Chapter 14.7, Problem 132RP

Atmospheric air enters an air-conditioning system at 30°C and 70 percent relative humidity with a volume flow rate of 4 m³/min and is cooled to 20°C and 20 percent relative humidity at a pressure of 1 atm. The system uses refrigerant-134a as the cooling fluid that enters the cooling section at 350 kPa with a quality of 20 percent and leaves as a saturated vapor. Show the process on the psychrometric chart. What is the heat transfer from the air to the cooling coils, in kW? If any water is condensed from the air, how much water will be condensed from the atmospheric air per min? Determine the mass flow rate of the refrigerant, in kg/min.

FIGURE P14–132

To determine

Show the process on the psychrometric chart; find the heat transfer from the air to the cooling coils, how much water will be condensed from the atmospheric air per min and the mass flow rate of the refrigerant.

Answer to Problem 132RP

The process on the psychrometric chart is shown below in Figure (1), the heat transfer from the air to the cooling coils is $\text{3}\text{.727}\text{kW}$, the amount of water condensed from the atmospheric air per min is $\text{0}\text{.07196}\text{kg}/\text{min}$ and the mass flow rate of the refrigerant is

$\text{1}\text{.435}\text{kg}/\text{min}$.

Explanation of Solution

As the process is a steady flow and thus the mass flow rate of dry air remains constant during the entire process.

$\begin{array}{c}{\dot{m}}_{a1}={\dot{m}}_{a2}\\ ={\dot{m}}_a\end{array}$

Here, the mass flow rate of air at inlet is ${\dot{m}}_{a1}$, mass flow rate of dry air at exit is ${\dot{m}}_{a2}$ and mass flow rate of dry air is ${\dot{m}}_a$.

Express the mass flow rate of dry air.

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

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

Express the mass flow rate of vapor at inlet.

${\dot{m}}_{v1}={\omega }_1{\dot{m}}_a$ (II)

Here, specific humidity at state 1 is ${\omega }_1$.

Express the mass flow rate of vapor at exit.

${\dot{m}}_{v2}={\omega }_2{\dot{m}}_a$ (III)

Here, specific humidity at state 2 is ${\omega }_2$.

Express the rate of condensation of water.

${\dot{m}}_w={\dot{m}}_{v1}-{\dot{m}}_{v2}$ (IV)

Express the enthalpy of condensate water.

$h_{w2}=h_{f@20\text{°C}}$ (V)

Here, enthalpy of saturation liquid at temperature of $\text{20°C}$ is $h_{f@20\text{°C}}$.

Express the rate required heat transfer rate from the atmospheric air to the evaporator fluid from an energy balance on the control volume.

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

Here, enthalpy at state 1 and 2 is $h_1\text{and}{h}_2$ respectively, and the rate required heat transfer rate is ${\dot{Q}}_{\text{out}}$.

Express enthalpy of refrigerant at inlet.

$h_{R1}=\left(1-x_{R1}\right)h_{f@350\text{kPa}}+x_{R1}{h}_{g@350\text{kPa}}$ (VII)

Here, quality of refrigerant at inlet is $x_{R1}$, enthalpy of saturation liquid at pressure of $\text{350}\text{kPa}$ is $h_{f@350\text{kPa}}$ and the enthalpy of saturation vapor at pressure of $\text{350}\text{kPa}$ is $h_{g@350\text{kPa}}$.

Express enthalpy of refrigerant at exit.

$h_{R2}=\left(1-x_{R2}\right)h_{f@350\text{kPa}}+x_{R2}{h}_{g@350\text{kPa}}$ (VIII)

Here, quality of refrigerant at exit is $x_{R2}$.

Express the mass flow rate of the refrigerant.

${\dot{m}}_R=\frac{{\dot{Q}}_{\text{out}}}{h_{R2}-h_{R1}}$ (IX)

Here, enthalpy of refrigerant at inlet and exit is $h_{R1}\text{and}{h}_{R2}$ respectively.

Conclusion:

Show the psychrometric diagram as in Figure (1).

Hence, the psychrometric diagram is shown in Figure (1).

