Electric motor- pump heat gain

[GeoffKZ]

ASHRAE Classifies heat gain into (3) as follows:

A = Motor in Driven Equip in. Btu/hr Heat gain = (BHP x 2545)/Eff

B = Motor out, Driven Equip in. Btu/hr Heat Gain = BHP x 2545

C = Motor in, Driven Equip out. Btu/hr Heat Gain = ((1-EFF)/Eff)) x BHP x 2545
My situation is as follows:
I have a 500 hp motor driving a water pump within closed room, pump is moving water at 120F, pump efficiency is 70%, motor efficiency is 95.6%,
Which equation to use to calculate heat load (heat generated) by this combo. water pumped is coming and going to and from different areas of the plant, pipe enters and exists through the wall. Need to know heat generated by both to design cooler.
Utilizing equation A generates heat load of 1,331,000 BTU/HR is this correct?
When I consider motor only Equation C I get 58,600 BTU/HR
Equation A does not use any pump info, however it generates huge load as compare to equation C.
If I will treat them separately and calculate motor load using C and how do I calculated pump load?
I know that maximum pump casing can reach 180F

 

[lukaiENG]

Calculate the electrical power into the room (your motor BHP)
Calculate the power transferred to the fluid using the pump/motor efficiency.
The difference is the heat gain to the room.
Also, don't forget if you are pumping 120degF water then the piping will contribute to the heat gain to the room, even if insulated.

 

[LittleInch]

Geoff KZ

You need to follow the energy and also be clear about where your power number apply.

As Lukai says, start with electrical power in ( Volts x amps) for the shaft power that the pump actually requires/consumes. This cannot be higher than your 500 hp, but could be a lot less.

Then subtract the motor losses. You don't say, but we assume that this is an air cooled motor, hence approx. 4.5% of the electrical power being consumed is lost as heat to the room.

Your pump data sheet will tell you how much shaft power the pump requires to do it's duty.
Approximately 30% of that shaft power will turn into heat and heat up the water ?( probably by a couple of degrees). The rest of the power input to the water will gradually turn into heat but this is spread around the water system due to frictional losses.

What the three equations are giving you is designed more for an air driven system.

Equation A is basically electrical power in and just works on the basis that the motor and its fan is within a duct and hence ALL the energy going into the motor at some point will add to the heat energy in the air
Equation B is saying that the motor is outside the duct and hence cooled by something else and the heat input into the system is the absorbed power of the fan / pump
Equation C would appear to the heat energy given off by the motor alone,

ALL the equations use the pump information because the BHP figure used is the shaft power of the pump NOT the rated power of the electric motor. These two figures can be the same or close, but the absorbed shaft power can be significantly lower than the maximum rated power of the motor.

Makes sense?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[GeoffKZ]

Thank you very much to both of you for replying to my question.
Looks like I have to only consider piping temperature (120F) transferring heat to the room which must stay below 115F. This obviously in addition to motor load and load due to cooling system and all the usual elements like walls, lights, wires etc.
Thank you again.