Chapter 11.10, Problem 115RP

An air conditioner with refrigerant-134a as the working fluid is used to keep a room at 26°C by rejecting the waste heat to the outside air at 34°C. The room is gaining heat through the walls and the windows at a rate of 250 kJ/min while the heat generated by the computer, TV, and lights amounts to 900 W. An unknown amount of heat is also generated by the people in the room. The condenser and evaporator pressures are 1200 and 500 kPa, respectively. The refrigerant is saturated liquid at the condenser exit and saturated vapor at the compressor inlet. If the refrigerant enters the compressor at a rate of 100 L/min and the isentropic efficiency of the compressor is 75 percent, determine (a) the temperature of the refrigerant at the compressor exit, (b) the rate of heat generation by the people in the room, (c) the COP of the air conditioner, and (d) the minimum volume flow rate of the refrigerant at the compressor inlet for the same compressor inlet and exit conditions.

FIGURE P11–115

To determine

The temperature of the refrigerant at the compressor exit.

Answer to Problem 115RP

The temperature of the refrigerant at the compressor exit is $\text{54}\text{.5°C}$.

Explanation of Solution

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

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

$h_3\cong h_4$ (I)

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

Express the specific enthalpy at state 2.

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

Here, specific enthalpy at state 2s is $h_{2s}$, isentropic efficiency is ${\eta }_C$ and specific enthalpy at state 1 and 2 is $h_1\text{and}{h}_2$ respectively.

Conclusion:

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

$\begin{array}{c}{P}_1=500\text{kPa}\\ =500\text{kPa}\left[\frac{\text{MPa}}{1000\text{kPa}}\right]\\ =0.5\text{MPa}\end{array}$

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to pressure of $\text{0}\text{.5}\text{MPa}$.

$\begin{array}{l}{h}_1=259.36\text{kJ}/\text{kg}\\ v_1=0.04117{\text{m}}^3/\text{kg}\\ s_1=0.9242\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, specific enthalpy, volume and entropy is $h_1,v_1\text{and}{s}_1$ respectively.

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

$\begin{array}{c}{P}_2=1200\text{kPa}\\ =1200\text{kPa}\left[\frac{\text{MPa}}{1000\text{kPa}}\right]\\ =1.2\text{MPa}\end{array}$

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the specific enthalpy at state 3 corresponding to pressure at state 3 of $\text{1200}\text{kPa}$.

$\begin{array}{c}{h}_3=h_f\\ =117.79\text{kJ}/\text{kg}\end{array}$

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

Substitute $117.79\text{kJ}/\text{kg}$ for $h_3$ in Equation (I).

$h_4=117.79\text{kJ}/\text{kg}$

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of $1.2\text{MPa}$ and specific entropy at state 2 $\left(s_2=s_1\right)$ of $0.9242\text{kJ}/\text{kg}\cdot \text{K}$ using interpolation method.

Write the formula of interpolation method of two variables.

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

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 $h_2\left(\text{kJ}/\text{kg}\right)$
0.9132 $\left(x_1\right)$	273.92 $\left(y_1\right)$
0.9242 $\left(x_2\right)$	$\left(y_2=?\right)$
0.9268 $\left(x_3\right)$	278.28 $\left(y_3\right)$

Substitute $\text{0}\text{.9132}\text{kJ}/\text{kg}\cdot \text{K}\text{,}0.9242\text{kJ}/\text{kg}\cdot \text{K}\text{and}\text{0}\text{.9268}\text{kJ}/\text{kg}\cdot \text{K}$ for $x_1,x_2\text{and}{x}_3$ respectively, $273.92\text{kJ}/\text{kg}$ for $y_1$ and $278.28\text{kJ}/\text{kg}$ for $y_3$ in Equation (III).

$\begin{array}{c}{y}_2=\frac{\left[\left(\text{0}\text{.9242}-\text{0}\text{.9132}\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\left[\left(278.28-273.92\right)\text{kJ}/\text{kg}\right]}{\left(\text{0}\text{.9268}-\text{0}\text{.9132}\right)\text{kJ}/\text{kg}\cdot \text{K}}+273.92\text{kJ}/\text{kg}\\ =277.45\text{kJ}/\text{kg}\\ =h_{2s}\end{array}$

Thus, the specific enthalpy at state 2s is,

$h_{2s}=277.45\text{kJ}/\text{kg}$

Substitute $0.75$ for ${\eta }_C$, $277.45\text{kJ}/\text{kg}$ for $h_{2s}$ and $259.36\text{kJ}/\text{kg}$ for $h_1$ in Equation (II).

