E5

MAPÚA UNIVERSITY Muralla St. Intramuros, Manila School of Mechanical and Manufacturing Engineering

EXPERIMENT NO. 5 CENTRIFUGAL FAN

12 MAHMUD, Ali R. 2015151413 ME144L – A1 Group No. 2

Date of Performance: Sep , 2018 Date of Submission: Sep 28, 2018

GRADE Engr. Teodulo A. Valle Instructor

TABLE OF CONTENTS

Objectives Theory and Principle List Of Apparatus Procedure Set-up of Apparatus Final Data Sheet Sample Computations Discussion of Result Questions and Answers Conclusion Reference Preliminary Data Sheet

Page 1 Page 1 Page 4 Page 5 Page 8 Page 9 Page 10 Page 12 Page 13 Page 15 Page 16 Page 17

OBJECTIVES 1. To be familiar with the operating procedure and principle 2. To determine the fan efficiency when subjected to varying discharge orifice THEORY AND PRINCIPLE Fans and blowers provide air for ventilation and industrial process requirements. Fans generate a pressure to move air (or gases) against a resistance caused by ducts, dampers, or other components in a fan system. The fan rotor receives energy from a rotating shaft and transmits it to the air. Fans, blowers and compressors are differentiated by the method used to move the air, and by the system pressure they must operate against. As per American Society of Mechanical Engineers (ASME) the specific ratio, the ratio of the discharge pressure over the suction pressure, is used for defining the fans, blowers and compressors.

Fig.1. Differences between fans, blowers, and compressors Fan and blower selection depends on the volume flow rate, pressure, type of material handled, space limitations, and efficiency. Fan efficiencies differ from design to design and also by types. Typical ranges of fan efficiencies are given in table below. Fans fall into two general categories: centrifugal flow and axial flow.

Fig.2. Fan Efficiencies

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In centrifugal flow, airflow changes direction twice - once when entering and second when leaving (forward curved, backward curved or inclined, radial) while in axial flow, air enters and leaves the fan with no change in direction (propeller, tube axial, vane axial).

Fig.3. Centrifugal Fan (left) and Axial Fan (right) The major types of centrifugal fan are: radial, forward curved and backward curved. Radial fans are industrial workhorses because of their high static pressures (up to 1400 mm WC) and ability to handle heavily contaminated airstreams. Because of their simple design, radial fans are well suited for high temperatures and medium blade tip speeds. Forward-curved fans are used in clean environments and operate at lower temperatures. They are well suited for low tip speed and high-airflow work - they are best suited for moving large volumes of air against relatively low pressures. Backward-inclined fans are more efficient than forward-curved fans. Backwardinclined fans reach their peak power consumption and then power demand drops off well within their useable airflow range. Backward-inclined fans are known as "non-overloading" because changes in static pressure do not overload the motor.

Fig.4. Types of Centrifugal Fan The major types of axial flow fans are: tube axial, vane axial and propeller. Tube axial fans have a wheel inside a cylindrical housing, with close clearance between blade and housing to improve airflow efficiency. The wheel turn faster than propeller fans, enabling operation under high-pressures 250 – 400 mm WC. The efficiency is up to 65%. Vane axial fans are similar to tube axial fans, but with addition of guide vanes that improve efficiency by directing and straightening 2

the flow. As a result, they have a higher static pressure with less dependence on the duct static pressure. Such fans are used generally for pressures up to 500 mm WC. Vane axial fans are typically the most energy-efficient fans available and should be used whenever possible. Propeller fans usually run at low speeds and moderate temperatures. They experience a large change in airflow with small changes in static pressure. They handle large volumes of air at low pressure or free delivery. Propeller fans are often used indoors as exhaust fans. Outdoor applications include air-cooled condensers and cooling towers. Efficiency is low – approximately 50% or less.

Fig.5. Types of Radial Fan In the experiment, the researcher focused on the performance of the centrifugal fan. Below are the formulas needed to compute for the unknown data. Solving for the density of air: 𝑃𝑉 = 𝑚𝑅𝑇 →

