Branch Circuit Design Calculations – Part One: Course Ee-1: Beginners' Electrical Design Course

Branch Circuit Design Calculations – Part One In " Course EE-1: Beginners' Electrical Design Course " which is a preparation for beginners in electrical design, I explained the following subjects:

1- The basic Concepts in electrical design (that every electrical designer must be familiar with it). 2- Electrical design philosophy for all building types. 3- Electrical load classification which include: 1. Lighting loads. 2. Appliances loads. 3. Power loads:  HVAC Loads,  Elevator loads,  Motor and pump loads. 4- Electrical design stages. 5- Electrical design criteria. Also, I listed the steps for electrical design process in the following two articles:  Electrical design process – part one  Electrical design process – part Two Also, in " Course EE-2: Basic Electrical Design Course – Level I " which is the second level for beginners in electrical design, I explained the following subjects: 1- How to determine the electrical requirements, needs and characteristics for any building/project. 2- How to develop different schematic diagrams with different connection configurations. 3- How to select optimal power sources and distribution networks. 4- How to select the equipment from the manufacturer’s catalogs 5- Check list foe electrical distribution design. 6- Solved examples. 7- ID-Spec software.

8- Preliminary electrical load estimation methods:  Space by space method,  Building method,  Area method. 9- Types of electrical diagrams. 10- How to draw and read the single line diagram

Important!!! Before proceeding with this new course, it is highly recommended to review the previous two courses mentioned above.

Today, I will begin explaining the Third Course in electrical design series which is " Course EE-3: Basic Electrical design course – Level II ", you can review the list of contents of this course by click on its link. I will begin with the first point; the branch circuit design calculations as follows.

Branch circuit design calculations

1. Introduction In low voltage distribution system, the power is transferred in the following manner: 1. From electrical power sources (Transformers, UDS, Generators and renewable energy sources) to service equipments ( MVSG, MLVSG or Main service Disconnect). 2. Then to (MLVSG, MCC or distribution center). 3. Then to (Switchboards and Panels). 4. Then to Final User Loads (Lighting, Small Appliances, Power Outlets, HVAC equipments, Motors, etc.). Regardless of the wiring methods used for transferring electrical power, the conductors carrying the power fall into one of two categories: 1. Feeders, 2. Branch-circuit conductors.

2. Difference between Service Conductors, feeders and Branch-circuit conductors So, let’s know the difference between branch circuit and feeder as follows:

A- Service Conductors: The conductors between the terminals of the electrical utility service equipment and the main disconnect switch of the main distribution in the premises

The service conductor can be divided to (3) parts based on the service point as follows: 1- Service Lateral: The underground conductors between the electrical utility service equipment and the service point. 2- Underground Service Conductors: The underground conductors between the service point and the main disconnect switch of the main distribution in the premises. 3- Service Drop: The overhead conductors between the utility electric supply system and the service point. Noting that The service point can be described as the point of demarcation between where the serving utility ends and the premises wiring begins. The serving utility generally specifies the location of the service point based on the conditions of service. B- Feeder:

It is A conductor that originates at the main distribution or main disconnect device and terminates at another distribution center, panelboard, or load center. C- Branch circuit: It is the portion of the wiring system extending past the final overcurrent device. These circuits usually originate at a panel and transfer power to user load devices. So, any circuit that extends beyond the final overcurrent protective device is called a branch circuit (Some branch circuits originate at safety switches (disconnects)). Example: Conductors from a 20A/1P circuit breaker in a panelboard connected to a receptacle in a room.

3. Essential definitions: Some other definitions are essential to understanding branch circuits and feeders, which are: 3.1 Service conductors: These conductors extend from the power company terminals to the main service disconnect. 3.2 Sub-feeders: These conductors originate at distribution centers other than the main distribution center and extend to panelboards, load centers, and disconnect switches that supply branch circuits. 3.3 Panelboard: This can be a single panel or multiple panels containing switches, fuses, and circuit breakers for switching, controlling, and protecting circuits.

3.4 Demand and Diversity factors: A- Demand factor: it is the ratio of the maximum demand of a system, or part of a system, to the total connected load on the system, or part of the system under consideration. So, Demand factor is always less than one.

B- Diversity factor: it is the ratio of the sum of the individual maximum demands of the various subdivisions of a system, or part of a system, to the maximum demand of the whole system, or part of the system, under consideration. So, Diversity factor is usually more than one. The sum of the connected loads supplied by a feeder-circuit can be multiplied by the demand factor to determine the load used to size the components of the system. While, the sum of the maximum demand loads for two or more feeders is divided by the diversity factor for the feeders to derive the Maximum Demand Load. Example: Consider four individual feeder-circuits with connected loads of 250 kVA, 200 kVA, 150 kVA and 400 kVA and demand factors of 90%, 80%, 75% and 85% respectively. Use a diversity factor of 1.5. calculate the Max demand load for the supply transformer.

Solution: Calculating demand for feeder-circuits 250 kVA x 90% = 225 kVA 200 kVA x 80% = 160 kVA 150 kVA x 75% = 112.5 kVA 400 kVA x 85% = 340 kVA The sum of the individual demands = 837.5 kVA If the main feeder-circuit were sized at unity diversity: kVA = 837.5 kVA ÷ 1.00 = 837.5 kVA The main feeder-circuit would have to be supplied by an 850 kVA transformer. However, using the diversity factor of 1.5, the kVA = 837.5 kVA ÷ 1.5 = 558 kVA for the main feeder. For diversity factor of 1.5, a 600 kVA transformer could be used. This means that a 600 kVA transformer can be used instead of an 850 kVA when applying the 1.5 diversity factor.

3.5 Electrical Load: it is the power drawn from the supply circuit by a particular piece of equipment, appliance, or luminaire (lighting fixture). A load can best be depicted as voltage drop.

Wherever there is voltage drop in an electrical circuit, there is a load being served. The electrical load for and device/equipment can be expressed by (3) different ways as follows: 1. Connected Load, 2. Calculated/Estimated Load, 3. Demand Load.

A- Connected Load: The load, measured in amperes or watts (volt-amperes), that is actually required by a load to operate properly. Connected load is the nameplate load rating of an appliance or other type of electrical utilization equipment. Example: A 60-watt light bulb connected to a 120- volt circuit will draw 1⁄2 ampere from the voltage source. This is the actual level of current flow obtained by dividing the 60 watts by the 120 volts (60 watts/120 volts = .5 ampere) as required by the power formula. There is a load of 60 watts, or 1⁄2 ampere, on this circuit due to this one load. The Total Connected Load: Many and various types of loads may be connected to the same circuit and the total load for this circuit is determined by adding together all of the connected loads.

B- Calculated/Estimated Load: It is The load of utilization equipment or outlets as determined by procedures allowed or required to be employed by the NEC. General lighting and general-use receptacle loads are examples of calculated loads. Examples: 1. The general lighting load is calculated from the total floor area of a building. 2. A motor load is determined to be 125% of the motor’s FLA rating. C- Demand Load: The load remaining after the demand factors allowed by the NEC for a number of like

or similar loads have been employed. Example: A demand load reduction is allowed for the total load for the ranges of an apartment building on the theory that not all ranges will be in use at the same time. Demand loads apply when a load that is available to many areas of the structure will not be fully loaded at the same time. Example: It is recognized by the NEC that not all of the outlets on all of the general use circuits will be in use at the same time. Therefore, for dealing with feeders, the NEC allows for de-rating of those loads to some lower level i.e. demand factors will apply only for a group of loads originate from same distribution center.

Also, Electrical Loads can be classified according duty time into: A- Continuous loads: loads where the maximum current is expected to continue for 3 hours or more. B- Non-continuous loads: other loads than Continuous Loads.

For more essential definitions, please view our Course " EC-1: Understanding NFPA 70 (National electrical code) course ".

4. Branch circuits types 1- According To Number of Loads:

A- Individual branch circuit: A branch circuit that supplies a single load. If this load is an appliance, it will be called Appliance branch circuit. B- Multi outlet branch circuit: A branch circuit with multiple loads which may be one of two cases as follows:  If these loads are appliances and lighting loads, it will be called General purpose branch circuit.  If these loads are appliances only, it will be called appliance branch circuit.

So, Appliance branch circuit: A branch circuit that supplies energy to one or more outlets to which appliances are to be connected and that has no permanently connected luminaires that are not a part of an appliance. And, General purpose branch circuit: A branch circuit that supplies two or more receptacles or outlets for lighting and appliances.

2- According To Number of Ungrounded Wires: A- Two wire branch circuit: A branch circuit with one ungrounded conductors and one grounded conductor. B- Multi-wire branch circuit: A branch circuit with two or more ungrounded conductors and one grounded conductor. A multi-wire circuit can be an individual circuit or a multi-outlet circuit.

3- According To Type of Load Branch-circuit loads are classified According To Type of Load into five categories: 1. Branch-circuit Lighting loads. 2. Branch-circuit Receptacle loads. 3. Branch-circuit Equipment loads. 4. Branch-circuit Heating and cooling loads. 5. Branch-circuit Motor loads.

Branch Circuit Design Calculations – Part Two In the previous article " Branch Circuit Design Calculations – Part One " in our new course " Course EE-3: Basic Electrical design course – Level II ", I explained some essential definitions and I listed different types of Branch circuits. Today I will list additional types of branch circuits and explain the basic features for any branch circuit as follows.

4.A Additional Branch circuits types 1- According to location

A- Indoor branch circuit: It includes All branch circuits inside the premises for lighting, receptacles, equipments, HVAC or other outlets. B- Outdoor Branch Circuits It is the branch circuits run on or between buildings, structures, or poles on the premises; and electrical equipment and wiring for the supply of utilization equipment that is located on or attached to the outside of buildings, structures, or poles. Examples for Outdoor Branch Circuits are listed in table 225.3 in below.

2- According to purpose of the Load A- General-Purpose Branch Circuits It included in Chapters 1, 2, 3, and 4 and not listed under Specific-Purpose Branch Circuits tables 210.2, 220.3 and chapters 5, 6&7.

B- Specific-Purpose Branch Circuits The provisions for branch circuits supplying equipment listed in Table 210.2 amend or supplement the provisions in this article and shall apply to branch circuits referred to therein.

Also table 220.3 list the calculation of loads in specialized applications that are in addition to, or modifications of, those Of general rules.

NEC code Chapters 5, 6, and 7 apply to special occupancies, special equipment, or other special conditions. These chapters supplement or modify the general rules.

3- According to type of building A- Dwelling Building A building providing complete and independent living facilities for one or more persons, including permanent provisions for living, sleeping, cooking, and sanitation. Dwelling buildings include the following types:  Dwelling, One-Family: A building that consists solely of one dwelling unit.  Dwelling, Two-Family: A building that consists solely of two dwelling units.

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Dwelling, Multifamily: A building that contains three or more dwelling units.

B- Non-dwelling buildings other buildings than dwelling like schools, hotels, factories, etc. 4- According to building’s construction status A- Branch circuits for Existing building. B- Branch circuits for New building.

5- According to Branch circuit voltage A- Branch Circuits Not More Than 600 Volts It is divided to (4) categories as follows: a- Branch Circuits not exceed 120 volts In residences and hotel rooms, circuits supplying lighting fixtures and small receptacle loads. b- Branch Circuits 120 volts and less it may be used to supply lampholders, auxiliary equipment of electric-discharge lamps, receptacles, and permanently wired equipment. c- Branch circuits 120 volts - 277 volts It may supply mogul-base screw-shell lampholders, ballasts for fluorescent lighting, ballasts for electric-discharge lighting, plug-connected appliances, and hard-wired appliances. Incandescent lighting operating over 150 volts is permitted in commercial construction. d- Branch circuits 277 volts - 600 volts it can supply mercury-vapor and fluorescent lighting, provided the lighting units are installed at heights not less that 22 feet above grade and in tunnels at heights no less than 18 feet.

B- Branch Circuits over 600 Volts It used for some special uses in non-dwelling buildings.

6- According to Branch circuit rating: A- 15 and 20 Ampere Branch Circuits It shall be permitted to supply lighting units or other utilization equipment, or a combination of both.

B- 30 Ampere Branch Circuits It shall be permitted to supply fixed lighting units with heavy-duty lampholders in other than a dwelling unit(s) or utilization equipment in any occupancy.

C- 40 and 50 Ampere Branch Circuits It shall be permitted to supply cooking appliances that are fastened in place in any occupancy. In other than dwelling units, such circuits shall be permitted to supply fixed lighting units with heavy-duty lampholders, infrared heating units, or other utilization equipment.

D- Branch Circuits Larger Than 50 Amperes It supply only non-lighting outlet loads.

5. Branch circuit features 5.1 Branch circuit Voltages

The voltage rating of electrical equipment shall not be less than the nominal voltage of a circuit to which it is connected.  Unless other voltages are specified, for purposes of calculating branch-circuit and feeder loads, nominal system voltages of 120, 120/240, 208Y/120, 240, 347, 480Y/277, 480, 600Y/347, and 600 volts shall be used.  Branch-circuit voltage limits are provided by NEC Code. These limits are based on the equipment being supplied by the circuit. 

5.2 Branch Circuit Conductors Conductors normally used to carry current shall be of copper unless otherwise noted. Conductors size unit: The American Wire Gage (AWG) for size identification, which is the same as the Brown and Sharpe (BS) Gage. The notation circular mils is used for Conductors larger than 4/0 AWG which will be equal to 250,000 circular mils. the notation MCM is used instead of circular mils, where MCM = 1000 circular mils. so, a 250,000-circular-mil conductor was labeled 250 MCM. the notation kcmil is used by both UL standards and IEEE standards instead of MCM. where kcmil = MCM = 1000 circular mils.

Example The circular mil area of a conductor is equal to its diameter in mils squared (1 in. = 1000 mils). What is the circular mil area of an 8 AWG solid conductor that has a 0.1285-in. diameter? Solution 0.1285 in. x 1000 = 128.5 mils 128.5 x 128.5 = 16,512.25 circular mils

5.3 Branch Circuit Rating The rating of any branch circuit will be the maximum permitted ampere rating or setting of the overcurrent device protecting this branch circuit. 

Branch circuits serving only one device can have any rating, while a circuit supplying more that one load is limited to ratings of 15, 20, 30, 40, or 50 amps. 

Example: A branch circuit wired with 10 AWG copper conductors has an allowable ampacity of at least 30 amperes, If this branch-circuit overcurrent protective device is a 20-ampere circuit breaker or fuse, what is the rating of this branch circuit?. Solution: the rating of this branch circuit is 20 amperes, based on the size or rating of the overcurrent protective device.

5.4 Number of Branch Circuits The minimum number of branch circuits shall be determined from the total calculated load and the size or rating of the circuits used. In all installations, the number of circuits shall be sufficient to supply the load served.

I will explain the branch circuit design calculations as per the classification of branch circuit according to type of load mentioned in Previous Article " Branch Circuit Design Calculations – Part Two ". First: Lighting Branch circuits In the broad sense, lighting loads may be categorized as follows: 1. General lighting. 2. Show-window lighting. 3. Track lighting. 4. Sign and outline lighting. 5. Other lighting.

Each lighting load is computed separately and then combined to determine the total lighting load.

Branch Circuit Design Calculations – Part Three

In the previous article " Branch Circuit Design Calculations – Part Two " in our new course " Course EE-3: Basic Electrical design course – Level II ", I explained some Additional definitions and features of Branch circuits. Also, I listed the (5) categories of Lighting Branch circuits, which were: 1. 2. 3. 4. 5.

General lighting. Show-window lighting. Track lighting. Sign and outline lighting. Other lighting.

Today I will explain the design calculations for the general lighting branch circuits as follows. You can review the following previous articles for more information: 

Branch Circuit Design Calculations – Part One

1- General lighting branch circuit 1.1 Definition: General lighting outlets are those Outlets intended for general use for fixed-in-place luminaires (lighting fixtures). They are only used for lighting for the normal use of the occupants and Its intensity should be adequate for any type of work performed in the area.

1.2 Lighting fixtures not included in this category: Specialized task lighting (Show-window lighting, Track lighting, accent, specialty, or display lighting).  Any special lighting for workshops, photography labs, or studios that may be located in the dwelling. 

1.3 Calculation Method Determining the general lighting load as per NEC will be based on the load per area

method as follows:

The NEC cod introduce minimum general lighting loads (in VA/ft2) for various types of buildings in Table 220.12.  Within the same building, there are normally several different types of areas like storage, office, hallways, and cafeterias, these areas must be considered separately if their (VA/ft2) values are available in table 220.12.  The general lighting load is calculated by multiplying the floor area (in ft2) of a building by its unit load (in VA/ft2) derived from the above table.  If the load is continuous (as in Most commercial structures), the calculated load is multiplied by 1.25 (the inverse of 80%) to determine the circuit requirements. (Please review the definition for the term ―continuous loads‖ in Article) 

How to calculate the Floor area? The floor area for each floor shall be calculated from the outside dimensions of the building, dwelling unit, or other area involved.  For dwelling units, the calculated floor area shall not include open porches, garages, or unused or unfinished spaces not adaptable for future use (like some attics, cellars, and crawl spaces). 

Example#1: A 25,000 ft2 office building is being designed. What is the general lighting load and what load does the circuit need to supply? Solution: From Table 220.12, the unit load for an office building is 3.5 VA/ft2. The general lighting load is determined by multiplying this value by the square footage of the building: The general lighting load = 3.5 VA/ft2 x 25,000 ft2 = 87,500 VA So, the general lighting load is 87,500 volt-amperes. However, the load is continuous and can only be 80% of the load supplied by the circuit. This value must be multiplied by 1.25 to determine the circuit requirements: the circuit rating = 87,500 VA x 1.25 = 109,375 VA so, The circuit is designed to supply 109.375 KVA

1.4 Notes for table 220.12:

Important!!! Don’t apply the values of table 220.12 before reviewing the following notes.

The unit values herein are based on minimum load conditions and 100 percent power factor (i.e. Load in VA = Load in Watt) and may not provide sufficient lighting for the installation contemplated. So, the designer can choose a higher value based on the existing design conditions.  Under any conditions, don’t use values less than that specified in table 220.12, there are no exceptions.  The general lighting load unit values specified in table 220.12 includes the following loads: 

1. All general-use receptacle outlets of 20-ampere rating or less, including receptacles connected to Bathroom Branch Circuits, 2. The outdoor receptacle outlets, 3. general-use receptacle Outlets used in Basements, Garages, and Accessory Buildings. 4. Wall lighting outlet used in Habitable Rooms, 5. Wall lighting outlets used in hallways, stairways, attached garages, and detached garages, 6. Wall lighting outlet used in Storage or Equipment Spaces (like attics, under-floor spaces, utility rooms, and basements), 7. Wall lighting outlet used in Guest Rooms or Guest Suites In hotels, motels, or similar occupancies. So, no need to add the above outlets in load calculations per NEC method. The NEC method and table 220.12 are applied for any Additions to Existing Installations for both dwelling and non-dwelling installations.  Energy saving–type calculations (which used to reduce the connected lighting load and actual power consumption) are not permitted to be used to determine the minimum calculated lighting load if they produce loads less than the load calculated according to 220.12. 

Example#2: A (2) floors (basement and main) dwelling unit have dimensions as show in below image. Calculate the total general lighting load for this unit and the Minimum Number of General Lighting Branch Circuits.

Solution: Step#1: Calculate the total area of the dwelling unit as follows: a- Basement area: area ―A‖ = 17 FT 9 IN. X (4 FT 4 IN. + 18 FT 6 IN. + 6 FT 0 IN.) = 17 FT 9 IN. X 28 FT 10 IN. = 512.0 FT2 Note that as mentioned in notes for table 220.12, the crawl space and garage areas are not considered as they are included in the NEC method calculation. b- Main floor area: AREA ―A‖= 6 FT 0 IN. X (11 FT 0 IN. + 17 FT 1 IN.) = 6 FT 0 IN. X 28 FT 1 IN. = 168.5 FT2 AREA ―B‖= 18 FT 6 IN. X (6 FT 3 IN. + 3 FT 8 IN. + 11 FT 0 IN. + 17 FT 1 IN.) = 18 FT 6 IN. X 38 FT 0 IN. = 730.0 FT2 AREA ―C‖= 4 FT 4 IN. X (3 FT 8 IN. + 11 FT 0 IN. + 17 FT 1 IN.) = 4 FT 4 IN. X 31 FT 9 IN. = 137.5 FT2 AREA ―D‖= 20 FT 2 IN. X (3 FT 8 IN. + 11 FT 0 IN.) = 20 FT 2 IN. X 14 FT 8 IN. = 295.5 FT2 Total main floor area (IN FT2) = 1304.5 FT2

So, Total area of the dwelling unit = 512.0 FT2 + 1304.5 FT2 = 1816.5 ft2

Step#2: The unit load listed in Table 220.12 shows 3 volt-amperes per ft2 for dwelling units. The general lighting load for the dwelling unit = 1816.5 ft2 x 3 watts per ft2 = 5449.5 watts

1.5 Notes for NEC method for calculation of lighting branch circuit load

Important!!! The NEC doesn’t introduce a procedure for calculating the actual full load for the individual lighting fixtures in a general lighting branch circuit.

If the required information for calculating the actual full load for every individual lighting fixture in the circuit is available, in this time, you can calculate the actual load for the lighting branch circuit by summing of actual full load for its individual lighting fixtures.  In this case, you need to compare the values obtained from NEC method with that obtained from actual load method and select the greater load value to be used in the design.  But actually, The NEC method for calculation of lighting load is not required if the actual full load for every individual lighting fixture in the circuit is determined.  Methods for determining the actual full load for every individual lighting fixture in the circuit is explained in our course " Advanced Course for Lighting Design - Level I " and I recommend reviewing these methods very well.  Section 220.18 (b) states that For circuits supplying lighting units that have ballasts, transformers, autotransformers, or LED drivers, the calculated load shall be based on the total ampere ratings of such units and not on the total watts of the lamps. This means that we must take into account the losses in light fixture switchgear (ballast, internal wiring, etc.) so, you must use the current rating of the ballast, not the tube wattage. 

Example#3: A fluorescent lighting fixtures with 4 numbers 2 feet lamps, 18 watt/ lamp. Calculate the actual load for this lighting fixture.

Solution: The actual total load of fixture = 4 lamps x 18 watt/lamp + losses So, we can’t know the actual losses, we will use the same equation in another form The actual total load of fixture = 4 ballast x watt/ballast = 4 x 20 w = 80 watt

1.6 Determining the Minimum Number of General Lighting Branch Circuits Example#4: For the dwelling unit in example# 2 above, Calculate the Minimum Number of General Lighting Branch Circuits. Solution: From example#2, the general lighting load for the dwelling unit = 5449.5 watts However, the load is continuous and can only be 80% of the total load. This value must be multiplied by 1.25 to determine the circuits requirements: the Load Rating = 5449.5 watts x 1.25 = 6811.9 watts The total ampere = 6811.9 watts / 120 V = 56.77 A (note that , NEC method assumed that PF=1) If we design 15-ampere circuits: The number of the general lighting circuits = 56.77/ 15 = 3.78 circuits. 

If we design 20-ampere circuits: The number of the general lighting circuits = 56.77/ 20 = 2.84 circuits 

Because it is not possible to have a partially powered electrical circuit that operates properly, and because these are minimum units, the total number of branch circuits with use of 15-ampere circuits is (4) circuits (3.78 circuits rounded up to 4 circuits) and (3) circuits with use of 20-ampere branch circuits (2.84 circuits rounded up to 3 circuits).

1.7 Determination of maximum Permissible number of lighting fixtures on a general lighting branch circuit You need to review the definition for branch circuit rating explained in the previous article " Branch Circuit Design Calculations – Part Two ". Rules to be applied are as follows: Section 210.23 states that: 1. In no case shall the load exceed the branch-circuit ampere rating. 2. An individual branch circuit shall be permitted to supply any load (any number of outlets) for which it is rated.  Section 210.23(A)(2) permits a 15- or 20-ampere branch circuit supplying lighting outlets to also supply utilization equipment fastened in place, such as appliances or an air conditioner. Under the conditions specified in this requirement, the utilization equipment load must not exceed 50 % of the branch-circuit ampere rating (7.5 amperes on a 15-ampere circuit and 10 amperes on a 20-ampere circuit). 

Note: according to 210.52(B), such fastened-in-place equipment is not permitted on the small-appliance branch circuits required in a kitchen, dining room, and so on.

Example#5: A 4 feet long, two-lamp fluorescent fixture ballast draws 0.7 amps at 120 volts. How many of these fixtures can be connected on a 20-amp circuit? Solution: This is a continuous load, so the current used by the lights can only be 80% of the circuit current rating: Allowable current = 20 A x 0.80 = 16 A By dividing the allowable load by the load of each lamp, the total number of Fixtures is determined: the total number of Fixtures = 16 A /0.7 A = 22.8 fixtures So, the maximum number of fixtures on the circuit is 22.

Example#6:

In example#5, If some utilization equipment fastened in place will be added to the circuit How many of these fixtures can be connected to the circuit? Solution: This is a continuous load, so the current used by the lights can only be 80% of the circuit current rating: Allowable current = 20 A x 0.80 = 16 A By following section 210.23(A)(2), the Allowable current = 16 A x 50% = 8 A By dividing the allowable load by the load of each lamp, the total number of lamps is determined: the total number of Fixtures = 8 A /0.7 A = 11.4 fixtures So, the maximum number of fixtures on the circuit is 11.

Important!! For good design, usually assume that the general lighting branch circuits will have some utilization equipment fastened in place and must be derated to 50% of its rating.

Special Rules In non-dwelling buildings: Lighting branch circuits that supply fixed lighting units with heavy-duty Lampholders can have 30 A or 40 A ratings.  Lighting Outlets for heavy-duty Lampholders shall be calculated at a minimum of 600 volt-amperes.  Use The above rules when it is applicable. 

Branch Circuit Design Calculations – Part Four In the previous article " Branch Circuit Design Calculations – Part Two " in our new course " Course EE-3: Basic Electrical design course – Level II " , I listed the (5) categories of Lighting Branch circuits, which were: 1. 2. 3.

General lighting. Show-window lighting. Track lighting.

4. 5.

Sign and outline lighting. Other lighting.

Also, in the previous Article " Branch Circuit Design Calculations – Part Three ", I explained the following points: The design calculation method for general lighting,  The calculation method for determination of the Minimum Number of General Lighting Branch Circuits in a building,  The calculation method for Determination of maximum Permissible number of lighting fixtures on a general lighting branch circuit. 

Today I will explain the design calculations for other categories of lighting branch circuits as follows. You can review the following previous articles for more information:

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Branch Circuit Design Calculations – Part One

2- Show-window lighting Branch Circuit

2.1 Definition From NEC article 100 the definition for Show-window is as follows: Show Window: Any window used or designed to be used for the display of goods or advertising material, whether it is fully or partly enclosed or entirely open at the rear and whether or not it has a platform raised higher than the street floor level. The Show-window lighting branch circuits are often used in commercial buildings.

2.2 NEC rules applied for Show-window lighting The Code includes several sections applicable to show-window lighting:   

Section 210.62 Show Windows Section 220.14(G) Load computation Section 220.43(A) Show-window lighting.

2.3 Design calculations for show window Load A- Determination of number of receptacles need for a show window: As per section 210.62, At least one receptacle outlet shall be installed within 450 mm (18 in.) of the top of a show window for each 3.7 linear m (12 linear ft) or major fraction thereof of show window area measured horizontally at its maximum width.

Notes to section 210.62: 1- To reduce the use of extension cords and floor receptacles which may cause physical injury, the required receptacle outlet(s) installed within 18 in. of the top of the show window. 2- Meaning of (12 linear ft) or major fraction is as follows: If we have a show window with total length = 12 feet then, we will need one receptacle,  If we have show window with total length = 13 feet (> 12) = 12 +1 then, we will need (2) receptacles  If we have show window with total length = 5 feet (<12) then, we will need one receptacle.  If we have show window with total length = 25 feet (> 12) = 2 x 12 +1 then, we will need (3) receptacles 

B- Determination of show window load: As per Sections 220.14(G) and 220.43(A), two options are permitted for the load calculations for branch circuits serving show windows: Option#1: If the receptacle(s) required according to 210.62 supplies the show window lighting load then the show window lighting load = 180 volt-amperes x number of receptacles Option#2: If the receptacle(s) required according to 210.62 don’t supplies the show window lighting load then the show window lighting load = 200 volt-amperes x show-window length (in feet) as shown in below image. And in this case: The show window total load = load of receptacles + load of lighting = (180 VA X number of receptacles) + (200 VA x show-window length (in feet)

Important!!! If the maximum volt-ampere rating of the equipment and lights used for show window lighting is known, then, calculate the show widow load as explained in the above two options. then, you must select the greater load value to be used for design of branch circuits and feeder. Don’t forget … The show window load is a continuous load and its value must be multiplied by 1.25 for determination of the circuit load requirements.

Important!! If you need to determine the minimum number of feet of show window allowed on a single branch circuit or/and maximum number of fixtures for a show window, please follow the same procedure explained in previous article " Branch Circuit Design Calculations – Part Two " for general lighting branch circuit.

Example#1: A department store has two lighted show windows, one 25 feet long and the other 20 feet long. What are the branch-circuit requirements for the show window load? Solution: Compute the load based on linear feet of show window: Total length = 25 + 20 = 45 feet Show-window load = 45 × 200 VA/ft = 9000 VA The lighting is a continuous load, so the show window load is multiplied by 1.25 to determine the circuit load requirements: Show-window load = 9000 VA × 1.25 = 11,250 VA The circuits supplying power for the show-window lighting must have a minimum capacity of 11,250 volt-amperes. In addition, receptacles are required for every 12 feet of show window. A total of five receptacles (two for the 20 feet window and three for the 25 feet window) are needed. The receptacle load = 5 × 180 VA = 900 VA

The show window total load = 11,250 VA + 900 VA = 12,150 VA

3- Track Lighting

3.1 Definition: From NEC section 410.2 the definition for track lighting is as follows: Lighting track is a manufactured assembly designed to support and energize luminaires (lighting fixtures) that are capable of being readily re-positioned on the track. Its length can be altered by addition or subtraction of sections of track. Track lighting is often used in commercial buildings for accent lighting.

3.2 NEC rules applied for Track lighting: The Code includes several sections applicable to Track lighting:  

Section 220.43(B) Track Lighting Section 410.151 (B) Connected Load.

3.3 Design calculations for Track Lighting Load As per NEC section 220.43(B), an additional load of 150 volt-amperes shall be included for every 600 mm (2 ft) of lighting track or fraction thereof.

Important!!! As per Section 220.43(B) ,This method for calculations for Track Lighting Load is applied for commercial buildings but it will not be applied for dwelling units or guest rooms or guest suites of hotels or motels) because in these buildings the track lighting load will not be added to the service load.

Example#2: 28 feet of track lighting is shown on the floor plans of a one-family dwelling. How much additional load will this track lighting add to the service? Solution: In accordance with 220.43(B), no additional load is required for track lighting installed in a dwelling unit.

Notes to section 220.43(B): 1. Where multi-circuit track is installed, the load shall be considered to be divided equally between the track circuits. 2. Meaning of mm (2 ft) of lighting track or fraction is that if there is a faction of track lighting the load computed for this faction is the same as a complete one.

Notes to section 410.151(B):

1. The maximum load on the track cannot exceed the rating of the branch circuit supplying the track. 2. Also, the track must be supplied by a branch circuit that has a rating not exceeding the rating of the track. (Please, review our previous article " Branch Circuit Design Calculations – Part One " for definition of branch circuit rating). Design calculations steps: Step#1: Determine the total length of track lighting,  Step#2: Divide the total length of the track by two,  Step#3: If, there is a faction, Round up the faction to one,  Step#4: Multiply the result from step#2 or 3(if there is a faction) by 150 voltamperes to get the total track load. 

Don’t forget … If The Track lighting load is considered as a continuous load then its value must be multiplied by 1.25 for determination of the circuit load requirements. Important!!! The actual track lighting fixtures are owner supplied, so, neither the quantity of track lighting fixtures nor the lamp wattage is specified. If the quantity of track lighting fixtures and the lamp wattage are known, then calculate the total load of the track = quantity of track lighting fixtures x number of lamps/fixture x lamp wattage, then, you must select the greater load value to be used for design of branch circuits and feeder. Important!! The above design calculation method as per Section 220.43(B) is not intended to determine the number of feet of track allowed on a single branch circuit and/or the maximum number of fixtures on an individual track. If you need to do this, please follow the same procedure explained in previous article " Branch Circuit Design Calculations – Part Two " for general lighting branch circuit.

Example#3: A lighting plan shows 62.5 linear ft of single-circuit track lighting for a small

department store featuring clothing. What is the minimum calculated load associated with the track lighting that must be added to the service or feeder supplying this store?

Solution: According to 220.43(B), the minimum calculated load to be added to the service or feeder supplying this track light installation is calculated as follows: Step#1: the total length of track lighting = 62.5 feet Step#2: 62.5 ft / 2 ft = 31.25, Step#3: Round up the length fraction, than rounded up to 32 Step#4: the total track load = 32 x 150 VA = 4800 VA The minimum load for the lighting track that is added to the service and/or feeder calculation is 4800 volt-amperes.

Special case of calculations for Track Lighting Load As per 220.43(B) exception, If the track lighting is supplied through a device that limits the current to the track (a supplementary over-current protective device that may be fuse or circuit breaker), the load shall be permitted to be calculated based on the rating of the device used to limit the current. I.e. the track lighting total load = voltage x device rating (Amps) Example#4: For department store in example#2, the entire length of the used track will be supplied by a single 20-ampere, 120-volt branch circuit. What is the minimum calculated load associated with the track lighting that must be added to the service or feeder supplying this store?

Solution: According to the exception to 220.43(B), the minimum calculated load to be added to the service or feeder supplying this track light installation is calculated as follows: The

track lighting total load = 120 V x 20 A= 2400 VA The minimum load for the lighting track that is added to the service and/or feeder calculation is 2400 volt-amperes.

Branch Circuit Design Calculations – Part Five In the previous article in our new course " Branch Circuit Design Calculations – Part Two ", I listed the (5) categories of Lighting Branch circuits, which were: 1. 2. 3. 4. 5.

General lighting. Show-window lighting. Track lighting. Sign and outline lighting. Other lighting.

Also, in the previous Article " Branch Circuit Design Calculations – Part Three ", I explained the following points: 1. The design calculation method for general lighting, 2. The calculation method for determination of the Minimum Number of General Lighting Branch Circuits in a building, 3. The calculation method for Determination of maximum Permissible number of lighting fixtures on a general lighting branch circuit. I explained how to design the show window and track lighting loads in the previous article " Branch Circuit Design Calculations – Part Four". Today I will explain the design calculations for sign and outline lighting branch circuits as follows. You can review the following previous articles for more information: 

Branch Circuit Design Calculations – Part One

4- Sign and outline lighting 4.1 Definition Sign and Outline Lighting: An arrangement of incandescent lamps, electric-discharge lighting, or other electrically powered light sources to outline or call attention to certain features such

as the shape of a building or the decoration of a window.

Fig.1: Section Sign Sign and outline lighting illumination systems include, but are not limited to:     

cold cathode neon tubing, high-intensity discharge lamps (HID), fluorescent or incandescent lamps, light-emitting diodes (LEDs), electroluminescent and inductance lighting.

Sometimes due to size restrictions or other logistical factors, the electric sign can be constructed in multiple factory-wired subassemblies (Section sign) that can be assembled at the sign-installation location. The definition for section sign will be as follows:

Section Sign: (see fig.1)

A sign or outline lighting system, shipped as sub-assemblies that requires fieldinstalled wiring between the sub-assemblies to complete the overall sign. The subassemblies are either physically joined to form a single sign unit or are installed as separate remote parts of an overall sign.

4.2 NEC rules applied for Sign and Outline lighting: The Code includes several sections applicable to sign and outline lighting:  

220.14 (F) Sign and Outline Lighting. Article 600.

2.3 Design calculations for Sign and Outline lighting: A- Number of required sign and outline branch circuits As per section 600.5 (A), At least one dedicated sign and outline branch circuit must be provided for each commercial Space at each entrance of a commercial building / occupancy accessible to pedestrians.

Important!!! Sometimes, the designer may assign A group of branch circuits supplied by a dedicated feeder for one sign lighting only, if there is need for installation of aLarge sign with higher loads than that specified by section 220.14 (F) in below and that usually be a multi section sign as described in the definitions in above. Notes for section 600.5(A): The term ―dedicated‖ in above means that the branch circuit for sign and outline lighting will not supply any other load.  Service hallways or corridors shall not be considered accessible to pedestrians, no sign and outline lighting branch circuits will be added there.  The loads for branch circuits of sign and outline lighting shall be added in the design stage of new commercial buildings for each commercial space as per 600.5 (A), because it is not convenient to add it later when the space is occupied or when a new occupant moves into an existing space. 

B- Sign and outline lighting branch circuit rating

As per section 600.5(B), the following rules are applied: Branch circuits that supply neon tubing installations shall not be rated in excess of 30 amperes.  Branch circuits that supply all other signs and outline lighting systems shall be rated not to exceed 20 amperes. 

Important!!! Large signs often have load requirements that exceed the ratings permitted bysection 600.5(B) will be divided to group of branch circuits supplied by a feeder which is not limited by the rating specified in 600.5(B). Important!!! In some cases, particularly for signs installed along highways, a utility servicededicated to the sign is provided. The rating of the feeder or service is not limited by the rating specified in 600.5(B).

C- Load Value for sign and outline lighting As per NEC section 220.14 (F) , Sign and outline lighting outlets shall be calculated at a minimum Load of 1200 volt-amperes for each required branch circuit specified in 600.5(A).

Important!!! Section 220.14(F) assigns 1200 volt-amperes as a minimum circuit load for the signs and outline lighting branch circuit, but If the actual load for a sign and outline lighting unit is known to be larger than 1200 VA, then the actual load is used for calculation purposes.

Fig.2 Important!!! Sign and outline lighting shall be considered to be continuous loads and must be multiplied by 1.25 for the purposes of feeder and protection calculations. As per section 600.6, each sign and outline lighting unit must be controlled by an externally operable and dedicated disconnecting means (switch or circuit breaker) that opens all ungrounded Conductors simultaneously as shown in Fig(2)above.

D- Location and Rating of sign and outline unit Disconnecting Means a- Location:

As per section 600.6(A), the following conditions shall be applied:

Fig.3 as in example#1 & 2 in Fig (3) above, The disconnecting means shall be within sight of the sign or outline lighting system that it controls. and as in example#3 in Fig (3) above, Where the disconnecting means is out of the line of sight from any section that is able to be energized, the disconnecting means shall be capable of being locked in the open position. 

For signs or outline lighting systems operated by mechanical or electromechanical controllers located external to the sign, the disconnecting means is required to be located within sightof (as in example# 2&3 in Fig (3) above) or in the same enclosure as the controller and must be capable of being locked in the open position. 

b- Rating: As per 600.6(B), The switch or flasher (not a circuit breaker) required to control the primary circuit of a transformer supplying a luminous gas tube, must have a current rating that is at least twice the current rating of this transformer.

Important!!! A disconnecting means shall not be required for cord-connected signs with an attachment plug.

Example#1:

A hardware store measures 80 feet × 120 feet. A portion of the building (80 feet × 40 feet) is used for storage. The remainder of the building is used as a showroom. There is a total of 45 feet of show windows, and there is one outdoor sign. What is the total lighting load for this building?

Solution: Each type of lighting load is computed separately and then combined to determine the total lighting load. First, the general lighting loads We have two areas in the store: Storage area = 80 feet × 40 feet = 3200 ft2 Showroom area = 80 feet × 80 feet = 6400 ft2

Table 220.3(A) included in the previous article " Branch Circuit Design Calculations – Part Three " lists unit loads for storage and showroom as 1/4 VA/ft2 and 3 VA/ft2, respectively, then: General lighting load (storage) = 3200 ft2 × 1/4 VA/ft2 = 800 VA General lighting load (showroom) = 6400 ft2 × 3 VA/ft2 = 19,200 VA Total General lighting load = 800 VA + 19,200 VA = 20,000 VA

Second: The show-window lighting load The show-window lighting load is based on 200 VA per linear foot: Show-window lighting load = 45 feet × 200 VA/ft = 9000 VA

Third: sign lighting The minimum load for the sign lighting is used: Sign lighting = 1200 VA Now the total lighting load of the building can be calculated by adding the parts together: Total lighting load of the building = 20,000 VA + 9000 VA + 1200 VA = 30,200 VA These loads are all continuous, so the total load is multiplied by a factor of 1.25 to determine the circuit requirements. So, Total lighting load of the building = 30,200 VA × 1.25 = 37,750 VA

5- Additional lighting loads 5.1 Definition The additional lighting loads are those loads not covered by NEC Code such as: 1. 2. 3. 4. 5. 6.

Security lighting, Parking area lighting, Sidewalk lighting, Roadway lighting, Stadium lighting, Tunnel Lighting.

5.2 Rules applied for Calculation of additional lighting loads Unfortunately, The NEC code don’t provide calculation rules for lighting types included under these additional lighting loads. So, all additional lighting loads are calculated using the actual load designed by professional lighting programs and methods as explained in our course " Advanced Course for Lighting Design - Level I ".

Important!!! All Additional lighting loads should be computed separately from the general lighting load and then added to the general lighting load.

Don’t forget… These additional lighting loads are considered continuous loads where appropriate and must be multiplied by 1.25 for feeder and overcurrent protection calculations.

Receptacle Branch Circuit Design Calculations – Part One In the previous article " Branch Circuit Design Calculations – Part One " in our new course " Course EE-3: Basic Electrical design course – Level II ", I indicated that Branch-circuit are classified According To Type of Load into five categories: 1. 2. 3. 4. 5.

Lighting Branch circuit, Receptacle Branch circuit, Equipment Branch circuit, Heating and cooling loads Branch circuit, Motor Branch circuit.

I explained the design calculations for the first type: lighting branch circuits in the following Previous Articles:     

Branch Branch Branch Branch Branch

Circuit Design Calculations Circuit Design Calculations Circuit Design Calculations Circuit Design Calculations Circuit Design Calculations

– – – – –

Part One Part Two Part Three Part Four Part Five

Today I will explain the design calculations for the second type: Receptacle branch circuits as follows.

Second: Receptacles branch circuit Receptacles are one of the most popular terms in the NEC code because of its importance especially in dwelling units and also in non-dwelling units as it is the device responsible for feeding power to most of equipments. NEC code includes large number of rules that are assigning the receptacle requirements for special locations, special loads, special installation methods and etc. So, I think we need to start with an introduction to receptacles before going to detailed explanation of NEC rules controlling the receptacles requirements and also, before explanation of design calculation steps for receptacle branch circuits.

1- Introduction 1.1 Essential Definitions Receptacle: A receptacle is a contact device installed at the outlet for the connection of an attachment plug. Receptacle Outlet: An outlet where one or more receptacles are installed.

Important!!! The number of receptacle Outlets my not equal to the number of receptacles installed on these outlets. For example, a duplex receptacle has two contact devices on the same receptacle outlet. In this case number of receptacles = 2 while the number of receptacle outlets = 1. Attachment Plug (Plug Cap) (Plug): A device that, by insertion in a receptacle, establishes a connection between the conductors of the attached flexible cord and the conductors connected permanently to the receptacle.

Note that… Attachment plug contact blades have specific shapes, sizes, and configurations so that a receptacle or cord connector will not accept an attachment plug of a voltage or current rating different from that for which the device is intended.

Multi-outlet Assembly: are metal or nonmetallic raceways that are usually surface, flush, or freestanding mounted designed to hold branch circuit conductors and receptacles, assembled in the field or at the factory.

The definition of multi-outlet assembly includes a reference to a freestanding assembly with multiple outlets, commonly called a power pole as shown in below image.

1.2 Types of Receptacles There are many types of Receptacles that had been classified according to many

factors such as:

Important!!! These Receptacle classifications are made as per NEC code, there will be other receptacle classifications as per other codes like IEC and BS but in this course I will explain that based on the NEC code only. 1- According to Type of Receptacle load a- Receptacle for dwelling loads: receptacles used for normal loads in dwelling units like general receptacles, Electric Dryers and Electric Cooking Appliances. These receptacles will have a maximum ampere of 30 A and voltage less than 600 V. b- Receptacle for Non-dwelling loads: receptacles used for special loads for non dwelling units like welding machines, heavy lamp holders, sign and outline lighting, Show Windows and Fixed Multi-outlet Assemblies (power poles). These receptacles will have a maximum ampere of 60 A and voltage less than 600 V.

2- According to Installation location a- Indoor Receptacles: Receptacles that will be used inside the building where there are no abnormal weather conditions, it is usually general purpose receptacles. b- Outdoor Receptacles: Receptacles that will be used outside the building or in opening areas inside the buildings, it is usually weather proof receptacles and GFCI type. c- Hazardous locations Receptacles: Receptacles that will be used in hazardous locations as defined by the NEC code, it is explosion proof receptacles type. d- Damp and wet location Receptacles: Receptacles that will be used in damp and wet locations like in kitchen and indoor swimming pools, it is GFCI type.

3- According to Method of installation

a- Pendant: A cord connector that is supplied by a permanently connected cord pendant shall be considered a receptacle outlet. b- Wall mounted, flush Receptacles. c- Wall mounted, Surface Receptacles. d- Wall mounted, Recessed Receptacles. e- Floor mounted Receptacles: receptacles installed in under floor boxes as a separate power source or as a part of FCC system (Flat conductor cable).

4- According to grounding connection

a- Un-grounded Receptacles: These Receptacles are similar to standard duplex receptacles, in that they accommodate two-prong plugs, but they are missing the ushaped grounding hole. b- Grounded Receptacles: these receptacles have the U shaped grounding hole.

5- According to polarization feature a- Polarized Receptacles: Receptacles that have wider neutral slot and narrower hot slot, the wider prong on the polarized plug will permit it to be plugged in only with the correct polarity. The narrower prong is the "hot" lead and the switch to the appliance is placed in that lead, guaranteeing that no voltage will reach the appliance when it is switched off. A non-polarized plug may have the switch in the neutral leg and thus be a shock hazard even when it is switched off as shown in the below image.

b- Non-polarized Receptacles: These outlets have two slots (hot and neutral) of the same width.

6- According to locking feature a- Locking Receptacles: Receptacles designed to lock an inserted plug with a matching blade configuration when the plug is rotated in a clockwise direction. The plug can only be removed by first turning it in a counter-clockwise direction.

b- Non-locking Receptacles: Receptacles without a locking feature, the plug can be removed by drawing it in the straight direction from the receptacle.

7- According to Mechanism of inserting and removing Plugs a- Twisted-Locking Receptacles: Twist-locking Receptacles types are used for heavy industrial and commercial equipment, where increased protection against accidental disconnection is required. Numbers prefixed by L are curved-blade, twist-locking connectors; others are straight blade and non-locking b- Straight blade Receptacles: A non-locking receptacle into which mating plugs are inserted at a right angle to the plane of the receptacle face.

8- According to Number of gangs on the same yoke

a- Single Receptacle: A single receptacle is a single contact device with no other contact device on the same yoke. b- Multiple Receptacle: A multiple receptacle is two or more contact devices on the same yoke.

c- Multiple Receptacle, Duplex: Two receptacles built with a common body and mounting means on the same yoke and accept two plugs. d- Multiple Receptacle, Triplex: A receptacle with a common mounting means on the same yoke which accepts three plugs. Multiple Receptacles, Quad (four-plex-type): A receptacle in a common housing that accepts up to four plugs. Four-In-One receptacles can be installed in place of duplex receptacles mounted in a single-gang box, providing a convenient means of adding receptacles without rewiring.

Important!!! Don’t be confused between duplex and twins receptacles, the duplex receptacles are two receptacles installed on the same outlet/yoke but the twins receptacles are two receptacles installed on two different outlets/yokes i.e. twin duplex receptacles = double duplex receptacle.

9- According to Ampere rating

15 A- Receptacles

20 A- Receptacles

50 A- Receptacles

60 A- Receptacles

30 A- Receptacles

Important!!! In dwelling units, the maximum rating of Receptacles will be 30 A while in Nondwelling it will be 60A.

10- According to Number of Receptacle poles

2- Poles Receptacle

3- Poles Receptacle

4- Poles Receptacle

11- According to Number of phases

Single phase Receptacle

Three phase Receptacle

12- According to Number of wires connected to the Receptacle

2- Wire Receptacle

3- Wire Receptacle

4- Wire Receptacle

5- Wire Receptacle

13- According to Receptacle Voltage rating

Single phase Receptacle

Three phase Receptacle

Single/Three phase Receptacle

125

250

120/208

250

480

125/250

277

600

277/480

347

347/600

480 600

14- According to Receptacle Configurations (NEMA Configurations) As per the following two images:

Note: the above two images summarize also the categories # 9 to # 13 explained in above. The NEMA Nomenclature is explained in the following image:

15- According to Receptacle Combination with other devices a- Non-Combined Receptacle: Receptacles that will be the only device installed on a yoke i.e. it will not share the same yoke with any other devices. b- Combined Receptacle: combined Receptacles are space-saving designs that provide two features in one device, such as an outlet with a guide light, a GFCI Receptacle with a switch or a switch with a Receptacle.

Combined Receptacles will include the following types: 1- GFCI Receptacle: A receptacle with a built in circuit that will detect leakage current to ground on the load side of the device. When the GFCI detects leakage current to ground, it will interrupt power to the load side of the device, preventing a hazardous ground fault condition. They can be used in indoor and outdoor locations.

GFCI Receptacle

Important!!! The use of GFCI Receptacles can be used as replacements for Un-grounded receptacles where a grounding means does not exist. Important!!! A receptacle supplying only a permanently installed fire alarm or burglar alarm system shall not be required to have ground-fault circuit-interrupter protection.

2- AFCI Receptacle: A receptacle with a built in device intended to provide protection from the effects of arc faults by recognizing characteristics unique to arcing and by functioning to de-energize the circuit when an arc fault is detected.

Important!!! AFCI Receptacle must be installed to be the first receptacle outlet of the branch circuit.

3- TVSS (SPD) Surge suppression Receptacle: A receptacle with built-in circuitry designed to protect its load side from high-voltage transients and surges like surge protectors. The circuitry will limit transient voltage peaks to help protect sensitive electronic equipment such as PC’s, modems, audio/video equipment, etc. without the need for power strips.

4- Switched Receptacle: A receptacle with built-in switch designed to turn on and off the power for the connected equipment to that Receptacle.

16- According to Receptacle Usage a- General purpose/Use Receptacles: These receptacles are normally used to supply lighting and general-purpose electrical equipment inside the buildings or outside the dwelling buildings only.

Important!!! general use - outdoor receptacle outlets are not required for commercial, institutional, and industrial occupancies, but as exception, outdoor receptacle outlets shall be installed at an accessible location for the servicing of heating, air-conditioning, and refrigeration equipment as per NEC 210.63 or at the discretion of the designer or owner. This receptacle is required to be provided with ground-fault circuit-interrupter protection. While an outdoor receptacle outlet shall not be required at one and two-family dwellings for the service of evaporative coolers.

b- Special purpose Receptacles: It includes the following types: 1- Pin and sleeve (industrial) Receptacle: A Receptacle with round pin-type contacts intended to mate with a connector having hollow cylindrical female contacts.

2- Hospital grade Receptacle: A receptacle designed to meet the performance requirements of high-abuse areas typically found in health care facilities. These receptacles are tested to the Hospital Grade requirements of Underwriters Laboratories Inc. Standard 498. 3- Isolated ground Receptacle: Receptacles intended for use in an Isolated Grounding system where the ground path is isolated from the facility grounding system. The grounding connection on these receptacles is isolated from the mounting strap. 4- Factory installed Receptacles: Receptacles installed on the equipment like electric baseboard heaters and provided as a separate assembly by the manufacturer. 5- Manufactured wiring system receptacles: special receptacles used only with Manufactured wiring system as shown in below image.

6- Shore power receptacles: Single locking- and grounding-type Receptacles intended to supply shore power to boats, it shall be housed in marine power enclosure listed as marina power enclosure or listed for set locations as shown in below image.

7- Tamper resistant Receptacle: A receptacle specially constructed so that access to its energized contacts is limited. It is designed to help protect children from electrical injury, they have a built-in shutter mechanism that blocks insertion of most small objects. The shutters only open when a properly rated plug is inserted. Once installed, they are permanent, offering continuous protection unlike plastic outlet caps that can be removed. 8- Weather-resistant Receptacle: These outlets are required by the NEC in damp or wet locations, such as patios, decks and pool areas, or any other residential outdoor location. The 15-amp and 20-amp weather-resistant outlets are built with UV stabilized thermoplastic and corrosion-resistant metals for superior performance outdoors, including cold impact resistance. You can choose from combined weather/tamper-resistant outlets or weather-resistant GFCIs with or without tamperresistance.

Note: Outlets in damp or wet locations should always be installed with weatherresistant covers. 9- Rotating receptacles: These Receptacles can be positioned to accommodate more than one of the large transformer-type plugs from cell phone chargers, hairdryers, cordless appliances, MP3 players, night lights and more, eliminating the need for power strips.

10- Split circuit receptacles: A split receptacle has two outlets with each wired on a

different circuit or with one outlet live and the other switched. With a split receptacle you can have a wall switch to turn a light on or off or remotely control one plug-in location but not the other. Most duplex receptacles provide break-off tabs that allow them to be converted into split-circuit receptacles.

11- Explosion Proof Receptacle: A receptacle constructed to meet the requirements of hazardous locations as defined by the NEC Code.

