Chapter 9, Problem 9.9P

(a)

Interpretation Introduction

Interpretation:

Considering vapor-compression refrigeration cycle and different thermodynamic processes taking place between two stages, the circulation rate of refrigerant, the heat transfer rate in the condenser, the power requirement, the coefficient of performance of the cycle and the coefficient of performance of a Carnot refrigeration cycle are to be calculated.

Concept introduction:

Below shown diagram represents vapor-compression refrigeration cycle on a $TS$ diagram which include four steps of the cycle. Line $1\to 2$ shows a liquid which is evaporating at constant pressure and providing a means for heat absorption at constant low temperature. The vapor thus produced is then compressed to a higher pressure. It is then cooled and condensed at higher temperature by heat rejection. By an expansion process, the liquid from the condenser is returned to its original pressure. This process is carried out by throttling through a partly open valve. Due to the fluid friction in the valve there is pressure drop in this irreversible process.

The line $4\to 1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2\to 3\text{'}$ and the actual compression is shown by the line $2\to 3$ where the direction of slope is in increasing enthalpy which reflects inherent irreversibility.

The equations used to calculate the heat absorbed in evaporator and the heat rejected in condenser are:

$\left|Q_C\right|=H_2-H_1$ ...... (1)

  $\left|Q_H\right|=H_3-H_4$ ...... (2)

The work of compression is:

$W=H_3-H_2$ ...... (3)

The coefficient of performance is:

$\omega =\frac{H_2-H_1}{H_3-H_2}$ ...... (4)

The rate of circulation of refrigerant, $\dot{m}$ is determined from the rate of heat absorption in the evaporator given by the equation:

$\dot{m}=\frac{\left|{\dot{Q}}_C\right|}{H_2-H_1}$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $\omega$ is attained at the given values of $T_C\text{ and }{T}_H$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $\omega$ is obtained.

Answer to Problem 9.9P

  $\begin{array}{c}\dot{m}=3.689\text{ kg/s}\\ {\dot{Q}}_H=-849.28\text{ kJ/s}\\ \dot{W}=249.34\text{ kW}\\ \omega =2.406\\ {\omega }_{\text{Carnot}}=10.51\end{array}$

Explanation of Solution

Given information:

The refrigerant is tetrafluoroethene.

Evaporation $T=0^{\circ }\text{C}$

Condensation $T={26}^{\circ }\text{C}$

$\eta =0.79$ for compressor

Refrigeration rate is $600\text{ kJ/s}$

The given data for the vapor-compression cycle are:

  $\begin{array}{c}{T}_C=0^{\circ }\text{C}=273.15\text{ K}\\ T_H={26}^{\circ }\text{C}=299.15\text{ K}\\ \eta =0.79\\ {\dot{Q}}_C=600\text{ kJ/s}\end{array}$

From table 9.1, the values of $H_2,\text{ and }{S}_2$ for the saturated tetrafluoroethene are:

  $\begin{array}{c}{H}_2=398.60\text{ kJ/kg}\\ S_2=1.727\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)\end{array}$

For isentropic compression,

  ${S^{\prime }}_3=S_2=1.727\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)$

The saturation pressure at point 4 is the pressure at which the vapor condenses which is $T_H={26}^{\circ }\text{C}$ and $H_4$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}=6.854\text{ bar}\\ H_4=235.97\text{ kJ/kg}\end{array}$

At point $3^{\prime }$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.727\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)$ to get the enthalpy ${H^{\prime }}_3$ at pressure $6.854\text{ bar}$ as:

  ${H^{\prime }}_3=452\text{ kJ/kg}$

Calculate the value of $\Delta H_{23}$ as:

  $\begin{array}{c}\Delta H_{23}=\frac{{H^{\prime }}_3-H_2}{\eta }\\ =\frac{452-398.60}{0.79}\\ =67.59\text{ kJ/kg}\end{array}$

Now, calculate $H_3$ as:

  $\begin{array}{c}{H}_3=H_2+\Delta H_{23}\\ =398.60+67.59\\ =466.19\text{ kJ/kg}\end{array}$

