Calculating Motor Temperature Rise

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W@ttson

W@ttson

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Hello,

Is there a way to calculate motor temperature rise based on a per unit of the full load current?

For example, here is this website that can tell you the expected life of the motor based on the per unit value of the full load current:

I would like to calculate the motor temperature rise for operating a motor for 30 seconds at 200% FLT. Then by comparing it to the allowable temperature for the insulation class I can see if that duration will have a negative effect on the motor.
 
You can make rough estimates.

For short duration transients, you ignore cooling to the environment and just assume any heat generated remains in the motor.

The copper losses are essentially I^2R, so double current means these losses go up by a factor of 4. The total additional watt seconds that the motor absorbs is 3*<normal losses>*30.

Next you need the heat capacity of the motor. Unless you know the mass of the individual motor parts, you are limited to <total motor mass> * 400 J/kg as a rough average for the materialsin the motor.

Total added losses / heat capacity gives you the approximate temperature rise.

Jon
 
winnie said:
You can make rough estimates.

For short duration transients, you ignore cooling to the environment and just assume any heat generated remains in the motor.

The copper losses are essentially I^2R, so double current means these losses go up by a factor of 4. The total additional watt seconds that the motor absorbs is 3*<normal losses>*30.

Next you need the heat capacity of the motor. Unless you know the mass of the individual motor parts, you are limited to <total motor mass> * 400 J/kg as a rough average for the materialsin the motor.

Total added losses / heat capacity gives you the approximate temperature rise.

Jon
Click to expand...
In your equation for the additional watt seconds that the motor absorbs, is that meant to be a “3” or should it be a “4”?
 
I assumed you were starting from the motor being at full load and already heated up to equilibrium.

Running at 200% current means 4x heat produced, minus 1x for the heat already being dissipated. Thus 3x additional heating.

Jon
 
How would you go about figuring out the copper losses (I^2R)? Would it just be as simple as measuring to the stator leads? I guess you could ask the motor manufacturer. Or perhaps looking at its efficiency. Its been a while since I cracked open my energy conversion text book.

I think qualitatively looking at things, if the motor is say rated for 60 min, that means that at full load you can run it for 60 min and it won't reach its maximum insulation temps. 30 seconds is 0.83% of the total rated time. Even if you have 3 times that additional heat, we can look at it eating into the time rating of the motor (At least I think). So for running at 30 seconds at full load would be more so like running for 120 seconds (30 seconds normal + 30 + 30 + 30).
 
There might be other design issues to consider besides heat.

Is the shaft designed for that much stress? Will it put more stress on the bearings? Will there be magnetic saturation of the core similar to a transformer?
 
As I said, this is a _very_ rough estimate. If you want serious accuracy, you need the internal design of the motor.

I'm assuming that total losses at 200% rated current will be approximately 400% of rated losses. As I said, very very crude.

You never specified the type of motor. as @steve66 suggests, there are lots of other design issues to consider.

-Jon
 
Steve/Jon,

It would be a 50HP motor Design B, TENV.

I used 200% as a round number, in actuality it would be closer to 140% - 150%.

Looking at a design B BDT is about 200%. So if it is 150% or 200% I see that the motor would be able to produce that when coupled to an oversized drive.
 
W@ttson said:
Steve/Jon,

It would be a 50HP motor Design B, TENV.

I used 200% as a round number, in actuality it would be closer to 140% - 150%.

Looking at a design B BDT is about 200%. So if it is 150% or 200% I see that the motor would be able to produce that when coupled to an oversized drive.
Click to expand...
Do you use BDT or LRT? In Design A and B motors, the BDT is higher that the motor LRT. Only Design C or D has an LRT that is expected to be higher than the breakdown torque.
 
topgone said:
Do you use BDT or LRT? In Design A and B motors, the BDT is higher that the motor LRT. Only Design C or D has an LRT that is expected to be higher than the breakdown torque.
Click to expand...
its coupled to the drive, you can produce any torque up to BDT at 0 speed (closed loop vector). As long as the transistor output stage of the drive can handle it.
 
I don't presume to be an expert in this area, but here's my take on it:
As the heat from I2R losses is generated, the rate of temperature rise will initially be determined by the specific heat capacity of copper windings. Then as the temperature of the windings rises, the heat transfer rate through the insulation to the rest of the motor will increase until it eventually reaches a thermal equilibrium and the conductor temperature stabilizes at some level. There will be multiple time constants involved as the heat passes from the conductor through the rest of the motor. A worst case conductor temperature rise estimate could be made by assuming that 30 seconds is significantly less than any of these time constants, and so the heat capacity of the copper is the primary factor in play. I don't know if this would be a valid assumption in the OP's case, but if this worst case result turns out to be acceptable for the insulation then it seems like you're good to go.
Of course, if there are multiple succesive 30 sec intervals like this at a higher current then that would have to be figured in using a thermal model that can predict the transient behavior.
 
If P=S*η*p.f. then η=P/S/p.f. where:
P= output power[kW] S=apparent power[kVA] p.f. =power factor and η=efficiency
S=sqrt(3)*Irated*Vrated=sqrt(3)*337*400/1000=233.48 kVA
Then η=200/233.48/0.87=0.9846
According to ABB The Motor Guide 2005 Second Edition 4.Electrical design, if ambient temperature is 50 degrees C the maximum permitted output % from rated will be 93%. That means instead of 200 kW 200*0.93=186 kW.
If we may consider the same efficiency for 93% we get :
S=186/0.9846=188.9 and Inew=S/sqrt(3)/Vrat=272.65 A [It has to be less than 292 A as you need]
 

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The 'ABB Motor Guide' linked by @Julius Right is a tremendously useful reference.

I did a search for it, and a quick read through lead me to the concept of 'standard motor duty types' (S1...S10). Searching on duty type S9 calculations, I hit on:

Starting at page 9 is directly relevant to the OP's question. Essentially up to 200% of normal rated current you can treat motor heating as an exponential. So what you can do is run the motor at full rated current while monitoring temperature until the temperature stabilizes. Then based on the heating curve you can figure out the thermal time constant of the motor. Then knowing the thermal time constant you calculate the heating curve scaled by I^2 and that will tell you the temperature rise rate.

Specifically go to page 15 where there is discussion of calculating the short term ratings of motors where you know the continuous rating. Note that the thermal time constants being used in the examples are all in the range of 1-1.5 _hours_. A motor with a time constant of 1 hour, starting cold, could reasonably be expected to operate for 15 _minutes_ at 200% current, without overheating.

For a 50 Hp 4 pole motor I think you will find a 30 second 200% overload causes a small blip in temperature, but I have no specific data to support that guess.

-Jon
 
Last edited:
I am sorry! My mistake:
S=186/0.9846/0.87=217.14 and Inew=S/sqrt(3)/Vrat=313,4 A [It has to be MORE than 292 A as you need]
 
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