Known A liquid water stream enters a cooling tower from a condenser at 38°C with a known mass flow rate. A stream of cooled water is returned to the condenser at 30°C and the same flow rate. Makeup water is added at 20°C. Atmospheric air enters the tower at 25°C and \phi=35 \%. Moist air exits the tower at 35°C and \phi=90 \%.
Find Determine the mass flow rates of the dry air and the makeup water, in kg/h.
Schematic and Given Data:
Engineering Model
1. The control volume shown in the accompanying figure operates at steady state. Heat transfer with the surroundings can be neglected, as can changes in kinetic and potential energy; also \dot{W}_{ cv }=0.
2. To evaluate specific enthalpies, each liquid stream is regarded as a saturated liquid at the corresponding specified temperature.
3. The moist air streams are regarded as ideal gas mixtures adhering to the Dalton model.
4. The pressure is constant throughout at 1 atm.
Analysis The required mass flow rates can be found from mass and energy rate balances. Mass balances for the dry air and water individually reduce at steady state to
\dot{m}_{ a 3}=\dot{m}_{ a 4} (dry air)
\dot{m}_{1}+\dot{m}_{5}+\dot{m}_{ v 3}=\dot{m}_{2}+\dot{m}_{ v 4} (water)
The common mass flow rate of the dry air is denoted as \dot{m}_{ a }. Since \dot{m}_{1}=\dot{m}_{2}, the second of these equations becomes
\dot{m}_{5}=\dot{m}_{ v 4}-\dot{m}_{ v 3}
With \dot{m}_{ v 3}=\omega_{3} \dot{m}_{ a } \text { and } \dot{m}_{ v 4}=\omega_{4} \dot{m}_{ a }
\dot{m}_{5}=\dot{m}_{ a }\left(\omega_{4}-\omega_{3}\right)
Accordingly, the two required mass flow rates, \dot{m}_{ a } \text { and } \dot{m}_{5}, are related by this equation. Another equation relating the flow rates is provided by the energy rate balance.
Reducing the energy rate balance with assumption 1 results in
0=\dot{m}_{1} h_{ w 1}+\left(\dot{m}_{ a } h_{ a 3}+\dot{m}_{ v 3} h_{ v 3}\right)+\dot{m}_{5} h_{ w 5}-\dot{m}_{2} h_{ w 2}-\left(\dot{m}_{ a } h_{ a 4}+\dot{m}_{ v 4} h_{ v 4}\right)
Evaluating the enthalpies of the water vapor as the saturated vapor values at the respective temperatures and the enthalpy of each liquid stream as the saturated liquid enthalpy at each respective temperature, the energy rate equation becomes
0=\dot{m}_{1} h_{ f 1}+\left(\dot{m}_{ a } h_{ a 3}+\dot{m}_{ v 3} h_{ g 3}\right)+\dot{m}_{5} h_{ f 5}-\dot{m}_{2} h_{ f 2}-\left(\dot{m}_{ a } h_{ a 4}+\dot{m}_{ v 4} h_{ g 4}\right)
Introducing \dot{m}_{1}=\dot{m}_{2}, \dot{m}_{5}=\dot{m}_{ a }\left(\omega_{4}-\omega_{3}\right), \dot{m}_{ v 3}=\omega_{3} \dot{m}_{ a }, \text { and } \dot{m}_{ v 4}=\omega_{4} \dot{m}_{ a } \text { and solving for } \dot{m}_{ a }
\dot{m}_{ a }=\frac{\dot{m}_{1}\left(h_{ f 1}-h_{ f 2}\right)}{h_{ a 4}-h_{ a 3}+\omega_{4} h_{ g 4}-\omega_{3} h_{ g 3}-\left(\omega_{4}-\omega_{3}\right) h_{ f 5}} (a)
The humidity ratios \omega_{3} \text { and } \omega_{4} required by this expression can be determined from Eq. 12.43, using the partial pressure of the water vapor obtained with the respective relative humidity. Thus, \omega_{3}=0.00688 kg(vapor)/kg(dry air) and \omega_{4}=0.0327 kg(vapor)/
kg(dry air).
\omega=0.622 \frac{p_{ v }}{p-p_{ v }} (12.43)
With enthalpies from Tables A-2 and A-22, as appropriate, and the known values for \omega_{3}, \omega_{4}, \text { and } \dot{m}_{1}, the expression for \dot{m}_{ a } becomes
\dot{m}_{ a }=\frac{\left(4.5 \times 10^{7}\right)(159.21-125.79)}{(308.2-298.2)+(0.0327)(2565.3)}
-(0.00688)(2547.2)-(0.0258)(83.96)
=2.03 \times 10^{7} kg / h
Finally, inserting known values into the expression for \dot{m}_{5} results in
\dot{m}_{5}=\left(2.03 \times 10^{7}\right)(0.0327-0.00688)=5.24 \times 10^{5} kg / h
Alternative Psychrometric Chart Solution Equation (a) can be rearranged to read
\dot{m}=\frac{\dot{m}_{1}\left(h_{ f 1}-h_{ f 2}\right)}{\underline{\left(h_{ a 4}+\omega_{4} h_{ g 4}\right)}-\underline{\left(h_{ a 3}+\omega_{3} h_{ g 3}\right)}-\left(\omega_{4}-\omega_{3}\right) h_{ f 5}}
The specific enthalpy terms h_{ f 1}, h_{ f 2}, \text { and } h_{ f 5} are obtained from Table A-2, as above. The underlined terms and \omega_{3} \text { and } \omega_{4} can be obtained by inspection of a psychrometric chart from the engineering literature providing data at states 3 and 4. Figure A-9 does not suffice in this application at state 4. The details are left as an exercise.
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, determine the partial pressure of the water vapor in the entering moist air stream, p_{ v 3}, in bar. Ans. 0.0111 bar.
