Chapter 14.7, Problem 110P

a)

To determine

The volume flow rate of air into the cooling tower.

Answer to Problem 110P

The volume flow rate of air into the cooling tower is $11.2{\text{m}}^3/\text{s}$.

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 volume flow rate of air into the cooling tower $\left({\dot{\nu }}_1\right)$.

${\dot{\nu }}_1={\dot{m}}_a{v}_1$ (III)

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

$\begin{array}{c}{P}_{v1}={\varphi }_1{P}_{g1}\\ ={\varphi }_1{P}_{\text{sat}@30\text{°C}}\end{array}$ (IV)

Here, saturation pressure of water at $\text{30°C}$ is $P_{g1}$ and relative humidity at state 1 is ${\varphi }_1$.

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

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

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

Write the expression to obtain the specific volume of duct $\left(v_1\right)$.

$v_1=\frac{R_a{T}_1}{P_{a1}}$ (VI)

Here, inlet temperature is $T_1$ and gas constant of air is $R_a$.

Write the expression to obtain the specific humidity $\left({\omega }_1\right)$ of air incoming.

${\omega }_1=\frac{0.622P_{v1}}{P_1-P_{v1}}$ (VII)

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

Write the expression to obtain the enthalpy at state 1 $\left(h_1\right)$.

$h_1=c_p{T}_1+{\omega }_1{h}_{g1}$ (VIII)

Here, specific heat of air is $c_p$, initial temperature is $T_1$, and initial condition of enthalpy at saturation vapor is $h_{g1}$.

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

$\begin{array}{c}{P}_{v2}={\varphi }_2{P}_{g2}\\ ={\varphi }_2{P}_{\text{sat}@20\text{°C}}\end{array}$ (IX)

Here, saturation pressure of water at $\text{20°C}$ is $P_{g2}$ and relative humidity at state 2 is ${\varphi }_2$.

Write the expression to obtain the specific humidity $\left({\omega }_2\right)$ of air incoming from second duct.

${\omega }_2=\frac{0.622P_{v2}}{P_2-P_{v2}}$ (X)

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

Write the expression to obtain the enthalpy at state 2 $\left(h_2\right)$.

$h_2=c_p{T}_2+{\omega }_2{h}_{g2}$ (XI)

Here, initial condition of enthalpy at saturation vapor at state 2 is $h_{g2}$.

Conclusion:

Refer Table A-4, “Saturated water – Temperature table”, obtain the properties of water at a temperature of $20°\text{C}$.

$\begin{array}{l}{P}_{sat@20°\text{C}}=2.3392\text{ kPa}\\ h_{g1}=2,537.4\text{kJ}/\text{kg}\end{array}$

Substitute 2.3392 kPa for $P_{\text{sat}@30\text{°C}}$ and 0.7 for ${\varphi }_1$ in Equation (IV).

$\begin{array}{c}{P}_{v1}=\left(0.7\right)\left(2.3392\text{kPa}\right)\\ =1.637\text{kPa}\end{array}$

Rewrite Equation (V) and substitute 96 kPa for $P_1$ and 1.637 kPa for $P_{v1}$.

$\begin{array}{c}{P}_{a1}=P_1-P_{v1}\\ =96\text{kPa}-1.637\text{kPa}\\ =94.363\text{kPa}\end{array}$

Convert the unit of $T_1$ from $\text{°C}$ to $\text{K}$.

$\begin{array}{c}{T}_1=20\text{°C}\\ =\left(20+273\right)\text{K}\\ =293\text{K}\end{array}$

Refer Table A-2, “Ideal gas specific heats of various common gases”, obtain the value of $R_a$ for air as $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ and $c_p$ for air as $1.005\text{kJ}/\text{kg}\cdot °\text{C}$.

Substitute $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R_a$, 293 K for $T_1$, and 94.363 kPa for $P_{a1}$ in Equation (VI).

$\begin{array}{c}{v}_1=\frac{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(293\text{K}\right)}{\left(94.363\text{kPa}\right)}\\ =0.891{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute 1.637 kPa for $P_{v1}$ and 96 kPa for $P_1$ in Equation (VII).

$\begin{array}{c}{\omega }_1=\frac{0.622\left(1.637\text{kPa}\right)}{96\text{kPa}-1.637\text{kPa}}\\ =0.0108\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $\text{20°C}$ for $T_1$, $0.0108\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$, and $2,537.4\text{kJ}/\text{kg}$ for $h_{g1}$ in Equation (VIII).

