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 T S diagram which include four steps of the cycle. Line 1 → 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 → 1 represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line 2 → 3 ' and the actual compression is shown by the line 2 → 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: | Q C | = H 2 − H 1 ...... (1) | Q H | = H 3 − H 4 ...... (2) The work of compression is: W = H 3 − H 2 ...... (3) The coefficient of performance is: ω = H 2 − H 1 H 3 − H 2 ...... (4) The rate of circulation of refrigerant, m ˙ is determined from the rate of heat absorption in the evaporator given by the equation: m ˙ = | Q ˙ C | H 2 − H 1 ...... (5) For Carnot refrigeration cycle, highest possible value of ω is attained at the given values of T C and T H . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of ω is obtained.

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Answer 1

Question

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(a)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $9.049$ .

The coefficient of performance for condensation temperature of $28∘C$ is $5.729$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the isentropic compression of vapor are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

Assume that the compressor efficiency is $η=1.00$ , so,

$H3=H′3$

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(16∘C)$ at pressure $5.043 bar$ as:

$H3(16∘C)=410.2 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87410.2−391.46=9.049$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(28∘C)$ at pressure $7.269 bar$ as:

$H3(28∘C)=418.1 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84418.1−391.46=5.729$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(40∘C)$ at pressure $10.166 bar$ as:

$H3(40∘C)=427.6 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41427.6−391.46=3.737$

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(b)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $6.786$ .

The coefficient of performance for condensation temperature of $28∘C$ is $4.297$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the compressor efficiency of $75%$ are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

The compressor efficiency is given as, $η=0.75$ .

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(16∘C)$ at pressure $5.043 bar$ as:

$H′3(16∘C)=410.2 kJ/kg$

Calculate $ΔH23(16∘C)$ as:

$ΔH23( 16∘C)= H ′3 ( 16 ∘ C)−H2η=410.2−391.460.75=24.99 kJ/kg$

Now, calculate $H3(16∘C)$ as:

$H3( 16∘C)=H2+ΔH23( 16∘C)=391.46+24.99=416.45 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87416.45−391.46=6.786$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(28∘C)$ at pressure $7.269 bar$ as:

$H′3(28∘C)=418.1 kJ/kg$

Calculate $ΔH23(28∘C)$ as:

$ΔH23( 28∘C)= H ′3 ( 28 ∘ C)−H2η=418.1−391.460.75=35.52 kJ/kg$

Now, calculate $H3(28∘C)$ as:

$H3( 28∘C)=H2+ΔH23( 28∘C)=391.46+35.52=426.98 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84426.98−391.46=4.297$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(40∘C)$ at pressure $10.166 bar$ as:

$H′3(40∘C)=427.6 kJ/kg$

Calculate $ΔH23(40∘C)$ as:

$ΔH23( 40∘C)= H ′3 ( 40 ∘ C)−H2η=427.6−391.460.75=48.19 kJ/kg$

Now, calculate $H3(40∘C)$ as:

$H3( 40∘C)=H2+ΔH23( 40∘C)=391.46+48.19=439.65 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41439.65−391.46=2.802$

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

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Answer 2

Question

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(a)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $9.049$ .

The coefficient of performance for condensation temperature of $28∘C$ is $5.729$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the isentropic compression of vapor are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

Assume that the compressor efficiency is $η=1.00$ , so,

$H3=H′3$

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(16∘C)$ at pressure $5.043 bar$ as:

$H3(16∘C)=410.2 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87410.2−391.46=9.049$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(28∘C)$ at pressure $7.269 bar$ as:

$H3(28∘C)=418.1 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84418.1−391.46=5.729$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(40∘C)$ at pressure $10.166 bar$ as:

$H3(40∘C)=427.6 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41427.6−391.46=3.737$

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(b)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $6.786$ .

The coefficient of performance for condensation temperature of $28∘C$ is $4.297$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the compressor efficiency of $75%$ are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

The compressor efficiency is given as, $η=0.75$ .

