DC 2-wire Feeder Analysis and Copper Saving

This problem explores the relationship between transmission voltage, current, and the amount of copper required for a DC 2-wire feeder when the power transmitted remains constant. The goal is to calculate the percentage saving in copper when the sending-end voltage is doubled.

Understanding the Initial and New Conditions

• Initial Condition: A DC 2-wire feeder has a sending-end voltage of \(V_1 = 220 \, \text{V}\). Let the power transmitted be \(P\) and the initial current be \(I_1\). The cross-sectional area of the copper conductor is \(A_1\).
• New Condition: The sending-end voltage is doubled to \(V_2 = 2 \times V_1 = 2 \times 220 = 440 \, \text{V}\). The power transmitted \(P\) remains constant. Let the new current be \(I_2\) and the new cross-sectional area of the copper conductor be \(A_2\).

Relationship Between Power, Voltage, and Current

For a DC circuit, the power transmitted (\(P\)) is the product of voltage (\(V\)) and current (\(I\)):

\[P = V \times I\]

Since the power transmitted (\(P\)) is constant in both scenarios, we can write:

\[V_1 I_1 = V_2 I_2\]

Calculating the Change in Current

We know that \(V_2 = 2V_1\). Substituting this into the equation above:

\[V_1 I_1 = (2V_1) I_2\]

Dividing both sides by \(V_1\) (since \(V_1 \neq 0\)):

\[I_1 = 2I_2\]

This implies:

\[I_2 = \frac{I_1}{2}\]

So, when the sending-end voltage is doubled while transmitting the same power, the current flowing through the feeder is halved.

Copper Saving Based on Conductor Area

The saving in copper is directly related to the reduction in the cross-sectional area of the conductor required. In many practical applications, particularly when designing for thermal limits or maintaining a consistent current density, the required cross-sectional area of a conductor is considered directly proportional to the current it needs to carry.

Thus, we assume that the cross-sectional area (\(A\)) is proportional to the current (\(I\)):

\[A \propto I\]

This means the ratio of the new area to the old area will be equal to the ratio of the new current to the old current:

\[\frac{A_2}{A_1} = \frac{I_2}{I_1}\]

Calculating the New Cross-sectional Area

Substitute the relationship \(I_2 = \frac{I_1}{2}\) into the area ratio equation:

\[\frac{A_2}{A_1} = \frac{I_1/2}{I_1}\]

\[\frac{A_2}{A_1} = \frac{1}{2}\]

Therefore, the new cross-sectional area \(A_2\) is half of the original cross-sectional area \(A_1\):

\[A_2 = \frac{A_1}{2}\]

Calculating the Percentage Saving in Copper

The saving in copper is the difference between the initial amount of copper and the new amount of copper required. Since the length of the feeder remains the same, the amount of copper (volume or weight) is directly proportional to the cross-sectional area.

Amount of copper saved \( \propto A_1 - A_2 \)

Substitute \(A_2 = \frac{A_1}{2}\):

Amount of copper saved \( \propto A_1 - \frac{A_1}{2} \)

Amount of copper saved \( \propto \frac{A_1}{2} \)

To find the percentage saving, we divide the amount saved by the initial amount and multiply by 100%:

\[\text{Percentage Saving} = \frac{\text{Amount of copper saved}}{\text{Initial amount of copper}} \times 100\%\]

\[\text{Percentage Saving} = \frac{A_1/2}{A_1} \times 100\%\]

\[\text{Percentage Saving} = \frac{1}{2} \times 100\%\]

\[\text{Percentage Saving} = 50\%\]

Thus, by doubling the sending-end voltage while keeping the transmitted power constant, there is a 50% saving in copper required for the feeder, assuming the conductor area scales directly with current.