Distribution System And Transformers

Energy Saving in Electrical Utilities K. R. GOVINDAN Kavoori Consultants 22, Janakiraman Street, West Mambalam, Chennai, 600 033.

Distribution System and Transformers

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About the faculty CEO of Kavoori consultants: services offered:  Energy audit, electrical safety and installation audit, relay protection and coordination studies, maintenance, technical training of executives and technicians of all trades, in-house as well as open seminars,  Technical trouble shooting 

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ATTITUDE Half full or half empty?

Two liters container; 1 liter liquid. Energy management

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ENERGY MANAGEMENT 

More precisely,

EFFICIENT MANAGEMENT OF ENERGY, THE VITAL RESOURCE. What is efficient management? Energy is utilized to do work; Use only the required minimum Or optimum requirement To perform a particular work.

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In the Present day context of:

  

Depleting energy sources Spiraling costs pollution of environment to alarming levels.

Energy management assumes top priority

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ENERGY AUDIT PRE – REQUISITE FOR AN ENERGY MANAGEMENT PROGRAMME BY ITSELF DOES NOT SAVE ENERGY HELPS MANAGEMENT IDENTIFY AREAS OF HIGHEST SAVINGS POTENTIAL

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FUNCTIONS OF AUDIT Assesses various forms of energy use  Compares with estimated minimum  Provides inputs for budgetary control 

MOST IMPORTANT FOR OLD PLANTS

? SET UP WHEN FUEL COST WAS VERY LOW NO CONCERN FOR ENERGY EFFICIENCY

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Good energy management is 

Increasing utilization efficiency or reducing losses Or  CONSERVATION OF ENERGY



Let us consider the electrical energy Energy management

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THE STORY OF ENERGY 300 BILLION YEARS AGO, ENTIRE ATMOSPHERE OF OUR EARTH- UNFIT FOR LIFE SUPPORT SLOWLY ALGEY AND LIKE PLANTS APPEARED, CONVERTED CO2 INTO O2 Received and STORED energy from the sun BY PHOTOSYNTHESIS Then, animals appeared Were living on plants

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SEEMS 63 MILLION YEARS AGO ALL LIVING AND NON LIVING THINGS SUDDENLY BURIED THE CAUSE MAY BE A DELUGE OR THE FALL OF AN ASTEROID UNDER HIGH PRESSURE FOR LONG TIME BECAME FOSSILS – THE ENERGY STORED IN THEM IS THE FUEL WE ARE ENJOYING NOW!

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STORY OF DEPLETING ENERGY!

Why energy conservation? We burn them, exhaust them; May be after some decades, no fossil fuel will be available. We convert atmosphere to CO2 and other pollutants May be after a few hundred years earth may become the old self and not suitable for any living being

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Energy saved and generated One kWhr electrical energy saved is equivalent to a saving of fuel for the generation of 5kWhrs!  How? 



Power from generating stations to the utilization point passes thro many equipments like transformers, transmission lines, cable feeders etc. Thermal efficiency of a turbo generator is only 30%!and other equipments efficiencies are also involved.

Hence the high figure! Energy management

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Power generation, transmission and distribution

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Typical industrial power distribution SLD

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One M.W. used a day Cnsumes 17 M.T. of coal, pollutes atmosphere by 3.4 tons of coal dust, 0.13 tons of SO2 and 0.18 Tons of oxides per day!

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Control of Atmospheric pollution Burning of fossil fuels generates  sulphuric, carbonic, and nitric acids  They fall on Earth as acid rain, affecting both

natural areas and the built environment.  Monuments and sculptures made from marble and limestone are vulnerable, as the acids dissolve calcium carbonate. A liter of petrol, diesel, kerosene used in a vehicle causes approximately 2.3 kg of CO2 emissions. Energy management

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Control of pollution Energy conserved reduces fuel consumption  Fossil fuels burnt generates green house gasses  Also causes acid rain etc.  Some of the solar radiation is reflected back by the earth and atmosphere and they escape to the space. 

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REMEMBER! 1.

GRID POWER EACH KW SAVED RESULTS IN REDUCING 6.4 TONNES OF CO2 EMISSIONS/ YEAR

2. DIESEL GENERATORS EACH KW SAVED RESULTS IN REDUCING 7.2 TONNES OF CO2 EMISSIONS/ YEAR Energy management

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Pollution – green house gasses 

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Effects of global warming Will melt polar ice caps and rise the sea levels  there will be about half to one meter increase in sea level by 2020  at the present levels of global warming  Coastal cities such as Mumbai, Kolkata and Chennai could go under sea by 2020  could make at least one billion people homeless between now and 2050  say scientists. 

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DO NOT MAKE THE ENTIRE EARTH LOOK LIKE THIS! PLEASE GIVE A GOOD EARTH TO OUR CHILDREN!

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We have a social responsibility for the future generation Leave the world, a wonderful place, as it isfor the future generation

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It's true that we don't know what we've got until we lose it!

Conserve the fast depleting conventional energy

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ENERGY CONSERVATION

Most urgent, top priority Depleting sources Spiraling cost Cannot have the luxury of unproductive usage and high demands

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ENERGY CONSERVATION OPPORTUNITIES IN  NO TWO IDENTICAL FACTORIES ARE ALIKE 

Scientific approach is needed to tackle unique problems of each industry An energy audit 

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CONCLUSION Audit helps in identifying energy conservation opportunities,  Not an one time function;  A continuous activity  Initial phase may provide plenty of opportunities; but  May taper down as the activity continues. 

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MISSING THE OBVIOUS

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PARETTO ANALYSIS

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ENERGY CONSERVATION

First let us look at:  What is power,  What is energy and  The sources of energy

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WHAT IS ELECTRICITY? AMPERES? VOLTS? WATTS? FLOW OF CURRENT WITH POTENTIAL DIFFERENCE ACROSS A RESISTANCE FLOW OF CURRENT GIVES POWER POWER FLOWING FOR A PERIOD

AMPERES VOLTS OHMS WATTS ENERGY

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Simple circuit

4 AMPS

960W

240 V

60  (Heater) 4 AMPS

4AMPS = 240VOLTS/60 OHMS VOLTAGE MAKES CURRENT FLOW THROUGH A RESISTANCE.

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POWER, ENERGY Power – rate of doing work  Energy – quantity of work done Electrical: Kilo Watt, Kilo Watt Hour Mechanical: Horse power, foot pound force (ft lbf) THERMAL: British thermal units (BTU) Joule Calorie 

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Energy equivalents: 

1 kilowatt hour =



3.6 10^6 Joules (J) or 3600000



859.85*10^3 k Calories (kcal) or 859850 cal



2.65 10^6 3412



(J)

foot pound force (ft lbf)or 2650000 British thermal units (BTU)

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ENERGY FORMS Coal Oil Gas Electricity Steam Compressed air Vacuum

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ENERGY SOURCES or FUELS Material capable of releasing energy  When chemical or physical structure changed or converted.  Releases energy either by chemical means -burning,  or by nuclear means, like nuclear fission or nuclear fusion.

