Chapter 8, Problem 167RQ

a)

To determine

The exit temperature of air and the entropy generated during the process.

Explanation of Solution

Given:

The volume flow rate of the air $\left({\dot{V}}_3\right)$ is $6{\text{m}}^{\text{3}}\text{/min}$.

The initial pressure of the air $\left(P_1\right)$ is $100\text{ kPa}$.

The initial temperature of the refrigerant-134a $\left(T_1\right)$ is $27°\text{C}$.

The initial pressure of the refrigerant-134a $\left(P_1\right)$ is $100\text{ kPa}$.

The quality of the refrigerant-134a is0.3

The mass flow rate of refrigerant-134a $\left({\dot{m}}_R\right)$ is 2 kg/min.

The surrounding temperature $\left(T_{surr}\right)$ is $32°\text{C}$.

The rate of heat gain from the surrounding $\left({\dot{Q}}_{in}\right)$ is $30\text{ kJ/min}$.

Calculation:

Refer Table A-12, “Saturated refrigerant-134a-Pressure table”, Obtain the following properties at saturated pressure of 120 kPa

Saturated liquid enthalpy, $h_f=22.47\text{kJ}/\text{kg}$

Evaporated enthalpy, $h_{fg}=214.52\text{kJ}/\text{kg}$

Saturated vapor enthalpy, $h_2=h_{g@120\text{kPa}}=236.99\text{kJ}/\text{kg}$

Saturated vapor entropy, $s_2=s_{g@120\text{kPa}}=0.9479\text{kJ}/\text{kg}\cdot \text{K}$

Saturated liquid entropy, $s_f=0.09269\text{kJ}/\text{kg}\cdot \text{K}$

Evaporated entropy, $s_{fg}=0.85520\text{kJ}/\text{kg}\cdot \text{K}$

Refrigerant –134a enters and leaves at the same pressure. Hence, $P_2=P_1$

Calculate the initial enthalpy of the refrigerant $\left(h_1\right)$.

  $\begin{array}{c}{h}_1=h_f+x_1{h}_{fg}\\ h_1=22.47\text{kJ}/\text{kg}+\left(0.3\right)\left(214.52\text{kJ}/\text{kg}\right)\\ =86.83\text{kJ}/\text{kg}\end{array}$

Calculate the initial entropy of the refrigerant.

  $\begin{array}{c}{s}_1=s_f+x_1{s}_{fg}\\ s_1=0.09269\text{kJ}/\text{kg}\cdot \text{K}+\left(0.3\right)\left(0.85520\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =0.3492\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-1E, “the molar mass, gas constant and critical–point properties table”,

The gas constant of air at room temperature as $0.287\text{kJ}/\text{kg}\cdot \text{K}$.

Calculate the mass flow rate of air $\left({\dot{m}}_{air}\right)$ .

  $\begin{array}{c}{\dot{m}}_{air}=\frac{P_3{\dot{V}}_3}{R{T}_3}\\ {\dot{m}}_{air}==\frac{\left(100\text{kPa}\right)\left(6{\text{m}}^3/\text{min}\right)}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)27°\text{C}}\\ =\frac{\left(100\text{kPa}\right)\left(6{\text{m}}^3/\text{min}\right)}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\frac{\text{1}\text{kPa}\cdot {\text{m}}^3}{1\text{kJ}}\right)\left(27+273\right)\text{K}}\\ =6.968\text{kg}/\text{min}\end{array}$

Write the expression for the mass balance of the system.

  ${\dot{m}}_{in}-{\dot{m}}_{out}=\Delta {\dot{m}}_{system}$        (I)

Here, mass flow rate into the control system is ${\dot{m}}_{in}$, mass flow rate exit from the control volume is ${\dot{m}}_{out}$ and mass flow rate change in the system is $\Delta {\dot{m}}_{system}$.

Substitute ${\dot{m}}_{in}={\dot{m}}_3$, ${\dot{m}}_{out}={\dot{m}}_4$, and $\Delta {\dot{m}}_{system}=0$ in the Equation (I).