17- According to Receptacle’s conductor marking a- Receptacles having no conductor marking: these receptacles will be used only

with copper conductors or copper-clad conductors. b- Receptacles marked “AL-CU”: these receptacles will be used only with copper conductors or copper-clad conductors. c- Receptacles marked “CO/ALR”: these receptacles will be used with UL-labeled aluminum, copper or copper-clad aluminum conductors.

18- According to Type of Receptacle branch circuit a- Branch Circuit, General-Purpose: A branch circuit that supplies two or more receptacles or outlets for lighting and appliances. b- Branch Circuit, Individual: A branch circuit that supplies only one utilization equipment via a single receptacle.

Important!!! A branch circuit supplying one duplex receptacle that supplies two cord-andplug-connected appliances or similar equipment is not an individual branch circuit.

c- Branch Circuit, Multiwire: A branch circuit that consists of two or more ungrounded conductors that have a voltage between them, and a grounded conductor that has equal voltage between it and each ungrounded conductor of the circuit and

that is connected to the neutral or grounded conductor of the system.

Receptacle Branch Circuit Design Calculations – Part Two In the previous article " Receptacle Branch Circuit Design Calculations – Part One " in our new course " Course EE-3: Basic Electrical design course – Level II ", I explained the essential definitions for receptacles branch circuit and the different types of receptacles. Today I will explain some basic principles and features of Receptacle branch circuits before proceeding with the design calculations of such type of circuits.

1- The difference between Multiple and Multiwire branch circuits

Multiple Branch Circuits: are two or more branch circuits supply devices or equipment on the same yoke. while, Branch Circuit, Multiwire: A branch circuit that consists of two or more ungrounded conductors that have a voltage between them, and a grounded conductor that has equal voltage between it and each ungrounded conductor of the circuit and that is connected to the neutral or grounded conductor of the system.

What do you notice from the two definitions?  If we have two devices to be fed from two different receptacle outlets installed on the same yoke, so we will need two different branch circuits, then we have 4-wires(2 hot and 2 grounded) provided to this yoke. In this case we have a multiple branch circuit.  But if we supply the same with a multiwire branch circuit, we will have 3 wires only (2 hot, one common grounded)

Don't Forget…. There is difference between receptacle device and receptacle outlet as explained before in the previous article.

1.1 Advantages of using multiwire instead of multiple branch circuit A multiwire circuit shall be permitted to be considered as multiple circuits. But the multiwire branch circuit has some advantages more than the Multiple circuit as follows:    

Using three wires to do the work of four (in place of two 2-wire circuits), Less raceway fill, Easier balancing and phasing of a system, Less voltage drop.

1.2 Constraints for using multiwire branch circuits: 1- All conductors of a multiple branch circuit shall originate from the same panelboard or similar distribution equipment. 2- The simultaneous opening of all ―hot‖ conductors at the panelboard effectively protects personnel from inadvertent contact with an energized conductor or device terminal during servicing. For examples:  For a single-phase installation, the disconnecting means could be two singlepole circuit breakers with an identified handle tie or a two-pole circuit breaker, as shown in Fig.1 (top), or by a 2-pole switch, as shown in Fig.1 (bottom).

Fig (1) For a 3-phase installation, a 3-pole circuit breaker, three single-pole circuit breakers with an identified handle tie, or a three-pole switch provides the required simultaneous opening of the ungrounded conductors. 

3- It is not permitted that the wiring terminals of a device, such as a receptacle, to be the means of maintaining the continuity of the grounded conductor between devices in a multiwire branch circuit.

A splice is made and a jumper is connected to the terminal, unless the neutral is looped; that is, a receptacle or lampholder could be replaced without interrupting the continuity of energized downstream line-to-neutral loads Opening the neutral could cause unbalanced voltages, and a considerably higher voltage would be impressed on one part of a multiwire branch circuit, especially if the downstream line to- neutral loads were appreciably unbalanced.

1.3 Uses of multiwire Branch circuits: It can supply Appliances that have both line-to-line and line-to-neutral connected loads, such as electric ranges and clothes dryers.  It can supply Loads that are line-to-neutral connected only, such as the splitwired combination device as shown in Fig.1 (bottom).  It can supply A device with a 250-volt receptacle (line-to-line) and a 125-volt receptacle (line-to-neutral), as shown in Fig.2, provided the branch-circuit overcurrent device simultaneously opens both of the ungrounded conductors. 

Fig (2)

2- GFCI protection basics As I mentioned before in previous article " Receptacle Branch Circuit Design Calculations – Part One " that Combined Receptacle are space-saving designs that provide two features in one device, such as an outlet with a guide light, a GFCI Receptacle with a switch or a switch with a Receptacle.

Important!!! As per NEC 406.5(G), A receptacle shall not be combined in enclosures with other receptacles, snap switches, or similar devices, unless they are arranged so that the voltage between adjacent devices does not exceed 300 volts or unless they are installed in enclosures equipped with identified, securely installed barriers between adjacent devices.

2.1 Definition: Ground-Fault Circuit Interrupter (GFCI) is A device intended for the protection of personnel that functions to de-energize a circuit or portion thereof within an established period of time when a current to ground exceeds the values established for a Class A device.

2.2 Theory of operation of GFCI Device: As in image below that shows a typical circuit arrangement of a GFCI which operates in the following procedure:

The line conductors are passed through a sensor and are connected to a shunttrip device.  As long as the current in the conductors is equal, the device remains in a closed position.  If one of the conductors comes in contact with a grounded object, either directly or through a person’s body, some of the current returns by an alternative path, resulting in an unbalanced current.  The toroidal coil senses the unbalanced current, and a circuit is established to the shunt-trip mechanism that reacts and opens the circuit. 

Important!!! The use of GFCI Receptacles can be used as replacements for Un-grounded receptacles where a grounding means does not exist. Important!!! A receptacle supplying only a permanently installed fire alarm or burglar alarm system shall not be required to have ground-fault circuit-interrupter protection.

Important!!! GFCIs do not protect persons from shock hazards where contact is between phase and neutral or between phase-to-phase conductors.

2.3 Types of GFCI devices:

A variety of GFCIs are available, including: 1. 2. 3. 4. 5.

portable types, plug-in types, circuit-breaker types, types built into attachment plug caps, receptacle types.

Each type has a test switch so that units can be checked periodically to ensure proper operation.

Important!!! All combined GFCI with receptacles must be readily accessible to be periodically tested, by pressing the test button on the device and verify that all receptacles protected through that GFCI device are de-energized.

3- AFCI Protection Basics 3.1 Definitions: Arc-Fault Circuit Interrupter (AFCI) is a device intended to provide protection from the effects of arc faults by recognizing characteristics unique to arcing and by functioning to de-energize the circuit when an arc fault is detected.

AFCI Receptacle: A receptacle with a built in device intended to provide protection from the effects of arc faults by recognizing characteristics unique to arcing and by functioning to de-energize the circuit when an arc fault is detected.

3.2 AFCI device uses: From the above definition, the uses for such devices are for: 1. De-energizing the branch circuit when an arc fault is detected.

Also, AFCI devices may have the capability to perform other functions such as: 1. 2. 3.

Overcurrent protection, Ground-fault circuit interruption Surge suppression.

Important!!! AFCI devices must be tested to verify that arc detection is not unduly inhibited by the presence of loads and circuit characteristics that may mask the hazardous arcing condition. Important!!! AFCI devices must be tested for unwanted tripping due to the presence of arcing that occurs in control and utilization equipment under normal operating conditions or to a loading condition that closely like to an arcing fault, such as a solid-state electronic ballast or a dimmed load.

3.3 AFCI device types: UL 1699 currently recognizes five types of arc fault circuit interrupters: 1. 2. 3. 4. 5.

Branch/feeder AFCI, Combination AFCI, Cord AFCI, Outlet AFCI, Portable AFCI.

The product standard requires specific marking on AFCI devices to indicate the type of protection provided.

Important!!! To provide the full range of AFCI protection covered in the UL standard for the branch-circuit conductors and at the outlets supplied by the branch circuit, use the combination- type AFCI devices not the Cord AFCI devices.

NEC Rules for using AFCI devices:

AS per NEC section 210.12(A), AFCI can be used with the following branch circuits: 

1. For all 15- and 20-ampere, 120-volt branch circuits that supply outlets including receptacle, lighting, and other outlets located throughout a dwelling unit except those outlets installed in kitchens, bathrooms, unfinished basements, garages, and outdoors (in this case GFCI will be used). 2. For shared circuits like circuit that feed a bedroom and other areas such as closets and hallways.

Important!!! There is no prohibition against using AFCI protection on other circuits or locations other than those specified in 210.12(A). 

As per 210.12(B), the branch circuit shall be protected by one of the following: 1. A listed combination-type AFCI located at the origin of the branch circuit. 2. A listed outlet branch-circuit type AFCI located at the first receptacle outlet of the existing branch circuit.

Receptacle Branch Circuit Design Calculations – Part Three In the previous article " Receptacle Branch Circuit Design Calculations – Part One ", I explained the essential definitions for receptacles branch circuit and the different types of receptacles. Also, in the previous article " Receptacle Branch Circuit Design Calculations – Part Two ", I explained basic principles, GFCI protection and AFCI protection for receptacle branch circuits. Today, I will explain the Receptacle Branch circuit calculations for both Dwelling and Non-dwelling buildings as follows.

First: Receptacle Branch circuit calculations in dwelling buildings 1- Essential definitions: As I mentioned before in previous article " Branch Circuit Design Calculations – Part Two " that the Dwelling Building is A single unit, providing complete and independent living facilities for one or more persons, including permanent provisions for living, sleeping, cooking, and sanitation. Dwelling buildings include the following types:   

Dwelling, One-Family: A building that consists solely of one dwelling unit. Dwelling, Two-Family: A building that consists solely of two dwelling units. Dwelling, Multifamily: A building that contains three or more dwelling units.

2- Receptacle Branch circuit ratings and permissible loads In no case shall the load exceed the branch-circuit ampere rating. The following are the permissible Receptacle Branch circuit ratings in dwelling buildings: A) 15- and 20-Ampere Branch Circuits 15- and 20-Ampere Branch Circuits shall be permitted to supply: 1- Only lighting units: this case explained before in previous article 2- Only utilization equipment: with condition that the combined load for all utilization equipment must not exceed the branch circuit rating. 3- Combination of both: in the case the permissible rating of the utilization equipment will depend on its type as follows: If it is not fastened-in-place, it can have a rating of up to 80 % of the branch circuit rating as in TABLE 210.21(B)(2).  If it is fastened-in-place, other than luminaires, it shall not exceed 50 % of the branch-circuit ampere rating. 

Important!!! Using fastened-in-place equipment is not permitted for the small-appliance branch circuits, laundry branch circuits, and bathroom branch circuits required in a dwelling unit and these branch circuits shall supply only their receptacle outlets required by the code.

B) 30-Ampere Branch Circuits A 30-ampere branch circuit shall be permitted to supply utilization equipment in any occupancy. A rating of any one cord-and-plug-connected utilization equipment shall not exceed 80 percent of the branch-circuit ampere rating.

Important!!! A single receptacle installed on an individual branch circuit shall have an ampere rating not less than that of the branch circuit.

C) 40- and 50-Ampere Branch Circuits A 40- or 50-ampere branch circuit shall be permitted to supply cooking appliances that are fastened in place in any occupancy.

3- Selecting Receptacle rating for a branch circuit A- Receptacle rating, general Where connected to a branch circuit supplying two or more receptacles or outlets, receptacle ratings shall conform to the values listed in Table 210.21(B) (3), or, where rated higher than 50 amperes, the receptacle rating shall not be less than the branchcircuit rating.

B- Single Receptacle on an Individual Branch Circuit (see below image) A single receptacle installed on an individual branch circuit shall have an ampere rating not less than that of the branch circuit. For example, a single receptacle on a 20-ampere individual branch circuit must be rated at 20 amperes.

C- Receptacle supplying Total Cord-and-Plug-Connected Load a Receptacle shall not supply a total cord-and plug- connected load in excess of the maximum specified in Table 210.21(B)(2).

4- Voltage ratings for Receptacle Branch circuit In dwelling units the voltage rating for Receptacle Branch circuit shall not exceed the following: A) 120 volts, nominal, between conductors For Receptacle Branch circuit that supplies Cord-and-plug-connected loads 1440 voltamperes, nominal, or less or less than 1⁄4 hp. B) 208-volt or 240-volt circuit, nominal, between conductors For Receptacle Branch circuit that supplies High-wattage cord-and-plug-connected loads, such as electric ranges, clothes dryers, and some window air conditioners.

5- The Maximum allowable number of receptacles on a branch circuit As per NEC section 220.14(I), Receptacle outlets load (see below image) shall be calculated at not less than:   

180 volt-amperes for each single receptacle, 180 volt-amperes for each multiple receptacle (duplex or triplex) on one yoke, 90 volt- amperes per receptacle for multiple receptacles (four or more).

Important!!! If a receptacle is dedicated for a specific device, then the actual load is used and If this dedicated load is continuous, then the 125% overrate is appropriate. To calculate the Maximum allowable number of receptacles on a branch circuit, make the following steps: Step#1: Determine the maximum circuit power by Multiply the branch circuit voltage and amperage.  Step#2: Then divide by 180 volt-amperes. 

The result from Step#2 = the Max. Allowable single, duplex or triplex receptacles or a combination of them on a branch circuit.

Example#1: How many receptacles can be placed on a 120-volt, 20-amp circuit? How many can be placed on a 120-volt, 15-amp circuit?

Solution: Step#1: Determine the maximum circuit power (for 20-amp circuit) = 120 V × 20 A = 2400 VA Determine the maximum circuit power (for 15-amp circuit) 120 V × 15 A = 1800 VA Step#2: Then divide the power by the load per receptacle For 20-amp circuit: Maximum allowable number of receptacles = 2400 VA / 180 VA = 13.3 For 15-amp circuit: Maximum allowable number of receptacles = 1800 VA / 180 VA = 10 So, A 120-volt, 20-amp circuit can supply 13 receptacles. A 120-volt, 15-amp circuit can supply 10 receptacles.

6- The Minimum number of receptacle branch circuits for bank or office buildings

As per NEC section 220.14(J), if the number of receptacles is unknown so for bank and office buildings, we can calculate the receptacles load by multiplying the area in ft2 by the unit value (1 VA/ft2). To get the required number of receptacle branch circuits for bank or office buildings make the following steps: Step#1: Calculate the total receptacles load for the whole building as explained above.  Step#2: Calculate the total receptacle circuit load in VA by multiplying its voltage by its amperage as follows: 1. For 15-A Branch circuits = 120 V × 15 A = 1800 VA 2. For 20-A Branch circuits = 120 V × 20 A = 2400 VA  Step#3: Divide the total receptacle load for the whole building by the maximum load per circuit to determine the minimum number of circuits. 

Example#2: Determine the total receptacle load for an 80 ft × 120 ft office building? And determine the number of 15-amp circuits needed to supply the load. Noting that the number of receptacles is unknown.

Solution: Step#1: The number of receptacles is unknown, so a receptacle load of 1 VA/ft2 can be calculated: Area = 80 ft × 120 ft = 9600 ft2 Total Receptacle load = 1 VA/ft2 × 9600 ft2 = 9600 VA Step#2: To determine the number of circuits required, first calculate the allowable load for a single circuit: The allowable load for a single circuit = 120 V × 15 A = 1800 VA Step#3: Divide the total receptacle load by the maximum load per circuit to determine the minimum number of circuits: Minimum number of 15-A receptacle Circuits = 9600 VA / 1800 VA = 5.33 This is the minimum number, so round up to six circuits.

Important!!! As per NEC section 220.14(J), limit using the unit value (1 VA/ft2) for banks and office buildings but it can be used also (as approximate) for other types of buildings including dwelling ones.

7- Receptacle branch circuits Calculations in dwelling buildings In the broad sense, Receptacle in dwelling units may serve one of the following loads: 1. 2. 3. 4. 5. 6. 7. 8.

General-use Receptacle Loads, Small appliance Loads, Laundry Load, Fastened-in-place Appliance loads, Cloth dryer Load, Household cooking appliances load, Heating and air conditioning loads, Motor loads.

In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code. 1- General-use Receptacle Loads 1.1 locations of General-use Receptacle Loads as per NEC section 210.52

Rule#1: Main rule (see below image) Receptacles shall be installed such that no point measured horizontally along the floor line of any wall space is more than 1.8 m (6 ft) from a receptacle outlet.

The wall space is a wall unbroken along the floor line by doorways, fireplaces, archways, and similar openings and may include two or more walls of a room (around corners), as illustrated in Exhibit 210.27. The Minimum length for a wall space is 2 ft.

Important!!! The wall space behind the swing of a door is included in the measurement. This does not mean that the receptacle outlet has to be located in that space, only that the space is included in the wall-line measurement.

Rule#2: electric baseboard heaters built-in receptacle (see below image)

If there is a permanently installed electric baseboard heaters (longer than 12 ft.) equipped with factory-installed receptacle outlets or outlets provided as a separate assembly by the manufacturer shall be permitted as the required outlet or outlets for the wall space utilized by such permanently installed heaters (see image below) . Such receptacle outlets shall not be connected to the heater circuits.

Rule#3: Receptacle designed for intended use Receptacle designed for intended utilization equipment or practical room use may be placed in corners, may be grouped, or may be placed in a convenient location. For example, receptacles in a living room and family

room that are intended to serve home entertainment equipment or home office equipment.

Rule#4: floor Receptacles (see below image) Receptacle outlets in floors shall not be counted as part of the required number of receptacle outlets unless located within 450 mm (18 in.) of the wall.

Rule#5: kitchen and dining areas counters receptacles other than that used for small appliances (see above image) Receptacle outlets installed to serve kitchen or dining area counters (for small appliances) cannot also be used as general-use receptacles for an adjacent wall space and in this case general-use receptacle branch circuits must be added to serve such locations.

Rule#6: receptacles not under main rule The following receptacles can't be used as general-use receptacles ( rule#1 will not apply for it): A receptacle that is Part of a luminaire or appliance, or A receptacle that is controlled by a wall switch, or A receptacle that is Located within cabinets or cupboards, or A receptacle that is located more than 1.7 m (5.5 ft) above the floor.

Important!!! A receptacle controlled by a switch may result in the occupant using an extension cord, run from an outlet or device that is not controlled by a switch, to supply appliances or equipment that require continuous power, such as an electric clock. So, this receptacle not considered as general-use receptacle.

Receptacle Branch Circuit Design Calculations – Part Four In the previous article " Receptacle Branch Circuit Design Calculations – Part Three ", I explained the following points: 1. Receptacle Branch circuit ratings and permissible loads, 2. How to select Receptacle rating for a branch circuit, 3. Voltage ratings for Receptacle Branch circuit, 4. How to calculate The Maximum allowable number of receptacles on a branch circuit, 5. How to calculate The Minimum number of receptacle branch circuits for bank or office buildings. Also, I stated that a Receptacle in dwelling units may serve one of the following loads: 1. 2. 3. 4. 5.

General-use Receptacle Loads, Small appliance Loads, Laundry Load, Fastened-in-place Appliance loads, Cloth dryer Load,

6. 7. 8.

Household cooking appliances load, Heating and air conditioning loads, Motor loads.

You can review the following previous articles for more information:  

Receptacle Branch Circuit Design Calculations – Part One Receptacle Branch Circuit Design Calculations – Part Two

In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code.

1- General-use Receptacle Loads 1.1 locations of General-use Receptacle Loads

the locations of General-use Receptacle Loads as per NEC section 210.52 is controlled by eleven (11) Rules as follows:

Important!!! The primary objective of the requirements covering any of the dwelling areas included in 210.52 is to minimize the need to use extension cords to supply utilization equipment. Rule#1: Main rule (see below image) Receptacles shall be installed such that no point measured horizontally along the floor line of any wall space is more than 1.8 m (6 ft) from a receptacle outlet.

Definition: The wall space is a wall unbroken along the floor line by doorways, fireplaces, archways, and similar openings and may include two or more walls of a room (around corners), as illustrated in Exhibit 210.27. The Minimum length for a wall space is 2 ft. Important!!! The wall space behind the swing of a door is included in the measurement. This does not mean that the receptacle outlet has to be located in that space, only that the space is included in the wall-line measurement.

Rule#2: electric baseboard heaters built-in receptacle (see below image) If there is a permanently installed electric baseboard heaters (longer than 12 ft.) equipped with factory-installed receptacle outlets or outlets provided as a separate assembly by the manufacturer shall be permitted as the required outlet or outlets for the wall space utilized by such permanently installed heaters (see image below) . Such receptacle outlets shall not be connected to the heater circuits.

Rule#3: Receptacle designed for intended use Receptacle designed for intended utilization equipment or practical room use may be placed in corners, may be grouped, or may be placed in a convenient location. For example, receptacles in a living room and family

room that are intended to serve home entertainment equipment or home office equipment. Rule#4: floor Receptacles (see below image) Receptacle outlets in floors shall not be counted as part of the required number of receptacle outlets unless located within 450 mm (18 in.) of the wall.

Rule#5: kitchen and dining areas counters receptacles other than that used for small appliances (see above image) Receptacle outlets installed to serve kitchen or dining area counters (for small appliances) cannot also be used as general-use receptacles for an adjacent wall space and in this case general-use receptacle branch circuits must be added to serve such locations.

Rule#6: receptacles not under main rule The following receptacles can't be used as general-use receptacles ( rule#1 will not apply for it): A receptacle that is Part of a luminaire or appliance, or A receptacle that is controlled by a wall switch, or A receptacle that is Located within cabinets or cupboards, or A receptacle that is located more than 1.7 m (5.5 ft) above the floor. Important!!! A receptacle controlled by a switch may result in the occupant using an extension cord, run from an outlet or device that is not controlled by a switch, to supply appliances or equipment that require continuous power, such as an electric clock. So, this receptacle not considered as general-use receptacle.

Rule#7: hallways As per NEC section 210.52(H), In dwelling units, hallways of 3.0 m (10 ft) or more in length shall have at least one receptacle outlet. The hallway length shall be considered the length along the centerline of the hallway without passing through a doorway. Important!!! Rule#7 does not apply to common hallways of hotels, motels, apartment buildings, condominiums, and similar occupancies.

Rule#8: foyers

As per NEC section 210.52(I), Foyers that are not part of a hallway in accordance with 210.52(H) and that have an area that is greater than 5.6 m2 (60 ft2) shall have a receptacle(s) located in each wall space 900 mm (3 ft) or more in width and unbroken by doorways, floor-to-ceiling windows, and similar openings.

Rule#9: Bathroom As per NEC Section 210.52(D), in dwelling units, at least one receptacle outlet shall be installed in bathrooms in one of the following locations: 1. Within 900 mm (3 ft) (36 inch) of the outside edge of each basin. 2. on the side or face of the basin cabinet not more than 300 mm (12 in.) below the basin countertop.

special cases for Rule#9 are as follows: 1. If there is more than one basin, a receptacle outlet is required adjacent to each basin location. 2. If the basins are in close proximity, one receptacle outlet installed between the two basins can be used to satisfy this requirement as shown in below image (top).

Definition: A bathroom is defined in Article 100 as ―an area including a basin with one or more of the following: a toilet, a urinal, a tub, a shower, a bidet, or similar plumbing fixtures.‖ Important!!! As per NEC section 406.9(C), it is prohibited to install a receptacle within or directly over a bathtub or inside a shower stall even if the receptacles are installed in a weatherproof enclosure.

Important!!! As per NEC section 210.11(C)(3), At least one 20-ampere branch circuit shall be provided to supply bathroom receptacle outlet(s). This circuit is permitted to supply the required receptacles in more than one bathroom as show in above image (bottom). Important!!! If the circuit supplies the required receptacle outlet in only one bathroom, it is allowed to also supply fastened-in-place equipment (lighting and an exhaust fan) in that bathroom, provided the lighting and fan load does not exceed 50 % of the branch-circuit ampere rating as per NEC section 210.23(A)(2).

Rule#10: Outdoors (see below image) As per NEC section 210.52(E), Outdoor receptacle outlets shall be installed in accordance with the following: 1- for One-Family and Two-Family Dwellings For a one family dwelling and each unit of a two-family dwelling that is at ground level, at least one receptacle outlet accessible while standing at grade level and located not more than 2.0 m (6.5 ft) above grade shall be installed at the front and at the back of each dwelling. 2- for Multifamily Dwellings For each dwelling unit of a multifamily dwelling where the dwelling unit is located at grade level and provided with individual exterior entrance/ egress, at least one receptacle outlet accessible from grade level and not more than 2.0 m (6.5 ft) above grade shall be installed. 3- for Balconies, Decks, and Porches Balconies, decks, and porches that are accessible from inside the dwelling unit and having an overall area of 20 ft2 or more must have at least one receptacle outlet installed within its perimeter. This receptacle shall not be located more than 2.0 m (6.5 ft) above the balcony, deck, or porch surface.

Important!!! Where outdoor heating, air conditioning, or refrigeration (HACR) equipment is located at grade level , the receptacle outlets required by this section ―which shall be located on the same level‖ can be used to comply with the receptacle outlet requirement of 210.63, provided that at least one of the outlets is located within 25 ft of the HACR equipment. Important!!! The receptacle outlet required for Balconies, Decks, and Porches can also be used to meet the above outdoor receptacle requirements of One-Family, Two-Family and Multifamily Dwellings.

Rule#11: Basements, Garages, and Accessory Buildings (see below image) As per NEC section 210.52(G), For a one-family dwelling, the following provisions shall apply: At least one receptacle outlet, in addition to those for specific equipment (Laundry for example), shall be installed in each basement, in each attached garage, and in each detached garage or accessory building with electric power.

Where a portion of the basement is finished into one or more habitable rooms, each separate unfinished portion shall have a receptacle outlet installed in accordance with this section.

Important!!! Where detached garages are not supplied with electricity, receptacle outlets do not have to be installed.

1.2 Calculation of general-use receptacles load A- For feeder and service calculation purposes As I mentioned before in previous article that In one-family, two-family, and multifamily dwellings and in guest rooms or guest suites of hotels and motels, The general lighting load unit values specified in table 220.12 (which was 3VA/ft2) will include the following receptacle loads: 1. All general-use receptacle outlets of 20-ampere rating or less, including receptacles connected to Bathroom Branch Circuits, 2. The outdoor receptacle outlets, 3. General-use receptacle Outlets used in Basements, Garages, and Accessory Buildings.

So, No additional load calculation is required for general-use receptacle loads i.e. general use receptacle loads (after applying the general lighting load unit value from table 220.12) = Zero. Example#1: A homeowner wants to add 10 numbers general-purpose Receptacles, what is the additional load to the service? Solution:

No additional load calculation is required for these general-use receptacle loads, the additional load to be added = Zero.

B- For branch circuit requirements (conductor ampacity & size and over-current protection) calculation This will be explained in coming articles, however you can review the calculation for maximum number of allowable receptacles on a branch circuit and the minimum required number of branch circuits for a dwelling building which were discussed before in previous article " Receptacle Branch Circuit Design Calculations – Part Three ".

1.3 How to specify the required Type of general-use receptacles for each area? 1- Grounding type: As per NEC section 406.4, Receptacles installed on 15- and 20-ampere branch circuits shall be of the grounding type. 2- GFCI type: As per NEC section 210.8 (A), all 125-volt, single-phase, 15- and 20-ampere receptacles are GFCI for the following locations: A- Bathrooms (see image below Rule#9) There are no exceptions to the bathroom GFCI requirement. For example, if a washing machine is located in the bathroom, the 15- or 20-ampere, 125-volt receptacle that is required to be supplied from the laundry branch circuit must be GFCI protected. B- Garages and accessory buildings (see below image) It is the locations that have a floor located at or below grade level not intended as habitable rooms and limited to storage areas, work areas, and areas of similar use.

There are no exceptions to the GFCI requirements in Garages and accessory buildings, and this ensures that all 125-volt, single-phase, 15- and 20-ampere receptacles installed in garages are GFCI type regardless of where the receptacle is located in the garage.

C- Unfinished basements Unfinished basements are defined as portions or areas of the basement not intended as habitable rooms and limited to storage areas, work areas, and the like. All The receptacles in a work area of a basement, as shown in image below Rule#11, must be GFCI type.

Important!!! If the basement has some finished areas such as sleeping rooms or family rooms, then all receptacles in such areas will not required to be GFCI

type as shown in image below Rule#11.

D- Crawl spaces All receptacles installed in crawl spaces (at or below ground level) must be GFCI type. E- Kitchens I explained in above rule#5 that a general-use receptacle branch circuits must be added to kitchens to serve Receptacles installed for disposals, dishwashers, and trash compactors. These receptacles are not required to be GFCI type. The same is for receptacle(s) installed behind a refrigerator to supply that appliance, not the countertop, and will not be GFCI type.

F- Sinks located in areas other than kitchens (see below image) All 125-volt, 15- and 20-ampere receptacles that are within 6 ft of any point along the outside edge of the sink located in areas other than kitchens is required to be GFCI type.

G- Outdoors All outdoor receptacles are GFCI type with the condition that it is readily accessible. If it is not, it will be exempt from the GFCI requirements as shown in below image.

The dwelling unit shown has four outdoor receptacles, illustrates the requirement of 210.8(A) (3). The three receptacles with direct grade-level access must be provided with GFCI protection. The fourth receptacle, located adjacent to the gutter for the roof-mounted snow-melting cable, is not readily accessible and, therefore, is exempt from the GFCI requirements of 210.8(A)(3) because of its dedicated function to supply the deicing equipment.

H- Boathouses Boathouse is A shed adjacent to a house at the edge of a river or lake used for housing boats.

3- Weatherproof type As per NEC section 406.9, the weatherproof feature is allowed for the following locations: A- Damp locations All 15- and 20-ampere, 125- and 250-volt receptacles installed in a damp location shall have an enclosure that is weatherproof whether or not the attachment plug cap is inserted.

B- Wet locations All 15- and 20-ampere, 125- and 250-volt receptacles installed in wet locations shall have an enclosure for the receptacle that is weatherproof. A receptacle suitable for wet locations shall also be considered suitable for damp locations.

Important!!! Receptacle, located under roofed open porches, canopies, marquees, and the like, and will not be subjected to a beating rain or water runoff, shall be not weatherproof type. the below image summarize the weatherproof features in all cases as per NEC section 406.9.

4- Tamper-Resistant type Tamper-Resistant receptacles are intended to increase safety for children. All nonlocking type 125- volt, 15- and 20-ampere receptacles shall be listed tamper resistant receptacles. the below image show the method of operation for TamperResistant feature:

The Receptacles in the following locations shall not be required to be tamperresistant: 1. Receptacles located more than 1.7 m (51⁄2 ft) above the floor (these receptacles are not accessible and well out of reach of small children). 2. Receptacles that are part of a luminaire or appliance. 3. A single receptacle or a duplex receptacle located within dedicated space for appliance that, in normal use, is not easily moved from one place to another ( like dishwashers, refrigerators, washing machines, and the like )

5- AFCI type All 15- and 20-ampere, 120-volt branch circuits that supply outlets including receptacle, lighting, and other outlets located throughout a dwelling unit family rooms, dining rooms, living rooms, parlors, libraries, dens, bedrooms, sunrooms, recreation rooms, closets, hallways, or similar rooms or areas shall be AFCI type.Other outlets installed in kitchens, bathrooms, unfinished basements, garages, and outdoors will not be AFCI type.

Please, note the following:  Shared circuits between a bedroom and other areas such as closets and hallways must have AFCI protection.  There is no prohibition against using AFCI protection on other circuits or locations other than those specified in 210.12(A). The AFCI device to be located at the first outlet supplied by a branch circuit in the following cases:

1. If the wiring method from the branch-circuit overcurrent protective device to the outlet is metallic conduit or tubing. 2. If the wiring method from the branch-circuit overcurrent protective device to the outlet is Nonmetallic conduit or tubing that is encased in not less than 2 inch of concrete.

Important!!! Another method for applying AFCI protection for branch circuit that require such protection is by installing A listed combination-type AFCI located at the origin of the branch circuit (in panelboard for example).

The following table will summarize the required receptacle type for each location in dwelling units:

Specifying a Receptacle in Dwelling Buildings Receptacle Type Location GFCI AFCI** family rooms X √ dining rooms X √ living rooms X √ parlors X √ libraries X √ dens X √ bedrooms X √ sunrooms X √ recreation rooms X √ closets X √ hallways X √ kitchens, counter receptacles

√

X

kitchens, general-use receptacles

X

√

bathrooms unfinished basements garages outdoors

X √ √ √

X X X X

Boathouses √ X Sinks located in areas other than √ X kitchens Crawl spaces √ X All 15- and 20-ampere, 125- and 250-volt receptacles installed in dwelling unit are grounding and non-locking type. All 15- and 20-ampere, 125- and 250-volt receptacles installed in a damp and Wet Locations are Weatherproof type (see exception above). All non-locking type 125- volt, 15- and 20-ampere receptacles installed in dwelling units are Tamper-Resistant type (see exceptions above) ** in branch circuits that need AFCI protection only the first Receptacle will be AFCI type and other receptacles (on the same branch circuit) will be normal type.

Receptacle Branch Circuit Design Calculations – Part Five In the previous article " Receptacle Branch Circuit Design Calculations – Part Four ", I explained General-use Receptacle branch circuits. Also, in previous article " Receptacle Branch Circuit Design Calculations – Part Three ", I stated that a Receptacle in dwelling units may serve one of the following loads: 1. 2. 3. 4. 5. 6. 7. 8.

General-use Receptacle Loads, Small appliance Loads, Laundry Load, Fastened-in-place Appliance loads, Cloth dryer Load, Household cooking appliances load, Heating and air conditioning loads, Motor loads.

In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code.

You can review the following previous articles for more information:  

Receptacle Branch Circuit Design Calculations – Part One Receptacle Branch Circuit Design Calculations – Part Two

2- Small Appliances Branch circuits 2.1 Applied NEC Rules for small appliances branch circuits: There are many NEC rules that control the number, location and type of small appliances branch circuits including: 210.8 A(6) Ground-Fault Circuit-Interrupter Protection for Personnel in Kitchens,  210.11(C)(1) Required Branch Circuits in Dwelling Units for Small-Appliance Branch Circuits,  210.52(A)(4) Dwelling Unit Receptacle Outlets for Countertop Receptacles,  210.52(B) Dwelling Unit Receptacle Outlets for Small Appliances,  210.52(C) Dwelling Unit Receptacle Outlets for Countertops,  220.52 (A) Small-Appliance Circuit Load in Dwelling Unit. 

2.2 Rules controlling small appliances Branch Circuits:

Rule#1: number of small appliance branch circuits In each dwelling unit, two or more 20-ampere small-appliance branch circuits shall be provided.

Rule#2: areas served by small appliances branch circuits All Receptacle Outlets In the kitchen, pantry, breakfast room, dining room, or similar area of a dwelling unit which serve Portable appliances used at a kitchen counter, such as toasters, coffee makers, skillets, mixers, and the like and after applying exceptions made by rules#6.

Rule#3: Receptacle Outlets served by small appliances branch circuits (see fig.1) The following receptacle outlets are permitted to be served by small appliance branch circuits: 1. All wall and floor receptacle outlets covered by 210.52(A) after applying exceptions made by rules#6, 2. All countertop outlets covered by 210.52(C), 3. Receptacle outlets for refrigeration equipment (see Rle#4). 4. Electric clock in any of the areas specified in Rule#2, 5. Supplemental equipment and lighting on gas-fired ranges, ovens, or counter-mounted cooking units.

The small appliance branch circuits are not permitted to serve any other outlets, such as might be connected to exhaust hoods or fans, disposals, or dishwashers.

Rule#4: Receptacle outlet for refrigeration equipment (see fig.1) The receptacle outlet for refrigeration equipment located in a kitchen or similar area shall be permitted to be supplied from one of the following circuits: 1. 20-A small appliance branch circuit, 2. An individual branch circuit rated 15 amperes or greater especially, when the receptacle outlet for the refrigerator is located so that it cannot be used to serve countertop surfaces.

Fig.1

Rule# 5: Distribution of small-appliance branch circuits (see fig.1) Small-appliance branch circuits must supply the required receptacles by rule#3 in the following way: The counter area receptacle outlets in the kitchen are required to be supplied by minimum two small-appliance branch circuits.  The wall receptacle outlets in the kitchen and dining room are permitted to be supplied by one or both of the small-appliance branch circuits that supply the counter area.  If there are many kitchens in same dwelling unit, no small-appliance branch circuit shall serve more than one Kitchen. 

Rule#6: floor Receptacles (see fig.1) Receptacle outlets in floors shall not be counted as part of the required

number of general-purpose receptacle outlets unless located within 450 mm (18 in.) of the wall. Otherwise, these Floor Receptacles will be counted as part of the small appliance branch circuits.

Rule#7: kitchen and dining areas counters receptacles other than that used for small appliances (see fig.1) Receptacle outlets installed to serve kitchen or dining area counters (for small appliances) cannot also be used as general-use receptacles for an adjacent wall space and in this case general-use receptacle branch circuits must be added to serve such locations.

Rule#8:general-purpose Switched receptacles (see fig.1) Switched receptacles supplied from general-purpose 15-ampere branch circuits are permitted to be located in kitchens, pantries, breakfast rooms, and similar areas.

2.3 rules controlling distribution of receptacle outlets of countertop spaces Countertops may be found In kitchens, pantries, breakfast rooms, dining rooms, and similar areas of dwelling units, these countertops have receptacle outlets which will be distributed in the countertop space as per the following rules:

Rule#9: spacing of Countertop receptacles (see figs.2,3) A receptacle outlet shall be installed at each wall countertop space that is 300 mm (12 in.) or wider And spaced so that no point along the wall line is more than 600 mm (24 in.) measured horizontally from adjacent receptacle outlet in that space.

Fig.2

Rule#10: a sink, range, or counter-mounted cooking unit installed in Countertop (see figs.2,3) Receptacle outlets shall not be required on a countertop wall directly behind a range, counter-mounted cooking unit, or sink.

Fig.3

Rule#11: dividing a Countertop (see figs.2,3) If the space depth behind a sink, range, or counter-mounted cooking unit is 12 in. or more or 18 in. or more (depending on the counter configuration), the countertop is still one space, and in this case the wall length behind a sink, range, or counter-mounted cooking unit must be included in measuring the total wall counter space to calculate the number of required Receptacle(s) for this countertop. But if the space depth behind a sink, range, or counter-mounted cooking unit is less than 300 mm (12 in.), countertop will be considered as two separate standalone countertops and rules #1 will be applied for each separate countertop.

Important!!! Rule#11 will apply for island and peninsular countertop (see fig.4) with a short dimension of at least 12 in. and a long dimension of at least 24 in. Noting that the measurement of a peninsular- type countertop is from the edge connecting to the non-peninsular counter. Peninsular countertop definition: kitchen is simply an island anchored to a wall or line of cabinets. It may be used as a breakfast bar, seating area or just extra countertop and storage space.

Fig.4

Rule#12: Receptacle locations on or above Countertop Receptacle outlets shall be located on or above, but not more than 500 mm (20 in.) above, the countertop.

Rule#13: Receptacle locations below Countertop Receptacle outlets shall be permitted to be mounted not more than 300 mm (12 in.) below the countertop with the following conditions: 1. There are no means to mount a receptacle within 500 mm (20 in.) above the countertop, such as an overhead cabinet. 2. Receptacle will be Accessible for use, 3. The countertop edge (beside the wall) extends less than 150 mm (6 in.) beyond its support base. 4. The countertop is flat across its entire surface (no backsplashes, dividers, etc.)

Important!!! Multi-outlet assemblies (see fig.5) shall be permitted to be to be used as the required countertop receptacle outlet(s) and it will be installed as one of the following ways: 1. Installed directly on the countertop 2. Installed on the bottom of the upper cabinets

Fig.5

Rule#14: Guest Rooms and Guest Suites in dwelling units and that are provided with permanent provisions for cooking (see fig.6) Guest rooms and guest suites that are provided with permanent provisions for cooking shall have branch circuits installed in the cooking area to meet the above rules#1 to #13.

Fig.6

2.4 Calculation of general-use receptacles load A- For feeder and service calculation purposes: As per NEC section 220.52(A), in each dwelling unit, the load shall be calculated at 1500 volt-amperes for each 2-wire small-appliance branch circuit. this load will be added for other loads to perform the feeder and service calculations.

B- For branch circuit requirements (conductor ampacity & size and over-current protection) calculation This will be explained in coming articles, however you can review the calculation for maximum number of allowable receptacles on a branch circuit which was discussed before in previous article " Receptacle Branch Circuit Design Calculations – Part Three ".

2.5 How to specify the required Type of receptacles for small appliances branch circuits? 1- GFCI Type As per NEC section 210.8 A (6), where the receptacles are installed to serve the countertop kitchen appliances around a kitchen sink must be GFCI Type. See fig.7.

Fig.7

Important!!! I explained in rule#5 in previous article ―Receptacle Branch Circuit Design Calculations – Part Five― that a general-use receptacle branch circuits must be added to kitchens to serve Receptacles installed for disposals, dishwashers, and trash compactors. These receptacles are not required to

be GFCI type. The same is for receptacle(s) installed behind a refrigerator to supply that appliance, not the countertop, and will not be GFCI type.

Important!!! According to 406.5(E), receptacles installed to serve countertops cannot be installed in the countertop in the face-up position because liquid, dirt, and other foreign material can enter the receptacle.

2- Tamper-Resistant type Tamper-Resistant receptacles are intended to increase safety for children. All nonlocking type 125- volt, 15- and 20-ampere receptacles shall be listed tamper resistant receptacles. The Receptacles in the following locations shall not be required to be tamper-resistant:

Receptacles located more than 1.7 m (51⁄2 ft) above the floor (these receptacles are not accessible and well out of reach of small children).  

Receptacles that are part of a luminaire or appliance.

A single receptacle or a duplex receptacle located within dedicated space for appliance that, in normal use, is not easily moved from one place to another ( like dishwashers, refrigerators, washing machines, and the like ) 

Receptacle Branch Circuit Design Calculations – Part Six In article " Receptacle Branch Circuit Design Calculations – Part Three ", I stated that a Receptacle in dwelling units may serve one of the following loads: 1. General-use Receptacle Loads, 2. Small appliance Loads, 3. Laundry Load, 4. Cloth dryer Load, 5. Fastened-in-place Appliance loads, 6. Household cooking appliances load, 7. Heating and air conditioning loads, 8. Motor loads.

I explained the first two types in the following articles:  

Receptacle Branch Circuit Design Calculations – Part Four Receptacle Branch Circuit Design Calculations – Part Five

In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code. You can review the following articles for more information:  Receptacle Branch Circuit Design Calculations – Part One 

Receptacle Branch Circuit Design Calculations – Part Two

3- Laundry Load Branch circuits

Definition: As per NEC section 550.2 Laundry Area is An area containing or designed to contain a laundry tray, clothes washer, or a clothes dryer. 3.1 Applied NEC Rules for Laundry branch circuit There are many NEC rules that control the location and load of Laundry branch circuit including:     

210.11 (C)(2) Required Laundry Branch Circuits in Dwelling Units 210.50 (C) Appliance Receptacle Outlets 210.52 (F) Laundry Areas 220.52 (B) Small-Appliance and Laundry Loads in Dwelling Unit 210.23 (A) Permissible Loads for 15- and 20-Ampere Branch Circuits

3.2 Rules controlling Laundry Branch Circuit

Rule#1: Number of Laundry branch circuits In each dwelling unit, At least one 20-ampere branch circuit shall be

provided to supply the laundry receptacle outlet(s).

Rule#2: Dedicated usage of Laundry branch circuits The laundry branch circuits required in a dwelling unit(s) by shall supply only the receptacle outlets specified for Laundry use and this circuit shall have no other outlets supplying other loads.(see below image)

Rule#3: Multifamily dwellings, Laundry in public area

As per NEC section 210.25(A), Branch circuits in each dwelling unit shall supply only loads within that dwelling unit or loads associated only with that dwelling unit. So, In multifamily dwellings building, if Laundry facilities are provided on the premises and are available to all building tenants (as a common usage). the laundry branch circuit will not be added to each individual dwelling unit. This Laundry branch circuits for public or common areas are required to be supplied from a separate ―house load‖ panelboard. This requirement permits access to the branch-circuit disconnecting means without the need to enter the space of any tenants. The requirement also prevents a tenant from turning off important circuits that may affect other tenants.

Rule#4: special case, no laundry facilities permitted As per Rule#3, In other than one-family dwellings where laundry facilities are not to be installed or permitted, a laundry receptacle shall not be required.

Rule#5: Location of receptacle outlet used for Laundry Appliance receptacle outlets installed in a dwelling unit for specific appliances, such as laundry equipment, shall be installed within 1.8 m (6 ft) of the intended location of the appliance. 3.3 Calculation of Laundry load A- For feeder and service calculation purposes:

Rule#6: Laundry branch circuit load In each dwelling unit, a load of not less than 1500 volt-amperes shall be included for each 2-wire laundry branch circuit installed. Where additional laundry branch circuits are provided, they also are calculated at 1500 volt-amperes per circuit. Important!!!

If Rule#3 above is applicable, then no Laundry load will be added for feeder calculation of each dwelling unit and the Laundry branch circuits load will be added to the ―house load‖ panelboard.

B- For branch circuit requirements (conductor ampacity & size and over-current protection) calculation This will be explained in coming articles, however you can review the calculation for maximum number of allowable receptacles on a branch circuit which was discussed before in previous article " Receptacle Branch Circuit Design Calculations – Part Three ".

3.4 How to specify the required Type of receptacles for Laundry branch circuits? A- GFCI Type The type of 15- or 20-ampere, 125-volt receptacle that is required to be supplied from the laundry branch circuit will depend on the washing machine location as follows: aIf the washing machine is located in the bathroom: As per NEC section 210.8 A (1), There are no exceptions to the bathroom GFCI requirement. So, the 15- or 20-ampere, 125-volt receptacle must be GFCI protected. b- If the washing machine is located in the Laundry Area or in a kitchen: The 15- or 20-ampere, 125-volt receptacle will not be GFCI protected.

B- Tamper-Resistant type Tamper-Resistant receptacles are intended to increase safety for children. A single receptacle or a duplex receptacle located within dedicated space for appliance that, in normal use, is not easily moved from one place to another ( like dishwashers, refrigerators, washing machines, and the like ) shall not be required to be tamper-resistant.

4- Clothes Dryer Load 4.1 Applied NEC Rules for Clothes Dryer Load

There are many NEC rules that control the location and load of Clothes Dryer including:   

210.50 (C) Appliance Receptacle Outlets 220.54 Electric Clothes Dryers in Dwelling Unit(s) 220.53 Appliance Load — Dwelling Unit(s)

4.2 Rules controlling Clothes Dryer Load

Rule#1: Location of receptacle outlet used for Clothes Dryer Appliance receptacle outlets installed in a dwelling unit for specific appliances, such as laundry equipment, shall be installed within 1.8 m (6 ft) of the intended location of the appliance.

Rule#2: Circuit ratings for Clothes Dryer High-wattage cord-and-plug-connected loads, such as electric ranges, clothes dryers, and some window air conditioners, may be connected to a 208-volt or 240-volt 3 phase, 4 wire circuit. Same can be done by using single phase multi-wire circuits that can supply both line-to-line and line-to-neutral connected loads as in below image.

Rule#3: Not Fastened-in-Place Appliances As per NEC section 220.53, electric ranges, clothes dryers, space-heating equipment or air conditioning equipment must not be included with the number of appliances that are fastened in place. Also, All portable small Appliances for kitchen and others are not Fastenedin-Place Appliances.

4.3 Calculation of Clothes dryer load A- For feeder and service calculation purposes:

Important!!! If a dwelling unit doesn’t include a clothes dryer, no Load for clothes dryer will be added for feeder calculation of this dwelling unit.

Important!!! A combination of clothes washer and clothes dryer (see below image) will be handled in calculations as it is a clothes dryer. But the branch circuit and 125V receptacle outlet for laundry is still required i.e. minimum Laundry requirements is mandatory while clothes dryer requirements is not.

Important!!! Kilovolt-amperes (kVA) shall be considered equivalent to kilowatts (kW) for clothes dryer loads. First: As per NEC Standard calculation method:

Rule#4: clothes dryers load in NEC Standard calculation method As per NEC section 220.54, the load for household electric clothes dryers in a dwelling unit(s) shall be either 5000 watts (volt-amperes) or the nameplate rating, whichever is larger, for each dryer served.

Important!!! The NEC does not prohibit applying the full load of all dryers to a service and/or feeder calculation especially when all the clothes dryers are in a common Laundry area in a multi-family dwelling because It is possible that all dryers will be operating at one time in a laundry facility. But when each tenant has separate clothes dryer, it is unlikely that all dryers will be in operation simultaneously. So, Table 220.54 Demand Factors for Household Electric Clothes Dryers will be used.

In multifamily dwelling, where two or more single-phase dryers are supplied by a 3-phase, 4-wire feeder or service, follow the below steps: Step#1: Assign the maximum number of dryers connected between any two phases.  Step#2: The total number of clothes dryers between two phases = 2 x the maximum number of clothes dryers connected between any two phases ---- (Result#1)  Step#3: 2-phase Total ―connected load‖ for clothes = totals of all the clothes dryers loads x (Result#1) ----- (Result#2)  Step#4: Based on The total number of clothes dryers between two phases (Result#1), assign the demand factor from Table 220.54.  Step#5: 2-phase Total ―demand load‖ for clothes dryers = (Result#2) X Demand factor ----- (Result#3)  Step#6: service 3-phase Total ―demand load‖ = (Result#3) X 3/2 

Example#1: What is the minimum service load for a 4.7 KW combination clothes washer and dryer? Solution: A combination of clothes washer and clothes dryer will be handled in calculations as it is a clothes dryer. So, the minimum service load for a 4.7 KW combination clothes washer and dryer = 5000 watts = 5 KW

Example#2: In a multi-family dwelling with 10 dwelling units, assuming the load of 10 single-phase dryers is connected as evenly as possible to the 3-phase system (3 dryers connected between phases A and B, 3 dryers connected between phases B and C, and 4 dryers connected between phases A and C), as in below image.

Solution: 3-phase Dryer load without applying the 220.54 demand factor = 5500 VA x 10 = 55,000 VA Step#1: The maximum number of dryers connected between any two phases is 4 Step#2: Twice the maximum number of dryers connected between any two phases is used as the basis for calculating the demand load = 4 dryers x 2 = 8 dryers Step#3: 2-phase Total ―connected load‖= 8 x 5500 VA = 44,000 VA Step#4: from Table 220.54, for 8 dryers, demand factor = 60% Step#5: 2-phase Total ―demand load‖= 44,000 VA x 0.6 = 26,400 VA (connected between two phases of the 3-phase system) Step#6: service 3-phase Total ―demand load‖ = 26,400 VA x 3/2 = 39,600 VA

Second: As per NEC Optional calculation method:

Rule#5: Application of NEC Optional calculation method NEC Optional calculation method will be used if the following condition is verified: 1. The service-entrance or feeder conductors have an ampacity of at least 100 amperes.

Important!!! If the service-entrance ampacity calculated by the optional method is less than 100A, re-calculate with using the standard method.

Important!!! In NEC Optional calculation method, for a multifamily dwelling, Table 220.84 ―Optional Calculations — Demand Factors for Three or More Multifamily Dwelling Units‖ will be used if the following conditions are verified: 1. If No dwelling unit is supplied by more than one feeder 2. Each dwelling unit is equipped with electric cooking equipment. 3. Each dwelling unit is equipped with either electric space heating or air conditioning, or both.

Important!!! The optional calculation can be used, provided all of the conditions for using table 220.84 listed above are met. Otherwise, the calculation for the multifamily dwelling is performed by using standard calculation method.

Important!!! In Optional calculation method, note the following: The dryer load is not subject to an individual demand factor as required by

table 220.54 for NEC Standard calculation method. The nameplate of the clothes dryer will be used, even if it is less than 5000 watts.

In a multi-family dwelling, for Optional calculation method, follow the below steps: Step#1: Total ―connected load‖ for clothes dryers = sum of all nameplate rating of clothes dryers loads in each dwelling unit---- (Result#1).  Step#2: Based on the total number of dwelling units in this building, assign the demand factor from Table 220.84.  Step#3: 3-phase service Total ―demand load‖ for clothes dryers = (Result#1) X Demand factor. 

Example#3: What is the one-family dwelling load contribution for a 4.5KW clothed dryer? Solution: The minimum load using the NEC standard calculation method for a clothes dryer = 5000 watts The load using the NEC optional calculation method for a clothes dryer = nameplate rating = 4500 watts

Example#4: Find the solution for the same multi-family dwelling building in example#1 by using the NEC optional calculation method. Solution: Step#1: Total ―connected load‖ for clothes dryers =5500 VA X 10 = 55,000 VA Step#2: from Table 220.84, For 10 dwelling units, the demand factor = 43 %. Step#3: 3-phase service Total ―demand load‖ for clothes dryers= 55,000 VA x 0.43 = 23,650 VA Service ampacity = 23,650VA /(208 V X1.732) = 66 A < 100 A, So the calculation must be repeated using NEC standard calculation method.

B- For branch circuit requirements (conductor ampacity & size and over-current protection) calculation This will be explained in coming articles, however you can review the calculation for maximum number of allowable receptacles on a branch circuit which was discussed before in previous article " Receptacle Branch Circuit Design Calculations – Part Three ".

4.4 How to specify the required Type of receptacles for Clothes dryer load? Same as that in paragraph 3.5 for Laundry receptacles.

Receptacle Branch Circuit Design Calculations – Part Seven In article " Receptacle Branch Circuit Design Calculations – Part Three ", I stated that a Receptacle in dwelling units may serve one of the following loads: 1. 2. 3. 4. 5. 6. 7. 8.

