Varun answered 4 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.
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Berin answered 4 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 4 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.
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