𝑚 𝑃 =𝜌= 𝑉 𝑅𝑇

Equation 1

Solving for the velocity head of air: 𝜌𝑎𝑖𝑟 𝑔ℎ𝑎𝑖𝑟 = 𝜌𝑤𝑎𝑡𝑒𝑟 𝑔ℎ𝑤𝑎𝑡𝑒𝑟 → ℎ𝑎𝑖𝑟 =

3

𝜌𝑤𝑎𝑡𝑒𝑟 ℎ𝑤𝑎𝑡𝑒𝑟 𝜌𝑎𝑖𝑟

Equation 2

Solving for the velocity of air: ℎ𝑎𝑖𝑟

𝑣𝑎𝑖𝑟 2 = → 𝑣𝑎𝑖𝑟 = √2𝑔ℎ𝑎𝑖𝑟 2𝑔

Equation 3

Solving for the volume flow rate of air: 𝑄 = 𝐴𝑜𝑟𝑖𝑓𝑖𝑐𝑒 𝑣𝑎𝑖𝑟

Equation 4

ℎ𝑇𝑤𝑎𝑡𝑒𝑟 = ℎ𝑠𝑡𝑎𝑡𝑖𝑐 + ℎ𝑣

Equation 5

𝜌𝑤𝑎𝑡𝑒𝑟 ℎ𝑇𝑤𝑎𝑡𝑒𝑟 𝜌𝑎𝑖𝑟

Equation 6

Solving for the total head of water:

Solving for the total head of air: ℎ𝑇𝑎𝑖𝑟 = Solving for the air power/ output power: 𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝛾𝑎𝑖𝑟 𝑄ℎ𝑇𝑎𝑖𝑟 = 𝜌𝑎𝑖𝑟 𝑔𝑄ℎ𝑇𝑎𝑖𝑟

Equation 7

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝐸𝐼𝑐𝑜𝑠(∅)

Equation 8

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 × 100% 𝑛𝑡𝑟𝑎𝑛𝑠 𝑛𝑚𝑜𝑡𝑜𝑟 𝑃𝑖𝑛𝑝𝑢𝑡

Equation 9

Solving for the input power:

Solving for the fan efficiency: 𝑛𝑓𝑎𝑛 =

4

LIST OF APPARATUS 1. Centrifugal fan

2. Electric motor 3. H2O Manometer

4. Pitot Tube

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5. Tachometer

6. Amprobe

7. Set of Orifice

6

8. Thermometer

9. Stop watch

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PROCEDURES 1. Prepare the apparatus needed for the experiment. 2. Insert the orifice with diameter 2.5 inches into the discharge casing of the centrifugal fan.

3. Start the motor so that the centrifugal fan will now able to intake air from the atmosphere. 4. Obtain the suction temperature by placing the thermometer directly to the suction side of the centrifugal fan. Perform this procedure for at least 3 minutes. 5. Determine the static and velocity head of the air using a manometer. In getting the velocity head, use a pitot tube and place it inside the discharge casing of the fan while for the static head, insert the tube of the manometer in the static section. Observe and record the deflections.

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6. Use a tachometer to determine the speed of the centrifugal fan. Make at least 3 trials and obtain the average which will be recorded in the data sheet. 7. Determine the current flowing into the motor using an amprobe.

8. Use the thermometer to determine the discharge temperature and do it at least 3 mins.

9. Make 2 more trials by replacing the orifice diameter into 4.5 inches and 7 inches.

10. Record all the data in the data sheet and calculate for the necessary data. Return the materials after the experiment. 9

SET-UP OF APPARATUS

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FINAL DATA SHEET

Trial

1 2 3

Orifice Speed Orifice Pressure Temperature Current Input Output Eff. Diameter (rpm) Static Velocity Suction Discharge (Amperes) (HP) (HP) (%) (inches) (inH2O) (inH2O) (℉) (℉) 2.5 1651 87.8 88.7 3.51 .87 .0739 8.34 1 1⁄2 1 1⁄8 4.5 1692 87.8 89.42 3.63 .91 .2289 25.16 1 3⁄8 1 3⁄8 3⁄ 7⁄ 7.0 1710 88.7 88.7 3.75 .98 .26 27.81 8 4

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SAMPLE COMPUTATIONS Trial 3 •

Solving for the density of air: 𝑙𝑏𝑓 12𝑖𝑛 2 )( ) 𝑚 𝑃 𝑙𝑏𝑚 𝑖𝑛2 1𝑓𝑡 𝑃𝑉 = 𝑚𝑅𝑇 → = 𝜌 = = = 0.07244 3 𝑓𝑡 − 𝑙𝑏𝑓 𝑉 𝑅𝑇 𝑓𝑡 (53.34 ) (87.8 + 460)𝑅 𝑙𝑏𝑚 − 𝑅 (14.7

•

Solving for the velocity head of air:

