Question

"A continuous electric current of 2,000 amps is to be transmitted from a generator to a transformer located 260 feet away. A copper conductor can be installed for $4.2 per pound, will have an estimated life of 23 years, and can be salvaged for $1.1 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 6.377/A kilowatt, where A is the cross-sectional area of the conductor in square inches. The cost of energy is $0.088 per kilowatt-hour, the interest rate is 12.8%, 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."

Answer 1

This is the simple question based on the Kelvin's law:

We will start this question by understanding Kelvin's Electricity law:

C1+C2A+C3/A=C

C is total cost

C2 is cost dependent on area

C3 is also inversly dependent on area

C1 is constant cost like cost of one time installation

Now , for optimum area A

dC/dA =0

Hence , above eqn becomes :

C2-C3/A² =0

A is the optimal area .

We get C2= $4.2 *23 years*365*24 -$1.1

C3=6.377 +$0.0088+12.8% of $0.0088

Solving we , get

A= 2.44 square inches.