Chapter 9, Problem 9.11P

(a)

Interpretation Introduction

Interpretation:

The circulation rate of refrigerant for the given system is 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.

(b)

Interpretation Introduction

Interpretation:

The reduction in the circulation rate is to be calculated if throttle valve is replaced by a turbine where the refrigerant expands isentropically.

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.

(c)

Interpretation Introduction

Interpretation:

If the cycle of (a) is modified, from the condenser if liquid enters the exchanger at ${26}^{\circ }\text{C}$ and from the evaporator if vapor enters at $-{25}^{\circ }\text{C}$ and leaves at ${20}^{\circ }\text{C}$ then the circulation rate of refrigerant for the given system is 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.

(d)

Interpretation Introduction

Interpretation:

The coefficient of performance for all the parts (a), (b), and (c) are to be calculated for isentropic compression.

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.