Chapter 14.7, Problem 138RP

a)

To determine

The mass flow rate of the cooling water.

Answer to Problem 138RP

The mass flow rate of the cooling water is $1337\text{kg}/\text{s}$.

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.

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

${\dot{m}}_3-{\dot{m}}_4={\dot{m}}_{\text{makeup}}$ (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)

Apply steady flow energy balance equation on the cooling water.

${\dot{Q}}_{\text{waste}}={\dot{m}}_3{h}_3-\left({\dot{m}}_3-{\dot{m}}_{\text{makeup}}\right)h_4$

${\dot{Q}}_{\text{waste}}={\dot{m}}_3{h}_3-\left({\dot{m}}_3-{\dot{m}}_a\left({\omega }_2-{\omega }_1\right)\right)h_4$ (III)

Here, rate of heat removal is ${\dot{Q}}_{\text{waste}}$.

Conclusion:

Refer Table A-4, “Saturated water – Temperature table”, obtain the enthalpy $\left(h_{f@30°\text{C}}=h_4\right)$ as $125.74\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ at a temperature of $30°\text{C}$.

Refer Table A-4, “Saturated water – Temperature table”, obtain the enthalpy $\left(h_{f@42°\text{C}}=h_3\right)$ as $175.90\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ at a temperature of $42°\text{C}$.

Refer Fig A-31, “Psychrometric chart at 1 atm total pressure”, at dry bulb temperature $\left(T_{\text{db}}=23°\text{C}\right)$, and wet bulb temperature $\left(T_{\text{wb}}=16°\text{C}\right)$, read the value of inlet enthalpy $\left(h_1\right)$ as $44.67\text{kJ}/\text{kg}\text{dry air}$, inlet specific humidity $\left({\omega }_1\right)$ as $0.008462\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$, and inlet specific volume of dry air $\left(v_1\right)$ as $0.8504{\text{m}}^3/\text{kg}\text{dry}\text{air}$.

Refer Fig A-31, “Psychrometric chart at 1 atm total pressure”, at temperature $\left(T_2=32°\text{C}\right)$, read the value of exit enthalpy $\left(h_2\right)$ as $110.69\text{kJ}/\text{kg}\text{dry air}$, and exit specific humidity $\left({\omega }_2\right)$ as $0.03065\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$.

Substitute $125.74\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ for $h_4$, $175.90\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ for $h_3$, $44.67\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ for $h_1$, $110.69\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ for $h_2$, $0.008462\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$, and $0.03065\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (II).

${\dot{m}}_a=\frac{{\dot{m}}_3\left(175.90\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}-125.74\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}\right)}{\left(\begin{array}{l}110.69\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}\\ -44.67\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}\end{array}\right)-\left[\begin{array}{l}\left(\begin{array}{l}0.03065\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\\ -0.008462\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\end{array}\right)\\ 125.74\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}\end{array}\right]}$

${\dot{m}}_a=0.7933{\dot{m}}_3$ (IV)

Substitute 70 MW for ${\dot{Q}}_{\text{waste}}$, $0.7933{\dot{m}}_3$ for ${\dot{m}}_a$, $125.74\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ for $h_4$, $175.90\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ for $h_3$, $0.008462\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$, and $0.03065\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (III).

$\begin{array}{l}70\text{ MW}\left(\frac{{10}^3\text{kJ}/\text{s}}{1\text{ MW}}\right)=\left\{\begin{array}{l}{\dot{m}}_3\left(175.90\text{kJ}/\text{kg}\right)\\ -\left({\dot{m}}_3-0.7933{\dot{m}}_3\left(\begin{array}{l}0.03065\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\\ -0.008462\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\end{array}\right)\right)\\ 125.74\text{kJ}/\text{kg}\end{array}\right\}\\ 70000=175.9{\dot{m}}_3-125.74{\dot{m}}_3+2.21{\dot{m}}_3\\ {\dot{m}}_3=1337\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the cooling water is $1337\text{kg}/\text{s}$.

b)

To determine

The volume flow rate of air into the cooling tower.

Answer to Problem 138RP

The volume flow rate of air into the cooling tower is $902{\text{m}}^3/\text{s}$.

Explanation of Solution

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

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

Conclusion:

Substitute $1337\text{kg}/\text{s}$ for ${\dot{m}}_3$ in Equation (IV).

$\begin{array}{c}{\dot{m}}_a=0.7933\left(1337\text{kg}/\text{s}\right)\\ =1061\text{kg}/\text{s}\end{array}$

Substitute $1061\text{kg}/\text{s}$ for ${\dot{m}}_a$ and $0.8504{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (V).

$\begin{array}{c}{\dot{\nu }}_1=1061\text{kg}/\text{s}\left(0.8504{\text{m}}^{\text{3}}/\text{kg}\right)\\ =902{\text{m}}^3/\text{s}\end{array}$

Thus, the volume flow rate of air into the cooling tower is $902{\text{m}}^3/\text{s}$.

c)

To determine

The mass flow rate of required makeup water.

Answer to Problem 138RP

The mass flow rate of required makeup water is $23.5\text{kg}/\text{s}$.

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).

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

Conclusion:

Substitute $1061\text{kg}/\text{s}$ for ${\dot{m}}_a$, $0.008462\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $0.03065\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (VI).

$\begin{array}{c}{\dot{m}}_{\text{makeup}}=1061\text{kg}/\text{s}\left(\begin{array}{l}0.03065\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\\ -0.008462\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\end{array}\right)\\ =23.5\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of required makeup water is $23.5\text{kg}/\text{s}$.