Chapter 7.13, Problem 206RP

a)

To determine

The average temperature of the room after 30 min.

Answer to Problem 206RP

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

Explanation of Solution

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 $\Delta E_{system}$.

Write the expression to calculate the final vapor quality.

$x_2=\frac{v_2-v_f}{v_g-v_f}$ (II)

Here, final vapor quality is $x_2$, final molar volume is $v_2$, saturated liquid molar volume is $v_f$ and saturated vapor molar volume is $v_g$.

Write the expression to calculate the final internal energy of the system.

$u_2=u_f+x_2{u}_{fg}$ (III)

Here, final internal energy of the system is $u_2$, saturated liquid internal energy is $u_f$, final vapor quality is $x_2$, and evaporation internal energy is $u_{fg}$.

Write the expression to calculate the final entropy of the system.

$s_2=s_f+x_2{s}_{fg}$ (IV)

Here, final entropy of the system is $s_2$, saturated liquid entropy is $s_f$, final vapor quality is $x_2$, and evaporation entropy is $s_{fg}$.

Write the expression to calculate the mass of the steam.

$m=\frac{{\nu }_1}{v_1}$ (V)

Here, mass of the steam is m, volume of the steam is ${\nu }_1$ and initial specific volume of steam is $v_1$.

Write the expression to calculate the ideal gas equation, to find the mass of the air.

$m_{air}=\frac{P_1{\nu }_1}{R{T}_1}$ (VI)

Here, mass of the air is $m_{air}$, initial pressure is $P_1$, initial volume is ${\nu }_1$, gas constant is R and initial temperature is $T_1$.

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

$W_{fan,in}={\dot{W}}_{fan,in}\left(\Delta t\right)$ (VII)

Here, rate of work done by fan is ${\dot{W}}_{fan,in}$ and time is $\Delta t$.

Conclusion:

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

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

Here, change in internal energy of system is $\Delta U$, heat transfer out is $Q_{out}$, mass of the steam is m, initial internal energy is $u_1$ and final internal energy is $u_2$.

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

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

From 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 $v_{\gamma }$.

Substitute $1.0805{\text{m}}^3/\text{kg}$ for $v_2$, $0.001043{\text{m}}^3/\text{kg}$ for $v_f$ and $1.6941{\text{m}}^3/\text{kg}$ for $v_g$ in Equation (II).

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

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

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

Here, Saturated liquid internal energy is $u_f$ and Evaporation internal energy is $u_{fg}$.

Substitute $417.40\text{kJ}/\text{kg}$ for $u_f$, 0.6376 for $x_2$ and $2088.2\text{kJ}/\text{kg}$ for $u_{fg}$ in Equation (III).

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

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

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

Here, Saturated liquid entropy is $s_f$ and evaporation entropy is $s_{fg}$.

Substitute $1.3028\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, 0.6376 for $x_2$ and $6.0562\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in Equation (IV).

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

Substitute $0.015{\text{m}}^3$ for ${\nu }_1$ and $1.0805{\text{m}}^3/\text{kg}$ for $v_1$ in Equation (V).

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

Substitute 0.01388 kg for m, $2654.4\text{kJ}/\text{kg}$ for $u_1$ and $1748.7\text{kJ}/\text{kg}$ for $u_2$ in Equation (VIII).

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

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

$\begin{array}{l}{Q}_{in}+W_{fan,in}-W_{b,out}=\Delta U\\ Q_{in}+W_{fan,in}=\Delta U+W_{b,out}\\ Q_{in}+W_{fan,in}=\Delta H\\ Q_{in}+W_{fan,in}=m{c}_p\left(T_2-T_1\right)\end{array}$ (IX)

Here, heat transfer in is $Q_{in}$, work done by the fan is $W_{fan,in}$, mass of the air is m, specific heat at constant pressure is $c_p$, initial temperature is $T_1$ and final temperature is $T_2$.

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

Substitute $\text{100}\text{kPa}$ for $P_1$, $\left(4\text{m}\times 4\text{m}\times 5\text{m}\right)$ for ${\nu }_1$, $0.2870\text{kJ}/\text{kg}\cdot \text{K}$ for R and $10 °\text{C}$ for $T_1$ in Equation (VI).

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

Substitute $0.120\text{kJ}/\text{s}$ for ${\dot{W}}_{fan,in}$ and 30 min for $\Delta t$ in Equation (VII).

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

From 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 12.6 kJ for $Q_{in}$, 216 kJ for $W_{fan,in}$, 98.5 kg for m, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ and $10 °\text{C}$ for $T_1$ in Equation (IX).

$\begin{array}{l}12.6\text{kJ}+216\text{kJ}=\left(98.5\text{ kg}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(T_2-10°\text{C}\right)\\ T_2=12.3\text{°C}\end{array}$

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

b)

To determine

The entropy change of the steam.

Answer to Problem 206RP

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

Explanation of Solution

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

$\Delta S_{steam}=m\left(s_2-s_1\right)$ (X)

Here, mass of the steam is m, final entropy is $s_2$ and initial entropy is $s_1$.

Conclusion:

Substitute 0.01388 kg for m, $5.1642\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$ and $7.5081\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$ in Equation (X).

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

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.

Answer to Problem 206RP

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

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, $P_2=P_1$.

$\Delta S_{air}=m{c}_p\ln \frac{T_2}{T_1}$ (XI)

Conclusion:

Substitute 98.5 kg for m, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $12.3 °\text{C}$ for $T_2$ and $10 °\text{C}$ for $T_1$ in Equation (XI).

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

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

d)

To determine

The entropy generation in the turbine.

Answer to Problem 206RP

The entropy generation in the turbine is $4.01\text{kW}/\text{K}.$

Explanation of Solution

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

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

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 $\Delta S_{system}$

Conclusion:

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

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

Here, entropy change of air is $\Delta S_{air}$ and entropy change of steam is $\Delta S_{stean}$.

Substitute $0.8013\text{kJ}/\text{K}$ for $\Delta S_{air}$ and $-0.032\text{5}\text{kJ}/\text{K}$ for $\Delta S_{steam}$ in Equation (XIII).

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

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