$\begin{array}{l}0.75=\frac{277.45\text{kJ}/\text{kg}-259.36\text{kJ}/\text{kg}}{h_2-259.36\text{kJ}/\text{kg}}\\ h_2=283.48\text{kJ}/\text{kg}\end{array}$

Refer Table A-13, “superheated refrigerant 134a”, and write the temperature at state 2 corresponding to pressure at state 2 of $1.2\text{MPa}$ and specific enthalpy at state 2 of $283.48\text{kJ}/\text{kg}$ using interpolation method.

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

Specific enthalpy at state 2s $h_2\left(\text{kJ}/\text{kg}\right)$	Temperature at state 2 $T_2\left(\text{°C}\right)$
278.28 $\left(x_1\right)$	50 $\left(y_1\right)$
283.48 $\left(x_2\right)$	$\left(y_2=?\right)$
289.66 $\left(x_3\right)$	60 $\left(y_3\right)$

Use excels and tabulates the values form Table (2) in Equation (III) to get,

$T_2=54.5\text{°C}$

Hence, the temperature of the refrigerant at the compressor exit is $\text{54}\text{.5°C}$.

To determine

The rate of heat generation by the people in the room.

Answer to Problem 115RP

The rate of heat generation by the people in the room is $\text{0}\text{.665}\text{kW}$.

Explanation of Solution

Express the mass flow rate of the refrigerant.

$\dot{m}=\frac{{\dot{\nu }}_1}{v_1}$ (IV)

Here, volume flow rate at state 1 is ${\dot{\nu }}_1$ and specific volume at state 1 is $v_1$.

Express the refrigeration load.

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

Express the rate of heat generation by the people in the room.

${\dot{Q}}_{\text{people}}={\dot{Q}}_L-{\dot{Q}}_{\text{heat}}-{\dot{Q}}_{\text{equip}}$ (VI)

Here, rate of heat generated is ${\dot{Q}}_{\text{heat}}$ and rate of heat generated by equipment is ${\dot{Q}}_{\text{equip}}$.

Conclusion:

Substitute $\text{100}\text{L}/\text{min}$ for ${\dot{\nu }}_1$ and $0.04117{\text{m}}^3/\text{kg}$ for $v_1$ in Equation (IV).

$\begin{array}{c}\dot{m}=\frac{\text{100}\text{L}/\text{min}}{0.04117{\text{m}}^3/\text{kg}}\\ =\frac{\text{100}\text{L}/\text{min}\left[\frac{{\text{m}}^{\text{3}}}{1000\text{L}}\right]\left[\frac{\min }{60\text{s}}\right]}{0.04117{\text{m}}^3/\text{kg}}\\ =0.04048\text{kg}/\text{s}\end{array}$

Substitute $0.04048\text{kg}/\text{s}$ for $\dot{m}$, $117.79\text{kJ}/\text{kg}$ for $h_4$ and $259.36\text{kJ}/\text{kg}$ for $h_1$ in Equation (V).

$\begin{array}{c}{\dot{Q}}_L=0.04048\text{kg}/\text{s}\left(259.36\text{kJ}/\text{kg}-117.79\text{kJ}/\text{kg}\right)\\ =5.731\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =5.731\text{kW}\end{array}$

Substitute $5.731\text{kW}$ for ${\dot{Q}}_L$, $\text{250}\text{kJ}/\text{min}$ for ${\dot{Q}}_{\text{heat}}$ and $\text{900}\text{W}$ for ${\dot{Q}}_{\text{equip}}$ in Equation (VI).

$\begin{array}{c}{\dot{Q}}_{\text{people}}=5.731\text{kW}-\text{250}\text{kJ}/\text{min}-\text{900}\text{W}\\ =5.731\text{kW}-\left[\left(\text{250}\text{kJ}/\text{min}\right)\frac{\min }{60\text{s}}\right]-\left[\left(\text{900}\text{W}\right)\frac{\text{kW}}{\text{1000}\text{W}}\right]\\ =5.731\text{kW}-\left(4.167\text{kJ}/\text{s}\right)\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]-0.9\text{kW}\end{array}$

$\begin{array}{c}=5.731\text{kW}-4.167\text{kW}-0.9\text{kW}\\ {\dot{Q}}_{\text{people}}=0.665\text{kW}\end{array}$

Hence, the rate of heat generation by the people in the room is $\text{0}\text{.665}\text{kW}$.

To determine

The COP of the air conditioner.

Answer to Problem 115RP

The COP of the air conditioner is $\text{5}\text{.87}$.

Explanation of Solution

Express the rate of work input.

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

Express the coefficient of performance of the air conditioner.