  $\begin{array}{l}{\dot{m}}_3-{\dot{m}}_4=0\\ {\dot{m}}_3={\dot{m}}_4={\dot{m}}_{air}\end{array}$

Refer the Table A-2, “Ideal-gas specific heats of various common gases”, select the value of the specific heat at constant pressure value of air as $1.005\text{kJ}/\text{kg}\cdot \text{K}$.

Write the expression for the energy balance equation for closed system.

  ${\dot{E}}_{in}-{\dot{E}}_{out}=\Delta {\dot{E}}_{system}$        (II)

Here, rate of energy transfer into the control volume is ${\dot{E}}_{in}$, rate of energy transfer exit from the control volume is ${\dot{E}}_{out}$ and rate of change in internal energy of system is $\Delta {\dot{E}}_{system}$

Substitute ${\dot{E}}_{in}={\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3$, ${\dot{E}}_{out}={\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4$ and  $\Delta {\dot{E}}_{system}=0$ in the Equation (II).

  $\begin{array}{l}{\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3-\left({\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4\right)=0\\ {\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3={\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4\end{array}$

  $\begin{array}{l}{\dot{m}}_R\left(h_2-h_1\right)={\dot{m}}_{air}\left(h_3-h_4\right)\\ {\dot{m}}_R\left(h_2-h_1\right)={\dot{m}}_{air}{c}_p\left(T_3-T_4\right)\\ T_4=T_3-\frac{{\dot{m}}_R\left(h_2-h_1\right)}{{\dot{m}}_{air}{c}_p}\end{array}$

  $\begin{array}{c}{T}_4=27°\text{C}-\frac{\left(2\text{kg}/\text{min}\right)\left(236.99\text{kJ}/\text{kg}-86.83\text{kJ}/\text{kg}\right)}{\left(6.968\text{kg}/\text{min}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =-15.9°\text{C}\end{array}$

Thus, the exit temperature of air is $-15.9°\text{C}$.

Write the expression for the rate of entropy balance for the system.

  ${\dot{S}}_{in}-{\dot{S}}_{out}+{\dot{S}}_{gen}=\Delta {\dot{S}}_{system}$        (III)

Here, rate of entropy in the system is ${\dot{S}}_{in}$, rate of entropy exit from the system is ${\dot{S}}_{out}$, rate of entropy generated is ${\dot{S}}_{gen}$ and rate of change of entropy in the system is $\Delta {\dot{S}}_{system}$.

For the steady flow system, change of entropy in the system is zero.

Substitute ${\dot{S}}_{in}={\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3$, ${\dot{S}}_{out}={\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4$, $\Delta {\dot{S}}_{system}=0$ in Equation (III).

  $\begin{array}{l}{\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3-\left({\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4\right)+{\dot{S}}_{gen}=0\\ {\dot{S}}_{gen}={\dot{m}}_R\left(s_2-s_1\right)+{\dot{m}}_{air}\left(s_4-s_3\right)\\ ={\dot{m}}_R\left(s_2-s_1\right)+{\dot{m}}_{air}\left(c_p\ln \left(\frac{T_4}{T_3}\right)-R\ln \left(\frac{P_4}{P_3}\right)\right)\\ {\dot{S}}_{gen}=\left\{\begin{array}{l}2\text{kg}/\text{min}\left(0.9479\text{kJ}/\text{kg}\cdot \text{K}-0.3492\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +6.968\text{kg}/\text{min}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{-15.9°\text{C}}{27°\text{C}}\right)-R\ln \left(\frac{P_4}{P_4}\right)\right)\end{array}\right\}\end{array}$

  $\begin{array}{c}{\dot{S}}_{gen}=\left\{\begin{array}{l}2\text{kg}/\text{min}\left(0.9479\text{kJ}/\text{kg}\cdot \text{K}-0.3492\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +6.968\text{kg}/\text{min}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{\left(-15.9+273\right)\text{K}}{\left(27+273\right)\text{K}}\right)-R\ln \left(\frac{P_4}{P_4}\right)\right)\end{array}\right\}\\ =1.1974-\left(6.968\right)0.1551\\ =0.1175\text{kJ}/\text{min}\cdot \text{K}\left(\frac{1\min }{60\sec }\right)\\ =0.00196\text{kW}/\text{K}\end{array}$

Thus, the entropy generated during the process is $0.00196\text{kW}/\text{K}$.

b)

To determine

The exit temperature of air and the entropy generated during the process.

