Answer
$351.3\text{ kJ/kg}$
Work Step by Step
$$
\begin{aligned}
s_1 & =s_2=s_{f @ 50 \mathrm{kPa}}=1.0912 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
s_3 & =s_4=6.5412 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
q_{\text {in }} & =2650.72 \mathrm{~kJ} / \mathrm{kg} \\
q_{\mathrm{out}} & =1931.8 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Processes 1-2 and 3-4 are isentropic. Thus, $i_{12}=\mathbf{0}$ and $i_{34}=\mathbf{0}$. Also, $$
\begin{aligned}
& x_{\text {destroyed } 23}=T_0\left(s_3-s_2+\frac{q_{R, 23}}{T_R}\right)=(290 \mathrm{~K})\left(6.5412-1.0912+\frac{-2650.8 \mathrm{~kJ} / \mathrm{kg}}{1500 \mathrm{~K}}\right)=1068 \mathrm{~kJ} / \mathrm{kg} \\
& x_{\text {destroyed } 41}=T_0\left(s_1-s_4+\frac{q_{R, 41}}{T_R}\right)=(290 \mathrm{~K})\left(1.0912-6.5412+\frac{1931.8 \mathrm{~kJ} / \mathrm{kg}}{290 \mathrm{~K}}\right)=351.3 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$
