Known Air at 100°F and \phi=10 \% enters an evaporative cooler with a volumetric flow rate of 5000 ft ^{3} / min. Moist air exits the cooler at 70°F. Water is added to the soaked pad of the cooler at 70°F.
Find Determine the mass flow rate of the water to the soaked pad, in lb/h, and the relative humidity of the moist air at the exit of the cooler.
Schematic and Given Data:
Engineering Model
1. The control volume shown in the accompanying figure operates at steady state. Changes in kinetic and potential energy can be neglected and \dot{W}_{ cv }=0.
2. There is no heat transfer with the surroundings.
3. The water added to the soaked pad enters as a liquid and evaporates fully into the moist air.
4. The pressure remains constant throughout at 1 atm.
5. The moist air streams are regarded as ideal gas mixtures adhering to the Dalton model.
Analysis
a. Applying conservation of mass to the dry air and water individually as in previous examples gives
\dot{m}_{ w }=\dot{m}_{ a }\left(\omega_{2}-\omega_{1}\right).
where \dot{m}_{ W } is the mass flow rate of the water to the soaked pad. To find \dot{m}_{ W } requires \omega_{1}, \dot{m}_{ a } \text {, and } \omega_{2}. These will now be evaluated in turn. The humidity ratio \omega_{1} can be found from Eq. 12.43, which requires p_{ v 1}, the partial pressure of the moist air entering the control volume. Using the given relative humidity \phi_{1} \text { and } p_{ g } \text { at } T_{1} from Table A-2E, we have p_{ v 1}=\phi_{1} p_{ g 1}=0.095 lbf / in .{ }^{2} \text { With this, } \omega_{1}=0.00405 lb(vapor)/lb(dry air).
\omega=0.622 \frac{p_{ v }}{p-p_{ v }} (12.43)
The mass flow rate of the dry air \dot{m}_{ a } can be found as in previous examples using the volumetric flow rate and specific volume of the dry air. Thus,
\dot{m}_{ a }=\frac{( AV )_{1}}{v_{ a 1}}
The specific volume of the dry air can be evaluated from the ideal gas equation of state. The result is v_{ a 1}=14.2 ft ^{3} / lb (dry air). Inserting values, the mass flow rate of the dry air is
\dot{m}_{ a }=\frac{5000 ft ^{3} / \min }{14.2 ft ^{3} / lb ( \text {dry air })}=352.1 \frac{ lb (\text { dry air })}{\min }
To find the humidity ratio \omega_{2}, reduce the steady-state forms of the mass and energy rate balances using assumption 1 to obtain
0=\left(\dot{m}_{ a } h_{ a 1}+\dot{m}_{ v 1} h_{ v 1}\right)+\dot{m}_{ w } h_{ w }-\left(\dot{m}_{ a } h_{ a 2}+\dot{m}_{ v 2} h_{ v 2}\right)
With the same reasoning as in previous examples, this can be expressed as the following special form of Eq. 12.55:
0=\dot{Q}_{ cv }+\dot{m}_{ a }\left[\underline{\left(h_{ a 1}-h_{ a 2}\right)}+\underline{\omega_{1} h_{ g 1}+\left(\omega_{2}-\omega_{1}\right) h_{ w }-\omega_{2} h_{ g 2}}\right] (12.55)
0=\left(h_{ a }+\omega h_{ g }\right)_{1}+\underline{\left(\omega_{2}-\omega_{1}\right) h_{ f }}-\left(h_{ a }+\omega h_{ g }\right)_{2} (a)
where h_{ f } denotes the specific enthalpy of the water entering the control volume at 70°F. Solving for \omega_{2}
1 \omega_{2}=\frac{h_{ a 1}-h_{ a 2}+\omega_{1}\left(h_{ g 1}-h_{ f }\right)}{h_{ g 2}-h_{ f }}=\frac{c_{p a }\left(T_{1}-T_{2}\right)+\omega_{1}\left(h_{ g 1}-h_{ f }\right)}{h_{ g 2}-h_{ f }}
where c_{p a }=0.24 Btu / lb \cdot{ }^{\circ} R \text {. With } h_{ f }, h_{ g 1}, \text { and } h_{ g 2} from Table A-2E
\omega_{2}=\frac{0.24(100-70)+0.00405(1105-38.1)}{(1092-38.1)}
=0.0109 \frac{l b(\text { vapor })}{l b (\text { dry air })}
Substituting values for \dot{m}_{ a }, \omega_{1} \text {, and } \omega_{2} into the expression for \dot{m}_{ w }
\dot{m}_{ w }=\left[352.1 \frac{ lb \text{( dry air )}}{\min }\left|\frac{60 \min }{1 h }\right|\right](0.0109-0.00405) \frac{ lb (\text{water})}{ lb \text{( dry air )}}
=144.7 \frac{ lb (\text{water})}{ h }
b. The relative humidity of the moist air at the exit can be determined using Eq. 12.44. The partial pressure of the water vapor required by this expression can be found by solving Eq. 12.43 to obtain
\left.\phi=\frac{p_{ v }}{p_{ g }}\right)_{T, p} (12.44)
p_{ v 2}=\frac{\omega_{2} p}{\omega_{2}+0.622}
Inserting values
p_{ v 2}=\frac{(0.0109)\left(14.696 lbf / in \cdot^{2}\right)}{(0.0109+0.622)}=0.253 lbf / in .{ }^{2}
At 70°F, the saturation pressure is 0.3632 lbf / in .^{2} Thus, the relative humidity at the exit is
\phi_{2}=\frac{0.253}{0.3632}=0.697(69.7 \%)
Alternative Psychrometric Chart Solution Since the underlined term in Eq. (a) is much smaller than either of the moist air enthalpies, the enthalpy of the moist air remains nearly constant, and thus evaporative cooling takes place at a nearly constant wetbulb temperature. See Fig. 12.13b and the accompanying discussion. Using this approach with the psychrometric chart, Fig. A-9E, determine humidity ratio and relative humidity at the exit, and compare with the previously determined values. The details are left as an exercise.
1 A constant value of the specific heat c_{p a} has been used here to evaluate the term \left(h_{ a 1}-h_{ a 2}\right). As shown in previous examples, this term can be evaluated alternatively using the ideal gas table for air.
Skills Developed
Ability to…
• apply psychrometric terminology and principles.
• apply mass and energy balances for a cooling tower process in a control volume at steady state.
• retrieve property data for dry air and water.
Quick Quiz
Using steam table data, what is the dew point temperature at the exit, in °F? Ans. 59.6°F.
