Chapter 14.7, Problem 105P

To determine

The rate of entropy generation for this process.

Answer to Problem 105P

The rate of entropy generation for this process is $\underset{\_}{3.402\times {10}^{-4}\text{Btu}/\text{s}\cdot \text{R}}$.

Explanation of Solution

Express the mass flow rate of dry air at state 1.

${\dot{m}}_{a1}=\frac{{\dot{V}}_1}{v_1}$ (I)

Here, volume flow rate at state 1 is ${\dot{V}}_1$ and specific volume at state 1 is $v_1$.

Express the mass flow rate of dry air at state 2.

${\dot{m}}_{a2}=\frac{{\dot{V}}_2}{v_2}$ (II)

Here, volume flow rate at state 2 is ${\dot{V}}_2$ and specific volume at state 2 is $v_2$.

Express the mass flow rate of dry air at state 3 from the conservation of mass.

${\dot{m}}_{a3}={\dot{m}}_{a1}+{\dot{m}}_{a2}$ (III)

Express enthalpy at state 3.

$\frac{{\dot{m}}_{a1}}{{\dot{m}}_{a2}}=\frac{h_2-h_3}{h_3-h_1}$ (IV)

Here, enthalpy at state 1 and 2 is $h_1\text{and}{h}_2$ respectively.

Express specific humidity at state 3.

$\frac{{\dot{m}}_{a1}}{{\dot{m}}_{a2}}=\frac{{\omega }_2-{\omega }_3}{{\omega }_3-{\omega }_1}$ (V)

Here, specific humidity at state 1 and 2 is ${\omega }_1\text{and}{\omega }_2$ respectively.

Determine the entropy balance on the mixing chamber for water.

$\Delta {\dot{S}}_{\text{w}}={\dot{m}}_{a3}{\omega }_3{s}_3-{\dot{m}}_{a1}{\omega }_1{s}_1-{\dot{m}}_{a2}{\omega }_2{s}_2$ (VI)

Determine the partial pressure of water vapour at state 1 for air steam.

$\begin{array}{c}{P}_{vap,1}={\varphi }_1\times P_{g1}\\ ={\varphi }_1\times P_{\text{sat@100°F}}\end{array}$ (VII)

Determine the partial pressure of dry air at state 1 for air steam.

$P_{a1}=P_1-P_{vap,1}$ (VIII)

Here, the pressure at the state 1 is $P_1$.

Determine the partial pressure of water vapour at state 2 for air steam.

$\begin{array}{c}{P}_{vap,2}={\varphi }_2\times P_{g2}\\ ={\varphi }_2\times P_{\text{sat@50°F}}\end{array}$ (IX)

Determine the partial pressure of dry air at state 2 for air steam.

$P_{a2}=P_2-P_{vap,2}$ (X)

Here, the pressure at the state 2 is $P_2$.

Determine the partial pressure of water vapour at state 3 for air steam.

$\begin{array}{c}{P}_{vap,3}={\varphi }_3\times P_{g3}\\ ={\varphi }_3\times P_{\text{sat@86}\text{.7°C}}\end{array}$ (XI)

Determine the partial pressure of dry air at state 3 for air steam.

$P_{a3}=P_3-P_{vap,3}$ (XII)

Here, the pressure at the state 3 is $P_3$.

Determine the entropy balance on the mixing chamber for the dry air.

$\begin{array}{c}\Delta {\dot{S}}_a={\dot{m}}_{a1}\left(s_3-s_1\right)+{\dot{m}}_{a2}\left(s_3-s_2\right)\\ =\left[\begin{array}{l}{\dot{m}}_{a1}\left(c_p\times \ln \frac{T_3}{T_1}-R\times \ln \frac{P_{a,3}}{P_{a,1}}\right)\\ +{\dot{m}}_{a2}\left(c_p\times \ln \frac{T_3}{T_2}-R\times \ln \frac{P_{a,3}}{P_{a,2}}\right)\end{array}\right]\end{array}$ (XIII)

Here, the coefficient of constant pressure is $c_p$, the temperature at the state 3 is $T_3$, the temperature at the state 1 is $T_1$, the temperature at the state 2 is $T_2$, and the universal gas constant is $R$.

Determine the rate of entropy generation.

${\dot{S}}_{\text{gen}}=\Delta {\dot{S}}_a+\Delta {\dot{S}}_w$ . (XIV)

Conclusion:

Refer Figure A-31E, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to dry bulb temperature of $\text{100°F}$ and relative humidity of $\text{90\%}$.

