Chapter 11.10, Problem 117RP

To determine

The greatest amount of exergy destroyed among the processes of system.

Answer to Problem 117RP

The greatest amount of exergy destroyed among the processes of system is $30.94\text{kW}$.

Explanation of Solution

Sketch the schematic layout of the two-stage compression refrigeration system as in Figure (1).

Sketch the $T-s$ diagram of the two-stage compression refrigeration as in Figure (2).

Write the relation between the specific enthalpies at the inlet and exit of throttling.

$h_i=h_e$ (I)

Here, specific enthalpy of the refrigerant at the inlet of throttling is $h_i$ and specific enthalpy of the refrigerant at the exit of throttling is $h_e$.

Write the energy balance equation for the separator.

${\dot{m}}_6\left(h_8-h_5\right)={\dot{m}}_2\left(h_1-h_4\right)$

${\dot{m}}_6={\dot{m}}_2\frac{h_1-h_4}{h_8-h_5}$ (II)

Here, mass flow rate of the refrigerant at the end of high pressure compressor is ${\dot{m}}_2$, mass flow rate at the inlet of evaporator is ${\dot{m}}_6$, specific enthalpy of refrigerant at the compressor inlet of high pressure cycle is $h_1$, specific enthalpy at the exit of expansion valve at high pressure cycle is $h_4$, specific enthalpy at the exit of compressor at the low pressure cycle is $h_8$, and the specific enthalpy at the inlet of expansion valve at the low pressure cycle is $h_5$.

Write the expression to calculate the quality of the refrigerant $\left(x\right)$ from specific enthalpy of saturated refrigerant.

$x=\frac{h-h_f}{h_{fg}}$ (III)

Here, specific enthalpy of the saturated liquid is $h_f$ and the specific enthalpy of the saturated refrigerant is $h_{fg}$.

Write the expression to calculate the specific entropy for saturated refrigerant $\left(s\right)$ from quality.

$s=s_f+x{h}_{fg}$ (IV)

Here, specific entropy of the saturated liquid is $s_f$ and the specific entropy of the saturated refrigerant is $s_{fg}$.

Write the expression to calculate the rate of cooling produced $\left(q_L\right)$ by the two-stage compression refrigeration.

$q_L=h_7-h_6$ (V)

Here, specific enthalpy at the inlet of low pressure compressor is $h_7$ and the specific enthalpy at the inlet evaporator is $h_6$.

Write the expression to calculate the rate of heat rejected from the system $\left(q_H\right)$.

$q_H=h_2-h_3$ (VI)

Here, specific enthalpy at the exit of high pressure compressor is $h_2$, and specific enthalpy at high pressure throttle inlet is $h_3$.

Write the general expression to calculate the exergy destruction $\left({\dot{X}}_{\text{dest}}\right)$ during a process.

${\dot{X}}_{\text{dest}}=\dot{m}{T}_0{s}_{gen}$

${\dot{X}}_{\text{dest}}=\dot{m}{T}_0\left(s_e-s_i-\frac{q_{\text{in}}}{T_{\text{source}}}+\frac{q_{\text{out}}}{T_{\text{sink}}}\right)$ (VII)

Here, mass flow rate is $\dot{m}$, surrounding temperature is $T_0$, specific entropy generated is $s_{\text{gen}}$, specific entropy at the exit is $s_e$, specific entropy at the inlet is $s_i$, heat input to the process is $q_{\text{in}}$, source temperature is $T_{\text{source}}$, heat output from the process is $q_{\text{out}}$, and the sink temperature is $T_{\text{sink}}$.

Conclusion:

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at high pressure compressor inlet temperature $\left(T_1\right)$ of $8.9°\text{C}$ as follows:

$\begin{array}{l}{h}_1=h_g=255.60\text{kJ}/\text{kg}\\ s_1=s_g=0.92711\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of high pressure compression exit at pressure $\left(P_2\right)$ of $1400\text{kPa}$ and specific entropy $\left(s_2\right)$ of $0.92711\text{kJ}/\text{kg}\cdot \text{K}$ as $281.56\text{kJ}/\text{kg}$.