Refer Figure A-31, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to dry bulb temperature of $\text{30°C}$ and relative humidity of $\text{70\%}$.

$\begin{array}{l}{h}_1=78.24\text{kJ}/\text{kg}\text{dry}\text{air}\\ {\omega }_1=0.01880\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ {\nu }_1=0.8847{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Figure A-31, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to dry bulb temperature of $\text{20°C}$ and relative humidity of $\text{20\%}$.

$\begin{array}{l}{h}_2=27.45\text{kJ}/\text{kg}\text{dry}\text{air}\\ {\omega }_2=0.002885\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $\text{4}{\text{m}}^{\text{3}}/\text{min}$ for ${\dot{\nu }}_1$ and $0.8847{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for ${\nu }_1$ in Equation (I).

$\begin{array}{c}{\dot{m}}_a=\frac{\text{4}{\text{m}}^{\text{3}}/\text{min}}{0.8847{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}}\\ =4.521\text{kg}/\text{min}\end{array}$

Substitute $4.521\text{kg}/\text{min}$ for ${\dot{m}}_a$ and $0.01880\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{v1}=\left(0.01880\right)\left(4.521\text{kg}/\text{min}\right)\\ =0.0850\text{kg}/\text{min}\end{array}$

Substitute $4.521\text{kg}/\text{min}$ for ${\dot{m}}_a$ and $0.002885\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (III).

$\begin{array}{c}{\dot{m}}_{v2}=\left(0.002885\right)\left(4.521\text{kg}/\text{min}\right)\\ =0.01304\text{kg}/\text{min}\end{array}$

Substitute $0.0850\text{kg}/\text{min}$ for ${\dot{m}}_{v1}$ and $0.01304\text{kg}/\text{min}$ for ${\dot{m}}_{v2}$ in Equation (IV).

$\begin{array}{c}{\dot{m}}_w=0.0850\text{kg}/\text{min}-0.01304\text{kg}/\text{min}\\ =\text{0}\text{.07196}\text{kg}/\text{min}\end{array}$

Hence, the amount of water condensed from the atmospheric air per min is $\text{0}\text{.07196}\text{kg}/\text{min}$.

Refer Table A-4, “saturated water-temperature table”, and write the enthalpy of condensate water at temperature of $\text{20°C}$.

$\begin{array}{c}{h}_{w_2}=h_{f@20\text{°C}}\\ =83.91\text{kJ}/\text{kg}\end{array}$

Here, enthalpy of saturation liquid is $h_f$.

Substitute $83.91\text{kJ}/\text{kg}$ for $h_{f@20\text{°C}}$ in Equation (V).

$h_{w2}=83.91\text{kJ}/\text{kg}$

Substitute $4.521\text{kg}/\text{min}$ for ${\dot{m}}_a$, $78.24\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_1$, $27.45\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$, $\text{0}\text{.07196}\text{kg}/\text{min}$ for ${\dot{m}}_w$ and $83.91\text{kJ}/\text{kg}$ for $h_{w2}$ in Equation (VI).

$\left(4.521\text{kg}/\text{min}\right)\left(78.24\text{kJ}/\text{kg}\right)=\left[\begin{array}{l}{\dot{Q}}_{\text{out}}+\left(4.521\text{kg}/\text{min}\right)\left(27.45\text{kJ}/\text{kg}\right)\\ +\left(\text{0}\text{.07196}\text{kg}/\text{min}\right)\left(83.91\text{kJ}/\text{kg}\right)\end{array}\right]$

$\begin{array}{c}{\dot{Q}}_{\text{out}}=223.6\text{kJ}/\text{min}\left[\frac{\text{s}}{60\min }\right]\\ =3.727\text{kJ}/\text{s}\left[\frac{\text{kW}}{60\text{kJ}/\text{s}}\right]\\ =3.727\text{kW}\end{array}$

Hence, the heat transfer from the air to the cooling coils is $\text{3}\text{.727}\text{kW}$.

Refer Table A-12, saturated refrigerant-134a-presure table”, and write enthalpy of saturation liquid at pressure of $\text{350}\text{kPa}$ using an interpolation method.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y pressure and enthalpy of saturation liquid respectively.

