Chapter 8.8, Problem 66P

a)

To determine

The rate of exergy destroyed during the process and the exit temperature $\left(T_4\right)$ of the air when the outer surface of air conditioner is insulated.

Answer to Problem 66P

The rate of exergy destroyed during the process is $\underset{\_}{0.594\text{kW}}$.

The exit temperature $\left(T_4\right)$ of the air is $\underset{\_}{\text{257}\text{.2}\text{K}\text{or}\left(-15.8\text{°C}\right)}$.

Explanation of Solution

Draw the schematic diagram of the flow of refrigerant-134a through evaporator section as shown in Figure (1).

Write the expression for the mass balances equation for the heat exchanger.

${\dot{m}}_{\text{in}}-{\dot{m}}_{\text{out}}=\Delta {\dot{m}}_{\text{system}}$

Here, mass flow rate of refrigerant at inlet is ${\dot{m}}_{\text{in}}$, mass flow rate of refrigerant at outlet is ${\dot{m}}_{\text{out}}$, and net mass flow rate of refrigerant through system is $\Delta {\dot{m}}_{\text{system}}$.

Since net mass flow rate of refrigerant-134a and air through system is 0.

${\dot{m}}_{\text{in}}-{\dot{m}}_{\text{out}}=0$

From Figure (1), the mass flow rate of refrigerant-134a at $\text{state}\text{1}\text{and}\text{2}$ is equal.

$\begin{array}{c}{\dot{m}}_1={\dot{m}}_2\\ ={\dot{m}}_{\text{R}}\end{array}$ (I)

Here, initial and final mass flow rate of refrigerant at $\text{state}\text{1}\text{and}\text{2}$ is ${\dot{m}}_1\text{and}{\dot{m}}_2$ respectively, and mass flow rate of refrigerant is ${\dot{m}}_{\text{R}}$.

From Figure (1), the mass flow rate of air at $\text{state}\text{3}\text{and}\text{4}$ is equal.

$\begin{array}{c}{\dot{m}}_3={\dot{m}}_4\\ ={\dot{m}}_{\text{a}}\end{array}$ (II)

Here, mass flow rate of air at $\text{state}3\text{and}\text{4}$ is ${\dot{m}}_3\text{and}{\dot{m}}_4$ respectively, and mass flow rate of air is ${\dot{m}}_{\text{a}}$.

Write the expression for the enthalpy at state 1 $\left(h_1\right)$.

$h_1=h_f+x_1{h}_{fg}$ (III)

Write the expression for the entropy at state 1$\left(s_1\right)$.

$s_1=s_f+x_1{s}_{fg}$ (IV)

Write the expression for the mass flow rate of air $\left({\dot{m}}_{\text{a}}\right)$.

${\dot{m}}_{\text{a}}=\frac{P_3{\dot{V}}_3}{R{T}_3}$ (V)

Here, gas constant of air is $R$, temperature at state 3 is $T_3$, pressure at state 3 is $P_3$, and volume flow rate of air at state 3 is ${\dot{V}}_3$.

Write the expression for energy balance for the heat exchanger

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}$ (VI)

Here, rate of net energy transfer in to the control volume is ${\dot{E}}_{\text{in}}$, rate of net energy transfer exit from the control volume is ${\dot{E}}_{\text{out}}$ and rate of change in internal energy of system is $\Delta {\dot{E}}_{\text{system}}$.

Substitute 0 for $\Delta {\dot{E}}_{\text{system}}$ in Equation (VI), and re-write the energy balance as follows:

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3={\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4\\ {\dot{m}}_2{h}_2-{\dot{m}}_1{h}_1={\dot{m}}_3{h}_3-{\dot{m}}_4{h}_4\end{array}$

$\begin{array}{l}{\dot{m}}_2\left(h_2-h_1\right)={\dot{m}}_3\left(h_3-h_4\right)\\ {\dot{m}}_{\text{R}}\left(h_2-h_1\right)={\dot{m}}_{\text{a}}\left(h_3-h_4\right)\\ ={\dot{m}}_{\text{a}}{c}_p\left(T_3-T_4\right)\\ T_4=T_3-\frac{{\dot{m}}_{\text{R}}\left(h_2-h_1\right)}{{\dot{m}}_{\text{a}}{c}_p}\end{array}$ (VII)

Here, mass flow rate at $\text{state}\text{1,2,3}\text{and}\text{4}$ is ${\dot{m}}_1,{\dot{m}}_2,{\dot{m}}_3\text{and}{\dot{m}}_4$ respectively, enthalpy at $\text{state}\text{1,2,3}\text{and}\text{4}$ is $h_1,h_2,h_3\text{and}{h}_4$ respectively, constant pressure specific heat of air at room temperature is $c_p$, temperature at $\text{state}\text{3}\text{and}\text{4}$ is $T_3\text{and}{T}_4$ respectively.

