Chapter 8.8, Problem 105RP

a)

To determine

The exit temperature

Answer to Problem 105RP

The exit temperature is $\underset{\_}{12.3\text{°C}}$.

Explanation of Solution

Write energy balance equation for a closed system of steam.

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}\\ -Q_{\text{out}}=\Delta U\\ Q_{\text{out}}=m\left(u_1-u_2\right)\end{array}$ (I)

Here, temperature at inlet and outlet condition is $T_1\text{and}{T}_2$ respectively, change in internal energy of a system is $\Delta U$, change in energy transfer of a system is $\Delta E_{\text{system}}$, internal energy at state 1 and 2 is $u_1\text{and}{u}_2$, energy transfer at inlet and outlet condition is $E_{\text{in}}\text{and}{E}_{\text{out}}$ respectively, heat transfer leaving from the steam radiator is $Q_{\text{out}}$, and total mass of air is m.

Write the expression for the dryness fraction at state 2$\left(x_2\right)$.

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

Here, $v_2=v_1$.

Write the expression of the internal energy at state 2$\left(u_2\right)$.

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

Write the expression of the entropy at state 2$\left(s_2\right)$.

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

Write the expression of the mass of steam $\left(m\right)$.

$m=\frac{V_1}{v_1}$ (V)

Here, initial volume of steam is $V_1$.

Write the expression of the mass of air $\left(m_{\text{air}}\right)$.

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

Here, initial temperature is $T_1$ and volume of air is $V_1$.

Write the expression of the amount of fan work done in 24 min.

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

Here, change in time is $\Delta t$.

Write the expression of energy balance equation for a closed system of air.

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}\\ Q_{\text{in}}+W_{\text{fan,in}}-W_{\text{b,out}}=\Delta U\\ Q_{\text{in}}+W_{\text{fan,in}}=\Delta H\\ Q_{\text{in}}+W_{\text{fan,in}}=m_{\text{air}}{c}_p\left(T_2-T_1\right)\end{array}$ (VIII)

Here, amount of heat transfer injected to the steam radiator is $Q_{\text{in}}$, boundary work output is $W_{\text{b,out}}$, and work input of fan is $W_{\text{fan,in}}$.

Conclusion:

From Table A-1, “molar mass, gas constant, and critical point properties”, Obtain the gas constant $\left(R\right)$ of air as $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$.

From Table A-3, “properties of common liquids, solids, and foods”, Obtain the specific heat $\left(c_p\right)$ of the air as $1.005\text{kJ}/\text{kg}\cdot \text{K}$.

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

$\begin{array}{l}{v}_1=1.0805{\text{m}}^{\text{3}}/\text{kg}\\ u_1=2654.6\text{kJ}/\text{kg}\\ s_1=7.5081\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From to Table A-5, “saturated water – pressure table”, obtain the following properties at the pressure of $100\text{kPa}$.

$\begin{array}{l}{v}_f=0.001043{\text{m}}^{\text{3}}/\text{kg}\\ v_g=1.6941{\text{m}}^{\text{3}}/\text{kg}\\ u_f=417.40\text{kJ}/\text{kg}\\ u_{fg}=2088.2\text{kJ}/\text{kg}\end{array}$

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

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

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

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

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

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

$\begin{array}{c}{s}_2=1.3028\text{kJ}/\text{kg}\cdot \text{K}+0.6376\left(6.0562\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =5.1639\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $0.015{\text{m}}^{\text{3}}$ for $V_1$ and $1.0805{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in equation (V).

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

0.01388 kg for m, $2654.6\text{kJ}/\text{kg}$ for $u_1$, and $1748.7\text{kJ}/\text{kg}$ for $u_2$ in equation (I) calculate $Q_{\text{out}}$.

$\begin{array}{c}{Q}_{\text{out}}=0.01388\text{kg}\left(2654.6\text{kJ}/\text{kg}-1748.7\text{kJ}/\text{kg}\right)\\ =12.58\text{kJ}\end{array}$

Calculate the volume of air.

$\begin{array}{c}{V}_1=4\text{m}\times 4\text{m}\times 5\text{m}\\ =80{\text{m}}^{\text{3}}\end{array}$

Substitute 283 K for $T_1$, 100 kPa for $P_1$, $0.2870\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for R, and $80{\text{m}}^{\text{3}}$ for $V_1$ in equation (VI).

$\begin{array}{c}{m}_{\text{air}}=\frac{\left(100\text{kPa}\right)\left(80{\text{m}}^{\text{3}}\right)}{0.2870\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\left(10°\text{C}\right)}\\ =\frac{\left(100\text{kPa}\right)\left(80{\text{m}}^{\text{3}}\right)}{0.2870\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\left(10+273\text{K}\right)}\\ =\frac{\left(100\text{kPa}\right)\left(80{\text{m}}^{\text{3}}\right)}{0.2870\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\left(283\text{K}\right)}\\ =98.5\text{kg}\end{array}$

Substitute $0.150\text{kJ}/\text{s}$ for ${\dot{W}}_{\text{fan,in}}$ and 24 min for $\Delta t$ in equation (VII).

