Heat Gain from Electric Motors
Heat Gain from Electric Motors
(OP)
Does anyone have a reasonable rule of thumb, or a reference that gives an idea of the heat gain you should expect based on horsepower from an electric motor? I'm sizing an exhaust fan in a pump room that has 4-40hp electric motors.





RE: Heat Gain from Electric Motors
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 -1) x BHP x 2545
RE: Heat Gain from Electric Motors
(40*2545)/0.84 = 121,190 Btu/hr
This seems a little high to me, am I using the formula wrong or what? I can't imagine having to provide 40 tons of cooling in this small room.
RE: Heat Gain from Electric Motors
You are correct if I understand what you the question is.
The energy conversion of 2545 BTU/HP is for useful work done by the motor. For example, if the motor drove a pump, the motor is doing useful work driving the impeller pumping a liquid to a higher pressure; this developes heat. The heat in this example is normally dispersed through the normal flow of the liquid; only if the pump is deadheaded (no flow) is a lot of heat generated that can damage the pump.
In your case, useful work is being done to forcefully flow air in a duct which will impart heat to the air flowing in the duct - but again that is normally dispersed without a significant increase in air temperature due to the large amount of air flowing and the fact that fresh air is probably being drawn in at 20 to 100% of the total air flow.
There is no useful conversion of energy between the air in the room and the motor. You only need to determine what the normal motor surface temperature is for full load amps; and then calculate the heat released to the atmosphere via radiation and conduction. It will not be anything near the complete energy conversion of 2545 BTU/Hp for the motor.
The more you learn, the less you are certain of.
RE: Heat Gain from Electric Motors
RE: Heat Gain from Electric Motors
RE: Heat Gain from Electric Motors
4 x 40 x (1-.84) x 2545 = 65152 BTUH or 5.43 Tons
Again you may not need this much refrigeration since some of this may be offset by fresh air intake through infiltration or intake ductwork.
The rest of the heat would be loss to the air driven by the fan 4x40x0.84x2545; assuming this leaves the room, it should not be a problem.
The more you learn, the less you are certain of.
RE: Heat Gain from Electric Motors
Btu/hr = 4 x (1/.84 - 1) x 40 x 2545 = 77562
Note this is higher than CHDO because input = bhp output/eff
In the ASHRAE equation of my previous post A = Input = B + C
B is the motor output energy that goes to the driven equipment or fluid,
C is the difference between input & output and is what goes to the space.
You should also add heat gain from lights and transmission from the exterior envelope (for quick estimate use equiv solar/air temperture of 130°F outdoors).
Typical indoor design for ventilation w/ OA is 104°F because this is typical max rating for motors & transformers. If the design outdoor air temperature is 95°F, then required CFM OA = (Btu/hr total heat gain)/(1.1 x (104-95)).
RE: Heat Gain from Electric Motors
RE: Heat Gain from Electric Motors
0 to 2 hp: 190 watts/hp
3 to 20 hp: 110 watts/hp
25 to 200 hp: 75 watts/hp
> 250 hp: 60 watts/hp
from HVAC Equations, Data, and Rules of Thumb by Arthur Bell, Jr. page 107
RE: Heat Gain from Electric Motors
Using the ASHRAE Table he references, motor Hp is the Nameplate HP, and motor efficiency accounts for all losses. The only adustment required is if the motor is overloaded in which case you are suppose to multiply factor by the max sdrvice factor for the motor.
Illiput: They don't say in the reference I have, but I wonder if we should account separately for the actual power factor at any load? I assume they just assume a power factor for the table?
The more you learn, the less you are certain of.
RE: Heat Gain from Electric Motors