Question

the kvar rating required for improving the power factor of a load operating at 500 kw and 0.85 power factor to 0.95 is

Answer 1

The pressure of air inside a soap bubble of diameter 0.7 millimetre is 8 millimetre of water above the atmospheric pressure determine the surface...

Answer 2

Answer:

kVAR rating to improve PF from 0.85 to 0.95 is 48.5kVAR.

Explanation:

To improve PF from 0.85 to 0.95  is 0.398

Required Capacitor kVAR to improve P.F from 0.85 to 0.95

Required Capacitor kVAR = kW x Table 1 multiplier  of 0.85 and 0.95

                                            = 500kW x 0.291

                                              = 145.5 kVAR

And Rating of Capacitors connected in each Phase

                                               = 145.5kVAR / 3

                                                = 48.5 kVAR

Solution # 2 (Classic Calculation Method)

Motor input = P = 500 kW

Original P.F = Cosθ1 = 0.85

Final P.F = Cosθ2 = 0.95

Required Capacitor kVAR = P (Tan θ1 – Tan θ2)

$\theta1 = Cos^{-1} (0.85) = 31.7^{0} ; Tan\theta_{1} = Tan (31.7^{0} ) = 0.617\\\theta_{2} = Cos^{-1} (0.95) = 18.19^{0} ; Tan\theta_{1} = Tan (18.19^{0} ) =0.328\\\\Required Capacitor kVAR = P (Tan\theta_{1} -Tan\theta_{2} ) = 500kW (0.617- 0.328) = 144.5 kVAR$

And Rating of Capacitors connected in each Phase

                                    144.5 kVAR / 3 = 48.1 kVAR.