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(16∘C)$ at pressure $5.043 bar$ as:

$H′3(16∘C)=410.2 kJ/kg$

Calculate $ΔH23(16∘C)$ as:

$ΔH23( 16∘C)= H ′3 ( 16 ∘ C)−H2η=410.2−391.460.75=24.99 kJ/kg$

Now, calculate $H3(16∘C)$ as:

$H3( 16∘C)=H2+ΔH23( 16∘C)=391.46+24.99=416.45 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87416.45−391.46=6.786$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(28∘C)$ at pressure $7.269 bar$ as:

$H′3(28∘C)=418.1 kJ/kg$

Calculate $ΔH23(28∘C)$ as:

$ΔH23( 28∘C)= H ′3 ( 28 ∘ C)−H2η=418.1−391.460.75=35.52 kJ/kg$

Now, calculate $H3(28∘C)$ as:

$H3( 28∘C)=H2+ΔH23( 28∘C)=391.46+35.52=426.98 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84426.98−391.46=4.297$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(40∘C)$ at pressure $10.166 bar$ as:

$H′3(40∘C)=427.6 kJ/kg$

Calculate $ΔH23(40∘C)$ as:

$ΔH23( 40∘C)= H ′3 ( 40 ∘ C)−H2η=427.6−391.460.75=48.19 kJ/kg$

Now, calculate $H3(40∘C)$ as:

$H3( 40∘C)=H2+ΔH23( 40∘C)=391.46+48.19=439.65 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41439.65−391.46=2.802$

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

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Answer 3

Question

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(a)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $9.049$ .

The coefficient of performance for condensation temperature of $28∘C$ is $5.729$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the isentropic compression of vapor are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

Assume that the compressor efficiency is $η=1.00$ , so,

$H3=H′3$

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(16∘C)$ at pressure $5.043 bar$ as:

$H3(16∘C)=410.2 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87410.2−391.46=9.049$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(28∘C)$ at pressure $7.269 bar$ as:

$H3(28∘C)=418.1 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84418.1−391.46=5.729$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(40∘C)$ at pressure $10.166 bar$ as:

$H3(40∘C)=427.6 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41427.6−391.46=3.737$

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(b)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $6.786$ .

The coefficient of performance for condensation temperature of $28∘C$ is $4.297$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the compressor efficiency of $75%$ are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

The compressor efficiency is given as, $η=0.75$ .

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(16∘C)$ at pressure $5.043 bar$ as:

$H′3(16∘C)=410.2 kJ/kg$

Calculate $ΔH23(16∘C)$ as:

$ΔH23( 16∘C)= H ′3 ( 16 ∘ C)−H2η=410.2−391.460.75=24.99 kJ/kg$

Now, calculate $H3(16∘C)$ as:

$H3( 16∘C)=H2+ΔH23( 16∘C)=391.46+24.99=416.45 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87416.45−391.46=6.786$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(28∘C)$ at pressure $7.269 bar$ as:

$H′3(28∘C)=418.1 kJ/kg$

Calculate $ΔH23(28∘C)$ as:

$ΔH23( 28∘C)= H ′3 ( 28 ∘ C)−H2η=418.1−391.460.75=35.52 kJ/kg$

Now, calculate $H3(28∘C)$ as:

$H3( 28∘C)=H2+ΔH23( 28∘C)=391.46+35.52=426.98 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84426.98−391.46=4.297$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(40∘C)$ at pressure $10.166 bar$ as:

$H′3(40∘C)=427.6 kJ/kg$

Calculate $ΔH23(40∘C)$ as:

$ΔH23( 40∘C)= H ′3 ( 40 ∘ C)−H2η=427.6−391.460.75=48.19 kJ/kg$

Now, calculate $H3(40∘C)$ as:

$H3( 40∘C)=H2+ΔH23( 40∘C)=391.46+48.19=439.65 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41439.65−391.46=2.802$

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

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Answer 4

Question

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(a)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $9.049$ .