𝜌𝑎𝑖𝑟 𝑔ℎ𝑎𝑖𝑟 = 𝜌𝑤𝑎𝑡𝑒𝑟 𝑔ℎ𝑤𝑎𝑡𝑒𝑟 → ℎ𝑎𝑖𝑟

1𝑓𝑡 3 𝑙𝑏𝑚 𝜌𝑤𝑎𝑡𝑒𝑟 ℎ𝑤𝑎𝑡𝑒𝑟 (1 8 𝑖𝑛)(12𝑖𝑛)(62.4 𝑓𝑡 3 ) = = 𝑙𝑏𝑚 𝜌𝑎𝑖𝑟 0.07244 3 𝑓𝑡

= 98.7024 𝑓𝑡 •

Solving for the velocity of air: ℎ𝑎𝑖𝑟 =

•

𝑣𝑎𝑖𝑟 2 𝑓𝑡 𝑓𝑡 → 𝑣𝑎𝑖𝑟 = √2𝑔ℎ𝑎𝑖𝑟 = √2 (32.2 2 ) (98.7024 𝑓𝑡) = 79.73 2𝑔 𝑠 𝑠

Solving for the air flow rate: 𝑄 = 𝐴𝑜𝑟𝑖𝑓𝑖𝑐𝑒 𝑣𝑎𝑖𝑟

•

𝜋 𝑓𝑡 8.8056𝑓𝑡 3 2 = (4.5𝑖𝑛) (79.73 ) = 4 𝑠 𝑠

Solving for the total head of air: 1 1 ℎ𝑇𝑤𝑎𝑡𝑒𝑟 = ℎ𝑠𝑡𝑎𝑡𝑖𝑐 + ℎ𝑣 = 1 𝑖𝑛 + 1 𝑖𝑛 = 2.625𝑖𝑛 2 8

ℎ𝑇𝑎𝑖𝑟 = •

𝜌𝑤𝑎𝑡𝑒𝑟 ℎ𝑇𝑤𝑎𝑡𝑒𝑟 = 𝜌𝑎𝑖𝑟

(62.4

1𝑓𝑡 𝑙𝑏𝑚 ) (1.625𝑖𝑛) (12𝑖𝑛) 3 𝑓𝑡 = 197.4047 𝑓𝑡 𝑙𝑏𝑚 0.07244 3 𝑓𝑡

Solving for the Output Power: (0.07244

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝛾𝑎𝑖𝑟 𝑄ℎ𝑇𝑎𝑖𝑟 = 𝜌𝑎𝑖𝑟 𝑔𝑄ℎ𝑇𝑎𝑖𝑟 =

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𝑓𝑡 8.8056𝑓𝑡 3 𝑙𝑏𝑚 ) (32.2 )( ) (197.4047 𝑓𝑡) 𝑠 𝑓𝑡 3 𝑠2 𝑙𝑏𝑚 − 𝑓𝑡 (32.2 ) 𝑙𝑏𝑓 − 𝑠 2

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = (125.92 •

𝑓𝑡 − 𝑙𝑏𝑓 1𝐻𝑃 )( ) = 0.2289 𝐻𝑃 550𝑓𝑡 − 𝑙𝑏𝑓 𝑠 𝑠

Solving for Input Power: 𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝐸𝐼𝑐𝑜𝑠(∅) = (220𝑉)(3.63𝐴)(1.00) (

•

1𝐻𝑃 ) = 1.0705 𝐻𝑃 746𝑊

Solving for fan efficiency (mechanical efficiency): 𝑛𝑚𝑜𝑡𝑜𝑟 = 𝑛𝑡𝑟𝑎𝑛𝑠 =

𝑀𝑜𝑡𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 → 𝑀𝑜𝑡𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 = 𝑛𝑚𝑜𝑡𝑜𝑟 𝑃𝑖𝑛𝑝𝑢𝑡 𝑃𝑖𝑛𝑝𝑢𝑡

𝐹𝑎𝑛 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 → 𝐹𝑎𝑛 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 = 𝑛𝑡𝑟𝑎𝑛𝑠 𝑀𝑜𝑡𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 𝑀𝑜𝑡𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 𝑛𝑓𝑎𝑛 =

=

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝐹𝑎𝑛 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 𝑛𝑡𝑟𝑎𝑛𝑠 𝑀𝑜𝑡𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 0.2289 𝐻𝑃 = × 100% = 25.1566% 𝑛𝑡𝑟𝑎𝑛𝑠 𝑛𝑚𝑜𝑡𝑜𝑟 𝑃𝑖𝑛𝑝𝑢𝑡 (1.0)(0.85)(1.0705 𝐻𝑃)