$\begin{array}{l}{h}_1=66.7\text{Btu}/\text{lbm}\text{dry}\text{air}\\ {\omega }_1=0.0386\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\\ v_1=14.98{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}\end{array}$

Here, initial enthalpy is $h_1$ and initial specific humidity is ${\omega }_1$.

Refer Figure A-31E, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to dry bulb temperature of $\text{50°C}$ and relative humidity of $\text{30\%}$.

$\begin{array}{l}{h}_2=14.5\text{Btu}/\text{lbm}\text{dry}\text{air}\\ {\omega }_2=0.0023\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\\ v_2=12.90{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}\end{array}$

Here, enthalpy at state 2 is $h_2$ and specific humidity at state 2 is ${\omega }_2$.

Refer Table A-4E, “saturated water-temperature table”, and write the entropy at state 1 and 2 at temperature of $100°\text{F}\text{and}50°\text{F}$.

$\begin{array}{c}{s}_{g1}=s_{g@100°\text{F}}\\ =1.9819\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

$\begin{array}{c}{s}_{g2}=s_{g@50°\text{F}}\\ =2.1256\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $\text{3}{\text{ft}}^{\text{3}}/\text{s}$ for ${\dot{\nu }}_1$ and $14.98{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$ for $v_1$ in Equation (I).

$\begin{array}{c}{\dot{m}}_{a1}=\frac{\text{3}{\text{ft}}^{\text{3}}/\text{s}}{14.98{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}}\\ =0.2002\text{lbm}/\text{s}\end{array}$

Substitute $\text{1}{\text{ft}}^{\text{3}}/\text{s}$ for ${\dot{\nu }}_2$ and $12.90{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$ for $v_2$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{a2}=\frac{1{\text{ft}}^{\text{3}}/\text{s}}{12.90{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}}\\ =0.07755\text{lbm}/\text{s}\end{array}$

Substitute $0.2002\text{lbm}/\text{s}$ for ${\dot{m}}_{a1}$ and $0.07755\text{lbm}/\text{s}$ for ${\dot{m}}_{a2}$ in Equation (III).

$\begin{array}{c}{\dot{m}}_{a3}=0.2002\text{lbm}/\text{s}+0.07755\text{lbm}/\text{s}\\ =0.2778\text{lbm}/\text{s}\end{array}$

Substitute $0.2002\text{lbm}/\text{s}$ for ${\dot{m}}_{a1}$, $0.07755\text{lbm}/\text{s}$ for ${\dot{m}}_{a2}$, $14.5\text{Btu}/\text{lbm}\text{dry}\text{air}$ for $h_2$ and $66.7\text{Btu}/\text{lbm}\text{dry}\text{air}$ for $h_1$ in Equation (IV).

$\begin{array}{l}\frac{0.2002\text{lbm}/\text{s}}{0.07755\text{lbm}/\text{s}}=\frac{14.5\text{Btu}/\text{lbm}\text{dry}\text{air}-h_3}{h_3-66.7\text{Btu}/\text{lbm}\text{dry}\text{air}}\\ 2.5815=\frac{14.5\text{Btu}/\text{lbm}\text{dry}\text{air}-h_3}{h_3-66.7\text{Btu}/\text{lbm}\text{dry}\text{air}}\\ h_3=52.1\text{Btu}/\text{lbm}\text{dry}\text{air}\end{array}$

Substitute $0.2002\text{lbm}/\text{s}$ for ${\dot{m}}_{a1}$, $0.07755\text{lbm}/\text{s}$ for ${\dot{m}}_{a2}$, $0.0023\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_2$ and $0.0386\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_1$ in Equation (V).

$\begin{array}{l}\frac{0.2002\text{lbm}/\text{s}}{0.07755\text{lbm}/\text{s}}=\frac{0.0023\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}-{\omega }_3}{{\omega }_3-0.0386\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}}\\ 2.5815=\frac{0.0023\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}-{\omega }_3}{{\omega }_3-0.0386\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}}\\ {\omega }_3=0.0284\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\end{array}$

Refer Figure A-31E, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to ${\omega }_3=0.0284\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ and $h_3=52.1\text{Btu}/\text{lbm}\text{dry}\text{air}$.

$\begin{array}{l}{T}_3=86.7\text{°F}\\ {\varphi }_3=1\left(100\%\right)\end{array}$

Refer Table A-4E, “saturated water-temperature table”, and write the entropy at state 3 at temperature of $\text{86}\text{.7°F}$ using an interpolation method.

