Capacitor Bank Calculation

Power Factor Improvement refers to the actions taken to increase the efficiency of an electrical system by minimizing the phase difference between voltage and current.

Here Total Load = 1245 KW
pf                         = 0.85
Required pf       = 0.93

Cos x = 0.85
Cos y = 0.95
Required capacitor KVAR = Kw * (tan x – tan y)

y = cos-1 0.85
= 31.8
x = cos-1 0.93
= 21.5

Required KVAR = 1245 (tan 31.8 – tan 21.5)
= 1245 (0.62 – 0.39)
= 1245 * 0.23
= 285 KVAR

So we choose 3 banks of 100 kVAR.
To find Capacitance, C = KVAR / 2 pi f v2

see below analysis!

Significant Role

1. Voltage Support
2. Reduced Transmission Losses
3. Increased System Capacity
4. Cost Savings
5. Improved System Stability
6. Enhancement of Equipment Lifespan
7. Flicker and Voltage Fluctuations Reduction