Answer to Question #194786 in Electrical Engineering for hazem

Question

The load on an installation is 800 kW. 0.8 lagging PF which works for 3000 hours per annum. The tariff is 1 $ per kVA plus 0.2 Cents per kWh. If the power factor is improved to 0.9 lagging by means of loss-free capacitors costing 0.6 $ per kVAR, calculated the annual saving effected. Allow 10% per annum for interest and depreciation on capacitors.

Answer 1

Leading kVAR taken by the capacitor $P(tan \phi_1 -tan \phi_2)=800(0.75 -0.4843)=212.56$

Annual cost before P.F correction

kVA demand =$\frac{800}{0.8}=1000$

kVA demand charges $=\$1 \times 1000= \$1000$

Units consumed 1 year $= 800 \times 3000 =2400000kWh$

Energy charges 1 year =$0.002 \times 2400000=\$4800$

Total annual cost $= \$ (1000+4800)=\$5800$

Annual cost after correction.

kVA demand =$\frac{800}{0.9}=888.89$

kVA demand charges $=\$1 \times 1000= \$888.89$

Energy charges = same as before i.e $4800

Capital cost of capacitors =$\$0.6 \times212.56=\$127.536$

Annual interest and depreciation =$\$0.1 \times 127.536=\$12.7536$

Total annual cost =$\$(888.89+4800+12.7536)\$5701.6436$

Annual saving =$5800-5701.6436=\$98.3564$