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Thread: Wattage and Amperage Calculations

  1. #1
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    Wattage and Amperage Calculations

    Hello,

    I have a question regarding finding loads for a building that has a 120/208V 3 Phase 4 wire 200 amp service. I am trying to find the wattage and amperage calculations for the following pieces of equipment and am looking for assistance in doing so. The pieces of equipment are as followed....

    One 15 kw 480v HVAC Unit
    One 10 kw 480v water heater
    16 100 watt 277v LED light fixtures
    4 40 watt LED 480v parking lot light fixtures
    2 460v 3 phase squirrel cage 25 horsepower motors
    2 277v specialty receptacles rates at 30 amps each
    6 277v specialty receptacles rated at 600 watts each

    Thanks for any help provided!

  2. #2
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    For the most part you just need ohms' law..

    For your motors you would need to look at the NEC motor Tables, NEC 430-248-252 and obtain your current then apply ohms law.

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    Since you are a student, work on converting the loads to wattage and show us what you find.
    You will get more out of it that way than us giving you the answers.
    Last edited by augie47; 04-16-18 at 08:18 PM.
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  3. #3
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    I understand that I have to use OHMs law to convert to and between watts and amps. I think I may be overthinking this in some ways. When I'm calculating for the wattage, I have 15,000W, 10000W. 1600W, 160W, and 3600W given to me. Therefore, I would need to calculate the wattage for the 2 specialty receptacles and for the squirrel cage motor. I have found the amperage of the motor to be 34 amps. However, what trips me up is the 3 phase service. In my problem it states that all the equipment is purely resistive. Therefore, that would indicate a power factor of 1. The equation used for 3 phase power is P=VI*1.73 so if everything is purely resistive then I would not multiple by the 1.73, correct? Also, when I'm calculating for wattage I am confused about whether I am using the service voltage in these calculations or the rated voltage of the equipment. I guess what I'm trying to say is I understand the equations and how to do the calculations, I'm just unsure about what the correct format should be before I do the calculations. I hope that makes sense.

  4. #4
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    sorry I thought you were looking for Amps
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  5. #5
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    Quote Originally Posted by LtsDeuce- View Post
    The equation used for 3 phase power is P=VI*1.73 so if everything is purely resistive then I would not multiple by the 1.73, correct?
    The 1.73 is capturing the fact that it is three phase power. sqrt(3)

    If you are finding current for a 240 volt single phase device, it would be VA / 240. If you are finding current for 208 volt three phase, it would be VA / ( 208 * sqrt(3) )
    Last edited by charlie b; 04-17-18 at 06:12 PM. Reason: corrected formula (208 versus 230)

  6. #6
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    Quote Originally Posted by LtsDeuce- View Post
    However, what trips me up is the 3 phase service. In my problem it states that all the equipment is purely resistive. Therefore, that would indicate a power factor of 1. The equation used for 3 phase power is P=VI*1.73 so if everything is purely resistive then I would not multiple by the 1.73, correct?
    That is the equation for 3 phase "apparent power." The equation used for 3 phase "real power" is P=VI*1.73* power factor. The types of problems you are dealing with should always work in "apparent power." If you are not yet familiar with the differences, this would be a good topic to review.

    Anything resistive would have a power factor of 1, but a motor will not. However, the good news is that the NEC table (from which you correctly determined that the current is 34 amps) is treating things as though the power factor was already figured into the tabulated value. Thus, the power assigned to each motor will be (460) * (34) * (1.732), or about 27,100 VA.

    Charles E. Beck, P.E., Seattle
    Comments based on 2017 NEC unless otherwise noted.

  7. #7
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    Quote Originally Posted by LtsDeuce- View Post
    I would need to calculate the wattage for the 2 specialty receptacles and . . . .
    The one piece of advice I will give you here is to always calculate things in terms of VA, and convert to amps only at the very end. For this part of your problem, you have 2 items, each drawing 30 amps at 277 volts. The voltage value tells me that this is a single phase load. Therefore, the VA is calculated as (2) * (30) * (277), or 16,620 VA. Since it is single phase, the value 1.732 (the square root of 3) does not enter into the equation.

    Charles E. Beck, P.E., Seattle
    Comments based on 2017 NEC unless otherwise noted.

  8. #8
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    Quote Originally Posted by LtsDeuce- View Post
    Hello,

    I have a question regarding finding loads for a building that has a 120/208V 3 Phase 4 wire 200 amp service. I am trying to find the wattage and amperage calculations for the following pieces of equipment and am looking for assistance in doing so. The pieces of equipment are as followed....

    One 15 kw 480v HVAC Unit
    One 10 kw 480v water heater
    16 100 watt 277v LED light fixtures
    4 40 watt LED 480v parking lot light fixtures
    2 460v 3 phase squirrel cage 25 horsepower motors
    2 277v specialty receptacles rates at 30 amps each
    6 277v specialty receptacles rated at 600 watts each

    Thanks for any help provided!
    you need a few basic equations
    1 and 3 ph: v = i r
    1 ph: p = v i
    3 ph: 1.732 v i
    we will ignore power factor since they do not give it in the problem
    technically it is in va not watts, but in this case considered equivilent

    the easiest way is to calc p for each device and add them up
    v and i do not matter if p is given

    15 kw (given)
    10 kw (given)
    16 x 100 w = 1600 w = 1.6 kw (1 kw = 1000 w)
    you can use w or kw, but all must be the same
    4 x 40 = 160 w = 0.160 kw
    you know it is 3 ph, to calc power you need i from the nec table, 34 A
    2 x (1.732 x 480 x 34) = 56534 w = 56.534 kw
    the next 2 are 1 ph
    2 x 277 x 30 = 16620 w = 16.62 kw
    they give watts so
    6 x 600 = 3600 w = 3.6 kw
    total p = 103.514 kw = 103,514 w

    you know p = 103514 w
    you know it must be 3 ph (due to 3 ph motors) so you know v = 480
    you need to find i
    you know p = 103514 = 1.732 x v x i = 1.732 x 480 x i
    so i = 103514 / (1.732 x 480) = 124.5 A

    you need a service > 125 A

    sometimes seeing the whole process helps
    Last edited by Ingenieur; 04-17-18 at 07:10 PM.

  9. #9
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    That makes sense now. Thanks for your help guys.

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