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. That is, it is assumed that
\(T_{H}^{*}=T_{H}\) and \(T_{L}^{*}=T_{L}\) so that there is no external
irreversibility. In that case, the thermal efficiency of the Carnot engine is
\(\eta_{C}=1-T_{L} / T_{H}\) In reality, however, we must maintain a reasonable
temperature difference between the two heat transfer media in order to have an
acceptable heat transfer rate through a finite heat exchanger surface area.
The heat transfer rates in that case can be expressed as
$$\begin{array}{l}
\dot{Q}_{H}=\left(h_{A}\right)_{H}\left(T_{H}-T_{H}^{*}\right) \\
\dot{Q}_{L}=(h A)_{L}\left(T_{L}^{*}-T_{L}\right)
\end{array}$$
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,
show that the power output will be a maximum when
$$\frac{T_{L}^{*}}{T_{H}^{*}}=\left(\frac{T_{L}}{T_{H}}\right)^{1 / 2}$$
Also, show that the maximum net power output in this case is
$$\dot{W}_{C, \max }=\frac{(h A)_{H} T_{H}}{1+(h A)_{H} /(h
A)_{L}}\left[1-\left(\frac{T_{L}}{T_{H}}\right)^{1 / 2}\right]^{2}$$
