Chapter 14.7, Problem 112P

To determine

The exergy destruction per unit mass of dry air.

Answer to Problem 112P

The exergy destruction per unit mass of dry air is $0.726\text{Btu}/\text{bm dry air}$.

Explanation of Solution

Apply the dry air mass balance on the cooling tower.

$\begin{array}{c}{\dot{m}}_{a_1}={\dot{m}}_{a_2}\\ ={\dot{m}}_a\end{array}$

Here, mass flow rate of air at inlet and outlet is ${\dot{m}}_{a_1}\text{and}{\dot{m}}_{a_2}$, and mass flow rate of liquid water is ${\dot{m}}_a$.

Apply the water mass balance on the cooling tower.

$\begin{array}{l}{\dot{m}}_3+{\dot{m}}_{a_1}{\omega }_1={\dot{m}}_4+{\dot{m}}_{a_2}{\omega }_2\\ {\dot{m}}_3-{\dot{m}}_4={\dot{m}}_a\left({\omega }_2-{\omega }_1\right)\\ {\dot{m}}_3-{\dot{m}}_4={\dot{m}}_{\text{makeup}}\end{array}$ (I)

Here, mass flow rate of water at state 3 and 4 is ${\dot{m}}_3$ and ${\dot{m}}_4$, and mass flow rate of required makeup water is ${\dot{m}}_{\text{makeup}}$.

Apply the energy balance on the cooling tower.

$\begin{array}{l}{\displaystyle \sum _{\text{in}}\dot{m}h}={\displaystyle \sum _{\text{out}}\dot{m}h}\\ {\dot{m}}_{a_1}{h}_1+{\dot{m}}_3{h}_3={\dot{m}}_{a_1}{h}_1+{\dot{m}}_4{h}_4\\ {\dot{m}}_3{h}_3={\dot{m}}_a\left(h_2-h_1\right)+\left({\dot{m}}_3-{\dot{m}}_{\text{makeup}}\right)h_4\\ {\dot{m}}_a=\frac{{\dot{m}}_3\left(h_3-h_4\right)}{\left(h_2-h_1\right)-\left({\omega }_2-{\omega }_1\right)h_4}\end{array}$ (II)

Write the expression to obtain the mass of water stream at state 3 $\left(m_3\right)$.

$m_3=\frac{{\dot{m}}_3}{{\dot{m}}_a}$ (III)

Write the expression to obtain the mass of water stream at state 4 $\left(m_4\right)$.

$m_4=m_3-\left({\omega }_2-{\omega }_1\right)$ (IV)

Write the expression to obtain the entropy change of water stream $\left(\Delta s_{\text{water}}\right)$.

$\Delta s_{\text{water}}=m_4{s}_4-m_3{s}_3$ (V)

Write the expression to obtain the entropy change of water vapor in the air stream $\left(\Delta s_{\text{vapor}}\right)$.

$\Delta s_{\text{vapor}}={\omega }_2{s}_{g2}-{\omega }_1{s}_{g1}$ (VI)

Write the expression to obtain the vapor pressure at inlet conditions $\left(P_{v1}\right)$.

$\begin{array}{c}{P}_{v1}={\varphi }_1{P}_{g1}\\ ={\varphi }_1{P}_{\text{sat}@65\text{°F}}\end{array}$ (VII)

Here, saturation pressure of water at $\text{65}°\text{F}$ is $P_{g1}$ and relative humidity at state 1 is ${\varphi }_1$.

Write the expression of the atmospheric pressure of an ideal gas mixture $\left(P_1\right)$.

$P_1=P_{a1}+P_{v1}$ (VIII)

Here, dry air partial pressure at state 1 is $P_{a1}$.

Write the expression to obtain the vapor pressure at second inlet conditions $\left(P_{v2}\right)$.

$\begin{array}{c}{P}_{v2}={\varphi }_2{P}_{g2}\\ ={\varphi }_2{P}_{\text{sat}@75\text{°F}}\end{array}$ (IX)

Here, saturation pressure of water at $\text{75°F}$ is $P_{g2}$ and relative humidity at state 2 is ${\varphi }_2$.

Write the expression of the atmospheric pressure of an ideal gas mixture $\left(P_2\right)$.

$P_1=P_{a1}+P_{v1}$ (X)

Here, dry air partial pressure at state 2 is $P_{a2}$.

Write the expression to obtain the entropy change of dry air $\left(\Delta s_a\right)$

$\Delta s_a=s_2-s_1$

$\Delta s_a=c_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{P_{a2}}{P_{a1}}\right)$ (XI)

Here, temperature at state 2 is $T_2$ and temperature at state 1 is $T_1$.

Write the expression for entropy generation in the cooling tower $\left(s_{\text{gen}}\right)$.

$s_{\text{gen}}=\Delta s_{\text{water}}+\Delta s_{\text{vapor}}+\Delta s_a$ (XII)

Write the exergy destruction per unit mass of dry air $\left(x_{\text{dest}}\right)$.