General-use Receptacle Loads, Small appliance Loads, Laundry Load, Cloth dryer Load, Household cooking appliances load, Fastened-in-place Appliance loads, Heating and air conditioning loads, Motor loads.

I explained the first four types in the following articles:



Receptacle Branch Circuit Design Calculations – Part Four Receptacle Branch Circuit Design Calculations – Part Five



Receptacle Branch Circuit Design Calculations – Part Six



In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code. You can review the following articles for more information: 

Receptacle Branch Circuit Design Calculations – Part One



Receptacle Branch Circuit Design Calculations – Part Two

5- Household cooking appliances load

Definition: Cooking Unit, Counter-Mounted is a cooking appliance designed for mounting in or on a counter and consisting of one or more heating elements, internal wiring, and built-in or mountable controls.(see below image)

Wall-mounted ovens and counter-mounted cooking units complete shall be permitted to be permanently connected or, only for ease in servicing or for installation, cord-and-plug-connected. 

A portable cooking appliance (e.g., cord-and-plug-connected microwave oven or hot plate) is not a permanent cooking facility. 

5.1 Applied NEC Rules for Household cooking appliances load There are many NEC rules that control the location and load of Household cooking appliances including: NEC section 220.53  210.18 Guest Rooms and Guest Suites - dwelling units  210.19 (A) (3) Minimum Ampacity and Size of conductors for Branch Circuits Not More Than 600 Volts supplying Household Ranges and Cooking Appliances  210.50 General (C) Appliance Receptacle Outlets  210.52(B) Small Appliances (2) No Other Outlets Exception No. 2  210.23 Permissible Loads (C) 40- and 50-Ampere Branch Circuits  220.55 Electric Ranges and Other Cooking Appliances — Dwelling Unit(s) 

Rule#1: Location of receptacle outlet used for Electric Ranges Appliance receptacle outlets installed in a dwelling unit for specific appliances, such as laundry equipment, shall be installed within 1.8 m (6 ft) of the intended location of the appliance. Receptacle outlets for Wall-mounted ovens and counter-mounted cooking units shall be located on or above, but not more than 500 mm (20 in.) above, the countertop.

Rule#2: Guest Rooms and Guest Suites in dwelling units and that are provided with permanent provisions for cooking(see below image) Guest rooms and guest suites that are provided with permanent provisions for cooking shall have branch circuits installed in the cooking area to meet the all of the requirements for dwelling units contained in Parts I, II, and III of Article 210.

Important!!! A portable cooking appliance (e.g., cord-and-plug-connected microwave oven or hot plate) is not a permanent cooking facility.

Rule#3: Not Fastened-in-Place Appliances As per NEC section 220.53, electric ranges, clothes dryers, space-heating equipment or air conditioning equipment must not be included with the number of appliances that are fastened in place. Also, All portable small Appliances for kitchen and others are not Fastenedin-Place Appliances.

Rule#4: Supplying Supplemental equipment and lighting Receptacle outlets of small appliance branch circuits are permitted to serve Supplemental equipment and lighting on gas-fired ranges, ovens, or countermounted cooking units.

Important!!! In most dwellings the countertop receptacle outlets supply more of the portable cooking appliances than the wall receptacles in the kitchen and dining areas, hence the requirement for the counter areas to be supplied by no fewer than two small-appliance branch circuits.

5.2 Calculation of Household Cooking Appliances load A- For feeder and service calculation purposes First: As per NEC Standard calculation method

Rule#5: Code Requirement for Household Cooking Appliances Household Cooking Appliances (ranges, wall mounted ovens, countermounted cooking appliances, etc.) are not required in a NEC load calculation. We can skip the calculation of Household Cooking Appliances Load if there are no cooking appliances rated over 1.75 KW.

Important!!! Kilovolt-amperes (kVA) shall be considered equivalent to kilowatts (kW) for Electric Ranges and Other Cooking Appliances.

Important!!! When the kilowatt rating fraction is 0.5 or more, it must be rounded up to the next whole kilowatt rating i.e. 14.5 KW up to a 15 KW and When the

fraction is less than 0.5, it can be dropped i.e. 14.4 KW dropped to 14 KW.

Rule#6: calculating household cooking appliances load in accordance with Table 220.55. As per NEC section 220.55, The load for household electric ranges, wallmounted ovens, counter-mounted cooking units, and other household cooking appliances individually rated in excess of 1.75 kW shall be permitted to be calculated in accordance with Table 220.55.

Important!!! The table 220.55 is not applicable for ranges rated more than 27 kW

because ranges rated more than 27 kW would not be considered household ranges. We have (5) cases for using table 220.55 as follows: Case#1: Individual Appliance not over 12 KW The service load for one household cooking appliance is permitted to be computed by using either the nameplate rating of the appliance or Table 220.55. In this case, Procedure for calculating the service demand load will be as follows: Step#1: Look in Left column of Table 220.55 for the number of appliances = 1 Step#2: Follow the row (for number of appliances = 1) across to the appropriate column A, B or C.  Column A is used where the rating of the appliance is less than 3½ kW  Column B is used where the rating of the appliance is 3½ kW through 8¾ Kw  Column C is used where the rating of the appliance not over 12-kW. Step#3: Service demand load calculation If Column A or B is used, Service demand load = Load of the Individual Appliance x Demand factor  If Column C is used, Service demand load = Max. Demand Load in column C 

Example#1: 12-kW electric range will be installed in a one-family dwelling, what is the service demand load for this range? Solution:

Step#1: Look in the left column of Table 220.55 for the number of appliances=1. Step#2: Follow the row across to the appropriate column which will be Column C Step#3: In accordance with Column C, a 12-kW range has a maximum demand of 8 kW

Example#2: What is the service demand load for a 9-kW range? Solution: Step#1: Look in the left column of Table 220.55 for the number of appliances=1. Step#2: Follow the row across to the appropriate column which will be Column C Step#3: In accordance with Column C, The maximum load required for one 9-kW range when calculating a service or feeder is 8 kW.

Example#3: What is the service demand load for a 3-kW, wall-mounted oven?

Solution:

Step#1: Look in the left column of Table 220.55 for the number of appliances=1. Step#2: Follow the row across to the appropriate column which will be Column A Step#3: In accordance with Column A, The demand factor percent = 80% So, the service demand Load= 3 KW X 0.8 = 2.4 KW

Case#2: group of Appliances with equal (same) ratings not over 12 KW The procedure is the same as in case#1 for individual Appliance as follows: Step#1: Look in Left column of Table 220.55 for the number of appliances, Step#2: Follow the row (for number of appliances given) across to the appropriate column A, B or C. Column A is used where the rating of the appliances (have same rating) is less than 3½ kW.  Column B is used where the rating of the appliances (have same rating) is 3½ kW through 8¾ KW  Column C is used where the rating of the appliances (have same rating) is not over 12-kW. 

Step#3: Service demand load calculation If Column A or B is used, Service demand load = Number of Appliances x Load of Individual appliance x Demand factor  If Column C is used, Service demand load = Max. Demand Load in column C 

Example#4: A 10-unit apartment building will have a 12-kW range in each apartment. What is the service demand load these ranges? Solution:

Step#1: Look in the left column of Table 220.55 for the number of appliances=10. Step#2: Follow the row across to the appropriate column which will be Column C Step#3: In accordance with Column C, for the number of appliances=10, the maximum service demand will be 25 kW

Example#5:

What is the service demand load for 40 wall-mounted ovens with same 3-kW rating? Solution:

Step#1: Look in the left column of Table 220.55 for the number of appliances=25. Step#2: Follow the row across to the appropriate column which will be Column A Step#3: In accordance with Column A, The demand factor percent = 30% So, the service demand Load= Number of Appliances x Load of Individual appliance x Demand factor = 25 x 3 KW X 0.3 = 22.5 KW

Example#6: What is the service demand load for 40 ranges with same 10-kW rating? Solution:

Step#1: Look in the left column of Table 220.55 for the number of appliances=40. Step#2: Follow the row across to the appropriate column which will be Column C Step#3: In accordance with Column C, for the number of appliances=40, the maximum service demand will be 15 kW + 1 kW for each range. So, the maximum service demand = 15 kW + number of appliances = 15 + 40 = 55 KW

Example#7: What is the service demand load for 60 ranges with same 12-kW rating? Solution:

Step#1: Look in the left column of Table 220.55 for the number of appliances=60. Step#2: Follow the row across to the appropriate column which will be Column C Step#3: In accordance with Column C, for the number of appliances=60, the maximum service demand will be 15 kW + 0.75 kW for each range. So, the maximum service demand = 25 kW + number of appliances X 0.75 KW = 25+ 60 X 0.75 = 70 KW

Case#3: group of Appliances with unequal ratings not over 12 KW The procedure is the same as in case#2 for individual Appliance but Steps#1, 2 and 3 will be repeated for each Appliance as follows: Step#1: Look in Left column of Table 220.55 for the number of appliances, Step#2: Follow the row (for number of appliances given) across to the appropriate column A, B or C. Column A is used where the rating of the appliances (have same rating) is less than 3½ kW.  Column B is used where the rating of the appliances (have same rating) is 3½ kW through 8¾ KW  Column C is used where the rating of the appliances (have same rating) is not over 12-kW. 

Step#3: Service demand load calculation If Column A or B is used, Service demand load = Number of Appliances x Load of Individual appliance x Demand factor  If Column C is used, Service demand load = Max. Demand Load in column C 

Step#4: Sum individual service demand loads to get the total service demand load.

Example#8: What is the service demand load for ten 2.5-kW ovens, ten 3-kW ovens, ten 5-kW cooktops and ten 6-kW cooktops? Solution: First: for 20 units that are within the limits of Column A Step#1:The 2.5-kW ovens have a rating = 10 × 2½ = 25 KW. The 3-kW ovens have a rating = 10 × 3 = 30KW. The ovens have a total rating = 25 + 30 = 55 KW Step#2: Because 20 ovens fall within the limits of Column A, the demand factor is 35 %. Step#3: The calculated demand for the ovens = 55 × 35% = 19.25 KW Second: for 20 units that are within the limits of Column B Step#1: The 5-kW cooktops have a rating = 10 × 5 = 50 KW The 6-kW cooktops have a rating = 10 × 6 = 60 KW The cooktops have a total rating = 50 + 60 = 110 KW Step#2: Because 20 ovens fall within the limits of Column B, the demand factor is 28 %. Step#3: The calculated demand for the cooktops = 110 × 28% = 30.8 KW Step#4: Now add the results to find the total demand load = 19.25 + 30.8 = 50.05 KW

Example#9: What is the service demand load for five 3.5-kW wall-mounted ovens, five 5-kW counter-mounted cooking units and ten 8-kW ranges? Solution: Step#1: the total number of units (5 + 5 + 10 = 20). Step#2: Because none of the appliances are rated less than 3½ kW or more than 8¾ kW, the demand factor will come from Column B. So, the demand factor percent across from 20 units (28 percent). Step#3: The ovens have a total rating of 17½ kW = 5 × 3.5 = 17.5KW The cooktops have total rating of 25 kW = 5 × 5 = 25 KW The ranges have a total rating of 80 kW = 10 × 8 = 80 KW Step#4: The combined rating of all the appliances = 17.5 + 25 + 80 = 122.5 KW The service demand load = 122.5 KW × 28% = 34.3 KW

Optional Calculation for Cases# 2 and 3 and appliances ratings Over 1 3 ⁄4 kW through 8 3⁄4 kW If appliances Ratings falling under Column A or B and C (ratings Over 1 3 ⁄4 kW through 8 3⁄4 kW), the calculation procedure will be as follows: Step#1: Calculate the service demand Load for such appliances whether falling under Column A or B Step#2: Calculate the same using Column C, Step#3: Choose the lowest value between steps# 1 and 2 to be the service demand Load.

Example#10: What is the service demand load for 6 ranges with same 8-kW rating? Solution: Using the optional method for household cooking appliances with ratings Over 1 3 ⁄4 kW through 8 3⁄4 kW Step#1: Calculate the service demand Load for such appliances whether falling under Column A or B In accordance with column B, the demand factor for 6 nos. appliances is 43% , so the service demand load = 6 x 8 x 0.43 = 20.64 KW Step#2: Calculate the same using Column C, In accordance with column C, for 6 nos. appliances, the Max. demand load = 21 KW Step#3: Choose the lowest value between steps# 1 and 2 to be the service demand Load. So, the service demand Load = the lowest value between steps# 1 and 2 = 20.64 KW

Case#4: Individual / group of Ranges with equal (same) ratings Over 12 kW through 27 KW For ranges individually rated more than 12 kW but not more than 27 kW, the calculation procedure will be as follows: Step#1: Calculate the maximum demand for the given number of ranges in Column C. Step#2: Find the KW Load increase over 12 KW = KW of one range – 12 Step#3: Find the increased demand amount = Step#1 x Step#2 x 0.05 Step#4: Total service demand load= step#1 +step#3

Example#11:

What is the service demand load for one 14-kW household electric range? Solution: Step#1: the maximum demand for one ranges in Column C = 8 KW Step#2: Load increase over 12 KW = 14 – 12 = 2 KW Step#3: the increased demand amount = Step#1 x Step#2 x 0.05 = 8x2x0.05 = 0.8 KW Step#4: Total service demand load= step#1 +step#3 = 8 + 0.8 = 8.8 kW

Case#5: group of Ranges with unequal ratings Over 8 3⁄4 kW through 27 kW For ranges individually rated more than 8 3⁄4 kW and of different ratings, but none exceeding 27 kW, the calculation procedure will be as follows: Step#1: Calculate the maximum demand for the given number of ranges in Column C. Step#2: Calculate the Average Rating = sum of all appliances rating (using 12 kW for any range rated less than 12 kW) / number of appliances Step#3: Find the KW Load increase over 12 KW = average load KW – 12 Step#4: Find the increased demand amount = Step#1 x Step#3 x 0.05 Step#5: Total service demand load= step#1 +step#4

Example#12: What is the service demand load for five 13-kW, five 15-kW and five 17-kW household electric ranges? Solution: Step#1: the maximum demand for 15 ranges in Column C = 30 KW Step#2: the Average Rating = [(5 × 13) + (5 × 15) + (5 × 17) / 15 = 15 KW Step#3: Load increase over 12 KW = 15 – 12 = 3 KW Step#4: the increased demand amount = Step#1 x Step#3 x 0.05 = 30x3x0.05 = 4.5 KW Step#5: Total service demand load= step#1 +step#3 = 30 + 4.5 = 34.5 kW

Example#13:What is the service demand load for eight 9-kW, six 13-kW and six 15-kW household electric ranges? Solution: Step#1: the maximum demand for 20 ranges in Column C = 35 KW Step#2: the Average Rating = [(8 × 12) + (6× 13) + (6 × 15) / 20 = 13.2 KW Notes:  

Use 12 kW for any range rated less than 12 kW 13.2 KW will be dropped to 13 KW

Step#3: Load increase over 12 KW = 13 – 12 = 1 KW Step#4: the increased demand amount = Step#1 x Step#3 x 0.05 = 35x1x0.05 = 1.75 KW Step#5: Total service demand load= step#1 +step#3 = 35 + 1.75 = 36.75 kW

Second: As per NEC Optional calculation method:

Rule#7: Application of NEC Optional calculation method NEC Optional calculation method will be used if the following condition is verified: 1. The service-entrance or feeder conductors have an ampacity of at least 100 amperes.

Important!!! If the service-entrance ampacity calculated by the optional method is less than 100A, re-calculate with using the standard method.

Important!!! In NEC Optional calculation method, for a multifamily dwelling, Table 220.84 ―Optional Calculations — Demand Factors for Three or More

Multifamily Dwelling Units‖ will be used if the following conditions are verified: 1. If No dwelling unit is supplied by more than one feeder 2. Each dwelling unit is equipped with electric cooking equipment. 3. Each dwelling unit is equipped with either electric space heating or air conditioning, or both.

Rule#8: For multifamily dwellings without electric cooking Range For multifamily dwellings without electric cooking range (using gas cooking ranges or other and may use electrical range in future) do the following: 1. Calculate the load of multifamily dwellings as per NEC standard method 2. Calculate the load of multifamily dwellings as per NEC optional method with adding electric cooking (based on 8 kW per unit) 3. Select the lowest load.

Important!!! The optional calculation can be used, provided all of the conditions for using table 220.84 listed above are met. Otherwise, the calculation for the multifamily dwelling is performed by using standard calculation method. Rule#9: Calculation of Household cooking appliances as per NEC optional method In Optional calculation method, note the following: 1. The Household cooking appliances load is not subject to an individual demand factor as required by table 220.55 for NEC Standard calculation method. 2. The nameplate load of the Household cooking appliances (fastened in place or permanently connected) will be used.

Example#14:

A multifamily dwelling has 20 dwelling units, One-half of the dwelling units are equipped with electric ranges not exceeding 12 kW each and other half of ranges are gas ranges. What is the service demand load for these ranges? Solution: 1- Calculated Load for Each Dwelling Unit In accordance with column C in table 220.55, Max. demand load for One Electric range = 8KW 2- Calculate the service demand load Step#1: Look in the left column of Table 220.55 for the number of appliances=10. Step#2: Follow the row across to the appropriate column which will be Column C Step#3: The maximum service demand load = 25 KW

B- For branch circuit requirements calculation 1- Determining Branch-Circuit Load

Important!!! The branch-circuit load for one household cooking appliance is permitted to be computed by using either the nameplate rating of the appliance or Table 220.55.

Important!!! Where counter-mounted cooking appliances like the one pictured in below image are used with a separate wall oven, it is permissible to run a single branch circuit, sized according to Note 4 to Table 220.55, to the kitchen and supply each with branch-circuit tap conductors installed as specified in 210.19(A)(3), Exception No. 1.

Important!!! If a single branch circuit supplies a counter-mounted cooking unit and not more than two wall-mounted ovens, all of which are located in the same room, the nameplate ratings of these appliances can be added and the total treated as the equivalent of one range, according to Note 4 of Table 220.55.

Rule#10: Circuit ratings for Electric Ranges

High-wattage cord-and-plug-connected loads, such as electric ranges, clothes dryers, and some window air conditioners, may be connected to a 208-volt or 240-volt 3 phase, 4 wire circuit. Same can be done by using single phase multi-wire circuits that can supply both line-to-line and line-to-neutral connected loads as in below image.

Important!!! Where two or more single-phase ranges are supplied by a 3-phase, 4-wire feeder or service, the total load shall be calculated on the basis of twice

the maximum number connected between any two phases.

Rule#11: household cooking appliances branch circuit rating As per 210.19(A)(3) Branch circuit conductors supplying household ranges, wall-mounted ovens, counter-mounted cooking units, and other household cooking appliances shall have an ampacity not less than the rating of the branch circuit and not less than the maximum load to be served. For ranges of 8 3⁄4 kW or more rating, the minimum branch-circuit rating shall be 40 amperes. the permissible rating For Wall-mounted ovens and counter-mounted cooking units to be connected to 15- and 20-Ampere Branch Circuits is as follows: 1. If it is not fastened-in-place, it can have a rating of up to 80 % of the branch circuit rating as in TABLE 210.21(B)(2). 2. If it is fastened-in-place, other than luminaires, it shall not exceed 50 % of the branch-circuit ampere rating.

Important!!! Exception to Rule#9, Conductors tapped from a 50-ampere branch circuit supplying electric ranges, wall-mounted electric ovens, and countermounted electric cooking units shall have an ampacity of not less than 20 amperes and shall be sufficient for the load to be served. As illustrated in fig. , this exception permits a 20-ampere tap if the following four conditions are met: 1. The taps are not longer than necessary to service or permit access to the junction box. 2. The taps to each unit are properly spliced. 3. The junction box is adjacent to each unit. 4. The taps are of sufficient size for the load to be served.

Household cooking appliances branch circuit conductor ampacity & size and overcurrent protectionwill be explained in coming articles, however you can review the calculation for maximum number of allowable receptacles on a branch circuit which was discussed before in previous article " Receptacle Branch Circuit Design Calculations – Part Three ".

5.4 How to specify the required Type and Rating of receptacles for household cooking appliances? A- Range Receptacle Rating The ampere rating of a range receptacle shall be permitted to be based on a single range demand load as specified in Table 220.55. B- GFCI Type As I explained in Rule#4 that a Receptacle outlet connected to small appliance branch circuits is permitted to serve Supplemental equipment and lighting on gas-fired ranges, ovens, or counter-mounted cooking units. As per NEC section 210.8 A (6), where the receptacles are installed to serve the countertop kitchen appliances (counter-mounted cooking units) around a kitchen sink must be GFCI Type. See below image.

But the receptacle serves Supplemental equipment and lighting on gas-fired ranges and ovens will not be GFCI type as in above image. According to 406.5(E), receptacles installed to serve countertops cannot be installed in the countertop in the face-up position because liquid, dirt, and other foreign material can enter the receptacle C- Tamper-Resistant type

Tamper-Resistant receptacles are intended to increase safety for children. The Receptacles in the following locations shall not be required to be tamperresistant: Receptacles located more than 1.7 m (51⁄2 ft) above the floor (these receptacles are not accessible and well out of reach of small children).  A single receptacle or a duplex receptacle located within dedicated space for appliance that, in normal use, is not easily moved from one place to another ( like dishwashers, refrigerators, washing machines, and the like ). 

Branch Circuit Design Calculations – Part Eight 3RT5QY4V42SU

In article " Receptacle Branch Circuit Design Calculations – Part Three ", I stated that a Receptacle in dwelling units may serve one of the following loads: 1. 2. 3. 4. 5. 6. 7. 8.

General-use Receptacle Loads, Small appliance Loads, Laundry Load, Cloth dryer Load, Household cooking appliances load, Fastened-in-place Appliance loads, Heating and air conditioning loads, Motor loads.

I explained the first five types in the following articles:



Receptacle Branch Circuit Design Calculations – Part Four Receptacle Branch Circuit Design Calculations – Part Five



Receptacle Branch Circuit Design Calculations – Part Six



Receptacle Branch Circuit Design Calculations – Part Seven



In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code. You can review the following articles for more information: 

Receptacle Branch Circuit Design Calculations – Part One



Receptacle Branch Circuit Design Calculations – Part Two

6- Fastened-in-place Appliances branch circuits

Definition: The term appliance designates utilization equipment commonly built in standardized types and sizes and installed as a unit to perform specific function(S) such as clothes washing, air conditioning, food mixing, deep frying, etc. Some examples of appliances that are fastened in place whether direct wired or cord and plug connected include dishwashers, kitchen-waste disposers, trash compactors, attic fans and water heaters (see below image).

Definitions:  Appliance, Fixed is An appliance that is fastened or otherwise secured at a specific location.  Appliance, Portable is An appliance that is actually moved or can easily be moved from one place to another in normal use.  Appliance, Stationary is An appliance that is not easily moved from one place to another in normal use.

Rule#1: Not Fastened-in-Place Appliances (see below image) As per NEC section 220.53, electric ranges, clothes dryers, space-heating equipment or air conditioning equipment must not be included with the number of appliances that are fastened in place. Also, All portable small Appliances for kitchen and others are not Fastenedin-Place Appliances.

Rule#2: Location of receptacle outlet used for Electric Appliances Appliance receptacle outlets installed in a dwelling unit for specific appliances, such as laundry equipment, shall be installed within 1.8 m (6 ft) of the intended location of the appliance.

Important!!! The load for an appliance is simply the ampere rating of that appliance which may be indicated on its nameplate and will be used in branch, feeder and service load calculations.

6.1 Applied NEC Rules for Fastened-in-place Appliances load

There are many NEC rules that control the location and load of Fastened-inplace appliances including: 210.23(A) Permissible Loads for 15- and 20-Ampere Branch Circuits  220.14 (A) Specific Appliances or Loads in All Occupancies  220.53 Appliance Load in Dwelling Unit  220.82(B)(3) Optional Feeder and Service Load Calculations of General Loads in Dwelling Unit 

6.2 Calculation of Fastened-in-place Appliances load A- For feeder and service calculation purposes First: As per NEC Standard calculation method:

Important!!! Kilovolt-amperes (kVA) shall be considered equivalent to kilowatts (kW) for Fastened-in-place Appliances.

Rule#3: Calculation of Fastened-in-place Appliances load As per NEC section, 220.53, It shall be permissible to apply a demand factor of 75 % to the nameplate rating load of four or more appliances fastened-inplace, that are served by the same feeder or service in a one-family, twofamily, or multifamily dwelling.

Important!!! No derating is allowed when there are only one, two, three fastened-inplace appliances.

Example#1: A one-family dwelling will have two feeders supplied by the service. One feeder will

provide power to a water heater and two attic ventilation fans. The other feeder will provide power to a dishwasher, a kitchen-waste disposer and a trash compactor. Where the 75 % demand factor to be applied? Solution: Since only three appliances are on each feeder, it is not permissible to apply the 75 % demand factor to either feeder load calculation. The demand factor can be applied to the service load calculation because it will supply power to six fastened-in-place appliances.

Example#2: What is the appliance load contribution, for a house that has a 1.2 KVA dishwasher, a 0.76 KVA garbage disposer, a 1.1 KVA trash compactor and a 4.5 KVA water heater? Solution: Step#1: find the total load or the Fastened-in-place Appliances Total load = 1.2+0.76+1.1+4.5 = 7.56 KVA Step#2: if the number of the Fastened-in-place Appliances is four or more, apply the 75% demand factor. Demand load = 7.56 x 75% = 5.67 KVA

Rule#4: Calculation of Fastened-in-place Appliances load For Multifamily Dwelling In a multifamily dwelling, the four or more fastened-in-place appliances do not have to be on the same feeder for each dwelling unit. In this case, the 75% demand factor will not apply to the feeder for each dwelling unit but it must be applied to the multifamily dwelling service.

Example#3: A 4.5-kilowatt (kW), 240-volt (V) water heater will be installed in each unit of a four-

unit multifamily dwelling. The water heaters are the only fastened-in-place appliance in this multifamily dwelling. One service will supply all four units. Using the standard method, what is the service load (after applying the demand factor) for fastened-inplace appliances? Solution: Each unit has only one fastened-in-place appliance, the 75% demand factor will not apply to the feeder for each dwelling unit. But there will be four fastened-in-place appliances on the multifamily dwelling service. Therefore, applying the 75 percent demand factor is permissible. The service load for these water heaters = 4x4.5x0.75 = 13.5 KW

Second: As per NEC Optional calculation method:

Rule#5: Application of NEC Optional calculation method NEC Optional calculation method will be used if the following condition is verified: 1. The service-entrance or feeder conductors have an ampacity of at least 100 amperes.

Important!!! If the service-entrance ampacity calculated by the optional method is less than 100A, re-calculate with using the standard method.

Important!!! In NEC Optional calculation method, for a multifamily dwelling, Table 220.84 ―Optional Calculations — Demand Factors for Three or More Multifamily Dwelling Units‖ will be used if the following conditions are verified: 1. If No dwelling unit is supplied by more than one feeder 2. Each dwelling unit is equipped with electric cooking equipment.

3. Each dwelling unit is equipped with either electric space heating or air conditioning, or both.

Important!!! The optional calculation can be used, provided all of the conditions for using table 220.84 listed above are met. Otherwise, the calculation for the multifamily dwelling is performed by using standard calculation method.

Rule#6: Calculation of Fastened-in-place Appliances load as per NEC Optional calculation method In Optional calculation method, note the following: 1. The Fastened-in-place Appliances load, regardless of the number, will not be derated as required per NEC section 220.53. 2. The nameplate of the Fastened-in-place Appliances that are fastened in place, permanently connected or located to be on a specific circuit will be used.

Important!!! The fastened-in-place appliances Load including water heater is given, do not include water heat load to the general load as per 220.82(B)(3)d . But if the water heater was not included with the fastened-in-place appliances, add the nameplate rating of the water heater to the general loads covered in 220.82(B). Example#4: what is the optional method service load calculation for a one-family dwelling with the following fastened-in-place appliances: a dishwasher rated 10 amperes (A) at 120 volts (V); a horsepower, 120V kitchen waste disposer; a trash compactor rated 7.5A at 120V; and a 4.5 kilowatts (kW), 240V water heater? Solution: The loads in KVA will be as follows:

The dishwasher load = 10A x120V = 1,200 VA = 1.2 KVA Kitchen waste disposer load = 9.8A x 120V = 1,176 VA = 1.176 KVA The compactor load = 7.5A x 120V = 900 VA = 0.9 KVA The water heater load = 4,500W = 4,500 VA = 4.5 KVA The optional method service load = 1.2 + 1.176 + 0.9 + 4.5 = 7.776 KVA

Example#5: Find the optional method service load calculation for a 12-unit multifamily dwelling with each unit containing the following fastened-in-place appliances: a dishwasher rated 10 amperes (A) at 120V; 9.8 A, 120V kitchen waste disposer, and a 4.5 kW, 240V water heater. Solution: The loads in KVA will be as follows: The dishwasher load = 10A x 120V = 1,200 VA. The kitchen waste disposer load = 9.8A x 120V = 1,176 VA. Water heater load = 4.5 KW x1,000 = 4,500 W = 4500 VA The fastened-in-place appliance load for each unit = 1,200 + 1,176 + 4,500 = 6,876 VA Since there are 12 units, multiply the appliance load by 12 The fastened-in-place appliance service load for the building = 12 x 6,876 VA = 82,512 VA Apply the Table 220.84 (in below image) for demand factor, of Multifamily dwelling, since there are 12 units, the demand factor will be 41%. The optional method demand load for this building = 82,512 VA x 41% = 33,830 VA

B- For branch circuit requirements (conductor ampacity & size and over-current protection) calculation

Rule#7: Combination of Fastened-in-place Appliances with other loads on a branch circuit

As per NEC section 210.23(A), The total rating of utilization equipment fastened in place, other than luminaires, shall not exceed 50 percent of the 15- and 20-Ampere Branch Circuits branch-circuit rating where lighting units, cord-and-plug-connected utilization equipment not fastened in place, or both, are also supplied.

Rule#8: Rating of Fastened-in-place Appliances Outlet As per NEC section 220.14 (A), an outlet for a specific appliance or other load not covered in 220.14(B) through (L) shall be calculated based on the ampere rating of the appliance or load served.

Example#6: What is the branch-circuit load for a 10-ampere, 120-volt dishwasher in a dwelling? Solution: Since the rating on the nameplate is 10 amperes, the branch-circuit load for this dishwasher is 10 amperes.

Important!!! Using fastened-in-place equipment is not permitted for the small-appliance branch circuits, laundry branch circuits, and bathroom branch circuits required in a dwelling unit and these branch circuits shall supply only their receptacle outlets required by the code.

Important!!! Small loads, such as those of 1440 volt-amperes or less and motors of less than 1⁄4 horsepower, are limited to 120- volt circuits.

Branch Circuit Design Calculations – Part Nine In article " Receptacle Branch Circuit Design Calculations – Part Three ", I stated that a Receptacle in dwelling units may serve one of the following loads:

1. 2. 3. 4. 5. 6. 7. 8.

General-use Receptacle Loads, Small appliance Loads, Laundry Load, Cloth dryer Load, Household cooking appliances load, Fastened-in-place Appliance loads, Heating and air conditioning loads, Motor loads.

I explained the first six types in the following articles:



Receptacle Branch Circuit Design Calculations – Part Four Receptacle Branch Circuit Design Calculations – Part Five



Receptacle Branch Circuit Design Calculations – Part Six



Receptacle Branch Circuit Design Calculations – Part Seven Branch Circuit Design Calculations – Part Eight





In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code. You can review the following articles for more information: 

Receptacle Branch Circuit Design Calculations – Part One



Receptacle Branch Circuit Design Calculations – Part Two

7- Heating and air conditioning loads

Definitions: A heat pump (see below image) is a device that acts as an air conditioner in the summer and as a heater in the winter. Heat pumps look and function exactly like an air conditioner except it has a reversible cycle.

7.1 Applied NEC Rules for Heating and air conditioning loads

There are many NEC rules that control the Heating and air conditioning loads including:       

220.50 Motors 220.51 Fixed Electric Space Heating 220.60 Noncoincident Loads 220.82(C) Heating and air conditioning loads in Dwelling Unit 220.83 Existing Dwelling Unit 220.85 two family dwelling 430.24 Several Motors or a Motor(s) and other Load(s)

 

440.3 Other Articles 440.6 Ampacity and Rating

7.2 Calculation of Heating and air conditioning loads A- For feeder and service calculation purposes

Important!!! Most service load calculations will include heating and/or air conditioning equipment, but not all feeder load calculations will include these types of loads. If the feeder will not supply power to heating and air conditioning equipment, calculate just the general loads on this feeder. If a service will not supply heating equipment calculate only the service for air condition only. If a service will not supply power to heating and air conditioning equipment, ignore this load in service load calculation.

First: As per NEC Standard calculation method

Rule#1: Service load for Room air conditioners The load for Room air conditioners shall be calculated at 100 % of its ampere rating which may be indicated on its nameplate and will be used in branch, feeder and service load calculations.

Rule#2: Service load for Fixed electric space-heating loads As per NEC section 220.51, Fixed electric space-heating loads shall be calculated at 100 percent of the total connected load. However, in no case shall a feeder or service load current rating be less than the rating of the largest branch circuit supplied.

Example#1:

A dwelling unit has seven wall heaters; each heater is rated 3,000 watts at 240 volts. How much load will these heaters add to a 240-volt, single-phase service? Assuming that The air conditioning load will be less than the heating load. Solution: In accordance with 220.51, calculate the heaters service load at 100 %= 7 × 3,000 = 21,000 watts Since the service voltage is known, the total current draw of the heaters can be calculated by dividing the total watts by 240 volts The total current draw of the heaters = 21,000 ÷ 240 = 87.5 A = 88 A

Rule#3: Central air conditioning and heating system Load Central air conditioning and heating system Load shall be calculated at 100 % of its nameplate and will be a Noncoincident Load.

Rule#4: Noncoincident Loads As per NEC section 220.60, where it is unlikely that two or more noncoincident loads will be in use simultaneously, it shall be permissible to use only the largest load(s) that will be used at one time for calculating the total load of a feeder or service. Important!!! Depending on the design, the heating system and the air conditioning system might be noncoincident loads.

Example#2: What is the service load contribution for a dwelling that will have electric space heaters and window air conditioners? The heat will consist of two rooms with space heaters rated 1,500 watts each and three rooms with space heaters rated 2,250 watts each.

The air conditioning will consist of four rooms with window air conditioning units rated 11.5 amperes at 240 volts. Assume space-heating watts are equivalent to volt-amperes (VA). Solution: In this dwelling, the heat load and the air conditioning load are noncoincident loads. The heat and the air conditioning will not be energized at the same time. Therefore, compare the total heat load to the total air conditioning load and omit the smaller of the two loads. First: calculate the heat load The total heat load = 1,500 + 1,500 + 2,250 + 2,250 + 2,250 = 9,750 VA. Second: calculate the air conditioning load Start by finding the volt-amperes of each unit (VA = E × I) (assume power factor or PF = 1.0). The load of each air conditioner = 240 × 11.5 = 2,760 VA The total air conditioner load = 2,760 VA × 4 = 11,040 VA Since the air conditioner load is larger than the heat load, omit the heat load. The service load contribution for the heating and air conditioning loads in this dwelling is 11,040 VA If the air conditioner compressors are the largest motors in this dwelling, multiply the load of one compressor by 25 percent and add it to the service load calculation.

Rule#5: the air handler (or blower motor) is not a noncoincident load Although the heating and air conditioning in package units and split systems are noncoincident loads, the air handler (or blower motor) (or evaporator motor) is not. Since the blower motor works with both the heating and air conditioning system, it must be included in both calculations. Example#3: A heating/cooling package unit will be installed in a one-family dwelling. The electric heater is rated 9.6 kW at 240 volts. The blower motor inside the package unit that circulates the air will be a ½ horsepower, 240-volt motor. How much load will this package unit add to a 240-volt, single-phase service? Assuming that the air conditioning load will be less than the heating load. Solution:

The heat load= 9.6 KW × 1,000 = 9,600 watts The full-load current in amperes of a ½ hp, single-phase, 240-volt motor (from Table 430.248) = 4.9 A The motor load = 4.9 A × 240 V = 1,176 VA The service load for this package unit = 1,176 + 9,600 = 10,776 watts

Example#4: A package unit has electric heat and air conditioning. The unit contains a compressor, a blower motor, a condenser fan motor and electric heat. The compressor draws 23 amperes at 240 volts, the blower motor draws 5 amperes at 240 volts and the condenser fan motor draws 2 amperes at 240 volts. The rating of the heat is 10 kW. Assume heating kilowatt rating is equivalent to kilovolt-amperes. What is the service load contribution for this heating/cooling package unit? Solution: The load of the air conditioner compressor =23 A × 240 V= 5,520 VA The load of the blower motor = 5 A × 240 V = 1,200 VA The load of the condenser fan motor = 2 A × 240 V = 480 VA The total air conditioning load = 5,520 + 1,200 + 480 = 7,200 VA Since the blower motor also works with the heat, the load of the blower motor must be added to the heat load. So, the heat load = 10 KW× 1,000 = 10,000 + 1,200 = 11,200 VA Since the heat load is larger than the air conditioner load, omit the air conditioner load. The service load contribution for this package unit is 11,200 VA.

Rule#6: Largest Motor in the feeder or service load calculation As per NEC sections 220.50 and 430.24, when calculating a feeder or service, the largest motor must be multiplied by 125 percent.

Important!!! Unless it is the largest motor in the feeder or service load calculation, do not multiply the full-load current of the motor by 125 percent.

Example#5:

In example#3, the blower motor in the heating/cooling package unit in the last example will be the largest motor in the one-family dwelling. How much load will this package unit add to a 240-volt, single-phase service? Solution: If the ½ hp blower motor is the largest motor in the calculation for the service, the ampacity must not be less than 125 percent of the full-load current rating plus the calculated load of the electric heat. Multiply the motor’s full-load current by 125 percent before adding it to the electric heat service load The motor load = 4.9 A × 240 V = 1,176 VA The service load for this package unit = 1,176 x 1.25 + 9,600 = 11,070 watts

Rule#7: A heat pump with supplementary heat is not a noncoincident load With a heat pump, the compressor (and accompanying motors) and some or all of the electric heat can be on at the same time. The load contribution of a heat pump is the air conditioning system load plus the maximum amount of heat that can be on while the air conditioner compressor is on.

Example#6: What is the service load contribution for a heat pump with supplementary heat? The compressor draws 26.4 amperes at 240 volts, the blower motor draws 6.5 amperes at 240 volts and the condenser fan motor draws 3 amperes at 240 volts. The electric heat in this unit has a rating of 15 kW. Assume heating kilowatt rating is equivalent to kilovolt-amperes.

Solution: The compressor and all of the heat in this heat pump can be energized at the same time. The load of the air conditioner compressor = 26.4 × 240 = 6,336 VA The load of the blower motor = 6.5 × 240 = 1,560 VA The load of the condenser fan motor = 3 × 240 = 720 The total air conditioning load = 6,336 + 1,560 + 720 = 8,616 VA

Since the air conditioning system and all of the heat can be on at the same time, add the two together to get the service load contribution. The service load contribution for this heat pump = 8,616 + 15,000 = 23,616 VA

Important!!! Noncoincident loads are not limited to heating and air conditioning systems.

Example#7: What is the feeder load contribution for two 5-hp, 230-volt, single phase pump motors? The motors will be wired so only one motor is in operation at any time. Solution: Because these two motors are noncoincident loads, it is permissible to omit the load of one motor. Start by finding the full-load current (FLC) of one motor. Full-load currents for single-phase, alternating-current motors are in Table 430.248. From Table 430.248, The FLC of this motor is 28 amperes. The load of one motor = 28 A × 230 V = 6,440VA Although there are two motors, they will never operate at the same time. Therefore, only one motor must be added to the feeder calculation. The feeder load contribution = 6,440 VA If this motor is the highest-rated motor on this feeder, multiply the load of this motor by 25 percent, and add it to the feeder load calculation.

Second: As per NEC Optional calculation method

Rule#8: Application of NEC Optional calculation method NEC Optional calculation method will be used if the following condition is verified: 1. The service-entrance or feeder conductors have an ampacity of at least 100 amperes.

Important!!! If the service-entrance ampacity calculated by the optional method is less than 100A, re-calculate with using the standard method.

Important!!! In NEC Optional calculation method, for a multifamily dwelling, Table 220.84 ―Optional Calculations — Demand Factors for Three or More Multifamily Dwelling Units‖ will be used if the following conditions are verified: 1. If No dwelling unit is supplied by more than one feeder 2. Each dwelling unit is equipped with electric cooking equipment. 3. Each dwelling unit is equipped with either electric space heating or air conditioning, or both.

Important!!! The optional calculation can be used, provided all of the conditions for using table 220.84 listed above are met. Otherwise, the calculation for the multifamily dwelling is performed by using standard calculation method.

Rule#9: Application of NEC Optional calculation method NEC Optional calculation method is applicable only for a single dwelling unit, an existing dwelling unit, a multifamily dwelling, two dwelling units, a school, an existing installation and a new restaurant.

Rule#10: Application of NEC Optional calculation method NEC Optional calculation method is applicable only for a single dwelling unit, an existing dwelling unit, a multifamily dwelling, two dwelling units, a school, an existing installation and a new restaurant.

I will explain the calculation of Heating and Air-Conditioning service and feeder loads for dwelling units according to the type of dwelling unit as follows: 1. 2. 3. 4.

A single dwelling unit, A multifamily dwelling, A two dwelling units, An existing dwelling unit.

First: for single dwelling units

Rule#11: Heating and Air-Conditioning Load as per NEC Optional calculation method As per NEC section 220.82 (C), for Heating and Air-Conditioning Load, The largest of the following six selections (load in kVA) shall be included: 1. 100 percent of the nameplate rating(s) of the air conditioning and cooling. 2. 100 percent of the nameplate rating(s) of the heat pump when the heat pump is used without any supplemental electric heating. 3. 100 percent of the nameplate rating(s) of the heat pump compressor and 65 percent of the supplemental electric heating for central electric space-heating systems. If the heat pump compressor is prevented from operating at the same time as the supplementary heat, it does not need to be added to the supplementary heat for the total central space heating load. 4. 65 percent of the nameplate rating(s) of electric space heating if less than four separately controlled units. 5. 40 percent of the nameplate rating(s) of electric space heating if four or more separately controlled units. 6. 100 percent of the nameplate ratings of electric thermal storage (ETS) and other heating systems where the usual load is expected to be continuous at the full nameplate value.

Case#1: If a dwelling has some type of heat other than electric

In this case, calculate the air conditioning load only at 100 percent of the nameplate rating. 

Example#8: A central heating and air conditioning unit will be installed in a one-family dwelling. The air conditioning system is electric, and the heating system is gas. The air conditioner compressor has a rating of 16.6 amperes (A) at 230 volts (V). The condenser fan motor has a rating of 2A at 115V and the air-handler (blower motor) has a rating of 3.2A at 115V. Calculating by the optional method, how much air conditioning load will be added to the general loads?

Solution: First: calculating the volt-ampere (VA) load for each motor. The compressor load =16.6 A x 230 V = 3,818 VA The condenser fan load = 2 Ax 115 V = 230VA The air handler load = 3.2 A x 115 V = 368VA Second: add the air conditioning loads to find the total The service load = 3,818 + 230 + 368 = 4,416 VA Because air conditioning loads only exist, it will be calculated at 100 percent.

Important!!! Where there is more than one air conditioning unit, the calculation method is the same as it is for one air conditioning unit.

Example#9: Four window air conditioners will be installed in a one-family dwelling. The heating system is not electric. Two units are rated 12.5A at 230V and the other two are rated 8.5 amperes at 230 volts. Calculating by the optional method, how much air conditioning load will be added to the general loads? Solution:

Each of the larger air conditioners has a load = 12.5A x 230 V = 2,875 VA Each of the smaller window units has a load = 8.5A x 230 V = 1,955 VA The total load for all four units = 2,875 + 2,875 + 1,955 + 1,955 = 9,660 VA The multiplication factor for four air conditioning units is the same as it is for one unit at 100% The service load for four air conditioning units = 9,660 VA x 100% = 9,660 VA

Case #2: Heat pumps equipped with or without electric supplemental heat Supplemental heat is sometimes referred to as auxiliary, backup or even emergency heat. A dual-fuel heat pump is an electric heat pump and a gas furnace all in one. Dual fuel heat pumps can be fueled with natural gas or propane. Because geothermal heat pumps (sometimes referred to as geo-exchange, earth-coupled, ground-source or water-source heat pumps) do not depend on the temperature of the outside air, they may or may not be equipped with supplemental heat.  Heat pumps not equipped with supplemental electric heat are calculated exactly the same as the air conditioning equipment specified in case#1. 

Case #3: When a heat pump is used with supplemental electric heat In this case, multiply the nameplate rating(s) of the heat pump compressor by 100 percent and multiply the supplemental electric heating for central electric spaceheating systems by 65 percent. 

Example#10: A heat pump with supplemental electric heat will be installed in a one-family dwelling. The-2 1/2 ton heat pump system has nameplate rating of 24A at 240V. This heat pump is equipped with 15 kilowatts (kW) of backup heat. Calculating by the optional method, how much heating and air conditioning load will be added to the general loads?

Solution: The heat pump load =24A x 240V = 5,760VA The backup or supplemental heat= 15 kW x 1,000 = 15,000 VA Multiply the backup heat by 65 percent. The backup or supplemental heat= 15,000 x

65% = 9,750 VA The total heating and air conditioning load = 5,760 + 9,750 = 15,510 VA

Important!!! Some heat pumps are designed so that only part of the electric heat will operate while the compressor is in operation. If the heat pump compressor is prevented from operating at the same time as the supplementary heat, it does not need to be added to the supplementary heat for the total central space-heating load.

Example#11: A heat pump with supplemental electric heat will be installed in a one-family dwelling. The 3-ton heat pump system has nameplate rating of 30A at 240V. Although this heat pump is equipped with 15 kW of backup heat, it is designed so that only 10 kW can operate while the compressor is running. The other 5 kW of electric heat will only operate when the compressor is in the off position. Calculating by the optional method, how much heating and air conditioning load will be added to the general loads?

Solution: The heat pump load = 30A x 240 V = 7,200 VA Because only part of the backup heat can operate while the compressor is running, multiply the load of this part by 65 percent. The backup heat load = 10,000VA x 65% = 6,500 VA The total heating and air conditioning load = 7,200 + 6,500 = 13,700 VA

Case#4: electric space heating by less than four separately controlled units In this case, 65 percent of the nameplate rating(s) of electric space heating if less than four separately controlled units. 

Example#12:

A total of three electric wall heaters will be installed in a one-family dwelling. Two of the units are rated 3,000 watts (W) at 240 volts (V), and one unit is rated 4,800W at 240V. Each wall heater is separately controlled. This dwelling does not have an air conditioning load. Calculating by the optional method and after applying the appropriate demand factors, what is the heating load for these three electric space heaters? Solution: The total rating for all three heaters = 3,000 + 3,000 + 4,800 = 10,800 VA Because there are less than four separately controlled space heating units, multiply the total rating by 65 percent. The total rating for all three heaters = 10,800VA x 65% = 7,020 VA

Case#5: electric space heating by four or more separately controlled units In this case, 40 percent of the nameplate rating(s) of electric space heating if four or more separately controlled units. 

Example#13: Six separately controlled electric wall heaters will be installed in a one-family dwelling. Two heaters are rated 1,500W at 120V, one heater is rated 2,000W at 240V, two heaters are rated 3,000W at 240V and one heater is rated 4,000W at 240V. This dwelling does not have an air conditioning load. Calculating by the optional method and after applying the appropriate demand factors, what is the heating load for these six electric space heaters? Solution: The total rating for all six heaters = 1,500 + 1,500 + 2,000 + 3,000 + 3,000 + 4,000 = 15,000 VA Because there are at least four separately controlled space heating units, multiply the total rating by 40 percent. The total rating for all six heaters = 15,000 x 40% = 6,000 VA

Case#6: using electric thermal storage (ETS) and other heat systems Some electric utilities offer a discounted rate for kilowatt-hours used during certain hours of the day and night. During off-peak hours, customers pay rates that are less than rates during peak hours. During off-peak hours, electric elements are used to heat ceramic bricks. The stored heat is then released throughout the day when rates are higher. Many installations include multiple units placed throughout the house.  Since all of the electric thermal storage units could be heating bricks at the same time, no demand factor can be applied to the units.  In this case, 100 percent of the nameplate ratings of electric thermal storage (ETS) and other heat systems where the usual load is expected to be continuous at the full nameplate value. 

Example#14: Four electric thermal storage units will be installed in a one-family dwelling. Two units are rated 5.4 kilowatts (kW) at 240V and two units are rated 7.2 kW at 240V. This dwelling does not have an air conditioning load. Calculating by the optional method and after applying the appropriate demand factors, what is the heating load for this dwelling? Solution: The total rating for all four units = 5.4 + 5.4 + 7.2 + 7.2 = 25.2 KVA Because electric thermal storage system loads are calculated at 100 percent, 25.2 kVA will be added to service load.

Rule#12: Noncoincident Loads As per NEC section 220.60, where it is unlikely that two or more noncoincident loads will be in use simultaneously, it shall be permissible to use only the largest load(s) that will be used at one time for calculating the total load of a feeder or service.

Important!!! Depending on the design, the heating system and the air conditioning

system might be noncoincident loads.

Important!!! Similarly, 220.82(C) requires that only the largest of the six choices needs to be included in the feeder or service calculation.

Example#15: Three window air conditioners and five separately controlled electric wall heaters will be installed in a one-family dwelling. One window unit is rated 11.5 amperes at 240V and the other two are rated 9 amperes at 240V. Two wall heaters are rated 2,000W at 240V, two heaters are rated 3,000W at 240V and one heater is rated 4,000W at 240V. Calculating by the optional method, how much heating and air conditioning load will be added to the general loads? Solution: The larger air conditioner has a rating =11.5A x 240 V= 2,760 VA Each smaller window unit has a rating = 9A x 240 V = 2,160 VA The total load for all three units = 2,760 + 2,160 + 2,160 = 7,080 VA Since the air conditioning load is calculated at 100 percent, the total air conditioning load = 7,080 x 100% = 7,080 VA The total rating for all five heaters = 2,000 + 2,000 + 3,000 + 3,000 + 4,000 = 14,000 VA Because there are at least four separately controlled space heating units, multiply the total rating by 40 percent. The total heating load = 14,000 x 40% = 5,600 VA Because the heating and air conditioning systems in this example will not run simultaneously, it is only required to include the larger of the two loads. The larger load, after applying demand factors, is the air conditioning load. Add service load = air conditioning load = 7,080 VA

Branch Circuit Design Calculations – Part Ten In article " Receptacle Branch Circuit Design Calculations – Part Three " I stated that a Receptacle in dwelling units may serve one of the following loads: 1. 2.

General-use Receptacle Loads, Small appliance Loads,

3. 4. 5. 6. 7. 8.

Laundry Load, Cloth dryer Load, Household cooking appliances load, Fastened-in-place Appliance loads, Heating and air conditioning loads, Motor loads.

I explained the first six types in the following articles:



Receptacle Branch Circuit Design Calculations – Part Four Receptacle Branch Circuit Design Calculations – Part Five



Receptacle Branch Circuit Design Calculations – Part Six



Receptacle Branch Circuit Design Calculations – Part Seven Branch Circuit Design Calculations – Part Eight





In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code. You can review the following articles for more information: 

Receptacle Branch Circuit Design Calculations – Part One



Receptacle Branch Circuit Design Calculations – Part Two

7- Heating and air conditioning loads – Part Two

In the previous Article " Heating and air conditioning loads – Part One ", I explained the following points:

 

1. Applied NEC Rules for Heating and air conditioning loads 2. Feeder and service calculation of Heating and air conditioning loads by two methods: First: As per NEC Standard calculation method Second: As per NEC Optional calculation method

The rules applied in second method differ from a dwelling unit type to another. I explained the rules for first dwelling type ―A single dwelling unit‖ in above previous Article.

Today, I will continue explaining the rules for feeder and service calculation of Heating and air conditioning loads as per NEC Optional calculation method for other dwelling unit types which are: 1. 2. 3.

A multifamily dwelling, A two dwelling units, An existing dwelling unit,

Second: Multifamily Dwelling

Rule#1: Table 220.84 for Multifamily Dwelling demand factors As per NEC section 220.84, for Multifamily Dwelling, the demand factors of Table 220.84 shall be applied to the larger of the air-conditioning load or the fixed electric space-heating load.

Rule#2: Calculation of feeder and service loads for air conditioning load in Multifamily Dwelling With the optional method multifamily dwelling load calculation, the air conditioning load is calculated at 100 % of the nameplate rating.

Important!!! The air conditioning load is calculated the same way in multifamily dwellings as it is in one-family dwellings. In both types, the air conditioning load is calculated at 100 percent of the nameplate rating.

Rule#3: Calculation of feeder and service loads for electric space heating load in Multifamily Dwelling With the optional method multifamily dwelling load calculation, the load for space heating units must be added to the calculation at the nameplate rating.

Important!!! The electric space heating load is not calculated the same way in multifamily dwellings as it is in one-family dwellings. With the optional method one-family dwelling load calculation, it is permissible to apply a demand factor to space heating units. The demand factor depends on the number of units.

Don’t Forget… When calculating a multifamily dwelling by the optional method, use the larger of the air conditioning loads or the fixed electric space-heating load.

