Chapter 14.7, Problem 134RP

An automobile air conditioner uses refrigerant-134a as the cooling fluid. The evaporator operates at 100 kPa gage and the condenser operates at 1.5 MPa gage. The compressor requires a power input of 6 kW and has an isentropic efficiency of 85 percent. Atmospheric air at 25°C and 60 percent relative humidity enters the evaporator and leaves at 8°C and 90 percent relative humidity. Determine the volume flow rate of the atmospheric air entering the evaporator of the air conditioner, in m³/min.

To determine

The volume flow rate of the air at the inlet of the evaporator.

Answer to Problem 134RP

The volume flow rate of the air at the inlet of the evaporator is $\underset{\_}{20.5{\text{m}}^3/\min }$.

Explanation of Solution

Express the mass flow rate of dry air at state 1.

${\dot{m}}_{a1}=\frac{{\dot{V}}_1}{v_1}$ (I)

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

Write the formula for mass flow rate of condensate water.

${\dot{m}}_w={\dot{m}}_{a1}\left({\omega }_1-{\omega }_2\right)$ (II)

Here, specific humidity at state 1 and 2 is ${\omega }_1\text{and}{\omega }_2$ respectively

Write the formula for an energy balance on the control volume.

${\dot{m}}_a{h}_1={\dot{Q}}_{\text{in}}+{\dot{m}}_a{h}_2+{\dot{m}}_w{h}_{w2}$ (III)

Here, the rate of heat absorbed by the R-134a is ${\dot{Q}}_{\text{in}}$, the enthalpy at the state of 2 is $h_2$, the enthalpy of the state of 1 in condensate water is $h_1$

Write the formula of the mass flow rate of the refrigerant.

${\dot{W}}_{\text{C}}={\dot{m}}_R\frac{h_{R2,s}-h_{R1}}{{\eta }_C}$ (IV)

Here, the mass flow rate of refrigerant is ${\dot{m}}_R$, the enthalpy of the refrigerant state 2 is $h_{R2,s}$, the enthalpy of the state 1 in the refrigerant is $h_{R1}$, and the efficiency of the condenser is ${\eta }_C$.

Write the formula for the rate of heat absorbed by the R-134ain the evaporator.

${\dot{Q}}_{R,\text{in}}={\dot{m}}_R\left(h_{R1}-h_{R4}\right)$ (V)

Here, the enthalpy of the state 4 in the refrigerant is $h_{R4}$.

Conclusion:

Refer Figure A-31, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to dry bulb pressure of $\text{100}\text{kPa}$ and relative humidity of $\text{90\%}$.

$\begin{array}{l}{h}_1=55.88\text{kJ}/\text{kg}\text{dry}\text{air}\\ {\omega }_1=0.01206\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ v_1=0.8724{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\\ h_2=23.31\text{kJ}/\text{kg}\text{dry}\text{air}\\ {\omega }_2=0.006064\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.8724{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (I).

${\dot{m}}_{a1}=\frac{{\dot{V}}_1}{\left(0.8724{\text{m}}^{\text{3}}/\text{kg}\right)}$

Substitute $\frac{{\dot{V}}_1}{\left(0.8724{\text{m}}^{\text{3}}/\text{kg}\right)}$ for ${\dot{m}}_{a1}$, $0.01206\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$, and $0.006064\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (II).

$\begin{array}{c}{\dot{m}}_w=\frac{{\dot{V}}_1}{\left(0.8724{\text{m}}^{\text{3}}/\text{kg}\right)}\times \left(\begin{array}{l}0.01206\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ -0.006064\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}\right)\\ =\frac{{\dot{V}}_1}{\left(0.8724{\text{m}}^{\text{3}}/\text{kg}\right)}\times \left(0.005996\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\right)\\ =0.006873{\dot{V}}_1\end{array}$

Refer Table A-4, “saturated water-temperature table”, and write the specific enthalpy of condensate water at state 2 at temperature of $\text{8°C}$ using an interpolation method.

$h_{w2}=h_{f@8\text{°C}}$ (VI)

Here, entropy of saturation liquid at temperature of $\text{8°C}$ is $h_{f@8\text{°C}}$.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y is temperature and specific enthalpy of condensate water at state 2 respectively.

Show the specific enthalpy of condensate water at state 2 corresponding to temperature as in Table (1).

Temperature $T\left(\text{°C}\right)$	Specific enthalpy at state 2 $h_{2@h_f}\left(\text{kJ}/\text{kg}\right)$
5 $\left(x_1\right)$	21.02 $\left(y_1\right)$
8 $\left(x_2\right)$	$\left(y_2=?\right)$
10 $\left(x_3\right)$	42.022 $\left(y_3\right)$

Substitute $\text{5°C,}\text{8°C}\text{and}\text{10°C}$ for $x_1,x_2\text{and}{x}_3$ respectively, $21.02\text{kJ}/\text{kg}$ for $y_1$ and $42.022\text{kJ}/\text{kg}$ for $y_3$ in Equation (VII).

