Chapter 11.10, Problem 57P

Repeat Prob. 11–56 for a flash chamber pressure of 0.6 MPa.

To determine

The fraction of the refrigerant that evaporates as it is throttled to the flash chamber.

Answer to Problem 57P

The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is $0.2528$.

Explanation of Solution

Show the T-s diagram for compression refrigeration cycle as in Figure (1).

From Figure (1), write the specific enthalpy at state 5 is equal to state 6 due to throttling process.

$h_5\cong h_6$ (I)

Here, specific enthalpy at state 5 and 6 is $h_5\text{and}{h}_6$ respectively.

From Figure (1), write the specific enthalpy at state 7 is equal to state 8 due to throttling process.

$h_7\cong h_8$ (II)

Here, specific enthalpy at state 7 and 8 is $h_7\text{and}{h}_8$ respectively.

Express the fraction of the refrigerant that evaporates as it is throttled to the flash chamber

$x_6=\frac{h_6-h_8}{h_{fg@600\text{kPa}}}$ (III)

Here, specific enthalpy at saturated vapor is $h_g$ and specific enthalpy at evaporation and pressure of $\text{600}\text{kPa}\left(\text{0}\text{.6}\text{MPa}\right)$ is $h_{fg@600\text{kPa}}$.

Conclusion:

Perform unit conversion of pressure at state 1 from $\text{kPa}\text{to}\text{MPa}$.

$\begin{array}{c}{P}_1=0.1\text{MPa}\left[\frac{\text{1000}\text{kPa}}{\text{MPa}}\right]\\ =100\text{kPa}\end{array}$

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure at state 1 $\left(P_1\right)$ of $\text{100}\text{kPa}$.

$\begin{array}{c}{h}_1=h_g\\ =234.46\text{kJ}/\text{kg}\\ s_1=s_g\\ =0.9519\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, specific entropy and enthalpy at state 1 is $s_1\text{and}{h}_1$ respectively,  specific enthalpy and entropy at saturated vapor is $h_g\text{and}{s}_g$ respectively.

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2 corresponding to pressure at state 2 of $0.6\text{MPa}$ and specific entropy at state 2 $\left(s_2=s_1\right)$ of $0.9519\text{kJ}/\text{kg}\cdot \text{K}$ using interpolation method.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (IV)

Here, the variables denote by x and y is specific entropy at state 2 and specific enthalpy at state 2 respectively.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (1).

Specific entropy at state 2 $s_2\left(\text{kJ}/\text{kg}\cdot \text{K}\right)$	Specific enthalpy at state 2 $h_2\left(\text{kJ}/\text{kg}\right)$
0.9500 $\left(x_1\right)$	270.83 $\left(y_1\right)$
0.9519 $\left(x_2\right)$	$\left(y_2=?\right)$
0.9817 $\left(x_3\right)$	280.60 $\left(y_3\right)$

Substitute $\text{0}\text{.9500}\text{kJ}/\text{kg}\cdot \text{K}\text{,}0.9519\text{kJ}/\text{kg}\cdot \text{K}\text{and}\text{0}\text{.9817}\text{kJ}/\text{kg}\cdot \text{K}$ for $x_1,x_2\text{and}{x}_3$ respectively, $270.83\text{kJ}/\text{kg}$ for $y_1$ and $280.60\text{kJ}/\text{kg}$ for $y_3$ in Equation (IV).

$\begin{array}{c}{y}_2=\frac{\left[\left(\text{0}\text{.9519}-\text{9500}\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\left[\left(280.60-270.83\right)\text{kJ}/\text{kg}\right]}{\left(\text{0}\text{.9817}-\text{0}\text{.9500}\right)\text{kJ}/\text{kg}\cdot \text{K}}+270.83\text{kJ}/\text{kg}\\ =271.42\text{kJ}/\text{kg}\\ =h_2\end{array}$

Thus, the specific enthalpy at state 2 is,

$h_2=271.42\text{kJ}/\text{kg}$

Perform unit conversion of pressure at state 3 from $\text{kPa}\text{to}\text{MPa}$.

$\begin{array}{c}{P}_3=0.6\text{MPa}\left[\frac{\text{1000}\text{kPa}}{\text{MPa}}\right]\\ =600\text{kPa}\end{array}$

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 3 $\left(P_3\right)$ of $\text{600}\text{kPa}$.

