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How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor Improvement  (Easiest way ever)

Example: 1

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?

Solution #1 (By Simple Table Method)

Motor Input = 5kW

From Table, Multiplier to improve PF from 0.75 to 0.90 is .398

Required Capacitor kVAR to improve P.F from 0.75 to 0.90

Required Capacitor kVAR = kW x Table 1 Multiplier of 0.75 and 0.90

= 5kW x .398

= 1.99 kVAR

And Rating of Capacitors connected in each Phase

1.99/3 = 0.663 kVAR

Solution # 2 (Classical Calculation Method)

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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