# Answer to Question #194786 in Electrical Engineering for hazem

Question #194786

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.

1
2021-05-19T04:37:02-0400

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"

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