Chapter 8, Problem 135P

To determine

The rate of entropy generation during the process.

Explanation of Solution

Given:

The inlet pressure of the water $\left(P_1\right)$ is 20 psia.

The inlet temperature of the water $\left(T_1\right)$ is $50°\text{F}$.

The mass flow rate of water $\left({\dot{m}}_1\right)$ is $300\text{lbm}/\text{min}$.

The inlet pressure of the steam $\left(P_2\right)$ is 20 psia.

The inlet temperature of the steam $\left(T_2\right)$ is $240°\text{F}$.

The heat lost to the surroundings $\left({\dot{Q}}_{out}\right)$ is $180\text{Btu}/\text{min}$.

The surrounding temperature of the air $\left(T_{surr}\right)$ is $70°\text{F}$.

Calculation:

Refer to Table A-4E, “Saturated water—Temperature table”, obtain the below properties at the pressure of $20\text{psia}$ and temperature of $50°\text{F}$.

The fluid entropy $\left(s_f=s_1\right)$ is $0.03609\text{Btu}/\text{lbm}\cdot \text{R}$.

The fluid enthalpy $\left(h_f=h_1\right)$ is $18.07\text{Btu}/\text{lbm}$.

Refer to Table A-4E, “Saturated water—Temperature table”, obtain the below properties at the pressure of $20\text{psia}$ and temperature of $240°\text{F}$.

The entropy at inlet 2 $\left(s_2\right)$ is $1.7406\text{Btu}/\text{lbm}\cdot \text{R}$.

The enthalpy at inlet 2 $\left(h_2\right)$ is $1162.3\text{Btu}/\text{lbm}$.

Refer to Table A-4E, “Saturated water—Temperature table”, obtain the below properties at the pressure of $20\text{psia}$ and temperature of $130°\text{F}$.

The entropy at exit 3 $\left(s_3\right)$ is $0.18174\text{Btu}/\text{lbm}\cdot \text{R}$.

The enthalpy at exit 3 $\left(h_3\right)$ is $97.99\text{Btu}/\text{lbm}$.

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

  ${\dot{E}}_{in}-{\dot{E}}_{out}=\Delta {\dot{E}}_{system}$                              …… (I).

Here, rate of net energy transfer in to the control volume is ${\dot{E}}_{in}$, rate of net 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 0 for $\Delta {\dot{E}}_{system}$ in Equation (I).

  $\begin{array}{l}{\dot{E}}_{in}-{\dot{E}}_{out}=0\\ {\dot{E}}_{in}={\dot{E}}_{out}\\ {\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2={\dot{Q}}_{out}+{\dot{m}}_3{h}_3\\ {\dot{Q}}_{out}={\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2-{\dot{m}}_3{h}_3\end{array}$                           …… (II).

Here, mass flow rate at inlet 1 is ${\dot{m}}_1$, mass flow rate at inlet 2 is ${\dot{m}}_2$, mass flow rate at exit is ${\dot{m}}_3$.

Substitute ${\dot{m}}_3={\dot{m}}_1+{\dot{m}}_2$ in Equation (II).

  $\begin{array}{c}{\dot{Q}}_{out}={\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2-\left({\dot{m}}_1+{\dot{m}}_2\right)h_3\\ ={\dot{m}}_1\left(h_1-h_3\right)+{\dot{m}}_2\left(h_2-h_3\right)\end{array}$

  $\begin{array}{l}180\text{Btu}/\text{min}=\left[\begin{array}{l}300\text{lbm}/\text{min}\left(18.07\text{Btu}/\text{lbm}-97.99\text{Btu}/\text{lbm}\right)\\ +{\dot{m}}_2\left(1162.3\text{Btu}/\text{lbm}-97.99\text{Btu}/\text{lbm}\right)\end{array}\right]\\ 180\text{Btu}/\text{min}=\left(-23976\text{Btu}/\text{min}\right)+{\dot{m}}_2\left(1064.31\text{Btu}/\text{lbm}\right)\\ {\dot{m}}_2=22.7\text{lbm}/\text{min}\end{array}$

The rate of change in internal energy of the system is zero at steady state,

Write the expression for the mass balance of the system.

  ${\dot{m}}_{in}-{\dot{m}}_{out}=\Delta {\dot{m}}_{system}$                    …… (III).

Here, inlet mass flow rate is ${\dot{m}}_{in}$ and outlet mass flow rate is ${\dot{m}}_{out}$ and change in mass flow rate is $\Delta {\dot{m}}_{system}$.

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

  $\begin{array}{l}{\dot{m}}_1+{\dot{m}}_2-{\dot{m}}_3=0\\ {\dot{m}}_1+{\dot{m}}_2={\dot{m}}_3\end{array}$

  $\begin{array}{c}{\dot{m}}_3=\left(300\text{lbm}/\text{min}\right)+\left(22.7\text{lbm}/\text{min}\right)\\ =322.7\text{lbm}/\text{min}\end{array}$

Write the expression for the entropy balance during the process.

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

Here, rate of net input entropy is ${\dot{S}}_{in}$, rate of net output entropy is ${\dot{S}}_{out}$, rate of entropy generation is ${\dot{S}}_{gen}$, and rate of change of entropy of the system is $\Delta {\dot{S}}_{system}$.

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

  $\begin{array}{l}{\dot{m}}_1{s}_1+{\dot{m}}_2{s}_2-{\dot{m}}_3{s}_3-\frac{{\dot{Q}}_{out}}{T_{surr}}+{\dot{S}}_{gen}=0\\ {\dot{S}}_{gen}={\dot{m}}_3{s}_3-{\dot{m}}_1{s}_1-{\dot{m}}_2{s}_2+\frac{{\dot{Q}}_{out}}{T_{surr}}\end{array}$

  $\begin{array}{c}{\dot{S}}_{gen}=\left[\begin{array}{l}\left(322.7\text{lbm}/\text{min}\right)\left(0.18174\text{Btu}/\text{lbm}\cdot \text{R}\right)-\left(300\text{lbm}/\text{min}\right)\left(0.03609\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ -\left(22.7\text{lbm}/\text{min}\right)\left(1.7406\text{Btu}/\text{lbm}\cdot \text{R}\right)+\frac{180\text{Btu}/\text{min}}{70°\text{F}}\end{array}\right]\\ =\left[\begin{array}{l}\left(322.7\text{lbm}/\text{min}\right)\left(0.18174\text{Btu}/\text{lbm}\cdot \text{R}\right)-\left(300\text{lbm}/\text{min}\right)\left(0.03609\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ -\left(22.7\text{lbm}/\text{min}\right)\left(1.7406\text{Btu}/\text{lbm}\cdot \text{R}\right)+\frac{180\text{Btu}/\text{min}}{\left(70+460\right)\text{R}}\end{array}\right]\\ =8.65\text{Btu}/\text{min}\cdot \text{R}\end{array}$

Thus, the rate of entropy generation during the process is $8.65\text{Btu}/\text{min}\cdot \text{R}$.