$\begin{array}{c}{h}_3=h_g\\ =262.46\text{kJ}/\text{kg}\end{array}$

Perform unit conversion of pressure at state 5 from $\text{kPa}\text{to}\text{MPa}$.

$\begin{array}{c}{P}_5=1.4\text{MPa}\left[\frac{\text{1000}\text{kPa}}{\text{MPa}}\right]\\ =1400\text{kPa}\end{array}$

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 5 $\left(P_5\right)$ of $\text{1400}\text{kPa}$.

$\begin{array}{c}{h}_5=h_f\\ =127.25\text{kJ}/\text{kg}\end{array}$

Here, specific enthalpy at saturated liquid is $h_f$.

Substitute $127.25\text{kJ}/\text{kg}$ for $h_5$ in Equation (I).

$h_6=127.25\text{kJ}/\text{kg}$

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 8 $\left(P_8\right)$ of $\text{600}\text{kPa}$.

$\begin{array}{c}{h}_8=h_f\\ =81.50\text{kJ}/\text{kg}\end{array}$

Substitute $81.50\text{kJ}/\text{kg}$ for $h_8$ in Equation (II).

$h_7=81.50\text{kJ}/\text{kg}$

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the specific enthalpy at evaporation and pressure of $\text{400}\text{kPa}$.

$h_{fg@600\text{kPa}}=180.95\text{kJ}/\text{kg}$

Substitute $127.25\text{kJ}/\text{kg}$ for $h_6$, $81.50\text{kJ}/\text{kg}$ for $h_8$ and $180.95\text{kJ}/\text{kg}$ for $h_{fg@400\text{kPa}}$ in Equation (III).

$\begin{array}{c}{x}_6=\frac{127.25\text{kJ}/\text{kg}-81.50\text{kJ}/\text{kg}}{180.95\text{kJ}/\text{kg}}\\ =0.2528\end{array}$

Hence, the fraction of the refrigerant that evaporates as it is throttled to the flash chamber is $0.2528$.

To determine

The rate of heat removed from the refrigerated space.

Answer to Problem 57P

The rate of heat removed from the refrigerated space is $\text{28}\text{.57}\text{kW}$.

Explanation of Solution

Express the enthalpy at state 9 by using an energy balance on the mixing chamber.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}\\ {\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\displaystyle \sum {\dot{m}}_e{h}_e=}{\displaystyle \sum {\dot{m}}_i{h}_i}\end{array}$

$\begin{array}{l}{\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ \left(1\right)h_9=x_6{h}_3+\left(1-x_6\right)h_2\end{array}$ (V)

Here, the rate of total energy entering the system is ${\dot{E}}_{\text{in}}$, the rate of total energy leaving the system is ${\dot{E}}_{\text{out}}$, the rate of change in the total energy of the system is $\Delta {\dot{E}}_{\text{system}}$, mass flow rate at exit and inlet is ${\dot{m}}_e\text{and}{\dot{m}}_i$ respectively, and specific enthalpy at exit and inlet is $h_e\text{and}{h}_i$ respectively.

Express the mass flow rate through the flash chamber.

${\dot{m}}_B=\left(1-x_6\right){\dot{m}}_A$ (VI)

Here, mass flow rate through condenser is ${\dot{m}}_A$.

Express The rate of heat removed from the refrigerated space.

${\dot{Q}}_L={\dot{m}}_B\left(h_1-h_8\right)$ (VII)

Conclusion:

Substitute $0.2528$ for $x_6$, $\text{262}\text{.46}\text{kJ}/\text{kg}$ for $h_3$, and $\text{271}\text{.42}\text{kJ}/\text{kg}$ for $h_2$ in Equation (V).

$\begin{array}{l}\left(1\right)h_9=\left(0.2528\right)\left(\text{262}\text{.46}\text{kJ}/\text{kg}\right)+\left(1-0.2528\right)\left(\text{271}\text{.42}\text{kJ}/\text{kg}\right)\\ h_9=\text{269}\text{.15}\text{kJ}/\text{kg}\end{array}$

Substitute $0.2528$ for $x_6$ and $\text{0}\text{.25}\text{kg}/\text{s}$ for ${\dot{m}}_A$ in Equation (VI).

$\begin{array}{c}{\dot{m}}_B=\left(1-0.2528\right)\left(\text{0}\text{.25}\text{kg}/\text{s}\right)\\ =0.1868\text{kg}/\text{s}\end{array}$

Substitute $0.1868\text{kg}/\text{s}$ for ${\dot{m}}_B$, $\text{234}\text{.46}\text{kJ}/\text{kg}\text{and}\text{81}\text{.50}\text{kJ}/\text{kg}$ for $h_1\text{and}{h}_8$ respectively in Equation (VII).

