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Thread: Calculating Motor Operating Costs

  1. #1
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    Calculating Motor Operating Costs

    I was asked to give an estimated running cost of operating a 3 phase 75 HP motor for 80 hrs. I assumed at 480 Vac that the motor would have roughly 100 FLA.

    My equation then became KW = 480 V * 100 A * 1.732 / 1000 (ignored power factor and demand) and came to roughly 83 KW.

    I had a coworker check to verify my answer and he used a different method. He used 1 HP = .746 KW and came to roughly 56 KW. Does motor phase affect this equation?

    Which method is correct when solving for KW in a three phase system and why?

    Any insight would be apprecciated, Thanks.

  2. #2
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    430.252 says 96 amps for a 75 HP 3 phase 480 VAC MOTOR
    "Electricity is really just organized lightning." George Carlin


    Derék

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    Your "cost of operating" has the word "cost" in there. Cost of electricity is for kW, not Amps. Amps is essentially irrelevant.

    So yes, there are 746 watts/HP, that has nothing to do with amps, volts, phase etc. That is the power that is used. You are metered on kWH, (kilo Watt Hours), so to determine the cost of operating something electrical, simply take the watts (kW) x operating time (Hours).

    Now, without measuring anything all you can REALLY estimate is the MAXIMUM POSSIBLE cost of operating that motor, because all you have is the RATING of that motor, not a measurement of how much power it is actually consuming. So 75HP x .746 = 55.95kW, x 80 hours = 4,476 kWH, x whatever your electricity rate is will give you the worst case operating cost. It will probably be less than that however, you will only be able to tell the exact amount by measuring.

  4. #4
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    110107-2001 EST

    You need to know what is the motor load, how constant, and duty cycle.

    A useful method would be to measure the KWH for one typical 24 hour period.

    If you want a wild guess at full mechanical load, then maybe use 1 KW/HP as an estimate, or 75 KW. Then you need to estimate the number of full load KW per day, possibly 16 hours. At 16 hours per day the consumption would be 16*1*75 = 1200 KWH/day, and if cost is $0.10/KWH the daily cost is $120.

    You can change the cost all over the place based upon assumption you make.

    .

  5. #5
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    Note that the stated horsepower of a motor is normally the mechanical output of which the motor is capable.
    In this case it is stated to be 75 HP which is indeed about 56KW.
    The input must be more than this owing to the losses in the motor, in the absence of detailed information on the motor efficiency, I would allow approaching 10% for losses giving an input of about 61KW.
    Multipying 61 by the cost of power per KWH would give the hourly operating cost.

    This assumes that the motor is fully loaded, the actual loading may be much less and can only be determined by measurement.

  6. #6
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    Quote Originally Posted by gar View Post
    110107-2001 EST

    You need to know what is the motor load, how constant, and duty cycle.

    A useful method would be to measure the KWH for one typical 24 hour period.

    If you want a wild guess at full mechanical load, then maybe use 1 KW/HP as an estimate, or 75 KW.
    I agree with you points on motor loading and duty cycle. But think that taking 1kW for 1 hp is an unduly pessimistic assumption about motor efficiency at the rating being discussed here.

  7. #7
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    110108-0834 EST

    Besoeker:

    I agree that 1 KW is probably high for a motor of that size, but it is an easy number to use for a quick estimate.

    There are too many other undefined parameters that a high estimate won't make much difference. Actual KWH measurements over a reasonable time are needed to get good information.

    .

  8. #8
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    Quote Originally Posted by gar View Post
    110108-0834 EST

    Besoeker:

    I agree that 1 KW is probably high for a motor of that size, but it is an easy number to use for a quick estimate.
    Quite.
    Of course, if the motors were rated in kW instead of the archaic HP, that would then eliminate the need for conversion factors.......
    Runs for cover........



    Quote Originally Posted by gar View Post
    There are too many other undefined parameters that a high estimate won't make much difference. Actual KWH measurements over a reasonable time are needed to get good information.
    I agree. With just the motor rating alone, all you can calculate is the maximum energy that could be consumed over the 80 hour period.

  9. #9
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    Once again

    So everyone missed my point. I guess i will spell out my question.

    This is all theoretical. Only want Ballpark answers.

    A 75 HP motor at 480 find KW.

    If single phase 75HP * .746KW/HP = roughly 56 KW

    If three phase 75 HP * .746KW/HP = roughly 56 KW

    If three phase and assuming 100 FLA at 480V
    KW = Volts *Amperes * Power Factor * 1.732

    KW = 480 * 100 *Assume 1 (reality .8) *1.732 = 83 KW or (100 KW if power factor is used)


    All i want to know is if you generically use the formual for 1 hp = 746 watts do you need to account for phree phase situation by multiplying by 1.732.

    I understand you are only charged for real power usage (KW) unless you are on a commerial site where your rate structure can ding you for you KVA usage (Low power factor = high ding). Amps doesn't matter you just want KW (I was using a generic voltage and Current to find KW).

  10. #10
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    1 HP=746 Watts regardless of single phase or three phase.
    In many cases though the HP is the mechanical output, not the electrical input which must be greater since the motor must be less than 100% efficient.

    As a rough estimate I would add about 10% to the output in order to estimate the input.

    If the motor is not fully loaded then measurement by use of a KWH meter is the best option.

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