Answer
$\eta_{\text {th }}=85.0\%$
$Q_{\text {negen }} =635.6\text{ kJ}$
$W_{\text {net,out }} =465.5\text{ kJ}
$
Work Step by Step
(a) The thermal efficiency of this totally reversible cycle is determined from
$$
\eta_{\text {th }}=1-\frac{T_L}{T_H}=1-\frac{300 \mathrm{~K}}{2000 \mathrm{~K}}=\mathbf{8 5 . 0} \%
$$ (b) The amount of heat transferred in the regenerator is $$
\begin{aligned}
Q_{\text {negen }} & =Q_{41 \text { in }}=m\left(u_1-u_4\right)=m c_v\left(T_1-T_4\right) \\
& =(0.12 \mathrm{~kg})(3.1156 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(2000-300) \mathrm{K} \\
& =\mathbf{6 3 5 . 6} \mathrm{~kJ}
\end{aligned}
$$ (c) The net work output is determined from $$
\begin{aligned}
\frac{P_3 \boldsymbol{v}_3}{T_3} & =\frac{P_1 \boldsymbol{v}_1}{T_1} \longrightarrow \frac{\boldsymbol{v}_3}{\boldsymbol{v}_1}=\frac{T_3 P_1}{T_1 P_3}=\frac{(300 \mathrm{~K})(3000 \mathrm{kPa})}{(2000 \mathrm{~K})(150 \mathrm{kPa})}=3=\frac{\boldsymbol{v}_2}{\boldsymbol{v}_1} \\
s_2-s_1 & =c_v \ln \frac{T_2}{T_1} \quad+R \ln \frac{\boldsymbol{v}_2}{\boldsymbol{v}_1}=(2.0769 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln (3)\\&=2.282 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
Q_{\text {in }} & =m T_H\left(s_2-s_1\right)=(0.12 \mathrm{~kg})(2000 \mathrm{~K})(2.282 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\\&=547.6 \mathrm{~kJ} \\
W_{\text {net,out }} & =\eta_{\text {th }} Q_{\text {in }}=(0.85)(547.6 \mathrm{~kJ})=\mathbf{4 6 5 . 5} \mathrm{~kJ}
\end{aligned}
$$