Also,

  $\begin{array}{c}{H}_1=H_4\\ =235.97\text{ kJ/kg}\end{array}$

Use equation (5) to calculate the value of $\dot{m}$ as:

  $\begin{array}{c}\dot{m}=\frac{\left|{\dot{Q}}_C\right|}{H_2-H_1}\\ =\frac{\left|600\right|}{398.60-235.97}\\ =3.689\text{ kg/s}\end{array}$

Use equation (2) to calculate the value of ${\dot{Q}}_H$ as:

  $\begin{array}{c}{\dot{Q}}_H=\dot{m}\left(H_3-H_4\right)\\ =\left(3.689\right)\left(466.19-235.97\right)\\ =849.28\text{ kJ/s}\end{array}$

Since, ${\dot{Q}}_H$ is the heat rejected, negative sign is used. Thus,

  ${\dot{Q}}_H=-849.28\text{ kJ/s}$

The work done is calculated as:

  $\begin{array}{c}\dot{W}=\dot{m}\times \Delta H_{23}\\ =3.689\times 67.59\\ =249.34\text{ kW}\end{array}$

Use equation (5) to calculate the coefficient of performance for vapor-compression cycle as:

  $\begin{array}{c}\omega =\frac{H_2-H_1}{H_3-H_2}\\ =\frac{398.60-235.97}{466.19-398.60}\\ =2.406\end{array}$

For Carnot cycle, the coefficient of performance is calculated as:

  $\begin{array}{c}{\omega }_{\text{Carnot}}=\frac{T_C}{T_H-T_C}\\ =\frac{273.15}{299.15-273.15}\\ =10.51\end{array}$

(b)

Interpretation Introduction

Interpretation:

Considering vapor-compression refrigeration cycle and different thermodynamic processes taking place between two stages, the circulation rate of refrigerant, the heat transfer rate in the condenser, the power requirement, the coefficient of performance of the cycle and the coefficient of performance of a Carnot refrigeration cycle are to be calculated.

Concept introduction:

Below shown diagram represents vapor-compression refrigeration cycle on a $TS$ diagram which include four steps of the cycle. Line $1\to 2$ shows a liquid which is evaporating at constant pressure and providing a means for heat absorption at constant low temperature. The vapor thus produced is then compressed to a higher pressure. It is then cooled and condensed at higher temperature by heat rejection. By an expansion process, the liquid from the condenser is returned to its original pressure. This process is carried out by throttling through a partly open valve. Due to the fluid friction in the valve there is pressure drop in this irreversible process.

The line $4\to 1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2\to 3\text{'}$ and the actual compression is shown by the line $2\to 3$ where the direction of slope is in increasing enthalpy which reflects inherent irreversibility.

The equations used to calculate the heat absorbed in evaporator and the heat rejected in condenser are:

$\left|Q_C\right|=H_2-H_1$ ...... (1)

  $\left|Q_H\right|=H_3-H_4$ ...... (2)

The work of compression is:

$W=H_3-H_2$ ...... (3)

The coefficient of performance is:

$\omega =\frac{H_2-H_1}{H_3-H_2}$ ...... (4)

The rate of circulation of refrigerant, $\dot{m}$ is determined from the rate of heat absorption in the evaporator given by the equation:

$\dot{m}=\frac{\left|{\dot{Q}}_C\right|}{H_2-H_1}$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $\omega$ is attained at the given values of $T_C\text{ and }{T}_H$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $\omega$ is obtained.

Answer to Problem 9.9P

  $\begin{array}{c}\dot{m}=3.01\text{ kg/s}\\ {\dot{Q}}_H=-689.561\text{ kJ/s}\\ \dot{W}=189.63\text{ kW}\\ \omega =2.363\\ {\omega }_{\text{Carnot}}=13.958\end{array}$

Explanation of Solution

Given information:

The refrigerant is tetrafluoroethene.

