Answer
$ x_{\text {destroyod reheat }}=214.3\text{ kJ/kg}$
$ x_{\text {destroyed regen }}=47.8\text{ kJ/kg}$
Work Step by Step
From Problem 10-55 and the steam tables,$$
\begin{aligned}
y & =0.2016 \\
s_3 & =s_{f @ 0.8 \mathrm{~MPa}}=2.0457 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
s_5 & =s_6=6.7585 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
s_7 & =7.8692 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
s_1 & =s_2=s_{f @ 10 \mathrm{~kPa}}=0.6492 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
q_{\text {reheat }} & =h_7-h_6=3481.3-2812.7\\&=668.6 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Then the exergy destruction associated with reheat and regeneration processes are $$
\begin{aligned}
& x_{\text {destroyed reheat }}=T_0\left(s_7-s_6+\frac{q_{R, 67}}{T_R}\right) \\
& =(290 \mathrm{~K})\left(7.8692-6.7585+\frac{-668.6 \mathrm{~kJ} / \mathrm{kg}}{1800 \mathrm{~K}}\right) \\
& =214.3 \mathrm{~kJ} / \mathrm{kg} \\
& x_{\text {destroyed regen }}=T_0 s_{\text {gen }}=T_0\left(\sum m_e s_e-\sum m_i s_i+\frac{q_{surr}^0}{T_0}\right)=T_0\left(s_3-y s_6-(1-y) s_2\right) \\
& =(290 \mathrm{~K})[2.0457-(0.2016)(6.7585)-(1-0.2016)(0.6492)] \\
& =47.8 \mathrm{~kJ} / \mathrm{kg} \\
&
\end{aligned}
$$