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle inlet pressure of $1400\text{ kPa}$ as follows:

$\begin{array}{l}{h}_3=h_f=127.25\text{kJ}/\text{kg}\\ s_3=s_f=0.45325\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $127.25\text{kJ}/\text{kg}$ for $h_3$ in Equation (I).

$h_4=127.25\text{kJ}/\text{kg}$

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle exit temperature of $8.9°\text{C}$ as follows:

$\begin{array}{l}{h}_f=63.85\text{kJ}/\text{kg}\\ h_{fg}=191.69\text{kJ}/\text{kg}\\ s_f=0.24751\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=0.6796\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $63.85\text{kJ}/\text{kg}$ for $h_f$, $191.69\text{kJ}/\text{kg}$ for $h_{fg}$, and $127.25\text{kJ}/\text{kg}$ for $h_4$ in Equation (III).

$\begin{array}{l}{x}_4=\frac{127.25\text{kJ}/\text{kg}-63.85\text{kJ}/\text{kg}}{191.69\text{kJ}/\text{kg}}\\ x_4=0.3307\end{array}$

Substitute $0.24751\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, $0.6796\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$, and $0.3307$ for $x_4$ in Equation (IV).

$\begin{array}{l}{s}_4=0.24751\text{kJ}/\text{kg}\cdot \text{K}+0.3307\left(0.6796\text{kJ}/\text{kg}\cdot \text{K}\right)\\ s_4=0.47225\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure cycle throttle inlet temperature of $8.9°\text{C}$ as follows:

$\begin{array}{l}{P}_5=P_{\text{sat}}=400\text{kPa}\\ h_5=h_f=63.91\text{kJ}/\text{kg}\\ s_5=s_f=0.24752\text{kJ}/\text{kgK}\end{array}$

Here, pressure of refrigerant at the separator is $P_5$.

Substitute $63.91\text{kJ}/\text{kg}$ for $h_5$ in Equation (I).

$h_6=63.91\text{kJ}/\text{kg}$

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure expansion valve exit temperature $\left(T_6\right)$ of $-32°\text{C}$ as follows:

$\begin{array}{l}{h}_f=10.09\text{kJ}/\text{kg}\\ h_{fg}=220.83\text{kJ}/\text{kg}\\ s_f=0.04249\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=0.9157\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $63.91\text{kJ}/\text{kg}$ for $h_6$, $10.09\text{kJ}/\text{kg}$ for $h_f$, and $220.83\text{kJ}/\text{kg}$ for $h_{fg}$ in Equation (III).

$\begin{array}{c}{x}_6=\frac{63.91-10.09}{220.83}\\ =0.2437\end{array}$

Substitute 0.2437 for $x_6$, $0.04249\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, and $0.95819\text{kJ}/\text{kg}\cdot \text{K}$ for $s_g$ in Equation (IV).

$\begin{array}{c}{s}_6=0.04249\text{kJ}/\text{kg}\cdot \text{K}+\left(0.2437\right)0.9157\text{kJ}/\text{kg}\cdot \text{K}\\ =0.26559\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure compressor inlet temperature $\left(T_7\right)$ of $-32°\text{C}$ as follows:

$\begin{array}{l}{h}_7=h_g=230.93\text{kJ}/\text{kg}\\ s_7=s_g=0.95819\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of low pressure compression exit at pressure $\left(P_8\right)$ of $400\text{kPa}$ and specific entropy $\left(s_8\right)$  of $0.95819\text{kJ}/\text{kg}\cdot \text{K}$ as $264.51\text{kJ}/\text{kg}$.

Substitute $2\text{kg}/\text{s}$ for ${\dot{m}}_2$, $255.60\text{kJ}/\text{kg}$ for $h_1$, $127.25\text{kJ}/\text{kg}$ for $h_4$, $264.51\text{kJ}/\text{kg}$ for $h_8$, and $63.91\text{kJ}/\text{kg}$ for $h_5$ in Equation (II).

