Question

If the power output in the previous problem is 1000 MW and line resistance is 2.00 Ω, what were the old and new line losses?

Short answer

The old line loss is for 335 kV voltage output and the new line loss is for the voltage output of 750 kV.

Step-by-step solution

Definition of power

Power is defined as the rate at which one performs work. It's also known as the rate of energy consumption.

Given information and Formula to be used

The power output is$P_{Loss}=1000MW\left(\frac{{10}^{6}W}{1MW}\right)={10}^{9}W$

The resistance of the line is $R=2\Omega$

Voltage in the secondary coil of the transformer is $V_S^1=335kV\left(\frac{1000V}{1kV}\right)=3.35\times {10}^{5}V$and$V_S^2=750kV\left(\frac{1000V}{1kV}\right)=7.50\times {10}^{5}V$

The power loss in the circuit can be expressed as,

$P_{Loss}=I^{2}R$……………(1)

Where P, I, and V are the power, voltage, and current in the circuit, respectively.

What is the ratio of input to output current?

Substituting the given data in equation (1) can be expressed as,

When the value of output voltage is, $V_S^1=3.35\times {10}^{5}V$, then

\(\begin{align}{}{P_{{\rm{Loss }}}} &= {\left( {\frac{P}{{V_S^1}}} \right)^2}R\\ & = {\left( {\frac{{{{10}^9}{\rm{W}}}}{{3.35 \times {{10}^5}\;{\rm{V}}}}} \right)^2} \times 2\;\Omega \\ & = 1.782 \times {10^7}\;{\rm{W}}\left( {\frac{{1\;{\rm{MW}}}}{{{{10}^6}\;{\rm{W}}}}} \right)\\ & = 17.82\;{\rm{MW}}\end{align}\)

When the value of output voltage is, $V_S^2=7.50\times {10}^{5}V$, then

$$\begin{gathered} P_{Loss}={\left(\frac{P}{V_S^2}\right)}^{2}R \\ =\left(\frac{{10}^{9}W}{7.50\times {10}^{5}V}\right)\times 2\Omega \\ =3.56\times {10}^{6}W\left(\frac{1MW}{{10}^{6}W}\right) \\ =3.56MW \end{gathered}$$

Therefore, the old line loss is $17.82MW$for 335 kV voltage output and the new line loss is $3.56MW$for the voltage output of 750 kV.