Chapter 7.13, Problem 156P

Liquid water at 200 kPa and 15°C is heated in a chamber by mixing it with superheated steam at 200 kPa and 150°C. Liquid water enters the mixing chamber at a rate of 4.3 kg/s, and the chamber is estimated to lose heat to the surrounding air at 20°C at a rate of 1200 kJ/min. If the mixture leaves the mixing chamber at 200 kPa and 80°C, determine (a) the mass flow rate of the superheated steam and (b) the rate of entropy generation during this mixing process.

FIGURE P7–154

To determine

The mass flow rate of the superheated steam.

Answer to Problem 156P

The mass flow rate of the superheated steam is $0.4806\text{kg}/\text{s}$.

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

Write the expression to calculate 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 $\Delta {\dot{m}}_{system}$.

Conclusion:

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

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

Here, mass flow rate at entry 1 is ${\dot{m}}_1$, mass flow rate at entry 2 is ${\dot{m}}_2$, mass flow rate at exit is ${\dot{m}}_3$, enthalpy at entry 1 is $h_1$, enthalpy at entry 2 is $h_2$ and enthalpy at exit is $h_3$.

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

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

Rewrite the Equation (IV) to calculate the mass flow rate at entry 2.

${\dot{m}}_2=\frac{{\dot{Q}}_{out}-{\dot{m}}_1\left(h_1-h_3\right)}{\left(h_2-h_3\right)}$ (V)

From Table A-4, “the saturated water table”, select the initial enthalpy at entry 1 $\left(h_1\right)$, and initial entropy $\left(s_1\right)$ at the temperature of $15°\text{C}$ as $62.98\text{kJ}/\text{kg}$, and $0.2245\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

From Table A-6, “Superheated water”, select the initial enthalpy at entry 2 $\left(h_2\right)$, initial entropy $\left(s_2\right)$ and specific volume $\left(v_1\right)$ at the pressure of $\text{200}\text{kPa}$ and temperature of 150 C as $2769.1\text{kJ}/\text{kg}$, and $7.2810\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

From Table A-4, “the saturated water table”, select the enthalpy at exit $\left(h_3\right)$, and exit entropy $\left(s_3\right)$ at the temperature of 80 C as $335.02\text{kJ}/\text{kg}$, and $1.0756\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

Substitute $1200\text{kJ}/\text{min}$ for ${\dot{Q}}_{out}$, $4.3\text{kg}/\text{s}$ for ${\dot{m}}_1$, $62.98\text{kJ}/\text{kg}$ for $h_1$, $2769.1\text{kJ}/\text{kg}$ for $h_2$ and $335.02\text{kJ}/\text{kg}$ for $h_3$ in Equation (V).

$\begin{array}{c}{\dot{m}}_2=\frac{\left(1200\text{kJ}/\text{min}\right)-4.3\text{kg}/\text{s}\left(62.98\text{kJ}/\text{kg}-335.02\text{kJ}/\text{kg}\right)}{\left(2769.1\text{kJ}/\text{kg}-335.02\text{kJ}/\text{kg}\right)}\\ =\frac{\left(1200\text{kJ}/\text{min}\right)\left(\frac{1\min }{60\sec }\right)-4.3\text{kg}/\text{s}\left(62.98\text{kJ}/\text{kg}-335.02\text{kJ}/\text{kg}\right)}{\left(2769.1\text{kJ}/\text{kg}-335.02\text{kJ}/\text{kg}\right)}\\ =0.4806\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the superheated steam is $0.4806\text{kg}/\text{s}$.

To determine

The rate of heat entropy generation during the process.

Answer to Problem 156P

The rate of heat entropy generation during the process is $0.746\text{kW}/\text{K}$.

Explanation of Solution

Write the expression for the entropy balance during the process.

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

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 ${\dot{m}}_1{s}_1+{\dot{m}}_2{s}_2$ 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 (VII).

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

Here, entropy at entry 1 is $s_1$, entropy at entry 2 is $s_2$ and entropy at exit is $s_3$ and surrounding temperature is $T_{surr}$.

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).

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

Substitute $4.3\text{kg}/\text{s}$ for ${\dot{m}}_1$, and $0.4806\text{kg}/\text{s}$ for ${\dot{m}}_2$ in Equation (VIII).

$\begin{array}{c}{\dot{m}}_3=\left(4.3\text{kg}/\text{s}\right)+\left(0.4806\text{kg}/\text{s}\right)\\ =4.781\text{kg}/\text{s}\end{array}$

Substitute $4.781\text{kg}/\text{s}$ for ${\dot{m}}_3$, $1.0756\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$, $4.3\text{kg}/\text{s}$ for ${\dot{m}}_1$, $0.2245\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$, $0.4806\text{kg}/\text{s}$ for ${\dot{m}}_2$, $7.2810\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, $1,200\text{kJ}/\text{min}$ for ${\dot{Q}}_{out}$, and 20 C for $T_{surr}$ in Equation (VII).

$\begin{array}{c}{\dot{S}}_{gen}=\left\{\begin{array}{l}\left(4.781\text{kg}/\text{s}\right)\left(1.0756\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(4.3\text{kg}/\text{s}\right)\left(0.2245\text{kJ}/\text{kg}\cdot \text{K}\right)\\ -\left(0.4806\text{kg}/\text{s}\right)\left(7.2810\text{kJ}/\text{kg}\cdot \text{K}\right)+\frac{\left(1200\text{kJ}/\text{min}\right)}{20°\text{C}}\end{array}\right\}\\ =\left\{\begin{array}{l}\left(4.781\text{kg}/\text{s}\right)\left(1.0756\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(4.3\text{kg}/\text{s}\right)\left(0.2245\text{kJ}/\text{kg}\cdot \text{K}\right)\\ -\left(0.4806\text{kg}/\text{s}\right)\left(7.2810\text{kJ}/\text{kg}\cdot \text{K}\right)+\frac{\left(1200\text{kJ}/\text{min}\right)\left(\frac{1\min }{60\sec }\right)}{\left(20+273\right)\text{K}}\end{array}\right\}\\ =0.746\text{kW}/\text{K}\end{array}$

Thus, the rate of heat entropy generation during the process is $0.746\text{kW}/\text{K}$.