Chapter 8, Problem 152P

a)

To determine

The mass flow rate of the superheated steam.

Explanation of Solution

Given:

The temperature of the liquid water $\left(T_1\right)$ is $15°\text{C}$.

The mass flow rate of hot water $\left({\dot{m}}_1\right)$ is $4.3\text{kg/s}$.

The temperature of the steam $\left(T_2\right)$ is $150°\text{C}$.

The temperature of the mixture $\left(T_3\right)$ is $80°\text{C}$.

The constant pressure $\left(P\right)$ is $200\text{ kPa}$.

The heat lost to the surroundings $\left(Q_{out}\right)$ is $1200\text{ kJ/min}$.

Calculation:

Refer 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.

Refer 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.

Refer 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.

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

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

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

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

Rewrite the Equation (III) to obtain the mass flow rate at entry 2.

  $\begin{array}{c}{\dot{m}}_2=\frac{{\dot{Q}}_{out}-{\dot{m}}_1\left(h_1-h_3\right)}{\left(h_2-h_3\right)}\\ {\dot{m}}_2=\frac{\left(1200\text{kJ}/\text{min}\right)-4.3\text{kg}/\text{s}\left(62.98\text{kJ}/\text{kg}-335.02\text{kJ}/\text{kg}\right)}{\left(2769.1\text{kJ}/\text{kg}-335.02\text{kJ}/\text{kg}\right)}\\ =\frac{\left(1200\text{kJ}/\text{min}\right)\left(\frac{1\min }{60\sec }\right)-4.3\text{kg}/\text{s}\left(62.98\text{kJ}/\text{kg}-335.02\text{kJ}/\text{kg}\right)}{\left(2769.1\text{kJ}/\text{kg}-335.02\text{kJ}/\text{kg}\right)}\\ =0.4806\text{kg}/\text{s}\end{array}$

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

b)

To determine

The rate of heat entropy generation during the process.

Explanation of Solution

Write the expression to calculate the mass balance of the system.

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

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 (IV).

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

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

Write the expression for the entropy balance during the process.

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

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 (V).

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

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

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