Percentage Saving in Feeder Copper Calculation

To determine the percentage saving in feeder copper when the line voltage is increased, we need to analyze the relationships between power transmitted, current, power loss, resistance, and the physical dimensions (cross-sectional area) of the copper conductors, assuming the transmission distance and power loss remain constant.

Let the initial voltage be \(V_1 = 100\) V and the increased voltage be \(V_2 = 200\) V. Let \(P\) be the constant power transmitted and \(P_{loss}\) be the constant power loss in the feeder. Let \(I_1\) and \(I_2\) be the currents at voltages \(V_1\) and \(V_2\), respectively. Let \(R_1\) and \(R_2\) be the total resistances of the 2-wire feeder at voltages \(V_1\) and \(V_2\), respectively.

Relationship between Power, Voltage, and Current

The power transmitted \(P\) in a DC system is given by: \[P = V \times I\] Since \(P\) is constant: \[P = V_1 I_1 = V_2 I_2\] \[\frac{I_2}{I_1} = \frac{V_1}{V_2}\] Given \(V_1 = 100\) V and \(V_2 = 200\) V: \[\frac{I_2}{I_1} = \frac{100}{200} = \frac{1}{2}\] So, the new current \(I_2\) is half of the original current \(I_1\): \(I_2 = \frac{I_1}{2}\).

Relationship between Power Loss, Current, and Resistance

The power loss \(P_{loss}\) in the feeder resistance is given by: \[P_{loss} = I^2 R\] Since the power loss \(P_{loss}\) is constant: \[P_{loss} = I_1^2 R_1 = I_2^2 R_2\] Substitute \(I_2 = \frac{I_1}{2}\): \[I_1^2 R_1 = \left(\frac{I_1}{2}\right)^2 R_2\] \[I_1^2 R_1 = \frac{I_1^2}{4} R_2\] Assuming \(I_1 \neq 0\), we can divide both sides by \(I_1^2\): \[R_1 = \frac{R_2}{4}\] This means the new resistance \(R_2\) is four times the original resistance \(R_1\): \(R_2 = 4R_1\).

Relationship between Resistance and Conductor Area

The resistance \(R\) of a conductor is given by: \[R = \rho \frac{L}{A}\] where \(\rho\) is the resistivity of the material (copper), \(L\) is the length of the conductor, and \(A\) is its cross-sectional area. For a 2-wire system, \(L\) is the total length (out and return) and \(A\) is the area of each wire (assuming both are the same). Since the power distance is the same, the length \(L\) is constant. The resistivity \(\rho\) of copper is also constant. So, \(R\) is inversely proportional to \(A\): \[R_1 = \rho \frac{L}{A_1} \quad \text{and} \quad R_2 = \rho \frac{L}{A_2}\]

Finding the Relationship between Areas

We found that \(R_2 = 4R_1\). Substitute the expressions for \(R_1\) and \(R_2\): \[\rho \frac{L}{A_2} = 4 \left(\rho \frac{L}{A_1}\right)\] Since \(\rho\) and \(L\) are constant and non-zero, we can cancel them from both sides: \[\frac{1}{A_2} = \frac{4}{A_1}\] This implies \(A_1 = 4A_2\), or \(A_2 = \frac{A_1}{4}\). The new cross-sectional area \(A_2\) is one-fourth of the original area \(A_1\).

Calculating Percentage Saving in Copper

The amount (volume or weight) of copper in the feeder is directly proportional to its cross-sectional area, as the length is constant. Let the original amount of copper be proportional to \(A_1\). Let the new amount of copper be proportional to \(A_2\). The saving in copper is proportional to the difference in area: \(A_1 - A_2\). Saving \( = A_1 - \frac{A_1}{4} = \frac{4A_1 - A_1}{4} = \frac{3A_1}{4}\).

The percentage saving is calculated as: \[\text{Percentage Saving} = \frac{\text{Saving}}{\text{Original Amount}} \times 100\%\] \[\text{Percentage Saving} = \frac{\frac{3A_1}{4}}{A_1} \times 100\%\] \[\text{Percentage Saving} = \frac{3}{4} \times 100\%\] \[\text{Percentage Saving} = 0.75 \times 100\%\] \[\text{Percentage Saving} = 75\%\]

Therefore, the percentage saving in feeder copper is 75%.