Example#1: For a 30-unit multifamily dwelling that have electric space heaters and window air conditioners as follows: The heating for each unit will consist of two separately controlled wall heaters rated 2,250W each and one heater rated 1,500W. The air conditioning for each unit will consist of two window air conditioning units rated 11.5A at 240V What is the optional method service load calculation (before applying the Table 220.84 demand factor)? (Assume space-heating watts are equivalent to volt-amperes).

Solution: First: calculate the heat load for each unit by adding the nameplate rating of heat units The heat load for each unit = 2,250 + 2,250 + 1,500 = 6,000 VA So, the total heat load for 30 units = 6,000 VA x 30 = 180,000 VA Second: calculate the air conditioning load for each unit by Multiplying volts, amperes and the number of air conditioners in each unit The air conditioning load for each unit = 240V x 11.5A x 2 = 5,520 VA So, the total air conditioning load for 30 units = 5,520 VA x 30 = 165,600 VA Third: compare between the heating load and the air conditioning load for all units

Since the heating load is larger than the air conditioning load, omit the air conditioning load. The optional method service load calculation (before applying the Table 220.84 demand factor) for the heating and air conditioning loads in this 30-unit multifamily dwelling = 180,000 VA

Important!!! Do not assume the heating load will always be larger than the air conditioning load. The heating system could be gas or oil. The dwelling could also be located in a warm climate where the air conditioning load is larger than the heating load.

Example#2: For a 12-unit multifamily dwelling that has an identical central heating and air unit for each dwelling unit as follows: Each package unit will contain a compressor, a blower motor, a condenser fan motor and electric heat. The compressor draws 23A at 240V, the blower motor draws 5A at 240V, and the condenser fan motor draws 2A at 240V. The rating of the electric heat is 5 kW. What is the optional method service load calculation (before applying the Table 220.84 demand factor)? (Assume the heating kilowatt rating is equivalent to kilovoltamperes). Solution: First: calculate the air conditioning load for each unit and then for all units The load of each air conditioner compressor =23A x 240 V = 5,520VA The load of each blower motor =5 A x 240 V = 1,200 VA The load of each condenser fan motor = 2A x 240 V = 480VA So, the air conditioning load for each unit = 5,520 + 1,200 + 480 = 7,200 VA And the total air conditioning load for 12 units = 7,200 VA x 12 = 86,400 VA Second: calculate the heat load for each unit and then for all units Since the blower motor also works with the heat, add the load of the blower motor to

the heat load The load of blower motor = 5 x 1,000 VA = 5,000 VA So, the heating load for each unit = 5,000 + 1,200 = 6,200 VA And the total heating load for 12 units = 6,200 x 12 = 74,400 VA Third: compare between the heating load and the air conditioning load for all units In this example, the air conditioning load is larger than the heating load; therefore, omit the heating load. The optional method service load calculation (before applying the Table 220.84 demand factor) for the heating and air conditioning system in this 12-unit multifamily dwelling = 86,400 VA

Important!!! Fixed electric space heating is not limited to space heaters and electric strip heat that is part of a package unit or furnace. Fixed electric space heating could also be a heat pump with supplementary heat.

Rule#4: If the Fixed electric space heating is a heat pump with supplementary heat With a heat pump, the compressor (and accompanying motors) and some or all of the electric heat can be energized at the same time. So, The load contribution of a heat pump = the air conditioning system load + the maximum amount of heat that can be on while the air conditioner compressor is energized.

Example#3: For a six-unit multifamily dwelling that has an identical heat pump for each dwelling unit as follows: Each unit’s compressor draws 26.8A at 240V, the blower motor draws 5.8A at 240V, and the condenser fan motor draws 2.6A at 240V. The electric heat in this unit has a rating of 15 kW. The compressor and all of the heat in this heat pump can be energized at the same time. What is the optional method service load calculation (before applying the Table 220.84 demand factor)? Assume the heating kilowatt rating is equivalent to kilovolt-amperes.

Solution: First: calculate the air conditioning and heat loads for each heat pump The load of the air conditioner compressor = 26.8 A x 240 V= 6,432 VA The load of the blower motor = 5.8 A x 240 V= 1,392 VA The load of the condenser fan motor = 2.6A x 240 V = 624 VA So, the air conditioning load for each unit = 6,432 + 1,392 + 624 = 8,448 VA Since the air conditioning system and all of the heat can be on at the same time, add the two together The air conditioning and heat loads for each heat pump = 8,448 + 15,000 = 23,448 VA Second: calculate the air conditioning and heat loads for all heat pumps The total load for six units = 23,448VA x 6 = 140,688 VA The optional method service load calculation (before applying the Table 220.84 demand factor) for the heating and air conditioning system in this six-unit multifamily dwelling = 140,688 VA

Rule#5: house loads of multifamily dwellings Applying the Table 220.84 demand factor to house loads of multifamily dwellings is not permitted. House loads must be calculated as per NEC standard method.

Rule#6: multifamily dwelling buildings with multiple services and feeders Sometimes, in large multifamily dwelling buildings, multiple services and feeders may be installed to supply power to different floors or different buildings; in this case, it will be necessary to perform optional load calculation for each feeder. Generally, if the number of units on the feeder is not the same as the number on the service, it will be necessary to perform more than one load calculation.

Important!!! If the total service load for a multifamily dwelling is known, and there is a requirement to supply the building by multiple feeders, do not just divide the service load calculation by the number of feeders. Follow rule#6.

Example#4: If the calculated services load for a 24- unit multifamily dwelling is 311,556 VA and there is a requirement to supply power for it by using four feeders, each feeder will supply power to six units. Each unit in this multifamily dwelling will have 1,050 square feet of floor area, two 20- A small-appliance branch circuits, one 20A laundry branch circuit, fastened-inplace appliances rated 6,600 VA, a range rated 12,000 VA, and an electric clothes dryer rated 5,000 VA. The heating and air conditioning system in each unit will consist of a compressor rated 4,200 VA, a blower motor rated 840 VA, a condenser fan motor rated 360 VA and electric heat rated 5,000 VA What is the optional method feeder load calculation for six units? Solution: First: In accordance with 220.84(C)(1), The general lighting and general-use receptacle load for six units =3 VA x 150 ft2 x 6 = 18,900 VA Second: In accordance with 220.84(C)(2), The small-appliance branch circuit load for six units = 2 x 1,500VA x 6 = 18,000 VA Third: In accordance with 220.84(C)(3), use the nameplate ratings for the fastened-inplace appliances, ranges and clothes dryers will be used: The laundry branch circuit load for six units = 1,500VA x 6 = 9,000 VA. The calculated load for the fastened-in-place appliances for six units = 6,600 VA X 6 = 39,600 VA The calculated load for the ranges for six units = 12,000 VA X 6 = 72,000 VA The calculated load for the clothes dryers for six units = 5,000 VA X 6 = 30,000 VA Forth: In accordance with 220.84(C)(5), add to the load calculation the larger of the air conditioning load or the fixed electric space heating load The fixed electric space-heating load for six units = (electric heat load + blower motor load) X 6 = (5,000 + 840) X 6 = 35,040 VA The total air conditioning unit for six units = (compressor load + blower motor load + condenser fan motor load) x 6 = (4,200 VA + 840 VA +360 VA) x 6 = 32,400 VA The air conditioning load is omitted because the heating load was larger.

Fifth: find the total connected load of the feeder The total connected loads for a feeder supplying six units = 18,900 + 18,000 + 9,000 + 39,600 + 72,000 + 30,000 + 35,040 = 222,540 VA Sixth: apply Table 220.84 demand factor (in below image) for six units is 44 percent. The calculated load after applying the demand factor = 222,540 X 44% = 97,917.6 VA = 97,918 VA

Notes: 1. Not considering any house loads, the optional method feeder load calculation for six units is 97,918 VA. 2. Do not divide the service load calculation by the number of feeders to find the load of one feeder which will give incorrect answer as follows: The feeder load = service load / number of feeders = 311,556 VA / 4 = 77,889



VA While the correct calculated value for one feeder is 97,918 VA as calculated in example#4 above. 

Third: Two family dwelling (that are supplied by a single feeder)

Rule#6: service and feeder Calculation for Two family dwelling In accordance with 220.85, where two dwelling units are supplied by a single feeder and the calculated load as per NEC standard method exceeds that for three identical units calculated in accordance with NEC optional method, the lesser of the two loads shall be permitted.

Important!!! The optional method load calculation for two dwelling units involves using the optional method load calculation procedures for multifamily dwellings, but not for two dwelling units. Perform the load calculation procedures in 220.84, but calculate these two dwelling units as if there were three identical units.

Important!!! In accordance with 220.85, it is necessary to calculate by both methods and then select the lesser of the two loads.

Important!!!

Performing the optional method load calculation for two dwelling units without performing the standard method load calculation is permissible, but the result could be larger than the standard method load calculation.

Example#5: A single feeder will supply two dwelling units. Each unit in this two-family dwelling will have 1,800 square feet of floor area, two 20-ampere (A) small-appliance branch circuits, one 20A laundry branch circuit, four fastened-in-place appliances with a total rating of 9,156 VA, a range rated 12,000 VA and an electric clothes dryer rated 5,000 VA. The heating and air conditioning system in each unit will consist of a compressor rated 4,200 VA, a blower motor rated 1,176 VA, a condenser fan motor rated 360 VA and electric heat rated 10,000 VA. What is the optional method feeder load calculation for this two-family dwelling? Solution: First: Start by calculating the load as per NEC standard method as follows: 1- As per 220.12, Calculate the general lighting and general use receptacle load at 3 VA per square foot The general lighting and general use receptacle load for each unit = 1,800 ft2 x 3 VA/ft2 = 5,400 VA The general lighting and general use receptacle load for both units =5,400 VA X 2 = 10,800 VA 2- As per 220.52(A) and (B), calculate the small-appliance and laundry branch-circuit load at 1,500 VA for each circuit The small-appliance and laundry branch-circuit load for each unit = 1,500VA X 3 = 4,500 VA The small-appliance and laundry branch circuit load for both units = 4,500VA X 2 = 9,000 VA The total general lighting load, including small-appliance and laundry branch circuits= 10,800 + 9,000 = 19,800 VA 3- Apply the Table 220.42 demand factors (in below image) to the general lighting load. The first 3,000 VA remain 3,000VA X 100% = 3,000VA The remaining =19,800 – 3,000 = 16,800 VA at 35 percent = 16,800 VA X 35% = 5,880 VA So, the general lighting load for both units = 3,000 + 5,880 = 8,880 VA

4- The total fastened-in-place appliance load for both units = 9,156 x 2 = 18,312 VA Because there are more than three fastened-in-place appliances, it is permissible to apply a demand factor of 75 percent to this load. After applying the Section 220.53 demand factor, the fastened-in-place appliance load = 18,312 X 75% = 13,734 VA 5- In accordance with 220.54, the electric clothes dryer load = 2 X 5,000 X100% = 10,000 VA

6- The maximum demand for two 12,000 VA ranges from Table 220.55 is 11,000 VA. 7- The heating load, with the blower motor, = 10,000 + 1,176 = 11,176 VA This load is larger than the air conditioning load, and because of 220.60, it is permissible to use only the larger of the noncoincident loads. Therefore, the total heating load for both units = 11,176 VA X 2 = 22,352 VA 8- As required by 220.50 and 430.24, this calculation must include 25 percent of the largest motor. Since the compressor was omitted, the largest motor is the blower motor. 25% of the largest motor = 1,176 X 25% = 294 VA 9- After applying all demand factors, the standard method load calculation for these two dwelling units = 8,880 + 13,734 + 10,000 + 11,000 + 22,352 + 294 = 66,180 VA

Second: Perform the load calculation procedures in 220.84 (NEC optional method), but calculate these two dwelling units as if there were three identical units. 1- As per 220.84(C)(1), Calculate the general lighting and general use receptacle load at 3 VA per square foot The general lighting and general use receptacle load for each unit = 1,800 ft2 x 3 VA/ft2 = 5,400 VA The general lighting and general use receptacle load for three units = 5,400 VA X 3 = 16,200 VA Remember, this optional calculation is based on three units, not two. 2- As per 220.84(C)(2), calculate the small-appliance and laundry branch-circuit load at 1,500 VA for each circuit The small-appliance and laundry branch circuit load for each unit =1,500VA X 3 = 4,500 VA The small-appliance and laundry branch-circuit load for three units = 4,500 VA X 3 = 13,500 VA 3- In accordance with 220.84(C)(3), use the nameplate ratings for the fastened-inplace appliances, ranges and clothes dryers. The load for the fastened-in-place appliances, ranges and clothes dryers in each unit = 9,156 + 12,000 + 5,000 = 26,156 VA The load for the fastened-in-place appliances, ranges and clothes dryers for three units = 26,156 VA X 3 = 78,468 VA 4- In accordance with 220.84(C)(5), add to the load calculation the larger of the air

conditioning load or the fixed electric space heating load. In this example, the heating load is larger than the air conditioning load, therefore omit the air conditioning load. The heating load for each unit = 10,000 + 1,176 = 11,176 VA The heating load for three units = 11,176VA X 3 = 33,528 VA 5- The total connected load for three units = 16,200 + 13,500 + 78,468 + 33,528 = 141,696 VA 6- After finding the total connected loads, apply the Table 220.84 demand factor for the number of dwelling units. The Table 220.84 demand factor for three units is 45 percent. The calculated load after applying the demand factor =141,696 X 45% = 63,763 VA Third: Compare between the results of standard and optional (as 3 units) loads and select the lesser of the two loads. The result of the standard load calculation was 66,180 VA. The result of the Optional load calculation (as 3 units) was 63,763 VA So, the calculated feeder load for these two dwellings is 63,763 VA.

Branch Circuit Design Calculations – Part Eleven In article " Receptacle Branch Circuit Design Calculations – Part Three ", I stated that a Receptacle in dwelling units may serve one of the following loads: 1. 2. 3. 4. 5. 6. 7. 8.

General-use Receptacle Loads, Small appliance Loads, Laundry Load, Cloth dryer Load, Household cooking appliances load, Fastened-in-place Appliance loads, Heating and air conditioning loads, Motor loads.

I explained the first six types in the following articles:



Receptacle Branch Circuit Design Calculations – Part Four Receptacle Branch Circuit Design Calculations – Part Five



Receptacle Branch Circuit Design Calculations – Part Six



Receptacle Branch Circuit Design Calculations – Part Seven Branch Circuit Design Calculations – Part Eight





In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code. You can review the following articles for more information:  Receptacle Branch Circuit Design Calculations – Part One 

Receptacle Branch Circuit Design Calculations – Part Two

7- Heating and air conditioning loads – Part Three In the previous Article " Heating and air conditioning loads – Part One ", I explained the following points:

 

1. Applied NEC Rules for Heating and air conditioning loads 2. Feeder and service calculation of Heating and air conditioning loads by two methods: First: As per NEC Standard calculation method, Second: As per NEC Optional calculation method.

The rules applied in second method differ from a dwelling unit type to another. I explained the rules for first dwelling type ―A single dwelling unit‖ in Article " Heating and air conditioning loads – Part One ". Also, I explained the rules for dwelling unit type ―multifamily dwelling and two dwelling units‖ in Article " Heating and air conditioning loads – Part Two " Today, I will continue explaining the rules for feeder and service calculation of Heating and air conditioning loads as per NEC Optional calculation method for the forth case of dwelling unit which is existing dwelling unit. Forth: Existing Dwelling Unit

Important!!! The purpose of doing service and feeder calculations for an existing dwelling unit is to determine if the existing service or feeder is of sufficient capacity to serve a required additional loads or not.

Rule#1: Conditions Of adding new loads to An Existing Dwelling Unit

As per NEC section 220.87, Additional loads may be connected to existing services and feeders under the following conditions: 1. The maximum demand kVA data for a minimum 1-year period (or the 30-day alternative method from the exception) is available. 2. The maximum demand at 125 percent plus the new load does not exceed the ampacity of the feeder or rating of the service. 3. The feeder has overcurrent protection in accordance with 240.4, and the service has overload protection in accordance with 230.90.

Important!!! For condition#1 in Rule#1 above, If the maximum demand data for a 1year period is not available, the calculated load shall be permitted to be based on the maximum demand (measure of average power demand over a 15-minute period) continuously recorded over a minimum 30-day period using a recording ammeter or power meter connected to the highest loaded phase of the feeder or service, based on the initial loading at the start of the recording. The recording shall reflect the maximum demand of the feeder or service by being taken when the building or space is occupied and shall include by measurement or calculation the larger of the heating or cooling equipment load, and other loads that may be periodic in nature due to seasonal or similar conditions.

Important!!! For condition#2 in Rule#1 above, apply the NEC standard Calculation method to get the total load as follows: Total Load = Existing Load Value + New Load Where: Existing Load Value = Max demand Value for a 1-year period from 220.87(1) x 125% New Load = Continuous loads x 125 % + Non-continuous loads x 100 %

Important!!! If condition#2 in Rule#1 above is not verified, you need to increase ampacity of the feeder and/or rating of the service to be able to add new loads to an existing dwelling unit.

Example#1: A business owner wants to add equipment to an existing building. The new equipment will not replace any of the existing equipment. The existing service has a rating of 1,200 amperes (A). A peak-demand meter has been connected to the service for more than a year. The maximum demand data from the peak-demand meter shows a demand of 743A. The new equipment that has a calculated load (in accordance with standard load calculation procedures) of 251A. Does this existing service have an ampere rating high enough for the existing loads and the new loads?

Solution: Existing Load Value = Max demand Value for a 1-year period from 220.87(1) x 125% = 743 x 125% = 929 A New Load = 251 A New Service Load = Existing Load Value + New Load = 929 + 251 = 1,180 A Since the existing service is rated 1,200A, this installation will be Code-compliant

Rule#2: Conditions Of Application Of NEC Optional Calculation Method For An Existing Dwelling Unit As per NEC section 220.83, The NEC optional calculation method for an existing dwelling can only be used if the dwelling is supplied by a singlephase service. The service can be fed from a 120/240-volt (V) or 208Y/120V system but must be a 3-wire system.

For existing dwelling unit, it shall be permissible to calculate the total load by using the NEC optional calculation method in the following two cases:

1. Where Additional Air-Conditioning Equipment or Electric Space-Heating Equipment Is Not to Be Installed. 2. Where Additional Air-Conditioning Equipment or Electric Space-Heating Equipment Is to Be Installed.

A- Where Additional Air-Conditioning Equipment or Electric Space-Heating Equipment Is Not to Be Installed

Rule#3: Calculation of electrical load as per NEC Optional Calculation Method For An Existing Dwelling Unit – Case (A) In this case, the calculation is almost identical to the calculation method in 220.82(B) for single family dwelling. The only difference is the amount of load that is rated at 100 percent, in 220.82(B) for single family dwelling it was 10 KVA but in 220.83 for existing dwelling it will be 8 KVA.

The following percentages shall be used for existing and additional new loads:

Load (kVA) First 8 kVA of load at Remainder of load at

Percent of Load 100 40

Important!!! The NEC optional calculation method for an existing dwelling shall Include all of the existing loads and the new loads. Example#2: An addition will be built onto an existing one-family dwelling; the existing kitchen will also be renovated. The area is 1,500 square feet. The existing dwelling has the following loads:  Two small-appliance branch circuits;  One laundry branch circuit;  Two attic roof ventilators rated 506 VA each;

   

One motor rated 1,176 VA; A range rated 8,000 VA; A clothes dryer rated 5,000 VA; And a water heater rated 4,500 VA.

The service supplying this existing dwelling is 120/240V, single phase, and is rated 100 A. The addition built onto this dwelling will add 500 square feet of floor area. Therefore, the total calculated floor area will be 2,000 square feet. The existing range will be replaced by a new 12,000 VA range. The kitchen renovation will also include two additional appliance loads: a dishwasher rated 1,380 VA and a kitchen waste disposer rated 864 VA. Two more small-appliance branch circuits will be installed, thus making a total of four small-appliance branch circuits. No additional air conditioning equipment or electric space-heating equipment will be installed Is this 100A service of sufficient capacity to serve these additional loads? If not, what size service is required?

Solution: When calculating an existing dwelling unit, include all the existing loads and all the additional new loads. First: calculate the general lighting and general as specified in [220.83(A)(1)] The general lighting and general-use receptacle load = 2,000 ft2 x 3 VA /ft2= 6,000 VA Second: calculate the small-appliance branch circuit and laundry branch circuit load as specified in [220.83(A)(1)] The small-appliance and laundry branch circuits load = (four small-appliance branch circuits + one Laundry branch circuit) = 1,500 VA x 5 = 7,500 VA. Third: calculate The Fastened-in-place appliance load, as specified in 220.83(A)(3)(a) The Fastened-in-place appliance load= Loads of (Two attic roof ventilators + motor + a dishwasher + kitchen waste disposer) = 506 + 506 + 1,176 + 1,380 + 864 = 4,432 VA Forth: calculate other loads as specified in 220.83(A)(3)(b), (c), (d) Other loads = new range load + clothes dryer load + water heater load = 12,000 VA + 5,000 VA + 4,500 VA = 21,500 VA Fifth: Before applying the demand factor, the total for the existing loads and the new loads = 6,000 + 7,500 + 4,432 + 21,500 = 39,432 VA

Sixth: In accordance with 220.83(A), calculate the first 8 kVA (8,000 VA) of load at 100 percent and the remainder of load at 40 percent. Demand of existing loads and the new loads = 8,000 VA x 100 % + (39,432 – 8,000) x 40% = 20,573 VA Seventh: Find the minimum ampere rating by dividing volt-amperes by voltage. The minimum ampere rating required for this service = 20,573 VA ÷ 240 V = 85.7 A = 86 A. Since the service supplying this dwelling is rated 100A, it is of sufficient capacity to serve these additional loads.

B- Where Additional Air-Conditioning Equipment or Electric Space-Heating Equipment Is to Be Installed

Rule#4: Calculation of electrical load as per NEC Optional Calculation Method For An Existing Dwelling Unit – Case (B) In this case, the calculation is identical to that of case (A), But In this case, there are additional loads for of air-conditioning or space-heating, so the following will be applied for these loads: 1. The larger connected load of air-conditioning or space-heating, but not both, shall be used. 2. The following percentages shall be used for existing and additional new loads:   

Air-conditioning equipment at 100% Central electric space heating at %100 Less than four separately controlled space heating units at 100%

The percentages in below image shall be used for existing and additional new loads.

Important!!! The NEC optional calculation method for an existing dwelling shall Include all of the existing loads and the new loads.

Example#3: In example#2, before the job starts, the homeowners change their mind. They decide to install a central air conditioning system that is rated 7,200 VA. Is this 100A service of sufficient capacity to serve these additional loads plus the central air conditioning unit? If not, what size service is required? Solution: Use rules as specified in 220.83(B), for existing and additional new loads where additional air conditioning equipment or electric space-heating equipment is to be installed. From example#2, the Demand of existing loads and the new loads = 20,573 VA As per Rule#4 above, the air conditioning system load must be added to the calculation at 100 percent of nameplate rating So, the Demand of existing loads and the new loads = 20,573 + 7,200 = 27,773 VA The minimum ampere rating required for this service = 27,773 VA ÷ 240 V = 115.7 A = 116 A , which exceeds the 100 A old service rating In accordance with 240.6(A), the next standard ampere rating above 116 is 125 amperes. Therefore, the minimum size service for this dwelling is 125 amperes.

Branch Circuit Design Calculations – Part Twelve In article " Receptacle Branch Circuit Design Calculations – Part Three ", I stated

that a Receptacle in dwelling units may serve one of the following loads: 1. 2. 3. 4. 5. 6. 7. 8.

General-use Receptacle Loads, Small appliance Loads, Laundry Load, Cloth dryer Load, Household cooking appliances load, Fastened-in-place Appliance loads, Heating and air conditioning loads, Motor loads.

I explained the first seven types in the following articles:        

Receptacle Branch Circuit Design Calculations – Part Four Receptacle Branch Circuit Design Calculations – Part Five Receptacle Branch Circuit Design Calculations – Part Six Receptacle Branch Circuit Design Calculations – Part Seven Branch Circuit Design Calculations – Part Eight Branch Circuit Design Calculations – Part Nine Branch Circuit Design Calculations – Part Ten Branch Circuit Design Calculations – Part Eleven

In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code.

You can review the following articles for more information:  

Receptacle Branch Circuit Design Calculations – Part One Receptacle Branch Circuit Design Calculations – Part Two

8- Motor loads 8.1 Applied NEC Rules for Motor loads

There are many NEC rules that control the Motor Loads including: 220.14(C) Motor Loads for All Occupancies  220.18(A) Maximum Loads for circuits supplying Motor-Operated and Combination Loads.  220.50 Motors  422.62 Appliances Consisting of Motors and Other Loads  220.82 Optional Method - Dwelling Unit  220.84 Optional Method - Multifamily Dwelling  430.6 Ampacity and Motor Rating Determination  430.22 Single Motor 

8.2 Calculation of Motor loads

First: As per NEC Standard calculation method

The motors exist in dwelling units as fastened in place appliances or separate motors (air conditioning compressors, fan blower, etc.)

Rule#1: Motor Loads as per NEC standard method When calculating a feeder or service As per NEC Standard calculation method, the largest motor must be multiplied by 25 percent and add it to the service load calculation.

Important!!! Most electrical equipment is rated in volt-amperes (VA) or watt input. While motors traditionally have been rated in horsepower output (Some motors are available with their output ratings expressed in watts and kilowatts).

Rule#2: Motor Loads as per NEC standard method

As per NEC section 430.6(A)(1), Do not use the actual current rating marked on the nameplate. When calculating motor loads, use the values given in Tables 430.247 through 430.250.

Important!!! Exceptions to 430.6(A)(1) : 1. Motors built for low speeds (less than 1,200 rpm) or high torques for multispeed motors. 2. For equipment that employs a shaded-pole or permanent-split capacitor-type fan or blower motor that is marked with the motor type, use the full load current for such motor marked on the nameplate of the equipment in which the fan or blower motor is employed. 3. For a listed motor-operated appliance that is marked with both

motor horsepower and full-load current, use the motor full-load current marked on the nameplate of the appliance.

Important!!!  Full-load currents for 3-phase motors are in Table 430.250,  Full-Load Current for Two-Phase Alternating- Current Motors (4-Wire) are in Table 430.249,  Full-load currents for single-phase motors are in Table 430.248,  Full-Load Currents for Direct-Current Motors are in Table 430.247.

Important!!! Kilovolt-amperes (kVA) shall be considered equivalent to kilowatts (kW) for Fastened-in-place Appliances.

Important!!! If the motor is one of the fastened-in-place appliances , Apply NEC section, 220.53, which states that ― It shall be permissible to apply a demand factor of 75 % to the nameplate rating load of four or more appliances fastened-inplace, that are served by the same feeder or service in a one-family, twofamily, or multifamily dwelling‖.

Important!!! As per NEC section 220.53, electric ranges, clothes dryers, space-heating equipment or air conditioning equipment must not be included with the number of appliances that are fastened in place. Also, All portable small Appliances for kitchen and others are not Fastenedin-Place Appliances.

Important!!!

No derating is allowed when there are only one, two, three fastened-inplace appliances.

Example#1: The following appliances will be installed in a one-family dwelling:  A dishwasher rated 10 amperes at 120 volts;  A ½ hp, 120-volt kitchen-waste disposer;  A 4.5 kW, 240-volt water heater. What is the total service load for these appliances?

Solution: The load for these three appliances must be calculated at 100 percent. The dishwasher load = 10 A × 120 V = 1,200 VA. Since the kitchen-waste (or garbage) disposer is a motor, the load must be determined by multiplying the full-load current in Table 430.248 by the system voltage. In accordance with Table 430.248, the full-load current of a ½-hp, 120-volt motor = 9.8 A Note: Although the current value of 9.8 amperes is in the column titled ―115 volts,‖ the currents listed in this column are permitted for system voltage ranges of 110 to 120 volts (see the last sentence of the text in Table 430.248). The disposer load =9.8 A × 120 V = 1,176 VA The load of the water heater = 4.5 × 1,000 = 4,500 VA Since derating is not permissible for only three appliances, The calculated load at 100 % = 1,200 + 1,176 + 4,500 = 6,876 VA If the disposer motor is the largest motor, then The calculated service load = 6,876 + disposer load x 25% = 7,170 VA

Example#2: For the same case in example#, add a fourth appliance; a trash compactor rated 7.5 amperes at 120 volts, What is the total service load? Solution: The compactor load = 7.5 A × 120 V = 900 VA The total before applying the demand factor = 1,200 + 1,176 + 900 + 4,500 = 7,776 VA Next, because there are four appliances, multiply the load by the 75 percent demand factor. The calculated service load for these appliances = 7,776 × 0.75 = 5,832 VA If the disposer motor is the largest motor, then The calculated service load = 5,832 + disposer load x 25% = 6,126 VA

Important!!!  If the motor is air conditioning compressor, usually the air conditioning compressor is the largest motor in dwelling units. in this case, multiply the load of one compressor by 25 percent and add it to the service load calculation.  But if the heating load is larger than the air conditioning load, and because of 220.60 which states that‖ it is permissible to use only the larger of the noncoincident loads‖ the air conditioning load will be omitted and the air conditioning compressor will not be the largest motor in this case.

Example#3: A single feeder will supply two dwelling units. Each unit in this two-family dwelling will have 1,800 square feet of floor area, two 20-ampere (A) small-appliance branch circuits, one 20A laundry branch circuit, four fastened-in-place appliances with a total rating of 9,156 VA, a range rated 12,000 VA and an electric clothes dryer rated 5,000 VA. The heating and air conditioning system in each unit will consist of a compressor rated 4,200 VA, a blower motor rated 1,176 VA, a condenser fan motor rated 360 VA and electric heat rated 10,000 VA.

What is the feeder load calculation for this two-family dwelling?

Solution: Start by calculating the load as per NEC standard method as follows: 1- As per 220.12, Calculate the general lighting and general use receptacle load at 3 VA per square foot The general lighting and general use receptacle load for each unit = 1,800 ft2 x 3 VA/ft2 = 5,400 VA The general lighting and general use receptacle load for both units =5,400 VA X 2 = 10,800 VA 2- As per 220.52(A) and (B), calculate the small-appliance and laundry branch-circuit load at 1,500 VA for each circuit The small-appliance and laundry branch-circuit load for each unit = 1,500VA X 3 = 4,500 VA The small-appliance and laundry branch circuit load for both units = 4,500VA X 2 = 9,000 VA The total general lighting load, including small-appliance and laundry branch circuits= 10,800 + 9,000 = 19,800 VA 3- Apply the Table 220.42 demand factors to the general lighting load. The first 3,000 VA remain 3,000VA X 100% = 3,000VA The remaining =19,800 – 3,000 = 16,800 VA at 35 percent = 16,800 VA X 35% = 5,880 VA So, the general lighting load for both units = 3,000 + 5,880 = 8,880 VA 4- The total fastened-in-place appliance load for both units = 9,156 x 2 = 18,312 VA Because there are more than three fastened-in-place appliances, it is permissible to apply a demand factor of 75 percent to this load. After applying the Section 220.53 demand factor, the fastened-in-place appliance load = 18,312 X 75% = 13,734 VA 5- In accordance with 220.54, the electric clothes dryer load = 2 X 5,000 X100% = 10,000 VA 6- The maximum demand for two 12,000 VA ranges from Table 220.55 is 11,000 VA.

7- The heating load, with the blower motor, = 10,000 + 1,176 = 11,176 VA This load is larger than the air conditioning load, and because of 220.60, it is permissible to use only the larger of the noncoincident loads. Therefore, the total heating load for both units = 11,176 VA X 2 = 22,352 VA 8- As required by 220.50 and 430.24, this calculation must include 25 percent of the largest motor. Since the compressor was omitted, the largest motor is the blower motor. 25% of the largest motor = 1,176 X 25% = 294 VA 9- After applying all demand factors, the standard method load calculation for these two dwelling units = 8,880 + 13,734 + 10,000 + 11,000 + 22,352 + 294 = 66,180 VA

Don’t Forget… For example#3, NEC optional method for three identical units must be calculated, because In accordance with 220.85, where two dwelling units are supplied by a single feeder and the calculated load as per NEC standard method exceeds that for three identical units calculated in accordance with NEC optional method, the lesser of the two loads shall be permitted.

Example#4: A heating/cooling package unit will be installed in a one-family dwelling. The electric heater is rated 9.6 kW at 240 volts. The blower motor inside the package unit that circulates the air will be a ½ horsepower, 240-volt motor. How much load will this package unit add to a 240-volt, single-phase service? Assuming that the air conditioning load will be less than the heating load. Solution: The heat load= 9.6 KW × 1,000 = 9,600 watts The full-load current in amperes of a ½ hp, single-phase, 240-volt motor (from Table 430.248) = 4.9 A The motor load = 4.9 A x 240 V = 1,176 VA

The service load for this package unit = 1,176 + 9,600 = 10,776 watts Because the blower motor is the largest motor in the service load The service load for this package unit = 10,776 + 1,176 x 25% = 11,070 watts

Second: As per NEC Optional calculation method

1- For single dwelling unit

Rule#3 Permanently Connected Motors As per NEC section 220.82(B)(4), add the nameplate ampere (A) or kilovolt ampere (kVA) rating of all permanently connected motors to the general loads covered in 220.82(B). Include all motors that are not already included with the general loads covered in 220.82(B)

Important!!! Permanently connected motors must be added to the general loads when calculating a feeder or service when calculated by the NEC optional method. Although motors are listed separately, some or all of the motors may have already been added to the load calculation as fastened-in-place appliances. For example, a kitchen waste disposer could be added to an optional load calculation as a fastened-in-place appliance or as a permanently connected motor. It is not necessary to add motors that have already been added to the load calculation as fastened-in-place appliances.

Important!!! When calculating the feeder or service load of a dwelling unit by the optional method, do not multiply the full-load current of permanently connected motors by 125 percent.

Important!!! Do not include heating and air conditioning equipment in the list with permanently connected motors. Heating and air conditioning equipment is covered in 220.82(C) and explained before in articles:   

Branch Circuit Design Calculations – Part Nine Branch Circuit Design Calculations – Part Ten Branch Circuit Design Calculations – Part Eleven

Example#5: What is the optional method service load calculation (before applying the demand factor) for a one-family dwelling with the following permanently connected motors:  Two automatic power attic roof ventilators rated horsepower (hp) at 115 volts (V) each  A swimming pool pump rated 1 hp at 230V Noting that the ampere rating of each motor is not known. These three motors have not been included with the general loads covered in 220.82(B).

Solution: Step#1: find the full-load current rating in amperes for each motor. Since the ampere ratings for these motors are not known, use the values given in Table 430.248 (which provides full-load currents in amperes for single-phase alternating-current motors) The full-load current for each of the automatic power attic roof ventilators = 4.4 A. The full-load current for the swimming pool pump = 10 A Step#2: calculate volt-amperes (VA) for each motor. The load for each attic fan = 115 V x 4.4 A = 506 VA The load for the swimming pool pump = 230 V x 10 A = 2,300 VA The service load for these permanently connected motors (As per NEC optional

calculation method) = 506 + 506 + 2,300 = 3,312 VA

2- For Multi-Family Dwelling

Rule#4: Permanently Connected Motors As per NEC section 220.84(C)(4), add the nameplate ampere (A) or kilovolt ampere (kVA) rating of all permanently connected motors to other categories of loads included in 220.84(C). Include all motors that are not already included with the item 220.84(C)(3).

Important!!! For multifamily dwelling load calculation, if there are any permanently connected motors that have not been included, add the nameplate ampere or kilovolt-ampere rating of those motors to the calculation.

Important!!! Do not include motors that are part of the air conditioning system or motors that are part of the fixed electric space-heating system. These motors will be included as 220.84(C)(5) requires and as explained before in articles:   

Branch Circuit Design Calculations – Part Nine Branch Circuit Design Calculations – Part Ten Branch Circuit Design Calculations – Part Eleven



Important!!! Include only permanently connected motors that are supplied from dwelling units. Do not include permanently connected motors that are supplied by house power.

Don’t Forget… After calculation of feeder or service load calculation for a Multifamily dwelling, apply the demand factors of table 220.84.

Example#6: The load of a multifamily dwelling will be calculated by the optional method. Two swimming pool pump motors and two automatic power attic roof ventilators will be powered by the house panel. Can these permanently connected motors be included with the connected loads to which Table 220.84 apply?

Solution: Because these permanently connected motors are supplied by house power, do not add them to the loads that will be calculated in accordance with Table 220.84(Optional Calculations — Demand Factors for Three or More Multifamily Dwelling Units). These motors must be calculated in accordance with Part III of Article 220.

Non-Dwelling Buildings Load Calculations- Part One In course " EE-1: Beginners' Electrical design course " which is a preparation for beginners in electrical design, I explained the Electrical design philosophy for all building types and indicated the different types of buildings according to usage which were: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Industrial buildings Commercial buildings Residential buildings Agricultural buildings Educational buildings Transportation buildings Religious buildings Parking and storage Military buildings Governmental buildings Cultural buildings Other buildings

In the current course" EE-3: Basic Electrical design course – Level II ", in Sixteen Articles, I explained the design of the Dwelling type buildings; I explained Where and how to distribute each type of load in a dwelling unit as per NEC code and how to calculate its Demand load for feeder and service sizing calculations, and finally I summarize the steps of calculating electrical load for any dwelling unit as per NEC

standard & Optional Calculation methods in two Excel sheets/calculator. These Sixteen Articles for the electrical design of dwelling buildings are as follows: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

Branch Circuit Design Calculations – Part One Branch Circuit Design Calculations – Part Two Branch Circuit Design Calculations – Part Three

Receptacle Branch Circuit Design Calculations – Part One Receptacle Branch Circuit Design Calculations – Part Two Receptacle Branch Circuit Design Calculations – Part Three Receptacle Branch Circuit Design Calculations – Part Four Receptacle Branch Circuit Design Calculations – Part Five Receptacle Branch Circuit Design Calculations – Part Six Receptacle Branch Circuit Design Calculations – Part Seven Branch Circuit Design Calculations – Part Eight Branch Circuit Design Calculations – Part Nine Branch Circuit Design Calculations – Part Ten Branch Circuit Design Calculations – Part Eleven Electrical Load Calculator for Dwelling Units Optional Electrical Load Calculator for Dwelling Units

Today, I will begin explaining the electrical design of Non-Dwelling Buildings as per NEC Code.

List of ordinary Non-Dwelling Buildings Loads: Most of Non-Dwelling Buildings will have the following types of loads: 1. 2. 3. 4. 5. 6.

Lighting loads, Receptacles Loads, Kitchen Loads, Heating, Ventilation and air conditioning Loads (Non-Coincident Loads), Motor Loads, Other Loads.

Again, but for above Non-Dwelling Buildings Loads, I will explain the following points:

Where and how to distribute each type of load in a dwelling unit as per NEC



code? 

How to calculate its Demand load for feeder and service sizing calculations?

Important!!! All design Calculations for Non-dwelling Buildings will be as per NEC standard calculation method but I will explain design calculations as per NEC Optional calculation method only for the following Non-Dwelling buildings as permitted by NEC, Part IV. Optional Feeder and Service Load Calculations:  Schools  Existing Installations  New Restaurants

First: Lighting Loads 1- Lighting branch circuit ratings in non-dwelling units Lighting branch circuit ratings in non-dwelling units may have one of the following ratings according to the type and application used as follows: a) 15- and 20-Ampere Branch Circuits It shall be permitted to supply lighting units or other utilization equipment, or a combination of both. b) 30-Ampere Branch Circuits It shall be permitted to supply fixed lighting units with heavy-duty lampholders. c) 40- and 50-Ampere Branch Circuits It shall be permitted to supply fixed lighting units with heavy-duty lampholders.

2- Lighting load Calculations In the broad sense, lighting loads for Non-Dwelling buildings may be categorized as

follows: 1. 2. 3. 4. 5.

General lighting. Show-window lighting. Track lighting. Sign and outline lighting. Other lighting.

2.1 General lighting

Definition: General lighting outlets are those Outlets intended for general use for fixedin-place luminaires (lighting fixtures). They are only used for lighting for the normal use of the occupants and Its intensity should be adequate for any type of work performed in the area.

Important!!! The following Lighting fixtures are not included in General Lighting category: 1. Specialized task lighting (Show-window lighting, Track lighting, accent, specialty, or display lighting). 2. Any special lighting for workshops, photography labs, or studios that may be located in the dwelling.

2.1.a Calculation Method Determining the general lighting load as per NEC will be based on the load per area method as follows: Step#1: determine the general lighting load density (in VA/ft2) for the building occupancy under design.

Rule#1: Applying Table 220.12 The NEC code introduces minimum general lighting loads (in VA/ft2) for various types of buildings in Table 220.12. Within the same building, there are normally several different types of areas like storage, office, hallways, and cafeterias, these areas must be considered separately if their (VA/ft2) values are available in table 220.12.

Step#2: calculate the floor area in (ft2)

Rule#2: Applying Table 220.12 The floor area for each floor shall be calculated from the outside dimensions of the building, or other area involved. The calculated floor area shall not include open porches, garages, or unused or unfinished spaces not adaptable for future use (like some attics, cellars, and crawl spaces).

Step#3: calculate the general lighting load The general lighting load is calculated by multiplying the floor area in (ft2) of a building by its unit load (in VA/ft2) derived from table 220.12. General lighting load in (VA) = Area of Occupancy in (ft2) X general lighting load density in (VA/ft2)

Step#4: Apply lighting demand factor from table 220.42 for type of building under design.

Important!!! The demand factors of table 220.42 shall not apply to the calculated load of feeders or services supplying areas in hospitals, hotels, and motels where the entire lighting is likely to be used at one time, as in operating rooms, ballrooms, or dining rooms.

Important!!! For the purpose of determining the circuit requirements, If the General Lighting load is continuous (as in Most commercial structures), the calculated load is multiplied by 1.25 (the inverse of 80%).

Rule#3: Calculating Lighting Outlets load for heavy-duty Lampholders As per NEC section 210.21, If there is Lighting Outlets for heavy-duty Lampholders in the Non-Dwelling Building, It shall be calculated at a minimum of 600 volt-amperes.

Example#1: A 25,000 ft2 office building is being designed. What is the general lighting load and what load does the circuit need to supply?

Solution: From Table 220.12, the unit load for an office building is 3.5 VA/ft2. The general lighting load is determined by multiplying this value by the square footage of the building: 3.5 VA/ft2 x 25,000 ft2 = 87,500 VA So, the general lighting load is 87,500 volt-amperes. However, the load is continuous and can only be 80% of the load supplied by the circuit. This value must be multiplied by 1.25 to determine the circuit requirements: 87,500 VA x 1.25 = 109,375 VA The circuit is designed to supply 109.375 KVA

Notes for table 220.12

Important!!! Don’t apply the values of table 220.12 before reviewing the following notes: 1. The unit values herein are based on minimum load conditions and 100 percent power factor (i.e. Load in VA = Load in Watt) and may not provide sufficient lighting for the installation contemplated. So, the designer can choose a higher value based on the existing design conditions. 2. Under any conditions, don’t use values less than that specified in table 220.12, there are no exceptions.

Important!!! The general lighting load unit values specified in table 220.12 for guest rooms or guest suites of hotels and motels includes the following loads: 1. All general-use receptacle outlets of 20-ampere rating or less, including receptacles connected to Bathroom Branch Circuits, 2. The outdoor receptacle outlets, 3. general-use receptacle Outlets used in Basements, Garages, and Accessory Buildings. 4. Wall lighting outlet used in Habitable Rooms, 5. Wall lighting outlets used in hallways, stairways, attached garages, and detached garages, 6. Wall lighting outlet used in Storage or Equipment Spaces (like attics, under-floor spaces, utility rooms, and basements), 7. Wall lighting outlet used in Guest Rooms or Guest Suites In hotels, motels, or similar occupancies.

So, no need to add the above outlets in load calculations per NEC method.

Important!!! The NEC method and table 220.12 are applied for any Additions to Existing

Installations for non-dwelling installations.

Important!!! Energy saving–type calculations (which used to reduce the connected lighting load and actual power consumption) are not permitted to be used to determine the minimum calculated lighting load if they produce loads less than the load calculated according to 220.12.

2.1.b Notes for NEC method for calculation of lighting branch circuit load

Important!!! The NEC doesn’t introduce a procedure for calculating the actual full load for the individual lighting fixtures in a general lighting branch circuit.

Rule#4: NEC method for calculation of lighting branch circuit load If the required information for calculating the actual full load for every individual lighting fixture in the circuit is available, the following procedure will be applied: 1. Calculate the actual load for the lighting branch circuit by summing of actual full load for its individual lighting fixtures. 2. Compare the values obtained from NEC method with that obtained from actual load method and select the greater load value to be used in the design.

Important!!! Actually, The NEC method for calculation of lighting load is not required if the

actual full load for every individual lighting fixture in the circuit is determined. Methods for determining the actual full load for every individual lighting fixture in the circuit is explained in our course and I recommend reviewing these methods very well.

Rule#5: determining the actual load current of circuits supplying lighting units that have ballasts, transformers, autotransformers, or LED drivers As per NEC Section 220.18 (b) states , circuits supplying lighting units that have ballasts, transformers, autotransformers, or LED drivers, the calculated load shall be based on the total ampere ratings of such units and not on the total watts of the lamps. This means that the losses in light fixture switchgear (ballast, internal wiring, etc.) must be taken into account when calculating the actual full load of light fixtures and the current rating of the ballast, not the tube wattage, will be used.

Example#2: A fluorescent lighting fixtures with 4 numbers 2 feet lamps, 18 watt/ lamp. Calculate the actual load for this lighting fixture. Solution: The actual total load of fixture = 4 lamps x 18 watt/lamp + losses So, we can’t know the actual losses, we will use the same equation in another form The actual total load of fixture = 4 ballast x watt/ballast = 4 x 20 w = 80 watt 2.2 Show-window lighting

The calculation of this type of lighting is explained in Article " Branch Circuit Design Calculations – Part Four ".

2.3 Track lighting

The calculation of this type of lighting is explained in Article " Branch Circuit Design Calculations – Part Four ".

2.4 Sign and outline lighting

The calculation of this type of lighting is explained in Article " Branch Circuit Design Calculations – Part Five ".

2.5 Other lighting Other lighting includes but not limited to the following: 1. 2. 3. 4. 5. 6.

Security lighting, Parking area lighting, Sidewalk lighting, Roadway lighting, Stadium lighting, Tunnel Lighting.

2.5.a Rules applied for Calculation of Other lighting loads

Unfortunately, The NEC code don’t provide calculation rules for lighting types included under these additional lighting loads. So, all additional lighting loads are calculated using the actual load designed by professional lighting programs and methods as explained in our course.

Important!!! All Additional lighting loads should be computed separately from the general lighting load and then added to the general lighting load.

Don’t forget… These additional lighting loads are considered continuous loads where appropriate and must be multiplied by 1.25 for feeder and overcurrent protection calculations.

Rule#6: Calculating the total lighting load for buildings having more than one lighting category. lighting loads for Non-Dwelling buildings may be categorized as follows: 1. 2. 3. 4. 5.

General lighting. Show-window lighting. Track lighting. Sign and outline lighting. Other lighting.

If more than one of the above lighting categories is existing in the same building, Each lighting Category load is computed separately and then combined to other lighting categories load to determine the total lighting load.

Non-Dwelling Buildings Load Calculations- Part Two In this part of course " EE-3: Basic Electrical design course – Level II " which explains the electrical design of Non-Dwelling Buildings as per NEC Code.

In Article " Non-Dwelling Buildings Load Calculations- Part One ", I introduced a List for ordinary Non-Dwelling Buildings Loads which was as follows:

1.

Lighting loads,

2.

Receptacles Loads,

3.

Kitchen Loads,

4.

Heating, Ventilation and air conditioning Loads (Non-Coincident Loads),

5.

Motor Loads,

6.

Other Loads.

Again, but for above Non-Dwelling Buildings Loads, I will explain the following points: 1. Where and how to distribute each type of load in a dwelling unit as per NEC code? 2. How to calculate its Demand load for feeder and service sizing calculations?

Important!!! All design Calculations for Non-dwelling Buildings will be as per NEC standard calculation method but I will explain design calculations as per NEC Optional calculation method only for the following Non-Dwelling buildings as permitted by NEC, Part IV. Optional Feeder and Service Load Calculations: 1. 2. 3.

Schools Existing Installations New Restaurants

I explained the Design Calculation for first type of Non-Dwelling Building loads which is Lighting loads in Article. Today, I will explain the second type of Non-Dwelling Building Loads which is Receptacle Loads.

Second: Receptacle Loads – Part One 1- Introduction Before we go ahead in explaining the design calculation for Receptacles Loads in NonDwelling buildings, there are some parts explained previously in the design calculations of Receptacles Loads of Dwelling buildings and are applicable also in Nondwelling buildings, these parts need to be reviewed from the following Articles:     

Introduction and Definitions Types of Receptacles The difference between Multiple and Multiwire branch circuits GFCI protection basics AFCI Protection Basics

2- Receptacle Branch circuit ratings and permissible loads In no case shall the load exceed the branch-circuit ampere rating. The following are the permissible Receptacle Branch circuit ratings in dwelling buildings:

A) 15- and 20-Ampere Branch Circuits 15- and 20-Ampere Branch Circuits shall be permitted to supply: 1- Only lighting units: this case explained before in previous article 2- Only utilization equipment: with condition that the combined load for all utilization equipment must not exceed the branch circuit rating. 3- Combination of both: in the case the permissible rating of the utilization equipment will depend on its type as follows:

If it is not fastened-in-place, it can have a rating of up to 80 % of the branch circuit rating as in TABLE 210.21(B)(2). 

If it is fastened-in-place, other than luminaires, it shall not exceed 50 % of the branch-circuit ampere rating. 

B) 30-Ampere Branch Circuits A 30-ampere branch circuit shall be permitted to supply utilization equipment in any occupancy. A rating of any cord-and-plug-connected utilization equipment shall not exceed 80 percent of the branch-circuit ampere rating.

Important!!! A single receptacle installed on an individual branch circuit shall have an ampere rating not less than that of the branch circuit.

C) 40- and 50-Ampere Branch Circuits A 40- or 50-ampere branch circuit shall be permitted to supply cooking appliances that are fastened in place in any occupancy. In other than dwelling units, such circuits shall be permitted to supply fixed lighting units with heavy-duty lampholders, infrared heating units, or other utilization equipment.

D) Branch Circuits Larger Than 50 Amperes Branch circuits larger than 50 amperes shall supply only non-lighting outlet loads especially on industrial premises where conditions of maintenance and supervision ensure that only qualified persons service the equipment. Example for using Multioutlet branch circuits greater than 50 amperes: A common practice at industrial premises is to provide several single receptacles with ratings of 50 amperes or higher on a single branch circuit, to allow quick relocation of equipment for production or maintenance use, such as in the case of electric welders. Generally, only one piece of equipment at a time is supplied from this type of receptacle circuit. The type of receptacle used in this situation is generally a configuration known as a pin-and-sleeve receptacle, although the Code does not preclude the use of other configurations and designs. Pin-and-sleeve receptacles may or may not be horsepower rated.

3- Selecting Receptacle rating for a branch circuit 3.1 Receptacle rating, general Where connected to a branch circuit supplying two or more receptacles or outlets, receptacle ratings shall conform to the values listed in Table 210.21(B) (3), or, where rated higher than 50 amperes, the receptacle rating shall not be less than the branchcircuit rating.

Exceptions for above rule are as follows:

1. Receptacles for one or more cord-and-plug- connected arc welders shall be permitted to have ampere ratings not less than the minimum branch-circuit conductor ampacity permitted by 630.11(A) or (B), as applicable for arc welders. 2. The ampere rating of a receptacle installed for electric discharge lighting shall be permitted to be based on 410.62(C).

3.2 Single Receptacle on an Individual Branch Circuit

A single receptacle installed on an individual branch circuit shall have an ampere rating not less than that of the branch circuit. For example, a single receptacle on a 20-ampere individual branch circuit must be rated at 20 amperes. 3.3 Receptacle supplying Total Cord-and-Plug-Connected Load A receptacle shall not supply a total cord-and plug- connected load in excess of the maximum specified in Table 210.21(B)(2) mentioned above.

3.4 Range Receptacle Rating

The ampere rating of a range receptacle shall be permitted to be based on a single range demand load as specified in Table 220.55 in below.

4- Voltage ratings for Receptacle Branch circuit In dwelling units the voltage rating for Receptacle Branch circuit shall not exceed the following: A) 120 Volts between Conductors Circuits not exceeding 120 volts, nominal, between conductors shall be permitted to supply the following:

1. The terminals of lampholders applied within their voltage ratings Section 210.6(B)(1) allows lampholders to be used only within their voltage ratings. 2. Auxiliary equipment of electric-discharge lamps Auxiliary equipment includes ballasts and starting devices for fluorescent and high-intensitydischarge (e.g., mercury vapor, metal halide, and sodium) lamps. 3. Cord-and-plug-connected or permanently connected utilization equipment

Important!!! In guest rooms or guest suites of hotels, motels, and similar occupancies, the voltage shall not exceed 120 volts, nominal, between conductors that supply the terminals of the following: 1. Luminaires, 2. Cord-and-plug-connected loads 1440 volt-amperes, nominal, or less or less than 1⁄4 hp. The term similar occupancies refer to sleeping rooms in dormitories, fraternities, sororities, nursing homes, and other such facilities.

B) 208-volt or 240-volt circuit, nominal, between conductors For Receptacle Branch circuit that supplies High-wattage cord-and-plug-connected loads, such as electric ranges and some window air conditioners. Exceptions for (B) are as follows: 1. For lampholders of infrared industrial heating appliances as provided in 422.14. 2.