Evaporation $T=6^{\circ }\text{C}$

Condensation $T={26}^{\circ }\text{C}$

$\eta =0.78$ for compressor

Refrigeration rate is $500\text{ kJ/s}$

The given data for the vapor-compression cycle are:

  $\begin{array}{c}{T}_C=6^{\circ }\text{C}=279.15\text{ K}\\ T_H={26}^{\circ }\text{C}=299.15\text{ K}\\ \eta =0.78\\ {\dot{Q}}_C=500\text{ kJ/s}\end{array}$

From table 9.1, the values of $H_2,\text{ and }{S}_2$ for the saturated tetrafluoroethene are:

  $\begin{array}{c}{H}_2=402.06\text{ kJ/kg}\\ S_2=1.724\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)\end{array}$

For isentropic compression,

  ${S^{\prime }}_3=S_2=1.724\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)$

The saturation pressure at point 4 is the pressure at which the vapor condenses which is $T_H={26}^{\circ }\text{C}$ and $H_4$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}=6.854\text{ bar}\\ H_4=235.97\text{ kJ/kg}\end{array}$

At point $3^{\prime }$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.724\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)$ to get the enthalpy ${H^{\prime }}_3$ at pressure $6.854\text{ bar}$ as:

  ${H^{\prime }}_3=451.2\text{ kJ/kg}$

Calculate the value of $\Delta H_{23}$ as:

  $\begin{array}{c}\Delta H_{23}=\frac{{H^{\prime }}_3-H_2}{\eta }\\ =\frac{451.2-402.06}{0.78}\\ =63\text{ kJ/kg}\end{array}$

Now, calculate $H_3$ as:

  $\begin{array}{c}{H}_3=H_2+\Delta H_{23}\\ =402.06+63\\ =465.06\text{ kJ/kg}\end{array}$

Also,

  $\begin{array}{c}{H}_1=H_4\\ =235.97\text{ kJ/kg}\end{array}$

Use equation (5) to calculate the value of $\dot{m}$ as:

  $\begin{array}{c}\dot{m}=\frac{\left|{\dot{Q}}_C\right|}{H_2-H_1}\\ =\frac{\left|500\right|}{402.06-235.97}\\ =3.01\text{ kg/s}\end{array}$

Use equation (2) to calculate the value of ${\dot{Q}}_H$ as:

  $\begin{array}{c}{\dot{Q}}_H=\dot{m}\left(H_3-H_4\right)\\ =\left(3.01\right)\left(465.06-235.97\right)\\ =689.561\text{ kJ/s}\end{array}$

Since, ${\dot{Q}}_H$ is the heat rejected, negative sign is used. Thus,

  ${\dot{Q}}_H=-689.561\text{ kJ/s}$

The work done is calculated as:

  $\begin{array}{c}\dot{W}=\dot{m}\times \Delta H_{23}\\ =3.01\times 63\\ =189.63\text{ kW}\end{array}$

Use equation (5) to calculate the coefficient of performance for vapor-compression cycle as:

  $\begin{array}{c}\omega =\frac{H_2-H_1}{H_3-H_2}\\ =\frac{402.06-235.97}{465.06-402.06}\\ =2.636\end{array}$

For Carnot cycle, the coefficient of performance is calculated as:

  $\begin{array}{c}{\omega }_{\text{Carnot}}=\frac{T_C}{T_H-T_C}\\ =\frac{279.15}{299.15-279.15}\\ =13.958\end{array}$

(c)

Interpretation Introduction

Interpretation:

Considering vapor-compression refrigeration cycle and different thermodynamic processes taking place between two stages, the circulation rate of refrigerant, the heat transfer rate in the condenser, the power requirement, the coefficient of performance of the cycle and the coefficient of performance of a Carnot refrigeration cycle are to be calculated.

Concept introduction:

Below shown diagram represents vapor-compression refrigeration cycle on a $TS$ diagram which include four steps of the cycle. Line $1\to 2$ shows a liquid which is evaporating at constant pressure and providing a means for heat absorption at constant low temperature. The vapor thus produced is then compressed to a higher pressure. It is then cooled and condensed at higher temperature by heat rejection. By an expansion process, the liquid from the condenser is returned to its original pressure. This process is carried out by throttling through a partly open valve. Due to the fluid friction in the valve there is pressure drop in this irreversible process.