Show the enthalpy of saturation liquid corresponding to pressure as in Table (1).

Pressure $P\left(\text{kPa}\right)$	Enthalpy of saturation liquid $h_f\left(\text{kJ}/\text{kg}\right)$
320 $\left(x_1\right)$	55.14 $\left(y_1\right)$
350 $\left(x_2\right)$	$\left(y_2=?\right)$
360 $\left(x_3\right)$	59.70 $\left(y_3\right)$

Substitute $\text{320}\text{kPa,}\text{350}\text{kPa}\text{and}\text{360}\text{kPa}$ for $x_1,x_2\text{and}{x}_3$ respectively, $55.14\text{kJ}/\text{kg}$ for $y_1$ and $59.70\text{kJ}/\text{kg}$ for $y_3$ in Equation (X).

$\begin{array}{c}{y}_2=\frac{\left(\text{350}\text{kPa}-\text{320}\text{kPa}\right)\left(59.70\text{kJ}/\text{kg}-55.14\text{kJ}/\text{kg}\right)}{\left(\text{360}\text{kPa}-\text{320}\text{kPa}\right)}+55.14\text{kJ}/\text{kg}\\ =58.56\text{kJ}/\text{kg}\\ =h_{f@350\text{kPa}}\end{array}$

Thus, the enthalpy of saturation liquid at pressure of $\text{350}\text{kPa}$ is,

$h_{f@350\text{kPa}}=58.56\text{kJ}/\text{kg}$

Refer Table A-12, saturated refrigerant-134a-presure table”, and write enthalpy of saturation vapor at pressure of $\text{350}\text{kPa}$ using an interpolation method.

Show the enthalpy of saturation vapor corresponding to pressure as in Table (2).

Pressure $P\left(\text{kPa}\right)$	Enthalpy of saturation liquid $h_g\left(\text{kJ}/\text{kg}\right)$
320 $\left(x_1\right)$	251.93 $\left(y_1\right)$
350 $\left(x_2\right)$	$\left(y_2=?\right)$
360 $\left(x_3\right)$	253.86 $\left(y_3\right)$

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

$h_{g@350\text{kPa}}=253.3775\text{kJ}/\text{kg}$

Substitute $0.20$ for $x_{R1}$, $58.56\text{kJ}/\text{kg}$ for $h_{f@350\text{kPa}}$ and $253.3775\text{kJ}/\text{kg}$ for $h_{g@350\text{kPa}}$ in Equation (VII).

$\begin{array}{c}{h}_{R1}=\left(1-0.20\right)\left(58.56\text{kJ}/\text{kg}\right)+\left(0.20\right)\left(253.3775\text{kJ}/\text{kg}\right)\\ =46.848\text{kJ}/\text{kg}+50.6755\text{kJ}/\text{kg}\\ =97.52\text{kJ}/\text{kg}\end{array}$

Substitute $1$ for $x_{R2}$, $58.56\text{kJ}/\text{kg}$ for $h_{f@350\text{kPa}}$ and $253.3775\text{kJ}/\text{kg}$ for $h_{g@350\text{kPa}}$ in Equation (VIII).

$\begin{array}{c}{h}_{R2}=\left(1-1\right)\left(58.56\text{kJ}/\text{kg}\right)+\left(1\right)\left(253.3775\text{kJ}/\text{kg}\right)\\ =0+253.3775\text{kJ}/\text{kg}\\ =253.3775\text{kJ}/\text{kg}\end{array}$

Substitute $223.6\text{kJ}/\text{min}$ for ${\dot{Q}}_{\text{out}}$, $253.3775\text{kJ}/\text{kg}$ for $h_{R2}$, and $97.52\text{kJ}/\text{kg}$ for $h_{R1}$ in Equation (IX).

$\begin{array}{c}{\dot{m}}_R=\frac{223.6\text{kJ}/\text{min}}{253.3775\text{kJ}/\text{kg}-97.52\text{kJ}/\text{kg}}\\ =1.435\text{kg}/\text{min}\end{array}$

Hence, the mass flow rate of the refrigerant is $\text{1}\text{.435}\text{kg}/\text{min}$.