Fan/Pump Heat Load

[AshleyF]

Does anyone know how to calculate the heat load, into the fluid, of a pump or fan? I believe the load is close to 100% of the input power. The only basis I have for this is 1 set of test data that indicated 92%.

This test was on a Vane Axial fan flowing ~ 1820 CFM (128.5 lb/min) with 32 In H2O of head rise. The input power was ~ 12.25 kW and the temperature rise was measured at ~21oF (11.3 kW). The calculated overall efficiency of this fan is ~56%.

Others witin my organization insist that the heat is only due to the inefficiencies of the shaft & motor. Therefore, heat would only be 44% of the input power from the above example.

If anyone has reference material that clearly explains this it would be very helpful.

Thanks for the help.

 

[AbbyNormal]

The inefficiency of the fan will create heat but the power comes from the fan shaft.

The commercial cooling load software I use quotes the 1989 ASHRAE HOF, and uses a formula of

TDf=0.363 x PD/ Eff

Where TDf is the temperature rise in degrees F, PD is the Pressure in inches of water and Eff is the combined efficiency of the fan and motor.

If the motor is out of the air stream as is most likely the case in your scenario, Eff is the fan efficiency. With the motor in the air stream you would multiply the motor efficiency by the fan efficiency to determine "Eff".

Checking your data

21= .363 x 32/.56

21 = 20.74, pretty close

The 2005 ASHRAE HOF goes into fan power a bit on page 30.30 and will give equations for heat gain based on the motor and drive being inside or outside of the air stream.




Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[wfn217]

It seems that compressing the air that much would raise the temperature by 11.6 F.