POWER FACTOR IMPROVEMENT..

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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Why we are normally using cylindrical shape conductors?

Example: 

Design a suitable conductor to carry a current of 100A.

Solution:

Let us assume the material is copper

copper has a current carrying capacity of 6A/sq.mm.

We know that the property of current, it only flows through the surface of the conductor.

so 1sq.mm copper has a current carrying capacity of 6A

there fore, 1A current must be carried by 0.1667sq.mm copper conductor

there fore, 100A will carried by a conductor of 16.667sq.mm area.

Design the conductor for 20sq.mm

Let us assume one conductor is cylindrical shape and another one is square prism.

Surface area of cylinder = pi x d x L + pi x (d/2)2