$\begin{array}{c}{W}_{\text{fan,in}}=0.150\text{kJ}/\text{s}\left(24\min \right)\\ =0.150\text{kJ}/\text{s}\left(24\min \times \frac{\text{60}\text{s}}{1\min }\right)\\ =216\text{kJ}\end{array}$

Substitute 12.58 kJ for $Q_{\text{in}}$, 216 kJ for $W_{\text{fan,in}}$, 98.5 kg for $m_{\text{air}}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, and $10\text{°C}$ for $T_1$ in Equation (VIII).

$\begin{array}{l}12.58\text{kJ}+\text{216}\text{kJ}=\left(98.5\text{kg}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(T_2-10\text{°C}\right)\\ T_2=12.3\text{°C}\end{array}$

Thus, the exit temperature is $\underset{\_}{12.3\text{°C}}$.

b)

To determine

The entropy change of the steam.

Answer to Problem 105RP

The entropy change of the steam is $\underset{\_}{-0.0325\text{kJ}/\text{K}}$.

Explanation of Solution

Write the expression the entropy change of the steam.

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

Conclusion:

Substitute 0.01388 kg for m, $5.1639\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, and $7.5081\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$ in equation (IX)

$\begin{array}{c}\Delta S_{\text{steam}}=0.01388\text{kg}\left(5.1639\text{kJ}/\text{kg}\cdot \text{K}-7.5081\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =-0.0325\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the steam is $\underset{\_}{-0.0325\text{kJ}/\text{K}}$.

c)

To determine

The entropy change of the air

Answer to Problem 105RP

The entropy change of the air is $\underset{\_}{0.8012\text{kJ}/\text{K}}$.

Explanation of Solution

Write the expression for the entropy change of the air.

$\begin{array}{c}\Delta S_{\text{air}}=m{c}_p\ln \frac{T_2}{T_1}-mR\ln {\frac{P_2}{P_1}}^{↵0}\\ =m{c}_p\ln \frac{T_2}{T_1}\end{array}$ (X)

Conclusion:

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

$\begin{array}{c}\Delta S_{\text{air}}=\left(\text{98}\text{.5}\text{kg}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{12.3°\text{C}}{10°\text{C}}\\ =\left(\text{98}\text{.5}\text{kg}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{\left(12.3+273\right)\text{K}}{\left(10+273\right)\text{K}}\\ =\left(\text{98}\text{.5}\text{kg}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{285.3\text{K}}{283\text{K}}\\ =0.8012\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the air is $\underset{\_}{0.8012\text{kJ}/\text{K}}$.

d)

To determine

The energy destroyed during the process

Answer to Problem 105RP

The energy destroyed during the process is $\underset{\_}{218\text{kJ}}$.

Explanation of Solution

For a closed system, write the simplification rate form of the entropy balance for the room.

$\begin{array}{l}{S}_{\text{in}}-S_{\text{out}}+S_{\text{gen}}=\Delta S_{\text{system}}\\ 0+S_{\text{gen}}=\Delta S_{\text{steam}}+\Delta S_{\text{air}}\\ S_{\text{gen}}=\Delta S_{\text{steam}}+\Delta S_{\text{air}}\end{array}$ (XI)

Here, entropy generation is $S_{\text{gen}}$, change of entropy is $\Delta S_{\text{system}}$, entropy at inlet condition is $S_{\text{in}}$, entropy at outlet condition is $S_{\text{out}}$.

Calculate the energy destroyed during the process $\left(X_{\text{destroyed}}\right)$.

$X_{\text{destroyed}}=\left(T_0\right)\left(\Delta S_{\text{steam}}+\Delta S_{\text{air}}\right)$ (XII)

Here, dead state temperature is $T_0$.

Conclusion:

Substitute $0.8012\text{kJ}/\text{K}$ for $\Delta S_{\text{air}}$ and $-0.0325\text{kJ}/\text{K}$ for $\Delta S_{\text{steam}}$ in Equation (XI).

$\begin{array}{c}{S}_{\text{gen}}=-0.0325\text{kJ}/\text{K}+0.8012\text{kJ}/\text{K}\\ =0.7687\text{kJ}/\text{K}\end{array}$

Substitute 283 K for $T_0$, $0.8012\text{kJ}/\text{K}$ for $\Delta S_{\text{air}}$ and $-0.0325\text{kJ}/\text{K}$ for $\Delta S_{\text{steam}}$

$\begin{array}{c}{X}_{\text{destroyed}}=283\text{K}\left(-0.0325\text{kJ}/\text{K}+0.8012\text{kJ}/\text{K}\right)\\ =218\text{kJ}\end{array}$

Thus, the energy destroyed during the process is $\underset{\_}{218\text{kJ}}$.