Write the expression for the entropy balance for the steady flow system as;

${\dot{S}}_{\text{in}}-{\dot{S}}_{\text{out}}+{\dot{S}}_{\text{gen}}=\Delta {\dot{S}}_{\text{system}}$ (VIII)

Here, rate of entropy generation is ${\dot{S}}_{\text{gen}}$, rate of change in entropy is $\Delta {\dot{S}}_{\text{system}}$, rate of entropy at inlet condition is ${\dot{S}}_{\text{in}}$, and rate of entropy at outlet condition is ${\dot{S}}_{\text{out}}$.

At steady state, rate of change in entropy of the system is zero.

Substitute 0 for $\Delta {\dot{S}}_{\text{system}}$ in Equation (VIII), and re-write the entropy balance as follows:

$\begin{array}{l}{\dot{S}}_{\text{in}}-{\dot{S}}_{\text{out}}+{\dot{S}}_{\text{gen}}=0\\ {\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3-{\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4+{\dot{S}}_{\text{gen}}=0\\ {\dot{m}}_{\text{R}}{s}_1+{\dot{m}}_{\text{a}}{s}_3-{\dot{m}}_{\text{R}}{s}_2+{\dot{m}}_{\text{a}}{s}_4+{\dot{S}}_{\text{gen}}=0\\ {\dot{S}}_{\text{gen}}={\dot{m}}_{\text{R}}\left(s_2-s_1\right)+{\dot{m}}_{\text{a}}\left(s_4-s_3\right)\end{array}$ (IX)

Here, entropy at $\text{state}\text{1,2,3}\text{and}\text{4}$ is $s_1,s_2,s_3\text{and}{s}_4$ respectively.

Write the expression for the change between state 4 entropy $\left(s_4\right)$ and state 3 entropy $\left(s_3\right)$.

$s_4-s_3=c_p\ln \left(\frac{T_4}{T_3}\right)-R\ln \left(\frac{P_4}{P_3}\right)$ (X)

Here, temperature at state $\text{3}\text{and}\text{4}$ is $T_3\text{and}{T}_4$ and pressure at state $\text{3}\text{and}\text{4}$ is $P_4\text{and}{P}_3$.

Write the expression for the exergy destroyed rate during the process $\left({\dot{X}}_{\text{destroyed}}\right)$.

${\dot{X}}_{\text{destroyed}}=T_0{\dot{S}}_{\text{gen}}$ (XI)

Here, dead state temperature is $T_0$.

Conclusion:

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure $\left(P_1\right)$ of $\text{120}\text{kPa}$ and state 1 quality $\left(x_1\right)$ of 0.3 as;

$\begin{array}{l}{h}_f=\text{22}\text{.47}\text{kJ}/\text{kg}\\ h_{fg}=214.52\text{kJ}/\text{kg}\\ s_f=0.09269\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=0.85520\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, enthalpy of saturated liquid is $h_f$, enthalpy of evaporation is $h_{fg}$, entropy of saturated liquid is $s_f$ and entropy of evaporation is $s_{fg}$.

Substitute $\text{22}\text{.47}\text{kJ}/\text{kg}$ for $h_f$, $0.3$ for $x_1$ and $214.52\text{kJ}/\text{kg}$ for $h_{fg}$ in equation (III)

$\begin{array}{c}{h}_1=\left(\text{22}\text{.47}\text{kJ}/\text{kg}\right)+\left(0.3\right)\left(214.52\text{kJ}/\text{kg}\right)\\ =\text{22}\text{.47}\text{kJ}/\text{kg}+64.356\text{kJ}/\text{kg}\\ =86.83\text{kJ}/\text{kg}\end{array}$

Substitute $0.09269\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, $0.3$ for $x_1$ and $0.85520\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in equation (IV).