For railway properties as described in 110.19.

C) 277 Volts to Ground Circuits exceeding 120 volts, nominal, between conductors and not exceeding 277

volts, nominal, to ground shall be permitted to supply the following:

1.

Listed electric-discharge or listed LED type luminaires

2. Listed incandescent luminaires, where supplied at 120 volts or less from the output of a step-down auto-transformer that is an integral component of the luminaire and the outer shell terminal is electrically connected to a grounded conductor of the branch circuit 3.

Luminaires equipped with mogul-base screw shell lampholders

4. Lampholders, other than the screw shell type, applied within their voltage ratings 5.

Auxiliary equipment of electric-discharge lamps

6. Cord-and-plug-connected or permanently connected utilization equipment. Typical examples of the cord-and- are through-the-wall heating and air-conditioning units and restaurant deep fat fryers that operate at 480 volts, 3 phase, from a grounded wye system. The following image shows some examples of luminaires permitted to be connected to branch circuits.

Important!!! The requirements in (C) describes the voltage as ―volts, nominal, to ground,‖ whereas (A), (B), (D), and (E) describe voltage as ―volts, nominal, between conductors.‖ Luminaires listed for and connected to a 480-volt

source may be used in applications permitted by 210.6(C), provided the 480volt system is in fact a grounded wye system that contains a grounded conductor (thus limiting the system ―voltage to ground‖ to the 277-volt level).

Exceptions for (C) are as follows: 1. For lampholders of infrared industrial heating appliances as provided in 422.14. 2.

For railway properties as described in 110.19.

D) 600 Volts between Conductors Circuits exceeding 277 volts, nominal, to ground and not exceeding 600 volts, nominal, between conductors shall be permitted to supply the following: (A) The auxiliary equipment of electric-discharge lamps mounted in permanently installed luminaires where the luminaires are mounted in accordance with one of the following:

1. Not less than a height of 6.7 m (22 ft) on poles or similar structures for the illumination of outdoor areas such as highways, roads, bridges, athletic fields, or parking lots. 2. Not less than a height of 5.5 m (18 ft) on other structures such as tunnels

Important!!! For luminaire installations that are not on poles or in a tunnel, the branch circuit Voltage is limited to 277 volts to ground.

(B) Cord-and-plug-connected or permanently connected utilization equipment other than luminaires. (C) Luminaires powered from direct-current systems where the luminaire contains a listed dc-rated ballast that provides isolation between the dc power source and the lamp circuit and protection from electric shock when changing lamps.

Exceptions for (D) are as follows:

1. For lampholders of infrared industrial heating appliances as provided in 422.14. 2.

For railway properties as described in 110.19.

E) Over 600 Volts between Conductors. Circuits exceeding 600 volts, nominal, between conductors shall be permitted to supply utilization equipment in installations where conditions of maintenance and supervision ensure that only qualified persons service the installation.

5- The Maximum allowable number of receptacles on a branch circuit As per NEC section 220.14(I), Receptacle outlets load shall be calculated at not less than:



180 volt-amperes for each single receptacle



180 volt-amperes for each multiple receptacle (duplex or triplex) on one yoke.



90 volt- amperes per receptacle for multiple receptacles (four or more).

But, if a receptacle is dedicated for a specific device, then the actual load is used and If this dedicated load is continuous, then the 125% overrate is appropriate. To calculate the Maximum allowable number of receptacles on a branch circuit, make the following steps: 1.

Multiply the branch circuit voltage and amperage

2.

Then divide by 180 volt-amperes.

3. The result = the Max. Allowable single, duplex or triplex receptacles or a combination of them on a branch circuit.

Example#1:

How many receptacles can be placed on a 120-volt, 20-amp circuit? How many can be placed on a 120-volt, 15-amp circuit? Solution: Determine the maximum circuit power (for 20-amp circuit) = 120 V × 20 A = 2400 VA Determine the maximum circuit power (for 15-amp circuit) 120 V × 15 A = 1800 VA Then divide the power by the load per receptacle For 20-amp circuit: Maximum allowable number of receptacles = 2400 VA / 180 VA = 13.3 For 15-amp circuit: Maximum allowable number of receptacles = 1800 VA / 180 VA = 10 So, A 120-volt, 20-amp circuit can supply 13 receptacles. A 120-volt, 15-amp circuit can supply 10 receptacles.

6- The Minimum number of receptacle branch circuits for bank or office buildings As per NEC section 220.14(J), if the number of receptacles is unknown so for bank and office buildings, we can calculate the receptacles load by multiplying the area in ft2 by the unit value (1 VA/ft2). To get the required number of receptacle branch circuits for bank or office buildings make the following steps:



Calculate the total receptacles load for the whole building as explained above.

Calculate the total receptacle circuit load in VA by multiplying its voltage by its amperage as follows: 

1.

For 15-A Branch circuits = 120 V × 15 A = 1800 VA

2. For 20-A Branch circuits = 120 V × 20 A = 2400 VA  Divide the total receptacle load for the whole building by the maximum load per circuit to determine the minimum number of circuits.

Example#2: Determine the total receptacle load for an 80 ft × 120 ft office building? And determine the number of 15-amp circuits needed to supply the load. Noting that the number of receptacles is unknown. Solution: (a) The number of receptacles is unknown, so a receptacle load of 1 VA/ft2 can be calculated: Area = 80 ft × 120 ft = 9600 ft2 Total Receptacle load = 1 VA/ft2 × 9600 ft2 = 9600 VA (b) To determine the number of circuits required, first calculate the allowable load for a single circuit: The allowable load for a single circuit = 120 V × 15 A = 1800 VA Divide the total receptacle load by the maximum load per circuit to determine the minimum number of circuits: Minimum number of 15-A receptacle Circuits = 9600 VA / 1800 VA = 5.33 This is the minimum number, so round up to six circuits.

Important!!! As per NEC section 220.14(J), limit using the unit value (1 VA/ft2) for banks and office buildings but it can be used also (as approximate) for other types of buildings including dwelling ones.

Non-Dwelling Buildings Load Calculations- Part Three In Article " Non-Dwelling Buildings Load Calculations- Part One ", I introduced a List for ordinary Non-Dwelling Buildings Loads which was as follows: Lighting loads, Receptacles Loads, Kitchen Loads, Heating, Ventilation and air conditioning Loads (Non-Coincident Loads), Motor Loads, Other Loads.

1. 2. 3. 4. 5. 6.

Again, but for above Non-Dwelling Buildings Loads, I will explain the following points: 1. Where and how to distribute each type of load in a dwelling unit as per NEC code? 2. How to calculate its Demand load for feeder and service sizing calculations?

Important!!! All design Calculations for Non-dwelling Buildings will be as per NEC standard calculation method but I will explain design calculations as per NEC Optional calculation method only for the following Non-Dwelling buildings as permitted by NEC, Part IV. Optional Feeder and Service Load Calculations: 1. 2. 3.

Schools Existing Installations New Restaurants

I explained the Design Calculation for first type of Non-Dwelling Building loads which is Lighting loads in Article " Non-Dwelling Buildings Load Calculations- Part One ". Also, I explained part one of Design Calculation for second type; Receptacle loads for Non-Dwelling Buildings in Article " Non-Dwelling Buildings Load Calculations- Part Two ". Today, I will explain the Design Calculation of Receptacle loads - second part for Non-

Dwelling Building Loads which is Receptacle Loads.

Second: Receptacle Loads – Part Two 1- Locations of General-use Receptacle Loads in Non-dwelling Buildings

Important!!! In Non-dwelling buildings (commercial and industrial buildings), many 120volt receptacles may be necessary to accommodate any devices that don’t have special electrical requirements (General-use). But The NEC code doesn’t specify the quantity and location of General-use receptacles in Non-dwelling buildings, Unlike the dwelling buildings where the NEC code specifies the quantity and location of General-use receptacles as explained before in Article

Rule#1: Quantity and Location of General-use receptacles in Nondwelling buildings – General Rule Since, there is no receptacle placement provisions is specified by NEC code for Non-Dwelling Buildings, the following provisions can be used: The ― 6 ft rule‖ which is the maximum distance to a receptacle measured horizontally along the walls in dwelling units will not applied here, while The maximum advised distance between receptacles (12 ft rule) is a helpful guideline for general office spaces.  No general-use receptacles are needed in commercial bathrooms, but if installed, it must be GFCI type.  The designer adds one receptacle for each restroom.  The designer adds one receptacle for approximately 20 ft of wall space in hallways and corridors (to accommodate cleaning equipments such as vacuums.  As per NEC section 210.63, one 120 volt receptacle within 25 ft of any heating, air conditioning or refrigeration equipment, which may include any roof mounted air conditioning equipment 

Important!!! The quantity and location of these General-use receptacles will ultimately depend on customer requirements, so we have two cases: Case#1: If the final customer requirements are known during the design phase Case#2: If the final customer requirements are not known

Rule#2: Quantity and Location of General-use receptacles in Nondwelling buildings in Case#1 In this case, the following provisions will be done: The designer customizes the quantity and location of general purpose receptacles to meet the customer’s specific needs.  The designer might meet directly with the customer to discuss any specific requirements and alter the design accordingly.  The practical locations of these receptacles will be based also on the permanent furniture layout of the building, space, or area under design. 

Rule#3: Quantity and Location of General-use receptacles in Nondwelling buildings in Case#2 In this case, the designer must strive to create a design that provides for the average user’s electrical needs and allows for flexibility for future additions and expansions. For each area inside the building, the minimum number of receptacle outlets should be determined by assuming there is no furniture in that area and the following provisions will be done: The designer adheres to the NEC requirements for general purpose receptacles in dwelling buildings 210.52(A)(1) to provide a sufficient 

quantity of receptacles and meet any future needs.  For any other areas within the facility, the designer should consider how frequently the space is occupied and what is the primary usage of each space is to determine the proper quantity and placement of general use receptacle outlets.

Important!!! In case#2, after getting the permanent furniture layout of the building, space, or area under designs, the practical locations of general-use receptacles can be adjusted on this time.

1.1 Special Case: Locations and Minimum Quantity of General-use Receptacle Loads in Guest Rooms, Guest Suites, Dormitories, and Similar Occupancies

Definitions: Guest Room: An accommodation combining living, sleeping, sanitary, and storage facilities within a compartment. Guest Suite: An accommodation with two or more contiguous rooms comprising a compartment, with or without doors between such rooms, that provides living, sleeping, sanitary, and storage facilities.

Rule#4: Receptacles Placement for Guest rooms or guest suites in hotels, motels, sleeping rooms in dormitories, and similar occupancies Receptacles shall be installed such that no point measured horizontally along the floor line of any wall space is more than 1.8 m (6 ft) from a receptacle outlet.

Definition: The wall space is a wall unbroken along the floor line by doorways, fireplaces, archways, and similar openings and may include two or more walls of a room (around corners), as illustrated in Exhibit 210.27. The Minimum length for a wall space is 2 ft.

Important!!! The wall space behind the swing of a door is included in the measurement. This does not mean that the receptacle outlet has to be located in that space, only that the space is included in the wall-line measurement.

Rule#5: Minimum Number of Receptacles for Guest rooms or guest suites in hotels, motels, sleeping rooms in dormitories, and similar occupancies As per 210.60(B), At least two receptacle outlets must be readily accessible without requiring the movement of furniture to access those receptacles.

Important!!! Where receptacles are installed behind the bed, the receptacle shall be located to prevent the bed from contacting any attachment plug that may be installed or the receptacle shall be provided with a suitable guard. To reduce the risk of fire to bedding material, receptacles located behind beds must include guards if attachment plugs could contact the bed.

Rule#6: Bathroom Receptacles for Guest rooms or guest suites in hotels, motels, sleeping rooms in dormitories, and similar occupancies As per NEC Section 210.52(D), at least one receptacle outlet shall be installed in bathrooms in one of the following locations: Within 900 mm (3 ft) (36 inch) of the outside edge of each basin. on the side or face of the basin cabinet not more than 300 mm (12 in.) below the basin countertop. special cases for Rule#9 are as follows: 1. If there is more than one basin, a receptacle outlet is required adjacent to each basin location. 2. If the basins are in close proximity, one receptacle outlet installed between the two basins can be used to satisfy this requirement as shown in below image (top).

Rule#7: Guest rooms or guest suites provided with permanent provisions for cooking and countertop areas Extended-stay hotels and motels are often equipped with permanent provisions for cooking and countertop areas and shall have receptacle outlets installed in accordance with all of the applicable rules in 210.52. Note that, a portable microwave oven is not considered to be a permanently installed cooking appliance.

How to specify the required Type of general-use receptacles for each area? 1- Grounding type

As per NEC section 406.4, Receptacles installed on 15- and 20-ampere branch circuits shall be of the grounding type.

2- GFCI type

As per NEC section 210.8 (B), all 125-volt, single-phase, 15- and 20-ampere receptacles are GFCI for the following locations: A) Bathrooms Some motel and hotel bathrooms have the basin located outside the door to the room containing the tub, toilet, or another basin.

Definition: A bathroom is defined in Article 100 as ―an area including a basin with one or more of the following: a toilet, a urinal, a tub, a shower, a bidet, or similar plumbing fixtures.‖

Important!!! As per NEC section 406.9(C), it is prohibited to install a receptacle within or

directly over a bathtub or inside a shower stall even if the receptacles are installed in a weatherproof enclosure.

Important!!! As per NEC section 210.11(C)(3), At least one 20-ampere branch circuit shall be provided to supply bathroom receptacle outlet(s). This circuit is permitted to supply the required receptacles in more than one bathroom as show in above image (bottom).

B) Kitchens All 15- and 20-ampere, 125-volt kitchen receptacles in non-dwelling-type kitchens must be GFCI protected, whether or not the receptacle serves countertop areas.

Definition: Kitchen: is an area with a sink and permanent facilities for food preparation and cooking. Note that, A portable cooking appliance (e.g., cord-and-plug-connected microwave oven or hot plate) is not a permanent cooking facility. Kitchens, meeting the Article 100 definition above, in restaurants, hotels, schools, churches, dining halls, and similar facilities are examples of the types of kitchens covered by this requirement.

C) Rooftops As per NEC 210.63, a 125-volt, single-phase, 15- or 20-ampere-rated receptacle outlet shall be installed at an accessible location for the servicing of heating, airconditioning, and refrigeration equipment. The receptacle shall be located on the same level and within 7.5 m (25 ft) of the heating, air-conditioning, and refrigeration equipment. The receptacle outlet shall not be connected to the load side of the equipment disconnecting means. This receptacle improves worker safety by eliminating the need to employ makeshift methods of obtaining 125-volt power for servicing and troubleshooting. Exception for GFCI requirement for 210.63 receptacles is as follows:

1. Receptacles that are not readily accessible and are supplied by a branch circuit dedicated to electric snow-melting, deicing, or pipeline and vessel heating equipment shall be permitted to be installed in accordance with 426.28 or 427.22, as applicable.

D) Outdoors Other than the limited exclusions covered by the two exceptions to 210.8(B)(4), all 125-volt, single-phase, 15- and 20-ampere receptacles installed outdoors at commercial, institutional, and industrial occupancies are required to be provided with ground-fault circuit-interrupter protection. The two exceptions to 210.8(B)(4) are as follows: 1. Receptacles that are not readily accessible and are supplied by a branch circuit dedicated to electric snow-melting, deicing, or pipeline and vessel heating equipment shall be permitted to be installed in accordance with 426.28 or 427.22, as applicable. 2. In industrial establishments only, where the conditions of maintenance and supervision ensure that only qualified personnel are involved, an assured equipment grounding conductor program as specified in 590.6(B) (2) shall be permitted for only those receptacle outlets used to supply equipment that would create a greater hazard if power is interrupted or having a design that is not compatible with GFCI protection.

Important!!! Although commercial, institutional, and industrial occupancies are not required to have outdoor receptacle outlets installed for general use, there may be outdoor receptacle outlets installed to meet the requirement of 210.63 or at the discretion of the designer or owner.

E) Sinks Where receptacles are installed within 1.8 m (6 ft) of the outside edge of the sink. This requirement covers receptacles installed near sinks in lunchrooms, janitors’ closets, classrooms, and all other areas that are not covered by the bathroom and kitchen provisions of 210.8(B)(1) and (B)(2).

Two exceptions to above are as follows: 1. In industrial laboratories, receptacles used to supply equipment where removal of power would introduce a greater hazard shall be permitted to be installed without GFCI protection. 2. For receptacles located in patient bed locations of general care or critical care areas of health care facilities other than those covered under 210.8(B)(1), GFCI protection shall not be required.

F) Indoor wet locations G) Locker rooms with associated showering facilities H) Garages, service bays, and similar areas where electrical diagnostic equipment, electrical hand tools, or portable lighting equipment are to be used

3- Tamper-Resistant type

NEC Section 406.12 requires listed tamper-resistant receptacles in all areas specified in 210.52 for dwelling units to increase the level of safety for children. Since guest rooms or guest suites in hotels and motels are also likely to be occupied by children, this section requires the same level of protection.  As per NEC 406.13, all non-locking-type, 125-volt, 15- and 20-ampere receptacles located inguest rooms and guest suites shall be listed tamper-resistant receptacles.  As per NEC 406.14 In all child care facilities, all non-locking-type, 125-volt, 15and 20- ampere receptacles shall be listed tamper-resistant receptacles. 

Non-Dwelling Buildings Load Calculations- Part Four In Article " Non-Dwelling Buildings Load Calculations- Part One ", I introduced a List for ordinary Non-Dwelling Buildings Loads which was as follows: 1. 2. 3. 4. 5. 6.

Lighting loads, Receptacles Loads, Kitchen Loads, Heating, Ventilation and air conditioning Loads (Non-Coincident Loads), Motor Loads, Other Loads.

Again, but for above Non-Dwelling Buildings Loads, I will explain the following points: 1. Where and how to distribute each type of load in a dwelling unit as per NEC code? 2. How to calculate its Demand load for feeder and service sizing calculations?

Important!!! All design Calculations for Non-dwelling Buildings will be as per NEC standard calculation method but I will explain design calculations as per NEC Optional calculation method only for the following Non-Dwelling buildings as permitted by NEC, Part IV. Optional Feeder and Service Load Calculations: 1. 2. 3.

Schools, Existing Installations, New Restaurants.

I explained the Design Calculation for first type of Non-Dwelling Building loads which is Lighting loads in Article " Non-Dwelling Buildings Load Calculations- Part One ". Also, I explained parts one & Two of Design Calculation for second type; Receptacle loads for Non-Dwelling Buildings in Articles:

 

Non-Dwelling Buildings Load Calculations- Part Two Non-Dwelling Buildings Load Calculations- Part Three

Today, I will explain the Design Calculation of Receptacle loads - Third part for NonDwelling Building Loads which is Receptacle Loads.

Second: Receptacle Loads – Part Three 1- General –Use Receptacle Loads in Non-Dwelling Units

Important!!! The receptacle outlets in Non-Dwelling Buildings will be counted to calculate their load, unlike the Dwelling buildings where the General-Use Receptacles Load is calculated with the general lighting Load by multiplying Building Area in (ft2) with its power density in (VA/ft2) from table 220.42 in below image.

Important!!! General-Use Receptacles are not-continuous Loads, No need to multiply it by 1.25 to calculate their contribution in the total load. But, if a Receptacle is dedicated for a specific device, then the actual load is used and If this dedicated load is continuous, then the 125% overrate is appropriate.

Rule#1: General-Use Receptacle Load (see below image) As per NEC section 220.14(I), Receptacle outlets load shall be calculated at not less than: 180 volt-amperes for each single receptacle,  180 volt-amperes for each multiple receptacle (duplex or triplex) on one yoke/strap,  90 volt- amperes per receptacle for multiple receptacles (four or more). 

Important!!! The general lighting load unit values specified in table 220.42 for guest rooms or guest suites of hotels and motels includes All general-use receptacle outlets of 20-ampere rating or less, including receptacles connected to Bathroom Branch Circuits, So, no need to add the above outlets in load calculations per NEC method.

Example#1: For a commercial building having 75 duplex and 10 single receptacles, what is the calculated load for these receptacles?

Solution: As per NEC section 220.14(I), Receptacle outlets load shall be calculated at not less than: 180 volt-amperes for each single receptacle, 180 volt-amperes for each multiple receptacle (duplex or triplex) on one yoke/strap. So, Receptacle calculated load = (75+10) x 180 VA = 15,300 VA

Important!!! In Commercial and Industrial Buildings, it is common to use fixed multioutlet assemblies, which can be classified to two types according to method of usage as follows: 1. Light use: which means that not all the cord-connected equipment is expected to be used at the same time, as defined in 220.14(H)(1). An example of light use is a workbench area where one worker uses one electrical tool at a time. 2. Heavy use: which is characterized by all the cord-connected equipment generally operating at the same time, as defined in 220.14(H)(2). An example of heavy use is a retail outlet displaying

television sets, where most, if not all, sets are operating simultaneously.

Rule#2: Fixed Multioutlet Assemblies Load (see below image) As per NEC section 220.14(H), Fixed multioutlet assemblies used shall be calculated as follows: 1. For Light use, each 1.5 m (5 ft) or fraction thereof of each separate and continuous length shall be considered as one outlet of not less than 180 volt-amperes. 2. For heavy use, each 300 mm (1 ft) or fraction thereof shall be considered as an outlet of not less than 180 volt-amperes.

Important!!! Fixed multioutlet assemblies are not-continuous Loads, No need to multiply it by 1.25 to calculate their contribution in the total load. But, if Fixed multioutlet assemblies is dedicated for a specific device, then the actual load is used and If this dedicated load is continuous, then the 125% overrate is appropriate.

Important!!! Fixed multioutlet assemblies in guest rooms or guest suites of hotels and motels are included in The general lighting load unit values specified in table 220.12.

So, no need to add the Fixed multioutlet assemblies in load calculations for these locations.

Example#2: For a commercial building has 75 linear feet of fixed multioutlet assembly, with 15 feet of the assembly subject to simultaneous use. What is the calculated load for this fixed multioutlet assembly?

Solution: Length of fixed multioutlet assembly Portion for Non- simultaneous use (Light Use) = 75 – 15 = 60 linear feet. Calculated load for Light use portion = (60/5) x 180 VA = 2,160 VA Calculated load for Heavy use (simultaneous use) portion = 15 x 180 VA = 2,700 VA Total calculated load for this fixed multioutlet assembly = 2,160 VA + 2,700 VA = 4,860 VA

Rule#3: General-Use Receptacles and Fixed Multioutlet Assemblies Loads Demand Factors As per NEC section 220.44, General-Use Receptacles and Fixed Multioutlet Assemblies are subjected to demand factors by either of the following two methods: 1. If the occupancy is one of the types (other than dwelling units) listed in Table 220.42, the receptacle load could be added to the general lighting load and made subject to the demand factors in Table 220.42. 2. If the occupancy is not one of the types listed in Table 220.42 and receptacle loads greater than 10,000 volt-amperes, The receptacle loads are calculated (without the lighting load) with demand factors from Table 220.44 (in below image) applied.

Example#3: The general lighting load, before applying demand factors, for hospital patient rooms is 100,000 volt-amperes. The receptacle branch-circuit load, calculated in accordance with 220.14(H) and (I), is 144,000 volt-amperes. What is the lighting and receptacle load after demand factors? Solution: Since Hospital is one of the building types (other than dwelling units) listed in Table 220.42, the receptacle load could be added to the general lighting load and made subject to the demand factors in Table 220.42. Total load = lighting load + receptacles load = 100,000 VA+ 144,000 VA = 244,000 VA Then apply demand factors for hospital from table 220.42 as follows: Multiply the first 50,000 by 40 percent = 50,000 × 40% = 20,000 VA The Remaining = 244,000 – 50,000 = 194,000 VA Multiply the remaining by 20 percent = 194,000 × 20% = 38,800 VA The lighting and receptacle load, after applying Table 220.42 demand factors= 20,000 + 38,800 = 58,800 VA

Example#4: For a hospital building, the receptacle branch-circuit load is 144,000 volt-amperes. What is the receptacle load after applying the demand factors in Table 220.44?

Solution: Applying the demand factors in Table 220.44 The first 10 kVA is multiplied by 100 percent = 10,000 x 100% = 10.000 VA The remainder = 144,000 – 10,000 = 134,000 VA Multiply the remainder by 50 percent = 134,000 × 50% = 67,000 VA Then, the receptacle load, after applying Table 220.44 demand factors = 10,000 + 67,000 = 77,000 VA

Example#5: A store has a total of 80 duplex receptacles. What is the receptacle load after demand factors?

Solution: First, multiply the 80 receptacles by 180 volt-amperes. Receptacle load before applying demand factors = 80 × 180 VA= 14,400 VA Second, Applying the demand factors in Table 220.44 The first 10 kVA is multiplied by 100 percent = 10,000 x 100% = 10.000 VA The remainder =14,400 – 10,000 = 4,400 VA Multiply the remainder by 50 percent = 4,400 × 50% = 2,200 VA Then, the receptacle load, after applying Table 220.44 demand factors= 10,000 + 2200 = 12,200 VA

Rule#4: Special case#1: where the actual number of receptacle outlets in banks and office buildings is unkown

As per NEC section 220.14(K), in bank and office buildings, if the number of receptacles is unknown, we can calculate the receptacles load by multiplying the bank or office building area in ft2 by the unit value (1 VA/ft2).

Rule#5: Special case#2: where the actual number of receptacle outlets in banks and office buildings is kown As per NEC section 220.14(K), in bank and office buildings, if the number of receptacles is known, then we can calculate the receptacles load to be the larger of the following two loads: 1. 2.

Load computed as per Rule#1 above, Load computed as per Rule#4 above.

Example#6: In an 8,500 square foot bank, it is not known how many receptacles are there, what is the calculated load for receptacles?

Solution: As per NEC section 220.14(K), in bank and office buildings, if the number of receptacles is unknown, we can calculate the receptacles load by multiplying the bank or office building area in ft2 by the unit value (1 VA/ft2). The calculated load for receptacles = 8500 ft2 x 1 VA/ft2 = 8,500 VA

Non-Dwelling Buildings Load Calculations- Part Five In Article " Non-Dwelling Buildings Load Calculations- Part One ", I introduced a List for ordinary Non-Dwelling Buildings Loads which was as follows: 1. Lighting loads, 2. Receptacles Loads,

3. 4. 5. 6.

Kitchen Loads, Heating, Ventilation and air conditioning Loads (Non-Coincident Loads), Motor Loads, Other Loads.

Again, but for above Non-Dwelling Buildings Loads, I will explain the following points: 1. Where and how to distribute each type of load in a dwelling unit as per NEC code? 2. How to calculate its Demand load for feeder and service sizing calculations?

Important!!! All design Calculations for Non-dwelling Buildings will be as per NEC standard calculation method but I will explain design calculations as per NEC Optional calculation method only for the following Non-Dwelling buildings as permitted by NEC, Part IV. Optional Feeder and Service Load Calculations: Schools, Existing Installations, New Restaurants.

I explained the Design Calculation for first type of Non-Dwelling Building loads which is Lighting loads in Article " Non-Dwelling Buildings Load Calculations- Part One ". Also, I explained parts one & Two of Design Calculation for second type; Receptacle loads for Non-Dwelling Buildings in Articles: 

Non-Dwelling Buildings Load Calculations- Part Two



Non-Dwelling Buildings Load Calculations- Part Three Non-Dwelling Buildings Load Calculations- Part Four



Today, I will explain the Design Calculation of third type of loads in Non-Dwelling Building Loads which is Kitchen Load.

Third: Kitchen Load

Definition: Kitchen: An area with a sink and permanent provisions for food preparation and cooking.

Important!!! Kitchen equipment is not limited to restaurants; it can be a portion of a load calculation for a school.

Rule#1: commercial electric cooking equipment Load (see below image for kitchen) As per NEC section 220.56, it shall be permissible to calculate the load for commercial electric cooking equipment, dishwasher booster heaters, water heaters, and other kitchen equipment in accordance with Table 220.56 (see below image). These demand factors shall be applied to all equipment that: 1. 2.

Has thermostatic control or Intermittently used as kitchen equipment.

Important!!! Demand factors of table 220.56 shall not apply to space-heating, ventilating, or air-conditioning equipment.

Important!!! As per table 220.56, No derating is allowed for only one or two pieces of kitchen equipment.

Important!!! Because commercial electric cooking and other kitchen equipment is used more intensively and for longer periods of time than electric cooking equipment in a dwelling unit, the demand factors in Table 220.56 do not provide a comparable demand to that permitted for dwelling unit cooking appliances until there are six or more commercial appliances.

Important!!! Because commercial electric cooking and other kitchen equipment is used more intensively and for longer periods of time than electric cooking equipment in a dwelling unit, the demand factors in Table 220.56 do not provide a comparable demand to that permitted for dwelling unit cooking appliances until there are six or more commercial appliances.

Important!!! Unlike Table 220.55, Table 220.56 is not just for ranges and cooking appliances; it will apply for all equipment that has either thermostatic control or intermittent use as kitchen equipment (like commercial electric cooking equipment, dishwasher booster heaters, water heaters, and other kitchen equipment).

Rule#2: Comparing the demand load to the sum of the largest two kitchen equipment loads As per NEC section 220.56, the calculated demand load for the feeder or service must not be less than the sum of the largest two kitchen equipment loads. Two cases are available: 1. The demand load > the sum of the largest two kitchen equipment loads, in this case, it is permissible to use the demand load. 2. The demand load < the sum of the largest two kitchen equipment loads, in this case, the demand load must not be used and The minimum load will be the sum of the largest two kitchen equipment loads.

Example#1: What is the demand load for a restaurant with two 14-kW stoves, a 6-kW oven, a 5-kW dishwasher, a 3-kW booster heater, and a 4-kW food waste disposer?

Solution: The total number of units is six. The demand factor percent in Table 220.56 for six units is 65 percent. The total rating of the equipment = 14 + 14 + 6 + 5 + 3 + 4 = 46 kW The load after applying the Table 220.56 demand factor = 46 × 65% = 29.9 kW Comparing the demand load to the sum of the largest two kitchen equipment loads The absolute minimum rating is the total combined rating of the largest two kitchen equipment loads = 14 + 14 = 28 KW Since the demand load of 29.9 kW is higher than the sum of the two highest rated units, it is permissible to use the demand load.

Example#2: What is the demand load for the following commercial kitchen equipment: five 3-kW ovens, one 5-kW cooktop and two 20-kW ranges?

Solution: The total number of units is eight. The demand factor percent in Table 220.56 for eight units is 65 percent. T The total rating of the equipment = 3 + 3 + 3 + 3 + 3 + 5 + 20 + 20 = 60 kW The load after applying the demand factor = 60 × 65% = 39 kW The sum of the largest two kitchen equipment loads = 20 + 20 = 40 Since the demand load of 39 kW is lower than the sum of the two highest rated units, the demand load must not be used. The minimum load for these eight units of kitchen equipment is 40 kW

Important!!! As per NEC section 220.55, demand factors in Table 220.55 cannot be applied to commercial kitchen equipment.

Definition: An instructional program Building: a building in which a discipline and an organized sequence or grouping of courses leading to a defined objective such as a major, degree, certificate, license, the acquisition of selected knowledge or skills, or transfer to another institution of higher education. For example A high School Classroom.

Rule#3: Cooking Appliances Load in instructional programs Calculating the Cooking Appliances Loads in Instructional programs building will have the following cases: 1. If the cooking appliances rated over 1.75 KW and are household type then apply demand factors from table 220.55 as per note#5 on this table. 2. If the cooking appliances rated over 1.75 KW and are Commercial type then apply demand factors from table 220.56.

Important!!! Household electric cooking equipment installed in other than dwellings units and not in instructional programs must be calculated in accordance with Table 220.56.

Example#3:

A high school classroom will be converted into a culinary arts classroom. Five 12-kW, 240-volt household electric ranges will be installed. A separate panelboard will be installed to supply the ranges. What is the kilowatt-demand load for the feeder and panelboard supplying power to these ranges?

Solution: Since the ranges will be used in instructional programs and they are household electric ranges. So, table 220.55 note#5 will be applied. The feeder and panelboard could be sized on 100 percent of the equipment’s nameplate ratings, which is 60 kW, but this is not necessary. Look in the left column of Table 220.55 for five appliances. Because of the range ratings, the maximum demand is in Column C. The maximum demand required for five 12-kW, 240-volt house-hold electric ranges is 20 kW.

Example#4: A high school classroom will be converted into a culinary arts classroom. Five 12-kW, 240-volt commercial electric ranges will be installed. A separate panelboard will be installed to supply the ranges. What is the kilowatt-demand load for the feeder and panelboard supplying power to these ranges? Solution: Since the ranges will be used in instructional programs and they are not household (commercial) electric ranges. Therefore, it is not permissible to apply the demand factors from Table 220.55. Demand factors for commercial electric cooking equipment are in Table 220.56. Look in the left column of Table 220.56 for five units of equipment and follow across for the demand factor. The demand factor percent for five units is 70 percent. These five ranges have a total rating = 5 × 12 = 60 kW The load after applying the demand factor = 60 × 70% = 42 kW The feeder and panelboard supplying power to these five commercial ranges must be capable of carrying a load of at least 42 kW

Non-Dwelling Buildings Load Calculations- Part Six In Article " Non-Dwelling Buildings Load Calculations- Part One ", I introduced a List for ordinary Non-Dwelling Buildings Loads which was as follows:

1.

Lighting loads,

2.

Receptacles Loads,

3.

Kitchen Loads,

4.

Heating, Ventilation and air conditioning Loads (Non-Coincident Loads),

5.

Motor Loads,

6.

Other Loads.

Again, but for above Non-Dwelling Buildings Loads, I will explain the following points: 1. Where and how to distribute each type of load in a dwelling unit as per NEC code? 2. How to calculate its Demand load for feeder and service sizing calculations?

Important!!! All design Calculations for Non-dwelling Buildings will be as per NEC standard calculation method but I will explain design calculations as per NEC Optional calculation method only for the following Non-Dwelling buildings as permitted by NEC, Part IV. Optional Feeder and Service Load Calculations: 1. 2. 3.

Schools, Existing Installations, New Restaurants.

I explained the Design Calculation for first type of Non-Dwelling Building loads which is Lighting loads in Article " Non-Dwelling Buildings Load Calculations- Part One ",.

Also, I explained Design Calculation for second type; Receptacle loads for NonDwelling Buildings in Articles:



Non-Dwelling Buildings Load Calculations- Part Two



Non-Dwelling Buildings Load Calculations- Part Three



Non-Dwelling Buildings Load Calculations- Part Four

And I explained the third type of Non-dwelling Loads; Kitchen Loads in Article " NonDwelling Buildings Load Calculations- Part Five ". Today, I will explain the Design Calculation of Heating, Ventilation and Air conditioning Loads (Non-Coincident Loads) in Non-Dwelling Building. You can review our course " Introduction to Heating, Ventilation and Air Conditioning Systems " for more information.

Forth: Heating, Ventilation and Air Conditioning Loads (Non-Coincident Loads)

Important!!! Regardless of the type of structure—residential, commercial, or industrial, the Heating, Ventilation and Air Conditioning Systems will be calculated by NEC part III (standard method) in the following manner.

Definition: A heat pump (see below image): a device that acts as an air conditioner in the summer and as a heater in the winter. Heat pumps look and function exactly like an air conditioner except it has a reversible cycle.

Important!!! Most service load calculations will include heating and/or air conditioning equipment, but not all feeder load calculations will include these types of loads. If the feeder will not supply power to heating and air conditioning equipment, calculate just the general loads on this feeder. If a service will not supply heating equipment calculate only the service for air condition only. If a service will not supply power to heating and air conditioning equipment, ignore this load in service load calculation.

Rule#1: Service load for Room air conditioners

The load for Room air conditioners shall be calculated at 100 % of its ampere rating which may be indicated on its nameplate and will be used in branch, feeder and service load calculations.

Rule#2: Service load for Fixed electric space-heating loads As per NEC section 220.51, Fixed electric space-heating loads shall be calculated at 100 percent of the total connected load. However, in no case shall a feeder or service load current rating be less than the rating of the largest branch circuit supplied.

Important!!! Cord and plug space heaters are not a permanent fixed heaters, then it will not be included in load calculation in this step. Fixed heating equipment, such as central heating systems, boilers, heating cable, and unit heaters (baseboard, panel, and duct heaters) will be included.

Important!!! Fixed electric space heating shall be considered continuous load.

Example#1:

An Office has seven wall heaters; each heater is rated 3,000 watts at 240 volts. How much load will these heaters add to a 240-volt, single-phase service? Assuming that The air conditioning load will be less than the heating load.

Solution:

In accordance with 220.51, calculate the heaters service load at 100 %= 7 × 3,000 = 21,000 watts Since the service voltage is known, the total current draw of the heaters can be calculated by dividing the total watts by 240 volts The total current draw of the heaters = 21,000 ÷ 240 = 87.5 A = 88 A.

Rule#3: Central air conditioning and heating system Load Central air conditioning and heating system Load shall be calculated at 100 % of its nameplate and will be a Noncoincident Load.

Rule#4: Noncoincident Loads As per NEC section 220.60, where it is unlikely that two or more noncoincident loads will be in use simultaneously, it shall be permissible to use only the largest load(s) that will be used at one time for calculating the total load of a feeder or service.

Important!!! Depending on the design, the heating system and the air conditioning system might be noncoincident loads.

Example#2:

What is the service load contribution for an Office Building that will have electric space heaters and window air conditioners? The heat will consist of two offices with space heaters rated 1,500 watts each and three offices with space heaters rated 2,250 watts each. The air conditioning will consist of four offices with window air conditioning units rated 11.5 amperes at 240 volts. Assume space-heating watts are equivalent to voltamperes (VA).

Solution:

In this Office Building, the heat load and the air conditioning load are noncoincident loads. The heat and the air conditioning will not be energized at the same time. Therefore, compare the total heat load to the total air conditioning load and omit the smaller of the two loads. First: calculate the heat load The total heat load = 1,500 + 1,500 + 2,250 + 2,250 + 2,250 = 9,750 VA. Second: calculate the air conditioning load Start by finding the volt-amperes of each unit (VA = E × I) (assume power factor or PF = 1.0). The load of each air conditioner = 240 × 11.5 = 2,760 VA The total air conditioner load = 2,760 VA × 4 = 11,040 VA Since the air conditioner load is larger than the heat load, omit the heat load. The service load contribution for the heating and air conditioning loads in this Office Building is 11,040 VA If the air conditioner compressors are the largest motors in this Office Building, multiply the load of one compressor by 25 percent and add it to the service load calculation.

Rule#5: the air handler (or blower motor) is not a noncoincident load Although the heating and air conditioning in package units and split systems are noncoincident loads, the air handler (or blower motor) (or evaporator motor) is not. Since the blower motor works with both the heating and air conditioning system, it must be included in both calculations.

Example#3:

A heating/cooling package unit will be installed in an Office. The electric heater is rated 9.6 kW at 240 volts. The blower motor inside the package unit that circulates the air will be a ½ horsepower, 240-volt motor. How much load will this package unit add to a 240-volt, single-phase service? Assuming that the air conditioning load will be less than the heating load.

Solution:

The heat load= 9.6 KW × 1,000 = 9,600 watts The full-load current in amperes of a ½ hp, single-phase, 240-volt motor (from Table 430.248) = 4.9 A The motor load = 4.9 A × 240 V = 1,176 VA The service load for this package unit = 1,176 + 9,600 = 10,776 watts

Example#4:

A package unit has electric heat and air conditioning. The unit contains a compressor, a blower motor, a condenser fan motor and electric heat. The compressor draws 23 amperes at 240 volts, the blower motor draws 5 amperes at 240 volts and the condenser fan motor draws 2 amperes at 240 volts. The rating of the heat is 10 kW. Assume heating kilowatt rating is equivalent to kilovolt-amperes. What is the service load contribution for this heating/cooling package unit?

Solution:

The load of the air conditioner compressor =23 A × 240 V= 5,520 VA The load of the blower motor = 5 A × 240 V = 1,200 VA The load of the condenser fan motor = 2 A × 240 V = 480 VA The total air conditioning load = 5,520 + 1,200 + 480 = 7,200 VA Since the blower motor also works with the heat, the load of the blower motor must be added to the heat load. So, the heat load = 10 KW× 1,000 = 10,000 + 1,200 = 11,200 VA Since the heat load is larger than the air conditioner load, omit the air conditioner load. The service load contribution for this package unit is 11,200 VA.

Rule#6: Largest Motor in the feeder or service load calculation As per NEC sections 220.50 and 430.24, when calculating a feeder or service, the largest motor must be multiplied by 125 percent.

Important!!! Unless it is the largest motor in the feeder or service load calculation, do not multiply the full-load current of the motor by 125 percent.

Example#5:

In example#3, the blower motor in the heating/cooling package unit will be the largest motor. How much load will this package unit add to a 240-volt, single-phase service?

Solution:

If the ½ hp blower motor is the largest motor in the calculation for the service, the ampacity must not be less than 125 percent of the full-load current rating plus the calculated load of the electric heat. Multiply the motor’s full-load current by 125 percent before adding it to the electric heat service load The motor load = 4.9 A × 240 V = 1,176 VA The service load for this package unit = 1,176 x 1.25 + 9,600 = 11,070 watts

Rule#7: A heat pump with supplementary heat is not a noncoincident load With a heat pump, the compressor (and accompanying motors) and some or all of the electric heat can be on at the same time. The load contribution of a heat pump is the air conditioning system load plus the maximum amount of heat that can be on while the air conditioner compressor is on.

Example#6:

What is the service load contribution for a heat pump with supplementary heat? The compressor draws 26.4 amperes at 240 volts, the blower motor draws 6.5 amperes at 240 volts and the condenser fan motor draws 3 amperes at 240 volts. The electric heat in this unit has a rating of 15 kW. Assume heating kilowatt rating is equivalent to kilovolt-amperes.

Solution:

The compressor and all of the heat in this heat pump can be energized at the same time. The load of the air conditioner compressor = 26.4 × 240 = 6,336 VA The load of the blower motor = 6.5 × 240 = 1,560 VA The load of the condenser fan motor = 3 × 240 = 720 The total air conditioning load = 6,336 + 1,560 + 720 = 8,616 VA Since the air conditioning system and all of the heat can be on at the same time, add the two together to get the service load contribution. The service load contribution for this heat pump = 8,616 + 15,000 = 23,616 VA

Important!!! Noncoincident loads are not limited to heating and air conditioning systems.

Example#7:

What is the feeder load contribution for two 5-hp, 230-volt, single phase pump motors? The motors will be wired so only one motor is in operation at any time.

Solution:

Because these two motors are noncoincident loads, it is permissible to omit the load of one motor. Start by finding the full-load current (FLC) of one motor. Full-load currents for single-phase, alternating-current motors are in Table 430.248. From Table 430.248, The FLC of this motor is 28 amperes. The load of one motor = 28 A × 230 V = 6,440VA Although there are two motors, they will never operate at the same time. Therefore, only one motor must be added to the feeder calculation. The feeder load contribution = 6,440 VA If this motor is the highest-rated motor on this feeder, multiply the load of this motor by 25 percent, and add it to the feeder load calculation.

Non-Dwelling Buildings Load Calculations- Part Seven In Article " Non-Dwelling Buildings Load Calculations- Part One " I introduced a List for ordinary Non-Dwelling Buildings Loads which was as follows: 1. 2. 3. 4. 5. 6.

Lighting loads, Receptacles Loads, Kitchen Loads, Heating, Ventilation and air conditioning Loads (Non-Coincident Loads), Motor Loads, Other Loads.

Again, but for above Non-Dwelling Buildings Loads, I will explain the following points:

1. Where and how to distribute each type of load in a dwelling unit as per NEC code? 2. How to calculate its Demand load for feeder and service sizing calculations?

Important!!! All design Calculations for Non-dwelling Buildings will be as per NEC standard calculation method but I will explain design calculations as per NEC Optional calculation method only for the following Non-Dwelling buildings as permitted by NEC, Part IV. Optional Feeder and Service Load Calculations: 1. 2. 3.

Schools, Existing Installations, New Restaurants.

I explained the first four parts as follows: 1- Load Calculation for Lighting Loads of Non-Dwelling Building in Article " NonDwelling Buildings Load Calculations- Part One ". 2- Load Calculation for Receptacle loads of Non-Dwelling Buildings in Articles: 

Non-Dwelling Buildings Load Calculations- Part Two



Non-Dwelling Buildings Load Calculations- Part Three



Non-Dwelling Buildings Load Calculations- Part Four

3- Load Calculation for Kitchen Equipment Loads of Non-dwelling Building in Article " Non-Dwelling Buildings Load Calculations- Part Five ". 4- Load Calculation for Heating, Ventilation and Air Conditioning Systems Loads (NonCoincident Loads) of Non-dwelling Building in Article " Non-Dwelling Buildings Load Calculations- Part Six ". Today, I will explain the Load Calculation for Motor and Other Loads of Non-Dwelling Building.

Fifth: Motor Loads

Important!!! Regardless of the type of structure—residential, commercial, or industrial, the Motors Loads will be calculated by in the same manner. The motors can be fastened in place appliances or separate motors (air conditioning compressors, fan blower, etc.)

Rule#1: Largest Motor Load When calculating a feeder or service As per NEC Standard calculation method, the largest motor must be multiplied by 25 percent and add it to the service load calculation.

Important!!! Most electrical equipment is rated in volt-amperes (VA) or watt input. While motors traditionally have been rated in horsepower output (Some motors are available with their output ratings expressed in watts and kilowatts).

Rule#2: Motor Loads As per NEC section 430.6(A)(1), Do not use the actual current rating marked on the nameplate. When calculating motor loads, use the values given in Tables 430.247 through 430.250.

Important!!! Exceptions to 430.6(A)(1) : 1.

Motors built for low speeds (less than 1,200 rpm) or high

torques for multispeed motors. 2. For equipment that employs a shaded-pole or permanent-split capacitor-type fan or blower motor that is marked with the motor type, use the full load current for such motor marked on the nameplate of the equipment in which the fan or blower motor is employed. 3. For a listed motor-operated appliance that is marked with both motor horsepower and full-load current, use the motor full-load current marked on the nameplate of the appliance.

Important!!! Full-load currents for 3-phase motors are in Table 430.250, Full-Load Current for Two-Phase Alternating- Current Motors (4-Wire) are in Table 430.249, Full-load currents for single-phase motors are in Table 430.248, Full-Load Currents for Direct-Current Motors are in Table 430.247.

Important!!! Kilovolt-amperes (kVA) shall be considered equivalent to kilowatts (kW) for Fastened-in-place Appliances.

Important!!! If the Largest motor is air conditioning compressor, in this case, multiply the load of one compressor by 25 percent and add it to the service load calculation

But if the heating load is larger than the air conditioning load, and because of 220.60 which states that‖ it is permissible to use only the larger of the noncoincident loads‖ the air conditioning load will be omitted and the air conditioning compressor will not be the largest motor in this case.

Example#1: For example, what is the minimum rating in amperes for conductors supplying a 2 hp, 230-volt, single-phase motor?

Solution: Find the full-load current in amperes for the given motor. As shown in Table 430.248, the current for a 2 hp, 230-volt motor is 12 amperes.

Example#2: What is the minimum rating in amperes for conductors supplying a 10 hp, 208-volt, 3phase motor and a 7½ hp, 208-volt, 3-phase motor?

Solution: The full-load current from Table 430.250 for a 10 hp, 208-volt, 3-phase motor is 30.8 amperes. The full-load current from the same table for a 7½ hp, 208-volt, 3-phase motor is 24.2 amperes. Multiply the highest full-load current by 125 percent = 30.8 x 125 % = 38.5 A Now add to this value to the full-load current of the other motor= 38.5 + 24.2 = 62.7 A So, the minimum rating in amperes for conductors supplying above motors is 62.7 A

Example#3: what is the minimum rating in amperes for conductors supplying a 5 hp, 230-volt, 3phase motor; a 10 hp, 230-volt, 3-phase motor; a 15 hp, 230-volt, 3-phase motor; and a 10 hp, 230-volt, single-phase motor?

Solution: Full-load currents for 3-phase motors are in Table 430.250, and full-load currents for single-phase motors are in 430.248. The full-load current for a 5 hp, 230-volt, 3-phase motor is 15.2 amperes; The full-load current for a 10 hp, 230-volt, 3-phase motor is 28 amperes; The full-load current for a 15 hp, 230-volt, 3-phase motor is 42 amperes; And the full-load current for a 10 hp, 230-volt, single-phase motor is 50 amperes Although the 15 hp motor has the largest amount of horsepower, the motor with the highest full-load current rating is the single-phase motor. Multiply the motor with the highest full-load current rating by 125 percent = 50 × 125% = 62.5 A Now add to this value to the full-load currents of the other motors = 62.5 + 15.2 + 28 + 42 = 147.7 = 148 A So, the minimum rating in amperes for conductors supplying these motors is 148 A

Example#4: A heating/cooling package unit will be installed in an Office. The electric heater is rated 9.6 kW at 240 volts. The blower motor inside the package unit that circulates the air will be a ½ horsepower, 240-volt motor. How much load will this package unit add to a 240-volt, single-phase service? Assuming that the air conditioning load will be less than the heating load.

Solution: The heat load= 9.6 KW × 1,000 = 9,600 watts The full-load current in amperes of a ½ hp, single-phase, 240-volt motor (from Table 430.248) = 4.9 A The motor load = 4.9 A x 240 V = 1,176 VA The service load for this package unit = 1,176 + 9,600 = 10,776 watts

Because the blower motor is the largest motor in the service load The service load for this package unit = 10,776 + 1,176 x 25% = 11,070 watts

Rule#3: Motor Load Balance It is permissible to balance the motors as evenly as possible between phases before performing motor-load calculations.

Example#5: What is the minimum rating in amperes for conductors supplying a 10 hp, 208-volt, 3phase motor and three 3 hp, 120-volt, single-phase motors?

Solution: First, find the full-load currents for the motors. The full-load current for a 10 hp, 208-volt, 3-phase motor is 30.8 amperes [Table 430.250]. Full-load currents for single-phase motors are in 430.248. Note that the currents listed in the 115-volt column are permitted for system voltage ranges of 110 to 120 volts. The full-load current for a 3 hp, 120-volt motor is 34 amperes. Next, balance the motors as evenly as possible between phases. Connect the 3-phase motor to each of three ungrounded (hot) conductors.

Because the 3 hp motors are 120-volts, connect each motor to the grounded conductor and one ungrounded conductor. One motor will be on phase A, one motor on B and one on phase C. (see above image) Although there are four motors total, there are only two motors on each phase Because the motors are balanced between phases, the full-load current on each phase = 30.8 + 34 = 64.8A. Because the single phase motor is the largest motor, multiply its amperes by 125 percent = 34 × 125% = 42.5 A Now add to this value the full-load currents of the other motor on the same phase = 42.5 + 30.8 = 73.3 A = 73 A So, the minimum rating in amperes for conductors supplying these motors is 73 A

Definition: Continuous Load: A load where the maximum current is expected to continue for 3 hours or more.

Important!!! The loads under the following categories are continuous loads : 1. General Lighting,

2. 3. 4.

Track Lighting, Show Window Lighting, Sign and Outline Lighting.

While the Loads under the following categories are Not-continuous loads: 1. General-Use Receptacles, 2. Multioutlet Assemblies.

Rule#4: Continuous Loads When calculating a feeder or service As per NEC Standard calculation method, the Loads classified as Continuous Loads must be multiplied by 25 % and add it to the service load calculation.

Sixth: All other Loads

Definition: All Other Loads: are the loads that don’t falling under one of the following loads categories: 1. 2. 3. 4. 5. 6. 7. 8. 9.

General lighting, Track lighting, Show window lighting, Sign and outline lighting, General-use receptacles, Multioutlet assemblies, Kitchen equipment, HVAC loads, Motor loads.

Rule#5: Other Loads All other Loads will be calculated as follows:  All other Non-Continuous Loads will be added at 100%,  All other Continuous Loads will be added at 125%.

Example#6: A bank has the following loads: Parking- Lot Lighting at 57 A, 120 V,  Water Heater at 4 KW, 208 V. What is the Service Load contribution of these loads? 

Solution: The two loads; Parking- Lot Lighting and Water Heater are not falling under any category from the definition of All Other Loads. So, these Loads will be calculated as follows: Parking- Lot Lighting is a continuous load, multiply its VA by 125% = 57 x 120 x 125% = 8,550 VA The Water Heater is a Non-continuous load, multiply its VA by 100% = 4 KW = 4,000 VA The Service Load contribution of these loads = 8,550 + 4,000 VA = 12,550 VA

Rule#6: House (Common Area) Loads in Non-Dwelling Buildings As per NEC section 210.25(B), The systems, equipment, and lighting for public or common areas are required to be supplied from a separate ―house load‖ panelboard. This requirement permits access to the branch-circuit disconnecting means

without the need to enter the space of any tenants. The requirement also prevents a tenant from turning off important circuits that may affect other tenants.

Non-Dwelling Buildings Load Calculations- Part Eight I explained the Loads calculation for Non-Dwelling Buildings as per the NEC Standard Method in the following Articles:        

Non-Dwelling Buildings Load Calculations- Part One Non-Dwelling Buildings Load Calculations- Part Two Non-Dwelling Buildings Load Calculations- Part Three Non-Dwelling Buildings Load Calculations- Part Four Non-Dwelling Buildings Load Calculations- Part Five Non-Dwelling Buildings Load Calculations- Part Six Non-Dwelling Buildings Load Calculations- Part Seven Electrical Load Calculator for Non-Dwelling Buildings

I also indicated that the NEC Optional calculation method can be applied only for the following Non-Dwelling buildings as permitted by NEC, Part IV. Optional Feeder and Service Load Calculations: 1. 2. 3.

