Chapter 11.10, Problem 21P

A refrigerator uses refrigerant-134a as the working fluid and operates on the vapor-compression refrigeration cycle. The evaporator and condenser pressures are 200 kPa and 1400 kPa, respectively. The isentropic efficiency of the compressor is 88 percent. The refrigerant enters the compressor at a rate of 0.025 kg/s superheated by 10.1°C and leaves the condenser subcooled by 4.4°C. Determine (a) the rate of cooling provided by the evaporator, the power input, and the COP. Determine (b) the same parameters if the cycle operated on the ideal vapor-compression refrigeration cycle between the same pressure limits.

To determine

The rate of cooling provided by the evaporator, the power input and the COP.

Answer to Problem 21P

The rate of cooling provided by the evaporator is $\text{3}\text{.317}\text{kW}\text{and}\text{4}\text{.535}\text{kW}$ respectively, the power input is $\text{1}\text{.218}\text{kW}$ and the COP is $2.723$.

Explanation of Solution

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

Express specific enthalpy at state 2 by using the formula of Carnot efficiency.

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

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

Express rate of heat lost by the evaporator.

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

Here, mass flow rate of refrigerant is $\dot{m}$, specific enthalpy at state 4 is $h_4$.

Express rate of heat supplied to the evaporator.

${\dot{Q}}_H=\dot{m}\left(h_2-h_3\right)$ (III)

Here, specific enthalpy at state 3 is $h_3$.

Express power input.

${\dot{W}}_{\text{in}}=\dot{m}\left(h_2-h_1\right)$ (IV)

Express coefficient of performance.

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

Express initial temperature.

$T_1=T_{\text{sat@200}\text{kPa}}+T_{\text{IR}}$ (VI)

Here, saturated temperature at initial pressure of $\text{200}\text{kPa}$ is $T_{\text{sat@200}\text{kPa}}$ and initial temperature of refrigerant is $T_{\text{IR}}$.

Express temperature at state 3.

$T_3=T_{\text{sat@1400}\text{kPa}}-T_{\text{FR}}$ (VII)

Here, saturated temperature at pressure at state 3 of $\text{1400}\text{kPa}$ is $T_{\text{sat@1400}\text{kPa}}$ and final temperature of refrigerant is $T_{\text{FR}}$.

Conclusion:

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the saturated temperature at initial pressure of $\text{200}\text{kPa}$.

$T_{\text{sat@200}\text{kPa}}=-10.1\text{°C}$

Substitute $-10.1\text{°C}$ for $T_{\text{sat@200}\text{kPa}}$ and $-10.1\text{°C}$ for $T_{\text{IR}}$ in Equation (VI).

$\begin{array}{c}{T}_1=-10.1\text{°C}+\left(-10.1\text{°C}\right)\\ =0\text{°C}\end{array}$

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the saturated temperature at pressure at state 3 of $\text{1400}\text{kPa}$.

$T_{\text{sat@1400}\text{kPa}}=52.4\text{°C}$

Substitute $52.4\text{°C}$ for $T_{\text{sat@1400}\text{kPa}}$ and $4.4\text{°C}$ for $T_{\text{FR}}$ in Equation (VII).

$\begin{array}{c}{T}_3=52.4\text{°C}-4.4\text{°C}\\ =48\text{°C}\end{array}$

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to initial pressure $\left(P_1\right)$ of $\text{200}\text{kPa}$ and initial temperature of $\text{0°C}$.

$\begin{array}{l}{h}_1=253.07\text{kJ}/\text{kg}\\ s_1=0.9699\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, initial specific entropy is $s_1$.

From Figure (1), the initial specific entropy is equal to specific entropy at state 2.

$\begin{array}{c}{s}_1=s_2\\ =0.9699\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Perform unit conversion of pressure at state 2 from $\text{kPa}\text{to}\text{MPa}$.

$\begin{array}{c}{P}_2=1400\text{kPa}\left[\frac{\text{MPa}}{\text{1000}\text{kPa}}\right]\\ =1.4\text{MPa}\end{array}$

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of $\text{1}\text{.4}\text{MPa}$ and specific entropy at state 2 $\left(s_2=s_1\right)$ of $0.9699\text{kJ}/\text{kg}\cdot \text{K}$ using interpolation method.

Write the formula of interpolation method of two variables.

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

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 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 $h_{2s}\left(\text{kJ}/\text{kg}\right)$
0.9389 $\left(x_1\right)$	285.47 $\left(y_1\right)$
0.9699 $\left(x_2\right)$	$\left(y_2=?\right)$
0.9733 $\left(x_3\right)$	297.10 $\left(y_3\right)$

Substitute $\text{0}\text{.9389}\text{kJ}/\text{kg}\cdot \text{K}\text{,}0.9699\text{kJ}/\text{kg}\cdot \text{K}\text{and}\text{0}\text{.9733}\text{kJ}/\text{kg}\cdot \text{K}$ for $x_1,x_2\text{and}{x}_3$ respectively, $285.47\text{kJ}/\text{kg}$ for $y_1$ and $297.10\text{kJ}/\text{kg}$ for $y_3$ in Equation (VIII).