$\begin{array}{c}{s}_1=\left(0.09269\text{kJ}/\text{kg}\cdot \text{K}\right)+\left(0.3\right)\left(0.85520\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =0.09269\text{kJ}/\text{kg}\cdot \text{K}+0.25656\text{kJ}/\text{kg}\cdot \text{K}\\ =0.34925\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure $\left(P_1\right)$ of $\text{120}\text{kPa}$ and quality is saturated vapor as;

$\begin{array}{l}{h}_{2,h_g@120\text{kPa}}=\text{236}\text{.99}\text{kJ}/\text{kg}\\ s_{2,s_g@120\text{kPa}}=0.94789\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, enthalpy at state 2 is $h_2$, entropy at state 2 is $s_2$, enthalpy of saturated vapor is $h_g$ and entropy of saturated vapor is $s_g$.

From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the gas constant of air gas $\left(R\right)$ as $\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$.

Substitute $\text{100}\text{kPa}$ for $P_3$, $\text{6}{\text{m}}^{\text{3}}/\text{min}$ for ${\dot{V}}_3$, $\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R$ and $\text{27°C}$ for $T_3$ in equation (V).

$\begin{array}{c}{\dot{m}}_{\text{a}}=\frac{\left(\text{100}\text{kPa}\right)\left(\text{6}{\text{m}}^{\text{3}}/\text{min}\right)}{\left(\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{27°C}\right)}\\ =\frac{\left(\text{100}\text{kPa}\right)\left(\text{6}{\text{m}}^{\text{3}}/\text{min}\right)}{\left(\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{27}+273\right)\text{K}}\\ =\frac{\left(\text{100}\text{kPa}\right)\left(\text{6}{\text{m}}^{\text{3}}/\text{min}\right)}{\left(\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(300\text{K}\right)}\\ =6.969\text{kg}/\text{min}\end{array}$

At steady state, rate of change in internal energy of the system is zero.

From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the constant pressure specific heat $\left(c_p\right)$ of air as $\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}$.

Substitute $\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $\text{27°C}$ for $T_3$, $\text{236}\text{.99}\text{kJ}/\text{kg}$ for $h_2$, $86.83\text{kJ}/\text{kg}$ for $h_1$, and $6.969\text{kg}/\text{min}$ for ${\dot{m}}_{\text{a}}$ and $2\text{kg}/\text{min}$ for ${\dot{m}}_{\text{R}}$ in Equation (VII).

$\begin{array}{c}{T}_4=\left(\text{27°C}\right)-\frac{\left(2\text{kg}/\text{min}\right)\left(\text{236}\text{.99}-86.83\right)\text{kJ}/\text{kg}}{\left(6.969\text{kg}/\text{min}\right)\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}\right)}\\ =\text{27°C}-\frac{300.32\text{kJ}/\min }{7.0038\text{kJ}/\min \cdot °\text{C}}\\ =\text{27°C}-42.879°\text{C}\\ =-15.8°\text{C}\end{array}$

$\begin{array}{c}=\left(-15.8+273\right)\text{K}\\ =257.2\text{K}\end{array}$

Thus, the exit temperature $\left(T_4\right)$ of the air is $\underset{\_}{\text{257}\text{.2}\text{K}\text{or}\left(-15.8\text{°C}\right)}$.

Substitute $\text{1}\text{.005}\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $\text{257}\text{.2}\text{K}$ for $T_4$, $\text{27°C}$ for $T_3$, $\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R$, 0 for $P_4$, and $\text{100}\text{kPa}$ for $P_3$ in Equation (X).

$\begin{array}{c}{s}_4-s_3=\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\text{257}\text{.2}\text{K}}{\text{27°C}}\right)-\left(\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{0}{\text{100}\text{kPa}}\right)\\ =\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left[\frac{\text{257}\text{.2}\text{K}}{\left(\text{27}+273\right)\text{K}}\right]-0\\ =\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\ln \left(\frac{\text{257}\text{.2}\text{K}}{300\text{K}}\right)\right]\\ =-0.1550\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $2\text{kg}/\text{min}$ for ${\dot{m}}_{\text{R}}$, $6.969\text{kg}/\text{min}$ for ${\dot{m}}_{\text{a}}$, $-0.1550\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4-s_3$, $0.94789\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$ and $0.34925\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$ in Equation (IX) to get ${\dot{S}}_{\text{gen}}$.

