Chapter 11.10, Problem 59P

(a)

To determine

The mass flow rate of the refrigerant through the upper cycle.

Answer to Problem 59P

The mass flow rate of the refrigerant through the upper cycle is $\text{0}\text{.384}\text{kg}/\text{s}$.

Explanation of Solution

Express the specific enthalpy at state 2 using Carnot efficiency.

${\eta }_C=\frac{h_{2s}-h_1}{h_2-h_1}$ (I)

Here, specific enthalpy at state 1, 2 and 2s is $h_1,h_2\text{and}{h}_{2s}$ respectively and Carnot efficiency is ${\eta }_C$.

Express specific enthalpy at state 3.

$h_3=h_{f@500\text{kPa}}$ (II)

Here, specific enthalpy at saturated liquid and pressure of $\text{500}\text{kPa}$ is $h_{f@500\text{kPa}}$.

Write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

$h_3\cong h_4$ (III)

Here, specific enthalpy at state 3 and 4 is $h_3\text{and}{h}_4$ respectively.

Express specific enthalpy at state 5.

$h_5=h_{g@400\text{kPa}}$ (IV)

Here, specific enthalpy at saturated vapor and pressure of $\text{400}\text{kPa}$ is $h_{g@400\text{kPa}}$.

Express specific entropy at state 5.

$s_5=s_{g@400\text{kPa}}$ (V)

Here, specific entropy at saturated vapor and pressure of $\text{400}\text{kPa}$ is $s_{g@400\text{kPa}}$.

Express the specific enthalpy at state 6 using Carnot efficiency.

${\eta }_C=\frac{h_{6s}-h_5}{h_6-h_5}$ (VI)

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

Express specific enthalpy at state 7.

$h_7=h_{f@1400\text{kPa}}$ (VII)

Here, specific enthalpy at saturated liquid and pressure of $\text{1400}\text{kPa}$ is $h_{f@1400\text{kPa}}$.

Write the specific enthalpy at state 7 is equal to state 8 due to throttling process.

$h_7\cong h_8$ (VIII)

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

Express the mass flow rate of the refrigerant through the upper cycle.

${\dot{m}}_A\left(h_5-h_8\right)={\dot{m}}_B\left(h_2-h_3\right)$ (IX)

Here, mass flow rate of the refrigerant through the lower cycle is ${\dot{m}}_B$.

Conclusion:

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of $\text{160}\text{kPa}$.

$\begin{array}{c}{h}_1=h_g\\ =241.14\text{kJ}/\text{kg}\\ s_1=s_g\\ =0.420\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

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 2s corresponding to pressure at state 2 of $0.50\text{MPa}$ and specific entropy at state 2 $\left(s_2=s_1\right)$ of $0.9420\text{kJ}/\text{kg}\cdot \text{K}$ using interpolation method.

Write the formula of interpolation method of two variables.

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

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

Show the specific enthalpy at state 2s 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 2s $h_{2s}\left(\text{kJ}/\text{kg}\right)$
0.9384 $\left(x_1\right)$	263.48 $\left(y_1\right)$
0.9420 $\left(x_2\right)$	$\left(y_2=?\right)$
0.9704 $\left(x_3\right)$	273.03 $\left(y_3\right)$

Substitute $\text{0}\text{.9384}\text{kJ}/\text{kg}\cdot \text{K}\text{,}0.9420\text{kJ}/\text{kg}\cdot \text{K}\text{and}\text{0}\text{.9704}\text{kJ}/\text{kg}\cdot \text{K}$ for $x_1,x_2\text{and}{x}_3$ respectively, $263.48\text{kJ}/\text{kg}$ for $y_1$ and $273.03\text{kJ}/\text{kg}$ for $y_3$ in Equation (X).

$\begin{array}{c}{y}_2=\frac{\left[\left(\text{0}\text{.9420}-\text{0}\text{.9384}\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\left[\left(273.03-263.48\right)\text{kJ}/\text{kg}\right]}{\left(\text{0}\text{.9704}-\text{0}\text{.9384}\right)\text{kJ}/\text{kg}\cdot \text{K}}+263.48\text{kJ}/\text{kg}\\ =264.55\text{kJ}/\text{kg}\\ =h_{2s}\end{array}$

Thus, the specific enthalpy at state 2s is,

$h_{2s}=264.55\text{kJ}/\text{kg}$

Substitute $264.55\text{kJ}/\text{kg}$ for $h_{2s}$, $241.14\text{kJ}/\text{kg}$ for $h_1$ and $0.80$ for ${\eta }_C$ in Equation (I).