$\begin{array}{c}{\dot{Q}}_L=\left(0.1868\text{kg}/\text{s}\right)\left(\text{234}\text{.46}\text{kJ}/\text{kg}-\text{81}\text{.50}\text{kJ}/\text{kg}\right)\\ =28.57\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =28.57\text{kW}\end{array}$

Hence, the rate of heat removed from the refrigerated space is $\text{28}\text{.57}\text{kW}$.

To determine

The coefficient of performance.

Answer to Problem 57P

The coefficient of performance is $\text{2}\text{.50}$.

Explanation of Solution

Express compressor work input per unit mass.

${\dot{W}}_{\text{in}}={\dot{m}}_A\left(h_4-h_9\right)+{\dot{m}}_B\left(h_2-h_1\right)$ (VIII)

Express the coefficient of performance.

${\text{COP}}_{\text{R}}=\frac{{\dot{Q}}_L}{{\dot{W}}_{\text{in}}}$ (IX)

Express entropy at state 4.

$s_4=x_6{s}_3+\left(1-x_6\right)s_2$ (X)

Here, specific entropy at state 3 is $s_3$.

Conclusion:

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 3 $\left(P_3\right)$ of $\text{600}\text{kPa}$.

$\begin{array}{c}{s}_3=s_g\\ =0.92196\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, specific entropy at saturated vapor is $s_g$.

Substitute $0.2528$ for $x_6$, $0.92196\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$ and $0.9519\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$ in Equation (X).

$\begin{array}{c}{s}_4=\left(0.2528\right)\left(0.92196\text{kJ}/\text{kg}\cdot \text{K}\right)+\left(1-0.2528\right)\left(0.9519\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =0.23307\text{kJ}/\text{kg}\cdot \text{K}+0.71125\text{kJ}/\text{kg}\cdot \text{K}\\ =0.9444\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 4 corresponding to pressure at state 4 of $1.4\text{MPa}$ and specific entropy at state 4 of $0.9444\text{kJ}/\text{kg}\cdot \text{K}$ using interpolation method.

Show the specific enthalpy at state 4 corresponding to specific entropy as in Table (2).

Specific entropy at state 4 $s_4\left(\text{kJ}/\text{kg}\cdot \text{K}\right)$	Specific enthalpy at state 4 $h_4\left(\text{kJ}/\text{kg}\right)$
0.9389 $\left(x_1\right)$	285.47 $\left(y_1\right)$
0.9444 $\left(x_2\right)$	$\left(y_2=?\right)$
0.9733 $\left(x_3\right)$	297.10 $\left(y_3\right)$

Use excels and substitute value from Table (2) in Equation (IV) to get,

$h_4=287.31\text{kJ}/\text{kg}$

Substitute $\text{0}\text{.25}\text{kg}/\text{s}\text{and}\text{0}\text{.1868}\text{kg}/\text{s}$ for ${\dot{m}}_A\text{and}{\dot{m}}_B$ respectively, $\text{287}\text{.31}\text{kJ}/\text{kg}$ for $h_4$, $\text{269}\text{.15}\text{kJ}/\text{kg}$ for $h_9$, $\text{271}\text{.42}\text{kJ}/\text{kg}\text{and}\text{234}\text{.46}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_1$ respectively in Equation (VIII).

$\begin{array}{c}{\dot{W}}_{\text{in}}=\left(\text{0}\text{.25}\text{kg}/\text{s}\right)\left(287.31-269.15\right)\text{kJ}/\text{kg}+\left(\text{0}\text{.1868}\text{kg}/\text{s}\right)\left(271.42-234.46\right)\text{kJ}/\text{kg}\\ =11.44\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =11.44\text{kW}\end{array}$

Substitute $\text{11}\text{.44}\text{kW}$ for ${\dot{W}}_{\text{in}}$ and $\text{28}\text{.57}\text{kW}$ for ${\dot{Q}}_L$ in Equation (IX).

$\begin{array}{c}{\text{COP}}_{\text{R}}=\frac{\text{28}\text{.57}\text{kW}}{11.44\text{kW}}\\ =2.50\end{array}$

Hence, the coefficient of performance is $\text{2}\text{.50}$.