Answer
a) $T_4=278\text{ K}$
b) $\mathrm{COP}_{\mathrm{R}}=1.49$
c) $\dot{W}_{\text {net,in }}=108.2\text{ kW}$
Work Step by Step
(a) The temperature of the helium at the turbine inlet is determined from an energy balance on the regenerator,
$$
\begin{aligned}
\dot{E}_{\text {in }}-\dot{E}_{\text {out }} & =\Delta \dot{E}_{\text {system }}^{\text { 0(steady) }}=0 \\
\dot{E}_{\text {in }} & =\dot{E}_{\text {out }} \\
\sum \dot{m}_e h_e & =\sum \dot{m}_i h_i \longrightarrow \dot{m}\left(h_3-h_4\right)=\dot{m}\left(h_1-h_6\right)
\end{aligned}
$$ or, $$
\dot{m} c_p\left(T_3-T_4\right)=\dot{m} c_p\left(T_1-T_6\right) \longrightarrow T_3-T_4=T_1-T_6
$$ Thus, $$
T_4=T_3-T_1+T_6=20^{\circ} \mathrm{C}-\left(-10^{\circ} \mathrm{C}\right)+\left(-25^{\circ} \mathrm{C}\right)=5^{\circ} \mathrm{C}=278 \mathrm{~K}
$$ (b) From the isentropic relations, $$
\begin{aligned}
& T_2=T_1\left(\frac{P_2}{P_1}\right)^{(\mathrm{k}-1) / \mathrm{k}}=(263 \mathrm{~K})(3)^{0.667 / 1.667}=408.2 \mathrm{~K}=135.2^{\circ} \mathrm{C} \\
& T_5=T_4\left(\frac{P_5}{P_4}\right)^{(\mathrm{k}-1) / \mathrm{k}}=(278 \mathrm{~K})\left(\frac{1}{3}\right)^{0.667 / 1.667}=179.1 \mathrm{~K}=-93.9^{\circ} \mathrm{C}
\end{aligned}
$$ Then the COP of this ideal gas refrigeration cycle is determined from $$
\begin{aligned}
\mathrm{COP}_{\mathrm{R}} & =\frac{q_L}{w_{\text {net,in }}}=\frac{q_L}{w_{\text {comp,in }}-w_{\text {turb,out }}}=\frac{h_6-h_5}{\left(h_2-h_1\right)-\left(h_4-h_5\right)} \\
& =\frac{T_6-T_5}{\left(T_2-T_1\right)-\left(T_4-T_5\right)}=\frac{-25^{\circ} \mathrm{C}-\left(-93.9^{\circ} \mathrm{C}\right)}{[135.2-(-10)]^{\circ} \mathrm{C}-[5-(-93.9)]^{\circ} \mathrm{C}}=\mathbf{1 . 4 9}
\end{aligned}
$$ (c) The net power input is determined from $$
\begin{aligned}
\dot{W}_{\text {net,in }} & =\dot{W}_{\text {comp,in }}-\dot{W}_{\text {turb,out }}=\dot{m}\left[\left(h_2-h_1\right)-\left(h_4-h_5\right)\right] \\
& =\dot{m} c_p\left[\left(T_2-T_1\right)-\left(T_4-T_5\right)\right] \\
& \left.=(0.45 \mathrm{~kg} / \mathrm{s})\left(5.1926 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)([135.2-(-10)]-[5-(-93.9))]\right) \\
& =\mathbf{1 0 8 . 2} \mathbf{k W}
\end{aligned}
$$
