Chapter 11.10, Problem 44P

Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 50°C at a rate of 0.022 kg/s and leaves at 750 kPa subcooled by 3°C. The refrigerant enters the compressor at 200 kPa superheated by 4°C. Determine (a) the isentropic efficiency of the compressor, (b) the rate of heat supplied to the heated room, and (c) the COP of the heat pump. Also, determine (d) the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the pressure limits of 200 and 800 kPa.

FIGURE P11–42

To determine

The isentropic efficiency of the compressor.

Answer to Problem 44P

The isentropic efficiency of the compressor is $\text{0}\text{.757}$.

Explanation of Solution

Show the T-s diagram for process as in Figure (1).

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 and 4 is $h_3\text{and}{h}_4$ respectively.

Express isentropic efficiency of the compressor.

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

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

Express the temperature at state 3.

$T_3=T_{\text{sat@750}\text{kPa}}-T_{\text{sub-cooled}}$ (II)

Here, saturated temperature at pressure of $\text{750}\text{kPa}$ is $T_{\text{sat@750}\text{kPa}}$ and sub-cooled temperature is $T_{\text{sub-cooled}}$.

Express the temperature at state 1.

$T_1=T_{\text{sat@200}\text{kPa}}-T_{\text{superheated}}$ (III)

Here, saturated temperature at pressure of $\text{200}\text{kPa}$ is $T_{\text{sat@750}\text{kPa}}$ and superheated temperature is $T_{\text{superheated}}$.

Express quality at state 2s.

$s_2=s_{f\text{@800}\text{kPa}}+x_{2s}{s}_{fg\text{@800}\text{kPa}}$ (IV)

Here, specific entropy at saturated liquid and evaporation and $-10\text{°C}$ is $s_{f\text{@800}\text{kPa}}$ and $s_{fg\text{@800}\text{kPa}}$ respectively.

Express specific enthalpy at state 2s.

$h_{2s}=h_{f\text{@800}\text{kPa}}+x_{2s}{h}_{fg\text{@800}\text{kPa}}$ (V)

Here, specific enthalpy at saturated liquid and evaporation and $-10\text{°C}$ is $h_{f\text{@800}\text{kPa}}$ and $h_{fg\text{@800}\text{kPa}}$ respectively.

Conclusion:

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

$\begin{array}{c}{P}_2=800\text{kPa}\left[\frac{\text{MPa}}{\text{1000}\text{kPa}}\right]\\ =0.8\text{MPa}\end{array}$

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2 corresponding to pressure at state 2 of $0.8\text{MPa}$ and temperature at state 2 of $\text{50°C}$.

$h_2=286.71\text{kJ}/\text{kg}$

Refer Table A-12, “saturated refrigerant-134a-pressure table” and write saturated temperature at pressure of $\text{750}\text{kPa}$.

$T_{\text{sat@750}\text{kPa}}=29.06\text{°C}$

Substitute $29.06\text{°C}$ for $T_{\text{sat@750}\text{kPa}}$ and $\text{3°C}$ for $T_{\text{sub-cooled}}$ in Equation (II).

$\begin{array}{c}{T}_3=29.06\text{°C}-3\text{°C}\\ =\text{26}\text{.06°C}\end{array}$

Refer Table A-12, “saturated refrigerant-134a-pressure table” and write specific enthalpy at state 3 corresponding to pressure at state 3 of $\text{750}\text{kPa}$ and temperature at state 3 of $\text{26}\text{.06°C}$ using an interpolation method.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y is temperature at state 3 and specific enthalpy at state 3 respectively.

Show the specific enthalpy at state 3 corresponding to temperature as in Table (1).

Temperature at state 3 $T_3\left(\text{°C}\right)$	Specific enthalpy at state 3 $h_3\left(\text{kJ}/\text{kg}\right)$
26.69 $\left(x_1\right)$	88.82 $\left(y_1\right)$
26.06 $\left(x_2\right)$	$\left(y_2=?\right)$
29.06 $\left(x_3\right)$	92.22 $\left(y_3\right)$

Substitute $\text{26}\text{.69}°\text{C,}\text{26}\text{.06}°\text{C}\text{and}\text{29}\text{.06}°\text{C}$ for $x_1,x_2\text{and}{x}_3$ respectively, $88.82\text{kJ}/\text{kg}$ for $y_1$ and $92.22\text{kJ}/\text{kg}$ for $y_3$ in Equation (VI).

