Chapter 8, Problem 151P

(a)

To determine

The mass flow rate of the cold water

Explanation of Solution

Given:

The temperature of the hot water $\left(T_1\right)$ is $70°\text{C}$.

The mass flow rate of hot water $\left({\dot{m}}_1\right)$ is $3.6\text{kg/s}$.

The temperature of the cold water $\left(T_2\right)$ is $20°\text{C}$.

The temperature of the mixture $\left(T_3\right)$ is $42°\text{C}$.

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

Calculation:

Refer Table A-4, “Saturated water-Temperature table”, obtain the following properties of the water at the temperature of $70°\text{C}$.

The initial enthalpy at entry 1 $\left(h_1\right)$ is $293.07\text{kJ}/\text{kg}$.

The initial entropy at entry 1  $\left(s_1\right)$ is $0.9551\text{kJ}/\text{kg}\cdot \text{K}$.

Refer Table A-4, "Saturated water- Temperature table”, obtain the following properties of the water at the temperature of $20°\text{C}$.

The enthalpy at entry 2$\left(h_2\right)$ is $83.91\text{kJ}/\text{kg}$.

The entropy at entry 2 $\left(s_2\right)$ is $0.2965\text{kJ}/\text{kg}\cdot \text{K}$.

Refer Table A-4, “Saturated water- Temperature table”, for the temperature of $42°\text{C}$ to obtain the following properties of water using interpolation method.

Temperature in $\text{K}$	Enthalpy in $\text{kJ}/\text{kg}$	Entropy in $\text{kJ}/\text{kg}\cdot \text{K}$
40	167.53	0.5724
42	?
45	188.44	0.6386

Write the formula of interpolation method of two variables.

  $y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$        (I)

Substitute $x_1=40°\text{C}$, $x_2=42°\text{C}$, $x_3=45°\text{C}$, $y_1=167.53\text{kJ}/\text{kg}$, and $y_3=188.44\text{kJ}/\text{kg}$ in Equation (I).

  $\begin{array}{c}{y}_2=\frac{\left(42-40\right)\left(167.53-188.44\right)}{\left(45-40\right)}+188.44\\ =175.90\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

The enthalpy of the mixture $\left(h_3\right)$ is $\underset{\_}{175.90\text{kJ}/\text{kg}\cdot \text{K}}$.

Substitute $x_1=40°\text{C}$, $x_2=42°\text{C}$, $x_3=45°\text{C}$, $y_1=0.5724\text{kJ}/\text{kg}\cdot \text{K}$, and $y_3=0.6386\text{kJ}/\text{kg}\cdot \text{K}$ in Equation (I).

  $\begin{array}{c}{y}_2=\frac{\left(42-40\right)\left(0.6386-0.5724\right)}{\left(45-40\right)}+0.5724\\ =0.5990\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

The entropy of the mixture $\left(s_3\right)$ is $\underset{\_}{0.5990\text{kJ}/\text{kg}\cdot \text{K}}$.

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 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 (I).

  $\begin{array}{l}{\dot{E}}_{in}-{\dot{E}}_{out}=0\\ {\dot{E}}_{in}={\dot{E}}_{out}\\ {\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2={\dot{m}}_3{h}_3\\ {\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2=\left({\dot{m}}_1+{\dot{m}}_2\right)h_3\end{array}$        (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 $h_3$.

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

  $\begin{array}{c}{\dot{m}}_2=\frac{{\dot{m}}_1\left(h_1-h_3\right)}{\left(h_3-h_2\right)}\\ {\dot{m}}_2=\frac{3.6\text{kg}/\text{s}\left(293.07\text{kJ}/\text{kg}-175.90\text{kJ}/\text{kg}\right)}{\left(175.90\text{kJ}/\text{kg}-83.91\text{kJ}/\text{kg}\right)}\\ =4.586\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the superheated steam is $4.586\text{kg}/\text{s}$.

(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).

  $\begin{array}{c}{\dot{m}}_1+{\dot{m}}_2-{\dot{m}}_3=0\\ {\dot{m}}_3={\dot{m}}_1+{\dot{m}}_2\\ {\dot{m}}_3=\left(3.6\text{kg}/\text{s}\right)+\left(4.586\text{kg}/\text{s}\right)\\ =8.186\text{kg}/\text{s}\end{array}$

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 ${\dot{m}}_1{s}_1+{\dot{m}}_2{s}_2$ for ${\dot{S}}_{in}$, ${\dot{m}}_3{s}_3$ for ${\dot{S}}_{out}$ and 0 for $\Delta {\dot{S}}_{system}$ in Equation (V).

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

  $\begin{array}{c}{\dot{S}}_{gen}=\left\{\begin{array}{l}\left(8.186\text{kg}/\text{s}\right)\left(0.05990\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(4.586\text{kg}/\text{s}\right)\left(0.2965\text{kJ}/\text{kg}\cdot \text{K}\right)\\ -\left(3.6\text{kg}/\text{s}\right)\left(0.9551\text{kJ}/\text{kg}\cdot \text{K}\right)\end{array}\right\}\\ =0.1054\text{kW}/\text{K}\end{array}$

Thus, the rate of heat entropy generation during the process is $0.1054\text{kW}/\text{K}$.