active power filter
Practical calculation of reactive power
Jan 08, 2018

Please refer to the following calculation examples


Type of circuit

Apparent power

S (kVA)

Active power

P (kW)

Reactive power

Q (kVAr)

Single phase (Ph + N)

S = V x I P = V x I x cos φ Q = V x I x sin φ

Single phase (Ph + Ph)

S = U x I P = U x I x cos φ Q = U x I x sin φ
Example:                                   5 kw load Cos φ= 0.5 10 kVA 5 Kw 8.7 kVAr
Three-phase (3Ph or 3Ph+N) S = 3 x U x I P = 3 x U x I x cos φ Q = 3 x U x I x sin φ

Example of Motor with               Pn = 51kW

cos φ = 0.86                         efficiency = 0.91

65 kVA 56 kW 33 kVAr

Calculations in the three-phase example were as follows:
Pn = power supplied to the rotary axis = 51 kW
P = active consumed power = Pn/ρ = 56 kW
S = apparent power = P/cos φ = P/0.86 = 65 kVA
Hence:
Q = √(S² - P²) = √(65² - 56²) = 33 kVAr
The average power factor values for various loads are given below.


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