The line $4\to 1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2\to 3\text{'}$ and the actual compression is shown by the line $2\to 3$ where the direction of slope is in increasing enthalpy which reflects inherent irreversibility.

The equations used to calculate the heat absorbed in evaporator and the heat rejected in condenser are:

$\left|Q_C\right|=H_2-H_1$ ...... (1)

  $\left|Q_H\right|=H_3-H_4$ ...... (2)

The work of compression is:

$W=H_3-H_2$ ...... (3)

The coefficient of performance is:

$\omega =\frac{H_2-H_1}{H_3-H_2}$ ...... (4)

The rate of circulation of refrigerant, $\dot{m}$ is determined from the rate of heat absorption in the evaporator given by the equation:

$\dot{m}=\frac{\left|{\dot{Q}}_C\right|}{H_2-H_1}$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $\omega$ is attained at the given values of $T_C\text{ and }{T}_H$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $\omega$ is obtained.

Answer to Problem 9.9P

  $\begin{array}{c}\dot{m}=2.572\text{ kg/s}\\ {\dot{Q}}_H=-634.875\text{ kJ/s}\\ \dot{W}=234.954\text{ kW}\\ \omega =1.702\\ {\omega }_{\text{Carnot}}=6.87\end{array}$

Explanation of Solution

Given information:

The refrigerant is tetrafluoroethene.

Evaporation $T=-{12}^{\circ }\text{C}$

Condensation $T={26}^{\circ }\text{C}$

$\eta =0.77$ for compressor

Refrigeration rate is $400\text{ kJ/s}$

The given data for the vapor-compression cycle are:

  $\begin{array}{c}{T}_C=-{12}^{\circ }\text{C}=261.15\text{ K}\\ T_H={26}^{\circ }\text{C}=299.15\text{ K}\\ \eta =0.77\\ {\dot{Q}}_C=400\text{ kJ/s}\end{array}$

From table 9.1, the values of $H_2,\text{ and }{S}_2$ for the saturated tetrafluoroethene are:

  $\begin{array}{c}{H}_2=\text{391}\text{.46 kJ/kg}\\ S_2=1.735\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)\end{array}$

For isentropic compression,

  ${S^{\prime }}_3=S_2=1.735\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)$

The saturation pressure at point 4 is the pressure at which the vapor condenses which is $T_H={26}^{\circ }\text{C}$ and $H_4$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}=6.854\text{ bar}\\ H_4=235.97\text{ kJ/kg}\end{array}$

At point $3^{\prime }$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)$ to get the enthalpy ${H^{\prime }}_3$ at pressure $6.854\text{ bar}$ as:

  ${H^{\prime }}_3=461.8\text{ kJ/kg}$

Calculate the value of $\Delta H_{23}$ as:

  $\begin{array}{c}\Delta H_{23}=\frac{{H^{\prime }}_3-H_2}{\eta }\\ =\frac{461.8-391.46}{0.77}\\ =91.351\text{ kJ/kg}\end{array}$

Now, calculate $H_3$ as:

  $\begin{array}{c}{H}_3=H_2+\Delta H_{23}\\ =391.46+91.351\\ =482.811\text{ kJ/kg}\end{array}$

Also,

  $\begin{array}{c}{H}_1=H_4\\ =235.97\text{ kJ/kg}\end{array}$

Use equation (5) to calculate the value of $\dot{m}$ as:

  $\begin{array}{c}\dot{m}=\frac{\left|{\dot{Q}}_C\right|}{H_2-H_1}\\ =\frac{\left|400\right|}{391.46-235.97}\\ =2.572\text{ kg/s}\end{array}$

Use equation (2) to calculate the value of ${\dot{Q}}_H$ as:

  $\begin{array}{c}{\dot{Q}}_H=\dot{m}\left(H_3-H_4\right)\\ =\left(2.572\right)\left(482.811-235.97\right)\\ =634.875\text{ kJ/s}\end{array}$

