Varun answered 3 years ago Motor input = P = 5 kW
Original P.F = Cosθ1 = 0.75
Final P.F = Cosθ2 = 0.90
θ1 = Cos-1 = (0.75) = 41°.41; Tan θ1 = Tan (41°.41) = 0.8819
θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843
Required Capacitor kVAR to improve P.F from 0.75 to 0.90
Required Capacitor kVAR = P (Tan θ1 – Tan θ2)
= 5kW (0.8819 – 0.4843)
= 1.99 kVAR
And Rating of Capacitors connected in each Phase
1.99/3 = 0.663 kVAR.
You are welcome to read more Important Interview Questions and MCQs.
Berin answered 3 years ago A 3 Phase, 5 kW Induction Motor has a P.F (Power factor) of 0.75 lagging. What size of Capacitor in kVAR is required to improve the P.F (Power Factor) to 0.90?
Motor input = P = 5 kW
Original P.F = Cosθ1 = 0.75
Final P.F = Cosθ2 = 0.90
θ1 = Cos-1 = (0.75) = 41°.41; Tan θ1 = Tan (41°.41) = 0.8819
θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843
Required Capacitor kVAR to improve P.F from 0.75 to 0.90
Required Capacitor kVAR = P (Tan θ1 – Tan θ2)
= 5kW (0.8819 – 0.4843)
= 1.99 kVAR
And Rating of Capacitors connected in each Phase
1.99/3 = 0.663 kVAR
ravu answered 3 years ago Step 1 – Calculate Actual Load (kW)
(Load) Power kW = Volts V x √3 x Current I x Power factor Pf
Step 2 – Calculate Required Power Factor Correction (kVAr)
Power Factor Correction kVAr = Power kW (TanΦi – TanΦd)
Φi = Cos-1 Initial Power Factor Pf
Φd = Cos-1 Required Power Factor Pf
Step 3 – Calculate Actual Power Factor Correction [kAVrl
Actual Power Factor Correction Pf = Cos (Tan-1 (TanΦi – Correction kVAr/Power kW ))
Typical Power Factor Correction available in multiples of 25kVAR’s, for further details please Contact Blakley Electrics Technical Department.
You are welcome to read more Important Interview Questions and MCQs.