$\begin{array}{l}{\dot{m}}_6=2\text{kg}/\text{s}\times \frac{255.60\text{kJ}/\text{kg}-127.25\text{kJ}/\text{kg}}{264.51\text{kJ}/\text{kg}-63.91\text{kJ}/\text{kg}}\\ {\dot{m}}_6=1.280\text{kg}/\text{s}\end{array}$

Substitute $230.93\text{kJ}/\text{kg}$ for $h_7$, and $63.91\text{kJ}/\text{kg}$ for $h_6$ in Equation (V).

$\begin{array}{l}{q}_L=230.93\text{kJ}/\text{kg}-63.91\text{kJ}/\text{kg}\\ q_L=167.02\text{kJ}/\text{kg}\end{array}$

Substitute $127.25\text{kJ}/\text{kg}$ for $h_3$ and $281.56\text{kJ}/\text{kg}$ for $h_2$ in Equation (VI).

$\begin{array}{l}{q}_H=281.56\text{kJ}/\text{kg}-127.25\text{kJ}/\text{kg}\\ q_H=154.31\text{kJ}/\text{kg}\end{array}$

Rewrite the Equation (VII) for the process 2 to 3.

${\dot{X}}_{\text{dest,2-3}}={\dot{m}}_2{T}_0\left(s_3-s_2+\frac{q_H}{T_H}\right)$ (VIII)

Here, the temperature of the high temperature reservoir is $T_H$.

Substitute $2\text{kg}/\text{s}$ for ${\dot{m}}_2$, $20°\text{C}$ for $T_0$, $0.45325\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$, $0.92711\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, $154.31\text{kJ}/\text{kg}$ for $q_H$, and $20°\text{C}$ for $T_H$ in Equation (VIII).

$\begin{array}{c}{\dot{X}}_{\text{dest,2-3}}=2\text{kg}/\text{s}\times 20°\text{C}\times \left(\begin{array}{l}0.45325\text{kJ}/\text{kg}\cdot \text{K}-0.92711\text{kJ}/\text{kg}\cdot \text{K}\\ +\frac{154.31\text{kJ}/\text{kg}}{20°\text{C}}\end{array}\right)\\ =2\text{kg}/\text{s}\times \left(20+273\right)\text{K}\times \left(\begin{array}{l}0.45325\text{kJ}/\text{kg}\cdot \text{K}-0.92711\text{kJ}/\text{kg}\cdot \text{K}\\ +\frac{154.31\text{kJ}/\text{kg}}{\left(20+273\right)\text{K}}\end{array}\right)\\ =30.94\text{kW}\end{array}$

Rewrite the Equation (VII) for the process 3 to 4.

${\dot{X}}_{\text{dest,3-4}}={\dot{m}}_2{T}_0\left(s_4-s_3\right)$ (IX)

Substitute $2\text{kg}/\text{s}$ for ${\dot{m}}_2$, $20°\text{C}$ for $T_0$, $0.45325\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$, and $0.47225\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4$ in Equation (IX).

$\begin{array}{c}{\dot{X}}_{\text{dest,3-4}}=2\text{kg}/\text{s}\times 20°\text{C}\times \left(0.47225\text{kJ}/\text{kg}\cdot \text{K}-0.45325\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =2\text{kg}/\text{s}\times \left(20+273\right)\text{K}\times \left(0.47225\text{kJ}/\text{kg}\cdot \text{K}-0.45325\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =11.03\text{kW}\end{array}$

Rewrite the Equation (VII) for the process 5 to 6.

${\dot{X}}_{\text{dest,5-6}}={\dot{m}}_6{T}_0\left(s_6-s_5\right)$ (X)

Substitute $1.280\text{kg}/\text{s}$ for ${\dot{m}}_6$, $20°\text{C}$ for $T_0$, $0.26559\text{kJ}/\text{kg}\cdot \text{K}$ for $s_6$, and $0.24752\text{kJ}/\text{kgK}$ for $s_5$ in Equation (X).