$x_{\text{dest}}=T_0{s}_{\text{gen}}$ (XIII)

Conclusion:

Refer Table A-4E, “Saturated water – Temperature table”, obtain the enthalpy $\left(h_{f@95°\text{F}}=h_3\right)$ as $63.04\text{Btu}/\text{lbm}$ and entropy $\left(s_{f@95°\text{F}}=s_3\right)$ as $0.12065\text{Btu}/\text{lbm}\cdot \text{R}$ at a temperature of $95°\text{F}$.

Refer Table A-4E, “Saturated water – Temperature table”, obtain the enthalpy $\left(h_{f@80°\text{F}}=h_4\right)$ as $48.07\text{Btu}/\text{lbm}$ and entropy $\left(s_{f@80°\text{F}}=s_4\right)$ as $0.09328\text{Btu}/\text{lbm}\cdot \text{R}$ at a temperature of $80\text{°F}$.

Refer Fig A-31E, “Psychrometric chart at 1 atm total pressure”, at inlet temperature $\left(T_1=65\text{°F}\right)$, and relative humidity $\left({\varphi }_1=30\%\right)$, read the value of inlet enthalpy $\left(h_1\right)$ as $19.9\text{Btu}/\text{lbm}\text{dry air}$, inlet specific humidity $\left({\omega }_1\right)$ as $0.00391\text{lbm}{\text{H}}_2\text{O}/\text{lbm}\text{dry}\text{air}$, and inlet specific volume of dry air $\left(v_1\right)$ as $13.31{\text{ft}}^3/\text{lbm}\text{dry}\text{air}$.

Refer Fig A-31E, “Psychrometric chart at 1 atm total pressure”, at outlet temperature $\left(T_2=75\text{°F}\right)$ and relative humidity $\left({\varphi }_2=80\%\right)$, read the value of exit enthalpy $\left(h_2\right)$ as $34.3\text{Btu}/\text{lbm}\text{dry air}$, and exit specific humidity $\left({\omega }_2\right)$ as $0.0149\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$.

Substitute $3\text{lbm}/\text{s}$ for ${\dot{m}}_3$, $63.04\text{Btu}/\text{lbm}$ for $h_3$, $48.07\text{Btu}/\text{lbm}$ for $h_4$, $19.9\text{Btu}/\text{lbm}\text{dry air}$ for $h_1$, $34.3\text{Btu}/\text{lbm}\text{dry air}$ for $h_2$, $0.00391\text{lbm}{\text{H}}_2\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_1$, and $0.0149\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (II).

$\begin{array}{c}{\dot{m}}_a=\frac{3\text{lbm}/\text{s}\left(63.04\text{Btu}/\text{lbm}-48.07\text{Btu}/\text{lbm}\right)}{\left(\begin{array}{l}34.3\text{Btu}/\text{lbm}\text{dry air}\\ -19.9\text{Btu}/\text{lbm}\text{dry air}\end{array}\right)-\left[\begin{array}{l}\left(\begin{array}{l}0.0149\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\\ -0.00391\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\end{array}\right)\\ 48.07\text{Btu}/\text{lbm}\end{array}\right]}\\ =3.22\text{lbm}/\text{s}\end{array}$

Substitute $3\text{lbm water}/\text{s}$ for ${\dot{m}}_3$ and $3.22\text{lbm dry air}/\text{s}$ for ${\dot{m}}_a$ in Equation (III).

$\begin{array}{c}{m}_3=\frac{3\text{lbm water}/\text{s}}{3.22\text{lbm dry air}/\text{s}}\\ =0.9317\text{lbm water}/\text{lbm dry air}\end{array}$

Substitute $0.9317\text{lbm water}/\text{lbm dry air}$ for $m_3$, $0.00391\text{lbm}{\text{H}}_2\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_1$, and $0.0149\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (IV).

$\begin{array}{c}{m}_4=0.9317\text{lbm water}/\text{lbm dry air}-\left(0.0149-0.00391\right)\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\\ =0.9207\text{lbm}\text{water}/\text{lbm}\text{dry}\text{air}\end{array}$

Substitute $0.9317\text{lbm water}/\text{lbm dry air}$ for $m_3$, $0.9207\text{lbm water}/\text{lbm dry air}$ for $m_4$, $0.12065\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_3$ and $0.09328\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_4$ in Equation (V).

$\begin{array}{c} \Delta s_{\text{water}}=\left\{\begin{array}{l}\left(0.9207\text{lbm}\text{water}/\text{lbm}\text{dry}\text{air}\times 0.09328\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ -\left(0.9317\text{lbm}\text{water}/\text{lbm}\text{dry}\text{air}\times 0.12065\text{Btu}/\text{lbm}\cdot \text{R}\right)\end{array}\right\}\\ =-0.02653\text{Btu}/\text{R}\cdot \text{lbm}\text{dry}\text{air}\end{array}$

Refer Table A-4E, “Saturated water – Temperature table”, obtain the entropy $\left(s_{g@65°\text{F}}=s_{g1}\right)$ as $2.0788\text{Btu}/\text{lbm}\cdot \text{R}$ at a temperature of $65°\text{F}$.