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DISCUSSION OF RESULT

Based on the data gathered by the researcher, three trials were performed by the researcher and each trial, the orifice diameter was varied. Taking an observation of the speed of the fan, the data were precise to each other because the values were slightly different with each other and the speed has a value of roughly 1651 rpm. Focusing on the pressure at the orifice, the static and velocity head measured by manometer has a decreasing value when the orifice diameter was increased, particularly for the velocity head. The deflection at static head decreases throughout the trial were not quite far from each other but in the velocity head, the researcher observed that the measured head decreased by almost half because of the size of the orifice diameter. From the formula in the continuity equation, the velocity and area of a certain section are indirectly proportional to each other and the resulting velocity affects the velocity head. Thus, the total head was affected by the orifice diameter which is indirectly proportional to each other. For the suction temperature, the values were precise to each other and the ambient temperature at that time is roughly 87.8 degrees Fahrenheit and the discharge temperature were also precise to each other projecting a temperature of roughly 88.7 degrees Fahrenheit. It shows that there is an increase in temperature due to some energy loss present in the system. The input power of the motor was affected by the current which was directly measured by the researcher. Throughout the trials, the output power from the fan increases with the diameter of the orifice in the discharge of air because the amount of air that flows in the system increases as a result of having the orifice diameter increased. The output power was affected also by the total head of air which decreases when the orifice diameter was increased, but the volume flow rate is significantly larger which affects the value of output power. Thus, the volume flow rate shows direct proportionality with the output power. Also, the output power was affected by the efficiency of the motor due to some losses and the resulting efficiency of the centrifugal fan were computed and ranging from 5%-30%. The data gathered were affected by some errors that were made by the researcher like rounding off values in the calculations or the condition of the apparatus used in the experiment. Since there is no equipment that operates at 100% efficiency, we assumed that losses are present in the system.

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QUESTION AND ANSWERS 1. Air is flowing in a duct with velocity of 7.62 m/s and a static pressure of 2.16 cm water gage. The duct diameter is 1.22 m, the barometric pressure 99.4 kPa and the gauge fluid temperature and air temperature are 30 degrees Celsius. What is the total pressure of air against which the fan will operate in cm of water? 𝑚 2 (7.62 𝑠 ) 𝑣2 ℎ = ℎ𝑠 + ℎ𝑣 ; ℎ𝑣 = = = 2.959 𝑚 𝑜𝑓 𝑎𝑖𝑟 2𝑔 2 (9.81 𝑚 ) 𝑠2 ℎ𝑣 (1000) = (2.959)(𝜌𝑎𝑖𝑟 ); 𝜌𝑎𝑖𝑟 =

99.4 𝑘𝑃𝑎 𝑘𝑔 = 1.143 3 0.287(30 + 273) 𝑚

ℎ𝑣 (1000) = (2.959)(1.143) → ℎ𝑣 = 0.0034 𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 0.34 𝑐𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 ℎ = ℎ𝑠 + ℎ𝑣 = 2.16 + 0.34 = 2.50 𝑐𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 2. The forced draft fans in parallel with a capacity of 73.2 m3/s each supplying combustion air to a steam generator. Air inlet is at 43.33 degrees Celsius, a static pressure of 254 mm water gauge is developed and the fan speed is 1200 rpm. The fan input is 257 kW each. Determine the capacity of the fan for a speed increase of 20 percent. 𝑄2 𝑁2 𝑁2 (73.2)(1.2)(1200) = → 𝑄2 = 𝑄1 = = 87.84 𝑚3 /𝑠 𝑄1 𝑁1 𝑁1 1200 3. A fan draws 1.42 m3/s of air at a static pressure of 2.54 cm of water through a duct 300 mm diameter and discharges it through a duct of 275 mm diameter. Determine the static fan efficiency if the total fan mechanical efficiency is 70% and air is measured at 25 degrees Celsius and 760 mmHg. ℎ𝑠 𝑛𝑠 = 𝑛𝑚𝑒𝑐ℎ ( ) ℎ 𝜌𝑎𝑖𝑟 = ℎ𝑠 = 𝑣=

𝑃 101.325 = = 1.18 𝑘𝑔/𝑚3 𝑅𝑇 (0.287)(25 + 273)