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ENERGY FORMS 

Identify source or carriers: CARRIERS steam water air electricity?

pressure, heat potential,Velocity (k.e) pressure potential difference

DO NOT GET CONSUMED Energy imparted, carried and delivered.

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ENERGY FORMS Identify source or carriers:  Sources:  Inherent energy expended by irreversible chemical process - burning  Fuels OIL GAS COAL Gets consumed. 

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REDUCE – WHAT? OUTPUT? USEFUL WORK DONE

NO ! WORK DONE SAME INPUT? YES.

HOW? ENERGY INPUT = USEFUL WORK DONE + ENERGY LOST IN CONVERSION / TRANSMISSION.

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ENERGY INPUT = USEFUL WORK + LOSSES

OR USEFUL WORK + (LOSSES+WASTAGE+LOW EFFICIENCY)

TO MINIMIZE ENERGY USE: ~IDENTIFY AND MINIMIZE LOSSES. intrinsic to the system and equipments. Energy management

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LOSSES    

AVOIDABLE WASTAGE LOW EFFICIENCY UN EVEN DEMAND

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INCREASING POWER FACTOR



REDUCES DEMAND. OK, BUT, DOES IT REDUCE ENERGY LOSSES? IF YES, HOW?



AC CIRCUITS POWER NOT = VOLT * AMPS



A PHASE ANGLE EXISTS BETWEEN VOLTAGE AND CURRENT



POWER = INST VOLTAGE * INST CURRENT Energy management

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Power factor 

Components of Impedance (I) Resistance + Reactance (Vectorial sum) Reactance = Inductive reactance + Capacitive reactance (Vectorial sum) These two oppose each other I.e. 180 degrees apart

Almost all circuits, especially in industries – inductive I.e, have low lagging power factor.

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Because Load consists mainly of: 1. 2. 3. 4. 5. 6. 7.

Induction motors Static controls – thyristors etc, Power transformers and voltage regulators, Welding machines, Electric-arc and induction furnaces, Choke coils and magnetic systems, Neon signs and discharge lamps.

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Inductive loads 

Higher inductive load: Lower power factor and higher reactive current Line losses depend directly upon the square of the current immaterial of its power factor Losses proportional to Sq of current!

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Lesser current lesser losses! kW

kVA1 `

KVA2`

1 2 R KVA Capacitive reactance in RKVA



From the sketch: Inductive component of kVA1 = kW*Tan1 to be reduced to kW * Tan 2. Or to reduce 1 to 2; the demand kVA1 is reduced to kVA2 Possible by supplying a leading RKVA equal to (kW * Tan 1) – (kW* Tan 2) Or, the capacitance required in RKVA = kW * (Tan 1 – Tan 2) Energy management

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Capacitance required for power factor correction Capacitance required in kVAr = Avr. Demand * Avr P.F. * (Tan 1 – Tan 2) Or,



Cap. required in kVAr = M.D * Present P.F. * (Tan(Cos-1 Prsnt P.F) – TanCos-1 required P.F.)) Power

Factor correction by static capacitors: In most industrial cases, pay back less than 18 months. Energy management

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Selection of capacitors 

POINTS TO BE CONSIDERED: 1. 2. 3. 4. 5. 6.

Reliability of the equipment to be installed Probable life. Capital cost. Maintenance cost. Running costs. Space required and ease of installation.

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LOCATION OF CAPACITORS Nearest to inductive load or switch board: Reduce current and I2R loss INDIVIDUAL CORRECTION Better across motor terminals Preferably 7.5 kW and above Avoids providing separate control gears for capacitors Improves starting condition voltage drop reduced at start I.e. Drop across cables, transformers, buses Reduces I2 or losses

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INDIVIDUAL CORRECTION 



Caution 1. Protective equipment of feeders/ equipments should be properly set 2. Capacitor size dependent on motor magnetizing current. 3. Motor overload trip setting: OLTA = OLTA * P.F. without capacitors (With capacitors) (Without capacitors) power factor with capacitors

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INDIVIDUAL POWER FACTOR CORRECTION OF MOTORS: 

Care necessary in deciding kVAr capacitor in relation to the magnetizing kVA of the machine. If rating too high, damage to motor and capacitor. Motor, still revolving after disconnection from supply, may act as a generator by self – excitation; produce voltage higher than supply voltage. If motor switched on again before speed fallen to 80% normal speed, high voltage superimposed on supply circuits; risk of damaging other equipment connected in same circuit.

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Capacitor location for motors 

Location A Capacitor installed on incoming side of starter, on line side of O/L relay (a) Capacitor size dependent on motor magnetizing current. (b) Current to starter not reduced. (c) Motor overload trip setting same as without the capacitor.

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Capacitor location for motors 

Location B



Capacitor installed on load side of starter, line side of the O/L relay. (a) Current to the starter reduced. (b) Motor overload trip setting is the same as without capacitor. (3) Location D Capacitor installed on load side of both starter and motor O/L relay. (a) Current to starter reduced. Energy management

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Capacitance value 

Correct size capacitor in kVAr not to exceed 85% of noload magnetizing kVA of machine. If motor runs, even momentarily, with windings and capacitor forming a closed circuit, and disconnected from mains, over-excitation occur if capacitance too large. Happens when: 1. Switching off supply to motor. 2. Step – changing a star/delta or auto-transformer starter, 3. Breaker trips, or fuses blow on distribution system such that: Motors with individual capacitors, or Group of motors and line capacitor, form closed circuits.

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LOSSES IN A CAPACITOR 

Capacitor: two conductors, separated by a dielectric, energized at opposite polarity. (i) There is no prefect conductor (ii) There is no prefect dielectric All conductors have some resistance All dielectrics have some conductance

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LOSSES IN A CAPACITOR 

Current caused by conductance in capacitor draws a small power. Known as dielectric loss Quality of capacitors depends on the dielectric loss, generally known as Tan  loss. Power loss = VI Cos  or, = VI Tan  When angle between actual and the quadrature current is extremely small.

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LOSSES IN A CAPACITOR 

The conductance in the dielectric is also called as leakage resistance The current due to this will cause power flow – I.e I2R Loss This Dielectric loss = Capacitor rating in kVA *  Tan This should be kept at a minimum. The limits as per standards are: 660 V Capacitors: (i) Mixed dielectric and film capacitors 0.0025 (ii) Paper Dielectric capacitors 0.005 Above 660 V : not exceeding 0.001

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CASE STUDIES 

In a large electrochemical industry, P.F. correction capacitor 4000 kVAr Dielectric loss = Tan  = .002 Total loss = 4000 * 0.002 = 8 kW Annual energy lost = 7008 kWhr, Costing Rs. 315360/- (@Rs 4.5/ kWhr)

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CAUTION IN HANDLING CAPACITORS 

Some Capacitors may contain Polychlorinated Biphenyl (PCB) very dangerous to health May cause cancer Should be only buried for disposal

Some may contain Isopropyl biphenyl These may be disposed by incineration Always follow EPA (Environmental Protection Agency) requirements or Central/ State government regulations. Energy management

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Dielectric Losses In Power Cables 

The dielectric power factor of cables for voltage of 33 kV and above is of great importance should have very low value. The dielectric power factor is loss in dielectric (watt) volts * amp

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Dielectric Lose In Power Cables 

If cable dielectric is ‘perfect’, when voltage is applied, charging current is in leading quadrature. Should not have in phase component. But actually has small in phase component; causes dielectric loss, generating heat.