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

Conclusion:

Substitute $0.04048\text{kg}/\text{s}$ for $\dot{m}$, $283.48\text{kJ}/\text{kg}$ for $h_2$ and $259.36\text{kJ}/\text{kg}$ for $h_1$ in Equation (VII).

$\begin{array}{c}{\dot{W}}_{\text{in}}=0.04048\text{kg}/\text{s}\left(283.48\text{kJ}/\text{kg}-259.36\text{kJ}/\text{kg}\right)\\ =0.9764\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =0.9764\text{kW}\end{array}$

Substitute $0.9764\text{kW}$ for ${\dot{W}}_{\text{in}}$ and $5.731\text{kW}$ for ${\dot{Q}}_L$ in Equation (VIII).

$\begin{array}{c}\text{COP}=\frac{5.731\text{kW}}{0.9764\text{kW}}\\ =5.87\end{array}$

Hence, the COP of the air conditioner is $\text{5}\text{.87}$.

To determine

The minimum volume flow rate of the refrigerant at the compressor inlet.

Answer to Problem 115RP

The minimum volume flow rate of the refrigerant at the compressor inlet is $\text{15}\text{.7}\text{L}/\min$.

Explanation of Solution

Express the reversible coefficient of performance of the cycle.

${\text{COP}}_{\text{rev}}=\frac{1}{\frac{T_H}{T_L}-1}$ (IX)

Here, high and low temperature medium is $T_H\text{and}{T}_L$ respectively.

Express corresponding minimum power input.

${\dot{W}}_{\text{in,min}}=\frac{{\dot{Q}}_L}{{\text{COP}}_{\text{rev}}}$ (X)

Express the minimum mass flow rate.

${\dot{m}}_{\min }=\frac{{\dot{W}}_{\text{in,min}}}{h_2-h_1}$ (XI)

Express the minimum volume flow rate of the refrigerant at the compressor inlet

${\dot{\nu }}_{1,\min }={\dot{m}}_{\min }{v}_1$ (XII)

Conclusion:

Substitute $\text{34°C}$ for $T_H$ and $\text{26°C}$ for $T_L$ in Equation (IX).

$\begin{array}{c}{\text{COP}}_{\text{rev}}=\frac{1}{\frac{\text{34°C}}{\text{26°C}}-1}\\ =\frac{1}{\frac{\left(\text{34}+273\right)\text{K}}{\left(\text{26}+273\right)\text{K}}-1}\\ =\frac{1}{\frac{307\text{K}}{299\text{K}}-1}\\ =37.38\end{array}$

Substitute $5.731\text{kW}$ for ${\dot{Q}}_L$ and $37.38$ for ${\text{COP}}_{\text{rev}}$ in Equation (X).

$\begin{array}{c}{\dot{W}}_{\text{in,min}}=\frac{5.731\text{kW}}{37.38}\\ =0.1533\text{kW}\end{array}$

Substitute $0.1533\text{kW}$ for ${\dot{W}}_{\text{in,min}}$, $283.48\text{kJ}/\text{kg}$ for $h_2$ and $259.36\text{kJ}/\text{kg}$ for $h_1$ in Equation (XI).

$\begin{array}{c}{\dot{m}}_{\min }=\frac{0.1533\text{kW}}{283.48\text{kJ}/\text{kg}-259.36\text{kJ}/\text{kg}}\\ =\frac{0.1533\text{kW}\left[\frac{\text{kJ}/\text{s}}{\text{kW}}\right]}{283.48\text{kJ}/\text{kg}-259.36\text{kJ}/\text{kg}}\\ =\frac{0.1533\text{kJ}/\text{s}}{283.48\text{kJ}/\text{kg}-259.36\text{kJ}/\text{kg}}\\ =0.006358\text{kg}/\text{s}\end{array}$

Substitute $0.006358\text{kg}/\text{s}$ for ${\dot{m}}_{\min }$ and $0.04117{\text{m}}^3/\text{kg}$ for $v_1$ in Equation (XII).

$\begin{array}{c}{\dot{\nu }}_{1,\min }=\left(0.006358\text{kg}/\text{s}\right)\left(0.04117{\text{m}}^3/\text{kg}\right)\\ =0.0002618{\text{m}}^{\text{3}}/\text{s}\left[\frac{\text{1000}\text{L}}{\text{m3}}\right]\left[\frac{\text{60}\text{min}}{\text{s}}\right]\\ =\text{15}\text{.7}\text{L}/\min \end{array}$

Hence, the minimum volume flow rate of the refrigerant at the compressor inlet is $\text{15}\text{.7}\text{L}/\min$.