Chapter 9, Problem 9.13P

(a)

Interpretation Introduction

Interpretation:

The effect of condensation temperature on the coefficient of performance is to be determined for different condensation temperatures assuming isentropic compression of vapor.

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 9.13P

The coefficient of performance for condensation temperature of ${16}^{\circ }\text{C}$ is $9.049$ .

The coefficient of performance for condensation temperature of ${28}^{\circ }\text{C}$ is $5.729$ .

The coefficient of performance for condensation temperature of ${40}^{\circ }\text{C}$ is $3.737$ .

As the condensation temperature is increased, the coefficient of performance decreases.

Explanation of Solution

Given information:

In a refrigerator, tetrafluoroethene acts as a refrigerant and operates with an evaporation temperature of $-{12}^{\circ }\text{C}$ . Different condensation temperature for the isentropic compression of vapor are ${16}^{\circ }\text{C, }{28}^{\circ }\text{C},\text{ and }{40}^{\circ }\text{C}$ .

From table 9.1, the values of $H_2,\text{ and }{S}_2$ and the pressure for the saturated tetrafluoroethene at $-{12}^{\circ }\text{C}$ are:

  $\begin{array}{c}{H}_2=391.46\text{ kJ/kg}\\ S_2=1.735\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)\\ P_2=1.852\text{ bar}\end{array}$

Assume that the compressor efficiency is $\eta =1.00$ , so,

  $H_3={H^{\prime }}_3$

For condensation temperature of ${16}^{\circ }\text{C}$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H_4\left({16}^{\circ }\text{C}\right)$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}\left({16}^{\circ }\text{C}\right)=5.043\text{ bar}\\ H_4\left({16}^{\circ }\text{C}\right)=221.87\text{ kJ/kg}\end{array}$

Also, $H_1\left({16}^{\circ }\text{C}\right)=H_4\left({16}^{\circ }\text{C}\right)=221.87\text{ kJ/kg}$

For isentropic compression,

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

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 $H_3\left({16}^{\circ }\text{C}\right)$ at pressure $5.043\text{ bar}$ as:

  $H_3\left({16}^{\circ }\text{C}\right)=410.2\text{ kJ/kg}$

Using equation (4), the coefficient of performance for condensation temperature of ${16}^{\circ }\text{C}$ is calculated as:

  $\begin{array}{c}\omega \left({16}^{\circ }\text{C}\right)=\frac{H_2-H_1\left({16}^{\circ }\text{C}\right)}{H_3\left({16}^{\circ }\text{C}\right)-H_2}\\ =\frac{391.46-221.87}{410.2-391.46}\\ =9.049\end{array}$

For condensation temperature of ${28}^{\circ }\text{C}$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H_4\left({28}^{\circ }\text{C}\right)$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}\left({28}^{\circ }\text{C}\right)=7.269\text{ bar}\\ H_4\left({28}^{\circ }\text{C}\right)=238.84\text{ kJ/kg}\end{array}$

Also, $H_1\left({28}^{\circ }\text{C}\right)=H_4\left({28}^{\circ }\text{C}\right)=238.84\text{ kJ/kg}$

For isentropic compression,

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

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 $H_3\left({28}^{\circ }\text{C}\right)$ at pressure $7.269\text{ bar}$ as:

  $H_3\left({28}^{\circ }\text{C}\right)=418.1\text{ kJ/kg}$

Using equation (4), the coefficient of performance for condensation temperature of ${28}^{\circ }\text{C}$ is calculated as:

  $\begin{array}{c}\omega \left({28}^{\circ }\text{C}\right)=\frac{H_2-H_1\left({28}^{\circ }\text{C}\right)}{H_3\left({28}^{\circ }\text{C}\right)-H_2}\\ =\frac{391.46-238.84}{418.1-391.46}\\ =5.729\end{array}$

For condensation temperature of ${40}^{\circ }\text{C}$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H_4\left({40}^{\circ }\text{C}\right)$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}\left({40}^{\circ }\text{C}\right)=10.166\text{ bar}\\ H_4\left({40}^{\circ }\text{C}\right)=256.41\text{ kJ/kg}\end{array}$

