Answer
$172.3\text{ kJ/kg}$
Work Step by Step
$$
\begin{aligned}
s_1 & =s_2=s_{f @10\text{ kPa}}=0.6492 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
s_3 & =s_4=6.5995 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
q_{\text {in }} & =3173.2 \mathrm{~kJ} / \mathrm{kg} \\
q_{\text {out }} & =1897.9 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Processes 1-2 and 3-4 are isentropic. Thus, $i_{12}=0$ and $i_{44}=0$. Also, $$
\begin{aligned}
& x_{\text {destroyed 23 }}=T_0\left(s_3-s_2+\frac{q_{R, 23}}{T_R}\right)=(290 \mathrm{~K})\left(6.5995-0.6492+\frac{-3173.2 \mathrm{~kJ} / \mathrm{kg}}{1500 \mathrm{~K}}\right)=1112.1 \mathrm{~kJ} / \mathrm{kg} \\
& x_{\text {destroyed } 41}=T_0\left(s_1-s_4+\frac{q_{R, 41}}{T_R}\right)=(290 \mathrm{~K})\left(0.6492-6.5995+\frac{1897.9 \mathrm{~kJ} / \mathrm{kg}}{290 \mathrm{~K}}\right)=172.3 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$
