Chapter 8, Problem 189RQ

a)

To determine

The average temperature of the room after 30 min.

Explanation of Solution

Given:

The dimension of room $\left(v_{room}\right)$ is $4\text{ m}\times \text{4 m}\times \text{5 m}$.

The temperature of the room $\left(T_{room}\right)$ is $10°\text{C}$.

The volume of the radiator is 15 L.

The pressure of the super-heated vapor $\left(P_1\right)$ is $200\text{kPa}$.

The temperature of the superheated vapor $\left(T_1\right)$ is $200°\text{C}$.

The pressure of the steam $\left(P_2\right)$ is $100\text{ kPa}$.

The time duration $\left(\Delta t\right)$ is 30 min.

The pressure of the room $\left(P_{room}\right)$  is 100 kPa.

Calculation:

Refer the Table A-6, “Superheated water table”, obtain the following properties of steam at temperature of $200°\text{C}$ and pressure of $\text{200}\text{kPa}$.

  $\begin{array}{l}{v}_1=1.0805{\text{m}}^3/\text{kg}\\ u_1=2654.4\text{kJ}/\text{kg}\\ s_1=7.5081\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of $\text{100}\text{kPa}$.

  $\begin{array}{l}{v}_f=0.001043{\text{m}}^3/\text{kg}\\ v_g=1.6941{\text{m}}^3/\text{kg}\end{array}$

Here, Saturated liquid specific volume is $v_f$ and saturated vapor specific volume is $v_{\gamma }$.

Calculate the final vapor quality.

  $\begin{array}{c}{x}_2=\frac{v_2-v_f}{v_g-v_f}\\ x_2=\frac{1.0805{\text{m}}^3/\text{kg}-0.001043{\text{m}}^3/\text{kg}}{1.6941{\text{m}}^3/\text{kg}-0.001043{\text{m}}^3/\text{kg}}\\ =0.6376\end{array}$

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of $\text{100}\text{kPa}$.

  $\begin{array}{l}{u}_f=417.40\text{kJ}/\text{kg}\\ u_{fg}=2088.2\text{kJ}/\text{kg}\end{array}$

Calculate the final internal energy of the system.

  $u_2=u_f+x_2{u}_{fg}$

  $\begin{array}{c}{u}_2=417.40\text{kJ}/\text{kg}+\left(0.6376\right)\left(2088.2\text{kJ}/\text{kg}\right)\\ =1748.7\text{kJ}/\text{kg}\end{array}$

Refer the Table A-6, “Saturated water-Pressure table”, obtain the following properties of steam at pressure of $\text{100}\text{kPa}$.

  $\begin{array}{l}{s}_f=1.3028\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=6.0562\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the final entropy of the system.

  $\begin{array}{c}{s}_2=s_f+x_2{s}_{fg}\\ s_2=1.3028\text{kJ}/\text{kg}\cdot \text{K}+\left(0.6376\right)\left(6.0562\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =5.1642\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the mass of the steam.

  $\begin{array}{c}m=\frac{{\nu }_1}{v_1}\\ m=\frac{0.015{\text{m}}^3}{1.0805{\text{m}}^3/\text{kg}}\\ =0.01388\text{kg}\end{array}$

Write the expression for the energy balance equation for closed system without air in the room.

  $E_{in}-E_{out}=\Delta E_{system}$        (I)

Here, energy transfer into the control volume is $E_{in}$, energy transfer exit from the control volume is $E_{out}$ and change in internal energy of system is $\Delta E_{system}$.

Substitute $E_{in}=0$, $E_{out}=Q_{out}$, and $\Delta E_{system}=\Delta U$ in Equation (I).

  $\begin{array}{l}0-Q_{out}=\Delta U\\ -Q_{out}=m\left(u_2-u_1\right)\\ Q_{out}=m\left(u_1-u_2\right)\end{array}$

  $\begin{array}{c}{Q}_{out}=\left(0.01388\text{kg}\right)\left(2654.4\text{kJ}/\text{kg}-1748.7\text{kJ}/\text{kg}\right)\\ =12.6\text{kJ}\end{array}$

Refer the Table A-1, “the molar mass, gas constant and critical point properties table”, select the gas constant of air as $0.2870\text{kJ}/\text{kg}\cdot \text{K}$.

Calculate the mass of the air using the ideal gas equation.