The coefficient of performance for condensation temperature of $28∘C$ is $5.729$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the isentropic compression of vapor are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

Assume that the compressor efficiency is $η=1.00$ , so,

$H3=H′3$

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(16∘C)$ at pressure $5.043 bar$ as:

$H3(16∘C)=410.2 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87410.2−391.46=9.049$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(28∘C)$ at pressure $7.269 bar$ as:

$H3(28∘C)=418.1 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84418.1−391.46=5.729$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(40∘C)$ at pressure $10.166 bar$ as:

$H3(40∘C)=427.6 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41427.6−391.46=3.737$

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(b)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $6.786$ .

The coefficient of performance for condensation temperature of $28∘C$ is $4.297$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the compressor efficiency of $75%$ are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

The compressor efficiency is given as, $η=0.75$ .

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(16∘C)$ at pressure $5.043 bar$ as:

$H′3(16∘C)=410.2 kJ/kg$

Calculate $ΔH23(16∘C)$ as:

$ΔH23( 16∘C)= H ′3 ( 16 ∘ C)−H2η=410.2−391.460.75=24.99 kJ/kg$

Now, calculate $H3(16∘C)$ as:

$H3( 16∘C)=H2+ΔH23( 16∘C)=391.46+24.99=416.45 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87416.45−391.46=6.786$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(28∘C)$ at pressure $7.269 bar$ as:

$H′3(28∘C)=418.1 kJ/kg$

Calculate $ΔH23(28∘C)$ as:

$ΔH23( 28∘C)= H ′3 ( 28 ∘ C)−H2η=418.1−391.460.75=35.52 kJ/kg$

Now, calculate $H3(28∘C)$ as:

$H3( 28∘C)=H2+ΔH23( 28∘C)=391.46+35.52=426.98 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84426.98−391.46=4.297$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(40∘C)$ at pressure $10.166 bar$ as:

$H′3(40∘C)=427.6 kJ/kg$

Calculate $ΔH23(40∘C)$ as:

$ΔH23( 40∘C)= H ′3 ( 40 ∘ C)−H2η=427.6−391.460.75=48.19 kJ/kg$

Now, calculate $H3(40∘C)$ as:

$H3( 40∘C)=H2+ΔH23( 40∘C)=391.46+48.19=439.65 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41439.65−391.46=2.802$

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

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Answer 5

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(a)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $9.049$ .

The coefficient of performance for condensation temperature of $28∘C$ is $5.729$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the isentropic compression of vapor are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

Assume that the compressor efficiency is $η=1.00$ , so,

$H3=H′3$

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(16∘C)$ at pressure $5.043 bar$ as:

$H3(16∘C)=410.2 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87410.2−391.46=9.049$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(28∘C)$ at pressure $7.269 bar$ as:

$H3(28∘C)=418.1 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84418.1−391.46=5.729$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(40∘C)$ at pressure $10.166 bar$ as:

$H3(40∘C)=427.6 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41427.6−391.46=3.737$

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(b)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $6.786$ .

The coefficient of performance for condensation temperature of $28∘C$ is $4.297$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the compressor efficiency of $75%$ are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

The compressor efficiency is given as, $η=0.75$ .

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(16∘C)$ at pressure $5.043 bar$ as:

$H′3(16∘C)=410.2 kJ/kg$

Calculate $ΔH23(16∘C)$ as:

$ΔH23( 16∘C)= H ′3 ( 16 ∘ C)−H2η=410.2−391.460.75=24.99 kJ/kg$

Now, calculate $H3(16∘C)$ as:

$H3( 16∘C)=H2+ΔH23( 16∘C)=391.46+24.99=416.45 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87416.45−391.46=6.786$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(28∘C)$ at pressure $7.269 bar$ as:

$H′3(28∘C)=418.1 kJ/kg$

Calculate $ΔH23(28∘C)$ as:

$ΔH23( 28∘C)= H ′3 ( 28 ∘ C)−H2η=418.1−391.460.75=35.52 kJ/kg$

Now, calculate $H3(28∘C)$ as:

$H3( 28∘C)=H2+ΔH23( 28∘C)=391.46+35.52=426.98 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84426.98−391.46=4.297$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(40∘C)$ at pressure $10.166 bar$ as:

$H′3(40∘C)=427.6 kJ/kg$

Calculate $ΔH23(40∘C)$ as:

$ΔH23( 40∘C)= H ′3 ( 40 ∘ C)−H2η=427.6−391.460.75=48.19 kJ/kg$

Now, calculate $H3(40∘C)$ as:

$H3( 40∘C)=H2+ΔH23( 40∘C)=391.46+48.19=439.65 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41439.65−391.46=2.802$

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

Answer 6

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(a)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $9.049$ .