$\begin{array}{c}{y}_2=\frac{\left[\left(\text{26}\text{.06}°\text{C}-\text{26}\text{.69}°\text{C}\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\left[\left(92.22-88.82\right)\text{kJ}/\text{kg}\right]}{\left(\text{29}\text{.06}°\text{C}-\text{26}\text{.69}°\text{C}\right)}+88.82\text{kJ}/\text{kg}\\ =87.93\text{kJ}/\text{kg}\\ =h_3\end{array}$

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

$\begin{array}{c}{h}_3\cong h_4\\ =87.93\text{kJ}/\text{kg}\end{array}$

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

$T_{\text{sat@200}\text{kPa}}=-10.09\text{°C}$

Substitute $-10.09\text{°C}$ for $T_{\text{sat@200}\text{kPa}}$ and $\text{4°C}$ for $T_{\text{superheated}}$ in Equation (III).

$\begin{array}{c}{T}_1=-10.09\text{°C}+\text{4°C}\\ =-6.09\text{°C}\end{array}$

Refer Table A-12, “saturated refrigerant-134a-pressure table” and write specific enthalpy and entropy at state 1 corresponding to pressure at state 1 of $\text{200}\text{kPa}$ and temperature at state 1 of $-\text{6}\text{.09°C}$ using an interpolation method.

$\begin{array}{l}{h}_1=247.88\text{kJ}/\text{kg}\\ s_1=0.9507\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, specific entropy at state 1 is $s_1$.

The specific entropy at state 1 is equal to specific entropy at state 1.

$\begin{array}{c}{s}_1=s_2\\ =0.9507\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, specific entropy at state 2 is $s_2$.

Refer Table A-12, “saturated refrigerant-134a-pressure table” and write the properties corresponding to pressure at state 2 of $\text{800}\text{kPa}$

$\begin{array}{l}{s}_{f\text{@800}\text{kPa}}=0.35408\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg\text{@800}\text{kPa}}=0.56445\text{kJ}/\text{kg}\cdot \text{K}\\ h_{f\text{@800}\text{kPa}}=95.48\text{kJ}/\text{kg}\\ h_{fg\text{@800}\text{kPa}}=171.86\text{kJ}/\text{kg}\end{array}$

Substitute $0.9507\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, $0.35408\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{f\text{@800}\text{kPa}}$ and $0.56445\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg\text{@800}\text{kPa}}$ in Equation (IV).

$\begin{array}{l}0.9507\text{kJ}/\text{kg}\cdot \text{K}=0.35408\text{kJ}/\text{kg}\cdot \text{K}+x_{2s}\left(0.56445\text{kJ}/\text{kg}\cdot \text{K}\right)\\ x_{2s}=1.056\end{array}$

Substitute $1.056$ for $x_{2s}$, $95.48\text{kJ}/\text{kg}$ for $h_{f\text{@800}\text{kPa}}$ and $171.86\text{kJ}/\text{kg}$ for $h_{fg\text{@800}\text{kPa}}$ in Equation (V).

$\begin{array}{c}{h}_{2s}=95.48\text{kJ}/\text{kg}+\left(1.056\right)\left(171.86\text{kJ}/\text{kg}\right)\\ =277.28\text{kJ}/\text{kg}\end{array}$

Substitute $\text{277}\text{.28}\text{kJ}/\text{kg}\text{and}\text{247}\text{.88}\text{kJ}/\text{kg}$ for $h_{2s}\text{and}{h}_1$ respectively, and $286.71\text{kJ}/\text{kg}$ for $h_2$ in Equation (I).

$\begin{array}{c}{\eta }_C=\frac{\text{277}\text{.28}\text{kJ}/\text{kg}-\text{247}\text{.88}\text{kJ}/\text{kg}}{\text{286}\text{.71}\text{kJ}/\text{kg}-\text{247}\text{.88}\text{kJ}/\text{kg}}\\ =0.757\end{array}$

Hence, the isentropic efficiency of the compressor is $\text{0}\text{.757}$.

To determine

The rate of heat supplied to the heated room.

Answer to Problem 44P

The rate of heat supplied to the heated room is $\text{4}\text{.373}\text{kW}$.

Explanation of Solution

Express the rate of heat supplied to the heated room.

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

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

Conclusion:

Substitute $\text{0}\text{.022}\text{kg}/\text{s}$ for $\dot{m}$, $\text{286}\text{.71}\text{kJ}/\text{kg}\text{and}\text{87}\text{.93}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_3$ respectively in equation (VII).

