Calculate 3 phase power consumption?

Question

Hello, Our company is looking to purchase some industrial equipment. I am doing a cost/profit analysis and trying to figure out power usage. Both devices run off of 208v 3 phase. Total usage is 29.2 amps running. For example, To figure out kwH for single phase power you take 120(volts) x 10(amps) gets you 1200 watts, times 1(hour) divided by 1000 = 1.2kwH if I remember correctly. Now, my question, how do you figure out 3 phase kwH? do you multiply by 208 for 1 phase, or 624 for all 3 phase'?

I can figure out what our cost per kwH is, just need to know how to figure out to get kwH from a 3phase device.

Thank you,
Jason

Answer 1

The proposed 3Ø, 120/208V equipment draws 29.2 amps.

.208 kV * 29.2 A * √3 = 10.5 kVA

Assuming a power factor of .85, the real power is 10.5 * .85 = 8.94 kW

Once you've got kW (or kVA) it doesn't matter if the equipment is 1Ø or 3Ø, the billing calculations stay the same. Just multiply by the number of hours of operation to get kWh (or kVAh).

As you analyze your costs, keep in mind that many utilities also have a demand charge. This charge can be rather significant and is usually on top of the kWh or kVAh energy usage.

The demand charge is based on your highest kW or kVA usage over a 15 or 30 minute interval. If these two new pieces of equipment are going to run continuously, they will each add 9 kW or 10.5 kVA to your demand charge.

Answer 2

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RE:
Calculate 3 phase power consumption?
Hello, Our company is looking to purchase some industrial equipment. I am doing a cost/profit analysis and trying to figure out power usage. Both devices run off of 208v 3 phase. Total usage is 29.2 amps running. For example, To figure out kwH for single phase power you take 120(volts) x 10(amps)...

Answer 3

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Your calculations are almost correct. The startup current can be neglected, since it should only last less than a few seconds. You should use a power factor, which is dependent on the motor specifications and its load level. Typical values would be about 0.75 - 0.85 for this small motor. You don't mention if the voltage is measured line-to-neutral, but this is a reasonable assumption in the U.S. Your calculation of daily energy consumption sounds about right for a small motor (about 2.5 hp) running for only about a half hour per day.

Answer 4

Apparent power (which is what the utility is charging you for) is often more than the Active power (which is what your machine is using to turn into mechanical work).

For a three-phase system:
(Apparent power) = kVA = (Sqrt 3)*(V)*(A)/1000

(Active power) = kW = kVA * PF
... where PF is the machine's 'power factor', a number between 0 and 1.

In your case, you have 208 V and 29.2 A (and some unidentified power factor).

Apparent Power = (Sqrt 3)*(208)*(29.2)/1000 = 10.5198 kVA

Answer 5

you just see the equipment watts
eg: 100w bulb use for 1 hr it consume 1unit of electricity i.e. 1kw of electricity.
if you have 3ph equip you need 1kw equipment
this 1kw is divided to 3 phase coils
more u tell as we clear the things then we sort out what you exactly want.