Answer
a) $\dot{Q}_L=61,100\text{ kJ/h}$
$COP^{}=1.97$
b) $\dot{E} \dot{x}_{\dot{Q}_L}=0.463\text{ kW}$
c) $\dot{E} x_{\text {dest }}=8.14\text{ kW}$
Work Step by Step
(a) The rate of heat absorbed from the bananas is $$
\dot{Q}_L=\dot{m} c_p\left(T_1-T_2\right)=(1140 \mathrm{~kg} / \mathrm{h})\left(3.35 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)(28-12)^{\circ} \mathrm{C}=\mathbf{6 1}, \mathbf{1 0 0 k J} / \mathbf{h}
$$ The COP is $$
\mathrm{COP}=\frac{\dot{Q}_L}{\dot{W}_{\text {in }}}=\frac{(61,100 / 3600) \mathrm{kW}}{8.6 \mathrm{~kW}}=\frac{16.97 \mathrm{~kW}}{8.6 \mathrm{~kW}}=\mathbf{1 . 9 7}
$$ (b) Theminimum power input is equal to the exergy of the heat transferred from the low-temperature medium: $$
\dot{E} \dot{x}_{\dot{Q}_L}=-\dot{Q}_L\left(1-\frac{T_0}{T_L}\right)=-(16.97 \mathrm{~kW})\left(1-\frac{28+273}{20+273}\right)=\mathbf{0 . 4 6 3\ k W}
$$ where the dead state temperature is taken as the inlet temperature of the eggplants $\left(T_0=28^{\circ} \mathrm{C}\right)$ and the temperature of the low-temperature medium is taken as the average temperature of bananas $T=(12+28) / 2=20^{\circ} \mathrm{C}$.
(c) The second-law efficiency of the cycle is $$
\eta_{\mathrm{II}}=\frac{\dot{E} x_{\dot{Q}_L}}{\dot{W}_{\text {in }}}=\frac{0.463}{8.6}=0.0539=\mathbf{5 . 3 9 \%}
$$ The exergy destruction is the difference between the exergy expended (power input) and the exergy recovered (the exergy of the heat transferred from the low-temperature medium): $$
\dot{E} x_{\text {dest }}=\dot{W}_{\text {in }}-\dot{E} x_{\dot{Q}_L}=8.6-0.463=8.14\ \mathbf{k W}
$$