The dielectric loss in watts per kilometer per phase is: 2f*C*U02 tan  10-6 watt/km per phase For paper insulated cables the DLA depends on density of the paper and the contamination in the oil and paper.

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Properties of different type of capacitors Sl. No

Details

Mixed Dielectric

100% Polypropylene

1

Losses

< 2.5 W/ kVAr.

0.5 W/ kVAr

2

Running Costs

Higher

1/5th of MD

3

Life

10 to 15 years

Same

4

Temp Rise

More

Less

5

Reliability

Higher temp rise, lower

More reliable

6

Size

Very large

Much smaller

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Energy meter and leading power factor 

Most energy meters erratic for leading power factor

CASE STUDIES In a plant in South Madras 100 kVAr capacitor in circuit left Weekend with no load: Meter reads 150 –200 units per day Misleading; Capacitors suspected defective; Replaced; No improvement Removed capacitors tested OK Tariff meter should assure accuracy for leading power factor Unnecessarily consumer billed for energy not consumed but shown as consumption by erratic energy meter Energy management

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Capacitors and Consumer Problems 

Many plants – high connected load but power drawn very low Machines intended for different types of production Not all used at one time Low utilization factor around 0.3 to 0.45 EB insists capacitance value based on connected load and some thumb rule! Leads to low leading power factor. Penalty levied for low power factor! Field engineers to be educated with correct method for selection of capacitance, or Listen to the consumer Energy management

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CASE STUDIES 

Connected load about 100 HP Power drawn hardly 25 kW



EB insisted 50 kVAr capacitor Average power factor goes to 0. Lead Penalty levied per month for low power factor at 20% I.e Rs.12,496! While energy consumed is 13.620 kWhrs costing only Rs.54,905! Best way is to install automatic power factor correction relays and controls. Switches on only required capacitance. But quite expensive for small industries to afford.

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Diesel Generators and Power Factor. 



It is believed that average power factor for a DG to operate is 0.8. A technically erroneous conclusion. Alternators rated in Volt – Amperes (kVA). To specify maximum current an alternator can deliver. Power factor specified to specify engine rating; kW loading and current loading should not be exceeded. Hence, power factor of loads supplied can be improved closer to unity by capacitors.

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CASE STUDIES 

DG set rating: 3 phase, 415 V, 50 Hz, 500 kVA; used for 6000 hours/ Year. Average load 250 kW at 0.65 PF. Full load copper loss of the alternator = 12 kW

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What is the saving if PF is improved to 0.93? 

Energy conservation by improving power factor. Rated Current of Alternator = 695.60 A Current at 0.65 PF = 535 A Copper loss at this current = 7.1 kW Current at 0.93 PF = 374 A Copper loss at this current = 3.5 kW Saving in copper losses = 7.1 – 3.5 kW = 3.6 kW For 6000 hour operation = 3.6 * 6000 kWh or, 21,600 kWh! Energy management

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TO CONSERVE ENERGY:

1.

REDUCE LOSSES

2.

CUT DOWN WASTE OF ENERGY

3.

INCREASE EFFICIENCY OF EQUIPMENTS & SYSTEM

4.

REDUCE PEAKING DEMANDS

5.

INCREASE POWER FACTOR.

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1. Location of power factor correction capacitor banks ** to be near the load d.B, to reduce i2r loss of cables 2. Major power consuming sectors should be as close as possible to main sub station 3. Capacitor dielectric losses tan 

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REDUCE LOSSES A. Optimal selection of transformers * At least loading should be between 40% to 60% B. Selection of cable sizes

* Generous size to reduce i2r loss * Warm cable means energy loss C. Selection of piping sizes optimum reduce pressure losses

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REDUCE LOSSES D. Optimal selection of equipments to work at max. Efficiency

E. Location of compressors, boilers nearer to consumers. F. Avoid P.R.Vs, Bends & Unnecessary Circuitous Routes.

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Transformer application in transmission and distribution systems

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Step-Up Transformers common and vital electrical tools used in power transmission.  They are usually the first major transformer in a transmission system and are often used in various forms throughout the system. 

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Step-Up Transformers     

Based on the same formulas of other transformers but they step up voltages to higher levels while reducing amperage and reduces power loss which is proportional to the square of the current Step-Up Transformers ideal in long-distance power transmission use; by stepping up voltage and reducing current to reduce energy lost, which is proportional to the square of the current.

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Step-up transformer        

has more turns on the secondary coil than on the primary coil the voltage induced in the secondary coil is higher than the primary coil voltage. number of turns on the primary coil is NP and on the secondary coil is NS, and if the respective voltages are VP and VS, then NS/NP = VS/VP. Example: the primary coil 200 turns and secondary coil 2,000 turns the voltage induced in the secondary coil is ten times higher than the primary coil voltage Electrical training Consultants

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Generator step up transformers (GSU) 

In all nuclear, thermal or hydro electric power stations, generator transformers are step-up transformers with deltaconnected LV windings energized by the generator voltage, while star connected HV windings are connected to the transmission lines.

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Generator step up transformers (GSU)     

Subject to voltage changes either due to load rejection or switching operations, followed by generator over excitation, must maintain ability to withstand over-loads. High currents involved requires control of magnetic field inside the tank to avoid localized overheating of associated metallic parts. All of these situations are taken into account during the design process of the specific units and tested with state-of-the-art techniques.

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A typical generator step up transformer Type :Indoor use, gas cooled three phase on-load tap changing gas insulated transformer  Gas pressure0.5MPa (at 20 deg.C)VoltagePrimary275kV (tap range: +10% -10%,23taps)  Secondary66kV  Tertiary21kV, 90 MVA  CapacityPrimary300MVASecondary300MVA  Impedance voltage22% (at 300MVA BASE)  Noise85dB 

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Oil filled transformers Generally, transformers are filled with insulating oil, to provide insulation as the clearances in side the tank and windings are very small. also serves as a medium for cooling the windings and core Since oil provides electrical insulation between internal live parts, it must remain stable at high temperatures for an extended period. To improve cooling of transformers, the oil-filled tank have external radiators through which the oil circulates by natural convection. Electrical training Consultants

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Transformer oil properties The flash point (min) and pour point (max) are 140 °C and −6 °C respectively.  The dielectric strength of new untreated oil is 12 MV/m (RMS) and  after treatment it should be >24 MV/m (RMS).  The dielectric strength of air is:3 MV/m (RMS) 

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Transformers

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Heat removal from transformers 

When transformers are on line, considerable amount of heat is produced in the windings and cores due to: ◦ Copper loss in the windings, I2R loss



Magnetic losses: ◦ Eddy current losses in the magnetic core etc ◦ Hysterises loss in the magnetic core etc This raises the temperature of the transformer and is dissipated by various cooling methods Electrical training Consultants

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Magnetic loss due to eddy currents

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Eddy Current Losses in the Core 

 



Alternating flux induces an EMF in the core proportional to flux density and frequency resulting in circulating currents Depends inversely upon the resistivity of the material and directly upon the thickness of the core. The losses per unit mass of core material, vary with square of the flux density, frequency and thickness of the core laminations. By using a laminated core, (thin sheets of silicon steel instead of a solid core) path of the eddy current is broken up without increasing the reluctance of the magnetic circuit. A comparison of solid iron core and a laminated iron core is shown in the sketch.