$\begin{array}{l}0.80=\frac{264.55\text{kJ}/\text{kg}-241.14\text{kJ}/\text{kg}}{h_2-241.14\text{kJ}/\text{kg}}\\ 0.80\left(h_2-241.14\text{kJ}/\text{kg}\right)=23.41\text{kJ}/\text{kg}\\ 0.80h_2-192.912\text{kJ}/\text{kg}=23.41\text{kJ}/\text{kg}\\ h_2=270.41\text{kJ}/\text{kg}\end{array}$

Perform unit conversion of pressure at state 3 and 5 from $\text{MPa}\text{to}\text{kPa}$.

$\begin{array}{c}{P}_3=0.5\text{MPa}\\ =0.5\text{MPa}\left[\frac{\text{1000}\text{kPa}}{\text{MPa}}\right]\\ =500\text{kPa}\end{array}$

$\begin{array}{c}{P}_5=0.4\text{MPa}\\ =0.4\text{MPa}\left[\frac{\text{1000}\text{kPa}}{\text{MPa}}\right]\\ =400\text{kPa}\end{array}$

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 3 $\left(P_3\right)$ of $\text{500}\text{kPa}$.

$\begin{array}{c}{h}_3=h_{f@500\text{kPa}}\\ =73.32\text{kJ}/\text{kg}\end{array}$

Substitute $73.32\text{kJ}/\text{kg}$ for $h_{f@500\text{kPa}}$ in Equation (II).

$h_3=73.32\text{kJ}/\text{kg}$

Substitute $73.32\text{kJ}/\text{kg}$ for $h_4$ in Equation (III).

$h_4=73.32\text{kJ}/\text{kg}$

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of $\text{400}\text{kPa}$.

$\begin{array}{c}{h}_5=h_{g@400\text{kPa}}\\ =255.61\text{kJ}/\text{kg}\\ s_5=s_{g@400\text{kPa}}\\ =0.9271\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $255.61\text{kJ}/\text{kg}$ for $h_{g@400\text{kPa}}$ in Equation (IV).

$h_5=255.61\text{kJ}/\text{kg}$

Substitute $0.9271\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{g@400\text{kPa}}$ in Equation (V).

$s_5=0.9271\text{kJ}/\text{kg}\cdot \text{K}$

Perform unit conversion of pressure at state 6 from $\text{kPa}\text{to}\text{MPa}$.

$\begin{array}{c}{P}_6=1.4\text{MPa}\\ =1.4\text{MPa}\left[\frac{\text{1000}\text{kPa}}{\text{MPa}}\right]\\ =1400\text{kPa}\end{array}$

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 6s corresponding to pressure at state 6 of $\text{1}\text{.4}\text{MPa}$ and specific entropy at state 6 $\left(s_6=s_5\right)$ of $0.9271\text{kJ}/\text{kg}\cdot \text{K}$ using interpolation method.

Show the specific enthalpy at state 6s corresponding to specific entropy as in Table (2).

Specific entropy at state 2 $s_6\left(\text{kJ}/\text{kg}\cdot \text{K}\right)$	Specific enthalpy at state 6s $h_{6s}\left(\text{kJ}/\text{kg}\right)$
0.9107 $\left(x_1\right)$	276.17 $\left(y_1\right)$
0.9271 $\left(x_2\right)$	$\left(y_2=?\right)$
0.9389 $\left(x_3\right)$	285.47 $\left(y_3\right)$

Use excels and tabulates the values from Table (2) in Equation (X) to get,

$h_{6s}=281.56\text{kJ}/\text{kg}$

Substitute $281.56\text{kJ}/\text{kg}$ for $h_{6s}$, $255.61\text{kJ}/\text{kg}$ for $h_5$ and $0.80$ for ${\eta }_C$ in Equation (VI).

$\begin{array}{l}0.80=\frac{281.56\text{kJ}/\text{kg}-255.61\text{kJ}/\text{kg}}{h_6-255.61\text{kJ}/\text{kg}}\\ 0.80\left(h_6-255.61\text{kJ}/\text{kg}\right)=281.56\text{kJ}/\text{kg}-255.61\text{kJ}/\text{kg}\\ 0.80h_6-204.488\text{kJ}/\text{kg}=25.95\text{kJ}/\text{kg}\\ h_6=288.04\text{kJ}/\text{kg}\end{array}$

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 7 $\left(P_7\right)$ of $\text{1400}\text{kPa}$.

