A continuous electric current of 2,000 amps is to be transmitted from a generato

Question

A continuous electric current of 2,000 amps is to be transmitted from a generator to a transformer located 360 feet away. A copper conductor can be installed for $4.6 per pound, will have an estimated life of 26 years, and can be salvaged for $0.9 per pound. Power loss from the conductor will be inversely proportional to the cross-sectional area of the conductor. The power loss in one hour may be expressed as 6.306/A kilowatt, where A is the cross-sectional area of the conductor in square inches. The cost of energy is $0.0817 per kilowatt-hour, the interest rate is 12.6%, and the density of copper is 555 pounds per cubic foot. Calculate the optimum cross-sectional area of the conductor. You should assume the conductor operates 24 hours a day, 365 days per year

Explanation / Answer

Power loss per year = 6.306 x24x365 /A Kw = 55240 / A

Cost of power loss per year= 0.0817 x 55240 /A = $4513/A

Volume of the power cable = A x L where A is the area of cross section in in2 and L is the length in foot.

Volume = 360 / 144 A ft3

Weight of power cable = 360 /144 A x 555 = 199800 /144A pounds

Cost of laying power cable = 199800X4.6 A = $919080/144A

Value of $1 after 26 years at 12.6% interest rate =( 1.126)26 = 13.59

Cost of laying cable after 26 yrs Including salvage value = 13.59x919080A - 0.9x199800 A

= 199800/144 A ( 13.59x4.6 - 0.9) =12310477 /144A

Cost of power saved = 4513/A ( 1+ 1.126+(1.126)2+(1.126)3+...+( 1.126)26)

= 4513 [ ( 1-(1.126)26] / A ( 1-1.126)

= 4513 ( 1-13.59) / -0.126 A = 12.6 x4513 / 0.126A = 451300 /A

For optimum cross section of wire, cost of power saved should be equal to the cost of power cable

451300 / A = 85490A

A2 = 5,27 gives A = 2.29 ft2 or 2.29 x144 in2 = 330 in2

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