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Eddy Current Losses in the Core For reducing eddy current losses, higher resistivity core material and thinner (Typical thickness of laminations is 0.35 mm) lamination of core are employed.  This loss decreases very slightly with increase in temperature.  This variation is very small and is neglected for all practical purposes.  Eddy current losses contribute to about 50% of the core losses. 

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Hysterisis losses when a magnetic field is applied all the grains of the magnetic material will orient in the direction of magnetizing force.  In next half cycle this grains will orient in opposite direction in the direction of magnetizing force.  The energy required to change the orientation of the magnetic grains in the direction of the magnetic field is lost in the form of heat. This loss is called hysterisis loss. 

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Transformer magnetic core material

CRGO Steel Laminations

Cold Rolled Grain Oriented (CRGO) silicon steels are used for laminations of the Power Transformers magnetic core. Properties:  Maximum magnetic induction to obtain high induction amplitude in an alternating field  Core loss will be independent of the load  CRGO steel sheets core loss is low; result in reduction of the constant losses.  Low apparent power input (Low hysterisis loss) results in low no load current  High grade surface insulation  Good mechanical processing properties  Low magnetostriction: results in low noise level

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Typical Losses in a 10 MVA Transformer Losses in 10 000 kVA 110kV/ 7 kV transformer are  No load loss or Magnetic losses at rated voltage :10.5 kW  Load loss or copper loss at rated current at 75oC : 55 kW 

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CLASSIFICATION OF TRANSFORMERS According to cooling method and permissible temperature rise. OIL IMMERSED TRANSFORMERS. Type Oil Circulation Cooling method ONAN Natural Air Natural ONAF Thermal Air Blast OB ONWF Head Only Water OFAN Forced by Air Natural OFAF Pump Air Blast OFB OFWF Water COMBINATION: ON/OB

ON/OFN

Symbol ON OW OFN OFW

ON/OFB Electrical training Consultants

ON/OFW. Kavoori 90

Oil filled transformers Double rated transformers and very large or high-power transformers (with capacities of thousands of KVA) may also have  cooling fans, start and stop initiated by the winding temperature indicators  oil pumps, and  even oil-to-water heat exchangers.  Cooling water pumps 

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Forced air cooled Oil Natural

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Forced air cooled Oil Natural

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Transformers

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Heat removal from transformers 

When transformers are on line, considerable amount of heat is produced in the windings and cores due to: ◦ Copper loss in the windings, I2R loss



Magnetic losses: ◦ Eddy current losses in the magnetic core etc ◦ Hysterises loss in the magnetic core etc This raises the temperature of the transformer and is dissipated by various cooling methods Electrical training Consultants

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Transformer Cooling Methods Losses in the transformer around 0.5 to 1% of its full load kW rating, converted in to heat;  temperature of the windings, core, oil and the tank rises.  This heat dissipated from the transformer tank and the radiator in to the atmosphere.  cooling arrangements helps in maintaining the temperature rise of various parts within permissible limits.  Cooling provided by the circulation of the oil. 

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Typical losses of transformers Rated

Power (kVA)

No-load loss(kW)

On-load loss(kW)

6300

9.3

45

8000

11.2

54

10000

13.2

63

12500

15.6

74

16000

18.8

90

20000

22.2

106

25000

26.2

126

31500

31.2

149

40000

37.3

179

50000

44.1

213

63000

52.5

255

75000

59.8

291

68.8

333

90000

Voltage Combination (kV)

60~150

Electrical training Consultants

Kavoori 97

Losses comparison : Dry type or liquid filled

Energy management

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98

Comparison of Losses: Oil type and dry type (Oil Transformer) Losses

Dry Type Transformer Losses

KVA

Full Load (W)

KVA

Full Load (W)

500

4930

500

10000

750

7900

750

15000

1000

8720

1000

16400

1500

13880

1500

22500

2000

16310

2000

26400

Energy management

Kavoori Consultants

99

Liquid, resin caste and dry type Transformers loss comparison Liquid:

Cast:

Dry:

Load Losses (kW)

16.38

21.00

18.52

No Load Losses (kW)

2.66

7.00

7.55

Total Losses (kW)

19.04

26.07

28.00

Energy management

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100

Transformer Oil Forms a very significant part of the transformer insulation system:  Has the important functions of acting as an electrical insulation as well as  A coolant to dissipate heat losses.  For small rating transformers heat removed from the transformer by natural thermal convection. 

Electrical training Consultants

Kavoori 101

Transformer Cooling Methods     

For large rating transformers this is not sufficient; As size and rating increases, losses increase at a faster rate. oil is circulated by means of oil pumps. Within the tank the oil is made to flow through the space between the coils of the windings. Several different combination of natural, forced, air, oil cooling methods are employed choice of transformer cooling method depends on the rating, size, and location.

Electrical training Consultants

Kavoori 102

Directed oil flow thro windings

Electrical training Consultants

Kavoori 103

Power transformer: Name plate details Make : Hack bridge – Hewittic and Easun Ltd. Rated voltage : 110 kV/ 7 kV. Rated current : 52.55 A/ 825.76 A Rated kVA : 10 000 kVA Connection: Primary Delta; Secondary Star No load loss at rated voltage :10500W Load loss at rated current at 75oC : 55000 W Imp voltage at rated current at 750C: 8.35%

Electrical training Consultants

Kavoori 104

Allowable temperature Rise Component

Cooling

Winding (Measured by Resistance)

Temp Rise Ambient C ON,OB,OW 55 Max 45 OFN, OFB 60 Daily Average 30

OFW Oil All (Measured by Thermometer)

65 (Yearly average 30) 45

Electrical training Consultants

Kavoori 105

COOLING MEDIUM -LETTER SYMBOLS Cooling Medium Mineral Oil Synthetic insulation liquid Gas Water Air Solid Insulant Natural Forced

Symbol O L B W A S N F Electrical training Consultants

Kavoori 106

Gas Insulated Power Transformers 

   

Use SF6 Gas as the insulating and cooling medium instead of insulating oil. First units produced in 1967. Several thousand units now in service worldwide. Transformer applications: GSU, Distribution class units up to 400 MVA, 345 kV. Primarily used in substations located in urban areas (including inside buildings, underground) due to safety benefits.

