This section will calculate the kVAR rating of the capacitor required to increase the power factor to a more efficient value, while decreasing utility energy consumption.
Advantages of power factor improvement.
- Reduce energy bill.
- Increase of system capacity.
- Reduce energy loss in cable.
- Reduce cable size, transformer and switcher gear rating.
Methods of pf improvement.
- Provide shunt capacitors - for small and medium industries.
- Synchronous motors for large industries.
Capacitor bank design
Example:
Let us assume connected load 1328.25kW
Take diversity factor = 0.85
We know that,
Maximum demand = diversity factor x connected load
=> 0.85 x 1328.25 = 1129.01 kW
Consider the actual value of p.f. is 0.8 lagging
There fore, cosØ1 = 0.8
Ø1 = cos-1(0.8) = 36.870
Assume the required value of p.f. is 0.95 lagging
There fore, cosØ2 = 0.8
Ø2 = cos-1(0.95) = 18.200
kVAR Req = kVAR1 - kVAR2
= kW x tan Ø1 – kW x tan Ø2
= 1328.25 x tan (36.87) – 1325.25 x tan (18.2)
= 559.4 kVAR
There fore capacitor bank required is 600kVAR
tan (36.87) – tan (18.2) = 0.4211557
≈0.4212
So the multiplying factor is 0.4212
That is connected load x 0.4212 gives the required kVAR rating of capacitor bank.
The value of capacitor is depends on the power factor pf
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