$\begin{array}{c}{h}_7=h_{f@1400\text{kPa}}\\ =127.25\text{kJ}/\text{kg}\end{array}$

Substitute $127.25\text{kJ}/\text{kg}$ for $h_{f@1400\text{kPa}}$ in Equation (VII).

$h_7=127.25\text{kJ}/\text{kg}$

Substitute $127.25\text{kJ}/\text{kg}$ for $h_7$ in Equation (VIII).

$h_8=127.25\text{kJ}/\text{kg}$

Substitute $255.61\text{kJ}/\text{kg}$ for $h_5$, $127.25\text{kJ}/\text{kg}$ for $h_8$, $\text{0}\text{.25}\text{kg}/\text{s}$ for ${\dot{m}}_B$, $270.41\text{kJ}/\text{kg}$ for $h_2$ and $73.32\text{kJ}/\text{kg}$ for $h_3$ in Equation (IX).

$\begin{array}{l}{\dot{m}}_A\left(255.61-127.25\right)=\left(\text{0}\text{.25}\text{kg}/\text{s}\right)\left(270.41-73.32\right)\text{kJ}/\text{kg}\\ {\dot{m}}_A\left(128.36\text{kJ}/\text{kg}\right)=\left(\text{0}\text{.25}\text{kg}/\text{s}\right)\left(197.09\text{kJ}/\text{kg}\right)\\ {\dot{m}}_A=0.3839\text{kg}/\text{s}\\ \simeq 0.384\text{kg}/\text{s}\end{array}$

Hence, the mass flow rate of the refrigerant through the upper cycle is $\text{0}\text{.384}\text{kg}/\text{s}$.

(b)

To determine

The rate of heat removal from the refrigerated space.

Answer to Problem 59P

The rate of heat removal from the refrigerated space is $\text{42}\text{kW}$.

Explanation of Solution

Express the rate of heat removal from the refrigerated space.

${\dot{Q}}_L={\dot{m}}_B\left(h_1-h_4\right)$ (XI)

Conclusion:

Substitute $\text{0}\text{.25}\text{kg}/\text{s}$ for ${\dot{m}}_B$, $241.14\text{kJ}/\text{kg}$ for $h_1$ and $73.32\text{kJ}/\text{kg}$ for $h_4$ in Equation (XI).

$\begin{array}{c}{\dot{Q}}_L=\left(\text{0}\text{.25}\text{kg}/\text{s}\right)\left(241.14-73.32\right)\text{kJ}/\text{kg}\\ =41.96\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =41.96\text{kW}\\ \simeq 42\text{kW}\end{array}$

Hence, the rate of heat removal from the refrigerated space is $\text{42}\text{kW}$.

(c)

To determine

The COP of the refrigerator.

Answer to Problem 59P

The COP of the refrigerator is $\text{2}\text{.12}$.

Explanation of Solution

Express the power input.

${\dot{W}}_{\text{in}}={\dot{m}}_A\left(h_6-h_5\right)+{\dot{m}}_B\left(h_2-h_1\right)$ (XII)

Express the COP of the refrigerator.

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

Conclusion:

Substitute $\text{0}\text{.3839}\text{kg}/\text{s}\text{and}\text{0}\text{.25}\text{kg}/\text{s}$ for ${\dot{m}}_A\text{and}{\dot{m}}_B$ respectively, $\text{288}\text{.04}\text{kJ}/\text{kg}\text{and}\text{255}\text{.61}\text{kJ}/\text{kg}$ for $h_6\text{and}{h}_5$ respectively, and $\text{270}\text{.41}\text{kJ}/\text{kg}\text{and}\text{241}\text{.14}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_1$ respectively in Equation (XII).

$\begin{array}{c}{\dot{W}}_{\text{in}}=\left(\text{0}\text{.3839}\text{kg}/\text{s}\right)\left(\text{288}\text{.04}-\text{255}\text{.61}\right)\text{kJ}/\text{kg}+\left(\text{0}\text{.25}\text{kg}/\text{s}\right)\left(270.41-241.14\right)\text{kJ}/\text{kg}\\ =19.77\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =19.77\text{kW}\end{array}$

Substitute $\text{19}\text{.77}\text{kW}$ for ${\dot{W}}_{\text{in}}$ and $\text{41}\text{.96}\text{kW}$ for ${\dot{Q}}_L$ in Equation (XIII).

$\begin{array}{c}\text{COP}=\frac{\text{41}\text{.96}\text{kW}}{\text{19}\text{.77}\text{kW}}\\ =2.12\end{array}$

Hence, the COP of the refrigerator is $\text{2}\text{.12}$.