Electrical training Consultants

Kavoori 107

Gas insulated transformers Space is becoming an important consideration. This has resulted in:  large-scale substations to be tucked away underground in overpopulated urban areas  incombustible and non-explosive , large-capacity gas insulated transformers for accident prevention and compactness of equipment.  In line with this requirement, several types of largecapacity gas insulated transformer have been developed. 

Electrical training Consultants

Kavoori 108

Gas insulated transformers The gas-forced cooling type was available for up to approximately 60MVA,  gas insulated transformer with higher ratings are liquid cooled.  Disadvantage: complex structure for liquid cooling.  certain manufactures began development of gas forced cooling type transformer,  TOSHIBA has delivered 275kV-300MVA gas cooled and gas insulated transformer,  its structure is as simple as the oil immersed type and is the largest capacity gas insulated transformer in the world. 

Electrical training Consultants

Kavoori 109

Gas insulated transformers Since heat capacity of SF6 gas is much smaller than that of insulating oil, the following measures are taken into account.  1. Raise the SF6 gas pressure to 0.5MPa  2. Produce as large flow as possible by optimizing the layout of gas ducts in the windings  3. Develop high capacity gas blower with high reliability  4. Apply highly thermal-resistant insulating materials to raise the limit of winding temperature rise

Electrical training Consultants

Kavoori 110

Sulfur Hexa Fluorine Gas (SF6) Physical properties  About five times heavier than air, density 6.14kg /m3.  Colorless, odorless and non-toxic.  Speed of sound propagation about three times less than in air, at atmospheric pressure. Hence interruption of arc less loud in SF6 than in air.  Dielectric strength on average 2.5 times that of air,  Increasing pressure, increases the dielectric strength  Around 3.5 bar, SF6 has the same strength as transformer oil.  Becomes liquid at - 63.2°C and in which noise propagates badly.

Electrical training Consultants

Kavoori 111

Gas insulated transformer

Electrical training Consultants

Kavoori 112

Gas insulated substation    



Gas insulated transformer does not need conservator, Height of transformer room reduced. It has non-flammability and non tank-explosion characteristics No need for fire fighting equipment in transformer room. So gas insulated transformer, gas insulated shunt reactor and GIS control panels installed in the same room. The substation is a fully SF6 gas insulated substation

Electrical training Consultants

Kavoori 113

Natural-cooled type SF6 gas-insulated transformer

Electrical training Consultants

Kavoori 114

Forced-gas-circulated, natural-air-cooled type SF6 gas-insulated transformer

Electrical training Consultants

Kavoori 115

Forced-gas-circulated, forced-air-cooled type SF6 gas-insulated transformer

Electrical training Consultants

Kavoori 116

Energy conservation and transformers The transformer efficiency is maximum when loaded at 45-50% of its rated capacity  Selection of transformers for an industry  Select two transformers of each rating for the full load of the plant.  In normal times, run them in paralleleach will be loaded to its 50% capacity, ie. At its maximum efficiency area. 

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TRANSFORMER LOSSES Constant loss or no load loss- does not depend upon load condition : about 1kW per 500 kVA Copper losses - proportional to load condition During lean periods, one transformer can be cut out of service - saves about 24 units per day i.e. Rs. 48/- per day per 500 kVA capacity Diagram - transformer losses

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118

TRANSFORMER LOSSES 

The higher the transformer capacity, the higher the constant losses

The idle loss of a 5000 kVA transformer is 10 kW! By prudent switching of transformers, this loss can be reduced.

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119

TRANSFORMER LOSSES Constant loss or no load loss- does not depend upon load condition : about 1kW per 500 kVA Copper losses - proportional to load condition During lean periods, one transformer can be cut out of service - saves about 24 units per day i.e. Rs. 48/- per day per 500 kVA capacity Diagram - transformer losses

Energy management

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120

TRANSFORMER LOSSES Transformer Load Losses- Model calcuations. KAVOORI CONSULTANTS, CHENNAI. Energy audit M/s. *************************** Ltd Table No. Transformer Load Losses, at the present loading condition: Transformers with Off Load Tap Changer Make Bharat Bijlee k.V.A. H.V. L.V. Imp % ge Units Rating 2000 11000 433 6.25 No load loss 3.3 kWs Full lload loss at temperature, oC 75 19.8 kWs Full lload loss at Operating temperature, oC 31.9 17.05 Full load current, L.T. 2669.9 Amps Full load current, H.T. 175.16 Amps Cost of electrical energy 5.95 Rs. No of transformers in Parallel 2 Single transformer in service Two transformers in service Load Losses, in kW Losses, in kW %ge Load No Load Load Total No Load Load Total At an operating temperatur of 31 oC 10.00% 3.3 0.17 3.47 6.6 0.04 6.64 20.00% 3.3 0.68 3.98 6.6 0.17 6.77 30.00% 3.3 1.53 4.83 6.6 0.38 6.98 40.00% 3.3 2.73 6.03 6.6 0.68 7.28 50.00% 3.3 4.26 7.56 6.6 1.07 7.67 60.00% 3.3 6.14 9.44 6.6 1.53 8.13 70.00% 3.3 8.35 11.65 6.6 2.09 8.69 80.00% 3.3 10.91 14.21 6.6 2.73 9.33 90.00% 3.3 13.81 17.11 6.6 3.45 10.05 100.00% 3.3 17.05 20.35 6.6 4.26 10.86 o

C

At an operating temperature of 10.00% 20.00% 30.00% 40.00% 50.00% 60.00% 70.00% 80.00% 90.00% 100.00%

3.3 3.3 3.3 3.3 3.3 3.3 3.3 3.3 3.3 3.3

0.19 0.77 1.72 3.07 4.79 6.90 9.39 12.26 15.52 19.16

3.49 4.07 5.02 6.37 8.09 10.20 12.69 15.56 18.82 22.46

65.00

19.16

6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6

0.05 0.19 0.43 0.77 1.20 1.72 2.35 3.07 3.88 4.79

Energy management

6.65 6.79 7.03 7.37 7.80 8.32 8.95 9.67 10.48 11.39

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121

Transformers efficiency v.s. load

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122

TRANSFORMER LOSSES 2000 kVA, 6600 /433 Volts Transformer. Total Losses Single, Two in parallel operation. o (Operating temperature C) 55

35.00 30.00

Total losses in kW (Load + No Load)

25.00 20.00

Single Transformer

15.00

Two transformers parallel

10.00 5.00 0.00 1

2

3

4

5

6

7

8

9

10

11

12

Transformer load in fraction of full load.

