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.

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 -1) x BHP x 2545

Just to make sure, but according to the above formula. A 40hp motor, driving a blower, indoors. Will generate around:

(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.

ChrisConley,

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.

You should use C if the fan is ducted outside the room ventilated and the efficiency you will use will be the motor efficiency which will be about 89% if standard, more if high efficiency type. Where does the blower get air fron & where does it blow it. Describe extend of blower suction & discharge ductwork with regards to the room ventilated.

I never said that this was ductwork. The 3-40 hp blowers are blowing into a 12" steel pipe. This pipe feeds various aeration processes. The blowers have an piped intake which is exterior to the building, and then they discharge the air to a process which is outside of the building. I used .84 efficiency as these blowers are being reused from a previous application and that is the efficiency that they operate at. The room is then ventilated independantly of any process equipment with an exhaust fan and an intake louver.

Additional input:  I would take HP x (1-Eff) x 2545 as an approximation of max heat that could be released from motor surface to atmosphere in room.  

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.

Apply the ASHRAE Equation C (since the blower air is not in the ventilated space envelope) as follows:

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)).

If you can put the exhaust right over the heat producing equipment, you can reduce the amount of ventilation air by having a higher temperature rise. You can design for an average temperature of 104°F and say temperature of exhausted air = 113°F, therefore your temperature rise = 113-95 = 18 & you can cut the required air flow in half.

Here are the rules of thumb that I use for motor (only) heat gain:

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

illiput is correct, I was thinking of the classic definition of motor efficiency used for pumps where motor efficiency is defined as a measure of how effectively the motor turns electrical energy into mechanical energy - so did forget to include energy losses for KVA input versus electrical output.
  
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.

I prefer to use bhp than motor hp to take into account motors are not fully loaded. The power factor will have no effect on the kw energy used by the motor. It only affects the Volt amps required to be produced by the power generator.