Chapter 15, Problem 21P

To determine

To find: The exhaust heat discharge.

Answer to Problem 21P

  $1.6\times {10}^{13}J/hr$

Explanation of Solution

Given Data:

  $T_H=625°C$

  $T_L=350°C$

Output $=1.3$ GW $\eta =75\%$

Calculation:

The Carnot efficiency $e_c=1-\frac{T_L}{T_H}$

  $\begin{array}{l}{e}_c=1-\frac{(273+350)}{(273+625)}\\ e_c=1-\frac{623}{898}\\ e_c=0.306\end{array}$

The efficiency of plant

  $\begin{array}{l}e=75\%e\\ e=0.229\end{array}$

Now intake heat flow rate, $e=\frac{W}{Q_H}$

  $\begin{array}{l}{Q}_H=\frac{W}{e}\hfill \\ =\frac{1.3\times {10}^9\text{J}/\text{s}}{0.229}\times 3600\text{s}/\text{h}\hfill \\ Q_H=2.04\times {10}^{13}\text{J}/\text{hr}\hfill \end{array}$

Let’s calculate the exhausted energy

  $\begin{array}{l}\begin{array}{l}{Q}_L=Q_H-W\\ Q_L=\left(2.04\times {10}^{13}\right)-\left(1.3\times {10}^9\right)(3600)\end{array}\hfill \\ Q_L=1.6\times {10}^{13}J/hr\hfill \\ \text{Heat exhausted}{Q}_L=1.6\times {10}^{13}J/hr\hfill \end{array}$