Chapter 15, Problem 25P

To determine

To find: The electric heat discharge per second.

Answer to Problem 25P

  $900\times {10}^6\text{W}$

Explanation of Solution

Given Data:

Efficiency $=38\%$

Output $=550$ MW

Formula Used:

  $e=\frac{W}{Q_H}$

Calculation:

  $\begin{array}{l}e=\frac{W}{Q_H}\\ e=\frac{W}{W+Q_L}\\ \begin{array}{cc}{Q}_L& =W\left(\frac{1}{e}-1\right)\\ \frac{Q_L}{t}& \begin{array}{l}\\ \\ =\frac{W}{t}\left(\frac{1}{e}-1\right)\\ =550\text{MW}\left(\frac{1}{0.38}-1\right)\\ =900\times {10}^6\text{W}\end{array}\end{array}\end{array}$