Since, ${\dot{Q}}_H$ is the heat rejected, negative sign is used. Thus,

  ${\dot{Q}}_H=-634.875\text{ kJ/s}$

The work done is calculated as:

  $\begin{array}{c}\dot{W}=\dot{m}\times \Delta H_{23}\\ =2.572\times 91.351\\ =234.954\text{ kW}\end{array}$

Use equation (5) to calculate the coefficient of performance for vapor-compression cycle as:

  $\begin{array}{c}\omega =\frac{H_2-H_1}{H_3-H_2}\\ =\frac{391.46-235.97}{482.811-391.46}\\ =1.702\end{array}$

For Carnot cycle, the coefficient of performance is calculated as:

  $\begin{array}{c}{\omega }_{\text{Carnot}}=\frac{T_C}{T_H-T_C}\\ =\frac{261.15}{299.15-261.15}\\ =6.87\end{array}$

(d)

Interpretation Introduction

Interpretation:

Considering vapor-compression refrigeration cycle and different thermodynamic processes taking place between two stages, the circulation rate of refrigerant, the heat transfer rate in the condenser, the power requirement, the coefficient of performance of the cycle and the coefficient of performance of a Carnot refrigeration cycle are to be calculated.

Concept introduction:

Below shown diagram represents vapor-compression refrigeration cycle on a $TS$ diagram which include four steps of the cycle. Line $1\to 2$ shows a liquid which is evaporating at constant pressure and providing a means for heat absorption at constant low temperature. The vapor thus produced is then compressed to a higher pressure. It is then cooled and condensed at higher temperature by heat rejection. By an expansion process, the liquid from the condenser is returned to its original pressure. This process is carried out by throttling through a partly open valve. Due to the fluid friction in the valve there is pressure drop in this irreversible process.

The line $4\to 1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2\to 3\text{'}$ and the actual compression is shown by the line $2\to 3$ where the direction of slope is in increasing enthalpy which reflects inherent irreversibility.

The equations used to calculate the heat absorbed in evaporator and the heat rejected in condenser are:

$\left|Q_C\right|=H_2-H_1$ ...... (1)

  $\left|Q_H\right|=H_3-H_4$ ...... (2)

The work of compression is:

$W=H_3-H_2$ ...... (3)

The coefficient of performance is:

$\omega =\frac{H_2-H_1}{H_3-H_2}$ ...... (4)

The rate of circulation of refrigerant, $\dot{m}$ is determined from the rate of heat absorption in the evaporator given by the equation:

$\dot{m}=\frac{\left|{\dot{Q}}_C\right|}{H_2-H_1}$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $\omega$ is attained at the given values of $T_C\text{ and }{T}_H$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $\omega$ is obtained.

Answer to Problem 9.9P

  $\begin{array}{c}\dot{m}=1.976\text{ kg/s}\\ {\dot{Q}}_H=-509.08\text{ kJ/s}\\ \dot{W}=209.08\text{ kW}\\ \omega =1.43\\ {\omega }_{\text{Carnot}}=5.799\end{array}$

Explanation of Solution

Given information:

The refrigerant is tetrafluoroethene.

Evaporation $T=-{18}^{\circ }\text{C}$

Condensation $T={26}^{\circ }\text{C}$

$\eta =0.76$ for compressor

Refrigeration rate is $300\text{ kJ/s}$

The given data for the vapor-compression cycle are:

  $\begin{array}{c}{T}_C=-{18}^{\circ }\text{C}=255.15\text{ K}\\ T_H={26}^{\circ }\text{C}=299.15\text{ K}\\ \eta =0.76\\ {\dot{Q}}_C=300\text{ kJ/s}\end{array}$

From table 9.1, the values of $H_2,\text{ and }{S}_2$ for the saturated tetrafluoroethene are:

  $\begin{array}{c}{H}_2=387.79\text{ kJ/kg}\\ S_2=1.740\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)\end{array}$

For isentropic compression,

  ${S^{\prime }}_3=S_2=1.740\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)$

The saturation pressure at point 4 is the pressure at which the vapor condenses which is $T_H={26}^{\circ }\text{C}$ and $H_4$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}=6.854\text{ bar}\\ H_4=235.97\text{ kJ/kg}\end{array}$

