Chapter 15, Problem 28P

To determine

To find: That rate of coal burn.

Answer to Problem 28P

  $158.2\text{kg/s}$

Explanation of Solution

Given Data:

  

  $T_1=670,T_2=440,T_3=430,T_4=290$

Output $W1+W2=1100\text{MW}$

Efficiency of mechanic $=60\%$

Formula used:

  $W_1+W_2=eQ+e^{\prime }Q(1-e)$

Calculation:

  $\begin{array}{l}\begin{array}{cc}e& =65\text{\%}\left[1-\frac{713\text{K}}{943\text{K}}\right]\\ & =0.1463\end{array}\\ \begin{array}{cc}e\text{'}& =60\text{\%}\left[1-\frac{563\text{K}}{703\text{K}}\right]\\ & =0.1194\end{array}\end{array}$

Now,

  $\begin{array}{l}Q\left[e+e^{\prime }(1-e)\right]=1100\text{MW}\\ Q[0.1463+0.1194(1-0.1463)]=1100\text{MW}\\ Q[0.24823]=1100\times {10}^6\text{W}\\ Q=\frac{1100\times {10}^6\text{W}}{0.24823}\\ Q=4431.37\times {10}^6\text{W}\left(\frac{\text{J/s}}{\text{W}}\right)\end{array}$

Calculating Rate,

  $\begin{array}{l}R=\frac{Q}{\left(2.8\times {10}^7\text{J}/\text{kg}\right)}\\ R=\frac{4431.37\times {10}^6\text{J}/\text{s}}{2.8\times {10}^7\text{J}/\text{kg}}\\ =158.2\text{kg/s}\end{array}$