Question: Transmitted power remaining the same, if the supply voltage of DC 2 wire feeder is increased by 100%, what is the saving in copper percentage?

Calculation:

Assuming the resistance of the wire remains the same, the power P transmitted through the wire can be expressed as:

P = V^2/R

where V is the voltage and R is the resistance of the wire.

Since the transmitted power remains the same, we can write:

V1^2/R = V2^2/R

where V1 and V2 are the original and new supply voltages, respectively.

Solving for V2, we get:

V2 = sqrt(2) x V1

Therefore, the new voltage is 1.414 times the original voltage.

The resistance R of the wire is proportional to its length L and inversely proportional to its cross-sectional area A, i.e.,

R = rho x L/A

where rho is the resistivity of the wire material.

If the cross-sectional area of the wire is reduced by a factor of k, the resistance becomes:

R' = rho x L/kA = kR

Therefore, the saving in copper, expressed as a percentage, is:

Saving = [(A - kA)/A] x 100% = (1 - k) x 100%

where A is the original cross-sectional area of the wire and k is the factor by which it is reduced.

Substituting k = 1/1.414 = 0.707, we get:

Saving = (1 - 0.707) x 100% = 29.3%

Therefore, the saving in copper is approximately 29.3%.

Conclusion:

Increasing the supply voltage of a DC 2 wire feeder by 100% while keeping the transmitted power constant results in a saving in copper of approximately 29.3%. This is because the new voltage allows for a smaller cross-sectional area of wire to be used while maintaining the same resistance and transmitted power.

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