$\begin{array}{c}{\dot{S}}_{\text{gen}}=\left\{\begin{array}{c}\left(2\text{kg}/\text{min}\right)\left[\left(0.94789-0.34925\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\\ +\left(6.969\text{kg}/\text{min}\right)\left(-0.1550\text{kJ}/\text{kg}\cdot \text{K}\right)\end{array}\right\}\\ =1.19728\text{kJ}/\text{min}\cdot \text{K}-1.0801\text{kJ}/\text{min}\cdot \text{K}\\ =0.1170\text{kJ}/\text{min}\cdot \text{K}\end{array}$

Substitute $\text{32°C}$ for $T_0$ and $0.1170\text{kJ}/\text{min}\cdot \text{K}$ for ${\dot{S}}_{\text{gen}}$ in Equation (XI).

$\begin{array}{c}{\dot{X}}_{\text{destroyed}}=\left(\text{32°C}\right)\left(0.1170\text{kJ}/\text{min}\cdot \text{K}\right)\\ =\left(\text{32}+273\right)\text{K}\left(0.1170\text{kJ}/\text{min}\cdot \text{K}\right)\\ =\left(305\text{K}\right)\left(0.1170\text{kJ}/\text{min}\cdot \text{K}\right)\\ =35.685\text{kJ}/\text{min}\left(\frac{1\min }{60\text{s}}\right)\end{array}$

$\begin{array}{c}=0.594\text{kJ}/\text{s}\left(\frac{\text{kW}}{\text{kJ}/\text{s}}\right)\\ =0.594\text{kW}\end{array}$

Thus, the rate of exergy destroyed during the process is $\underset{\_}{0.594\text{kW}}$.

b)

To determine

The exit temperature of the air and the rate of exergy destroyed during the process without insulation.

Answer to Problem 66P

The exit temperature of the air without insulation is $\underset{\_}{261.4\text{K or }\left(-11.6°\text{C }\right)}$.

The rate of exergy destroyed during the process without insulation is $\underset{\_}{0.68\text{kW}}$.

Explanation of Solution

Write the expression for the state 4 temperature $\left(T_4\right)$ from steady-flow energy equation.

$\begin{array}{l}{\dot{Q}}_{\text{in}}={\dot{m}}_{\text{R}}\left(h_2-h_1\right)+{\dot{m}}_{\text{a}}{c}_p\left(T_4-T_3\right)\\ T_4=T_3+\frac{{\dot{Q}}_{\text{in}}-{\dot{m}}_{\text{R}}\left(h_2-h_1\right)}{{\dot{m}}_{\text{a}}{c}_p}\end{array}$ (XII)

Here, heat gain is from the surrounding ${\dot{Q}}_{\text{in}}$.

Write the expression for the entropy balance For an extended system as;

$\begin{array}{l}{\dot{S}}_{\text{in}}-{\dot{S}}_{\text{out}}+{\dot{S}}_{\text{gen}}=\Delta {\dot{S}}_{\text{system}}\\ \frac{{\dot{Q}}_{\text{in}}}{T_0}+{\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3-{\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4+{\dot{S}}_{\text{gen}}=0\\ \frac{{\dot{Q}}_{\text{in}}}{T_0}+{\dot{m}}_{\text{R}}{s}_1+{\dot{m}}_{\text{a}}{s}_3-{\dot{m}}_{\text{R}}{s}_2+{\dot{m}}_{\text{a}}{s}_4+{\dot{S}}_{\text{gen}}=0\\ {\dot{S}}_{\text{gen}}={\dot{m}}_{\text{R}}\left(s_2-s_1\right)+{\dot{m}}_{\text{a}}\left(s_4-s_3\right)-\frac{{\dot{Q}}_{\text{in}}}{T_0}\end{array}$ (XIII)

Conclusion:

Substitute $\text{27°C}$ for $T_3$, $\text{30}\text{kJ}/\text{min}$ for ${\dot{Q}}_{\text{in}}$, $\text{2}\text{kg}/\text{min}$ for ${\dot{m}}_{\text{R}}$, $\text{236}\text{.99}\text{kJ}/\text{kg}$ for $h_2$, $86.83\text{kJ}/\text{kg}$ for $h_1$, $6.969\text{kg}/\text{min}$ for ${\dot{m}}_{\text{a}}$ and $\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$ in Equation (XII) to get $T_4$.

