Chapter 5.5, Problem 81P

Refrigerant-134a at 1 MPa and 90°C is to be cooled to 1 MPa and 30°C in a condenser by air. The air enters at 100 kPa and 27°C with a volume flow rate of 600 m³/min and leaves at 95 kPa and 60°C. Determine the mass flow rate of the refrigerant.

FIGURE P5–81

To determine

The mass flow rate of the refrigerant.

Answer to Problem 81P

The mass flow rate of the refrigerant is $100\text{kg/min}$.

Explanation of Solution

Consider the system is in steady state. Hence, the inlet and exit mass flow rates are equal.

The mass flow rate of air $\left({\dot{m}}_a\right)$ is as follows.

${\dot{m}}_1={\dot{m}}_2={\dot{m}}_a$

The mass flow rate of refrigerant $\left({\dot{m}}_R\right)$ is as follows.

${\dot{m}}_3={\dot{m}}_4={\dot{m}}_R$

Write the energy rate balance equation for one inlet and one outlet system.

$\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}$ (I)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Neglect the work transfer, heat transfer to the surrounding, potential and kinetic energies.

The Equations (I) reduced as follows for air.

$\begin{array}{l}\left[0+0+{\dot{m}}_1\left(h_1+0+0\right)\right]-\left[0+0+{\dot{m}}_2\left(h_2+0+0\right)\right]=0\\ {\dot{m}}_1{h}_1-{\dot{m}}_2{h}_2=0\\ {\dot{m}}_1{h}_1={\dot{m}}_2{h}_2\end{array}$ (II)

The Equations (I) reduced as follows for refrigerant.

$\begin{array}{l}\left[0+0+{\dot{m}}_3\left(h_3+0+0\right)\right]-\left[0+0+{\dot{m}}_4\left(h_4+0+0\right)\right]=0\\ {\dot{m}}_3{h}_3-{\dot{m}}_4{h}_4=0\\ {\dot{m}}_3{h}_3={\dot{m}}_4{h}_4\end{array}$ (III)

Combining Equation (II) and (III).

$\begin{array}{l}{\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3={\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4\\ {\dot{m}}_2{h}_2-{\dot{m}}_1{h}_1={\dot{m}}_3{h}_3-{\dot{m}}_4{h}_4\end{array}$

Substitute ${\dot{m}}_a$ for ${\dot{m}}_1,{\dot{m}}_2$ and ${\dot{m}}_R$ for ${\dot{m}}_3,{\dot{m}}_4$.

$\begin{array}{l}{\dot{m}}_a{h}_2-{\dot{m}}_a{h}_1={\dot{m}}_R{h}_3-{\dot{m}}_R{h}_4\\ {\dot{m}}_a\left(h_2-h_1\right)={\dot{m}}_R\left(h_3-h_4\right)\\ {\dot{m}}_R=\frac{{\dot{m}}_a\left(h_2-h_1\right)}{h_3-h_4}\end{array}$ (IV)

Write the formula for change in enthalpy $\left(h_2-h_1\right)$ of air.

$h_2-h_1=c_{p,a}\left(T_2-T_1\right)$

Substitute $c_{p,a}\left(T_2-T_1\right)$ for $h_2-h_1$ in Equation (IV).

$\begin{array}{c}{\dot{m}}_R=\frac{{\dot{m}}_a\left[c_{p,a}\left(T_2-T_1\right)\right]}{h_3-h_4}\\ =\frac{{\dot{m}}_a{c}_{p,a}\left(T_2-T_1\right)}{h_3-h_4}\end{array}$ (V)

For refrigerant:

At inlet:

The refrigerant is at the state of superheated condition.

Refer Table A-13, “Superheated refrigerant-134a”.

Obtain the inlet enthalpy $\left(h_3\right)$ corresponding to the pressure of $1\text{MPa}$ and the temperature of $90°\text{C}$.

$h_3=324.66\text{kJ/kg}$

At exit:

The refrigerant is at the state of saturated liquid.

Refer Table A-11, “Saturated refrigerant-134a-Temperature table”.

Obtain the exit enthalpy $\left(h_4\right)$ corresponding to the temperature of $30°\text{C}$.

$h_f=h_4=93.58\text{kJ/kg}$

For air:

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant of air $\left(R\right)$ is $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$.

Refer Table A-2, “Ideal2gas specific heats of various common gases”.

The specific heat at constant pressure $\left(c_{p@300\text{K}}\right)$ at the temperature of $300\text{K}$ is $1.005\text{kJ/kg}\cdot °\text{C}$.

Write the formula for mass flow rate of air $\left({\dot{m}}_a\right)$.

${\dot{m}}_a=\frac{\dot{V}{P}_1}{R{T}_1}$ (VI)

Here, the volumetric flow rate of air is $\dot{V}$, the pressure is $P$, the gas constant of air is $R$ and the temperature is $T$; the suffix one indicates the inlet condition of air.

Conclusion:

Substitute $600{\text{m}}^3/\min$ for $\dot{V}$, $100\text{kPa}$ for $P_1$, $0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$ and $27\text{°C}$ for $T_1$ in Equation (VI).

$\begin{array}{c}{\dot{m}}_{air}=\frac{\left(600{\text{m}}^3/\text{min}\right)\left(100\text{kPa}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(27\text{°C}\right)}\\ =\frac{60000\text{kPa}\cdot {\text{m}}^3/\text{min}}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(27+273\right)\text{K}}\\ =\frac{60000\text{kPa}\cdot {\text{m}}^3/\text{min}}{\text{86}\text{.1}\text{kPa}\cdot {\text{m}}^3\text{/kg}}\\ =696.8641\text{kg/min}\end{array}$

Substitute $696.8641\text{kg/min}$ for ${\dot{m}}_a$, $1.005\text{kJ/kg}\cdot °\text{C}$ for $c_{p,a}$, $60°\text{C}$ for $T_2$, $27\text{°C}$ for $T_1$, $324.66\text{kJ/kg}$ for $h_3$, $93.58\text{kJ/kg}$ for $h_4$ in Equation (V).

$\begin{array}{c}{\dot{m}}_R=\frac{\left(696.8641\text{kg/min}\right)\left(1.005\text{kJ/kg}\cdot °\text{C}\right)\left(60°\text{C}-27\text{°C}\right)}{324.66\text{kJ/kg}-93.58\text{kJ/kg}}\\ =\frac{\left(696.8641\text{kg/min}\right)\left(1.005\text{kJ/kg}\cdot °\text{C}\right)\left(33\text{K}\right)}{231.08}\\ =\frac{23111.4979\text{kJ/min}}{231.08\text{kJ/kg}}\\ =100.0151\text{kg/min}\end{array}$

$=100\text{kg/min}$

Thus, the mass flow rate of the refrigerant is $100\text{kg/min}$.