$\begin{array}{c}{y}_2=\frac{\left[\left(\text{0}\text{.9699}-\text{0}\text{.9389}\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\left[\left(297.10-285.47\right)\text{kJ}/\text{kg}\right]}{\left(\text{0}\text{.9733}-\text{0}\text{.9389}\right)\text{kJ}/\text{kg}\cdot \text{K}}+285.47\text{kJ}/\text{kg}\\ =295.95\text{kJ}/\text{kg}\\ =h_{2s}\end{array}$

From Figure (1), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

$h_3\cong h_4$

Here, specific enthalpy at state 3 is $h_3$.

Refer Table A-11, “saturated refrigerant-134a-pressure table”, and write the specific enthalpy at state 3 corresponding to temperature at state 3 of $\text{48}°\text{C}$

$\begin{array}{c}{h}_3=h_4\\ =h_f\\ =120.41\text{kJ}/\text{kg}\end{array}$

Substitute $\text{295}\text{.95}\text{kJ}/\text{kg}$ for $h_{2s}$, $0.88$ for ${\eta }_C$, and $\text{253}\text{.07}\text{kJ}/\text{kg}$ for $h_1$ in Equation (I).

$\begin{array}{l}0.88=\frac{\left(\text{295}\text{.95}-253.07\right)\text{kJ}/\text{kg}}{h_2-253.07\text{kJ}/\text{kg}}\\ h_2-253.07\text{kJ}/\text{kg}=48.7272\text{kJ}/\text{kg}\\ h_2=301.80\text{kJ}/\text{kg}\end{array}$

Substitute $\text{0}\text{.025}\text{kg}/\text{s}$ for $\dot{m}$, $\text{253}\text{.07}\text{kJ}/\text{kg}\text{and}\text{120}\text{.41}\text{kJ}/\text{kg}$ for $h_1\text{and}{h}_4$ respectively in Equation (II).

$\begin{array}{c}{\dot{Q}}_L=\text{0}\text{.025}\text{kg}/\text{s}\left(\text{253}\text{.07}\text{kJ}/\text{kg}-\text{120}\text{.41}\text{kJ}/\text{kg}\right)\\ =3.317\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =3.317\text{kW}\end{array}$

Substitute $\text{0}\text{.025}\text{kg}/\text{s}$ for $\dot{m}$, $\text{301}\text{.80}\text{kJ}/\text{kg}\text{and}\text{120}\text{.41}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_3$ respectively in Equation (III).

$\begin{array}{c}{\dot{Q}}_H=\text{0}\text{.025}\text{kg}/\text{s}\left(301.80\text{kJ}/\text{kg}-120.41\text{kJ}/\text{kg}\right)\\ =4.535\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =4.535\text{kW}\end{array}$

Hence, the rate of cooling provided by the evaporator is $\text{3}\text{.317}\text{kW}\text{and}\text{4}\text{.535}\text{kW}$ respectively.

Substitute $\text{0}\text{.025}\text{kg}/\text{s}$ for $\dot{m}$, $\text{301}\text{.80}\text{kJ}/\text{kg}\text{and}\text{253}\text{.07}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_1$ respectively in Equation (IV).

$\begin{array}{c}{\dot{W}}_{\text{in}}=\text{0}\text{.025}\text{kg}/\text{s}\left(\text{301}\text{.80}\text{kJ}/\text{kg}-\text{253}\text{.07}\text{kJ}/\text{kg}\right)\\ =1.218\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =1.218\text{kW}\end{array}$

Hence, the power input is $\text{1}\text{.218}\text{kW}$.

Substitute $1.218\text{kW}$ for ${\dot{W}}_{\text{in}}$ and $\text{3}\text{.317}\text{kW}$ for ${\dot{Q}}_L$ in Equation (V).

$\begin{array}{c}\text{COP}=\frac{\text{3}\text{.317}\text{kW}}{1.218\text{kW}}\\ =2.723\end{array}$

Hence, the COP is $2.723$.

To determine

The rate of cooling provided by the evaporator, the power input and the COP.

Answer to Problem 21P

The rate of cooling provided by the evaporator is $\text{2}\text{.931}\text{kW}\text{and}\text{3}\text{.947}\text{kW}$ respectively, the power input is $\text{1}\text{.016}\text{kW}$ and the COP is $2.885$.

Explanation of Solution

Show the T-s diagram for the ideal vapor-compression refrigeration cycle as in Figure (2).

Express rate of heat lost by the evaporator.

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

Here, mass flow rate of refrigerant is $\dot{m}$, specific enthalpy at state 4 is $h_4$.

Express rate of heat supplied to the evaporator.

${\dot{Q}}_H=\dot{m}\left(h_2-h_3\right)$ (X)

Here, specific enthalpy at state 3 is $h_3$.

