Chapter 14.7, Problem 111P

(a)

To determine

The volume flow rate of air into the cooling tower.

Answer to Problem 111P

The volume flow rate of air into the cooling tower is $\text{7}\text{.58}{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

As the process is a steady flow and thus the mass flow rate of dry air remains constant during the entire process.

$\begin{array}{c}{\dot{m}}_{a1}={\dot{m}}_{a2}\\ ={\dot{m}}_a\end{array}$

Here, the mass flow rate of air at inlet is ${\dot{m}}_{a1}$, mass flow rate of dry air at exit is ${\dot{m}}_{a2}$ and mass flow rate of dry air is ${\dot{m}}_a$.

Express the water mass balance:

$\begin{array}{l}{\displaystyle \sum {\dot{m}}_{w,i}}={\displaystyle \sum {\dot{m}}_{w,e}}\\ {\dot{m}}_3+{\dot{m}}_{a1}{\omega }_1={\dot{m}}_4+{\dot{m}}_2{\omega }_2\\ {\dot{m}}_3-{\dot{m}}_4={\dot{m}}_a\left({\omega }_2-{\omega }_1\right)\end{array}$

Here, mass flow rate of water at inlet and exit is ${\dot{m}}_{w,i}\text{and}{\dot{m}}_{w,e}$ respectively, specific humidity at state 1 and 2 is ${\omega }_1\text{and}{\omega }_2$ respectively; mass flow rate at state 2, 3 and 4 is ${\dot{m}}_2,{\dot{m}}_3\text{and}{\dot{m}}_4$ respectively.

Express the energy balance.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}\\ {\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_i{h}_i=}{\displaystyle \sum {\dot{m}}_e{h}_e}\end{array}$

$\begin{array}{l}0={\displaystyle \sum {\dot{m}}_e{h}_e}-{\displaystyle \sum {\dot{m}}_i{h}_i}\\ 0={\dot{m}}_{a2}{h}_2+{\dot{m}}_4{h}_4-{\dot{m}}_{a1}{h}_1-{\dot{m}}_3{h}_3\\ {\dot{m}}_a=\frac{{\dot{m}}_3\left(h_3-h_4\right)}{\left(h_2-h_1\right)-\left({\omega }_2-{\omega }_1\right)h_4}\end{array}$ (I)

Here, the rate of total energy entering the system is ${\dot{E}}_{\text{in}}$, the rate of total energy leaving the system is ${\dot{E}}_{\text{out}}$, the rate of change in the total energy of the system is $\Delta {\dot{E}}_{\text{system}}$, initial and exit mass flow rate is ${\dot{m}}_i\text{and}{\dot{m}}_e$ respectively, enthalpy at inlet and exit is $h_i\text{and}{h}_e$ respectively and enthalpy at state 1, 2, 3 and 4 is $h_1,h_2,h_3\text{and}{h}_4$ respectively.

Express initial partial pressure.

$\begin{array}{c}{P}_{\nu 1}={\varphi }_1{P}_{g1}\\ ={\varphi }_1{P}_{\text{sat@20°C}}\end{array}$ (II)

Here, relative humidity at state 1 is ${\varphi }_1$, initial vapor pressure is $P_{g1}$ and saturation pressure at temperature of $\text{20°C}$ is $P_{\text{sat@20°C}}$.

Express partial pressure of air at state 1.

$P_{a1}=P_1-P_{\nu 1}$ (III)

Here, pressure at state 1 is $P_1$.

Express specific volume at state 1.

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

Here, gas constant of air is $R_a$ and temperature at state 1 is $T_1$.

Express initial humidity ratio.

${\omega }_1=\frac{0.622P_{\nu 1}}{P_1-P_{\nu 1}}$ (V)

Express initial enthalpy.

$h_1=c_p{T}_1+{\omega }_1{h}_{g1@20\text{°C}}$ (VI)

Here, specific heat at constant pressure is $c_p$ and initial specific enthalpy saturated vapor at temperature of $\text{20°C}$ is $h_{g1\text{@20°C}}$.

Express final partial pressure.

$\begin{array}{c}{P}_{\nu 2}={\varphi }_2{P}_{g2}\\ ={\varphi }_2{P}_{\text{sat@35°C}}\end{array}$ (VII)

Here, relative humidity at state 2 is ${\varphi }_2$, final vapor pressure is $P_{g2}$ and saturation pressure at temperature of $\text{35°C}$ is $P_{\text{sat@35°C}}$.

Express final humidity ratio.

${\omega }_2=\frac{0.622P_{\nu 2}}{P_2-P_{\nu 2}}$ (VIII)

Here, pressure at state 2 is $P_2$.

Express final enthalpy.

$h_2=c_p{T}_2+{\omega }_2{h}_{g2@35\text{°C}}$ (IX)

Here, final specific enthalpy saturated vapor at temperature of $\text{35°C}$ is $h_{g2\text{@35°C}}$.

Express the volume flow rate of air into the cooling tower.

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

Here, specific volume at inlet is $v_1$.

Conclusion:

Refer Table A-2, “ideal-gas specific heats of various common gases”, and write the properties of air.

$\begin{array}{l}{c}_p=1.005\text{kJ}/\text{kg}\cdot °\text{C}\\ R_a=0.2870\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure and initial specific enthalpy saturated vapor at temperature of $\text{20°C}$.

$\begin{array}{l}{P}_{\text{sat@20°C}}=2.3392\text{kPa}\\ h_{g1\text{@20°C}}=2537.4\text{kJ}/\text{kg}\end{array}$

Substitute $0.70$ for ${\varphi }_1$ and $2.3392\text{kPa}$ for $P_{\text{sat@20°C}}$ in Equation (II).

