Question

The capacitor size in kVAr required to improve power factor from 0.90 to unity for 900 kW Load will be? ​

Answer 1

Given:

Power (P) = 900kW

Original PF = cosθ₁ = 0.90

Final PF = cosθ₂ = 1

To find:

Required Capacitor kVAR to improve PF from 0.90 to 1

Solution:

cosθ₁ = 0.90

θ₁ = cos⁻¹(0.90) = 25.84°

tanθ₁ = tan(25.84) = 0.4843

cosθ₂ = 1

θ₂ = cos⁻¹(1) = 0°

tanθ₂ = tan(0) = 0

Required Capacitor kVAR to improve PF from 0.90 to 1

Required Capacitor kVAR = P (tanθ₁ – tanθ₂)

Required Capacitor kVAR = 900kW (0.4843 - 0)

Required Capacitor kVAR = 900kW × 0.4843

Required Capacitor kVAR = 435.78 kVAR

Therefore, the capacitor size inkVAR is 435.78 kVAR