If the current is 45 amperes under this condition, calculate the power factor, reactive power, and reactive kilovolt-amperes
You didn't state whether it was a 3-phase or single-phase motor. At 8.8kW input, the equivalent input horsepower HP is: 8.8kW/(.746kW/HP) = 11.8HP input equivalent. I say input equivalent because horsepower normally means mechanical output power. Around here, that would require three-phase power, so assuming it is operated at a nameplate voltage rating of 230V 3-phase, the following calculations can be made: input kW = 8.8kW (given) input kVA = 45A x .230kV x 3^(1/2) = 17.93kVA power factor PF = (input kW) / (input kVA), or PF = 8.8 kW / 17.93 kVA = 0.4908 = 49% The reactive factor RF = sin ( arccos ( PF ) ) RF = sin (arccos ( 0.4908) ) = 0.8713 = 87% Reactive power = RF * kVA = .08713 x 17.93kVA = 15.6kVAR Reactive kVA = 15.6 kVAR Check: kVAR = ( kVA^2 - kW^2 ) ^ (1/2) kVAR = ( (17.93kVA)^2 - (8.8kW)^2) ^(1/2) kVAR = 15.6 kVAR it checks.
If it is a 3-phase motor, the power factor is 0.49 calculated as shown in another answer. That is a rather low power factor, but not unreasonable, particularly if it is a low speed motor or is lightly loaded. If it is a single phase motor, the kva is simply 45 X 230 / 1000 = 10.35 and the power factor is 8.8 / 10.35 = 0.85. Assuming that it is a standard motor operating at full load, it is most likely a 7.5 kW (10 Hp) motor. In that case the efficiency would be 7.5/8.8 X 100 = 85.2%, a reasonable figure for that size motor. It would be unusual for a motor of that size to be a single phase motor, but single phase motors of that size are listed in some motor catalogs.
power factor is 0.797, reactive power is 6.6kvar, reactive power is given in VAR or KVAR only.It is same as reactive kilovolt amperes. The way is simple for a single phase motor power is VICos (phi).and reactive power is VI Sin(phi)
The current o 45 amperes is too high for an 8.8 kw motor.Normally it should be 38 amperes only.The motor could burn at 45 amperes.
