Feeder Transmission Efficiency Explained

Understanding the transmission efficiency of a DC 2-wire feeder is crucial in electrical engineering. This problem involves calculating the efficiency based on the voltage drop across the feeder conductors. A DC 2-wire feeder, as the name suggests, uses two conductors to transmit direct current power from the sending end to the receiving end.

Voltage Drop Analysis for a DC 2-Wire Feeder

In any power transmission system, there is always some voltage drop due to the resistance of the conductors. For a DC 2-wire feeder, power flows through two conductors: one for the supply (go wire) and one for the return path (return wire). The problem states that the voltage drop per feeder conductor is 2%.

• Let \(V_S\) be the sending end voltage of the feeder.
• The voltage drop across one feeder conductor is given as 2% of the sending end voltage.
• Voltage drop per conductor \( = 0.02 \times V_S \).

Since it's a 2-wire feeder, there are two conductors contributing to the total voltage drop from the sending end to the receiving end. The total voltage drop is the sum of the drops in both the go and return conductors.

• Total voltage drop in the feeder \( = \text{Voltage drop in go conductor} + \text{Voltage drop in return conductor} \).
• Total voltage drop \( = (0.02 \times V_S) + (0.02 \times V_S) \).
• Total voltage drop \( = 0.04 \times V_S \).

This means that 4% of the sending end voltage is lost as a drop within the feeder conductors.

Receiving End Voltage Determination

The voltage available at the receiving end of the feeder (\(V_R\)) will be the sending end voltage minus the total voltage drop along the feeder.

• Receiving end voltage \(V_R = V_S - \text{Total voltage drop}\).
• \(V_R = V_S - (0.04 \times V_S)\).
• \(V_R = V_S (1 - 0.04)\).
• \(V_R = 0.96 \times V_S\).

So, the receiving end voltage is 96% of the sending end voltage.

Calculating Feeder Transmission Efficiency

The transmission efficiency (\(\eta\)) of a feeder is defined as the ratio of the power delivered at the receiving end to the power supplied at the sending end, usually expressed as a percentage. For a DC system, power (\(P\)) is calculated as voltage (\(V\)) multiplied by current (\(I\)). Since the current (\(I\)) flowing through the feeder is constant from the sending to the receiving end (assuming no loads tapped off in between, which is typical for such problems), the efficiency can be simplified based on voltages.

• Sending end power \(P_S = V_S \times I\).
• Receiving end power \(P_R = V_R \times I\).
• Transmission efficiency \(\eta = \frac{P_R}{P_S} \times 100\%\).
• \(\eta = \frac{V_R \times I}{V_S \times I} \times 100\%\).
• \(\eta = \frac{V_R}{V_S} \times 100\%\).

Substituting the value of \(V_R\) we found:

• \(\eta = \frac{0.96 \times V_S}{V_S} \times 100\%\).
• \(\eta = 0.96 \times 100\%\).
• \(\eta = 96\%\).

Therefore, the transmission efficiency of the DC 2-wire feeder is 96%.