ℎ𝑤 𝜌𝑤 0.0254(1000) = = 21.52 𝑚 𝑜𝑓 𝑎𝑖𝑟 𝜌𝑎 1.18

𝑄 1.42 𝑚 1.42 𝑚 ; 𝑣𝑠 = 𝜋 = 20.09 ; 𝑣𝑑 = 𝜋 = 23.9 𝐴 𝑠 𝑠 (0.3)2 (0.275)2 4 4

𝑚 2 𝑚 2 𝑣𝑑 2 − 𝑣𝑠 2 (23.9 𝑠 ) − (20.09 𝑠 ) ℎ𝑣 = = = 8.54 𝑚 𝑜𝑓 𝑎𝑖𝑟 𝑚 2𝑔 2 (9.81 2 ) 𝑠 ℎ = ℎ𝑠 + ℎ𝑣 = 21.52 𝑚 𝑜𝑓 𝑎𝑖𝑟 + 8.54 𝑚 𝑜𝑓 𝑎𝑖𝑟 = 30.06 𝑚 𝑜𝑓 𝑎𝑖𝑟 15

21.52 𝑚 𝑜𝑓 𝑎𝑖𝑟 𝑛𝑠 = (0.70) ( ) = 50.11% 30.06 𝑚 𝑜𝑓 𝑎𝑖𝑟 4. A large forced-draft fan is handling air at 1 atm, 43.3 degrees Celsius under a total head of 26.6 cm water. The power input to the fan is 224 kW and the fan is 75% efficient. Compute the volume of air handled per minute. 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡 𝑡𝑜 𝐹𝑎𝑛 = 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 𝑃=

𝛾𝑄ℎ 9.81(𝑄)(0.0266) 𝑚3 60𝑠 𝑚3 → 224 = → 𝑄 = 64.38 × = 3862.87 𝑛𝑓𝑎𝑛 0.75 𝑠 1min 𝑚𝑖𝑛

5. A fan is listed as having the following performance with standard air: • • • •

Volume discharge = 120 m3/s Speed = 7 rps Static Pressure =310 mm water Brake Power = 620 kW

The system duct will remain the same and the fan will discharge the same volume of 120 cubic meter per second of air at 93 degrees Celsius and a barometric pressure of 735 mmHg when its speed is 7 rps. Find the brake power input and the static pressure required. 𝜌2 𝐵𝑟𝑎𝑘𝑒 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡 = 620 ( ) 𝜌1 101.325 735 ( 760 ) 𝑘𝑔 𝜌2 = = 0.9329 3 0.287(93 + 273) 𝑚 ℎ2 0.9329 = → ℎ2 = 241 𝑚𝑚 𝑤𝑎𝑡𝑒𝑟 310 1.2 0.9329 𝐵𝑟𝑎𝑘𝑒 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡 = 620 ( ) = 482 𝑘𝑊 1.2

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CONCLUSION

The researcher was able to be familiar with the parts of centrifugal fan as well as its operating procedures. The primary purpose of using a fan in an industrial process is to apply power to a gas particularly air in order to increase the energy content of the working fluid (air) which causes it to flow since it is a fluid. There are several types of fans but the researcher only investigated the performance of a centrifugal fan which was located at North Building of Mapua University. The researcher was also able to calculate the efficiency of the centrifugal fan when the diameter of the orifice was increased. It portrays that it has a fan efficiency of 10% - 20% because of the air power/ output power. The output power of a fan was affected by variables such as the volume flow rate of air as well as the total head of air. The efficiency of a fan was determined in order to examine the performance of the centrifugal fan in which factors such as losses were also considered by the researcher. These losses are present in the motors, transmitting element, and the centrifugal fan but since the fan is directly coupled with the motor, the transmission efficiency was set aside. Having a low efficiency decreases the capacity of the equipment so proper maintenance to the equipment must be performed in order to save energy. Also, proper lubrication must be applied to the joints of the centrifugal fan when needed so that friction losses can be minimized and the save power will be increased. Lastly, follow the instructions made by the instructor and analyze the procedure carefully in order to obtain data. As much as possible, the researcher should wear PPE since the researcher is dealing with equipment and any type of accidents can occur.

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REFERENCES 1. Fans and Blowers. (2018). Retrieved from https://www.saylor.org/site/wpcontent/uploads/2011/09/Chapter-3.5-Fans-Blowers.pdf 2. Difference between Fan and Blower. (2017). Retrieved from http://www.differencebetween.info/difference-between-fan-and-blower 3. Centrifugal Fans. (2018). Retrieved from https://www.ebmpapst.com/en/products/centrifugal-fans/centrifugal_fans.html 4. Centrifugal Fans and Blowers. (2018). Retrieved from https://www.nyb.com/centrifugalfans/ 5. Fan Efficiencies. (2018). Retrieved from http://www.greenheck.com/media/articles/Product_guide/CS104-13_FEG.pdf

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