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123

TRANSFORMER LOSSES Constant loss or no load loss- does not depend upon load condition : about 1kW per 500 kVA Copper losses - proportional to load condition During lean periods, one transformer can be cut out of service - saves about 24 units per day i.e. Rs. 48/- per day per 500 kVA capacity Diagram - transformer losses

Energy management

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124

Transformer efficiency VS. Load 100.00

99.50

Percentage efficiency

99.00

98.50 2000 kVA transformer

98.00 2500 kVA Transformer

23500 kVA Transformer

97.50

500 kVA Transformer

97.00

96.50 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

Load, fraction of the rating

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TYPICAL EDDY CURRENT LOSS FACTORS FOR OIL-FILLED TRANSFORMERS Transformer size oil filled transformer Up to 1 MVA

Eddy current loss factor 1%

1 MVA TO 5 MVA Greater than 5 MVA

1 to 5 % 9 to 15%

126

SELECTION OF CABLE SIZE 

CONSIDER, SAY, A BULK LOAD OF 207 HP + 5 kW CONNECTED BY 1000 M OF CABLE FROM THE SUB STATION. Cable selected was 3-1/2 * 240 sq.Mm aluminum conductor p.V.C insulated armored cables I2r loss in the cable = 7626 w

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SELECTION OF CABLE SIZE 

If a 3-1/2 * 300 sq.Mm cable is used, the loss will be only 6100 w Difference in loss of power = 1525 w Difference in loss of energy in one year = 13,360 units cost saved @ Rs. 3 = Rs. 40,000/- unit

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Power distribution systems 

Power Factor Improvement Capacitors – Location Assume a sectional load of 155 kW located at about 1000m from the main substation and connected by an aluminum cable of size d ½* 240 sq.mm cable.

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129

Power Factor Improvement Capacitors – Location 

DC resistance of the cable 0.125/km load of the remove section 155 kW power factor of the load = 0.8

Consider the power factor capacitor at this main substation bus.  Current drawn by the load at 0.8 pf = 240a Power loss in the cable = 2702 * 0.125 = 9.082 kW

Energy management

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130

Power Factor Improvement Capacitors – Location 

If the power factor correction capacitor is connected at the load section distribution board: For a corrected power factor (of say 0.97) The current drawn will be 222.3 a power loss in the cable for this current = 222.3 * 0.125 = 6.177 kW

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Power Factor Improvement Capacitors – Location Saving in power loss= 9.082 - 6.177= 2.905 kW or 3 kW

Saving in one year of operation = 3 * 24 * 365 = 26,680 kW Energy cost saved per year = 26,280 * 3 = Rs. 78,840/to minimize the power loss and save energy and its cost, always locate capacitors at the section using maximum power, as close as possible to the respective substation panel.

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Load location - Cable losses 

Suppose the sub-station is close by and only 100 m cable is used. Loss in cable = 610 w Energy saving in one year = 48,092 kWhr

Cost of energy saved @ Rs. 1.4/kWhr = Rs. 1,44,000/To minimize power loss and save energy and its cost always locate the section using maximum power as close as possible to the main substation.

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133

ENERGY UTILIZATION EFFICIENCY IN HARMONIC ENVIRONMENT

SHRI. K.R. GOVINDAN, KAVOORI CONSULTANTS, New No: 22, JANAKIRAM STREET, WEST MAMBALAM, CHENNAI – 600 033. PH:24846139. 134

POWER UTILIZATION 

 





All alternating current equipments and power distribution systems and elements Designed to work from a power source with voltages of 50 HZ frequency and a sinusoidal waveform Their behavior, energy utilization efficiency and other characteristics are much affected when supplied with distorted wave forms. Incandescent lamps, heaters, etc draw current proportional to the voltage following sinusoidal waveform Hence these loads are called linear loads 135

POWER DRAWN BY A LINEAR RESISTIVE LOAD      

Both current and voltage rise and fall together Hence current is in phase with the voltage The power drawn at any instant is I X V during a negative half cycle, voltage and current are negative Since the power is the product of voltage and current it becomes positive Hence a positive power is drawn thorough out the cycle 136

POWER UTILIZATION 

 





All alternating current equipments and power distribution systems and elements Designed to work from a power source with voltages of 50 HZ frequency and a sinusoidal waveform Their behavior, energy utilization efficiency and other characteristics are much affected when supplied with distorted wave forms. Incandescent lamps, heaters, etc draw current proportional to the voltage following sinusoidal waveform Hence these loads are called linear loads 137

Unity power factor Voltage, current and power wave forms

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POWER DRAWN BY

A LINEAR INDUCTIVE LOAD Induction motors also draw current proportional to voltage  Since current drawn is inductive, lags the voltagebut still follow sinusoidal waveform  Same hold good for capacitors, but current leads the voltage  In phase or out of phase, the current drawn is proportional to the voltage  Hence these are also linear loads 

139

POWER UTILIZATION INDUCTIVE LOAD







Though current follows voltage waveform, the peak and the zero value of the current is displaced by an angle from the peak and zero point of voltage waveform With respect to the instantaneous voltage value, the current value becomes a function of the Cos of the angle (between voltage and current) Hence power at any instant is equal to Voltage X Current X Cos of the angle between them

140

LINEAR INDUCTIVE LOAD

141

POWER UTILIZATION INDUCTIVE LOADS 



 

If the load is totally inductive like a reactor or a induction coil the current drawn lags the voltage by 90 degrees Since power is the product of instantaneous voltage and current, its frequency is double of the voltage frequency It also passes through the negative half of the cycle Since the negative half and the positive half of the waveform are identical I.e. positive power and negative power, total power drawn by the load is zero 142

POWER UTILIZATION INDUCTIVE LOADS 

No net power flows

143

POWER CONTROL In the past, a resistance or an auto transformer was employed to regulate power  It controls the peak value of the voltage applied  But still the voltage follows a sinusoidal waveform but with lesser amplitude  Since power is a product of voltage and current, the power follows sinusoidal waveform  With reduction in peak value the power drawn is also reduced  But, involves wastage of power in the controlling element 

144

LINEAR POWER CONTROL

1

2

1. Line voltage, 2. Controlled voltage 145

SOLID STATE POWER CONTROL      

To eliminate the losses in the controlling elements. Solid state or thyristor controls employed. These follow different technique to control power Chops off a portion of the wave so that the volume of power to the load is reduced Now the current is not following the voltage waveform; it is like interrupted impulses of current This is a non sinusoidal distorted waveform

146

SOLID STATE CONTROL OF POWER DISTORTS WAVEFORM

147

HARMONICS AND ENERGY LOSS    





Harmonic currents are just circulating in the network They do not contribute to the power delivered But causes I2R losses In addition the magnetic effect of harmonics creates other problems which also results in considerable losses Alternating current passing though a conductor sets up alternating magnetic field. Create varying magnetic field around the conductor

148

HARMONICS AND ENERGY LOSS SKIN EFFECT Center of the conductor enveloped by more varying magnetic flux than on the outside.  They push the current to the periphery of the conductor as the center is subjected to higher intensity of magnetic field  This concentration at surface is “the skin Effect” Increases conductor effective resistance  This is more pronounced if the conductors are associated with magnetic material as the flux density is much higher 

149

HARMONICS AND ENERGY LOSS CONDUCTORS, CABLES ETC. SKIN EFFECT These effects are proportional to the frequency of the alternating current  Hence very high for higher frequency harmonic currents  Since effective area of cross section is reduced, higher resistance offered to the current flow  Very high I2R losses are involved  For closely placed conductors another factor comes in to play –I.e.“Proximity Effects” 

150

HARMONICS AND ENERGY LOSS CONDUCTORS, CABLES ETC. PROXIMITY EFFECT

Conductor halves in close proximity cut by more Flux than the remote halves. Current distribution not even throughout the Cross-section, Greater portion carried by remote halves. When currents are in opposite directions,halves in closer proximity carry more current. Overall effect- increase in effective resistance.