Also, $H_1\left({40}^{\circ }\text{C}\right)=H_4\left({40}^{\circ }\text{C}\right)=256.41\text{ kJ/kg}$

For isentropic compression,

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

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 $H_3\left({40}^{\circ }\text{C}\right)$ at pressure $10.166\text{ bar}$ as:

  $H_3\left({40}^{\circ }\text{C}\right)=427.6\text{ kJ/kg}$

Using equation (4), the coefficient of performance for condensation temperature of ${40}^{\circ }\text{C}$ is calculated as:

  $\begin{array}{c}\omega \left({40}^{\circ }\text{C}\right)=\frac{H_2-H_1\left({40}^{\circ }\text{C}\right)}{H_3\left({40}^{\circ }\text{C}\right)-H_2}\\ =\frac{391.46-256.41}{427.6-391.46}\\ =3.737\end{array}$

As the condensation temperature is increased, the coefficient of performance decreases.

(b)

Interpretation Introduction

Interpretation:

The effect of condensation temperature on the coefficient of performance is to be determined for different condensation temperatures assuming compressor efficiency of $75\%$ .

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 9.13P

The coefficient of performance for condensation temperature of ${16}^{\circ }\text{C}$ is $6.786$ .

The coefficient of performance for condensation temperature of ${28}^{\circ }\text{C}$ is $4.297$ .

The coefficient of performance for condensation temperature of ${40}^{\circ }\text{C}$ is $2.802$ .

As the condensation temperature is increased, the coefficient of performance decreases.

Explanation of Solution

Given information:

In a refrigerator, tetrafluoroethene acts as a refrigerant and operates with an evaporation temperature of $-{12}^{\circ }\text{C}$ . Different condensation temperature for the compressor efficiency of $75\%$ are ${16}^{\circ }\text{C, }{28}^{\circ }\text{C},\text{ and }{40}^{\circ }\text{C}$ .

From table 9.1, the values of $H_2,\text{ and }{S}_2$ and the pressure for the saturated tetrafluoroethene at $-{12}^{\circ }\text{C}$ are:

  $\begin{array}{c}{H}_2=391.46\text{ kJ/kg}\\ S_2=1.735\text{ kJ/}\left(\text{kg}\cdot \text{K}\right)\\ P_2=1.852\text{ bar}\end{array}$

The compressor efficiency is given as, $\eta =0.75$ .

For condensation temperature of ${16}^{\circ }\text{C}$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H_4\left({16}^{\circ }\text{C}\right)$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}\left({16}^{\circ }\text{C}\right)=5.043\text{ bar}\\ H_4\left({16}^{\circ }\text{C}\right)=221.87\text{ kJ/kg}\end{array}$

Also, $H_1\left({16}^{\circ }\text{C}\right)=H_4\left({16}^{\circ }\text{C}\right)=221.87\text{ kJ/kg}$

For isentropic compression,

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

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 ${H^{\prime }}_3\left({16}^{\circ }\text{C}\right)$ at pressure $5.043\text{ bar}$ as:

  ${H^{\prime }}_3\left({16}^{\circ }\text{C}\right)=410.2\text{ kJ/kg}$

Calculate $\Delta H_{23}\left({16}^{\circ }\text{C}\right)$ as:

  $\begin{array}{c}\Delta H_{23}\left({16}^{\circ }\text{C}\right)=\frac{{H^{\prime }}_3 \left({16}^{\circ }\text{C}\right)-H_2}{\eta }\\ =\frac{410.2-391.46}{0.75}\\ =24.99\text{ kJ/kg}\end{array}$

Now, calculate $H_3\left({16}^{\circ }\text{C}\right)$ as:

  $\begin{array}{c}{H}_3\left({16}^{\circ }\text{C}\right)=H_2+\Delta H_{23}\left({16}^{\circ }\text{C}\right)\\ =391.46+24.99\\ =416.45\text{ kJ/kg}\end{array}$

Using equation (4), the coefficient of performance for condensation temperature of ${16}^{\circ }\text{C}$ is calculated as:

  $\begin{array}{c}\omega \left({16}^{\circ }\text{C}\right)=\frac{H_2-H_1\left({16}^{\circ }\text{C}\right)}{H_3\left({16}^{\circ }\text{C}\right)-H_2}\\ =\frac{391.46-221.87}{416.45-391.46}\\ =6.786\end{array}$