$\begin{array}{c}{P}_{\nu 1}=\left(0.70\right)\left(2.3392\text{kPa}\right)\\ =1.637\text{kPa}\end{array}$

Substitute $\text{96}\text{kPa}$ for $P_1$ and $1.637\text{kPa}$ for $P_{\nu 1}$ in Equation (III).

$\begin{array}{c}{P}_{a1}=\text{96}\text{kPa}-1.637\text{kPa}\\ =\text{94}\text{.363}\text{kPa}\end{array}$

Substitute $0.2870\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R_a$, $\text{20°C}$ for $T_1$ and $\text{94}\text{.363}\text{kPa}$ for $P_{a1}$ in Equation (IV).

$\begin{array}{c}{v}_1=\frac{\left(0.2870\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{20°C}\right)}{\text{94}\text{.363}\text{kPa}}\\ =\frac{\left(0.2870\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left[\left(\text{20}+273\right)\text{K}\right]}{\text{94}\text{.363}\text{kPa}}\\ =\frac{\left(0.2870\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{293}\text{K}\right)}{\text{94}\text{.363}\text{kPa}}\\ =0.8911{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $\text{1}\text{.637}\text{kPa}$ for $P_{\nu 1}$ and $\text{96}\text{kPa}$ for $P_1$ in Equation (V).

$\begin{array}{c}{\omega }_1=\frac{0.622\left(\text{1}\text{.637}\text{kPa}\right)}{\text{96}\text{kPa}-\text{1}\text{.637}\text{kPa}}\\ =0.01079\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $\text{20°C}$ for $T_1$, $0.01079\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$, and $2537.4\text{kJ}/\text{kg}$ for $h_{g1\text{@20°C}}$ in Equation (VI).

$\begin{array}{c}{h}_1=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{20°C}\right)+\left(0.01079\right)\left(2537.4\text{kJ}/\text{kg}\right)\\ =47.5\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure and final specific enthalpy saturated vapor at temperature of $\text{35°C}$.

$\begin{array}{l}{P}_{\text{sat@35°C}}=5.6291\text{kPa}\\ h_{g2\text{@35°C}}=2564.6\text{kJ}/\text{kg}\end{array}$

Substitute $1$ for ${\varphi }_2$ and $5.6291\text{kPa}$ for $P_{\text{sat@35°C}}$ in Equation (VII).

$\begin{array}{c}{P}_{\nu 2}=\left(1\right)\left(5.6291\text{kPa}\right)\\ =5.6291\text{kPa}\end{array}$

Substitute $5.6291\text{kPa}$ for $P_{\nu 2}$ and $\text{96}\text{kPa}$ for $P_2$ in Equation (VIII).

$\begin{array}{c}{\omega }_2=\frac{0.622\left(5.6291\text{kPa}\right)}{\text{96}\text{kPa}-5.6291\text{kPa}}\\ =0.03874\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $\text{35°C}$ for $T_2$, $0.03874\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$, and $2564.6\text{kJ}/\text{kg}$ for $h_{g2\text{@35°C}}$ in Equation (IX).

$\begin{array}{c}{h}_2=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{35°C}\right)+\left(0.03874\right)\left(2564.6\text{kJ}/\text{kg}\right)\\ =134.5\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Table A-4, “saturated water-temperature table”, and write the enthalpy at state 3 at temperature of $\text{40°C}$.

$\begin{array}{c}{h}_3=h_f\\ =167.53\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}\end{array}$

Here, enthalpy of saturation liquid is $h_f$.

Refer Table A-4, “saturated water-temperature table”, and write the enthalpy at state 4 at temperature of $\text{30°C}$.

$\begin{array}{c}{h}_4=h_f\\ =125.74\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}\end{array}$

Substitute $\text{17}\text{kg}/\text{s}$ for ${\dot{m}}_3$, $167.53\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ for $h_3$, $125.74\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ for $h_4$, $134.5\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$, $47.5\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_1$, $0.01079\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $0.03874\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (I).

$\begin{array}{c}{\dot{m}}_a=\frac{\left(\text{17}\text{kg}/\text{s}\right)\left(167.53-125.74\right)\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}}{\left(134.5-47.5\right)\text{kJ}/\text{kg}-\left(0.03874-0.01079\right)\left(125.74\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}\right)}\\ =8.507\text{kg}/\text{s}\end{array}$

Substitute $8.507\text{kg}/\text{s}$ for ${\dot{m}}_a$ and $0.8911{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for ${\nu }_1$ in Equation (X).

$\begin{array}{c}{\dot{\nu }}_1=\left(8.507\text{kg}/\text{s}\right)\left(0.8911{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\right)\\ =7.58{\text{m}}^{\text{3}}/\text{s}\end{array}$

Hence, the volume flow rate of air into the cooling tower is $\text{7}\text{.58}{\text{m}}^{\text{3}}/\text{s}$.

(b)

To determine

The mass flow rate of the required makeup water.

Answer to Problem 111P

The mass flow rate of the required makeup water is $\text{0}\text{.238}\text{kg}/\text{s}$.

Explanation of Solution

Express the mass flow rate of the required makeup water.

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

Conclusion:

Substitute $8.507\text{kg}/\text{s}$ for ${\dot{m}}_a$, $0.01079\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $0.03874\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (XI).

$\begin{array}{c}{\dot{m}}_{\text{makeup}}=8.507\text{kg}/\text{s}\left(0.03874-0.01079\right)\\ =\text{0}\text{.238}\text{kg}/\text{s}\end{array}$

Hence, the mass flow rate of the required makeup water is $\text{0}\text{.238}\text{kg}/\text{s}$.