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How to Calculate Three-Phase Power & kWh

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Three-phase power is used primarily in power distribution systems such as the power systems that power homes and businesses. Three-phase is referred to as three phase because alternating current flows along three separate conductors. Each current is slightly delayed or out-of-phase with the other. For example, if you assume conductor A as the lead, conductor B is delayed one-third of a cycle in comparison to A and conductor C is delayed two-thirds of a cycle in comparison to A. Together the conductors create the 3-phase circuit and the associated current, voltage and power levels.

Determine the phase voltage for each conductor. Connect a voltmeter between each conductor and neutral. Record the voltage. Do this for all three conductors. As an example, assume V1 = 300 V, V2 = 280V and V3 = 250 V

Determine the phase currents for each conductor. Connect an ammeter between each conductor and neutral. Record the current. Do this for all three conductors. As an example, assume I1 = 130 amps, I2 = 120 amps and I3 = 110 amps.

Calculate the power for each phase. Power is voltage times current or P = VI. Do this for each conductor. Using the examples above:

P1 = V1 x I1 = 300V x 130 amps = 39,000 VA or 39 KVA P2 = V2 x I2 = 280V x 120 amps = 33,600 VA or 33.6 KVA P3 = V3 x I3 = 250V x 110 amps = 27,500 VA or 27.5 KVA

Calculate the total 3-phase power, or "Ptotal," by adding the power of each phase together: Ptotal = P1 + P2 + P3. Using the above example:

Ptotal = 39KVA + 33.6 KVA + 27.5 KVA = 100.1 KVA

Convert Ptotal from KVA into Kilowatts using the formula: P(KW) = P (KVA) x power factor. Refer to operational specifications to find the power factor associated with the system. If we assume a power factor of 0.86 and apply the numbers from above:

P (KW) = P (KVA) x power factor = 100.1 KVA and 0.86 = 86KW

Determine Kilowatt-hours (kWh) associated with power usage using the formula: P(KW) x hours of use. If we assume 8 hours of use and continuing with the example:

kWh = P(KW) x hours of use = 86 KW x 8 hours = 688 kWh

About the Author

Dwight Chestnut has been a freelance business researcher and article writer for over 18 years. He has published several business articles online and written several business ebooks. Chestnut holds a bachelor's degree in electrical engineering from the University of Mississippi (1980) and a Master of Business Administration from University of Phoenix (2004).

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