$s_3=s_{g@86.7\text{°F}}$ (XV)

Here, entropy of saturation liquid at temperature of $\text{86}\text{.7°F}$ is $s_{g@86.7\text{°F}}$.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y is temperature and specific entropy at state 3 respectively.

Show the specific entropy at state 3 corresponding to temperature as in Table (1).

Temperature $T\left(\text{°F}\right)$	Specific entropy at state 3 $s_{3@s_g}\left(\text{Btu}/\text{lbm}\cdot \text{R}\right)$
85 $\left(x_1\right)$	2.0214 $\left(y_1\right)$
86.7 $\left(x_2\right)$	$\left(y_2=?\right)$
90 $\left(x_3\right)$	2.0079 $\left(y_3\right)$

Substitute $\text{80°F,}\text{86}\text{.7°F}\text{and}\text{90°F}$ for $x_1,x_2\text{and}{x}_3$ respectively, $2.0214\text{Btu}/\text{lbm}\cdot \text{R}$ for $y_1$ and $\text{2}\text{.0079}\text{Btu}/\text{lbm}\cdot \text{R}$ for $y_3$ in Equation (XIV).

$\begin{array}{c}{y}_2=\frac{\left(\text{86}\text{.7°F}-\text{85°F}\right)\left(\text{2}\text{.0079}\text{Btu}/\text{lbm}\cdot \text{R}-2.0214\text{Btu}/\text{lbm}\cdot \text{R}\right)}{\left(\text{90°F}-85\text{°F}\right)}+2.0214\text{Btu}/\text{lbm}\cdot \text{R}\\ =\text{2}\text{.0168}\text{Btu}/\text{lbm}\cdot \text{R}\\ =s_{g@86.7\text{°F}}\end{array}$

Substitute $\text{2}\text{.0168}\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_{g@86.7\text{°F}}$ in Equation (XV).

$s_{g3}=\text{2}\text{.0168}\text{Btu}/\text{lbm}\cdot \text{R}$

Substitute $0.2002\text{lbm}/\text{s}$ for ${\dot{m}}_{a1}$, $0.07755\text{lbm}/\text{s}$ for ${\dot{m}}_{a2}$, $0.2778\text{lbm}/\text{s}$ for ${\dot{m}}_{a3}$, $0.0023\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_2$, $0.0386\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_1$, $0.0284\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_3$, $\text{2}\text{.0168}\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_{g3}$, $1.9819\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_{g1}$, and $2.1256\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_{g2}$ in Equation (VI).

$\begin{array}{c}\Delta {\dot{S}}_{\text{w}}=\left[\begin{array}{l}\left(0.2778\text{lbm}/\text{s}\times 0.0284\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\times \text{2}\text{.0168}\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ -\left(0.2002\text{lbm}/\text{s}\times 0.0386\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\times 1.9819\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ -\left(0.07755\text{lbm}/\text{s}\times 0.0023\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\times 2.1256\text{Btu}/\text{lbm}\cdot \text{R}\right)\end{array}\right]\\ =\left[\left(0.015912\text{Btu}/\text{s}\cdot \text{R}\right)-\left(0.015316\text{Btu}/\text{s}\cdot \text{R}\right)-\left(0.000379\text{Btu}/\text{s}\cdot \text{R}\right)\right]\\ =0.000217\text{Btu}/\text{s}\cdot \text{R}\end{array}$

Substitute $0.90$ for ${\varphi }_1$ and $0.95052\text{psia}$ for $P_{sat@100°\text{F}}$ in Equation (VII).

$\begin{array}{c}{P}_{vap,1}=\left(0.90\right)\times \left(0.95052\text{psia}\right)\\ =0.8555\text{psia}\end{array}$

Substitute 1 atm for $P_1$ and $0.8555\text{psia}$ for $P_{vap,1}$ in Equation (VIII).

$\begin{array}{c}{P}_{a1}=1\text{atm}-\left(0.8555\text{psia}\right)\\ =1\text{atm}\times \left(\frac{\left(14.696\text{psia}\right)}{1\text{atm}}\right)-\left(0.8555\text{psia}\right)\\ =14.696\text{psia}-0.8555\text{psia}\\ =13.84\text{psia}\end{array}$

Substitute $0.30$ for ${\varphi }_2$ and $0.17812\text{psia}$ for $P_{sat@50°\text{F}}$ in Equation (IX).

