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I'm in the process of trying to build a small hobby farm from scratch. One of the initial steps is to get temp service on the property, and to do so I have to fill out an electric load form. I will mostly be operating pumps, table saws, etc. Single phase, and most of the time not at the same time. Part of the sheet requires you to fill in the KVA and the KW of the different electric loads. I've done a fail amount of electrical work, but honestly these are metrics that I'm not familiar with. Can someone provide a little guidance on how one would fill this out? I vaguely recall from engineering school that the difference between KVA and KW is one is absolute and one is apparent power, but what that means within this context I'm not sure.

Thanks for the help

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    \$\begingroup\$ "I've done a fail amount of electrical work ...". Haven't we all! \$\endgroup\$
    – Transistor
    Aug 7 '17 at 14:05
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    \$\begingroup\$ Ha! Freudian slip, I think I'll leave it \$\endgroup\$
    – user379468
    Aug 7 '17 at 14:11
  • \$\begingroup\$ I understand this is just temporary service (construction?) But I think they have to plan to provide adequate service to handle the worst case, momentary load. \$\endgroup\$
    – jonk
    Aug 7 '17 at 16:08
  • \$\begingroup\$ What other info do they want as the Motors line has a star against it..... \$\endgroup\$
    – Solar Mike
    Aug 7 '17 at 16:49
  • \$\begingroup\$ They want to know the starts per day if the motor is over 10 hp \$\endgroup\$
    – user379468
    Aug 7 '17 at 16:50
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kW = Volts x Amps x power factor / 1000 for single phase and Volts x Amps x power factor x square root of 3 / 1000 for three phase.

kVA is the same except without multiplying by power factor.

For everything except motors, assume that power factor = 1 and kVA = kW.

HVAC and refrigeration would be primarily motor loads.

For items that have watts listed in the description or marked on the product assume kW = kVA. For items that only have current given, calculate the kVA and leave the kW blank.

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  • \$\begingroup\$ I recommend checking which equipment may be needed later. A simple conveyor belt may need lots of power, depending on load. \$\endgroup\$
    – Janka
    Aug 7 '17 at 13:49
  • \$\begingroup\$ You should clarify with the electric utility company if they want an estimate of your ultimate usage or just what you need for temporary power. \$\endgroup\$ Aug 7 '17 at 14:03
  • \$\begingroup\$ PSE&G I described the situation, but they only have one load sheet apparently, even though she said temp service is possible ... \$\endgroup\$
    – user379468
    Aug 7 '17 at 14:12

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