151

EFFECTIVE AREA OF CONDUCTORS FOR HARMONIC CURRENTS Cross – sectional area of a round conductor available for conducting DC current “DC resistance”

Cross sectional area of the same conductor available for conducting normal-frequency AC “AC resistance”

Cross sectional area of the same conductor available for conducting high-frequency AC “AC resistance”

152

HARMONICS AND ENERGY LOSS CONDUCTORS, CABLES ETC.

Proximity effect decreases with increase In spacing between cables. 

At certain harmonics the combined effect results in twice the I2R loss A.C/D.C resistance ratio

Frequency

Harmonic of 50 Hz

1.01

50

1

1.21

250

5

1.35

350

7

1.65

550

11 153

HARMONICS AND INDUCTION MOTOR When the power supplied to the stator of the motor contains harmonics,  The stator winding affected by skin effect  The rotor is severely affected, as the conductors are subjected to magnetic field of varying frequencies.  1.5 Hz to 300 Hz.  In the motor the rotating magnetic field developed by the fundamental frequency voltage only develop necessary torque – delivers shaft power

154

HARMONICS AND INDUCTION MOTOR  

 

With motor designed for 3% slip, the rotor currents have a frequency of 1.5Hz; The rotor is designed to have the reactance and DC resistance nearly equal at this frequency to get optimum efficiency. But, different types of Rotating Magnetic fields are setup by individual harmonic currents While fields created by forward magnetic fields subtract on the rotor field, negative ones added up to the rotor field

155

HARMONICS AND INDUCTION MOTOR 5th harmonic creates 250 Hz frequency while 11th and 13th pair together to induce 500 Hz in the rotor  These high frequency harmonics snow balls the skin effect and the rotor I2R loss becomes very high  The rotor have currents at 6,12,18,12 etc times the stator frequency  High frequency means higher eddy current and hysterisis loss  The negative torques will affect the shaft horse power; some times create very bad vibration  At certain level the efficiency drops down about 10% 

156

HARMONICS AND INDUCTION MOTOR     

Harmonic fields rotating relative to each other produce torque pulsations Needs re-examination of torsional characteristics of entire shaft system Leakage flux set up in stator and rotor end windings added to the losses With skewed rotor bars, high frequency produce substantial iron loss; Depends upon amount of skew and iron loss characteristics

157

HARMONICS AND INDUCTION MOTOR

Case Study:  Test on a 15 kW motor at full out put  With 50 Hz fundamental sinusoidal voltage loss at full load = 1303 Watts  With Quasi-square wave voltage 1600 Watts  Losses up by 23%

158

HARMONICS AND TRANSFORMER

Transformers essentially comprises of current carrying conductors encircled by iron core  Hence harmonics effects results in:  Higher eddy current and hysterisis losses  Skin effects due to harmonic current  High copper losses  This effect more important for converter transformers  Filters do not neutralize harmonic current in these transformers; due to higher losses develop unexpected hot spot in tanks 

159

NO LOAD CURRENT OF A STAR/STAR TRANSFORMER HARMONIC RESOLUTION

Harmonic analysis of peaked no load current wave of i0 = 100 sin  + 31.5 sin 5+… 160

HARMONICS AND TRANSFORMER Third harmonics-Important for power transformers; circulation of triplen zero sequence current in delta windings  These extra currents over heat the windings  The RMS value of pure sine wave is 0.707 of peak value  340 V peak value has an RMS voltage of 240  But this ratio is not true for a distorted waveform  RMS value is the measure of the heat generated by an equivalent DC current  Hence, heat produced by harmonics are much higher 

161

THIRD HARMONICS IN PHASE WITH FUNDAMENTAL

162

THIRD HARMONICS OUT OF PHASE WITH FUNDAMENTAL

163

Third harmonics phase relation ship

164

HARMONICS AND POWER FACTOR 





 

Since harmonic currents are neither in phase nor follow supplied voltage they do not deliver any power In a pure sinusoidal waveform the displacement angle between the current and the voltage decides the power factor, known as displacement power factor or apparent power factor This does not hold good in case of harmonic currents as they do not have any such angular relation Hence power factor is kW/Volts X Ampere Actually this is the true power factor in a circuit which has harmonic currents 165

HARMONICS AND TRANSFORMER The losses in a transformer are a combination of 1. Excitation (No load loss) I.e. Eddy current, hysterisis, stray losses 2. Load losses mainly due to I2R loss in the conductor 3. Both the losses increase as the square of the frequency but does not contribute to the power transfer 4. Heats transformer; increases the temperature resulting in premature failure apart from wasting energy!

166

DERATING FACTOR FOR TRANSFORMERS

167

HARMONICS AND INSULATED CABLES A cable is essentially a conductor surrounded by an insulation These two components create losses; The conductor develops I2R loss due to the current flow If the current passing through contains higher harmonics this loss is increased due to the increased skin and proximity effects as shown earlier

168

HARMONICS AND INSULATED CABLES

   

The insulation is subjected to dielectric loss This loss is = 2 f C U02 tan  10 -6 (watt/km per phase) For a specified design, C and U02 are constant; therefore, loss is proportional to the frequency

Higher the harmonics higher the losses

169

BALANCED LOAD Neutral Current

5A

5A

0A

N

5A

5A

5A

R Y B

170

BALANCED LOAD WITH THIRD HARMONICS Neutral Current

5A

10A

15A

N

5A

5A

5A

R Y B

171

THIRD HARMONICS AND NEUTRAL CURRENTS

172

ELECTRICAL FAILURE MECHANISM • •

All protective systems are based on Current2 & Time Rarely – Mechanical Damage.

Resistance

Current 2

Power Loss

Energy Loss

Heat

Time Insulation Failure

Temperature





 



ELECTRICAL FAILURE Power loss is proportional to the square of the current; Immaterial, whether the current is in phase with voltage or of fundamental frequency Harmonic currents are no exception to this; They do not deliver power, but circulate in the system, contributing to energy loss. result: higher temperature

ELECTRICAL FAILURE



Most of the protective schemes are based on this, I.e. I2t, resistance being almost constant. But added disadvantage with harmonics is They increase the resistance also, by skin and proximity effects.



