Islandsparky
Electrical
Looking for info on calculating the heat produced by electric motors of various voltages.
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Heat produced by electric motors 2
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Skogsgurra
Electrical
Or in the book-shelf?
We are not like this. Not normally. But sometimes.
What have you found so far? What motors are you talking about? There are so many types. Are they DC, AC, small, large, asynchronous, synchronous, single phase, three-phase, open, enclosed, ventilated? You need to reformulate your question before you can expect a decent answer.
Gunnar Englund
We are not like this. Not normally. But sometimes.
What have you found so far? What motors are you talking about? There are so many types. Are they DC, AC, small, large, asynchronous, synchronous, single phase, three-phase, open, enclosed, ventilated? You need to reformulate your question before you can expect a decent answer.
Gunnar Englund
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Islandsparky
Electrical
The motors are all AC, h.p range from 1 to 20 h.p.totally enclosed fan cooled. The ones I am most instrested in calculating are 3 phaze but would like to learn how to calculate 1 phase as well. Most of the motors are Leesen motors (brand name) I assume should not make a difference.
These motors run about 8 to 10 hours a day.I was hoping there is a formula which would be a standard in heat calculations.
These motors run about 8 to 10 hours a day.I was hoping there is a formula which would be a standard in heat calculations.
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The issue is complicated by the fact that electric motor efficiencies can be all over the map, anywhere from 84 - 94% in that size range, depending on design. Here is a website that shows a chart on Figure 2 representing the range of efficiencies found in AC motors. BC Hydro Link
The only way to be sure is get the motor manufacturer's data on each specific motor. If that is impractical (and it usually is), the rule of thumb that my HVAC friends use is to assume 75% load on each motor, and 85% efficiency. This is a gross generalization of course, but if you are calculating heat losses into a building for AC load calcs, that represents a reasonable compromise. The motors that are running at higher loads will have slightly lower loss percentages but more overall power to begin with, and lightly loaded motors have higher loss percentages but lower overall power.
So how you would do that is to convert each motor HP to kW (746W/HP), multiply by .75 for loading, then again by .15 for losses (85% efficiency), which will give you watts loss (heat rejection) into the environment. From that you can calculate BTUs if necessary.
As a side note to this, if it is a new installation, chances are the motors will need to be energy efficient designs so the "guesstimate" efficiency at 75% load can be higher, probably around 87%.
Eng-Tips: Help for your job, not for your homework Read faq731-376![[pirate]](/data/assets/smilies/pirate.gif "[pirate] [pirate]")
The only way to be sure is get the motor manufacturer's data on each specific motor. If that is impractical (and it usually is), the rule of thumb that my HVAC friends use is to assume 75% load on each motor, and 85% efficiency. This is a gross generalization of course, but if you are calculating heat losses into a building for AC load calcs, that represents a reasonable compromise. The motors that are running at higher loads will have slightly lower loss percentages but more overall power to begin with, and lightly loaded motors have higher loss percentages but lower overall power.
So how you would do that is to convert each motor HP to kW (746W/HP), multiply by .75 for loading, then again by .15 for losses (85% efficiency), which will give you watts loss (heat rejection) into the environment. From that you can calculate BTUs if necessary.
As a side note to this, if it is a new installation, chances are the motors will need to be energy efficient designs so the "guesstimate" efficiency at 75% load can be higher, probably around 87%.
Eng-Tips: Help for your job, not for your homework Read faq731-376
Heat produced by electric motors, no matter what type (AC single or 3 ph, DC, Brushless DC, Reluctance, Stepper, Homopolar yada yada yada) is the power loss that has to be dissipated by the motor. If you can accurately measure the power supplied to the motor (electrical), and just as accurately measure the output power (mechanical) of the motor you will have the power (in watts) that is contributing to heat and not work. As already mentioned, I2R is often the main source of heat. In some cases eddy current, hysteresis, friction, windage etc. can be significant sources of heat. Some manufacturers provide thermal resistance (ºC/W) for their products and this can be of some use, but these values are always under predetermined conditions that usually do not relate to your application. I’m not exactly sure what you are asking, so I hope this will help. And remember, it's alway the last place you look.
motorspert
Electrical
If you want a quick and dirty formula, (given to me by someone older and much wiser than me) then.............
take the centre height (in mm)and divide by 100.
ie for a 710 centre height, expect 7.5kW dissipation.
Its the right ball park for medium machines - I've checked a couple some years ago. Not sure how it would translate to fractional horsepower machines though
Richard
take the centre height (in mm)and divide by 100.
ie for a 710 centre height, expect 7.5kW dissipation.
Its the right ball park for medium machines - I've checked a couple some years ago. Not sure how it would translate to fractional horsepower machines though
Richard
To motorspert:
This 'formula' is rather strange...
Motors of 710mm shaft axis height may have output of several MW, with total losses well over 100 kW (which mean heating of surroundings - unless the losses are carried away by cooling medium, eg. water).
This 'formula' is rather strange...
Motors of 710mm shaft axis height may have output of several MW, with total losses well over 100 kW (which mean heating of surroundings - unless the losses are carried away by cooling medium, eg. water).
More details about termal operation of electric motors you may find from Richard Welch, well-known motor designer and expert - see for example
motorspert
Electrical
rotman
yes you are correct, I should have stated that this is the heat from a TEWAC (water cooled) machine. I was eating my lunch when i responded and didnt fully absorb the question. Having taken blinkers off now,, i realise that my formula is not much use for totally enclosed machines as all the losses have to find their way through the frame. sorry to have confused you
yes you are correct, I should have stated that this is the heat from a TEWAC (water cooled) machine. I was eating my lunch when i responded and didnt fully absorb the question. Having taken blinkers off now,, i realise that my formula is not much use for totally enclosed machines as all the losses have to find their way through the frame. sorry to have confused you
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