  $\begin{array}{c}{m}_{air}=\frac{P_1{\nu }_1}{R{T}_1}\\ m_{air}=\frac{\left(100\text{ kPa}\right)\left(4\text{m}\times 4\text{m}\times 5\text{m}\right)}{\left(0.2870\text{kJ}/\text{kg}\cdot \text{K}\right)10°\text{C}}\\ =\frac{\left(100\text{ kPa}\right)\left(80{\text{m}}^{\text{3}}\right)}{\left(0.2870\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)\left(10+273\right)\text{K}}\\ =98.5\text{kg}\end{array}$

Write the expression to calculate the total work done by the fan.

  $\begin{array}{c}{W}_{fan,in}={\dot{W}}_{fan,in}\left(\Delta t\right)\\ W_{fan,in}=\left(0.120\text{kJ}/\text{s}\right)\left(30\min \right)\\ =\left(0.120\text{kJ}/\text{s}\right)\left(30\min \right)\left(\frac{60\sec }{1\min }\right)\\ =216\text{kJ}\end{array}$

Refer the Table A-2, “the ideal–gas equation specific heats of various common gases table”, select the specific heat at constant pressure for air as $1.005\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $E_{in}=Q_{in}+W_{fan,in}$, $E_{out}=W_{b,out}$ and $\Delta E_{system}=\Delta U$ in Equation (I).

$\begin{array}{l}{Q}_{in}+W_{fan,in}-W_{b,out}=\Delta U\\ Q_{in}+W_{fan,in}=\Delta U+W_{b,out}\\ Q_{in}+W_{fan,in}=\Delta H\\ Q_{in}+W_{fan,in}=m{c}_p\left(T_2-T_1\right)\end{array}$$\begin{array}{l}12.6\text{kJ}+216\text{kJ}=\left(98.5\text{ kg}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(T_2-10°\text{C}\right)\\ T_2=12.3\text{°C}\end{array}$

Thus, the average temperature of the room after 30 min is $12.3\text{°C}$.

b)

To determine

The entropy change of the steam.

Explanation of Solution

Write the expression to calculate the change of entropy of the steam.

  $\begin{array}{c}\Delta S_{steam}=m\left(s_2-s_1\right)\\ \Delta S_{steam}=\left(0.01388\text{ kg}\right)\left(5.1642\text{kJ}/\text{kg}\cdot \text{K}-7.5081\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =-0.032\text{5}\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the steam is $-0.032\text{5}\text{kJ}/\text{K}$.

c)

To determine

The entropy change of the air.

Explanation of Solution

Write the expression to calculate the entropy change of air.

  $\Delta S_{air}=m{c}_p\ln \frac{T_2}{T_1}-mR\ln \frac{P_2}{P_1}$

Since, $P_2=P_1$.

  $\begin{array}{c}\Delta S_{air}=m{c}_p\ln \frac{T_2}{T_1}\\ \Delta S_{air}=\left(98.5\text{ kg}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{12.3 °\text{C}}{10 °\text{C}}\\ =\left(98.5\text{ kg}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{\left(12.3+273\right)\text{K}}{\left(10+273\right)\text{K}}\\ =0.8013\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the air is $0.8013\text{kJ}/\text{K}$.

d)

To determine

The entropy generation in the turbine.

Explanation of Solution

Write the expression for the entropy balance equation of the system.

  $S_{in}-S_{out}+S_{gen}=\Delta S_{system}$        (II)

Here, rate of net entropy in is $S_{in}$, rate of net entropy out is $S_{out}$, rate of entropy generation is $S_{gen}$ and change of entropy of the system is $\Delta S_{system}$

Substitute $S_{in}=0$, $S_{out}=0$, and $\Delta S_{system}=\Delta S_{air}+\Delta S_{steam}$ in Equation (II).

  $\begin{array}{l}0-0+S_{gen}=\Delta S_{air}+\Delta S_{steam}\\ S_{gen}=\Delta S_{air}+\Delta S_{steam}\end{array}$

  $\begin{array}{c}{S}_{gen}=0.8013\text{kJ}/\text{K}+-0.032\text{5}\text{kJ}/\text{K}\\ =0.7688\text{kJ}/\text{K}\end{array}$

Thus, the total entropy generated during the process is $0.7688\text{kJ}/\text{K}$.