A dc transmission facility is modeled by the approximate circuit shown in Fig. 2.21. If the load voltage is known to be Vload = 458.3 kV. we wish to find the voltage at the sending end of the line and the power loss in the line.

Question

A dc transmission facility is modeled by the approximate circuit shown in Fig. 2.21. If the load voltage is known to be [latex]V_{load}[/latex] = 458.3 kV, we wish to find the voltage at the sending end of the line and the power loss in the line.

Accepted Answer

Knowing the load voltage and load resistance, we can obtain the line current using Ohm’s law:

[latex]I_{L}[/latex] = 458.3k/220
= 2.083 kA

The voltage drop across the line is

[latex]V_{line}[/latex] = [latex](I_{L})(R_{line})[/latex]
= 41.66 kV

Now, using KVL,

[latex]V_{S}\,=\,V_{line}\,+\,V_{load}[/latex]
= 500 kV

Note that since the network is simply a voltage-divider circuit, we could obtain [latex]V_{S}[/latex] immediately from our knowledge of [latex]R_{line},\,R_{load},[/latex] and [latex]V_{load}.[/latex] That is,

[latex]V_{load}\,=\,[\frac{R_{load}}{R_{load}\,+\,R_{line}}]V_{S}[/latex]

and [latex]V_{S}[/latex] is the only unknown in this equation.

The power absorbed by the line is

[latex]P_{line}\,=\,I^{2}_{L}R_{line}[/latex]
= 86.79 MW

Question 2.16: A dc transmission facility is modeled by the approximate cir...

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