Answer
$x_{\text {destroyed } 23}=1110\text{ kJ/kg}$
$x_{\text {destroyed } 45}=104.6\text{ kJ/kg}$
$x_{\text {destroyed } 61}=240.6\text{ kJ/kg}$
Work Step by Step
From Problem 10-31,
$$
\begin{aligned}
s_1 & =s_2=s_{f @ 20 \mathrm{kPa}}=0.8320 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
s_3 & =s_4=6.5432 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
s_5 & =s_6=7.1292 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
q_{23, \text { in }} & =3178.3-257.50=2920.8 \mathrm{~kJ} / \mathrm{kg} \\
q_{45 \text {, in }} & =3248.4-2901.0=347.3 \mathrm{~kJ} / \mathrm{kg} \\
q_{\text {out }} & =h_6-h_1=2349.7-251.42=2098.3 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Processes 1-2, 3-4, and 5-6 are isentropic. Thus, $i_{12}=i_{34}=i_{56}=\mathbf{0}$. Also,
$$
\begin{aligned}
& x_{\text {destroyed } 23}=T_0\left(s_3-s_2+\frac{q_{R, 23}}{T_R}\right)=(295 \mathrm{~K})\left(6.5432-0.8320+\frac{-2920.8 \mathrm{~kJ} / \mathrm{kg}}{1500 \mathrm{~K}}\right)=\mathbf{1 1 1 0 ~} \mathbf{k J} / \mathbf{k g} \\
& x_{\text {destroyed } 45}=T_0\left(s_5-s_4+\frac{q_{R, 45}}{T_R}\right)=(295 \mathrm{~K})\left(7.1292-6.5432+\frac{-347.3 \mathrm{~kJ} / \mathrm{kg}}{1500 \mathrm{~K}}\right)=\mathbf{1 0 4 . 6 ~ k J /} / \mathbf{k g} \\
& x_{\text {destroyed } 61}=T_0\left(s_1-s_6+\frac{q_{R, 61}}{T_R}\right)=(295 \mathrm{~K})\left(0.8320-7.1292+\frac{2098.3 \mathrm{~kJ} / \mathrm{kg}}{295 \mathrm{~K}}\right)=\mathbf{2 4 0 . 6} \mathbf{~ k J} / \mathbf{k g}
\end{aligned}
$$
