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PAYBACK CALCULATIONS FOR POWER FACTOR IMPROVEMENT at CHPP, Zambia.

A. Deduction in demand charges by improving power factorSystem present demand ( kVA1 ) = 2500 kVA

System present power factor (Cos 1) = 0.7

Power factor (Cos ) = kW/kVA

kW = kVA x Cos

at present power factor kW = kVA 1 x Cos 1 = 2500 x 0.7 = 1750

We are improving the power factor from 0.7 to 0.95 by installing the capacitor banks at CHPP MCC

Now system demand (kVA2) = kVA2

Power factor improvement (Cos2) = 0.95

Now at this power factor kW = kVA 2 x Cos 2

= kVA2 x 0.95

As at the same power required demands are different depending on the power factor

kVA 1 x Cos 1 = kVA 2 x Cos 2

1500 = kVA 2 x 0.95

kVA 2 = 1750/0.95 = 1842

after improving the power factor required demand at the same power is 1842 kVA instead of 2500 kVA

Therefore reduction in demand charges in energy bill due to reduction in maximum demand by

improving power factor from 0.7 to 0.9 shall be

Saving in demand charges per month = Rs. 100 x (2500 1842) = Rs. 1,84,200 per month

Net reduction per annum = 1,84,200 x 12 = 22,10,400

B. Cost investment for installation of power factor improving equipment There are two sets of 300 kVAr and one set of 250 kVAr capacitor banks for each MCC

Cost of capacitor banks = Rs. 1000000

Switching and associated equipment and installation = Rs. 200000

Total cost = Rs. 1200000

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C. Pay back calculationTotal cost of the capacitor banks = Rs. 1200000

Annual depreciation and interest @ 20% = 1200000 x 0.2 = Rs. 240000

Net annual saving = Rs. 2210400 240000 = Rs. 1970400

Net monthly saving = Rs. 1970400 / 12 = Rs. 164200

Therefore, payback period = 1200000 / 164200 = 7.3 months

This may be consider that as demand (kVA) got reduced by improving power factor, current on the

system also reduced further, I2 R losses also reduced. So there is further reduction in energy bill by

decreasing losses on the system

Advantages of improving power factor:

1. Reduced system losses, and the losses in the cables, lines, and feeder circuits and hence lowersizes could be opted.

2. Improved system voltages, thus enable maintaining rated voltage to motors, pumps and otherequipment. The voltage drop in supply conductors is a resistive loss, and wastes power heating

the conductors. A 5% drop in voltage means that 5% of your power is wasted as heat before it

even reaches the motor. Improving the power factor, especially at the motor terminals, can

improve your efficiency by reducing the line current and the line losses.

3. Improved voltage regulation.4. Increased system capacity, by release of kVA capacity of transformers and cables for the same

kW , thus permitting additional loading without immediate augmentation.