Chapter 6, Problem 74P

To determine

The mass flow rate of the refrigerant.

Explanation of Solution

Given:

The inlet pressure of Refrigerant $\left(P_1\right)$ is 1 MPa.

The inlet temperature of Refrigerant  $\left(T_1\right)$ is $90°\text{C}$.

The outlet pressure of Refrigerant  $\left(P_2\right)$ is 1 MPa.

The outlet temperature of Refrigerant  $\left(T_2\right)$ is $30°\text{C}$.

The inlet pressure of air $\left(P_3\right)$ is 100 kPa.

The inlet temperature of air $\left(T_3\right)$ is $27°\text{C}$.

The outlet pressure of air $\left(P_4\right)$ is 95 kPa.

The outlet temperature of air $\left(T_4\right)$ is $60°\text{C}$.

Calculation:

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.

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

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

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

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 refrigerant .

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

The Equations (I) reduced as follows for air.

  $\begin{array}{l}\left[0+0+{\dot{m}}_3\left(h_3+0+0\right)\right]-\left[0+0+{\dot{m}}_4\left(h_4+0+0\right)\right]=0\\ {\dot{m}}_3{h}_3-{\dot{m}}_4{h}_4=0\\ {\dot{m}}_3{h}_3={\dot{m}}_4{h}_4\end{array}$        (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 ${\dot{m}}_R={\dot{m}}_1,{\dot{m}}_2$ and ${\dot{m}}_a={\dot{m}}_3,{\dot{m}}_4$ in the above equation.

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

Write the formula for change in enthalpy $\left(h_4-h_3\right)$ of air.

  $h_4-h_3=c_{p,a}\left(T_4-T_3\right)$

Substitute $h_4-h_3=c_{p,a}\left(T_4-T_3\right)$ in Equation (IV).

  $\begin{array}{c}{\dot{m}}_R=\frac{{\dot{m}}_a\left[c_{p,a}\left(T_4-T_3\right)\right]}{h_1-h_2}\\ =\frac{{\dot{m}}_a{c}_{p,a}\left(T_4-T_3\right)}{h_1-h_2}\end{array}$        (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 $\left(h_1\right)$ corresponding to the pressure of $1\text{MPa}$ and the temperature of $90°\text{C}$.

  $h_1=324.66\text{kJ/kg}$

At exit:

The refrigerant is at the state of saturated liquid.

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

Obtain the exit enthalpy $\left(h_2\right)$ corresponding to the temperature of $30°\text{C}$.

  $h_f=h_2=93.58\text{kJ/kg}$

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 $\left({\dot{m}}_a\right)$.

  ${\dot{m}}_a=\frac{\dot{V}{P}_3}{R{T}_3}$        (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 $\dot{V}=600{\text{m}}^3/\min$, $P_3=100\text{kPa}$, $R=0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ and $T_3=27\text{°C}$ 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 ${\dot{m}}_a=696.8641\text{kg/min}$, $c_{p,a}=1.005\text{kJ/kg}\cdot °\text{C}$, $T_4=60°\text{C}$, $T_3=27\text{°C}$, $h_1=324.66\text{kJ/kg}$, $h_2=93.58\text{kJ/kg}$ 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 $100\text{kg/min}$.