$\begin{array}{c}{h}_1=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{20°C}\right)+\left[\begin{array}{l}\left(0.0108\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\right)\\ \left(2,537.4\text{kJ}/\text{kg}\right)\end{array}\right]\\ =47.5\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Table A-4, “Saturated water – Temperature table”, obtain the properties of water at a temperature of $35°\text{C}$.

$\begin{array}{l}{P}_{sat@35°\text{C}}=5.6291\text{ kPa}\\ h_{g2}=2,564.6\text{kJ}/\text{kg}\end{array}$

Substitute 5.6291 kPa for $P_{\text{sat}@35\text{°C}}$ and 1.00 for ${\varphi }_2$ in Equation (IX).

$\begin{array}{c}{P}_{v2}=\left(1.00\right)\left(5.6291\text{kPa}\right)\\ =5.6291\text{kPa}\end{array}$

Substitute 5.6291 kPa for $P_{v2}$ and 96 kPa for $P_2$ in Equation (X).

$\begin{array}{c}{\omega }_2=\frac{0.622\left(5.6291\text{kPa}\right)}{96\text{kPa}-5.6291\text{kPa}}\\ =0.0387\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $\text{35°C}$ for $T_2$, $0.0387\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$, and $2,564.6\text{kJ}/\text{kg}$ for $h_{g2}$ in Equation (XI).

$\begin{array}{c}{h}_2=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{35°C}\right)+\left[\begin{array}{l}\left(0.0387\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\right)\\ \left(2,564.6\text{kJ}/\text{kg}\right)\end{array}\right]\\ =134.4\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Table A-4, “Saturated water – Temperature table”, obtain the value of enthalpy at saturation liquid $\left(h_{f@40\text{°C}}\approx h_3\right)$ as $167.53\text{kJ}/\text{kg}{\text{H}}_2\text{O}$ at a temperature of $40°\text{C}$.

Refer Table A-4, “Saturated water – Temperature table”, obtain the value of enthalpy at saturation liquid $\left(h_{f@30\text{°C}}\approx h_4\right)$ as $125.74\text{kJ}/\text{kg}{\text{H}}_2\text{O}$ at a temperature of $30°\text{C}$.

Substitute $25\text{kg}/\text{s}$ for ${\dot{m}}_3$, $125.74\text{kJ}/\text{kg}{\text{H}}_2\text{O}$ for $h_4$, $167.53\text{kJ}/\text{kg}{\text{H}}_2\text{O}$ for $h_3$, $47.5\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_1$, $134.4\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$, $0.0108\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$, and $0.0387\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (II).

$\begin{array}{c}{\dot{m}}_a=\frac{25\text{kg}/\text{s}\left(167.53\text{kJ}/\text{kg}{\text{H}}_2\text{O}-125.74\text{kJ}/\text{kg}{\text{H}}_2\text{O}\right)}{\left(\begin{array}{l}134.4\text{kJ}/\text{kg}\text{dry}\text{air}\\ -47.5\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}\right)-\left[\begin{array}{l}\left(\begin{array}{l}0.0387\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\\ -0.0108\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\end{array}\right)\\ 125.74\text{kJ}/\text{kg}{\text{H}}_2\text{O}\end{array}\right]}\\ =12.53\text{kg}/\text{s}\end{array}$

Substitute $12.53\text{kg}/\text{s}$ for ${\dot{m}}_a$ and $13.76{\text{ft}}^3/\text{lbm}\text{dry}\text{air}$ for $v_1$ in Equation (III).

$\begin{array}{c}{\dot{\nu }}_1=12.53\text{kg}/\text{s}\left(0.891{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\right)\\ =11.2{\text{m}}^3/\text{s}\end{array}$

Thus, the volume flow rate of air into the cooling tower is $11.2{\text{m}}^3/\text{s}$.

b)

To determine

The mass flow rate of required makeup water.

Answer to Problem 110P

The mass flow rate of required makeup water is $0.35\text{kg}/\text{s}$.

Explanation of Solution

Write the expression to obtain the mass flow rate of required makeup water $\left({\dot{m}}_{\text{makeup}}\right)$ from Equation (I).

${\dot{m}}_{\text{makeup}}={\dot{m}}_a\left({\omega }_2-{\omega }_1\right)$ (XII)

Conclusion:

Substitute $12.53\text{kg}/\text{s}$ for ${\dot{m}}_a$, $0.0108\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $0.0387\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (XII).

$\begin{array}{c}{\dot{m}}_{\text{makeup}}=12.53\text{kg}/\text{s}\left(\begin{array}{l}0.0387\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\\ -0.0108\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\end{array}\right)\\ =0.35\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of required makeup water is $0.35\text{kg}/\text{s}$.