Refer Table A-4E, “Saturated water – Temperature table”, obtain the entropy $\left(s_{g@80°\text{F}}=s_{g2}\right)$ as $2.0352\text{Btu}/\text{lbm}\cdot \text{R}$ at a temperature of $80°\text{F}$.

Substitute $2.0352\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_{g2}$, $2.0788\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_{g1}$, $0.00391\text{lbm}{\text{H}}_2\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_1$, and $0.0149\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (VI).

$\begin{array}{c} \Delta s_{\text{vapor}}=\left\{\begin{array}{l}\left(0.0149\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\times 2.0352\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ -\left(0.00391\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\times 2.0788\text{Btu}/\text{lbm}\cdot \text{R}\right)\end{array}\right\}\\ =0.0222\text{Btu}/\text{R}\cdot \text{lbm dry air}\end{array}$

Refer Table A-4E, “Saturated water – Temperature table”, obtain the properties of water at a temperature of $65°\text{F}$.

$P_{sat@65°\text{F}}=0.30578\text{ psia}$

Substitute $0.30578\text{ psia}$ for $P_{sat@65°\text{F}}$ and 0.3 for ${\varphi }_1$ in Equation (VII).

$\begin{array}{c}{P}_{v1}=\left(0.3\right)\left(0.30578\text{ psia}\right)\\ =0.0917\text{ psia}\end{array}$

Rewrite Equation (VIII) and substitute 14.696 psia for $P_1$ and 0.0917 psia for $P_{v1}$.

$\begin{array}{c}{P}_{a1}=P_1-P_{v1}\\ =14.696\text{ psia}-0.0917\text{ psia}\\ =14.60\text{ psia}\end{array}$

Refer Table A-4E, “Saturated water – Temperature table”, obtain the properties of water at a temperature of $75°\text{F}$.

$P_{sat@75°\text{F}}=0.43016\text{ psia}$

Substitute $0.43016\text{ psia}$ for $P_{sat@75°\text{F}}$ and 0.80 for ${\varphi }_2$ in Equation (IX).

$\begin{array}{c}{P}_{v2}=\left(0.80\right)\left(0.43016\text{ psia}\right)\\ =0.3441\text{ psia}\end{array}$

Rewrite Equation (X) and substitute 14.696 psia for $P_2$ and 0.3441 psia for $P_{v2}$.

$\begin{array}{c}{P}_{a2}=P_2-P_{v2}\\ =14.696\text{ psia}-0.3441\text{ psia}\\ =14.35\text{ psia}\end{array}$

Refer Table A-2E, “Ideal gas specific heats of various common gases”, obtain the value of $c_p\text{ and }R$ for air as $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ and $0.06855\text{Btu}/\text{lbm}\cdot \text{R}$ respectively.

Substitute $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $0.06855\text{Btu}/\text{lbm}\cdot \text{R}$ for R, $75°\text{F}$ for $T_2$, $65°\text{F}$ for $T_1$, $14.35\text{ psia}$ for $P_{a2}$ and $14.60\text{ psia}$ for $P_{a1}$ in Equation (XI).

$\begin{array}{c}\Delta s_a=\left\{\begin{array}{l}0.240\text{Btu}/\text{lbm}\cdot \text{R}\ln \left(\frac{\left(75°\text{F}+460\right)\text{R}}{\left(65°\text{F}+460\right)\text{R}}\right)\\ -0.06855\text{Btu}/\text{lbm}\cdot \text{R}\ln \left(\frac{14.35\text{ psia}}{14.60\text{ psia}}\right)\end{array}\right\}\\ =0.005712\text{Btu}/\text{lbm}\cdot \text{R dry air}\end{array}$

Substitute $0.005712\text{Btu}/\text{R}\cdot \text{lbm dry air}$ for $\Delta s_a$, $0.0222\text{Btu}/\text{R}\cdot \text{lbm dry air}$ for $\Delta s_{\text{vapor}}$ and $-0.02653\text{Btu}/\text{R}\cdot \text{lbm}\text{dry}\text{air}$ for $\Delta s_{\text{water}}$ in Equation (XII).

$\begin{array}{c}{s}_{\text{gen}}=\left\{\begin{array}{c}-0.02653\text{Btu}/\text{R}\cdot \text{lbm dry air}+0.02220\text{Btu}/\text{R}\cdot \text{lbm dry air}\\ +0.005712\text{Btu}/\text{R}\cdot \text{lbm dry air}\end{array}\right\}\\ =0.001382\text{Btu}/\text{R}\cdot \text{lbm dry air}\end{array}$

Substitute $65°\text{F}$ for $T_0$, and $0.001382\text{Btu}/\text{R}\cdot \text{lbm dry air}$ for $s_{\text{gen}}$ in Equation (XIII).

$\begin{array}{c}{x}_{\text{dest}}=\left(65°\text{F}+460\right)\text{R}\times 0.001382\text{Btu}/\text{R}\cdot \text{lbm dry air}\\ =0.726\text{Btu}/\text{bm dry air}\end{array}$

Thus the exergy destruction per unit mass of dry air is $0.726\text{Btu}/\text{bm dry air}$.