Express power input.

${\dot{W}}_{\text{in}}=\dot{m}\left(h_2-h_1\right)$ (XI)

Express coefficient of performance.

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

Conclusion:

From Figure (2), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

$h_3\cong h_4$

Here, specific enthalpy at state 3 is $h_3$.

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure at state 1 $\left(P_1\right)$ of $\text{200}\text{kPa}$.

$\begin{array}{c}{h}_1=h_g\\ =244.50\text{kJ}/\text{kg}\cdot \\ s_1=s_g\\ =0.9378\text{kJ}/\text{kgK}\end{array}$

Here, specific entropy at state 1 is $s_1$, 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 $\text{1}\text{.4}\text{MPa}$ and specific entropy at state 2 $\left(s_2=s_1\right)$ of $0.9378\text{kJ}/\text{kgK}$ using interpolation method.

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

Specific entropy at state 2 $s_2\left(\text{kJ}/\text{kg}\cdot \text{K}\right)$	Specific enthalpy at state 2 $h_2\left(\text{kJ}/\text{kg}\right)$
0.9107 $\left(x_1\right)$	276.17 $\left(y_1\right)$
0.9378 $\left(x_2\right)$	$\left(y_2=?\right)$
0.9389 $\left(x_3\right)$	285.47 $\left(y_3\right)$

Substitute $\text{0}\text{.9107}\text{kJ}/\text{kg}\cdot \text{K}\text{,}0.9378\text{kJ}/\text{kg}\cdot \text{K}\text{and}\text{0}\text{.9389}\text{kJ}/\text{kg}\cdot \text{K}$ for $x_1,x_2\text{and}{x}_3$ respectively, $276.17\text{kJ}/\text{kg}$ for $y_1$ and $285.47\text{kJ}/\text{kg}$ for $y_3$ in Equation (VIII).

$\begin{array}{c}{y}_2=\frac{\left[\left(\text{0}\text{.9378}-\text{0}\text{.9107}\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\left[\left(285.47-276.17\right)\text{kJ}/\text{kg}\right]}{\left(\text{0}\text{.9389}-\text{0}\text{.9107}\right)\text{kJ}/\text{kg}\cdot \text{K}}+276.17\text{kJ}/\text{kg}\\ =285.13\text{kJ}/\text{kg}\\ =h_2\end{array}$

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the specific enthalpy at state 3 corresponding to pressure at state 3 $\left(P_3\right)$ of $\text{1400}\text{kPa}$.

$\begin{array}{c}{h}_3=h_4\\ =h_f\\ =127.25\text{kJ}/\text{kg}\end{array}$

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

Substitute $\text{0}\text{.025}\text{kg}/\text{s}$ for $\dot{m}$, $\text{244}\text{.50}\text{kJ}/\text{kg}\text{and}\text{127}\text{.25}\text{kJ}/\text{kg}$ for $h_1\text{and}{h}_4$ respectively in Equation (IX).

$\begin{array}{c}{\dot{Q}}_L=\text{0}\text{.025}\text{kg}/\text{s}\left(\text{244}\text{.50}\text{kJ}/\text{kg}-\text{127}\text{.25}\text{kJ}/\text{kg}\right)\\ =2.931\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =2.931\text{kW}\end{array}$

Substitute $\text{0}\text{.025}\text{kg}/\text{s}$ for $\dot{m}$, $285.13\text{kJ}/\text{kg}\text{and}\text{127}\text{.25}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_3$ respectively in Equation (X).

$\begin{array}{c}{\dot{Q}}_H=\text{0}\text{.025}\text{kg}/\text{s}\left(285.13\text{kJ}/\text{kg}-127.25\text{kJ}/\text{kg}\right)\\ =3.947\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =3.947\text{kW}\end{array}$

Hence, the rate of cooling provided by the evaporator is $\text{2}\text{.931}\text{kW}\text{and}\text{3}\text{.947}\text{kW}$ respectively.

Substitute $\text{0}\text{.025}\text{kg}/\text{s}$ for $\dot{m}$, $\text{285}\text{.13}\text{kJ}/\text{kg}\text{and}\text{244}\text{.50}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_1$ respectively in Equation (XI).

$\begin{array}{c}{\dot{W}}_{\text{in}}=\text{0}\text{.025}\text{kg}/\text{s}\left(\text{285}\text{.13}\text{kJ}/\text{kg}-\text{244}\text{.50}\text{kJ}/\text{kg}\right)\\ =1.016\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =1.016\text{kW}\end{array}$

Hence, the power input is $\text{1}\text{.016}\text{kW}$.

Substitute $1.016\text{kW}$ for ${\dot{W}}_{\text{in}}$ and $\text{2}\text{.931}\text{kW}$ for ${\dot{Q}}_L$ in Equation (XII).

$\begin{array}{c}\text{COP}=\frac{2.931\text{kW}}{1.016\text{kW}}\\ =2.885\end{array}$

Hence, the COP is $2.885$.