When discussing Carnot engines, it is assumed that the

engine is in thermal equilibrium with the source and the sink

during the heat addition and heat rejection processes,

respectively. It is assumed that $T_H^{**}=T_H$ and $T_L^{**}=T_L$ so that

there is no external irreversibility and the thermal efficiency of

the Carnot engine is ${}_C=1-\frac{T_L}{T_H}.$ In reality, we must

maintain a reasonable temperature difference between the two

heat transfer media in order to have an acceptable heat transfer

rate. The heat transfer rates can be expressed as

$Q_H^{}=(hA{)}_H(T_H-T_H^{**})$

$Q_L^{}=(hA{)}_L(T_L^{**}-T_L)$

where $h$ and A are the heat transfer coefficient and heat

transfer surface area, respectively. When the values of $h,$

$A,T_H,$ and $T_L$ are fixed,

$($a$)$ Show that the power output will be a maximum when

$\frac{T_L^{**}}{T_H^{**}}=(\frac{T_L}{T_H}{)}^{\frac{1}{2}}$

$($b$)$ Show that the maximum net power output is

$W_{C,\text{m}\text{a}\text{x}}^{}=\frac{(hA{)}_H{T}_H}{1+\frac{(hA{)}_H}{(hA{)}_L}}[1-(\frac{T_L}{T_H}{)}^{\frac{1}{2}}{]}^2$