Chapter 6.11, Problem 154RP

To determine

The maximum net power output.

Answer to Problem 154RP

The maximum net power output is ${\stackrel{.}{W}}_{\max }=\frac{{\left(hA\right)}_H{T}_H}{1+\raisebox{1ex}{${\left(hA\right)}_H$}\!\left/ \!\raisebox{-1ex}{${\left(hA\right)}_L$}\right.}{\left\{1-{\left(\frac{T_L}{T_H}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right\}}^2$

Explanation of Solution

Write the heat gain transfer rate equation $\left({\dot{Q}}_H\right)$.

${\dot{Q}}_H={\left(hA\right)}_H\left(T_H-T_H^{*}\right)$ (I)

Here, heat transfer coefficient with heat gain transfer per unit area for hot body is ${\left(hA\right)}_H$ and temperature of hot reservoir is $T_H^{*}$ or $T_H$.

Write the heat loss transfer rate equation $\left({\dot{Q}}_L\right)$.

${\dot{Q}}_L={\left(hA\right)}_L\left(T_L-T_L^{*}\right)$ (II)

Here, heat transfer coefficient with heat loss transfer per unit area for hot body is ${\left(hA\right)}_L$ and temperature of cold reservoir is $T_L^{*}$ or $T_L$.

Write the work done equation

$\dot{W}={\eta }_{th}{\dot{Q}}_H$ (III)

Here, efficiency of the heat engine is ${\eta }_{th}$.

Write the reversible cycle relation between the hot and cold temperature reservoir.

$\frac{T_H^{*}}{T_L^{*}}=\frac{{\dot{Q}}_H}{{\dot{Q}}_L}$ (IV)

Write the temperature relation between the reservoirs.

$\frac{T_L^{*}}{T_H}=\frac{T_L^{*}}{T_H^{*}}\frac{T_H^{*}}{T_H}=r\left(1-x\right)$ (V)

Here, variables are $r$ and $x$.

Conclusion:

Substitute ${\left(hA\right)}_H\left(T_H-T_H^{*}\right)$ for ${\dot{Q}}_H$ and $\left(1-\frac{T_L^{*}}{T_H^{*}}\right)$ for ${\eta }_{th}$ in Equation (III).

$\begin{array}{c}\dot{W}=\left(1-\frac{T_L^{*}}{T_H^{*}}\right){\left(hA\right)}_H\left(T_H-T_H^{*}\right)\\ =\left(1-\frac{T_L^{*}}{T_H^{*}}\right){\left(hA\right)}_H\left(1-\frac{T_H^{*}}{T_H}\right)T_H\\ ={\left(hA\right)}_H{T}_H\left(1-\frac{T_L^{*}}{T_H^{*}}\right)\left(1-\frac{T_H^{*}}{T_H}\right)\end{array}$ (VI)

Substitute $\raisebox{1ex}{$T_L^{*}$}\!\left/ \!\raisebox{-1ex}{$T_H^{*}$}\right.$ for $r$ and $1-\raisebox{1ex}{$T_H^{*}$}\!\left/ \!\raisebox{-1ex}{$T_H$}\right.$ for $x$ in Equation (VI).

$\dot{W}={\left(hA\right)}_H{T}_H\left(1-r\right)x$

Substitute $\raisebox{1ex}{$T_L^{*}$}\!\left/ \!\raisebox{-1ex}{$T_H^{*}$}\right.$ for $r$, ${\left(hA\right)}_H\left(T_H-T_H^{*}\right)$ for ${\dot{Q}}_H$, and ${\left(hA\right)}_L\left(T_L-T_L^{*}\right)$ for ${\dot{Q}}_L$ in Equation (V).

$\begin{array}{l}\frac{1}{r}=\frac{{\left(hA\right)}_H\left(T_H-T_H^{*}\right)}{{\left(hA\right)}_L\left(T_L-T_L^{*}\right)}\\ =\frac{{\left(hA\right)}_H{T}_H\left(1-\raisebox{1ex}{$T_H^{*}$}\!\left/ \!\raisebox{-1ex}{$T_H$}\right.\right)}{{\left(hA\right)}_L{T}_H\left(\raisebox{1ex}{$T_L^{*}$}\!\left/ \!\raisebox{-1ex}{$T_H$}\right.-\raisebox{1ex}{$T_L^{}$}\!\left/ \!\raisebox{-1ex}{$T_H$}\right.\right)}\end{array}$

Substitute $\raisebox{1ex}{$T_L^{*}$}\!\left/ \!\raisebox{-1ex}{$T_H^{*}$}\right.$ for $r$ and $1-\raisebox{1ex}{$T_H^{*}$}\!\left/ \!\raisebox{-1ex}{$T_H$}\right.$ for $x$ in Equation (IV).

$\frac{1}{r}=\frac{{\left(hA\right)}_{H}x}{{\left(hA\right)}_L\left[r\left(1-x\right)-\raisebox{1ex}{$T_L^{}$}\!\left/ \!\raisebox{-1ex}{$T_H$}\right.\right]}$

$x=\frac{r-T_L/T_H}{r\left[{\left(hA\right)}_H/{\left(hA\right)}_L+1\right]}$ (VII)

Substitute Equation (VII) in (IV).

$\frac{\stackrel{.}{W}}{{\left(hA\right)}_H{T}_H}={\left(hA\right)}_H{T}_H\left(1-r\right)\frac{r-\raisebox{1ex}{$T_L$}\!\left/ \!\raisebox{-1ex}{$T_H$}\right.}{r\left[\raisebox{1ex}{${\left(hA\right)}_H$}\!\left/ \!\raisebox{-1ex}{${\left(hA\right)}_L$}\right.+1\right]}$ (VIII)

Taking the partial derivative $\frac{\partial \stackrel{.}{W}}{\partial r}$ holding everything else constant and setting it equal to zero gives

$r=\frac{T_L^{*}}{T_H^{*}}={\left(\frac{T_L}{T_H}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ (IX)

Substitute Equation (IX) into (VIII).

${\stackrel{.}{W}}_{\max }=\frac{{\left(hA\right)}_H{T}_H}{1+\raisebox{1ex}{${\left(hA\right)}_H$}\!\left/ \!\raisebox{-1ex}{${\left(hA\right)}_L$}\right.}{\left\{1-{\left(\frac{T_L}{T_H}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right\}}^2$

Thus, the maximum net power output is ${\stackrel{.}{W}}_{\max }=\frac{{\left(hA\right)}_H{T}_H}{1+\raisebox{1ex}{${\left(hA\right)}_H$}\!\left/ \!\raisebox{-1ex}{${\left(hA\right)}_L$}\right.}{\left\{1-{\left(\frac{T_L}{T_H}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right\}}^2$.