Chapter 21, Problem 36P

To determine

To estimate the power saving if the voltage is stepped up from 120 V to 1200 V and then stepped down again.

Answer to Problem 36P

Solution:

The power saving is 54.97 MW.

Explanation of Solution

Given:

Output power, $P_0=2\times {10}^{6}W$

Output voltage, V_o = 120 V

Resistance, R = 0.100 W

Efficiency, h = 99%

Formula used:

We know that the output power is given by the relation,

${\text{I}}_{\text{i}}$

Calculation:

Substituting the values of $P_o$ and $V_o$, we obtain the line current I_l

$2\times {10}^6=120\times I_l$

$I_l=16.67\times {10}^3\text{ A}$

Now, the power lost in the two transmission lines

$P_{loss}=I_{{}_l}^2\times R_l$

$P_{loss}={\left(16.67\times {10}^3\right)}^2\times 0.100\times 2=564.5\times {10}^3\text{ W}$

Now, we calculate the generated power

$P_g=P+P_{loss}$

Substituting the values of $P_o$ and P, we get

$P_g=2+55.58=57.58\text{ MW}$

Let us consider the step up transformer.

Power at the end of the transmission line,

$P_l=0.99\left(V_l\times I_l\right)=V_o\times I_o$

By rearranging the terms, we get the line current ${\text{I}}_{\text{i}}$,

$I_l=\frac{V_o\times I_o}{0.99\times V_l}=\frac{120\times 16.67\times {10}^3}{0.99\times 1200}=1.68\text{ A}$

Now, considering the efficiency of the transformer, the total power will be the summation of the power from the step down transformer and the power loss. That is

$P_{total}=\frac{2\times {10}^6}{0.99}+564.5\times {10}^3=2.58\times {10}^6\text{ W}$

Therefore, the power that passes through the transmission lines

$=\frac{2.58\times {10}^6}{0.99}\text{ =2}\text{.61}\times {10}^6\text{W}$

Now, the power saving will be

$P_{saving}=P_g-P=57.58\times {10}^6-2.61\times {10}^6=54.97\times {10}^6\text{ W}$