Hastens failure, reduce useful life



CAPTIVE POWER GENERATORS AND HARMONICS Generators for large lighting installations:  discharge lamps with inductive chokes etc generate 30% 3rd harmonics  If generated voltage contains 3% harmonics, with harmonic loads, waveform may worsen  Even in a well balanced three phase lighting system 20% 3rd harmonic may exist in each phase.  3rd harmonics are of additive nature; in the neutral it will be 60%..  will heat up the machines and neutral conductors  but the fundamental current may be zero

CAPTIVE POWER GENERATORS AND HARMONICS 











Eg: A carefully balanced 250 kVA fluorescent lighting load in a warehouse. fed from the public utility - 13% of full load line current observed in the neutral fed from 320 kVA stand by generator - current increases to 250 Ampere (72% full load current) due to high third harmonic content in the generator output waveform The solution was to replace 11/12 pitch of the stator winding by 2/3 pitch The neutral current was below than the value when supplied from the utilities

CAPTIVE POWER GENERATORS AND HARMONICS Sizing generators for non linear loads:  Simple rules of the thumb is to oversize the standard generators for the load to be catered  Some allow 50% non linear loads  But, manufacturers should be given full information of non linear loads while ordering  The crux of the problem - one of the generating impedance  Current harmonics of non linear loads are constant – do not depend upon the power supply

CAPTIVE POWER GENERATORS AND HARMONICS 







 

But voltage distortion is a direct function of generating impedance The stator pitch configuration have varying reactance for each harmonics Hence evaluating the voltage distortion for all harmonics individually is necessary These distorted voltages affect the performance of AVRs affecting their stability PMG excitation system has improved this situation The power to the AVR is constant irrespective of generated output

CAPTIVE POWER GENERATORS AND HARMONICS 







Designing generators with specific winding pitches and low reactance is not quite commercially viable Hence practical solution is to derate the standard industrial generators Some reputed manufacturers select a 0.12 p.u. subtransient reactance as a good practical solution The basics; 6 pulse VFD motor drive with 26% current distortion

CAPTIVE POWER GENERATORS AND HARMONICS POWER FACTOR:  Conventional power factor is Watts/Volt amp is = Cosine of the angle between current and voltage  This is really a displacement power factor  But with harmonic currents, power factor as Cos  does not hold good  Because there are many harmonic currents flowing in the circuit  If the total RMS value of the current is taken into consideration, the power factor value may become worse

CAPTIVE POWER GENERATORS AND HARMONICS 









The power drawn is a function of the fundamental current only Harmonic current increase the total RMS current without increasing the power Discrepancy arise between ammeter reading and voltmeter reading Standard power factor meter measures displacement power factor only They may show a unity power factor while infact the real power factor may be as low as 0.70

ELECTRICAL FAILURE MECHANISM • •

All protective systems are based on Current2 & Time Rarely – Mechanical Damage.

Resistance

Current 2

Power Loss

Energy Loss

Heat

Time Insulation Failure

Temperature





 



ELECTRICAL FAILURE Power loss is proportional to the square of the current; Immaterial, whether the current is in phase with voltage or of fundamental frequency Harmonic currents are no exception to this; They do not deliver power, but circulate in the system, contributing to energy loss. result: higher temperature

ELECTRICAL FAILURE



Most of the protective schemes are based on this, I.e. I2t, resistance being almost constant. But added disadvantage with harmonics is They increase the resistance also, by skin and proximity effects.



Hastens failure, reduce useful life

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CAPTIVE POWER GENERATORS AND HARMONICS Generators for large lighting installations:  discharge lamps with inductive chokes etc generate 30% 3rd harmonics  If generated voltage contains 3% harmonics, with harmonic loads, waveform may worsen  Even in a well balanced three phase lighting system 20% 3rd harmonic may exist in each phase.  3rd harmonics are of additive nature; in the neutral it will be 60%..  will heat up the machines and neutral conductors  but the fundamental current may be zero

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Eg: A carefully balanced 250 kVA fluorescent lighting load in a warehouse. fed from the public utility - 13% of full load line current observed in the neutral fed from 320 kVA stand by generator - current increases to 250 Ampere (72% full load current) due to high third harmonic content in the generator output waveform The solution was to replace 11/12 pitch of the stator winding by 2/3 pitch The neutral current was below than the value when supplied from the utilities

CAPTIVE POWER GENERATORS AND HARMONICS Sizing generators for non linear loads:  Simple rules of the thumb is to oversize the standard generators for the load to be catered  Some allow 50% non linear loads  But, manufacturers should be given full information of non linear loads while ordering  The crux of the problem - one of the generating impedance  Current harmonics of non linear loads are constant – do not depend upon the power supply

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But voltage distortion is a direct function of generating impedance The stator pitch configuration have varying reactance for each harmonics Hence evaluating the voltage distortion for all harmonics individually is necessary These distorted voltages affect the performance of AVRs affecting their stability PMG excitation system has improved this situation The power to the AVR is constant irrespective of generated output

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Designing generators with specific winding pitches and low reactance is not quite commercially viable Hence practical solution is to derate the standard industrial generators Some reputed manufacturers select a 0.12 p.u. subtransient reactance as a good practical solution The basics; 6 pulse VFD motor drive with 26% current distortion

CAPTIVE POWER GENERATORS AND HARMONICS POWER FACTOR:  Conventional power factor is Watts/Volt amp is = Cosine of the angle between current and voltage  This is really a displacement power factor  But with harmonic currents, power factor as Cos  does not hold good  Because there are many harmonic currents flowing in the circuit  If the total RMS value of the current is taken into consideration, the power factor value may become worse

Lamp Characteristics: efficacy, life and colour rendering index. Lamp type

Previous coding

ILCOS coding

Lamp efficacy (lumens/ Watt)

Quoted lamp life (hours)

Colour rendering Index compared to Inc lamp

Tungsten filament

GLS

I

10 to 18

1000 to 2000

100

Tungsten halogen

TH

HS

15 to 25

2000 to 4000

100

High pressure mercury

MBF

QE

30 to 60

14000 to 25000

47

Low pressure mercury (fluorescent)

MCF

FD (tubular) FS (compact)

65 to 95 65 to 95

6000 to 15000 8000 to 10 000

11

Metal halide

MBI

M

65 to 85

6000 to 13000

Low pressure sodium

SOX

LS

70 to 150

11000 to 22000

High pressure sodium

SON

S

55 to 120

12000 to 26000

XF

70 to 80

60000

Induction

Energy management

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Kavoori Consultants

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The power drawn is a function of the fundamental current only Harmonic current increase the total RMS current without increasing the power Discrepancy arise between ammeter reading and voltmeter reading Standard power factor meter measures displacement power factor only They may show a unity power factor while infact the real power factor may be as low as 0.70

CONCLUSION Harmonics are created in a power system by the consumer and also by the supplier  But major portion by consumer  Harmonics creates lot of problems, destroys equipments  All energy efficient equipments essentially creates harmonics;  These result in added energy losses  Hence harmonics are to be limited  While selecting energy efficient equipments these points are to be given greater attention 

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