For condensation temperature of ${28}^{\circ }\text{C}$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H_4\left({28}^{\circ }\text{C}\right)$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}\left({28}^{\circ }\text{C}\right)=7.269\text{ bar}\\ H_4\left({28}^{\circ }\text{C}\right)=238.84\text{ kJ/kg}\end{array}$

Also, $H_1\left({28}^{\circ }\text{C}\right)=H_4\left({28}^{\circ }\text{C}\right)=238.84\text{ kJ/kg}$

For isentropic compression,

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

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 ${H^{\prime }}_3\left({28}^{\circ }\text{C}\right)$ at pressure $7.269\text{ bar}$ as:

  ${H^{\prime }}_3\left({28}^{\circ }\text{C}\right)=418.1\text{ kJ/kg}$

Calculate $\Delta H_{23}\left({28}^{\circ }\text{C}\right)$ as:

  $\begin{array}{c}\Delta H_{23}\left({28}^{\circ }\text{C}\right)=\frac{{H^{\prime }}_3 \left({28}^{\circ }\text{C}\right)-H_2}{\eta }\\ =\frac{418.1-391.46}{0.75}\\ =35.52\text{ kJ/kg}\end{array}$

Now, calculate $H_3\left({28}^{\circ }\text{C}\right)$ as:

  $\begin{array}{c}{H}_3\left({28}^{\circ }\text{C}\right)=H_2+\Delta H_{23}\left({28}^{\circ }\text{C}\right)\\ =391.46+35.52\\ =426.98\text{ kJ/kg}\end{array}$

Using equation (4), the coefficient of performance for condensation temperature of ${28}^{\circ }\text{C}$ is calculated as:

  $\begin{array}{c}\omega \left({28}^{\circ }\text{C}\right)=\frac{H_2-H_1\left({28}^{\circ }\text{C}\right)}{H_3\left({28}^{\circ }\text{C}\right)-H_2}\\ =\frac{391.46-238.84}{426.98-391.46}\\ =4.297\end{array}$

For condensation temperature of ${40}^{\circ }\text{C}$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H_4\left({40}^{\circ }\text{C}\right)$ at this point is:

  $\begin{array}{c}{P}^{\text{sat}}\left({40}^{\circ }\text{C}\right)=10.166\text{ bar}\\ H_4\left({40}^{\circ }\text{C}\right)=256.41\text{ kJ/kg}\end{array}$

Also, $H_1\left({40}^{\circ }\text{C}\right)=H_4\left({40}^{\circ }\text{C}\right)=256.41\text{ kJ/kg}$

For isentropic compression,

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

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 ${H^{\prime }}_3\left({40}^{\circ }\text{C}\right)$ at pressure $10.166\text{ bar}$ as:

  ${H^{\prime }}_3\left({40}^{\circ }\text{C}\right)=427.6\text{ kJ/kg}$

Calculate $\Delta H_{23}\left({40}^{\circ }\text{C}\right)$ as:

  $\begin{array}{c}\Delta H_{23}\left({40}^{\circ }\text{C}\right)=\frac{{H^{\prime }}_3 \left({40}^{\circ }\text{C}\right)-H_2}{\eta }\\ =\frac{427.6-391.46}{0.75}\\ =48.19\text{ kJ/kg}\end{array}$

Now, calculate $H_3\left({40}^{\circ }\text{C}\right)$ as:

  $\begin{array}{c}{H}_3\left({40}^{\circ }\text{C}\right)=H_2+\Delta H_{23}\left({40}^{\circ }\text{C}\right)\\ =391.46+48.19\\ =439.65\text{ kJ/kg}\end{array}$

Using equation (4), the coefficient of performance for condensation temperature of ${40}^{\circ }\text{C}$ is calculated as:

  $\begin{array}{c}\omega \left({40}^{\circ }\text{C}\right)=\frac{H_2-H_1\left({40}^{\circ }\text{C}\right)}{H_3\left({40}^{\circ }\text{C}\right)-H_2}\\ =\frac{391.46-256.41}{439.65-391.46}\\ =2.802\end{array}$

As the condensation temperature is increased, the coefficient of performance decreases.