Explanation of Solution

Substitute ${\dot{E}}_{in}={\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3+{\dot{Q}}_{in}$, ${\dot{E}}_{out}={\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4$ and $\Delta {\dot{E}}_{system}=0$ in Equation (II).

  $\begin{array}{l}{\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3+{\dot{Q}}_{in}-\left({\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4\right)=0\\ {\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3+{\dot{Q}}_{in}={\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4\\ {\dot{Q}}_{in}={\dot{m}}_R\left(h_2-h_1\right)+{\dot{m}}_{air}{c}_p\left(T_4-T_3\right)\\ T_4=T_3+\frac{{\dot{Q}}_{in}-{\dot{m}}_R\left(h_2-h_1\right)}{{\dot{m}}_{air}{c}_p}\end{array}$

  $\begin{array}{c}{T}_4=27°\text{C}+\frac{30\text{kJ}/\text{min}-\left(2\text{kg}/\text{min}\right)\left(236.99\text{kJ}/\text{kg}-86.83\text{kJ}/\text{kg}\right)}{\left(6.968\text{kg}/\text{min}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =-11.6°\text{C}\end{array}$

Thus, the exit temperature of air is $-11.6°\text{C}$.

For the steady flow system, change of entropy in the system is zero.

Substitute ${\dot{S}}_{in}={\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3+\frac{{\dot{Q}}_{in}}{T_{surr}}$, ${\dot{S}}_{out}={\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4$, $\Delta {\dot{S}}_{system}=0$ in Equation (III).

  $\begin{array}{l}{\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3+\frac{{\dot{Q}}_{in}}{T_{surr}}-\left({\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4\right)+{\dot{S}}_{gen}=0\\ {\dot{S}}_{gen}={\dot{m}}_R\left(s_2-s_1\right)+{\dot{m}}_{air}\left(s_4-s_3\right)-\frac{{\dot{Q}}_{in}}{T_{surr}}\\ {\dot{S}}_{gen}={\dot{m}}_R\left(s_2-s_1\right)+{\dot{m}}_{air}\left(c_p\ln \left(\frac{T_4}{T_3}\right)-R\ln \left(\frac{P_4}{P_3}\right)\right)-\frac{{\dot{Q}}_{in}}{T_{surr}}\end{array}$${\dot{S}}_{gen}=\left\{\begin{array}{l}2\text{kg}/\text{min}\left(0.9479\text{kJ}/\text{kg}\cdot \text{K}-0.3492\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +6.968\text{kg}/\text{min}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{-11.6°\text{C}}{27°\text{C}}\right)-R\ln \left(\frac{P_4}{P_4}\right)\right)\\ -\frac{30\text{kJ}/\text{min}}{32°\text{C}}\end{array}\right\}$

  $\begin{array}{c}{\dot{S}}_{gen}=\left\{\begin{array}{l}2\text{kg}/\text{min}\left(0.9479\text{kJ}/\text{kg}\cdot \text{K}-0.3492\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +6.968\text{kg}/\text{min}\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\ln \left(\frac{\left(-11.6+273\right)\text{K}}{\left(27+273\right)\text{K}}\right)-R\ln \left(\frac{P_4}{P_4}\right)\right)\\ -\frac{30\text{kJ}/\text{min}}{\left(32+273\right)\text{K}}\end{array}\right\}\\ =1.1974+\left(6.968\right)\left(-0.1384\right)-0.09836\\ =0.1348\text{kJ}/\text{min}\cdot \text{K}\left(\frac{1\min }{60\sec }\right)\\ =0.00225\text{kW}/\text{K}\end{array}$

Thus, the entropy generated during the process is $0.00225\text{kW}/\text{K}$.