At point $3^{\prime }$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.740\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)$ to get the enthalpy ${H^{\prime }}_3$ at pressure $6.854\text{ bar}$ as:

  ${H^{\prime }}_3=468.2\text{ kJ/kg}$

Calculate the value of $\Delta H_{23}$ as:

  $\begin{array}{c}\Delta H_{23}=\frac{{H^{\prime }}_3-H_2}{\eta }\\ =\frac{468.2-387.79}{0.76}\\ =105.81\text{ kJ/kg}\end{array}$

Now, calculate $H_3$ as:

  $\begin{array}{c}{H}_3=H_2+\Delta H_{23}\\ =387.79+105.81\\ =493.6\text{ kJ/kg}\end{array}$

Also,

  $\begin{array}{c}{H}_1=H_4\\ =235.97\text{ kJ/kg}\end{array}$

Use equation (5) to calculate the value of $\dot{m}$ as:

  $\begin{array}{c}\dot{m}=\frac{\left|{\dot{Q}}_C\right|}{H_2-H_1}\\ =\frac{\left|300\right|}{387.79-235.97}\\ =1.976\text{ kg/s}\end{array}$

Use equation (2) to calculate the value of ${\dot{Q}}_H$ as:

  $\begin{array}{c}{\dot{Q}}_H=\dot{m}\left(H_3-H_4\right)\\ =\left(1.976\right)\left(493.6-235.97\right)\\ =509.08\text{ kJ/s}\end{array}$

Since, ${\dot{Q}}_H$ is the heat rejected, negative sign is used. Thus,

  ${\dot{Q}}_H=-509.08\text{ kJ/s}$

The work done is calculated as:

  $\begin{array}{c}\dot{W}=\dot{m}\times \Delta H_{23}\\ =1.976\times 105.81\\ =209.08\text{ kW}\end{array}$

Use equation (5) to calculate the coefficient of performance for vapor-compression cycle as:

  $\begin{array}{c}\omega =\frac{H_2-H_1}{H_3-H_2}\\ =\frac{387.79-235.97}{493.6-387.79}\\ =1.43\end{array}$

For Carnot cycle, the coefficient of performance is calculated as:

  $\begin{array}{c}{\omega }_{\text{Carnot}}=\frac{T_C}{T_H-T_C}\\ =\frac{255.15}{299.15-255.15}\\ =5.799\end{array}$

(e)

Interpretation Introduction

Interpretation:

Considering vapor-compression refrigeration cycle and different thermodynamic processes taking place between two stages, the circulation rate of refrigerant, the heat transfer rate in the condenser, the power requirement, the coefficient of performance of the cycle and the coefficient of performance of a Carnot refrigeration cycle are to be calculated.

Concept introduction:

Below shown diagram represents vapor-compression refrigeration cycle on a $TS$ diagram which include four steps of the cycle. Line $1\to 2$ shows a liquid which is evaporating at constant pressure and providing a means for heat absorption at constant low temperature. The vapor thus produced is then compressed to a higher pressure. It is then cooled and condensed at higher temperature by heat rejection. By an expansion process, the liquid from the condenser is returned to its original pressure. This process is carried out by throttling through a partly open valve. Due to the fluid friction in the valve there is pressure drop in this irreversible process.

The line $4\to 1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2\to 3\text{'}$ and the actual compression is shown by the line $2\to 3$ where the direction of slope is in increasing enthalpy which reflects inherent irreversibility.

The equations used to calculate the heat absorbed in evaporator and the heat rejected in condenser are:

$\left|Q_C\right|=H_2-H_1$ ...... (1)

  $\left|Q_H\right|=H_3-H_4$ ...... (2)

The work of compression is:

$W=H_3-H_2$ ...... (3)

The coefficient of performance is:

$\omega =\frac{H_2-H_1}{H_3-H_2}$ ...... (4)

The rate of circulation of refrigerant, $\dot{m}$ is determined from the rate of heat absorption in the evaporator given by the equation:

$\dot{m}=\frac{\left|{\dot{Q}}_C\right|}{H_2-H_1}$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $\omega$ is attained at the given values of $T_C\text{ and }{T}_H$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $\omega$ is obtained.