$\begin{array}{c}{T}_4=\text{27°C}+\frac{\text{30}\text{kJ}/\text{min}-\left[\left(\text{2}\text{kg}/\text{min}\right)\left(\text{236}\text{.99}-86.83\right)\text{kJ}/\text{kg}\right]}{\left(6.969\text{kg}/\text{min}\right)\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}\right)}\\ =\text{27°C}+\frac{\text{30}\text{kJ}/\text{min}-300.32\text{kJ}/\text{min}}{7.0038\text{kg}/\text{min}\cdot °\text{C}}\\ =\text{27°C}-38.59\text{°C}\\ =-11.6\text{°C}\end{array}$

$\begin{array}{c}=\left(11.6+273\right)\text{K}\\ =261.4\text{K}\end{array}$

Thus, the exit temperature of the air is $\underset{\_}{261.4\text{K or }\left(-11.6°\text{C }\right)}$.

substitute $\text{1}\text{.005}\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $\text{261}\text{.4}\text{K}$ for $T_4$, $\text{27°C}$ for $T_3$, $\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R$, 0 for $P_4$, and $\text{100}\text{kPa}$ for $P_3$ in Equation (XIII).

$\begin{array}{c}{s}_4-s_3=\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\text{261}\text{.4}\text{K}}{\text{27°C}}\right)-\left(\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{0}{\text{100}\text{kPa}}\right)\\ =\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left[\frac{\text{261}\text{.4}\text{K}}{\left(\text{27}+273\right)\text{K}}\right]-0\\ =\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\ln \left(\frac{\text{261}\text{.4}\text{K}}{300\text{K}}\right)\right]\end{array}$

$=-0.1384\text{kJ}/\text{kg}\cdot \text{K}$

Substitute $\text{30}\text{kJ}/\text{min}$ for ${\dot{Q}}_{\text{in}}$, $\text{2}\text{kg}/\text{min}$ for ${\dot{m}}_{\text{R}}$, $6.969\text{kg}/\text{min}$ for ${\dot{m}}_{\text{a}}$, $0.94789\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$ and $0.34925\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$, $-0.1384\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4-s_3$, and $\text{27°C}$ for $T_0$ in Equation (XIII) to get ${\dot{S}}_{\text{gen}}$.

$\begin{array}{c}{\dot{S}}_{\text{gen}}=\left\{\begin{array}{c}\left(2\text{kg}/\text{min}\right)\left[\left(0.94789-0.34925\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\\ +\left(6.969\text{kg}/\text{min}\right)\left(-0.1384\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{\text{30}\text{kJ}/\text{min}}{27\text{°C}}\end{array}\right\}\\ =1.19728\text{kJ}/\text{min}\cdot \text{K}-0.9645\text{kJ}/\text{min}\cdot \text{K}-\frac{\text{30}\text{kJ}/\text{min}}{300\text{K}}\\ =1.19728\text{kJ}/\text{min}\cdot \text{K}-0.9645\text{kJ}/\text{min}\cdot \text{K}-0.1\text{kJ}/\text{min}\cdot \text{K}\\ =0.13278\text{kJ}/\text{min}\cdot \text{K}\end{array}$

Substitute $\text{32°C}$ for $T_0$ and $0.13278\text{kJ}/\text{min}\cdot \text{K}$ ${\dot{S}}_{\text{gen}}$ in Equation (XI).

$\begin{array}{c}{\dot{X}}_{\text{destroyed}}=T_0{\dot{S}}_{\text{gen}}\\ =\left(\text{32°C}\right)\left(0.13278\text{kJ}/\text{min}\cdot \text{K}\right)\\ =\left(\text{32}+273\right)\text{K}\left(0.13278\text{kJ}/\text{min}\cdot \text{K}\right)\\ =\left(305\text{K}\right)\left(0.13278\text{kJ}/\text{min}\cdot \text{K}\right)\end{array}$

$\begin{array}{c}=40.5\text{kJ}/\text{min}\left(\frac{1\min }{60\text{s}}\right)\\ =0.68\text{kJ}/\text{s}\left(\frac{\text{kW}}{\text{kJ}/\text{s}}\right)\\ =0.68\text{kW}\end{array}$

Thus, the rate of exergy destroyed during the process is $\underset{\_}{0.68\text{kW}}$.