Chapter 25, Problem 7STP

To determine

To find: the power that the power station can produce.

To find: the power if the line is replaced by one operator.

Explanation of Solution

Given:

The voltage, $V_1=440\text{V}$ ,

The output power, $P_{out}=88\text{ kW}=88000\text{ W }$

The new voltage, $V_2=12\text{ kV}$ .

Formula used:

The expression for power can be expressed as,

  $P_{out}=V\times I$    .......(1)

Here, $P_{out}$ is the output power, $V$ is the voltage and $I$ is current.

Total power can be expressed as,

  $\begin{array}{l}\text{Input power=Output power+Losses}\\ P_{\text{in}}=P_{\text{out}}+I^{2}R\end{array}$    .......(2)

Here, R is the resistance.

Calculation:

Equation (2) can be written as,

  ${\text{P}}_{in}={\text{P}}_{out}+\frac{V^2}{R}$    .......(3)

Substituting $440\text{V}$ in $V$ and $88000\text{ W}$ in $P_{out}$ in equation (1),

  $\begin{array}{l}I=\frac{80000}{440}\\ I=181.9\text{A}\end{array}$

Substitute $181.8\text{A}$ in $I$ and $0.2\Omega$ for $R$ in equation (2),

  $\begin{array}{l}{P}_{in}=80000+{181.8}^2\times 0.2\\ P_{in}=86.6\text{ kW}\end{array}$

Substituting 12kV in V in equation (1)

  $\begin{array}{l}I=\frac{80000}{12000}\\ I=6.67\text{A}\end{array}$

Substituting 6.67 A for I in equation (2)

  $\begin{array}{l}{P}_{in}=80000+{6.67}^2\times 0.2\\ P_{in}=80.008\text{ kW}\end{array}$

Conclusion:

Hence, the required power are $86.6\text{ kW}$ and $80.008\text{ kW}$ respectively.