Schools, Existing Installations, New Restaurants.

Today, I will explain the load calculations for these special Non-Dwelling Buildings (Schools, Existing Installations and New Restaurants) by using NEC optional method.

First: Schools

Important!!! As with dwellings, there are two methods for calculating services or feeders for schools as follows: 1. Standard NEC Method in Part III of Article 220, Feeder and Service Load Calculations, 2. Optional NEC Method Part IV of Article 220, Optional Feeder and Service Load Calculations.

Rule#1: Condition for Using NEC Optional Load Calculation Method for Schools In accordance with 220.86, using the optional calculation method is only permissible if the school is equipped with electric space heating, air conditioning or both.

Rule#2: Load Calculation for Feeders within the School Building If the School building or structure has been calculated by the optional method, it will be necessary to calculate the loads of feeders within the School building or structure by the NEC Standard Method. Calculating feeders within the School building or structure by the optional method shall not be permitted.

Important!!! If, by using the NEC Standard Method, the ampacity of an individual feeder is greater than the ampacity required for the entire building, it shall not be required that the ampere rating of the feeder be larger than the ampere rating of the service.

Rule#3: Load Calculation for Portable Classroom Buildings Although many schools have portable classroom buildings, using the NEC Optional Method to calculate loads to those portable classroom buildings shall not be permitted. It must be calculated by NEC Standard Method.

Rule#4: Neutral Load Calculation There is no optional method for calculating neutral loads; therefore, neutral loads must be determined in accordance with 220.61.

Procedure for School Load Calculation by NEC Optional Method This optional method calculation procedure for schools can be divided into four easy steps as follows: Step#1: Find the total connected load.

Important!!! The connected load shall include all of the interior and exterior lighting, power, water heating, cooking, other loads and the larger of the air conditioning load or space heating load within the building or structure.

Important!!! When calculating the total load of a school note the following:  Use only 100 percent of the load,  Do not multiply continuous loads by 125 percent,  Do not derate receptacle loads,  Do not derate cooking equipment,  Do not increase the largest motor by 25 percent.

Step#2: find the average volt-ampere per square foot (VA/ft2). Step#3: applying the Table 220.86 demand factors.

Step#4: multiplying the derated volt-ampere per square foot by the square-foot area of the school to get the new derated feeder or service load of the school.

Example#1: What is the service load calculation for a school with a total connected load of 750 kilovolt-amperes (kVA) and a floor area of 30,000 square feet? Solution: Step#1: Find the total connected load The total connected load is provided and is 750 kVA Step#2: find the average volt-ampere per square foot (VA/ft2).

To find volt-amperes per square foot, divide the total connected load by the total square-foot area of the school. The average volt-amperes per square foot = 750,000 VA ÷ 30,000 ft2 = 25 VA/ft2 Step#3: applying the Table 220.86 demand factors. Multiply the first 3 (VA/ft2) by 100% = 3 VA/ft2 The Reminder = 25 – 3 = 22 VA/ft2 Multiply the next 17 by 75% = 17 x 75% = 12.75 VA/ft2 The Reminder = 22- 17 = 5 VA/ft2 Multiply the reminder by 25% = 5 x 25% = 1.25 VA/ft2 After applying the demand factors in table 220.86, the VA/ft2 = 3 + 12.75 + 1.25 = 17 VA/ft2 Step#4: multiplying the derated volt-ampere per square foot by the square-foot area of the school to get the new derated feeder or service load of the school. derated service load of the school = 17 VA/ft2 x 30,000 ft2 = 510,000 VA = 510 KVA

Second: Determining Existing Loads for Existing Installations

Important!!! there are two methods for calculating services or feeders for Existing Installations as follows: 1. NEC Standard Method, Section 220.16 covers loads for additions to existing installations for both dwelling units and other than dwelling units. 2. NEC Optional Method, Article 220.87 covers optional method load calculation specifications for Existing Installations regardless of the type of occupancy. Noting that Article 220.83 covers optional method load calculation specifications for an existing dwelling unit.

Important!!! The purpose of doing service and feeder calculations for an existing dwelling unit is to determine if the existing service or feeder is of sufficient capacity to serve a required additional loads or not.

Rule#5: Conditions Of adding new loads to An Existing Dwelling Unit In accordance with 220.87, when calculating a feeder or service load for existing installations, regardless of the type of occupancy, it shall be permissible to use actual maximum demand to determine the existing load where meeting three conditions: 1. The maximum demand kVA data for a minimum 1-year period (or the 30-day alternative method from the exception) is available. 2. The maximum demand at 125 percent plus the new load does not exceed the ampacity of the feeder or rating of the service. 3. The feeder has overcurrent protection in accordance with 240.4, and the service has overload protection in accordance with 230.90.

Important!!! For condition#1 in Rule#5 above, If the maximum demand data for a 1year period is not available, the calculated load shall be permitted to be based on the maximum demand (measure of average power demand over a 15-minute period) continuously recorded over a minimum 30-day period using a recording ammeter or power meter connected to the highest loaded phase of the feeder or service, based on the initial loading at the start of the recording. The recording shall reflect the maximum demand of the feeder or service by being taken when the building or space is occupied and shall include by measurement or calculation the larger of the heating or cooling equipment load, and other loads that may be periodic in nature due to seasonal or similar conditions.

Important!!! For condition#2 in Rule#5 above, apply the NEC standard Calculation method to get the total load as follows: Total Load = Existing Load Value + New Load Where: Existing Load Value = Max demand Value for a 1-year period from 220.87(1) x 125% New Load = Continuous loads x 125 % + Non-continuous loads x 100 %

Important!!! If condition#2 in Rule#5 above is not verified, you need to increase ampacity of the feeder and/or rating of the service to be able to add new loads to an existing dwelling unit. Example#2: A business owner wants to add equipment to an existing building. The new equipment will not replace any of the existing equipment. The existing service has a rating of 1,200 amperes (A). A peak-demand meter has been connected to the service for more than a year. The maximum demand data from the peak-demand meter shows a demand of 743A. The new equipment that has a calculated load (in accordance with standard load calculation procedures) of 251A. Does this existing service have an ampere rating high enough for the existing loads and the new loads? Solution: Existing Load Value = Max demand Value for a 1-year period from 220.87(1) x 125% = 743 x 125% = 929 A New Load = 251 A New Service Load = Existing Load Value + New Load = 929 + 251 = 1,180 A Since the existing service is rated 1,200A, this installation will be Code-compliant

Third: New Restaurants

Important!!! As with dwellings and schools, there are two methods for calculating services or feeders for New Restaurants as follows: 1. Standard NEC Method in Part III of Article 220, Feeder and Service Load Calculations, 2. Optional NEC Method Part IV of Article 220, section 220.88, Optional Feeder and Service Load Calculations.

Rule#6: NEC Optional Method Load Calculation of service or feeders for New Restaurants As 220.88 states, instead of using the NEC Standard Method in Part III of Article 220, the calculation of a service or feeder load for a new restaurant shall be permitted by the NEC Optional Method in accordance with Table 220.88.

Important!!! In accordance with section 220.88, The feeder has overcurrent protection in

accordance with 240.4, and the service has overload protection in accordance with 230.90.

Important!!! In accordance with section 220.88, Feeder conductors shall not be required to be of greater ampacity than the service conductors.

Rule#7: Neutral Load Calculation There is no optional method for calculating neutral loads; therefore, neutral loads must be determined in accordance with 220.61.

Procedure for New Restaurants Load Calculation by NEC Optional Method This optional method calculation procedure for New Restaurants can be divided into four easy steps as follows: Step#1: Find the new restaurant’s total connected load in kilovolt-amperes (kVA).

Important!!! In accordance with the note under Table 220.88, the total connected load must include all electrical loads, including both heating and cooling loads.

Important!!! When calculating the total load of a new restaurant note the following:   

Use only 100 percent of the load, Do not multiply continuous loads by 125 percent, Do not derate receptacle loads,

 

Do not derate cooking equipment, Do not increase the largest motor by 25 percent.

Step#2: determine the type of the new restaurant: all electric and not all electric. If it is all electric, use the second Column of table 220.88. If it is not all electric, use the third column. Step#3: Once the new restaurant’s total connected load has been calculated, select the row with the range of numbers that includes the total connected load. Step#4: get the new derated feeder or service load of the new restaurant as follows: A- for All Electric new restaurants: If the connected load is no more than 200 kVA, multiply the calculated load by 80 percent.  If the total connected load is more than 200 kVA, an additional math step is required as per the appropriate row in table 220.88. 

B- for Not All Electric new restaurants:  If the connected load is no more than 200 kVA, the calculated load must be taken at 100 percent.  If the connected load is more than 200 kVA, an additional math step is required as per the appropriate row in table 220.88.

Example#3: What is the optional method service load calculation for a new restaurant with a total connected load of 180 kVA? This new restaurant will be all electric. Solution: Step#1: Find the new restaurant’s total connected load in kilovolt-amperes (kVA). The total connected load is given = 180 KVA Step#2: determine the type of the new restaurant: all electric and not all electric. The type of new restaurant is all electric Step#3: select the first row (0–200 KVA).

Step#4: get the new derated service load of the new restaurant as follows: The connected load is less than 200 kVA, multiply the connected load by 80 percent The new derated service load of the new restaurant = 180 x 80% = 144 kVA

Example#4: What is the optional method service load calculation for a new restaurant with a total connected load of 394 kVA? This new restaurant will be all electric.

Solution: Step#1: Find the new restaurant’s total connected load in kilovolt-amperes (kVA). The total connected load is given = 394 KVA Step#2: determine the type of the new restaurant: all electric and not all electric. The type of new restaurant is all electric Step#3: since the total connected load is given = 394 KVA, select the third row (326– 800 KVA). Step#4: get the new derated service load of the new restaurant as follows: Calculate the amount over 325= 394 – 325 = 69 KVA Now multiply the amount over 325 by 50 percent = 69 KVA x 50% = 34.5 KVA Finally, The optional method service load calculation for this new restaurant = 34.5 KVA + 172.5 KVA = 207 KVA

Example#5: What is the optional method service load calculation for a new restaurant with a total connected load of 455 kVA? This new restaurant will be supplied by natural gas and electricity.

Solution: Step#1: Find the new restaurant’s total connected load in kilovolt-amperes (kVA). The total connected load is given = 455 KVA Step#2: determine the type of the new restaurant: all electric and not all electric. The type of new restaurant is Not all electric Step#3: since the total connected load is given = 455 KVA, select the third row (326– 800 KVA). Step#4: get the new derated service load of the new restaurant as follows: Calculate the amount over 325= 455 – 325 = 130 KVA Now multiply the amount over 325 by 45 percent = 130 KVA x 45% = 58.5 KVA Finally, The optional method service load calculation for this new restaurant = 58.5 KVA + 262.5 KVA = 321 KVA

Electrical Load Calculations for Farms Now, we arrive to Part V of Article 220, where the Electrical Load Calculations for Farms are discussed, you may note that farm load calculations are not part of the optional method load calculations because Farms usually have a dwelling unit and one or more additional buildings or structures that are not dwelling units. Today, I will explain the load calculations for Farms as follows.

Farms Electrical Load Calculations

Important!!! Farms usually have a dwelling unit and one or more additional buildings or structures (Other than dwelling units). So, Total Farm load will include two loads; Farm Dwelling Load and Farm Loads.

Rule#1: Farms Electrical Load Calculations As per NEC Part V of Article 220, Farms Electrical Load Calculations will vary based on the existing case of the following: 

Case#1: If the Farm Dwelling load and Farm Loads are supplied by

separate services  Case#2: If the Farm Dwelling load and Farm Loads are supplied by one Common service.

Case#1: If the Farm Dwelling load and Farm Load are supplied by separate services

Rule#2: Electrical Load Calculation of Farm Dwelling Load As per NEC section 220.102(A), If the Farm Dwelling load and Farm Load are supplied by separate services, calculate the Farm dwelling Load by either Standard NEC Method in (Part III of Article 220) or Optional NEC Method (Part IV of Article 220).

Rule#3: Electrical Load Calculation of Farm Loads (Other Than Dwelling) As per NEC section 220.102(B), If the Farm Dwelling load and Farm Load are supplied by separate services, and Where a feeder or service supplies a farm building or other load that has two or more separate branch circuits, the load for feeders, service conductors and service equipment shall be calculated in accordance with demand factors not less than indicated in Table 220.102.

Procedures for calculating Farm Loads (Other Than Dwelling) with separate service Calculating Farm Load (Other Than Dwelling) as per Table 220.12 will be done in four steps as follows: Step#1: compare the three following loads and select the largest: 1. All the loads that are expected to operate simultaneously, or 2. 125 percent of the full load current of the largest motor, or 3. First 60 amperes of the load. Step#2: Multiply the next 60A of all other Loads by 50 percent. Step#3: Multiply the remaining loads by 25 percent. Step#4: the service or feeder Loads for Farm Load = Sum of all loads from steps#1,2 and 3.

Example#1: A farm dwelling and farm building will be supplied by separate services. What is the

minimum size service for the farm loads? The total ampere rating for all the farm loads (including simultaneously operated loads) will be 385 amperes (A). All the loads that are expected to operate simultaneously will have a total rating of 65A. The largest motor will be a 10 horsepower (hp), 240-volt (V), single-phase motor. The farm service will be supplied by a 120/240V, single-phase system.

Solution: Calculate the service for this farm equipment in accordance with Table 220.102. Step#1: compare the three following loads and select the largest: a- All the loads that are expected to operate simultaneously will be 65A. b- In accordance with Table 220.102, multiply the full-load current of the largest motor by 125 percent. The full-load current for the 10-hp motor (from Table 430.248) is 50A. After multiplying by 125 percent, the rating is = 50 A x 125% = 62.5 A c- The third load is the first 60A of the total load. After comparing the three, the largest is 65A. This load will be added to the calculation at 100 percent. The remaining farm loads = 385 A – 65 A = 320 A Step#2: Multiply the next 60A of all other Loads by 50 percent Multiply the next 60A by 50 percent = 60 x 50% = 30 A The remaining farm loads = 320 A – 60 A = 260 A. Step#3: Multiply the remaining loads by 25 percent. The remainder of all other loads = 260 x 25% = 65 A Step#4: the service or feeder Loads for Farm Load = Sum of all loads from steps#1, 2 and 3 The total rating after applying the demand factors = 65 + 30 + 65 = 160 A This rating is not a standard rating. In accordance with 240.6(A), the next standard ampere rating higher than 160 is 200. So, the minimum size service for the farm loads in this example is 200 A.

Case#2: If the Farm Dwelling load and Farm Load are supplied by one Common service.

Rule#4: Electrical Load Calculation of Farm Dwelling Load As per NEC section 220.102(A), If the Farm Dwelling load and Farm Load are supplied by separate services, calculate the Farm dwelling Load as per below Table and according to existing case.

Table for identifying used Method for Electrical Load Calculation of Farm Dwelling Load IF Farm Dwelling Load has Electric Heat No Electric Heat

IF Farm Load

has electric grain-drying systems

Standard NEC Method

Standard or Optional NEC Method

has No electric graindrying systems

Standard or Optional NEC Method

Standard or Optional NEC Method

Rule#5: Electrical Load Calculation of Farm Loads As per NEC section 220.103, If the Farm Dwelling load and Farm Load are supplied by one Common service, apply demand factor of Table 220.103 for calculating the total Farm load.

Important!!! For Rule#5 above, the largest Motor Load must be multiplied by 125% before applying demand factors of table 220.103.

Rule#6: Common Service Farm Load As per Note under Table 220.103, Common Service Farm Load = Total Farm Loads (calculated as per Rule#5 in above) + Farm Dwelling Load (calculated as per Rule#4 in above)

Example#2: One service will supply both a farm dwelling and farm equipment. What is the minimum size service for the following loads? All the loads that are expected to operate simultaneously will have a total rating of 60 amperes (A). The largest motor will be a 10 horsepower (hp), 240-volt (V), single-phase motor. Another load will be 50A. The remaining loads will be 92A. The total load for the dwelling, calculated in accordance with Part III in Article 220, is 110A. A 120/240V, single-phase system will

supply the farm service.

Solution: Multiply the full-load current of the largest motor by 125 percent. The full-load current for the 10-hp motor (from Table 430.248) is 50 A. After multiplying by 125 percent, the rating is = 50 A x 125% = 62.5 A. - The motor load is the largest load. This load will be added to the calculation at 100 percent as per table 220.103. The motor load = 62.5 A X 100% = 62.5 A - The next largest load (all the loads that are expected to operate simultaneously) is 60A. The second largest load after applying the demand factor of table 220.103 = 60 A x 75% = 45 A - The third largest load is 50A. The third largest load after applying the demand factor of table 220.103 = 50 A x 65% = 32.5 A - The remaining load, after applying the demand factor of table 220.103 = 92A x 50% = 46 A The total load, after applying the demand factors of table 220.103 = 62.5 + 45 + 32.5 + 46 = 186 A As specified in the note under Table 220.103, add the total farm loads (after demand factors) to the farm dwelling unit load. In this example, the farm dwelling unit was calculated in accordance with Part III in Article 220 and the load is 110A. The minimum ampere rating for the dwelling unit and farm equipment = 186 A+ 110 A= 296 A This minimum rating is not a standard rating in accordance with 240.6(A). The next standard ampere rating higher than 296 is 300. So, the minimum size service for the farm loads and dwelling unit load in this example is 300 A.

Example#3:

What is the minimum size service for a farm that will have the dwelling unit and four barns supplied by the one service? The total load for the dwelling, calculated in accordance with Part III in Article 220, is 147A. The barn loads are as follows: barn 1 = 103A; barn 2 = 84A; barn 3 = 72A; and barn 4 = 62A. A 120/240V, single-phase system will supply the farm service.

Solution: Calculate the barn loads separately from the dwelling unit load then multiply the barn loads by the demand factor percentages in Table 220.103. - The largest load is 103A The largest load after applying the demand factor of table 220.103 = 103 A x 100% = 103 A - The second largest load is 84A The second largest load after applying the demand factor of table 220.103 = 84 A x 75% = 63 A - The third largest load is 72 A The third largest load after applying the demand factor of table 220.103 = 72 A x 65% = 46.8 A - The fourth largest load is 62A The remaining load, after applying the demand factor of table 220.103 = 62A x 50% = 31 A The total load after applying the demand factors = 103 + 63 + 46.8 + 31 = 243.8 A Now add the total farm loads (after demand factors) to the farm dwelling unit load. The dwelling unit load is 147A. The minimum ampere rating for the dwelling unit and the four barns = 243.8 + 147 = 390.8 A = 391 A This minimum rating is not a standard rating in accordance with 240.6(A). The next standard ampere rating higher than 391 is 400. So, the minimum size service for the farm loads in this example is 400 A.

Load Calculations for Feeder and Service Neutral Load Calculations for Feeder and Service Neutral

Definitions:

Neutral Conductor: is The conductor connected to the neutral point of a system that is intended to carry current under normal conditions.

Neutral Point: is The common point on a wye-connection in a polyphase system or midpoint on a single-phase, 3-wire system, or midpoint of a single-phase portion of a 3-phase delta system, or a midpoint of a 3-wire, direct-current system.

Important!!! The neutral conductor is a current-carrying conductor. Many believe that, because the neutral conductor is a grounded conductor, it is safe to work on it while it is energized. This is a very dangerous practice that has led to many serious electric shocks.

Four examples of a neutral point in a system (see below image)

An example of the first system is a three-phase, four-wire wye-connected

transformer.

An example of the second system is a single-phase, three-wire system that supplies power to Most of one-family dwellings.

An example of the third system is an ungrounded delta system which is a type of three-phase delta system that is sometimes used in industrial facilities

An example of the fourth system is a direct current system used to supply power derived from batteries, power supplies, generators or other sources to the loads.

Important!!! At the neutral point of the system, the vectorial sum of the nominal voltages from all other phases within the system that utilize the neutral, with respect to the neutral point, is zero potential. In some type of delta system, not all phases utilize the neutral; as in below image where phase (B) does not utilize the neutral.

Important!!!

The neutral load must be calculated in accordance section 220.61 even if the feeder or service is calculated by the NEC standard method (Part III) or the NEC optional method (Part IV).

Rule#1: The feeder or service neutral load Basic Calculation

As per NEC section 220.61(A), The feeder or service neutral load shall be the maximum unbalance of the load determined by article 220. The maximum unbalanced load shall be the maximum net calculated load between the neutral conductor and any one ungrounded conductor.

Exception for Rule#1: For 3-wire, 2-phase or 5-wire, 2-phase systems, the maximum unbalanced load shall be the maximum net calculated load between the neutral conductor and any one ungrounded conductor multiplied by 140 percent.

Important!!! When calculating the neutral load, it is not necessary to include loads that do not contribute to the neutral current.

Example#1: A non-dwelling occupancy has a calculated service load of 92 amperes on one phase and 88 amperes on the other phase. The loads have been balanced as evenly as possible. A water heater has a current draw of 19 amperes at 240 volts. The voltage at this panelboard will be single-phase, 120/240 volts. What is the neutral load?

Solution: Since the water heater is 240 volts, it will not contribute to the neutral current. Therefore, 19 amperes can be omitted from the neutral load calculation. The neutral load = highest calculated load – water heater Load = 92 – 19 = 73 amperes Although the feeder conductors must have an ampere rating of at least 92 amperes, the conductor feeding the neutral must have an ampacity of at least 73 amperes.

Rule#2: Permitted Reduction in Neutral Load - 1

As per NEC section 220.61(B)(1), When calculating the feeder or service neutral load, it is permissible to apply a demand factor of 70 percent to household electric ranges, wall-mounted ovens, counter-mounted cooking units and electric dryers, where the maximum unbalanced load has been determined in accordance with Table 220.55 for ranges and Table 220.54 for dryers.

Example#2: What is the minimum neutral load on the service for 5,500-watt clothes dryer installed in a one-family dwelling?

Solution: When calculating feeder or service neutral loads for clothes dryers, start by finding the feeder or service load. As per Table 220.54, because there is only one dryer, the minimum service load (at 100 percent) is 5,500 volt-amperes Apply Rule#2 above; multiply the service load by 70 percent. So, the minimum neutral load on the service = 5,500 VA × 70% = 3,850 VA

Example#3: A 10-unit apartment building will have a 12-kW range in each apartment. What load will these ranges add to a service neutral calculation?

Solution: Start by finding the feeder or service load. As per column C of Table 220.55 for 10 appliances, the maximum demand for 10 nos. 12-kW ranges is 25 kW. Apply Rule#2 above; multiply the service load by 70 percent. The neutral load = 25 KW × 70% = 17.5 kW

Rule#3: Permitted Reduction in Neutral Load - 2

As per NEC section 220.61(B)(2), Where the calculated neutral current is more than 200 amperes, another reduction is permitted. Where the feeder or service is supplied from a three-wire DC or single-phase AC system; a four-wire, three-phase, three-wire, two-phase system; or a five-wire, two-phase system, it is permissible to apply a demand factor of 70 percent to that portion of the unbalanced load in excess of 200 amperes [220.61(B)(2)]. This demand factor is also in addition to any demand factors that may have already been applied.

Example#4:

After calculating the service load for an office building by the NEC Standard calculation, the neutral load is 216,000 volt-amperes. What is the neutral load after applying the demand factor from 220.61(B)(2)? The electrical service will be supplied by a 208Y/120-volt, three-phase, four-wire system.

Solution: First, convert volt-amperes to amperes. Since this is a 208-volt, three-phase system, the total voltage = 208 V × 1.732 = 360.256 V = 360 V The neutral Load in Amps = 216,000 VA ÷ 360 V = 600 A Apply Rule#3 above; multiply the amperes in excess of 200 amperes by 70 % The amperes in excess of 200 amperes = 600 A – 200 A = 400 A The demand neutral Load in excess of 200 amperes = 400 A × 70% = 280 A The total demand neutral load = 280 A + 200 A = 480 A

Rule#4: Prohibited Reduction in Neutral Load - 1

As per NEC section 220.61(C)(1), there shall be no reduction of the neutral or grounded conductor capacity applied to Any portion of a three-wire circuit consisting of two ungrounded (hot) conductors and the neutral conductor of a three-phase, four-wire, wye-connected system.

Important!!! As per Rule#2 above, it is permissible to apply an additional demand factor of 70 percent to a feeder or service neutral supplying household electric ranges. But, As per Rula#4, applying the 70 percent demand factor is not permissible if the ranges are supplied by a single-phase panel that is fed from a three-phase, four-wire, wye-connected system.

Definition: Nonlinear Load: A load where the wave shape of the steady-state current does not follow the wave shape of the applied voltage.

Electronic equipment, electronic/electric-discharge lighting, adjustable-speed drive systems, and similar equipment may be nonlinear loads.

Rule#5: Prohibited Reduction in Neutral Load - 2

As per NEC section 220.61(C)(2), there shall be no reduction of the neutral or grounded conductor capacity applied to That portion consisting of nonlinear loads supplied from a 4-wire, wye-connected, 3-phase system.

Important!!! As per Rule#3 above, where the feeder or service-neutral load exceeds 200 amperes, it is permissible to apply an additional demand factor of 70 percent to that portion of the unbalanced load in excess of 200 amperes. But, As per Rula#5, applying the 70 percent demand factor to reduce the neutral or grounded conductor’s capacity is not permissible for that portion consisting of nonlinear loads supplied from three-phase, 4-wire, wye-connected systems .

Example#5: After calculating the service load for an office building by the basic calculation, the neutral load is 216,000 volt-amperes. The major portion of the neutral load is fluorescent luminaires and information technology equipment. The electrical service will be supplied by a 208Y/120 volt, three-phase, 4-wire system. After demand factors, what is the neutral load in amperes?

Solution: First, convert volt-amperes to amperes. Since this is a 208-volt, three-phase system, the total voltage = 208 V × 1.732 = 360.256 V = 360 V The neutral Load in Amps = 216,000 VA ÷ 360 V = 600 A

Apply Rule#5 above; since the major portion of the neutral load consists of nonlinear loads, applying the additional demand factor of 70 percent is not permissible. This office building has a neutral demand load of 600 amperes

Important!!! If the neutral load calculated as per 220.61 is less than the calculated load of the ungrounded (hot) conductors, it might be possible to reduce the size of the feeder or service-neutral conductor to be smaller than the ungrounded conductors size, but this is not a must, The installed neutral conductor can be the same size as the ungrounded conductors.

Rule#6: Relation between Sizes of Neutral Conductor and Equipment-Grounding Conductor

Compare the minimum size required for the neutral calculated load with the minimum size equipment-grounding conductor from table 250.122 (in below image); then, select the larger of the two to be the size of the neutral conductor.

Example#6: a 400-ampere panelboard (protected by a 400-ampere breaker) will be installed in an area within an industrial plant. The panelboard will be supplied by a 208Y/120 volt, three-phase, 4-wire system. The size of the ungrounded feeder conductors will be 500 kcmil copper. The calculated neutral load is 67 amperes. What is the minimum size conductor required for the neutral?

Solution: Based on the calculated neutral load only, the minimum size conductor from Table 310.16 is 4 AWG. (See image for table 310.16)

The size equipment-grounding conductor specified in Table 250.122 for a 400-ampere

overcurrent device is 3 AWG copper. Therefore, the minimum size neutral conductor is 3 AWG copper.

Rule#6: Relation between Sizes of Neutral Conductor and Grounded Electrode Conductor

The neutral conductor must not be smaller than the required grounded electrode specified in table 250.66 (in below image).

Electrical Rules and Calculations for Air-Conditioning Systems – Part One This is the first Article in our new Course HVAC-2: Electrical Rules and Calculations for Air-Conditioning Systems, which will list, explain, and discuss with examples all the topics covering the Electrical Rules and Calculations for Air-Conditioning Systems which will include but not limited to the following points:

Introduction for Air-Conditioning Systems types,  Introduction for Types of Motors/compressors used for Air-Conditioning Systems,  Electrical Wiring for different Air-Conditioning Systems types,  Types and locations of Disconnecting means,  Sizing of Disconnecting means,  Types of motor-compressor controllers,  Sizing of motor-compressor controllers ratings,  Sizing feeder/branch circuit overcurrent protective devices (OCPDs),  Sizing feeder/branch circuit conductors,  Sizing motor-compressor overload protection. 

First: Introduction for Air-Conditioning Systems Types

1- Air-Conditioning Systems Types As we explained before in our course HVAC-1: An introduction to heating, ventilation and air conditioning (HVAC) systems that the HVAC system can be divided to (2) main systems: 1. 2.

Heating systems, Cooling systems.

Here, we are interested in the cooling systems or what we will called ― Air Conditioning systems‖ in this course, which in turn divided to (4) sub-main systems:    

De-centralized systems (individual room systems), Semi-centralized systems (packaged systems), Centralized systems (central Hydronic systems), Special systems.

1.1 De-Centralized Systems (Individual Room Systems) These systems include: 1. 2. 3.

Window air conditioning units, Split air conditioning units, Mini-heat pumps.

Window air conditioning units

Split air conditioning units

Mini-heat pumps

For more information about Decentralized Systems (Individual Room Systems), please review Article " Air Conditioning System Configurations – Part One ".

1.2 Semi-Centralized Systems (Packaged Systems) These systems include: A- Unitary packaged systems (one pipe systems) which can be divided to: 1. 2.

Packaged air conditioners with water cooled condensers, Packaged air conditioners with air cooled condensers.

Packaged air conditioners with water cooled condensers -1

Packaged air conditioners with water cooled condensers -2

Roof Top Packaged air conditioners with air cooled condensers

B- Ducted split systems (two pipes systems)

Ducted split systems For more information about Semi-centralized systems (packaged systems), please review Article "Air Conditioning System Configurations – Part Two ".

Also, the units used in packaged systems can be divided to: 1. 2. 3.

Rooftop packaged units, Indoor packaged units, Split units,

A- Roof Top Packaged Units: Roof top packaged units (RPU) are an all in one air cooled A/C unit that are installed on top of the roof and can provide cooling and heating. The condensing unit and the air handler reside inside a single housing. In order to connect the unit to the inside duct work, an opening on the roof for supply and return air is necessary. The duct work distributes the air evenly through-out the space. Heating can be provided by reversing the condensing unit (heat pump) or connecting an electric heater, gas heater, or steam coil. 

Roof Top Packaged Units

B- Indoor Packaged Units: Indoor package units (IPU) are similar to rooftop units, but because the unit is located inside, it requires a different way of cooling the condensing unit. The most common way, is air-cooled using louvered openings (horizontal 

angled slats that prevent outside weather to come inside, but allow air flow) to get fresh outside air. In situations where there is no access to fresh outside air, water cooled units are the best solution. Indoor units come in vertical, or space saving horizontal configurations. These units are the same as roof top units in the way they connect to duct work and have the same heating options as mention in the roof top packaged units.

Indoor Packaged Units C- Split Packaged Systems: Split systems are comprised of a condensing unit (a condensing coil and a compressor) located outside, and an air handler (pushes air across the evaporator coil) located inside. Refrigerant lines for high and low pressure are run from the condensing unit to the air handler. Since the high pressure lines get cold, they are covered with insulation to prevent condensation issues and having water accumulate inside the space. The air handler comes in different configurations to best fit the desired installation. Split Systems can be divided to: 1.

Ducted (Connected to duct work),

2. Ductless (wall mounted, ceiling suspended, ceiling cassettes, and floor standing). You may also have multiple air handlers connected to one condensing unit in computer controlled split systems.

Ducted Split Packaged Systems

Ductless Split Packaged Systems

1.3 Centralized Systems (Central Hydronic Systems) These systems include: 1. 2. 3.

Centralized duct ―All air‖ systems, Centralized fluid based hydronic systems ―all water systems‖, Combined (hybrid) water and air systems.

Centralized duct “All air” systems Concept

Centralized “all water systems” Concept

Combined (hybrid) water and air systems Concept

For more information about Centralized systems (central Hydronic systems), please review the following Articles:  

Air Conditioning System Configurations – Part Two. Air Conditioning System Configurations – Part Three.

1.4 Special Systems These systems include: 1. 2.

Evaporative cooling systems, Central air washer (central evaporative air) cooling systems.

Evaporative cooling systems

Fixed Evaporative cooler

Central air washer (central evaporative air) cooling systems

For more information about Special systems, please review Article " Air Conditioning System Configurations – Part Three ".

Second: Introduction for Types of Motors/Compressors used in AirConditioningSystems

2- Types of Motors/Compressors used in Air-Conditioning Systems The types of Motors/compressors used for Air-Conditioning systems can be divided to: A- Positive-Displacement Compressors: They physically compress the vaporized refrigerant into a smaller volume and higher pressure, and include reciprocating, rotary, and scroll types. These deliver a constant volume of gas under a constant speed. They can be divided to: 1. 2. 3.

Reciprocating compressors, Screw Compressor, Scroll Compressors.

B- Dynamic Compressors: They increase vaporized refrigerant pressure by the kinetic energy imparted on refrigerant by a rotating impeller. A centrifugal compressor is a dynamic compressor and is not a constant displacement type. This type includes: 1.

Centrifugal Compressors.

2.1 Reciprocating Compressors Reciprocating compressors are driven by a motor and use pistons, cylinders and valves to compress the refrigerant. Reciprocating compressors are usually used in smaller systems up to 100 tons.

These compressors are available in (2) configurations namely: 1. Hermetic Refrigerant Motor-Compressor (welded Hermetic), 2. Non- Hermetic Refrigerant Motor-Compressor, which can be divided to (2) sub-configurations:  

Semi-hermetic Refrigerant Motor-Compressor (Bolted Hermetic), Open type (Direct driven) Motor-Compressor.

Reciprocating Compressors Types

2.1.A Hermetic Refrigerant Motor-Compressor (welded Hermetic)

Hermetic Refrigerant Motor-Compressor

Construction: A combination consisting of a compressor and the driving motor, both of which are enclosed in the same welded steel casing (housing) and the two are connected by a common shaft with no external shaft or shaft seals. This makes the whole compressor and the motor a single compact and portable unit. But since the components (compressor and motor) are not accessible for repair, the entire compressor unit must be replaced if it fails. 

Usage: It is widely used for the refrigeration and air conditioning applications like the household refrigerators, deep freezers, window air conditioners, split air conditioners, most of the packaged air conditioners. 

Window air conditioners with Hermetic Refrigerant Motor-Compressor

Capacity: It is used with motor power requirements from 1/20 (for small refrigerator compressors)to to several hundred HP (for centrifugal chillers). With cooling capacity for one single unit about 10 ton but if more cooling capacity is needed, several hermetic motor-compressor can be installed in the same air conditioning unit.  The smaller motors up to about 3 HP are usually of single phase design, while larger motors are invariably three phase. 

Advantages of the hermetic motor-compressors are: They are smaller, more compact and have less vibration than the open compressor,  They have no external shaft; this eliminates mechanical problems with shafts, belts, and sheaves and concern of refrigerant leakage, 

In a hermetic unit the motor is located within the refrigerant atmosphere. So, the motor is continuously cooled by the refrigerant vapor flowing to the compressor section valves,  Lubrication is also simplified since both the motor and the compressor operate in the same closed space with the oil.  The shaft is shorter and more rigid, the bearing arrangement is simplified, the machine is quieter. 

The disadvantages of hermetic motor-compressors are: They are limited on capacity,  They are limited on speed because the compressor has to run at the motor speed (direct drive),  If the control of the related air conditioning system is poorly designed or ineptly operated, or if the compressor is allowed to start too frequently, there is a greater risk of motor burn-out than with open machines,  They are not a field serviceable. Therefore, if a motor burns out in a hermitic compressor, or any other internal problem occurs, the maintenance trend is for a complete replacement of the unit. Otherwise, the entire unit must be returned to the shop or factory to be dismantled and reconditioned. 

2.1.B Semi- Hermetic Refrigerant Motor-Compressor (bolted Hermetic)

Semi- Hermetic Refrigerant Motor-Compressor

Construction: Semi-hermetic Motor-Compressor is similar to the hermetic type but have a removable cover bolted on to the end of the casing to facilitate occasional maintenance. The motor is also part of the unit, however it is not sealed. So, Semi-hermetic compressor is almost completely accessible. 

Capacity and Usage: Welded, hermetic machines are used for the smaller duties, up to about 70 kW of refrigeration and bolted, semi-hermetics are used for larger loads. Packaged, water chilling units having multiple, semi-hermetic compressors are used with capacities up to about 700 kW of refrigeration. 

Advantage of Semi-Hermetic compressors over hermetic compressors:

Semi-Hermetic compressors can be taken apart to do maintenance work or re-built a number of times if necessary giving a much longer service life.  Semi- Hermetic Refrigerant compressors are direct drive and hermetic sealing. 

2.1.C Open Type (Direct Driven) Motor-Compressor

Open Type (Direct Driven) Motor-Compressor Construction: Open type of Motor-Compressor in which the compressor and the motor are different entities and the compressor is driven by external power source, such as an electric motor, an engine or a turbine through a direct coupling or a 

vee-belt. Power shaft protrudes through the compressor housing and seal is required to prevent refrigerant from leaking out of the compressor housing. Motor is cooled in a conventional manner by air that is drawn in from the surrounding.

Usage and Capacity: Open Type (direct driven) Motor-Compressors use as an older technology (The motor and compressor are separated by a flexible coupling), so they are not commonly used today.  The capacities for Open type of Motor-Compressors vary from a fractional ton to 400 ton in a single machine. 

Comparison between Open and Semi-hermetic / Hermetic Systems Open

Semi-hermetic / Hermetic

Open systems have motor separately

Any problem in motor affects

connected to compressor. Not

refrigerant charge and oil charge.

connected with refrigerant and oil charge in the system. Compressor being open design, can

Motor and compressor are in the

be inspected / opened by just closing

same casing. Maintenance at site in

the isolation valves. No botheration

case of hermetic is impossible and

of removal / loss of refrigerant gas.

very difficult for semihermetic machines. Some of the internationally well known brands recommend opening of semi-hermetic also only at the factory / service centre and not at site.

Since these can be attended to at

If the compressor has to be taken to

site, down time is very minimal.

factory, long unavoidable delays will

Since motors are outside the

result and disrupt complete cooling. Hermetic compressors are not a field serviceable.

of makes and type possible.

Semi-Hermetic compressors can be taken apart to do maintenance work or re-built a number of times if necessary giving a much longer service life.

Motor windings are cooled by

Motor windings are in the refrigerant

ambient air.

vapour surroundings.

refrigerant environment, wide choice

Open machines are costlier compared Hermetic machines were basically to hermetic.

introduced to reduce manufacturer's first cost. Motor is smaller and cheaper as it is cooled by the refrigerant. Copper and iron content are about half that of an open design.

Open motors do not require any

Hermetic motors consume cooling

refrigeration effect and thus do not

energy produced by the refrigerant.

reduce the capacity of the system

These vapours are to be compressed by the same compressor; Hence, effectively some cooling capacity is lost in motor cooling itself.

Open system is about 10% more

Net effectiveness of cooling capacity

efficient considering loss in cooling

and power consumption is about 10%

motor and penalty in power for

poorer due to hermetic / semi-

compressing vapour.

hermetic design.

Open motors are more efficient as

Hermetic motors are less efficient as

they are cooled / ventilated by

they are cooled by refrigerant and

ambient air.

are loaded beyond peak efficiency point. These motors are rotating in a much denser refrigerant atmosphere and have higher windage loss.

Power fluctuations are not highly

Power fluctuations and electrical

detrimental to open motors.

transients can produce a flash in the refrigerant atmosphere, which can break down into carbon, fluorine, chlorine etc. and will be carried into

the system. When combined with moisture, hydrochloric and hydrofluoric acids are formed which can cause large – scale contamination. Insurance premia charged for open

Hermetics will take very long before

systems is much less as the damages

they are put back into operation,

are not catastrophic and not for long. hence the insurance companies charge much higher premium as the refrigerant and oil invariably need to be replaced. Overall maintenance is very fast,

Heavy maintenance in case of

quick and without any problem on

burnouts such as cleaning, flushing,

refrigerant side.

vacuumising, pressure testing and recharging fresh refrigerant and oil.

The refrigerant NH3 is very cheap

HFC refrigerants are very costly and

(economical).

cannot return to soil if leaked. ODP and/or GWP problems are plenty.

Oil being heavier than refrigerant,

Oil is partially miscible and oil

can be drained easily even during

recovery and maintaining the oil level

running.

is a big problem.

2.2 Scroll Compressors

Scroll compressors features two involutes scrolls, one stationary and one orbiting around the first. This movement draws gas into the outer pocket and the gas is forced toward the center of the scroll, creating increasingly higher gas pressures. The upper limit of the refrigeration capacity of currently manufactured scroll compressors is 60 tons. 

2.3 Screw Compressors

Screw compressors are based on a mechanism made up of two threaded rotors (screws) that are coupled together. The gas is compressed due to the progressive overlapping of the lobes, causing a reduction in the volume occupied by the gas. Continuous and step-less capacity control is provided by moving a sliding valve toward the discharge port, which opens a shortcut re-circulating passage to the suction port.  The refrigeration capacity of twin-screw compressors is 50 to 1500 tons but is normally used in the 200 tons to 800 tons range. 

2.4 Centrifugal Compressors

Centrifugal compressors are made up of a rotor located inside a special chamber. The rotor is rotated at high speed, imparting high kinetic energy to the gas, which is forced through the narrow outlet opening, thus increasing its pressure. The characteristics of a centrifugal compressor make it ideal for air conditioning applications because it is suitable for variable loads, has few moving parts, and is economical to operate.  The available refrigeration capacity for centrifugal compressors ranges from 100 to 2,000 tons. 

Important Notes

1- Compressor Capacities: The size of refrigeration compressors is given in either of the following:



Motor input horse power (HP),



Motor input kilowatts (kW input),



Refrigeration cooling capacity (kW cooling),



British Thermal Units per hour (Btu/h) or

Tons of refrigeration (TR): A refrigeration ton is equal to heat extraction rate of 12,000 BTU's/hr; therefore a 3 TR chiller can remove 36,000 BTU's/hr. 

2- Usage Of Reciprocating Compressors hermetic Motor-Compressors is widely used for the refrigeration and air conditioning applications like the household refrigerators, deep freezers, window air conditioners, split air conditioners, most of the packaged air conditioners. 

Semi-hermetic Motor-Compressors are used with Packaged, water chilling units having multiple, semi-hermetic compressors with capacities up to about 700 kW of refrigeration. 

Open Type (direct driven) Motor-Compressors use as an older technology (The motor and compressor are separated by a flexible coupling), so they are not commonly used today. 

In Article " Electrical Rules and Calculations for Air-Conditioning Systems – Part One ", which was the first Article in our new Course HVAC-2: Electrical Rules and Calculations for Air-Conditioning Systems, I explained the following points:  

Introduction for Air-Conditioning Systems Types, Introduction for Types of Motors/Compressors used in Air-Conditioning Systems,

Today, I will explain Electrical Wiring for different Air-Conditioning Systems types.

Third: Electrical Wiring for Air-Conditioning Systems

1- Importance Of Electrical Wiring For Air Conditioning Systems In the detailed design phase, the electrical designer must size and select the wires/cables, conduits, starters, disconnects and switchgear necessary for supplying power and control to HVAC equipment. This information designed by the electrical designer will be and must appear on the electrical drawings for proper installation by the electrical contractor. So, to determine the electrical equipment and power supply required for the HVAC system proper operation, the electrical designer needs: Knowing the size of the HVAC system (equipment types, locations, …), Understanding how different HVAC equipment operates in a certain HVAC system.  

The above points can be fulfilled by understanding the electrical wiring diagram of individual HVAC equipment and of the whole system also. Note: Also the HVAC designer will need to know the size of the electrical loads to assess the impact of the heat generated by the electrical system on the HVAC load.

2- How to get the Electrical Wiring for Air Conditioning systems? Usually, the electrical wiring diagram of any HVAC equipment can be acquired from the manufacturer of this equipment who provides the electrical wiring diagram in the user's manual (seeFig.1) or sometimes on the equipment itself (see Fig.2).

Fig.1

Fig.2

3- Types of Electrical Wiring Diagrams For Air Conditioning Systems

There are three basic types of wiring diagrams used in the HVAC/R industry today, which are:

1. 2. 3.

The Ladder Diagram, The Line Diagram, The installation diagram.

3.1 The Ladder Diagram

It is the most common type of wiring Diagrams. It is called ladder because the symbols that are used to represent the components in the system have been placed on the rungs of a ladder. ladder diagrams will be referred to as ―schematic‖ diagrams, or simply ―schematics.‖ A typical schematic of a packaged air conditioner is shown in Fig.3.

Fig.3

In electrical schematics, the symbols stand for various components in the circuit, and the lines stand for the wires connecting them. The intention of the overall schematic is to show how the circuit functions, not how it actually looks.

Note:

A wiring schematic shows the condition of a piece of equipment when there is no power being applied to the unit. For example, if a switch is depicted as being normally open (N/O) or normally closed (N/C), remember that the position of the switch is shown as it appears when there is no power applied to that circuit. If there is any deviation from this practice, there will be an explanatory note on the schematic.

Before you begin looking at electrical schematic diagrams, though, remember that there are always five basic components to any schematic:

1. A power supply, 2. A path for the power, 3. A load or component that operates from the power, 4. A switch or component that interrupts the power to the load, 5. A legend (see Fig.4)or key that explains what the various symbols and abbreviations used in the wiring diagram.

Fig.4

3.2 The Line Diagram

It usually includes drawings that more closely resemble the components themselves, rather than symbols.  Fig.5 on the previous page is an example of a typical line diagram. CompareFig.3 and Fig.5 and note the differences in the way that motors, switches, and transformers are represented. Today it is not uncommon for some manufacturers to show both types of diagrams on their equipment. 

Fig.5

3.3 The Installation Diagram

This diagram is used primarily by the installing contractor. It normally shows only what the terminal board connections are, and very rarely will it include any internal wiring of the unit. Fig.6 is a typical installation diagram for a residential cooling system.

Fig.6

4- How to read Electrical Wiring Diagrams?

In order to read electrical Schematics, you need to be familiar with the following:

1. 2. 3.

Symbols Used In Schematics, Schematic Diagram Configurations, Schematic Diagram Locators.

4.1 Symbols Used In Schematics

The most important symbols used in electrical schematics are:

   

Power Supplies, Wiring, Switches, Loads.

A- Power Supplies: (see Fig.7)

Many different supply voltages are used in the HVAC/R industry, ranging from 575-V, three-phase power supplies to 24-V control circuit voltages. Power supplies may be indicated by solid lines or by dashed or dotted lines.

Fig.7

B- Wiring:

Most schematics use straight lines to represent the wires that connect components to each other.  If two wires are connected internally, the connection usually is shown as a dot (a solid black circle), as illustrated at those points marked ―A‖ in Fig.8.1. But note that there is no dot to indicate a junction or connection at point ―B.‖ This means that one wire simply crosses over the other wire. 

Fig.8

Now look at Fig.8.2, crossover wires are shown with half circles or loops that ―jump‖ over other wires (see those points marked ―A‖). Note also that in this type of diagram, junctions are shown without connection dots (see those points marked ―B‖). 

Notes:

Not all manufacturers follow the same schematic diagram practices and you will see several different styles of wiring diagrams.  If dots are used to show junctions, then intersecting lines without dots mean that the two wires cross without connecting. If loops or jumps are used to depict crossovers, then wires that meet—even without dots—are connected. 

Wiring identification:(see Fig.9)

Every manufacturer can identify the wires used in electrical diagrams by one of the following methods:

1. Using Different line thicknesses to represent different types of wires. 2. Using numbers or colors (or both) to help identify the various wires found in a unit .

Fig.9

Note:

The used wiring identification method should be clearly indicated in the legend that accompanies the drawing.

C- Switches:

A switch is a device that interrupts power to the load. It may be:

1. 2. 3.

Manually Operated Switch, Activated automatically by pressure or temperature (Control Switches), Electrically controlled switch (Relays and Contactors).

C.1 Manually Operated Switch:

The switch can be in the closed position (Normally closed) (N/C) or in the open position (Normally open) (N/O) (see Fig.10). You must note that in electrical wiring schematic the position of the switch is shown as it appears when there is no power applied to that circuit. If there is any deviation from this practice, there will be an explanatory note on the schematic.

Fig.10

A switch is characterized by : (see Fig.11)

1. The number of contacts (or poles): the number of poles can be considered as the number of circuits that the switch can control at one time or the number of contacts in the switch. 2. The number of positions (or throws) it has: the number of throws can be considered as the number of paths a single circuit can take.

Fig.11

Note:

The dashed line in the switch symbols represents the mechanical connection that makes the contacts move together, but these contacts are not connected electrically.

C.2 Activated automatically by pressure or temperature (Control Switches):

Pressure and temperature controls are switches; they also may be configured with various combinations of poles and throws. The position of the switch ―arm‖ in the schematic symbol indicates the operation of the control. for examples: (see Fig.12)

Fig.12

The temperature switch (RS-2) is shown with the arm above the contacts. This signifies that the switch opens on a rise in temperature and closes on a drop in temperature.  The pressure switch (AFS-2) is shown with the arm below the contacts. This signifies that the switch opens on a drop in pressure and closes on a rise in pressure.  For SPDT limit switch (LS), When there is an increase in temperature, the contacts ―C‖ to ―N/C‖ move to the ―N/O‖ position. When the temperature decreases, the contacts ―C‖ to ―N/O‖ move back to the ―N/C‖ position. 

C.3 Electrically Controlled Switch:

a- Relays:

Relays are electrically operated control switches. The schematic symbols used to represent relays are the same as those for manually operated switches, except that relay symbols often include a solenoid coil.  There are several possible ways of depicting the solenoid coil. Fig.13 shows two different schematic representations of a DPDT relay. 

Fig.13 Note that multiple-pole Relays, like multiple-pole switches, are connected mechanically but not electrically. 

b- Contactors:

A contactor is a type of heavy-duty relay that handles higher voltages and higher currents than a control relay. Contactors appear nearly identical to relays on schematic diagrams.  Some manufacturers employ contactors that use a single set of contacts. A ―bus bar‖ is placed over the connection where the other set would be, as shown in Fig.14. 

Fig.14

D- Loads:(see Fig.15)

Loads are devices that consume power and convert it to some other form of energy, such as motion or heat. They may be motors, heaters, lights, or other pieces of equipment. A transformer is a type of power-consuming device, but rather than converting energy, a transformer changes the voltage or current.

Fig.15

4.2 Schematic Diagram Configurations

There are two basic configurations used in schematics today to show the approximate placement of loads, switches, and different power or supply voltages. They are : (see Fig.16)

1.

Side-by-Side Arrangement,

2.

Up-and-Down Arrangement.

A- Side-by-Side Arrangement:

In this arrangement, Manufacturers usually place motors and other power-consuming components on the right side of the diagram. This is called the load side. The switches and other controllers are placed on the left side of the diagram. This is called the line side.

Fig.16

4.3 Schematic Diagram Locators

As in roadmaps, almost all mapmakers place numbers and/or letters along the vertical and horizontal edges of maps to help users find particular cities, towns, landmarks, or other locations. Electrical schematics utilize a similar system.

Take a look at Fig.18. This is the same schematic of a packaged air conditioner that you saw inFig.3, but notice that now a column of small numbers has been added, running down the left hand side of the diagram. These numbers are used to indicate the relative location of each horizontal line in the diagram.

Fig.18

Note:

If a line falls between two numbers, the number lower on the page generally is used as the location reference.

This type of line-numbering system can be very useful in helping the reader identify the location of a specific component on the schematic, as well as its controlling switch. For examples:(see Fig.19)

Fig.19

A- Numbers on Left hand side of the diagram:

In Fig.19A, ―C1‖ contacts are located on lines 7 and 11. Similarly, in Fig.19B, you can find the high voltage switches ―IFR‖ and ―C1‖ on lines 28 and 35, respectively. Now look at the lower portion of the wiring diagram in Fig.18 and locate the relay coils ―IFR‖ and ―C1‖ on lines 52 and 57.

B- Numbers on Right hand side of the diagram:

In Fig.19C, note that there are small numbers along the right hand side of the diagram as well. These numbers designate the line location of relay contacts. The small number 28 in the right-hand margin tells you the line location of the contacts associated with relay coil ―IFR.‖ Look back at line 28 in Fig.19B, and you will find the ―IFR‖ contacts. Likewise, the numbers 7, 11, and 35 in the right-hand margin of Fig.19C refer you to the lines where the contacts associated with relay coil ―C1‖ can be found.

C- Underlined Numbers:

Note that the 35 is underlined. An underlined number signifies a normally closed contact (and, conversely, a number that is not underlined signifies a normally open contact). Accordingly, you will find that the ―C1‖ contacts located on line 35 in Fig.19B are shown as normally closed, and that the ―C1‖ contacts on lines 7 and 11 in Fig.19A are shown as normally open.

Electrical Wiring Diagrams for Air Conditioning Systems – Part Two In Article " Electrical Rules and Calculations for Air-Conditioning Systems – Part One ", which was the first Article in our new Course HVAC-2: Electrical Rules and Calculations for Air-Conditioning Systems, I explained the following points: 

Introduction for Air-Conditioning Systems Types,



Introduction for Types of Motors/Compressors used in Air-Conditioning Systems.

And in Article " Electrical Wiring Diagrams for Air Conditioning Systems – Part One ", I explained the following points:

Importance of Electrical Wiring for Air Conditioning Systems, How to get the Electrical Wiring for Air Conditioning systems?, Types of Electrical Wiring Diagrams For Air Conditioning Systems, How to read Electrical Wiring Diagrams?