Answer to Problem 9.9P

  $\begin{array}{c}\dot{m}=1.356\text{ kg/s}\\ {\dot{Q}}_H=-269.63\text{ kJ/s}\\ \dot{W}=69.64\text{ kW}\\ \omega =2.87\\ {\omega }_{\text{Carnot}}=4.87\end{array}$

Explanation of Solution

Given information:

The refrigerant is tetrafluoroethene.

Evaporation $T=-{25}^{\circ }\text{C}$

Condensation $T={26}^{\circ }\text{C}$

$\eta =0.75$ for compressor

Refrigeration rate is $200\text{ kJ/s}$

The given data for the vapor-compression cycle are:

  $\begin{array}{c}{T}_C=-{25}^{\circ }\text{C}=248.15\text{ K}\\ T_H={26}^{\circ }\text{C}=299.15\text{ K}\\ \eta =0.75\\ {\dot{Q}}_C=200\text{ kJ/s}\end{array}$

From table 9.1, the values of $H_2,\text{ and }{S}_2$ for the saturated tetrafluoroethene are:

  $\begin{array}{c}{H}_2=383.45\text{ kJ/kg}\\ S_2=1.746\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)\end{array}$

For isentropic compression,

  ${S^{\prime }}_3=S_2=1.746\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)$

The saturation pressure at point 4 is the pressure at which the vapor condenses which is $T_H={26}^{\circ }\text{C}$ and $H_4$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}=6.854\text{ bar}\\ H_4=235.97\text{ kJ/kg}\end{array}$

At point $3^{\prime }$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.727\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)$ to get the enthalpy ${H^{\prime }}_3$ at pressure $6.854\text{ bar}$ as:

  ${H^{\prime }}_3=421.97\text{ kJ/kg}$

Calculate the value of $\Delta H_{23}$ as:

  $\begin{array}{c}\Delta H_{23}=\frac{{H^{\prime }}_3-H_2}{\eta }\\ =\frac{421.97-383.45}{0.75}\\ =51.36\text{ kJ/kg}\end{array}$

Now, calculate $H_3$ as:

  $\begin{array}{c}{H}_3=H_2+\Delta H_{23}\\ =383.45+51.36\\ =434.81\text{ kJ/kg}\end{array}$

Also,

  $\begin{array}{c}{H}_1=H_4\\ =235.97\text{ kJ/kg}\end{array}$

Use equation (5) to calculate the value of $\dot{m}$ as:

  $\begin{array}{c}\dot{m}=\frac{\left|{\dot{Q}}_C\right|}{H_2-H_1}\\ =\frac{\left|200\right|}{383.45-235.97}\\ =1.356\text{ kg/s}\end{array}$

Use equation (2) to calculate the value of ${\dot{Q}}_H$ as:

  $\begin{array}{c}{\dot{Q}}_H=\dot{m}\left(H_3-H_4\right)\\ =\left(1.356\right)\left(434.81-235.97\right)\\ =269.63\text{ kJ/s}\end{array}$

Since, ${\dot{Q}}_H$ is the heat rejected, negative sign is used. Thus,

  ${\dot{Q}}_H=-269.63\text{ kJ/s}$

The work done is calculated as:

  $\begin{array}{c}\dot{W}=\dot{m}\times \Delta H_{23}\\ =1.356\times 51.36\\ =69.64\text{ kW}\end{array}$

Use equation (5) to calculate the coefficient of performance for vapor-compression cycle as:

  $\begin{array}{c}\omega =\frac{H_2-H_1}{H_3-H_2}\\ =\frac{383.45-235.97}{434.81-383.45}\\ =2.87\end{array}$

For Carnot cycle, the coefficient of performance is calculated as:

  $\begin{array}{c}{\omega }_{\text{Carnot}}=\frac{T_C}{T_H-T_C}\\ =\frac{248.15}{299.15-248.15}\\ =4.87\end{array}$