$\begin{array}{c}{\dot{X}}_{\text{dest,5-6}}=1.280\text{kg}/\text{s}\times 20°\text{C}\times \left(0.26559\text{kJ}/\text{kg}\cdot \text{K}-0.24752\text{kJ}/\text{kgK}\right)\\ =1.280\text{kg}/\text{s}\times \left(20+273\right)\text{K}\times \left(0.26559\text{kJ}/\text{kg}\cdot \text{K}-0.24752\text{kJ}/\text{kgK}\right)\\ =6.793\text{kW}\end{array}$

Rewrite the Equation (VII) for the process 6 to 7.

${\dot{X}}_{\text{dest,6-7}}={\dot{m}}_6{T}_0\left(s_7-s_6-\frac{q_L}{T_L}\right)$ (XI)

Substitute $1.280\text{kg}/\text{s}$ for ${\dot{m}}_6$, $20°\text{C}$ for $T_0$, $0.26559\text{kJ}/\text{kg}\cdot \text{K}$ for $s_6$, and $0.95819\text{kJ}/\text{kgK}$ for $s_7$, $167.02\text{kJ}/\text{kg}$ for $q_L$, and $-22°\text{C}$ for $T_L$ in Equation (XI).

$\begin{array}{c}{\dot{X}}_{\text{dest,6-7}}=1.280\text{kg}/\text{s}\times 20°\text{C}\times \left(0.95819\text{kJ}/\text{kgK}-0.26559\text{kJ}/\text{kg}\cdot \text{K}-\frac{167.02\text{kJ}/\text{kg}}{\left(-22°\text{C}\right)}\right)\\ =1.280\text{kg}/\text{s}\times 20°\text{C}\times \left(0.95819\text{kJ}/\text{kgK}-0.26559\text{kJ}/\text{kg}\cdot \text{K}-\frac{167.02\text{kJ}/\text{kg}}{\left(-22+273\right)\text{K}}\right)\\ =10.18\text{kW}\end{array}$

Rewrite the Equation (VII) for the separator process.

${\dot{X}}_{\text{dest,separator}}=T_0\left[{\dot{m}}_6\left(s_5-s_8\right)-{\dot{m}}_2\left(s_1-s_4\right)\right]$ (XII)

Substitute $1.280\text{kg}/\text{s}$ for ${\dot{m}}_6$, $20°\text{C}$ for $T_0$, $0.24752\text{kJ}/\text{kgK}$ for $s_5$, $0.95819\text{kJ}/\text{kg}$ for $s_8$, $2\text{kg}/\text{s}$ for ${\dot{m}}_2$, $0.47225\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4$, and $0.92711\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$ in Equation (XII).

$\begin{array}{c}{\dot{X}}_{\text{dest,separator}}=20°\text{C}\left[\begin{array}{l}1.280\text{kg}/\text{s}\left(0.24752\text{kJ}/\text{kgK}-0.95819\text{kJ}/\text{kg}\right)\\ -2\text{kg}/\text{s}\left(0.92711\text{kJ}/\text{kg}\cdot \text{K}-0.47225\text{kJ}/\text{kg}\cdot \text{K}\right)\end{array}\right]\\ =\left(20+273\right)\text{K}\left[\begin{array}{l}1.280\text{kg}/\text{s}\left(0.24752\text{kJ}/\text{kgK}-0.95819\text{kJ}/\text{kg}\right)\\ -2\text{kg}/\text{s}\left(0.92711\text{kJ}/\text{kg}\cdot \text{K}-0.47225\text{kJ}/\text{kg}\cdot \text{K}\right)\end{array}\right]\\ =0.197\text{kW}\end{array}$

The exergy destruction for the isentropic processes is zero. Hence,

$\begin{array}{l}{\dot{X}}_{\text{dest,1-2}}=0\\ {\dot{X}}_{\text{dest,7-8}}=0\end{array}$

Thus, the greatest amount of exergy destroyed among the processes of system is $30.94\text{kW}$.