$\begin{array}{c}{P}_{vap,2}=\left(0.30\right)\times \left(0.17812\text{psia}\right)\\ =0.0534\text{psia}\end{array}$

Substitute 1 atm for $P_2$ and $0.0534\text{psia}$ for $P_{vap,2}$ in Equation (X).

$\begin{array}{c}{P}_{a2}=1\text{atm}-\left(0.0534\text{psia}\right)\\ =1\text{atm}\times \left(\frac{\left(14.696\text{psia}\right)}{1\text{atm}}\right)-\left(0.0534\text{psia}\right)\\ =14.696\text{psia}-0.0534\text{psia}\\ =14.64\text{psia}\end{array}$

Substitute $1.0$ for ${\varphi }_3$ and $0.6298\text{psia}$ for $P_{sat@86.7°\text{C}}$ in Equation (XI).

$\begin{array}{c}{P}_{vap,3}=\left(1.0\right)\times \left(0.6298\text{psia}\right)\\ =0.6298\text{psia}\end{array}$

Substitute 1 atm for $P_3$ and $0.6298\text{psia}$ for $P_{vap,3}$ in Equation (XII).

$\begin{array}{c}{P}_{a3}=1\text{atm}-\left(0.6298\text{psia}\right)\\ =1\text{atm}\times \left(\frac{\left(14.696\text{psia}\right)}{1\text{atm}}\right)-\left(0.6298\text{psia}\right)\\ =14.696\text{psia}-0.6298\text{psia}\\ =14.07\text{psia}\end{array}$

Substitute $0.2002\text{lbm}/\text{s}$ for ${\dot{m}}_{a1}$, $0.07755\text{lbm}/\text{s}$ for ${\dot{m}}_{a2}$, $0.06855\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $100°\text{F}$ for $T_1$, $50°\text{F}$ for $T_2$, $86.7°\text{C}$ for $T_3$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, 14.64 psia for $P_{a2}$, $13.84\text{psia}$ for $P_{a1}$, and 14.07 psia for $P_{a3}$ in Equation (XIII).

$\begin{array}{c}\Delta {\dot{S}}_a=\left[\begin{array}{l}\left(0.2002\text{lbm}/\text{s}\right)\left(\begin{array}{l}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\times \ln \frac{86.7°\text{C}}{100°\text{F}}\\ -\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\times \ln \frac{14.07\text{psia}}{13.84\text{psia}}\end{array}\right)\\ +\left(0.07755\text{lbm}/\text{s}\right)\left(\begin{array}{l}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\times \ln \frac{86.7°\text{C}}{50°\text{F}}\\ -\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\times \ln \frac{14.07\text{psia}}{14.64\text{psia}}\end{array}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(0.2002\text{lbm}/\text{s}\right)\left(\begin{array}{l}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\times \ln \frac{86.7°\text{C}}{100°\text{F+460}}\\ -\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\times \ln \frac{14.07\text{psia}}{13.84\text{psia}}\end{array}\right)\\ +\left(0.07755\text{lbm}/\text{s}\right)\left(\begin{array}{l}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\times \ln \frac{86.7°\text{C}}{50°\text{F+460}}\\ -\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\times \ln \frac{14.07\text{psia}}{14.64\text{psia}}\end{array}\right)\end{array}\right]\\ =\left(0.2002\text{lbm}/\text{s}\right)\left(-0.006899\text{Btu}/\text{lbm}\right)+\left(0.07755\text{lbm}/\text{s}\right)\left(0.01940\text{Btu}/\text{lbm}\right)\\ =0.0001233\text{Btu}/\text{s}\cdot \text{R}\end{array}$

Substitute $0.0001233\text{Btu}/\text{s}\cdot \text{R}$ for $\Delta {\dot{S}}_a$ and $0.0002169\text{Btu}/\text{s}\cdot \text{R}$ for $\Delta {\dot{S}}_w$ in Equation (XIV).

$\begin{array}{c}{\dot{S}}_{\text{gen}}=0.0001233\text{Btu}/\text{s}\cdot \text{R}+0.0002169\text{Btu}/\text{s}\cdot \text{R}\\ =0.0003402\text{Btu}/\text{s}\cdot \text{R}\\ \cong 3402\times {10}^{-4}\text{Btu}/\text{s}\cdot \text{R}\end{array}$

Thus, the rate of entropy generation for this process is $\underset{\_}{3.402\times {10}^{-4}\text{Btu}/\text{s}\cdot \text{R}}$.