Induction Motor Heat Generation Calculation

[hanksmith]

1st time having to do this and schooling is a little rusty and don't think we even touched this in schooling.

I have been asked to calculate how much heat a group of newly installed motors will generate into the surrounding room.

I figured I could just use the motor eff and then calculated the wasted wattage, this was shot down by my supervisor, he stated that the heat generated would be much lower then this and to try finding a thermal model of an induction motor,

If someone could point me towards an informative web site or even a technical paper on calculating heat losses in motors that would be great.

Thanks
Hank

 

[electricpete]

The thermal model is important for transients, but not for steady state.

In steady state, the main things you need to know about the motor are :
1 - motor load level
2 - motor efficiency at that load level

You might make determining motor efficiency a little more complicated if you have knowledge about voltage deviations or votlage unbalance, but that is a refinement.

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[dpc]

Your analysis is correct and your boss is wrong. All motor losses (kW) end up as heat. Ask him to explain where those lost kW go if not into heat.

 

[edison123]

dpc is right. All losses end up in heat. Efficiency is good enough to calculate the losses and hence the heat.

OEM's do temperature rise test for motors and generators, which is quite complicated and expensive. Hence, it is only done as a type test.

* Women are like the police. They can have all the evidence in the world and yet they still want a confession - Chris Rock *

 

[hanksmith]

Thanks guys,

If I get an explanation from him as to what he was thinking I will post it.

Hank

 

[jraef]

One thing to consider though, you need to get the efficiency curves for the motors, as it varies with loading. With those then, you will also need to calculate loading, which can be tricky. Power factor is a decent way of determining percent of load, but a kW meter is best. Unless you just want a worst case scenario, in which case you would pick the worst point on the curve, usually full load.

Here is a decent resource for you. On the 2nd page they describe methods of determining motor efficiency if you don't already know it. I suggest getting their MotorMaster software. It used to be free, I don't know if it still is, but I doubt it is expensive.

http://www1.eere.energy.gov/industr...estimate_motor_efficiency_motor_systemts2.pdf

This paper goes into more depth if you want to do it on your own.

http://www1.eere.energy.gov/industry/bestpractices/pdfs/10097517.pdf

 

[badservo]

unless the work done by the motors is outside or leaves the surrounding room, then the heat will be the total energy used by the motors.

 

[jraef]

Hmmm... machine tools, yes maybe. But if it's a pump, isn't energy put into moving the liquid, or a fan into moving the air, or a conveyor onto moving the belt and goods, or a compressor into the air or coolant going onto the pipes, etc. etc. etc.? I would think that most motors are expressing most of their energy to another location. I mean, 60% of all electric motors are on pumps. The entire closed system for which the pump is being used would need to be contained in the same room / building for that to be an issue. I can't think of an example off the top of my head.

 

[aolalde]

By definition an electric motor converts electric power into mechanical power. The motor losses is the part of electric energy that is converted to heat and it is proportional to the motor efficiency, with exception of the mechanical energy required by the self propelled motor air cooling fans.

 

[itsmoked]

Yes Jeff but any transmissions have losses. Possibly more loss than modern motors. A belt drive is what 3% loss? A geared transmission same or more?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[dpc]

There will certainly be losses in the pump and friction losses in the piping, but the motor knows nothing about this. The motor losses are all rejected as heat from the motor based on its efficiency.

I suppose some heat from the motor could be conducted through the shaft, into the pump and on to the fluid being pumped and then on to who knows where, but it is still starts as heat from the motor.

 

[gepman]

badservo is correct. This was probably a better question to ask in the HVACR forum since they are accounting for this all the time in their heat load calculations.

Multiply these numbers by the loading factor on the motor to get an approximate answer. The tricky part is determining where the work goes, inside or outside the room.

See attached.

 

[itsmoked]

hanksmith; What are these motors running?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[hanksmith]

Most are running slurry pumps, some are conveyors, either belt or screw, few are exhaust fans which I have ignored because I figured they would take care of their own heat.

Just need a rough calculation so that we know how many changes of air are required in the building

 

[gepman]

If the pumps are pumping the liquid out of the building then you can assume that they are "motor-in, driven equipment-out" (by my attachment in my previous post). If the conveyors are totally in the building then they are "motor-in, driven equipment-in". If the exhaust fans are mounted indoors (usually they are on the roof which wouldn't count), they are "motor-in, driven equipment-out" since they air is exhausted. Circulation fans are "motor-in, driven equipment-in".

Remember all the work winds up as heat also, just draw a box around your building or room and calculate the energy balance. If you are pumping or conveying outside the room then the work is going out. The exception to this would be a conveyor that is half in or half out of the room, then about 50% of the load would be in the room depending on where the motor was located. For pumps the inefficiency of the pump goes to heating the liquid which usually leaves the room (unless it is recirculated) so that energy does not stay in the room. The rest of the energy goes to overcoming friction which turns to heat and is usually carried away by the fluid. Energy used to lift the elevation of the fluid is stored and then released once it returns to its original elevation (if it does).

 

[jraef]

I think you guys misinterpreted what I was saying, but no matter. In this case, all of the motor LOSSES will be rejected into the room, but not all of the motor ENERGY will be.

 

[DickDV]

I would think that the only likely extra heating component would be a gearbox between the motor and the load, if it exists.

In that case, use jraef's analysis for the motor's heat input into the room and then consult the gearbox manufacturer for its losses at the speed and torque that you feel would be appropriate for its worst case continuous operation.

As for all this talk of energy being converted into heat in the load, it is technically correct but usually not a factor unless, as stated above, the energy is dissapated in the same enclosure as the motor/gearbox.