$\begin{array}{c}{y}_2=\frac{\left(\text{8°C}-\text{5°C}\right)\left(42.022\text{kJ}/\text{kg}-21.02\text{kJ}/\text{kg}\right)}{\left(\text{10°C}-5\text{°C}\right)}+21.02\text{kJ}/\text{kg}\\ =33.6212\text{kJ}/\text{kg}\\ =h_{2@h_f}\left(\text{kJ}/\text{kg}\right)\end{array}$

Substitute $33.6212\text{kJ}/\text{kg}$ for $h_{2@h_f}\left(\text{kJ}/\text{kg}\right)$ in Equation (VI).

$h_{w2}=33.6212\text{kJ}/\text{kg}$

Substitute $\frac{{\dot{V}}_1}{\left(0.8724{\text{m}}^{\text{3}}/\text{kg}\right)}$ for ${\dot{m}}_{a1}$, $55.88\text{kJ}/\text{kg}$ for $h_1$, $23.11\text{kJ}/\text{kg}$ for $h_2$, $33.63\text{kJ}/\text{kg}$ for $h_{w,2}$, and $0.006873{\dot{V}}_1$ for ${\dot{m}}_w$ in Equation (III).

$\frac{{\dot{V}}_1}{\left(0.8724{\text{m}}^{\text{3}}/\text{kg}\right)}\left(55.88\text{kJ}/\text{kg}\right)=\left[\begin{array}{l}{\dot{Q}}_{R,\text{in}}+\frac{{\dot{V}}_1}{\left(0.8724{\text{m}}^{\text{3}}/\text{kg}\right)}\\ \times \left(23.11\text{kJ}/\text{kg}\right)\\ +\left(0.006873{\dot{V}}_1\right)\left(33.63\text{kJ}/\text{kg}\right)\end{array}\right]$ (VIII)

From the Table A-12 “Saturated Refrigerant-134a-Pressure Table”, obtain the value of the specific enthalpy and entropy at the inlet of the compressor for isentropic process at 200 kPa of pressure and dryness fraction of 1 as

$\begin{array}{l}{h}_{R1}=244.50\text{kJ}/\text{kg}\\ s_{R1}=0.93788\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-12 “Saturated Refrigerant-134a-Pressure Table”, and write the specific enthalpy at the exit of the compressor for isentropic process at 1600 kPa of pressure and entropy of $0.93788\text{kJ}/\text{kg}\cdot \text{K}$ using an interpolation method Equation (VII).

$h_{R2,s}=287.96\text{kJ}/\text{kg}$

From the Table A-12 “Saturated Refrigerant-134a-Pressure Table”, obtain the value of the specific enthalpy at the inlet of the compressor for isentropic process at 1600 kPa of pressure as:

$\begin{array}{c}{h}_{R3}=h_{f@1600\text{kPa}}\\ =135.96\text{kJ}/\text{kg}\end{array}$

Here, the specific enthalpy at the state 4 and 3 are equal in the refrigerant.

Substitute 6 KW for ${\dot{W}}_C$, $287.96\text{kJ}/\text{kg}$ for $h_{R2,s}$, $244.50\text{kJ}/\text{kg}$ for $h_{R1}$, and 0.85 for ${\eta }_C$ in Equation (IV).

$\begin{array}{l}6\text{KW}={\dot{m}}_R\frac{\left(287.96-244.50\right)\text{kJ}/\text{kg}}{0.85}\\ 6\text{KW}={\dot{m}}_R\frac{43.46\text{kJ}/\text{kg}}{0.85}\\ {\dot{m}}_R=0.1173\text{kg}/\text{s}\\ {\dot{m}}_R=0.1173\text{kg}/\text{s}\times \left(\frac{1\text{kg}/\text{min}}{60\text{kg}/\text{s}}\right)\end{array}$

${\dot{m}}_R=7.041\text{kg}/\text{min}$

Substitute $7.041\text{kg}/\text{min}$ for ${\dot{m}}_R$, $244.50\text{kJ}/\text{kg}$ for $h_{R1}$, and $135.96\text{kJ}/\text{kg}$ for $h_{R4}$ in Equation (V).

$\begin{array}{c}{\dot{Q}}_{R,\text{in}}=\left(7.041\text{kg}/\text{min}\right)\times \left(244.50\text{kJ}/\text{kg}-135.96\text{kJ}/\text{kg}\right)\\ =\left(7.041\text{kg}/\text{min}\right)\times \left(108.54\text{kJ}/\text{kg}\right)\\ =764.2\text{kJ}/\text{min}\end{array}$

Substitute $764.2\text{kJ}/\text{min}$ for ${\dot{Q}}_{R,\text{in}}$ in Equation (VIII).

$\begin{array}{l}\frac{{\dot{V}}_1}{\left(0.8724{\text{m}}^{\text{3}}/\text{kg}\right)}\left(55.88\text{kJ}/\text{kg}\right)=\left[\begin{array}{l}\left(764.2\text{kJ}/\text{min}\right)+\frac{{\dot{V}}_1}{\left(0.8724{\text{m}}^{\text{3}}/\text{kg}\right)}\\ \times \left(23.11\text{kJ}/\text{kg}\right)\\ +\left(0.006873{\dot{V}}_1\right)\left(33.63\text{kJ}/\text{kg}\right)\end{array}\right]\\ {\dot{V}}_1=20.5{\text{m}}^{\text{3}}/\text{min}\end{array}$

Thus, the volume flow rate of the air at the inlet of the evaporator is $\underset{\_}{20.5{\text{m}}^3/\min }$.