Chapter 7.13, Problem 140P

Water at 20 psia and 50°F enters a mixing chamber at a rate of 300 lbm/min where it is mixed steadily with steam entering at 20 psia and 240°F. The mixture leaves the chamber at 20 psia and 130°F, and heat is lost to the surrounding air at 70°F at a rate of 180 Btu/min. Neglecting the changes in kinetic and potential energies, determine the rate of entropy generation during this process.

FIGURE P7–140E

To determine

The rate of entropy generation during the process.

Answer to Problem 140P

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

Explanation of Solution

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 $\Delta {\dot{E}}_{system}$.

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.

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

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 $\Delta {\dot{m}}_{system}$.

Write the expression for the entropy balance during the process.

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

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 $\Delta {\dot{S}}_{system}$.

Conclusion:

Substitute 0 for $\Delta {\dot{E}}_{system}$ in Equation (I)

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

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 ${\dot{m}}_3$, enthalpy at inlet 1 is $h_1$, enthalpy at inlet 2 is $h_2$ and enthalpy at exit is $h_3$.

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

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

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

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

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

$\begin{array}{l}{h}_f=h_1=18.07\text{Btu}/\text{lbm}\\ s_f=s_1=0.03609\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Here, fluid entropy is $s_f$ and fluid enthalpy is $h_f$.

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

$\begin{array}{l}{h}_2=1162.3\text{Btu}/\text{lbm}\\ s_2=1.7406\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Here, entropy at inlet 2 is $s_2$ and enthalpy at inlet 2 is $h_2$

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

$\begin{array}{l}{h}_3=97.99\text{Btu}/\text{lbm}\\ s_3=0.18174\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Here, entropy at exit 3 is $s_3$ and enthalpy at exit 3 is $h_3$.

Substitute $180\text{Btu}/\text{min}$ for ${\dot{Q}}_{out}$, $300\text{lbm}/\text{min}$ for ${\dot{m}}_1$, $18.07\text{Btu}/\text{lbm}$ for $h_1$$1162.3\text{Btu}/\text{lbm}$ for $h_2$ and $97.99\text{Btu}/\text{lbm}$ for $h_3$ in Equation (VI).

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

Substitute $300\text{lbm}/\text{min}$ for ${\dot{m}}_1$, and $22.7\text{lbm}/\text{min}$ for ${\dot{m}}_2$ in Equation (V).

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

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

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

Substitute $322.7\text{lbm}/\text{min}$ for ${\dot{m}}_3$, $0.18174\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_3$, $300\text{lbm}/\text{min}$ for ${\dot{m}}_1$, $0.03609\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_1$, $22.7\text{lbm}/\text{min}$ for ${\dot{m}}_2$, $1.7406\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_2$, $180\text{Btu}/\text{min}$ for ${\dot{Q}}_{out}$, and $70°\text{F}$ for $T_{surr}$ in Equation (VII).

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

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