   

Today, I will explain Electrical Wiring for different Air-Conditioning Systems Types and Equipment.

Third: Electrical Wiring Diagrams for Air-Conditioning Systems - Continued

The Electrical wiring diagrams for Typical Air conditioning equipment

The main types and equipments in common Air conditioning systems were:

           

Window air conditioning units, Split air conditioning units, Multi-Split air conditioning units, Mini-heat pumps, Split Packaged units, Unitary Packaged units, Chillers, Air Handling Units, Fan Coil Units, Pumps, VAV boxes, Dampers.

1- Window Air Conditioning Units

1.1 Window Air Conditioning Units Construction

Window air conditioning unit casing houses the following components: (see Fig.1)

Fig.1: Window Air Conditioning Units Construction

1. 2. 3. 4.

Condenser (outdoor coil), Condenser fan, Hermetic compressor, Evaporator (indoor conditioning coil),

5. Evaporator fan (blower), 6. Controls: The controls for window units are simple and inbuilt, it includes: (see Fig.2)

Fig.2: Window Air Conditioning Units Controls

A rotating selector/Mode switch marked with a hot-cold scale of five positions (off, high cool, low cool, high fan, low fan) with no temperature settings.  A rotating Thermostat switch work as on/off switch for the compressor, its status is depending on what temperature/cooling degree you set it at (usually there are 8 positions for cooling degree).  Louvers swing switch: it is on/off switch which controls the swing motor responsible for controlling the movement and direction angle in which the air be supplied from the louvers to the room. 

1.2 The power flow in Branch circuit of a Typical Window air conditioning unit

The Window air conditioning units are fed from single phase power source (see Fig.3), so its branch circuit and its main power cord consisting of 3 wires (The ground wire, hot wire and neutral wire). 

Fig.3: Window Air Conditioning Unit Power Circuit

The branch circuit will originate from one of the single pole Overcurrent protective device OCPD included in an electrical panel.  Then go through raceway system (conduits, ducts, …) to a disconnect means of some type suitable for the application .  Finally, the main power cord of the Window air conditioning unit is connected to this disconnecting means from one side, the other side enters the casing of the unit to be connected to the unit’s terminal box. 

1.3 Electrical wiring connections inside The Window air conditioning units

Here we are interested on how the main power cord is connected inside the unit and this can be explained as follow (see Fig.4):

Fig.4: Window Air Conditioning Unit Internal Electrical Wiring

A- Inside the unit the main power cord is split to: 1. The ground wire (either green or a bare wire) is screwed to the metal casing of the unit. 2. Hot wire 3. Neutral wire.

B- Hot wire goes to the selector switch on a window unit to feed power to the vital parts, compressor and fan motor as follows:  

Hot wire to selector switch to thermostat switch to compressor Hot wire to selector switch to fan motor.

C- Neutral wire will be connected to fan motor and compressor without goes through any switch. These connections are made on the wire connector in the back of the selector switch so, all neutral wires are common to each other because they are connected to the same point. some examples for the complete electrical wiring diagrams for Window Air Conditioning Unit are inFig.5.

Fig.5: Window Air Conditioning Unit Electrical Wiring Diagrams Also, in Fig.6, you can find examples for the complete wiring diagrams for Window Air Conditioning Unit which be mounted on the unit casing.

Fig.6: Window Air Conditioning Unit Electrical Wiring Diagrams - Factory builtin

also, you can find examples for the complete wiring diagrams for Window Air Conditioning Unit, touch and remote control type in Fig.7.

Fig.7: Window Air Conditioning Unit Electrical Wiring Diagrams - Touch and Remote Control Type

1.4 The power flow inside a Typical Window air conditioning unit in the cooling mode

When you turn the selector switch to cool mode, the power that came in from the cord that connected to the selector via hot wire goes to the fan so the fan operates.  The selector switch also sends the power via hot wire to the compressor, but the compressor will not operate until the thermostat comes to the on position, then the compressor will operates and the cooling cycle begin. 

2- Split Air Cooling Units

2.1 Split air Cooling Units Construction

The split systems are individual systems in which the two heat exchangers are separated (one outside, one inside) (see Fig.8). There are two main parts of the split air conditioner which are:

Fig.8: Split air Cooling Units Construction

1. 2.

Outdoor unit, Indoor unit.

1- Outdoor unit:

This unit is installed outside the room or office space which is to be cooled and houses important components of the air conditioner like:   

The compressor, Condenser Cooling Fan, Expansion Valve.

2- Indoor unit:

The most common type of the indoor unit is the wall mounted type though other types like ceiling mounted and floor mounted are also used. The indoor unit produces the cooling effect inside the room or the office and houses the following components:

     

The evaporator coil or the cooling coil, Cooling Fan or Blower, The drain pipe, Louvers or Fins, Air filter, Controls.

2.2 The power flow in Branch circuit of a Typical split air conditioning unit

The split air conditioning units are fed from either:

Single phase power source (see Fig.9 and Fig.11), so its branch circuit and its main power cord consisting of 3 wires (The ground wire, hot wire and neutral wire). 

Or Three phase power source (see Fig.12), so its branch circuit and its main power cord consisting of 5 wires (The ground wire, 3 hot wires and neutral wire). 

Fig.9: Split air Cooling Units - Single Phase - Indoor feed Outdoor

Fig.10: Split air Cooling Units - Single Phase - Electrical Wiring Diagram

Fig.11: Split air Cooling Units - Single Phase - Outdoor feed Indoor

Fig.12: Split air Cooling Units - Three Phase

Fig.13: Split air Cooling Units - Three Phase - Electrical Wiring Diagram

The branch circuit will originate from one of the single pole/three pole Overcurrent protective device OCPD included in an electrical panel.  Then go through raceway system (conduits, ducts, …) to a disconnect means of some type suitable for the application.  After that, the main power cord of the split air conditioning unit is connected to this disconnecting means from one side, the other side is connected to the terminal box in the indoor unit (see Fig.9) or in the outdoor unit (see Fig.10)according to the manufacturer’s recommendations and wiring diagrams. 

Note:

if power source connections made in indoor unit, an indoor disconnecting means is used and if power source connections made in outdoor unit, an outdoor disconnecting means (see Fig.14) with suitable identity of protection (IP) is used (review the manufacturer’s recommendations and wiring diagrams).

Fig.14: Outdoor Disconnecting Means

Finally, the power is transferred via 3-wire cable or 5-wire cable from the terminal box in indoor unit to the terminal box in outdoor unit or vice versa as indicated in the above point. 

Note:

There is a signal cable also connecting the control in the indoor unit with the control in the outdoor unit.

2.3 Electrical wiring connections inside The Split air conditioning units The electrical wiring inside both of indoor and outdoor units is more complicated than that of window air conditioning units. It is always factory wiring and from our point of view as electrical power engineers, it will not affect our work at all. However, we provide some examples for the electrical wiring diagrams including control wiring for reference as in below Fig.15.

Fig.15: Split air conditioning units - Internal Electrical wiring Diagram

3- Multi-Split Air Conditioners

3.1 The power wiring for multi-split air conditioners

These days, Multi-split air conditioners are also being used commonly (see Fig.16). In units for one outdoor unit there are two indoor units which can be placed in two different rooms or at two different locations inside a large room. 

Fig.16: Multi-split air conditioners



The power wiring for multi-split air conditioners will be as in Fig.17 below.

Fig.17: Multi-split air conditioners Power Wiring

in Fig.18, you can find examples for the complete wiring diagrams for Multi-split air conditioners.

Fig.18: Multi-split air conditioners Electrical Wiring Diagram

4- Mini- Heat Pumps

4.1 The power wiring of Mini- Heat Pumps

The power wiring of Mini- Heat Pumps will look like that of the Split air Cooling Units for far extent (seeFig.19).

Fig.19: Mini-Heat Pumps

However, you can find in below some examples for wiring diagrams for MiniHeat Pumps (see Fig.20) and you can compare them with that of Split air Cooling Units especially in the power (high voltage) wiring.

Fig.20: Mini-Heat Pumps Electrical Wiring Diagram

5- Split Packaged Units

5.1 Split Packaged Units Construction

A split system describes an air conditioning or heat pump system that is split into two sections (see Fig.21)which are:

1. 2.

Outdoor section, Indoor section.

Fig.21: Split Packaged Units Construction

1- Outdoor section:

The outdoor unit is located outside usually on the ground but sometimes on the roof. It houses the following components:

       

Compressor(s), Condenser coil(s), Condenser fan(s), Condenser's fan motor(s), Fan grille, Shut off valves, Reversing valve, Optional accessories (if any).

2- Indoor section:

The indoor section usually located in an interior closet or garage. It houses the

following components:

    

Blower(s), Evaporator coil, Thermal expansion valve(s) and distributor(s), Bearings and shaft, Optional accessories.

5.2 The Electrical wiring in Split Packaged units

The Electrical wiring in Split Packaged units consists of 3 main parts as follows:

1. 2. 3.

High voltage part (power part), High voltage control and motors part, Low voltage control part.

1- High voltage part (power part):(see Fig.22)

Fig.22: Electrical wiring of Split Packaged unit - High voltage part

The branch circuit will originate from one of the three pole Overcurrent protective device OCPD included in an electrical panel.

Then go through raceway system (conduits, ducts, …) to :

 

A disconnect means of indoor unit (Air handler unit), A disconnect means of outdoor unit (condenser/evaporator unit).

2- High voltage control and motors part:(see Fig.23)

Fig.23: Electrical wiring of Split Packaged unit - High voltage control and motors part

This includes the high voltage wiring inside air handler unit and inside condenser/evaporator unit.  Inside the air handler unit, the high voltage wiring powers the indoor fan, the heater and provide power for the transformer.  Inside the condenser/evaporator unit, the high voltage wiring powers the outside fan and the compressor. 

3- Low voltage control part:

This part has (2) mode for operation which are: 1. 2.

A/C Mode, Heat Mode.

A- In the A/C Mode:(see Fig.24)

Fig.24: Electrical wiring of Split Packaged unit - Low voltage

control part - A/C Mode

The thermostat send signal in (2) directions as follows:  

Via the Y wire to turn on the outside fan and the compressor, Via the G wire to turn on the indoor fan.

B- In the heat Mode:(see Fig.25)

Fig.25: Electrical wiring of Split Packaged unit - Low voltage control part - heat Mode

Also, the thermostat, in this mode, sends signal in (2) directions as follows:  

Via the G wire to turn on the indoor fan, Via the W wire to turn on the heater.

So, the complete wiring diagram will be as in Fig.26 below:

Fig.26: Electrical wiring of Split Packaged unit - Complete Circuit

Note:

The thermostat usually have (5) positions which are Off – Cold – Auto – Heat – on.

You can find in below some examples for wiring diagrams for split packaged units with different starting methods in Fig.27.

Fig.27: Electrical wiring of Split Packaged unit with different Starting Methods

6- Unitary Packaged Units

6.1 The power circuit for Unitary packaged units

Unitary packaged systems (see Fig.28) are by far the most commonly used air conditioning equipment in commercial buildings. A packaged air conditioning unit is a self-contained air conditioner. It provides the cooling, heating and the motion of the air. All the components needed for cooling, heating, and air movement are assembled in a steel casing. Most packaged units use semi hermetic compressors which mean that the motor and compressor unit are mounted in one housing. 

Fig.28: Rooftop packaged units Construction

Electrical Wiring Diagrams for Air Conditioning Systems – Part Three In Article " Electrical Rules and Calculations for Air-Conditioning Systems – Part One ", which was the first Article in our new Course HVAC-2: Electrical Rules and Calculations for Air-Conditioning Systems, I explained the following points: 

Introduction for Air-Conditioning Systems Types,



Introduction for Types of Motors/Compressors used in Air-Conditioning Systems.

And in Article " Electrical Wiring Diagrams for Air Conditioning Systems – Part One ", I explained the following points:  Importance of Electrical Wiring for Air Conditioning Systems,  How to get the Electrical Wiring for Air Conditioning systems?,  Types of Electrical Wiring Diagrams For Air Conditioning Systems,  How to read Electrical Wiring Diagrams?

Third: Electrical Wiring Diagrams for Air-Conditioning Systems Continued

The Electrical wiring diagrams for Typical Air conditioning equipment The main types and equipments in common Air conditioning systems were:          

Window air conditioning units, Split air conditioning units, Multi-Split air conditioning units, Mini-heat pumps, Split Packaged units, Unitary Packaged units, Chillers, Air Handling Units, Fan Coil Units, Pumps,

 

VAV boxes, Dampers.

In Article " Electrical Wiring Diagrams for Air Conditioning Systems – Part Two ", I explained the electrical wiring diagrams for some Typical Air conditioning equipments. Today, I will explain Electrical Wiring Diagrams for other Typical Air-Conditioning Equipment.

7- Chillers

7.1 Overview of Central HVAC Systems A heating, ventilating and air-conditioning (HVAC) system is a simple system of heating and cooling exchangers using water or refrigerant (direct expansion system) as the medium. Pumps move the heated or cooled water to the exchangers. Fans then move the warmed or cooled air created at the exchangers to the occupied building interiors.  So there are two stages to heating and cooling which are: 

1. Water stage: water is the most efficient and inexpensive medium that we can cool directly (through a chiller) or heat (through a boiler) 2. Air stage: air is the medium for heat exchange in the building as it can be cooled or heated through coils.

Fig.1 below illustrates a typical HVAC system showing water and air heat exchangers with the main components of the system which are:

Fig.1: Typical HVAC Central System

Fans: for air circulation and ventilation.  Chillers: for the production of chilled water for large buildings (note: for small buildings use the direct expansion cooling systems such as packaged airconditioners).  Boilers: for the production of hot water for Heating (note: it is often to use the electric heaters for zonal reheat).  Pumps: for the circulation of heating hot water, chilled water and condenser water.  Cooling towers: for heat rejection. The primary energy use is the cooling tower fan and pumps.  Controls: coordinate the operation of the mechanical components together as a system. 

7.2 Overview of a Water Chiller A chiller is a mechanical refrigeration device, like an air conditioner, except that it cools a fluid (usually water) instead of cooling air.  When a large air conditioner is required it is sometimes more feasible to use one large chiller instead of many small air conditioners.  Chillers are also used wherever there is a need for cooling a fluid such as a chemical process or for plastics molding.  To simplify the concept of a chiller you should compare it to a drinking fountain where you get cold ―chilled‖ water.  There are a variety of water chiller types (see Fig.2). Most commonly, they are absorption, centrifugal, helical rotary, and scroll. Some reciprocating chillers are also available. 

Fig.2: Types of Chillers

For more information about Chillers and their types, please review Article " HVAC Systems Main Equipment ".

7.3 The Major Components of a Water Chiller A chiller consists of a few major components as follows: 1- The condenser heat exchanger:

It can be either air-cooled, a coil or coils and a fan or fans, or water-cooled, another shell & tube heat exchanger cooled by cooling tower or other water. 1.1 Air cooled condenser: A type of condenser where refrigerant flows through the tubes and rejects heat into a flow of ambient air, most chiller units are fitted with air cooled condensers. 1.2 Water Cooled Condenser: A type of chiller condenser that uses water to remove heat from the refrigerant, this is normally a shell & tube type design. 2- Capillary Tube: A type of expansion device typically fitted on small capacity hire chillers, it comprises a long tube which reduces the pressure of the refrigerant. Centrifugal Fan – A type of fan fitted to an air cooled chiller allowing the fitting of ductwork onto the hot side of an air cooled condenser. The fan is designed to work against a static pressure. 3- Compressor: The main component in a chiller system, the compressor is used to increase the pressure & temperature of the refrigerant vapour. Compressors are usually reciprocating, scroll, centrifugal, or rotary screw types. 4- Condenser: The part of a chiller system where the refrigerant vapour is converted to liquid as it rejects heat. 5- Distributor: A device used to supply uniform gas supply through a submerged coil in a tank chiller evaporator. 6- Evaporator: The part of the chiller system where cool liquid refrigerant absorbs heat from the chilled water circuit. It is usually of shell & tube constructions and is the exchanger where chilled water would be produced. 6.1 Shell & Tube Evaporator:

A type of evaporator where refrigerant flows through the tubes & chilled water fills the surrounding shell 7- Expansion valve: A device used to maintain the pressure difference between the high pressure & low pressure sides of the chiller system. 8- Hot Gas Muffler: A device installed at the discharge side of the chiller compressor to reduce noise and vibration in reciprocating compressors. 9- Liquid Line Filter Drier: A device installed in the liquid line to remove moisture and foreign matter, designed to protect the chillers compressor. 10- Shut Off Valve: It is used to isolate one part of the chiller system from the rest. 11- Sub-Cooler: It is the lower portion of the chillers condenser that further cools the saturated liquid refrigerant. 12- Suction Header: A section of pipe within the chiller system used to collect the refrigerant vapour when it leaves the tubes of a submerged coil evaporator 13- Suction Line Filter: A devise installed into the chillers suction line to remove foreign matter from the refrigeration system 14- Cabinet: Cabinet is made from heavy gauge galvanized steel. Steel sheet panels are zinc coated and galvanized by hot dip process followed by air dry paint or backed on electrostatic polyester dry powder coat. 15- Control Panel: (see Fig.3)

Fig.3: Example of Chiller Control Panel The control panel design is equivalent to NEMA 4 (IP55) with hinged door for easy access ensuring dust and weatherproof construction. Internal power and control wiring is neatly routed, adequately anchored and all wires identified with cable markers as per standards applicable to HVAC industry. The electrical controls, used in the control panel, must be reliable in operation at high ambient conditions for a long period.

7.4 The Electrical Wiring Diagrams for a Water Chiller

Fig.4: Air cooled screw water chillers Let’s take an example for Water Chillers like Air cooled screw water chillers from Cooline Co. model # ASQ115B, the following electrical data for this model will be as in Fig.5.

Fig.5: Electrical Data for Cooline Co. model # ASQ115B

The model drawing and main components are indicated as in Fig.6.

Fig.6: Cooline Co. model # ASQ115B- Drawing and Main Components

And a Typical Schematic Wiring Diagram (Part Winding Start) for this model is indicated in Fig.7.

Fig.7: Cooline Co. model # ASQ115B -Typical Schematic Wiring Diagram you can note that Fig.7, Fig.6 and Fig.5 must be identical for:  the number of compressors = 2.  the number of condenser fans = 8.

Also, the Legend, Notes & wiring diagram for optional items used for this model is indicated in Fig.8.

Fig.8: Cooline Co. model # ASQ115B- Legend, Notes & wiring diagram

7.5 The Field Wiring For Chillers Internal power and control wiring is neatly routed, adequately anchored in manufactures' factory prior to delivery. The connections that must be made by the installer are as follows: A- Connection to the power network: (see Fig.9)

Fig.9: Chillers Connection to the power network Field power wiring is a single point connection; Main 3 phase power must be supplied from a single field supplied and mounted Disconnect switch, using dual element time delay fuse or circuit breaker with Rating as recommended by manufacturer. Also, electrical lugs for incoming power are provided by manufacturer.

B- Connection to the control and monitoring system: Interlocking connections are needed with water flow switch, pumps, remote monitoring system and two barrel units water temperature sensor, if any.

8- Air Handling Units, Fans, and Pumps

8.1 The Field Electrical Wiring for other Major Components in Central HVAC System Field power wiring for other Major Components in Central HVAC System like Pumps, Fans and AHUs are also a single point connection as in the case of Chillers but we must use some type of motor starters or/and Variable frequency drives (VFDs) as follows: If you will use DOL starters (see Fig.10), you must install a separate Disconnect switch, using dual element time delay fuse or circuit breaker (see Fig.11) with Rating as recommended by manufacturer. 

Fig.10: Examples for DOL Starters

Fig.11: The Field Electrical Wiring for Loads with DOL Starters and VFD

And if you use VFD which you must install a separate Disconnect switch, using dual element time delay fuse or circuit breaker (see Fig.11) with Rating as recommended by manufacturer. 

But if you use DOL combination starters (see Fig.12) which come with a Disconnect switch or circuit breaker built in the unit (see Fig.13). This will eliminate the cost of wiring between separate disconnect means and starters. 

Fig.12: Examples for DOL Combination Starters

Fig.13: The Field Electrical Wiring for Loads with DOL Combination Starters

Electrical Rules and Calculations for AirConditioning Systems – Part Two In Article " Electrical Rules and Calculations for Air-Conditioning Systems – Part One ", which was the first Article in our new Course HVAC-2: Electrical Rules and Calculations for Air-Conditioning Systems, I explained the following points:  

Introduction for Air-Conditioning Systems Types Introduction for Types of Motors/Compressors used in Air-Conditioning Systems

And in Article " Electrical Wiring Diagrams for Air Conditioning Systems – Part One ", I explained the following points: Importance of Electrical Wiring for Air Conditioning Systems, How to get the Electrical Wiring for Air Conditioning systems?, Types of Electrical Wiring Diagrams For Air Conditioning Systems, How to read Electrical Wiring Diagrams?

   

Also, I explained the electrical wiring diagrams for Typical Air conditioning equipments in the following Articles: Electrical Wiring Diagrams for Air Conditioning Systems – Part Two Electrical Wiring Diagrams for Air Conditioning Systems – Part Three

 

Today, I will explain in detail Different Types, Locations and Sizes of Disconnecting Means used for HVAC Systems.

Types of Disconnecting Means for Air-Conditioning Systems

In this Article and next Articles, you will learn the following: Types of Disconnecting Means used for Air-Conditioning Systems,  Where to locate the disconnecting means in the power circuit of AirConditioning Systems,  How to size these disconnecting means to be capable of disconnecting air-conditioning and refrigerating equipment, including motor-compressors and controllers from the circuit conductors. 

And before proceeding with the above points, let’s take a general overview for NEC Rules controlling motor loads of Air-Conditioning Systems.

1- General overview for NEC Rules controlling Motor loads of AirConditioning Systems

The NEC rules that control Motor loads of Air-Conditioning Systems depends on the type of the motor which may be one of the following types: 1. Only Non-Hermetic Refrigerant Motor-Compressor Loads, 2. Only Hermetic Refrigerant Motor-Compressor Loads (Airconditioning equipment, refrigerating equipment, or both), 3. Combination loads (motor(s) with other loads).

1- Only Non-Hermetic Refrigerant Motor-Compressor Loads:

Fig.1: Non-Hermetic Refrigerant Motor-Compressor Article 210 covers all branch circuits except for branch circuits that supply only Non-Hermetic Refrigerant Motor-Compressor Loads, which are covered in Article 430. 

Article 430 shall apply where a branch circuit supplies air-conditioning and refrigerating equipment that does not incorporate a hermetic refrigerant motor-compressor.  The rules of Articles 422, 424, or 430, as applicable, shall apply to airconditioning and refrigerating equipment that does not incorporate a hermetic refrigerant motor-compressor. This equipment includes devices that employ refrigeration compressors driven by conventional motors, furnaces with airconditioning evaporator coils installed, fan-coil units, remote forced air-cooled condensers, remote commercial refrigerators, and so forth. 

2- Only Hermetic Refrigerant Motor-Compressor Loads (Air-conditioning equipment, refrigerating equipment, or both):

Fig.2: Hermetic Refrigerant Motor-Compressor Article 440 shall apply Where a circuit supplies only air-conditioning equipment, refrigerating equipment, or both, that incorporate a hermetic refrigerant motor-compressor(s).  Article 440 does not apply unless a hermetic refrigerant motorcompressor is supplied.  Article 440 is in addition to or amendatory of the provisions of Article 430 and other applicable articles. Many requirements — for example, for disconnecting means, controllers, single or group installations and sizing of conductors— are the same as or similar to those applied in Article 430. So, 

Article 440 must be applied in conjunction with Article 430.  Equipment such as room air conditioners, household refrigerators and freezers, drinking water coolers, and beverage dispensers shall be considered appliances, and the provisions of Article 422 shall also apply.  Hermetic refrigerant motor-compressors, circuits, controllers, and equipment shall also comply with the applicable provisions of the Articles include in below table:

Equipment/Occupancy Capacitors Commercial garages, aircraft hangars, motor fuel dispensing facilities, bulk storage plants, spray application, dipping, and coating processes, and inhalation anesthetizing locations Hazardous (classified) locations

Article

Section 460.9

511, 513, 514, 515, 516, and 517 Part IV 500–503 and 505

Motion picture and television studios and similar locations

530

Resistors and reactors

470

3- Combination loads (motor(s) with other loads):

Fig.3: Combination loads (motor(s) with other loads) Clause 220.18A shall apply for circuits supplying loads consisting of motor-operated utilization equipment that is fastened in place and has a motor (Hermetic or non- Hermetic) larger than 1⁄8 hp in combination with other loads.  Note that as per Article 422, Equipment such as room air conditioners, household refrigerators and freezers, drinking water coolers, and beverage dispensers shall be considered appliances and can be fastened in place, in this case Clause 220.18A shall be applied. 

NEC 220.18A : For circuits supplying loads consisting of motor-operated

utilization equipment that is fastened in place and has a motor larger than 1⁄8 hp in combination with other loads, the total calculated load shall be based on 125 percent of the largest motor load plus the sum of the other loads. Article 430 apply to branch circuits with combination loads (NonHermetic Refrigerant Motor-Compressor(s) type with other loads).  Article 440 is applied when two or more hermetic refrigerant motorcompressors or one or more hermetic refrigerant motor-compressor with other motors or loads. 

2- Important Definitions 2.1 Disconnecting Means: A device, or group of devices, or other means by which the conductors of a circuit can be disconnected from their source of supply. 2.2 In Sight From (Within Sight From, Within Sight):

Fig.4: Disconnect Means In Sight From Equipment

Where one equipment is specified to be ―in sight from,‖ ―within sight from,‖ or ―within sight of,‖ and so forth, another equipment, the specified equipment is to be visible and not more than 15 m (50 ft) distant from the other.

Fig.5: Disconnect Means Not In Sight From Equipment/Motor 2.3 Branch-Circuit Selection Current (BCSC): When a branch-circuit selection current is marked on a nameplate, it must be used instead of the rated-load current to determine the size/Ratings of the disconnecting means, the controller, the motor branch circuit conductors, and the overcurrent protective devices for the branch-circuit conductors and the motor. The value of branch-circuit selection current is always greater than themarked rated-load current. 2.4 Rated-Load Current (RLC):

Fig.6:Air Conditioner Nameplate RLC, FLA The rated-load current for a hermetic refrigerant motor-compressor is the current resulting when the motor-compressor is operated at the rated load, rated voltage, and rated frequency of the equipment it serves.  The rated-load current in amperes of the motor-compressor shall be marked by the equipment manufacturer on either or both the motorcompressor nameplate and the nameplate of the equipment in which the motor-compressor is used. 

3- Types of disconnecting means Disconnecting means for air-conditioning and refrigerating equipment are used mainly to disconnect the main load(s) in these equipment, which is motorcompressors, from the power circuit. The specified types of disconnection means for this purpose as per NEC 430.109 will be as follows: 1- A listed Motor Circuit Switch: Ii must be a horsepower-rated switch capable of interrupting the maximum overload current of a motor. 2- A listed Molded Case Circuit Breaker. 3- A listed Molded Case Switch (non-automatic circuit interrupter):

It is a circuit-breaker like device without the overcurrent element and automatic trip mechanism. It is rated in amperes and is suitable for use as a motor circuit disconnect based on its ampere rating, as is a circuit breaker. 4- Instantaneous Trip Circuit Breaker that is part of a listed combination motor controller. 5- Listed Self-Protected Combination Controller:

Fig.7: Listed Self-Protected Combination Controller

6- Manual Motor Controller:

Fig.8: Manual Motor Controller Listed manual motor controllers additionally marked ―Suitable as Motor Disconnect‖ shall be permitted as a disconnecting means where installed between the final motor branch-circuit short-circuit protective device and the motor.  Listed manual motor controllers additionally marked ―Suitable as Motor Disconnect‖ shall be permitted as disconnecting means on the line side of the fuses permitted in lieu of devices listed in Table 430.52 for power electronic devices in a solid-state motor controller system. In this case, these fuses shall be considered supplementary fuses, and suitable branch-circuit short-circuit and ground fault protective devices shall be installed on the line side of the manual motor controller additionally marked ―Suitable as Motor Disconnect.‖ 

Table 430.52

For stationary motors rated at 2 hp or less and 300 volts or less, A listed manual motor controller having a horsepower rating not less than the rating of the motor and marked ―Suitable as Motor Disconnect‖ can be permitted as the disconnecting means for these motors. 

7- System Isolation Equipment SIE: System isolation equipment is ―a redundantly monitored, remotely operated contactor-isolating system, packaged to provide the disconnection/ isolation function.‖ SIE can be used as a means to disconnect and isolate separately operable parts of a large industrial machine. SIE is used where repeated operation of disconnecting means for maintenance or servicing is inherent to the process and the risk of injury to personnel is increased due to moving parts and multiple points of entry.  System isolation equipment shall be installed on the load side of the overcurrent protection and its disconnecting means. The disconnecting means shall be one of the following types: 

1. 2. 3.

A listed Motor Circuit Switch, A listed Molded Case Circuit Breaker, A listed Molded Case Switch (non-automatic circuit interrupter).

8- branch-circuit overcurrent device:

Fig.9: branch-circuit overcurrent device

It shall be permitted to serve as the disconnecting means for Stationary Motors of 1⁄8 Horsepower or Less. 9- A general-use switch: It can be used as a disconnecting means in the following cases: For stationary motor rated at 2 hp or less and 300 volts or less with the condition that the switch having an ampere rating not less than twice the fullload current rating of this stationary motor.  For Torque Motors.  For stationary motors rated at more than 40 hp dc or 100 hp ac, the general-use switch must be marked ―Do not operate under load.‖  For motors of over 2 hp to and including 100 hp with an autotransformertype controller where all of the following provisions are met: 

1. The motor drives a generator that is provided with overload protection. 2. The controller is capable of interrupting the locked rotor current of the motors, is provided with a no voltage release, and is provided with running overload protection not exceeding 125 percent of the motor full load current rating. 3. Separate fuses or an inverse time circuit breaker rated or set at not more than 150 percent of the motor full-load current is provided in the motor branch circuit.

10- A general-use snap switch suitable only for use on ac (not general-use ac–dc snap switches):

Fig.10: A general-use snap switch suitable only for use on ac It can be used as a disconnecting means for stationary motors rated at 2 hp or less and 300 volts or less and with the condition that the motor full-load current rating is not more than 80 percent of the ampere rating of the switch. 11- Isolating Switches:

Fig.11: Isolating Switches For stationary motors rated at more than 40 hp dc or 100 hp ac, the disconnecting means shall be permitted to be a isolating switch where plainly marked ―Do not operate under load.‖ 12- A Horsepower-Rated Attachment Plug and Receptacle :

Fig.12: Horsepower-Rated Attachment Plug and Receptacle For a cord-and plug- connected motor, a horsepower-rated attachment plug and receptacle, flanged surface inlet and cord connector, or attachment plug and cord connector having ratings no less than the motor ratings shall be permitted to serve as the disconnecting means. Note: Horsepower-rated attachment plugs, flanged surface inlets, receptacles, or cord connectors shall not be required for cord-and-plug-connected appliances in accordance with 422.33, room air conditioners in accordance with 440.63, or portable motors rated 1⁄3 hp or less. 13- An Attachment Plug and Receptacle:

Fig.13: An Attachment Plug and Receptacle For cord-and-plug-connected appliances, an accessible separable connector or an accessible plug and receptacle shall be permitted to serve as the disconnecting means.  Where the separable connector or plug and receptacle are not accessible, cord-and-plug-connected appliances shall be provided with disconnecting means as the branch-circuit overcurrent Device or switch according to the motor rating. The rating of a receptacle or of a separable connector shall not be less than the rating of any appliance connected thereto. 

Note: Demand factors authorized elsewhere in this Code shall be permitted to be applied to the rating of a receptacle or of a separable connector. An attachment plug and receptacle or cord connector shall be permitted to serve as the disconnecting means for a single- phase room air conditioner rated 250 volts or less if: 

1. the manual controls on the room air conditioner are readily accessible and located within 1.8 m (6 ft) of the floor, or 2. An approved manually operable disconnecting means is installed in a readily accessible location within sight from the room air conditioner.

Fig.14: Cord Connector

14- The branch-circuit switch or circuit breaker: For permanently connected appliances rated over 300 volt-amperes, the branch-circuit switch or circuit breaker shall be permitted to serve as the disconnecting means where the switch or circuit breaker is within sight from the appliance or is capable of being locked in the open position. The provision for locking or adding a lock to the disconnecting means shall be installed on or at the switch or circuit breaker used as the disconnecting means and shall remain in place with or without the lock installed. 

Fig.15: Disconnects with provisions for adding lock For permanently connected motor-operated appliances with motors rated over 1⁄8 horse power, the branch-circuit switch or circuit breaker shall be permitted to serve as the disconnecting means where the switch or circuit breaker is within sight from the appliance. The disconnecting means shall comply with 430.109 and 430.110. 

Note: If an appliance of more than 1⁄8 hp is provided with a unit switch in below paragraph, the switch or circuit breaker serving as the other disconnecting means shall be permitted to be out of sight from the appliance. 15- A unit switch(es) with a marked-off position that is a part of an appliance and disconnects all ungrounded conductors where other means for disconnection are provided in the occupancy: These switches are permitted as the disconnecting means where other means for disconnection are provided in the following occupancies: A- Multifamily Dwellings: In multifamily dwellings, the other disconnecting means shall be within the dwelling unit, or on the same floor as the dwelling unit in which the appliance is installed, and shall be permitted to control lamps and other appliances.

B- Two-Family Dwellings: In two-family dwellings, the other disconnecting means shall be permitted either inside or outside of the dwelling unit in which the appliance is installed. In this case, an individual switch or circuit breaker for the dwelling unit shall be permitted and shall also be permitted to control lamps and other appliances. C- One-Family Dwellings: In one-family dwellings, the service disconnecting means shall be permitted to be the other disconnecting means. D- Other Occupancies: In other occupancies, the branch circuit switch or circuit breaker, where readily accessible for servicing of the appliance, shall be permitted as the other disconnecting means.

4- Configuration Of Disconnection Means for Motor(S) in Multi-Wire Branch Circuit As per NEC 210.4 (B) Each multi-wire branch circuit shall be provided with a disconnecting means that will simultaneously disconnect all ungrounded conductors at the point where the branch circuit originates. 1- For a single-phase multi-wire branch circuit supplying only one utilization equipment: The disconnecting means could be one of the following:   

Two single-pole circuit breakers with an identified handle tie, Two-pole circuit breaker, 2-pole switch.

Fig.16: Disconnecting means for single-phase multi-wire branch circuit supplying only one utilization equipment In Fig.16, in which line-to-line loads are supplied from single-phase, identified handle ties or multi-pole common trip circuit breakers are permitted. 2- For a 3-phase installation:

The disconnecting means could be one of the following: 3-pole circuit breaker,  Three single-pole circuit breakers with an identified handle tie,  Three-pole switch provides the required simultaneous opening of the ungrounded conductors. 

Fig.17: Disconnecting means for Three-phase multi-wire branch circuit

In Fig.17A, where multi-pole common trip circuit breakers are required, handle ties are not permitted because the circuits are supplied from ungrounded systems.  In Fig.17B, where the supply systems are grounded, single-pole circuit breakers are permitted and handle ties or common trip operation is not required because the circuits supply line-to-neutral loads.  In Fig.17C, in which line-to-line loads are supplied from 4-wire, 3-phase systems, identified handle ties or multi-pole common trip circuit breakers are permitted. 

Electrical Rules and Calculations for AirConditioning Systems – Part Three In Article " Electrical Rules and Calculations for Air-Conditioning Systems – Part One ", which was the first Article in our new Course HVAC-2: Electrical Rules and Calculations for Air-Conditioning Systems, I explained the following points:  

Introduction for Air-Conditioning Systems Types Introduction for Types of Motors/Compressors used in Air-Conditioning Systems

And in Article " Electrical Wiring Diagrams for Air Conditioning Systems – Part One ", I explained the following points:    

Importance of Electrical Wiring for Air Conditioning Systems, How to get the Electrical Wiring for Air Conditioning systems?, Types of Electrical Wiring Diagrams For Air Conditioning Systems, How to read Electrical Wiring Diagrams?

Also, I explained the electrical wiring diagrams for Typical Air conditioning equipments in the following Articles:  Electrical Wiring Diagrams for Air Conditioning Systems – Part Two  Electrical Wiring Diagrams for Air Conditioning Systems – Part Three And in Article " Electrical Rules and Calculations for Air-Conditioning Systems – Part Two ", I explained Types of Disconnecting Means for Air-Conditioning Systems Today, I will explain in detail identification and Marking of Disconnecting Means used for Air Conditioning Systems.

1- Classification of Electrical Distribution Systems

according to Interrupting Ratings of their OCPD Combinations

All the disconnecting means (or generally all the overcurrent protective devices OCPD) in a building’s electrical distribution system are arranged, with respect to their short circuit ratings /interrupting ratings, in two combinations as follows: 1. 2.

Fully rated system, Series rated system.

Each system from the above can consist of all fuses, all circuit breakers, or a combination of fuses and circuit breakers.

1.1 Differences between Fully Rated And Series Rated Systems Fully Rated System Interrupting rating of overcurrent protective devices OCPD is equal to or greater than (typically 10% greater) the available fault level at the bus where the OCPD are installed. i.e.: interrupting rating of downstream and upstream OCPD ≥ fault level where OCPD are installed.

Interrupting rating of downstream OCPD≥ fault level where OCPD are installed. Fully rated systems can consist of

Series Rated System Interrupting rating of overcurrent protective devices OCPD is above the interrupting rating of the load side (protected) circuit breaker, but not above the interrupting rating of the line-side (protecting) device. i.e.: interrupting rating Rating of downstream OCPD < fault level where OCPD are installed < interrupting rating of main upstream OCPD. It allows installing of downstream OCPD with interrupting rating < fault level where OCPD are installed. A series rated combination can

all fuses, all circuit breakers, or a combination of fuses and circuit breakers. Fully rated systems are recommended and can be used everywhere, as long as individual interrupting ratings are in compliance with 110.9.

fully rated combinations can be selectively coordinated. Fully rated system is more expensive than series rated system.

consist of fuses protecting circuit breakers, or circuit breakers protecting circuit breakers. series rated combinations have limited applications (the only proper application of series rated combinations is for branch circuit, lighting panels) Series rated combinations can’t applied in health care systems, or emergency circuits like fire pumps and elevators. Series rated combinations have extra NEC requirements that must be met. Series rated combinations inherently can’t be selectively coordinated. Series rated combinations can lower the overall system cost.

1.2 Examples for Fully Rated And Series Rated Systems 1- Example for Fully Rated System: As In Fig.1 , fully rated system has protection devices tested to withstand the fault current available at their respective applied location.

Fig.1: Example for Fully Rated System 2- Example for Series Rated System In the event of a short circuit as in Fig.2, both downstream and upstream breakers will open simultaneously to clear the fault which results in unnecessarily blackout a portion of the electrical system.

Fig.2: Example for Series Rated System Note: Replacement of breakers and fuses of panelboards and switchboards in series rated systems must be done by using either: Breakers and fuses part number listed on the panelboard /switchboard label, or Tables of available fuse/circuit breaker series rated combinations published by panelboard and switchboard manufacturers (see Fig.3).

Fig.3: Example for Series Rated Combinations Published by Panelboard and Switchboard Manufacturers

1.3 Difference between Series Rated System and Selective Coordinated System

Fig.4: Difference between Series Rated System and Selective Coordinated System As shown in Fig.4, Series rated combinations can’t be selectively coordinated. In order to protect the load side circuit breaker, the line side (protecting) device must open in conjunction with the load side (protected) circuit breaker. This means that the entire panel can lose power because the device feeding the panel must open even under relatively low-level shortcircuit conditions.  But in selective coordinated system, only the load side circuit breaker 

will open in a case of fault without opening the line side (protecting) device.

2- Identification and Marking of Disconnecting Means

Based on the above discussion, any disconnect means will be used either in a fully Rated Combination System or in a Series Rated Combination System as described in above. First: Identification of Disconnecting Means used in a fully Rated Combination System For Identification of Disconnecting Means used in a fully Rated Combination System, the rules NEC 240.60 and 240.83 are applied as shown in below. As we stated above that a fully rated system consist of all fuses, all circuit breakers, or a combination of fuses and circuit breakers. For this reason, we will explain the identification/Marking of Fuses and Molded Case Circuit breakers that widely used as disconnecting means for Air conditioning systems.

1- Marking for Fuses Fuses shall be plainly marked, either by printing on the fuse barrel or by a label attached to the barrel showing the following: (see Fig.5)     

Ampere rating, Voltage rating, Interrupting rating where other than 10,000 amperes, Current limiting where applicable, The name or trademark of the manufacturer.

Fig.5: Fuse Marking

2- Marking for Molded Case Circuit breakers Circuit breakers shall have marking showing the following: A- General Marking

B- Position Indication

Type Designation,  On and Off (Open and Closed)  Manufacturer’s Name,  Trip and Reset,  Voltage Rating,  Electrical Operation (On and Off),  Ampere Rating,  Electrical  Line and Load Operation (Trip and Identification. Reset). 

D- Interrupting Ratings Interrupting Ratings. 

C- Interchangeable Trip Units  Manufacturer’s Name,  Ampere Rating,  Frame Designation,  Magnetic Settings.

E- Terminations  

Cu-Al Wire, Small Size Wire,

F- Adjustable Trip 

Trip,

Instantaneous

Tightening Torque,  Maximum Wire Size,  Multiple Conductor Connectors,  60/75°C Wire,  Separately Shipped Connectors,  Cable Connection Only,  Bus Bar Sizes. 

I- Special Markings Non-Conducting Enclosure,  Ventilated Enclosure,  40°C,  Current Limiting,  Class CTL,  Delta,  2-Pole — 3-Phase Rated,  3-Pole — 1-Phase Rated,  4-Pole — 3-Phase Rated,  Multi-Wire Circuit,  DC Rated 3-Pole,  100 Percent Continuous Rated,  SWD,  Independent Trip,  Special Characteristics, 



Type A and Type

B, Adjustable Controls. 

K- Circuit Breaker/Ground Fault Circuit Interrupter  Line and Load  ―Test‖ Function, Identification,  ―Class A‖  Identification of Marking, Fuses,  Instructions,  No Open Fuse  Terminal Tripping, Identification,  General Markings.  General Markings. J - Fused Circuit Breakers

For Replacement Not CTL,  Special Purpose Not General,  HID,  Remotely Operated Circuit Breaker. 

L- Circuit Breaker/Equipment Ground Fault Protection  ―Test‖ Function,  Trip Level Marking,  Instructions,  Terminal Identification,  Use Marking,  General Markings.

M- Circuit Breaker Surge-Protective Device Types,  Voltage Protection Rating,  Maximum Continuous Operating Voltage Rating (MCOV),  Nominal Discharge Current (In) Rating,  Short-circuit Current Rating (SCCR),  General Markings. 

P- Circuit Breaker Adapters  Ratings,  Type Designation,  Shunt Trip,  Manufacturer’s Name,  Separately Shipped,  Terminations,  External Dropping  Circuit Breaker, Resistor.  Instructions.

N- High-Fault Protectors And Accessory High-Fault Modules  Type Designation,  Manufacturer’s Name,  Terminations,  Circuit Breaker,  Interrupting Rating.

O- Accessories

R- Circuit Breakers For Use In Communications Equipment

S- Molded Case Circuit Breakers Also Listed As Combination Type ArcFault

Q- Circuit Protectors 

Manufacturer’s

Name, Voltage Rating,  Ampere Rating,  Reset Instructions. 

T- Molded Case Circuit Breakers Also Listed As Branch/Feeder Type Arc

Circuit Interrupters Ambient Operating Temperature,  Wire Insulation Temperature Rating,  Same Polarity,  General Markings.



U- Classified Molded Case Circuit Breakers For Use In Specified Equipment  Classified Only and Compatibility List,  Classified and Listed Compatibility List,  Compatibility List,  Classification and Listing Mark,  General Markings.

V- Molded Case Circuit Breakers For Use In Photvoltaic (Pv) Systems  Voltage Rating,  PV marking,  Multi-pole PV Circuit Breakers,  Temperature Rating,  Wire Range and Type,  General Markings.



  

Device Identifier, ―TEST‖ Function, Instructions, General Markings.

   

Fault Circuit Interrupters Device Identifier, ―TEST‖ Function, Instructions, General Markings.

W- Marking Location Codes           

A B C D E F G H I J K

Examples for important markings of Molded Case Circuit breakers: Example#1: As shown in Fig.6, the numbered marking will be as follows:

Fig.6 1. ON and OFF: The ON and OFF (closed and open) positions of the handle must be marked (NEC 240.81). These positions may also be marked with the internationally recognized ―I‖ and ―O‖ symbols, although this is not a UL requirement. If these markings are not visible when a motor operator is installed over the circuit breaker markings, then they must appear on the motor operator. Motor operators may be found in applications where remote or automatic operation of a circuit breaker is required. 2. Ampere Rating (if 100 A or less): The ampere rating may be located on the handle escutcheon or on the handle itself (NEC 240.83(B)). Circuit breakers that are rated more than 100 A may have their ampere rating marked in a position that is not visible with trims or covers in place. 3. HACR type:This marking indicates the circuit breaker is suitable for use with the group motor installations typically found in heating, air conditioning and refrigeration equipment. The NEC 2005 no longer has this marking requirement. The electrical industry determined that circuit breakers are considered suitable for use with such equipment without any further testing, therefore, the HACR marking is no longer required on air conditioning and refrigeration equipment or on circuit breakers for use in these applications. The requirement for this marking has also been removed from the UL 1995 product standard for HVAC equipment. Example#2: As shown in Fig.7, the numbered marking will be as follows:

Fig.7 1. Manufacturer’s Name: This marking may be the manufacturer’s name, trademark or other recognized means to identifying the company that made the circuit breaker. 2. Type Designation: All circuit breakers are marked with a type designation, which may be a catalog number prefix or a separate designation. Equipment labels, such as on panelboards, will list the circuit breaker types suitable for use. Note that the word ―type‖ may or may not be used on the circuit breaker or equipment labels. It is important to review the markings on the equipment, such as a panelboard, to make sure the circuit breaker designations on the equipment match the marking on the circuit breaker. 3. Voltage Rating: All circuit breakers must be marked with a voltage rating. If the rating is not marked ―ac‖ or ―dc,‖ then it is suitable for both. 120/240 V rated circuit breakers are suitable for use on single and three-phase 4-wire systems where the line-to-ground voltage does not exceed 120 V. Wye rated circuit breakers such as those rated 480Y/277 V, are suitable for use on three-phase 4-wire systems where the voltage to ground does not exceed 277 V. Special attention needs to be given to high leg or corner-grounded delta systems to insure that the circuit breaker has the appropriate rating. A review to see that the circuit breakers installed have a voltage rating suitable for the application is paramount for a code-compliant installation (NEC 240.83(E)). 4. SWD: 15- or 20-A circuit breakers rated 347 V or less may be marked ―SWD,‖ meaning that they are suitable for switching fluorescent lighting loads on a regular basis

(NEC 240.83(D)). These circuit breakers are evaluated for high endurance use, since they will be used similar to a light switch. 5. HID: 50 A or less circuit breakers rated 480 V or less may be marked ―HID,‖ meaning they are suitable for switching high intensity discharge or fluorescent lighting loads on a regular basis. These circuit breakers may employ a different construction than a standard SWD circuit breaker in order to address the high inrush current resulting from the lower power factor created by the HID lighting (NEC 240.83(D)). These circuit breakers also undergo additional endurance evaluation to demonstrate their ability to perform the switching duty. 6. Trip and Reset: Circuit breaker handles typically assume an intermediate position when tripped. This position must either be marked on the circuit breaker or on the equipment into which it is to be installed. If these markings are not visible when a motor operator is installed, then a ―tripped‖ marking may appear on the motor operator. Example#3: As shown in Fig.8, the numbered marking will be as follows:

Fig.8 1. Line and Load Designation: Circuit breakers marked with ―line‖ and ―load‖ designations are not suitable for reverse connection. Circuit breakers with interchangeable trip units must be marked ―line‖ and ―load‖ unless there is no risk of shock when changing the trip unit. 2. Interrupting Ratings: All circuit breakers with an interrupting rating more than 5000 A must be

marked with an interrupting rating (NEC 240.83(C)). Interrupting ratings are stated in RMS symmetrical amperes. If the short-circuit current rating of the equipment in which the circuit breaker is installed is less than the interrupting rating of the circuit breaker, then the lesser rating applies. Circuit breakers should be reviewed after installation to ensure they have an interrupting rating suitable for the application. This marking may be found in any location except the back of circuit breakers that are 1-½ inches wide per pole or less due to the size constraint. 3. Ampere rating (if more than 100 A): The ampere rating of a circuit breaker larger than 100 A may be found in a location that is visible after the cover or trim is removed. This marking requirement also applies to interchangeable trip units (NEC 240.83(A)). 4. 40°C:This marking indicates the maximum ambient temperature in which the circuit breaker can be applied at its marked ampere rating without rerating the ampacity of the circuit breaker. This marking is required for thermal- magnetic circuit breakers and is optional for electronic trip circuit breakers unless they are only suitable for a 25°C ambient, in which case they must be marked 25°C. When the ambient temperature rises above 40°C, the designer may need to consult the manufacturer to obtain rerating information. 5. Terminations (Cu-Al wire): Circuit breakers must be marked with the type material (Cu-Al) and size of wire for which their terminals are suitable for use. If suitable for use with only copper or only aluminum, then the word ―only‖ must be used. The abbreviations ―CU‖ and ―AL‖ are generally found on circuit breakers as permitted by the product standard. If only solid 10–14 AWG wire can be used, then that information must be noted. The number of wires per terminal will also be noted if more than one wire per terminal is permitted (NEC 110.14(A)). 6. Tightening Torque: The nominal torque for all field-wiring terminals must be marked. If the width of the circuit breaker is 1-½ inches per pole or less, then this marking may be found in any location except on the back. 7. Wire Temperature Ratings: Circuit breakers rated 125 A or less may be marked as suitable for use with 60°C, 60/75°C or 75°C only wire. Circuit breakers rated more than 125 A are rated for use with 75°C wire; the marking is optional. It is always permissible to use wire with a higher temperature rating, but it must be sized in accordance with the temperature marking on the circuit breaker and NEC Table 310.16. If the width of the circuit breaker is 1-½ inches per pole or less, then this marking may be in any location except on the back (NEC 110.14(C)1). 8. Separately Shipped Connectors:If connectors are not factory installed on a

circuit breaker, then it must be marked with the proper connectors or terminal kits required in any location except the back. Example#4: As shown in Fig.9, the numbered marking will be as follows:

Fig.9 1. “1 – 3” Marking: A 2-pole circuit breaker used to protect a 3-phase load on a corner-grounded delta system must be rated and marked for such an installation. Circuit breakers marked ―1-phase – 3-phase‖ or ―1 – 3 ―are suitable for use on 3-phase corner-grounded delta or single-phase circuits (NEC 240.85). 100% Rated Marking: 100 percent continuous rated – Circuit breakers are typically intended for use at not more than 80% of rated current where the load is considered continuous, or will continue for 3 hours or more (NEC 210.20). However, some circuit breakers are rated for continuous use at 100% of their current rating. These circuit breakers must be so marked in any location except on the back. Enclosure information such as a specific type or specific volume must also be marked. A requirement for the use of 90°C insulated wire sized to the 75°C column in NEC Table 310.16 and specific ventilation requirements may also be marked on the circuit breaker or equipment (NEC 210.20(A) and 215.3).

Notes for Marking of Fuses and Circuit breakers Fuses or circuit breakers used for supplementary overcurrent protection of fluorescent fixtures, semiconductor rectifiers, motor-operated appliances, and so on, need not be marked for Interrupting Current IC.  Circuit Breakers with interrupting rating 5000 amperes will not have Interrupting rating marking.  Fuses with interrupting rating 10,000 amperes will not have Interrupting rating marking. 

Second: Identification Disconnecting Means used in a Series Rated Combination System For Identification of Disconnecting Means used in a Series Rated Combination System, the rules NEC 240.86 and 110.22 are applied as shown in below. The arrangement of protective components in a series rated system is determined by two methods as follows: 1. A licensed professional engineer and in this case the system is called Engineered Series Combination Systems and It can be applied For Existing installations only. 2. The system manufacturer and in this case the system is called Tested Series Combination System and It can be applied For Existing and new installations.

The marking of a series rated system will differ according to the two above method used for arrangement of protective components as follows: 1- Marking for Engineered Series Combination Systems: Equipment enclosures for circuit breakers or fuses shall be legibly marked in the field as directed by the engineer and the marking shall be readily visible. 2- Marking for Tested Series Combination Systems: (see Fig.10)

Fig.10: Marking for Tested Series Combination Systems

The enclosures of circuit breakers or fuses must be legibly marked by the equipment manufacturer and the marking shall be readily visible.  When the equipment is installed in the field, the equipment must have an additional label to indicate that the equipment has been applied with a series combination rating. 

Example for series rated system marking: A- In case of Engineered Series Combination Systems, The marking can be as follows: CAUTION — ENGINEERED SERIES COMBINATION SYSTEM RATED ………. AMPERES. IDENTIFIED REPLACEMENT COMPONENTS REQUIRED.

B- In case of Tested Series Combination Systems, the field marking can be as follows: CAUTION — SERIES COMBINATION SYSTEM RATED ………. AMPERES. IDENTIFIED REPLACEMENT COMPONENTS REQUIRED.