The coefficient of performance for condensation temperature of $28∘C$ is $5.729$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the isentropic compression of vapor are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

Assume that the compressor efficiency is $η=1.00$ , so,

$H3=H′3$

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(16∘C)$ at pressure $5.043 bar$ as:

$H3(16∘C)=410.2 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87410.2−391.46=9.049$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(28∘C)$ at pressure $7.269 bar$ as:

$H3(28∘C)=418.1 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84418.1−391.46=5.729$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(40∘C)$ at pressure $10.166 bar$ as:

$H3(40∘C)=427.6 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41427.6−391.46=3.737$

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

Answer 7

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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

Answer 8

(a)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $9.049$ .

The coefficient of performance for condensation temperature of $28∘C$ is $5.729$ .

The coefficient of performance for condensation temperature of $40∘C$ is $3.737$ .

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

In a refrigerator, tetrafluoroethene acts as a refrigerant and operates with an evaporation temperature of $−12∘C$ . Different condensation temperature for the isentropic compression of vapor are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

Assume that the compressor efficiency is $η=1.00$ , so,

$H3=H′3$

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(16∘C)$ at pressure $5.043 bar$ as:

$H3(16∘C)=410.2 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87410.2−391.46=9.049$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(28∘C)$ at pressure $7.269 bar$ as:

$H3(28∘C)=418.1 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84418.1−391.46=5.729$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H3(40∘C)$ at pressure $10.166 bar$ as:

$H3(40∘C)=427.6 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41427.6−391.46=3.737$

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

Answer 13

(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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

(b)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $6.786$ .

The coefficient of performance for condensation temperature of $28∘C$ is $4.297$ .

The coefficient of performance for condensation temperature of $40∘C$ 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 $−12∘C$ . Different condensation temperature for the compressor efficiency of $75%$ are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

The compressor efficiency is given as, $η=0.75$ .

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(16∘C)$ at pressure $5.043 bar$ as:

$H′3(16∘C)=410.2 kJ/kg$

Calculate $ΔH23(16∘C)$ as:

$ΔH23( 16∘C)= H ′3 ( 16 ∘ C)−H2η=410.2−391.460.75=24.99 kJ/kg$

Now, calculate $H3(16∘C)$ as:

$H3( 16∘C)=H2+ΔH23( 16∘C)=391.46+24.99=416.45 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87416.45−391.46=6.786$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(28∘C)$ at pressure $7.269 bar$ as:

$H′3(28∘C)=418.1 kJ/kg$

Calculate $ΔH23(28∘C)$ as:

$ΔH23( 28∘C)= H ′3 ( 28 ∘ C)−H2η=418.1−391.460.75=35.52 kJ/kg$

Now, calculate $H3(28∘C)$ as:

$H3( 28∘C)=H2+ΔH23( 28∘C)=391.46+35.52=426.98 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84426.98−391.46=4.297$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(40∘C)$ at pressure $10.166 bar$ as:

$H′3(40∘C)=427.6 kJ/kg$

Calculate $ΔH23(40∘C)$ as:

$ΔH23( 40∘C)= H ′3 ( 40 ∘ C)−H2η=427.6−391.460.75=48.19 kJ/kg$

Now, calculate $H3(40∘C)$ as:

$H3( 40∘C)=H2+ΔH23( 40∘C)=391.46+48.19=439.65 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $40∘C$ is calculated as:

$ω( 40∘C)=H2−H1( 40 ∘ C)H3( 40 ∘ C)−H2=391.46−256.41439.65−391.46=2.802$

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

Answer 14

(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 $1→2$ 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 $4→1$ represents throttling process which occurs at constant enthalpy. The path of isentropic compression is shown by line $2→3'$ and the actual compression is shown by the line $2→3$ 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:

$|QC|=H2−H1$ ...... (1)

$|QH|=H3−H4$ ...... (2)

The work of compression is:

$W=H3−H2$ ...... (3)

The coefficient of performance is:

$ω=H2−H1H3−H2$ ...... (4)

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

$m˙=|Q˙C|H2−H1$ ...... (5)

For Carnot refrigeration cycle, highest possible value of $ω$ is attained at the given values of $TC and TH$ . Due to irreversible expansion in a throttle valve and irreversible compression in the vapor-compression cycle, lower values of $ω$ is obtained.

Answer 15

(b)

Expert Solution

Answer to Problem 9.13P

The coefficient of performance for condensation temperature of $16∘C$ is $6.786$ .

The coefficient of performance for condensation temperature of $28∘C$ is $4.297$ .

The coefficient of performance for condensation temperature of $40∘C$ is $2.802$ .

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

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

In a refrigerator, tetrafluoroethene acts as a refrigerant and operates with an evaporation temperature of $−12∘C$ . Different condensation temperature for the compressor efficiency of $75%$ are $16∘C, 28∘C, and 40∘C$ .

From table 9.1, the values of $H2, and S2$ and the pressure for the saturated tetrafluoroethene at $−12∘C$ are:

$H2=391.46 kJ/kgS2=1.735 kJ/(kg⋅K)P2=1.852 bar$

The compressor efficiency is given as, $η=0.75$ .

For condensation temperature of $16∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(16∘C)$ at this point is:

$Psat( 16∘C)=5.043 barH4( 16∘C)=221.87 kJ/kg$

Also, $H1(16∘C)=H4(16∘C)=221.87 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(16∘C)$ at pressure $5.043 bar$ as:

$H′3(16∘C)=410.2 kJ/kg$

Calculate $ΔH23(16∘C)$ as:

$ΔH23( 16∘C)= H ′3 ( 16 ∘ C)−H2η=410.2−391.460.75=24.99 kJ/kg$

Now, calculate $H3(16∘C)$ as:

$H3( 16∘C)=H2+ΔH23( 16∘C)=391.46+24.99=416.45 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $16∘C$ is calculated as:

$ω( 16∘C)=H2−H1( 16 ∘ C)H3( 16 ∘ C)−H2=391.46−221.87416.45−391.46=6.786$

For condensation temperature of $28∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(28∘C)$ at this point is:

$Psat( 28∘C)=7.269 barH4( 28∘C)=238.84 kJ/kg$

Also, $H1(28∘C)=H4(28∘C)=238.84 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(28∘C)$ at pressure $7.269 bar$ as:

$H′3(28∘C)=418.1 kJ/kg$

Calculate $ΔH23(28∘C)$ as:

$ΔH23( 28∘C)= H ′3 ( 28 ∘ C)−H2η=418.1−391.460.75=35.52 kJ/kg$

Now, calculate $H3(28∘C)$ as:

$H3( 28∘C)=H2+ΔH23( 28∘C)=391.46+35.52=426.98 kJ/kg$

Using equation (4), the coefficient of performance for condensation temperature of $28∘C$ is calculated as:

$ω( 28∘C)=H2−H1( 28 ∘ C)H3( 28 ∘ C)−H2=391.46−238.84426.98−391.46=4.297$

For condensation temperature of $40∘C$ .

The saturation pressure at point 4 is the pressure at which the vapor condenses and $H4(40∘C)$ at this point is:

$Psat( 40∘C)=10.166 barH4( 40∘C)=256.41 kJ/kg$

Also, $H1(40∘C)=H4(40∘C)=256.41 kJ/kg$

For isentropic compression,

$S′3=S2=1.735 kJ/(kg⋅K)$

At point $3′$ , the thermodynamic state properties are defines using the saturation pressure. From figure F.2 of appendix F, follow the constant enthalpy curve of $1.735 kJ/(kg⋅K)$ to get the enthalpy $H′3(40∘C)$ at pressure $10.166 bar$ as: