Answer
$x_{\text {destroyed cycle }}=1155\text{ kJ/kg}$
Work Step by Step
From Problem 10-45, $q_{\text {in }}=2692.2 \mathrm{~kJ} / \mathrm{kg}$ and $q_{\text {out }}=1675.7 \mathrm{~kJ} / \mathrm{kg}$. Then the exergy destruction associated with this regenerative cycle is $$
x_{\text {destroyed cycle }}=T_0\left(\frac{q_{\text {out }}}{T_L}-\frac{q_{\text {in }}}{T_H}\right)=(290 \mathrm{~K})\left(\frac{1675.7 \mathrm{~kJ} / \mathrm{kg}}{290 \mathrm{~K}}-\frac{2692.2 \mathrm{~kJ} / \mathrm{kg}}{1500 \mathrm{~K}}\right)=1155 \mathrm{~kJ} / \mathrm{kg}
$$