General Notes For Disconnecting Means Markings The markings shall specifically identify the purpose of each piece of equipment unless located and arranged so the purpose is evident (seeFig.11). For example, the marking should not simply indicate ―disconnect means‖ but rather ―disconnect means, water pump‖ or not simply ―disconnect means‖ but rather ―disconnect means, front lobby.‖ 

Fig.11: Purpose/Location Specified Marking The marking shall be of sufficient durability to withstand the environment involved.  The markings shall not fade or wear off. 

Electrical Rules and Calculations for Air-Conditioning Systems – Part Four In Article " Electrical Rules and Calculations for Air-Conditioning Systems – Part One ", which was the first Article in our new Course HVAC-2: Electrical Rules and Calculations for Air-Conditioning Systems, I explained the following points: Introduction for Air-Conditioning Systems Types  Introduction for Types of Motors/Compressors used in Air-Conditioning Systems 

And in Article " Electrical Wiring Diagrams for Air Conditioning Systems – Part One ", I explained the following points:    

Importance of Electrical Wiring for Air Conditioning Systems, How to get the Electrical Wiring for Air Conditioning systems?, Types of Electrical Wiring Diagrams For Air Conditioning Systems, How to read Electrical Wiring Diagrams?

Also, I explained the electrical wiring diagrams for Typical Air conditioning equipments in the following Articles:  

Electrical Wiring Diagrams for Air Conditioning Systems – Part Two Electrical Wiring Diagrams for Air Conditioning Systems – Part Three

And in Article " Electrical Rules and Calculations for Air-Conditioning Systems – Part Two ", I explained Types of Disconnecting Means for Air-Conditioning Systems. Also, in Article " Electrical Rules and Calculations for Air-Conditioning Systems – Part Three ", I explained identification and Marking of Disconnecting Means for AirConditioning Systems. Today, I will explain in where to locate Disconnecting Means used for Air Conditioning Systems.

Location of Disconnecting Means with respect to Controller, Motor and Driven Machinery Location

As we stated in Article " Electrical Rules and Calculations for AirConditioning Systems – Part Two " that the definition of the term In Sight From (Within Sight From, Within Sight) was as follows: Where one equipment is specified to be ―in sight from,‖ ―within sight from,‖ or ―within sight of,‖ and so forth, another equipment, the specified equipment is to be visible and not more than 15 m (50 ft) distant from the other. (see Fig.1 & Fig.2)

Fig.1: Disconnect insight from an equipment

Fig.2: Disconnect out of site from motor

First: for Motors and Motor-operated Equipments

1- NEC Code “In Sight From” Rules There are several Rules in the NEC Code for a disconnecting means to be in sight from the equipment that it controls.

Rule#1 NEC 430.102(A) requires an individual disconnecting means to be provided for each controller and shall be in sight from the controller location. This means that the controller must be visible and not more than 50 ft (15 m) from the disconnecting means.

Fig.3: The disconnecting means within sight of the controller location

Rule#2 NEC 430.102(B) requires that Disconnect must be in sight of motor location and driven machinery. If the disconnect as required in 430.102(A) is in sight of the controller, the motor location, and driven machinery, then that disconnect meets the requirements of both 430.102(A) and 430.102(B). (seeFig.4)

Fig.4: The disconnecting means within sight of the motor, controller and driven machinery locations If, because of the nature of the installation, the disconnect is in sight of the controller, but not in sight of the motor location and the driven machinery, then, generally, another disconnect must be installed. (see Fig.5)

Fig.5: The disconnecting means within sight of the controller location but not in sight of motor and driven machinery locations Note: When providing more than one disconnecting means, At least one of them shall be readily accessible as per NEC 430.107.

Rule#3 NEC 430.102(A) Exception No. 2 requires that A single disconnecting means may be located adjacent to (or in sight from) a group of coordinated controllers, where the controllers are mounted on a multi-motor continuous process machine. (see Fig.6)

Fig.6: Disconnect means for multiple motors on a single machine Note: NEC Code uses the terms motor location and controller location instead of

motor and controller because in many instances, the motor or controller is inside an enclosure and is out of sight until an access panel is removed. If the terms motor and controller were to be used, then it would be mandatory to install the disconnect inside of the enclosure. This is not always practical. The Commercial Building's roof-top air-conditioning units are examples.

Rule#4 NEC 430.113 for Motor and motor-operated equipment with More Than One power Source shall be provided with disconnecting means from each source of electric energy immediately adjacent to the equipment served.  Where multiple disconnecting means are provided, a permanent warning sign shall be provided on or adjacent to each disconnecting means to warn the user that other power sources are present.  Motor and motor-operated equipment that require multiple separate sources of power to operate properly like: 1. Motor space heater, 2. Motor with a speed switch, 3. Synchronous motors commonly use dc power for excitation purposes. (see Fig.7) 

Fig.7: Synchronous motors with 3-phase power supply and dc

power supply for excitation.

Exceptions to Rule#4: Exception No. 1: Where a motor receives electric energy from more than one source, the disconnecting means for the main power supply to the motor shall not be required to be immediately adjacent to the motor, provided the controller disconnecting means is capable of being locked in the open position. Exception No. 2: A separate disconnecting means shall not be required for a Class 2 remote-control circuit conforming with Article 725, rated not more than 30 volts, and isolated and ungrounded.

2- Exceptions NEC Code “In Sight From” Rules Three exceptions permit disconnecting means to be located elsewhereand not in sight of a motor.

430.102(A) Exception No. 1 For motors over 600 volts, the controller disconnecting means may be out of sight of the controller, provided the controller has a warning label indicating the location and identification of the disconnecting means, which must be capable of being locked in the open position.

430.102(A) Exception No. 3

Fig.8: valve actuator motor (VAM) assemblies The disconnecting means shall not be required to be in sight from valve actuator motor (VAM) assemblies containing the controller (see Fig.8) where such a location introduces additional or increased hazards to persons or property and the following conditions are met: 1. The valve actuator motor assembly is marked with a warning label giving the location of the disconnecting means. 2. The provision for locking or adding a lock to the disconnecting means shall be installed on or at the switch or circuit breaker used as the disconnecting means and shall remain in place with or without the lock installed.

430.102(B) exception the disconnecting means may only be out of sight of the motor, if the disconnecting means complying with 430.102(A) is individually capable of being locked in the open position and meets the following cases: Case#1:

If locating the disconnecting means close to the motor location and driven machinery is impracticable due to the type of machinery, the type of facility, lack of space for locating large equipment such as disconnecting means rated over 600 volts, or any increased hazard to persons or property, the disconnecting means is permitted to be located remotely.(see Fig.9) 

Fig.9: disconnecting means out of site for Motors rated in excess of 100 hp Some examples of increased or additional hazards to persons or property include, but are not limited to the following: 1. Motors rated in excess of 100 hp, 2. Multi-motor equipment, 3. Submersible motors, 4. Motors associated with adjustable speed drives, 5. Motors located in hazardous (classified) locations. 

Case#2: In industrial installations, if all of the following conditions are verified: 

1. Industrial installations have written safety procedures. 2. The conditions of maintenance and supervision ensure only qualified persons service the equipment.

Notes: The provision for locking or attaching a lock to the disconnecting means must be part of the disconnect and a permanent component of the switch or circuit breaker. (see Fig.10) 

Fig.10: Locking Hardware as a part of a circuit breaker The provision for locking (or adding a lock to) the disconnecting means is on the switch or circuit breaker and remains in place whether or not the lock is installed i.e. the lock is not readily removable or transferable type. (seeFig.10)  Disconnect switches or circuit breakers that are located only behind the locked door of a panelboard or within locked rooms do not comply with the requirements of 430.102. 

Second: for Air Conditioning Equipment

1- NEC Code “In Sight From” Rule

NEC 440.14 The Disconnecting means shall be located within sight from and readily accessible from the air-conditioning or refrigerating equipment.  The disconnecting means shall be permitted to be installed on or within the air-conditioning or refrigerating equipment, but it can't be located in access panels or where it will obscure the equipment nameplate.  This requirement is amendatory even if there is also a remote disconnect capable of being locked in the ―open‖ position under the provision of the exception to 430.102(B). 

Note: The main purpose of above rule is providing protection for service personnel working on equipment located in attics, on roofs, or outside in a remote location where it is difficult to gain access to a remote lockable disconnect

2- Exceptions NEC Code “In Sight From” Rule

440.14 Exception 1 There is no need to place a disconnecting means within sight from the equipment, if all of the following conditions are verified: 1. The disconnecting means is capable of being individually locked in the open position. 2. The equipment is essential to an industrial process in a facility

that has written safety procedures. 3. The conditions of maintenance and supervision ensure only qualified persons service the equipment. 4. The provision for locking (or adding a lock to) the disconnecting means is on the switch or circuit breaker and remains in place whether or not the lock is installed i.e. the lock is not readily removable or transferable type.(see Fig.10)

Note: 440.14 Ex 1 is for special conditions such as very large process refrigeration equipment and is rarely applicable for the following reasons: 1. This equipment is very large ones, so rated disconnects may not be available, 2. This equipment may be in hazardous locations, and locating disconnecting means within sight of the motor may introduce additional hazards.

3- Exceptions NEC Code “Readily Accessible” Rule

440.14 Exception 2 An accessible attachment plug and receptacle can serve as the disconnecting means for cord-connected equipment as per 440.13. The receptacle for the attachment plug doesn't have to be readily accessible. 



Examples of cord-connected equipment included in 440.13: 1.

Room air conditioners,(see Fig.11)

2.

Household refrigerators,

3.

Freezers,

4.

Beverage dispersers,

5.

Drinking water coolers.

Fig.11: an attachment plug and receptacle as a disconnecting means for Room air conditioners

As per 440.63 for Room air conditioners, an attachment plug and receptacle or cord connector shall be permitted to serve as the disconnecting means for a single- phase room air conditioner rated 250 volts or less if:(see Fig.11) 

1. the manual controls on the room air conditioner are readily accessible and located within 1.8 m (6 ft) of the floor, or 2. An approved manually operable disconnecting means is installed in a readily accessible location within sight from the room air conditioner. You can’t, for example, run the cord through a sheet of plywood and plug it in on the other side.



And as per 440.64 , A flexible cord that supplies a room air conditioner

can’t be longer than: 1.

3.0 m (10 ft) for 120V units,

2.

1.8 m (6 ft) for 208V or 240V units.

Electrical Rules and Calculations for Air-Conditioning Systems – Part Five I explained some of the Electrical Rules and Calculations for Air-Conditioning Systems in the following Articles:

Topic

Article

Introduction for Air-Conditioning Systems Types, Electrical Rules and Calculations for  Introduction for Types of Motors/Compressors used in AirAir-Conditioning Systems – Part One Conditioning Systems. 

Types of Disconnecting Means for Air-Conditioning Systems 

identification and Marking of Disconnecting Means for AirConditioning Systems 

where to locate Disconnecting Means used for Air Conditioning Systems

Electrical Rules and Calculations for Air-Conditioning Systems – Part Two Electrical Rules and Calculations for Air-Conditioning Systems – Part Three



Electrical Rules and Calculations for Air-Conditioning Systems – Part Four

I explained the Electrical Wiring Diagrams for Air Conditioning Systems equipment in the following Articles:

Topic Importance of Electrical Wiring for Air Conditioning Systems. 

Article Electrical Wiring Diagrams for Air Conditioning Systems – Part One

How to get the Electrical Wiring for Air Conditioning systems?  Types of Electrical Wiring Diagrams For Air Conditioning Systems.  How to read Electrical Wiring Diagrams? 

The electrical wiring diagrams for:  Window air conditioning units,  Split air conditioning units,  Multi-Split air conditioning units,  Mini-heat pumps,  Split Packaged units,  Unitary Packaged units. The electrical wiring diagrams for:  Chillers,  Air Handling Units, Fans, and Pumps.

Electrical Wiring Diagrams for Air Conditioning Systems – Part Two

Electrical Wiring Diagrams for Air Conditioning Systems – Part Three

Today, I will explain how to size Disconnecting Means used for Air Conditioning Systems.

Sizing of Disconnecting means for Air conditioning equipment

Important Definitions 1.1 Full Load Current FLC (or FLA): Full load current is the current drawn by a motor at its rated voltage, speed, frequency and horsepower.  The term Full load current FLC is equivalent to the term Rated load current RLC. 

Note: Where there is change in the motor voltage, speed, frequency and horsepower from the rated values, the current drawn by the motor in this case will be called the motor running current where: 

Motor running current ≤ Motor Full Load Current FLC 1.2 Locked Rotor Current LRC (or LRA): Locked Rotor Current, LRC is the steady-state current taken from the power line with the rotor locked (stopped) and with the rated voltage and frequency applied. Steady-state means that the motor starts at rest or zero speed. Why the motor Locked Rotor Current LRC is important? 1- Knowing a motor's locked rotor amps (LRA) and acceleration time helps in selecting upstream breakers. We make sure the circuit breaker or fuse can supply the locked rotor amps long enough to get the motor load up to speed and that it will trip if locked rotor amps continues beyond the allowable stall time. Note: Compare the LRA of a motor to the instantaneous time current curve (not the continuous current rating) for a protective device to see if the motor will be allowed to start without causing a nuisance trip. 2- Locked Rotor Current is a good approximation of the short circuit current that a motor will add to a short circuit on the power system. When a short occurs, the motor acts like an induction generator and delivers current to the fault until the energy stored in the motor's magnetic field and spinning rotor is dissipated. The current does not last long, but it must be accounted for in short circuit calculations.

Introduction

As we stated before in Article Electrical Rules and Calculations for AirConditioning Systems – Part Two that NEC Article 440 is in addition to or amendatory of the provisions of Article 430 and other applicable articles. Many requirements — for example, for disconnecting means, controllers, single or group installations and sizing of conductors— are the same as or similar to those applied in Article 430. So, Article 440 must be applied in conjunction with Article 430. So, let’s first study the disconnecting means sizing requirements as required by NEC Article 430.

First: Sizing of Disconnecting means for Motor as per NEC Article 430

Rule#1: Disconnecting Means Ampere Rating for Motors As per NEC 430.110 Ampere, The disconnecting means for motor circuits rated 600 volts, nominal, or less shall have an ampere rating (AR dis) not less than 115 percent of the full-load current rating of the motor. 

AR dis = 1.15 x I

FL

of motor

The full-load current (IFL) equivalent to the horsepower rating of motor is obtained from Table 430.247, Table 430.248, Table 430.249, or Table 430.250.  In case the horsepower rating of the motor do not correspond to the horsepower ratings shown in Table 430.247, Table 430.248, Table 430.249, or Table 430.250, the horsepower rating corresponding to the next higher value shall be selected. 

Table 430.247

Table 430.248

Table 430.249

Table 430.250

Example#1: A 50 HP Motor, 3Ø, 460 V, Design B load. Determine the size of the disconnect required for this motor. Solution: Get the full-load current (IFL) equivalent to the horsepower rating of each motor from Table 430.250.

IFL = 65 A

Apply Rule#1 for calculating Ampere Rating of disconnecting means AR follows:

dis as

AR dis = 1.15 x I FL of motor = 1.15 x 65 A = 74.75 A So, the Ampere Rating of the disconnect must be at least 74.75 amps.

Determining Disconnecting Means equivalent horsepower Rating in case of Motors To determine the equivalent horsepower Rating of the Disconnecting Means, the horsepower rating shall be selected from Table 430.247, Table 430.248, Table 430.249, or Table 430.250 corresponding to the Disconnecting Means Ampere Rating, and also the horsepower rating from 

Table 430.251(A) or Table 430.251(B) corresponding to the locked-rotor current.  In case the Full Load current and locked-rotor current do not correspond to the currents shown in Table 430.247, Table 430.248, Table 430.249, or Table 430.250, Table 430.251(A), or Table 430.251(B), the equivalent horsepower rating corresponding to the next higher value shall be selected.  In case different horsepower ratings are obtained when applying these tables, a horsepower rating at least equal to the larger of the values obtained shall be selected.

Exceptions to Rule#1: Disconnecting Means Ampere Rating for Motors Exception#1: A listed unfused motor-circuit switch having a horsepower rating not less than the motor horsepower shall be permitted to have an ampere rating less than 115 percent of the full-load current rating of the motor. Exception#2: Disconnecting means for a torque motor shall have an ampere rating of at least 115 percent of the motor nameplate current.

Rule#2: Sizing of disconnect means for Motors in Combination with other Loads For Combination Loads which operate simultaneously by a single disconnecting means and have one of the following combinations: 1. 2.

Two or more motors are used together, One or more motors are used in combination with other loads,

such as resistance heaters.

The rating of the disconnecting means shall be determined based on the rating of the combined load which can be calculated as in the following (2) cases: Case#1: If Motors ratings are given in horsepower Step#1: Get the full-load current (IFL) equivalent to the horsepower rating of each motor from Table 430.247, Table 430.248, Table 430.249, or Table 430.250. Step#2: Calculate Equivalent full-load current for the combined load (I as follows:

CFL)

Equivalent full-load current for the combined load = Σ all full load currents, including resistance loads I

CFL

= Σ IFL of motors + Σ IFL of resistance loads

Step#3: Calculate the Min. ampere rating of disconnect switch (AR dis.min ) based on Equivalent full-load current for the combined load (I CFL ) as follows: Min. ampere rating of disconnect means = 1.15 x Equivalent full-load current for the combined load AR dis.min = 1.15 x I

CFL

Step#4: Get the locked-rotor current (I LR) equivalent to the horsepower rating of each motor from Table 430.251(A) or Table 430.251(B). (see note for small motors in below) Step#5: Calculate Equivalent locked-rotor current for the combined load (I CLR) as follows: Equivalent locked-rotor current for the combined load = Σ all Locked Rotor currents, including resistance loads I

CLR = Σ

I

LR

of motors + Σ IFL of resistance loads

Step#6: Select the disconnect means horsepower rating (HPR dis) that correspond to Equivalent locked-rotor current (I CLR) for the combined load

from Table 430.251(A) or Table 430.251(B). Step#7: get full load current from Table 430.247, Table 430.248, Table 430.249, or Table 430.250 that correspond to the disconnect means horsepower rating (HPR dis) of step#6 Step#8: calculate selected disconnect Ampere rating (AR follows: AR dis. Selected = 1.15 x I

FL

dis. Selected)

as

from step#7

Step#9: compare between disconnect means ampere rating (AR dis. Selected) from step#8 and Min. ampere rating of disconnect means (AR dis.min ) from step#3 as follows: if (AR dis. Selected) from step#8 ≥ (AR dis.min) from step#3, then the horse power rating of the disconnect means will be (HPR dis) from step#6.  if (AR dis. Selected) from step#8 <(AR dis.min) from step#3, repeat steps #7,8 and 9 with next higher value of (HPR dis) obtained from step#6. 

Note: Where two or more motors or other loads cannot be started simultaneously, then Equivalent locked-rotor current for the combined load (I CLR) will be calculated as follows: I

CLR = Σ

I

LR

of motors that can be started simultaneously + Σ IFL of other concurrent loads

Case#2: If Motors ratings are given in Ampere The Ampere Rating of the disconnecting means will be as follows: AR dis ≥ 1.15 x I

Where: I Exception:

CFL

CFL

= Σ IFL of motors + Σ IFL of resistance loads

A listed non-fused motor-circuit switch having a horsepower rating equal to or greater than the equivalent horsepower of the combined loads, determined by Case#1, shall be permitted to have an ampere rating less than 115 percent of the sum of all currents at the full-load condition.

Table 430.251(A)

Table 430.251(B)

Important Note for small motors For small motors not covered by Table 430.247, Table 430.248, Table 430.249, or Table 430.250, the locked-rotor current (I LR) shall be assumed to be six times the full-load current (I FL). I

LR

=6I

FL

Note: A general-use switch, circuit breaker, molded case switch (nonautomatic circuit interrupter), or attachment plug and receptacle used as a disconnecting means must have an ampere rating of not less than 115 

percent of the motor full-load current.  Listed circuit breakers and molded case switches are tested under overload conditions at six times their rating, to cover motor circuit applications, and are suitable for use as a motor disconnecting means.

Example#2: A combined load consists of one 5-hp, one 3-hp, and two 1⁄2-hp motors, plus a 10-kW, all rated 240 volts, 3-phase. All motors are Design B motors. Determine the size of the disconnect required for this combination load. Solution: Step#1: Get the full-load current (IFL) equivalent to the horsepower rating of each motor from Table 430.250. Load 5-hp motor 3-hp motor 1⁄2-hp motor 1⁄2-hp motor 10-kW motor

Table Used 430.250 430.250 430.250 430.250

Step#1

IFL (in Ampere) 15.2 9.6 2.2 2.2 (10 x1000) / ((240 x1.732) = 24.1

Step#2: Calculate Equivalent full-load current for the combined load (I I

CFL =

Σ IFL of motors + Σ IFL of resistance loads

I

CFL =

15.2 + 9.6 + 2.2 + 2.2 + 24.1 = 53.3 A

CFL)

Step#3: Calculate the Min. ampere rating of disconnect switch (AR dis.min ) based on Equivalent full-load current for the combined load (I CFL ) AR dis.min = 1.15 x I

CFL

= 1.15 x 53.3 = 61.3 A

Step#4: Get the locked-rotor current (I of each motor from Table 430.251(B). Load 5-hp motor 3-hp motor 1⁄2-hp motor 1⁄2-hp motor

LR)

equivalent to the horsepower rating

Table Used 430.251(B) 430.251(B) 430.251(B) 430.251(B)

10-kW motor

ILR (in Ampere) 92 64 20 20 (10 x1000) / ((240 x1.732) = 24.1

Step#4 Step#5: Calculate Equivalent locked-rotor current for the combined load (I I

CLR = Σ

I

CLR

I LR of motors + Σ IFL of resistance loads

= 92 + 64 + 20 + 20 + 24.1 = 220.1 A

CLR)

Step#6: Select the disconnect means horsepower rating (HPR dis) that correspond to Equivalent locked-rotor current (I CLR) for the combined load from Table 430.251(B). For LRA = I CLR = 220.1 A at 230 V, A 15-hp general-purpose or heavy-duty switch is suitable which has LRA = 232 A ≥ 220.1 A

Step#6 Step#7: get full load current from Table 430.250 that correspond to the disconnect means horsepower rating (HPR dis) of step#6 A 15-hp general-purpose or heavy-duty switch will has a full load current = 42 A

Step#7 Step#8: calculate selected disconnect Ampere rating (AR

dis. Selected)

as follows:

AR dis. Selected = 1.15 x I FL from step#7 = 1.15 x 42 = 48.3 A Step#9: compare between disconnect means ampere rating (AR dis. Selected) from step#8 and Min. ampere rating of disconnect means (AR dis.min ) from step#3 as follows: Since (AR dis. Selected) from step#8 = 48.3 A < (AR dis.min) from step#3 = 61.3A Then go to the next higher value of (HPR dis) obtained from step#6 which is 20 HP which has full load current = 54 A and its (AR dis. Selected) = 1.15 x 54 = 62.1 A > (AR dis.min) from step#3 = 61.3A

Step#9 So, a 20-hp general-purpose or heavy-duty switch is suitable for this load combination.

Second: Sizing of Disconnecting means for Air Conditioning Equipment as per NEC Article 440

Introduction Which current value should be used in determining the Ampere Rating of the disconnect means for Hermetic Refrigerant Motor-Compressor and fans? We have three ratings can be marked on the nameplate of air conditioning equipment including Hermetic Refrigerant Motor-Compressor and fans which are: 1.

The nameplate rated-load current or nameplate full load

current, 2. The nameplate branch-circuit selection current (BCSC), 3. The nameplate horsepower rating.

The used rating in determining the Ampere Rating of the disconnect means for Hermetic Refrigerant Motor-Compressor and fans will be as follows: As per NEC 440.6 (A), the rated-load current marked on the nameplate of the equipment in which the motor-compressor is employed shall be used in determining the rating or ampacity of the disconnecting means,  But if no rated-load current is shown on the equipment nameplate, the rated-load current shown on the compressor nameplate shall be used.  As per NEC 440.2, When a branch-circuit selection current (BCSC) is marked on a nameplate, it must be used instead of the rated-load current to determine the size/Ratings of the disconnecting means. The value ofbranch-circuit selection current is always greater than the marked ratedload current.  As per NEC 440.6 (B), For multimotor equipment employing a shadedpole or permanent split-capacitor-type fan or blower motor, the full-load current for such motor marked on the nameplate of the equipment in which the fan or blower motor is employed shall be used instead of the horsepower rating to determine the ampacity or rating of the disconnecting means. This marking on the equipment nameplate shall not be less than the current marked on the fan or blower motor nameplate. 

Rule#1: Disconnecting Means Ampere Rating for Hermetic Refrigerant Motor-Compressor As per NEC 440.12(A)(1), For a hermetic refrigerant motor compressor, the ampere rating of the disconnecting means must be at least 115% of the nameplate rated-load current or branch circuit selection current (whichever is greater). AR dis = 1.15 x Σ the greatest value from nameplate hermetic refrigerant motor-compressor rated-load current(s) or branch-circuit selection current(s)

Determining Disconnecting Means equivalent horsepower Rating in case ofHermetic Refrigerant Motor-Compressor As per NEC 440.12(A)(2), To determine the equivalent horsepower Rating of the Disconnecting Means, the horsepower rating shall be selected from Table 430.248, Table 430.249, or Table 430.250 corresponding to the rated load current or branch-circuit selection current, whichever is greater, and also the horsepower rating from Table 430.251(A) or Table 430.251(B) corresponding to the locked-rotor current.  In case the nameplate rated-load current or branch-circuit selection current and locked-rotor current do not correspond to the currents shown in Table 430.248, Table 430.249, Table 430.250, Table 430.251(A), or Table 430.251(B), the horsepower rating corresponding to the next higher value shall be selected.  In case different horsepower ratings are obtained when applying these tables, a horsepower rating at least equal to the larger of the values obtained shall be selected. 

Exceptions to Rule#1: Disconnecting Means Ampere Rating for Hermetic Refrigerant Motor-Compressor A listed unfused motor circuit switch, without fuseholders, having a horsepower rating not less than the equivalent horsepower determined in accordance with 440.12(A)(2)explained in above, shall be permitted to have an ampere rating less than 115 percent of the specified current.

Rule#2: Sizing of disconnect means for hermetic refrigerant motorcompressors inCombination with other Loads As per NEC 440.12(B), For Combination Loads which operate simultaneously by a single disconnecting means and have one of the following

combinations: 1. Two or more hermetic refrigerant motor-compressors are used together, 2. One or more hermetic refrigerant motor-compressor with other motors or loads such as resistance loads.

The rating of the disconnecting means shall be determined based on the rating of the combined load which can be calculated as in the following (2) cases: Case#1: If Motors ratings are given in horsepower Step#1: Get the full-load currents of all loads as follows: A- For motors and loads other than a hermetic refrigerant motorcompressor, and fan or blower motors: Get the full-load current (I FL) equivalent to the horsepower rating of each motor from Table 430.248, Table 430.249, or Table 430.250. Then sum these motors full load currents with other full load currents of resistance loads to get a combined full load current (I CFL-1) as follows: I

CFL-1 =

Σ IFL of motors + Σ IFL of resistance loads

B- For hermetic refrigerant motor- compressor (s): I

= Σ the greatest value from nameplate hermetic refrigerant motorcompressor rated-load current(s) or branch-circuit selection current(s)

FL-HRMC

C- For fan or blower motors: I

FL-FAN

= Σ The nameplate full load currents of fan(s)

Step#2: Calculate Equivalent Total full-load current for the combined load (I FL-TOTAL) as follows: I FL-TOTAL = I

CFL-1

+ I FL-HRMC + I

FL-FAN

Step#3: Calculate the Min. ampere rating of disconnect switch (AR

dis.min

)

based on Equivalent Total full-load current for the combined load (I as follows:

FL-TOTAL)

Min. ampere rating of disconnect means = 1.15 x Equivalent full-load current for the combined load AR dis.min = 1.15 x I

FL-TOTAL

Step#4: Get the locked-rotor currents of all loads as follows: A- For motors and loads other than a hermetic refrigerant motorcompressor, and fan or blower motors: Get the locked-rotor current (I LR) equivalent to the horsepower rating of each motor from Table 430.251(A) or Table 430.251(B). (see note for small motors in below) Then sum these motors locked-rotor currents with other full load currents of resistance loads to get a combined locked-rotor current current (I CLR-1) as follows: I

CLR-1 = Σ

I LR of motors + Σ IFL of resistance loads

B- For hermetic refrigerant motor- compressor (s): I

LR-HRMC

= Σ Nameplate Locked rotor current of hermetic refrigerant motorcompressor (s)

C- For fan or blower motors: I

LR-FAN

= Σ The nameplate Locked rotor currents of fan(s)

Step#5: Calculate Equivalent Total locked-rotor current for the combined load (ILR-TOTAL) as follows: ILR-TOTAL = I

CLR-1

+I

LR-HRMC

+I

LR-FAN

Step#6: Select the disconnect means horsepower rating (HPR dis) that correspond to Equivalent total locked-rotor current (ILR-TOTAL) for the combined load from Table 430.251(A) or Table 430.251(B). Step#7: get full load current from Table 430.248, Table 430.249, or Table 430.250 that correspond to the disconnect means horsepower rating (HPR dis) of step#6.

Step#8: calculate selected disconnect Ampere rating (AR follows: AR dis. Selected = 1.15 x I

FL

dis. Selected)

as

from step#7

Step#9: compare between disconnect means ampere rating (AR dis. Selected) from step#8 and Min. ampere rating of disconnect means (AR dis.min ) from step#3 as follows: if (AR dis. Selected) from step#8 ≥ (AR dis.min) from step#3, then the horse power rating of the disconnect means will be (HPR dis) from step#6.  if (AR dis. Selected) from step#8 <(AR dis.min) from step#3, repeat steps #7,8 and 9 with next higher value of (HPR dis) obtained from step#6. 

Note: Where two or more motors or other loads such as resistance heaters, or both, cannot be started simultaneously, then Equivalent locked-rotor current for the combined load (ILR-TOTAL) will be calculated as follows: ILR-TOTAL = current of simultaneously motors or fans loads + current of non-simultaneously motors or fans loads + Σ IFL of other concurrent loads Where: current of simultaneously motors or fans loads = Σ I LR of motors( other than hermetic refrigerant motor-compressor or fan) that can be started simultaneously + Σ Nameplate Locked rotor current of hermetic refrigerant motor-compressor (s) that can be started simultaneously + Σ The nameplate Locked rotor currents of fan(s) that can be started simultaneously current of non-simultaneously motors or fans loads = Σ the greatest value from nameplate of non- simultaneously hermetic refrigerant motorcompressor rated-load current(s) or branch-circuit selection current(s) + Σ The nameplate full load currents of non- simultaneously fan(s)

Case#2: If Motors ratings are given in Ampere The Ampere Rating of the disconnect means will be as follows: AR dis ≥ 1.15 x I

FL-TOTAL

Where: I

FL-TOTAL

=I

CFL-1

+I

FL-HRMC

+I

FL-FAN

Exception: A listed unfused motor circuit switch, without fuseholders, having a horsepower rating not less than the equivalent horsepower determined by case#1 shall be permitted to have an ampere rating less than 115 percent of the sum of all currents.

Important Note for small motors For small motor compressors not having the locked-rotor current marked on the nameplate, or for small motors not covered by Table 430.248, Table 430.249, or Table 430.250, the locked-rotor current (I LR) shall be assumed to be six times the full-load current (I FL). I

LR

=6I

FL

Note: A general-use switch, circuit breaker, molded case switch (nonautomatic circuit interrupter), or attachment plug and receptacle used as a disconnecting means must have an ampere rating of not less than 115 percent of the motor full-load current.  Listed circuit breakers and molded case switches are tested under overload conditions at six times their rating, to cover motor circuit applications, and are suitable for use as a motor disconnecting means. 

Important Note for Disconnecting Means Rated in Excess of 100 Horsepower Where the rated-load or locked-rotor current as determined by rules#1 or 2

would indicate a disconnecting means rated in excess of 100 hp, the disconnecting means shall be permitted to be a general-use or isolating switch where plainly marked ―Do not operate under load.‖

Electrical Water Heaters Power Rating Calculations – Part One Most Domestic and commercial buildings need a service hot water system. Depending upon the type of building, this system could range from as small as an under sink water heater for washing hands to a 10,000 gallon hot water storage tank system used in a hospital laundry. This Article is intended to help designers to choose the appropriate type and calculate the required power rating for thee chosen type of Electrical Water Heater. Before going on with the calculations, we need to give you a brief about the following points:   

Hot Water System Components, Different types of Water Heaters used in domestic and commercial buildings, How to choose the best type of water heater for any application?

With this brief, you will be familiar with the types and construction of common Water Heaters.

1-

Hot Water System Components

A Hot water heating system has (4) major components (see Fig.1): 1. 2. 3. 4.

Heat energy source, Heat transfer equipment, Distribution system, Terminal hot water usage devices.

Fig (1): Hot Water System Components The most important components are the first and second ones.

1.A- Heat energy sources, they may be: 1. 2. 3. 4.

Fuel combustion, Solar energy collection, Electrical conversion, Recovered waste heat.

1.B- Heat transfer equipment, they may be: Direct heat transfer (see Fig.2) is from the combustion of fuels or direct conversion of electrical energy into heat,

Fig (2): Direct Heat Transfer Indirect heat transfer (see Fig.3) uses heat energy originating from remote heat sources, such as boilers, solar heat collectors, cogeneration refrigeration or waste heat.

Fig (3): Indirect Heat Transfer

2- Types of Water Heaters in Domestic And Commercial Buildings

Common types of commercial and industrial water heating equipment include: 1. 2. 3.

Storage water Heaters (Tank-Type), Instantaneous water heaters (Tankless-Type), Hybrid water heaters.

2.1- Storage Water Heaters (Tank-Type) This type of heaters incorporates the burner, storage tank, outer jacket,

insulation and controls in a single unit and is normally installed without dependence on other hot water storage equipment (see Fig.4).

Fig (4): Typical Storage Water Heater (Tank-Type). They are available in electric, liquid propane (LP) and natural gas models. Natural gas and LP water heaters normally use less energy and are less expensive to operate than electric models of the same size. There are two types of Storage water heaters (Tank-Type) which are: 1. 2.

Volume Storage water heaters (Tank-Type), Small Storage water heaters (Tank-Type).

2.1.A Volume Storage Water Heaters (Tank-Type) They are typically vertical, cylindrical tanks, usually standing on the floor or on a platform raised a short distance above the floor (as in Fig.4). In houses they can be mounted in the ceiling space over laundry-utility rooms. Typical sizes for household use range from 75 to 400 liters (20 to 100 US gallons).

2.1.B Small Storage Water Heaters (Tank-Type) Small storage tank water heaters, known as point of use (POU), utility or mobile home water heaters, are good choices for adding hot water to small buildings, shops or garages (see fig.5).

Fig (5): Small Storage Water Heaters (Tank-Type) These water heaters usually range in size from 2.5 to 19 gallons. The largest of these miniature units can also be used to provide hot water to secondary bathrooms that may be situated far from your home's main water heater. Tiny point-of-use (POU) electric storage water heaters with capacities ranging from 8 to 32 liters (2 to 6 gallons) are made for installation in kitchen and bath cabinets or on the wall above a sink. They typically use low power heating elements, about 1 kW to 1.5 kW, and can provide hot water long enough for hand washing, or, if plumbed into an existing hot water line, until hot water arrives from a remote high capacity water heater.

Storage water heaters (Tank-Type) have a special type which is the Solar water heaters. 2.1.C Solar Water Heaters Solar powered water heaters have two main components (see Fig.6): 1. 2.

Solar collectors, Storage tank.

Fig (6): Solar Water Heater Components 1- Solar Collectors Solar collectors are installed outside dwellings, typically on the roof or walls or nearby, 2- Storage Tank The potable hot water storage tank is typically a pre-existing or new conventional water heater, or a water heater specifically designed for solar thermal. Types Of Solar Water Heaters Solar powered water heaters have two main types:

1. 2.

The direct-gain type, The Indirect or closed-loop type.

A- The Direct-Gain Type: In this type (see Fig.7), the potable water is directly sent into the collector. Many such systems are said to use integrated collector storage (ICS), as direct-gain systems typically have storage integrated within the collector.

Fig (7): Direct-Gain Solar Water Heater Heating water directly is inherently more efficient than heating it indirectly via heat exchangers, but such systems offer very limited freeze protection (if any), can easily heat water to temperatures unsafe for domestic use, and ICS systems suffer from severe heat loss on cold nights and cold, cloudy days. B- The Indirect Or Closed-Loop Type: This type does not allow potable water through the panels, but rather pump a heat transfer fluid (either water or a water/antifreeze mix) through the panels (see Fig.8). After collecting heat in the panels, the heat transfer fluid flows through a heat exchanger, transferring its heat to the potable hot water.

Fig (8): Closed-Loop Solar Water Heater When the panels are cooler than the storage tank or when the storage tank has already reached its maximum temperature, the controller in closed-loop systems will stop the circulation pumps.

2.2- Instantaneous Water Heaters (Tankless-Type) They can called also on-demand water heaters, this type of heaters has minimal storage capacity, they do not store hot water; rather they Heat water as it passes through a series of coils in the unit (see Fig.9). They are available in electric, LP and natural gas models. Most tankless units can provide up to 3.5 gallons of heated water per minute.

Fig (9): Instantaneous Water Heaters (Tankless-Type) They usually include a flow switch as part of the control system. Tankless, instantaneous water heaters are best used for a steady, continuous supply of hot water. Tankless heaters may be installed throughout a household at more than one point-of-use (POU), far from a central water heater, or larger centralized models may still be used to provide all the hot water requirements for an entire house. The main advantages of tankless water heaters are: 1. A continuous flow of hot water as compared to a limited flow of continuously heated hot water from conventional tank water heaters, 2. Since the unit only heats water as you use it, a tankless heater is usually more energy efficient than a traditional storage tank water heater.

The main disadvantages of tankless water heaters are: 1.

Tankless water heaters can provide an unlimited amount of hot

water, but it can only provide a limited number of POU, 2. Common types of water heaters can't supply hot water at more than two points at a time.

A famous type of Instantaneous water heaters (Tankless-Type) is Electric shower heads 2.2.A Electric Shower Heads An electric heating element is incorporated into such shower heads to instantly heat the water as it flows through (see Fig.10).

Fig (10): Electric Shower Heads Electric showers have a simple electric system, working like a coffee maker, but with a larger water flow. A flow switch turns on the device when water flows through it. Once the water is stopped, the device turns off automatically. An ordinary electric shower often has three heat settings: low (2.5 kW), high (5.5 kW) or cold (0 W) to use when a central heater system is available or in hot seasons.

2.3 Hybrid Water Heaters

A hybrid water heater (see Fig.11) is a water heating system that integrates technology traits from both the tank-type water heaters and the tankless water heaters. It Heats cold water via an electrical heating element and heat pump that pulls in ambient air and extracts the available heat.

Fig (11): Hybrid Water Heaters They are also called Heat Pump Water Heaters (HPWH), they have small storage tanks that temper incoming cold water; This means hybrids only have to increase water temperature from warm to hot as opposed to tankless which has to raise completely cold water to hot. The defining characteristics of a "hybrid water heater" are: 1. A combination of water flow of tank and efficiency of tankless of water heater, 2. Built-in small storage water reservoir as part of heat exchanger (typically between two gallons to 20 gallons), 3. Dual activation: flow sensing and thermostat control.

Hybrid water heaters can be gas-fired (natural gas or propane), or be electrically powered using a combination of heat pump and conventional electric heating element.

3- How to choose the best type of water heater?

Many factors determine which water heater is best for your home. The three main factors to consider when choosing your water heater are: 1. 2. 3. 4. 5. 6. 7.

Water Storage Capacity, Water Heating Method, water heating system structure, water heating system Fuel Type, Recovery Rate, Space Limitations, Energy Efficiency.

3.1 Water Storage Capacity (by gallons or liters) An undersized water heater will work harder and have a shorter lifespan. So make sure to select a hot water heater that provides enough hot water for your home. The typical capacities for different types of water heaters are as follows: Volume Storage water heaters (Tank-Type): Typical sizes for household use range from 75 to 400 litres (20 to 100 US gallons).  Small Storage water heaters (Tank-Type: range in size from 2.5 to 19 gallons.  Tiny point-of-use (POU) electric storage water heaters with capacities ranging from 8 to 32 liters (2 to 6 gallons). 

Instantaneous water heaters (Tankless-Type): Most tankless units can provide up to 3.5 gallons of heated water per minute.  Hybrid water heaters: Typical sizes between two gallons to 20 gallons. 

3.2 Water Heating Method According to the user requirements, the selected method of water heating method can be determined as follows: 1. 2. 3.

Storage water Heaters (Tank-Type), Instantaneous water heaters (Tankless-Type), Hybrid water heaters.

Each offers unique advantages, and you can compare features and benefits in the table below.

Type

Method of operation

Factors to be Considered

Conventional Stores hot water Tank regularly in a tank sized to suit the users requirements.



Tankless



They do not store hot water; rather they Heat water as it passes through a series of coils in the water heater.

Economical,  Can be positioned in closet, basement or garage  Capacity ranges from 20 to 100 gallons,  Efficiency varies between models, brands and  fuel sources.

Require a larger up-front investment,  best used for a steady, continuous supply of hot water,  A continuous flow of hot water,

Hang on wall and frees up floor space,  Excellent option for residences occupied part-time,  usually more energy efficient by as much as 30%,  can provide an unlimited amount of hot water, but for only limited number of POU,  Requires ventilation. 

Hybrid

Heats cold water via an electrical heating element and heat pump that pulls in ambient air and extracts the available heat.

Require a larger up-front investment,  Magnesium anode rod extends life of the tank,  Heat pump delivers more hot water, up to 33 percent faster than standard electric water heater. 

3.3 Water Heating System Structure Also, according to the user requirements, the structure or distribution of hot water system can be one of the following types: 1. 2.

Centralized (Whole House) system, Point of Use Systems.

Each type is explained in detail in above paragraphs.

3.4 Water Heating System Fuel Type As per the user requirements, the selected fuel for the water heating

system can be one of the following: 1. 2. 3. 4.

Natural gas, LP (liquid propane gas), Electric, Solar.

Most water heaters are fueled by gas or electricity. Refer to the table below for comparison: Type

Factors to be Considered Requires a slightly larger up-front investment,  Must be vented outdoors for safety,  Units with sealed combustion or power venting increase safety,  Cost less to operate,  Not affected by power outages (tank-style only). 

Gas

 

Electric

  

Generally cost less than gas models, Easy to maintain, Requires no combustibles or venting, Heats water quickly, Offer high energy factor ratings.

Requires a larger up-front investment,  Requires no combustibles or venting,  Heats water up to 33 percent quicker than standard electric models,  Lower operating costs saves hundreds annually,  8700 BTU/h compressor is the most powerful in its class. 

Hybrid (electric)

Requires a slightly larger up-front investment,  Must be vented outdoors for safety,  Units with sealed combustion or power venting increase safety,  Cost less to operate than electric models,  Not affected by power outages (tank-style only). 

Liquid Propane (gas)

relatively good payback period, in average between 5-10 years,  low maintenance costs,  relatively high upfront costs,  in most areas they will require electrical or gas or other fuel backup during the winter period,  they require excellent overheating and freeze protection,  Environment friendly. 

Solar Power

3.5 Recovery Rate It is the amount of gallons or liters of hot water that the water heater is capable of providing in a given period of time (hour or minute). So, The greater your demand for hot water, the higher recovery rate you need. The used units expressing the recovery rate for water heaters are: GPH (Gallons Per Hour): The amount of water, in gallons, that is used each hour by the plumbing fixtures and equipment, such as dish machines.  GPM (Gallons Per Minute): The amount of water, in gallons, flowing through a plumbing fixture or through an instantaneous water heater per minute. 

Common Recovery Rates: Common recovery rates for Electrical water heaters for given input power are as in below table:

Recovery Rates in Gallons per hour – Electrical Water Heater Common recovery rates for Gas water heaters for consumed BTU are as in below table:

Recovery Rates in Gallons per hour – Gas Water Heater BTU (British Thermal Unit): The quantity of heat required to raise the temperature of one pound of water one degree F.

3.6 Space Limitations Once you know the capacity for your water heater, remember to take the unit’s dimensions into consideration. You can follow the following recommendations: A new conventional storage replacement unit may be larger than the old one because more insulation is required to meet the latest strict federal 

energy standards. Keep this in mind where units are installed in closets or other close quarters,  If you are upgrading to a larger unit, you may need to have plumbing run to it if it has to be relocated. One way to avoid relocating the unit is to select a model in a non-standard size, such as a unit that is shorter but larger around, known as a ―low boy‖ hot water heater. Lowboys vary between 30 to 49 inches and hold up to 50 gallons of water,  Tall water heaters range from 50 to 76 inches and can hold up to 100 gallons of water. They're ideal for basements or garages where height isn't an issue,  If you are purchasing a tankless water heater, be sure the location you choose for installation meets ventilation requirements,  The ideal location for a tankless unit is on an exterior wall near a gas supply line, water supply line and electrical power source. This is also the easiest and most cost-effective way to run the venting,  The unit should have ½-inch clearance on the sides, 12 inches on the front and 18 inches off the floor,  Hybrids offer a narrow 21-inch diameter for access into smaller locations.

3.7 Energy Efficiency Whichever fuel source you use, a water heater can be the third largest energy user in your home, so you’ll want a unit that offers energy and cost savings. Fortunately, almost all water heaters offer increased efficiencies to meet increasingly strict federal energy standards. Energy factor EF and yearly operating costs can be found on the Energy Guide label on the unit (see Fig.12).

Fig (12): Energy Guide Label Energy factor EF measures how efficiently a unit converts energy into heat as well as how much heat is lost during storage. The higher the energy factor, the more energy efficient the water heater is. Look for EF ratings as close to 1 as possible. Electric heaters tend to have the highest EF ratings.

Electrical Water Heaters Power Rating Calculations – Part Two This is the second Article for helping designers to choose the appropriate type and calculate the required power rating for the chosen type of Electrical Water Heater. In the first Article " Electrical Water Heaters Power Rating Calculations – Part One", we gave you a brief about the following points: 

Hot Water System Components,

Different types of Water Heaters used in domestic and commercial buildings, How to choose the best type of water heater for any application?

 

With this brief, you will be familiar with the types and construction of common Water Heaters. Today, we will explain in detail the Sizing and Power Rating Calculations for Water Heaters.

1-

Design Methodology of Electrical Water Heaters for any Building

First let's start from the beginning, you are the electrical designer for a new building and you want to size and calculate the power rating for the electrical water heaters that will be used through this building, so what is the first information that you must know? Yes, it is the type of this building/ occupancy, is it a residential, business, hospital, hotel, office building … etc?  The second information that you must know or calculate is the number of plumbing facilities in this building; this means how many Water Closets WCs, Lavatories, Bathtubs, showers, drinking fountains and service sinks included in this building.  The third information that you must calculate is the total demand of water (in gallons or liters) used per hour or minute in units of GPH (gallon per hour) or GPM (gallon per minute), this can be done by using the standard tables of GPH or GPM for each type of plumbing facilities for each type of buildings provided by many international or local plumbing codes.  At the end, by following one of the calculation procedures explained in this course, you can calculate the size (volume) in gallons or liters and the power rating in KW for the required electrical water heaters for this building. 

Fig (1): Design Methodology of Electrical Water Heaters for any Building As a summary for the above design methodology which is represented in Fig.1, we are going to calculate the following: The Minimum number of plumbing facilities for a given type of building/occupancy, 

  

The total demand of Water in GPH or GPM, The size (volume) in gallons or liters for the required electrical water heaters, The power rating in KW for the required electrical water heaters.

Step#1: Determination of The Building/Occupancy Type

As we all know that there are different types of buildings/occupancies which they are differentiated as per their main usage to the following main categories: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Industrial buildings, Commercial buildings, Residential buildings, Agricultural buildings, Educational buildings, Transportation buildings, Religious buildings, Parking and storage, Military buildings, Governmental buildings, Cultural buildings, Other buildings.

For more information about the above types of buildings, please read our article " Electrical Design Philosophy for Major Types of Buildings ".

Step#2: Calculation of The Minimum Number of Plumbing Facilities for a Given Type of Building/Occupancy

The common Plumbing Facilities for included in any Type of the above Buildings are as follows: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Water Closets WCs, Lavatories, Bathtubs, Showers, Drinking fountains, Service sinks, Water coolers, Water dispensers, Dishwashers, Laundries.

And to know the Minimum Number of Plumbing Facilities for a Given Type of Building/Occupancy, you must use one of the two following methods: 1. 2.

Quantity survey, Calculation by using Excel Spreadsheets.

1- Quantity survey: In this method to get the number of Plumbing Facilities, A quantity survey will be done from the architectural drawings of the new constructed buildings or by a site visit for the existing/old buildings.

2- Calculation by using Excel sheets: This method will be used only in two cases: Case#1: Where you can't get the architectural drawings of the new constructed building or,  Case#2: In the preliminary design phase where preliminary electrical load calculations have to be done, In this case, we can Calculations Spreadsheet for Minimum Number of Plumbing Facilities Required as per International Plumbing Code (IPC). 

Calculations Spreadsheet For Minimum Number Of Plumbing Facilities Required

The standard data used in this excel sheet are from International Plumbing Code (IPC) that developed by the International Code Council (ICC). This Excel Spreadsheet uses IPC - Table 403.1 for Minimum Number of Fixtures Required (see Fig.2). The table specifying the minimum number of plumbing fixtures required based on the following factors:   

The Building Occupancy Type, The Occupancy Classification, The Number of Building Occupants.

Fig (2): Minimum Number of Fixtures Required (Table 403.1)

Advantages of using IPC - Table 403.1 for Minimum Number of Fixtures Required: The IPC - Table 403.1 is based on various studies and is intended to provide equal access to fixtures,  The IPC - Table 403.1 is very accurate since it is directly related to the referenced building codes, 

The IPC - Table 403.1 takes into consideration the female population and the waiting period for the female population. 

How to use the Calculations Spreadsheet for Minimum Number of Plumbing Facilities Required?

This excel sheet includes (4) sheets as follows: 1. 2. 3. 4.

Cover and General data sheet, Floor input data sheet, Floor Calculation results sheet, Standard data sheet.

1- Cover and General Data Sheet

Fig (3): Cover and General data sheet

This worksheet lists the other worksheets in an index as shown in above Fig.3 ; you can go to a specific worksheet by just clicking on its name.  You can email me directly by clicking on the phrase ―Designed by: Ali Hassan‖ for any inquiries or notes.  This worksheet includes also the general data about the project and the designer which must be filled for every new project designed by this excel sheet. 

2- Floor Input Data Sheet

Fig (4): Floor input data sheet

In this sheet (see Fig.4), you must Input the following data:

a- Floor Level: input Floor level, b- The Building Occupancy Type: Pick Building Classification from the dropdown menu which includes (4) nos. of basic types of Building Occupancy as follows: 1. 2. 3. 4.

Business, Assembly, Institutional, Residential.

c- The Occupancy Classification: Pick Occupancy type from the dropdown menu which includes (26) nos. of Classifications of Occupancies as follows:

Theaters

Factory & Industrial

Nightclubs

High Hazard (See Section 403.2 & 403.3)

Asylums, reformatories, Etc Mercantile (See Section 403.2, 403.3 & 403.5)

Restaurants

Residential Care

Hotels, Motels

Halls, Museums, Etc.

Hospitals

Lodges

Business (See Section 403.2, 403.3 & 403.5)

Ambulatory Nursing Home Patients Day Nurseries, Sanitariums, Nonambulatory Nursing Home Patients, Etc. Employees, other than Residential care Visitors, other than Residential care

Educational

Prisons

Coliseums, Arenas Churches Stadiums, Pools, Etc.

Multiple family Dormitories One & Two family Dwellings Storage (see Sections 403.2 & 403.4)

d- The Area of the building in square footage (floor wise): Input square footage for floor under design. Note: If the input cell is shaded in grey, so no input required.

e- The Number of Building Occupants: Input number of people planned to use this floor. Note: If the input cell is shaded in grey, so no input required.

f- Number of Cells in Housing Pod: Input number of cells within housing pod. Program assumes that cells are double bunked. Note: If the input cell is shaded in grey, so no input required.

g- Percentage of Males and Females: The required number of fixtures shall be distributed equally between the sexes based on the percentage of each sex anticipated load. The occupant load shall be composed of 50% of each sex, unless statistical data approved by the code official indicate a different distribution of the sexes. This condition will not occur in prisons or other similar type facilities.

3- Floor Calculation Results Sheet

Fig (5): Floor Calculation results sheet In this sheet (see Fig.5), the calculation of Minimum Number of Fixtures Required for the floor under design will be done, you will find the following results: 1. 2. 3. 4. 5.

Percentage of Males, Percentage of Females, Total number of people, Number of men, Number of women.

Details of calculated plumbing Facilities Required as follows: 1. 2. 3. 4. 5. 6.

Water closets, Urinals, Lavatories, Bathtubs/Showers, Drinking fountains, Service sinks.

In bottom of this sheet, you will find the Totals of Calculated Plumbing Facilities Required for Floor under design. Note: You must repeat the calculation for each floor to get the total Number of Plumbing Facilities Required for it. These results will be used to calculate the total demand of Water in GPH or GPM for the building floor under design. 

4- Standard Data Sheet

Fig (6): Standard data sheet This sheet (see Fig.6) includes the standard values of IPC - Table 403.1 for Minimum Number of Plumbing Facilities required based on the factors described above.

To download your copy of Calculations Spreadsheet for Minimum Number of Plumbing Facilities Required as per IPC code, please click on the link.

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