$\begin{array}{c}{\dot{Q}}_H=\left(\text{0}\text{.022}\text{kg}/\text{s}\right)\left(286.71-87.93\right)\text{kJ}/\text{kg}\\ =4.373\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =4.373\text{kW}\end{array}$

Hence, the rate of heat supplied to the heated room is $\text{4}\text{.373}\text{kW}$.

To determine

The COP of the heat pump.

Answer to Problem 44P

The COP of the heat pump is $\text{5}\text{.12}$.

Explanation of Solution

Express the rate of work input.

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

Express coefficient of performance of heat pump.

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

Conclusion:

Substitute $\text{0}\text{.022}\text{kg}/\text{s}$ for $\dot{m}$, $\text{286}\text{.71}\text{kJ}/\text{kg}\text{and}\text{247}\text{.88}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_1$ respectively in equation (VIII).

$\begin{array}{c}{\dot{W}}_{\text{in}}=\left(\text{0}\text{.022}\text{kg}/\text{s}\right)\left(286.71-247.88\right)\text{kJ}/\text{kg}\\ =0.8542\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =0.8542\text{kW}\end{array}$

Substitute $0.8542\text{kW}$ for ${\dot{W}}_{\text{in}}$ and $4.373\text{kW}$ for ${\dot{W}}_{\text{in}}$ in Equation (IX).

$\begin{array}{c}\text{COP}=\frac{4.373\text{kW}}{0.8542\text{kW}}\\ =5.12\end{array}$

Hence, the COP of the heat pump is $\text{5}\text{.12}$.

To determine

The COP and the rate of heat supplied to the heated room.

Answer to Problem 44P

The COP and the rate of heat supplied to the heated room is $\text{6}\text{.18}\text{and}\text{3}\text{.91}\text{kW}$ respectively.

Explanation of Solution

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

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

$h_3\cong h_4$

Express the coefficient of performance.

$\text{COP}=\frac{h_2-h_3}{h_2-h_1}$ (X)

Express the rate of heat supplied to the heated room.

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

Conclusion:

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

$\begin{array}{c}{h}_1=h_g\\ =244.50\text{kJ}/\text{kg}\\ s_1=s_g\\ =0.9379\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

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 $0.8\text{MPa}$ and specific entropy at state 2 $\left(s_2=s_1\right)$ of $0.9379\text{kJ}/\text{kg}\cdot \text{K}$ 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 $h_2\left(\text{kJ}/\text{kg}\right)$
0.9185 $\left(x_1\right)$	267.34 $\left(y_1\right)$
0.9379 $\left(x_2\right)$	$\left(y_2=?\right)$
0.9481 $\left(x_3\right)$	276.46 $\left(y_3\right)$

Use Excels and substitute the values from Table (2) in Equation (VI) to get,

$h_2=95.48\text{kJ}/\text{kg}$

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 $\text{800}\text{kPa}$.

$\begin{array}{c}{h}_3=h_4\\ =h_f\\ =95.48\text{kJ}/\text{kg}\end{array}$

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

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

$\begin{array}{c}{h}_3\cong h_4\\ =95.48\text{kJ}/\text{kg}\end{array}$

Substitute $\text{273}\text{.29}\text{kJ}/\text{kg}\text{and}\text{95}\text{.48}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_3$ respectively, and $244.50\text{kJ}/\text{kg}$ for $h_1$ in Equation (X).

$\begin{array}{c}\text{COP}=\frac{\text{273}\text{.29}\text{kJ}/\text{kg}-\text{95}\text{.48}\text{kJ}/\text{kg}}{\text{273}\text{.29}\text{kJ}/\text{kg}-\text{244}\text{.50}\text{kJ}/\text{kg}}\\ =6.18\end{array}$

Substitute $\text{0}\text{.022}\text{kg}/\text{s}$ for $\dot{m}$, $\text{273}\text{.29}\text{kJ}/\text{kg}\text{and}\text{95}\text{.48}\text{kJ}/\text{kg}$ for $h_2\text{and}{h}_3$ respectively in equation (XI).

$\begin{array}{c}{\dot{Q}}_H=\left(\text{0}\text{.022}\text{kg}/\text{s}\right)\left(273.29-95.48\right)\text{kJ}/\text{kg}\\ =3.91\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =3.91\text{kW}\end{array}$

Hence, the COP and the rate of heat supplied to the heated room is $\text{6}\text{.18}\text{and}\text{3}\text{.91}\text{kW}$ respectively.