Known Moist air enters a dehumidifier at 30°C and 50% relative humidity with a volumetric flow rate of 280 m ^{3} / min. Condensate and moist air exit in separate streams at 10°C.
Determine Find the mass flow rate of the dry air, in kg/min, the rate at which water is condensed, in kg per kg of dry air, and the required refrigerating capacity, in tons.
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 significant heat transfer to the surroundings.
3. The pressure remains constant throughout at 1.013 bar.
4. At location 2, the moist air is saturated. The condensate exits at location 3 as a saturated liquid at temperature T_{2}.
5. The moist air streams are regarded as ideal gas mixtures adhering to the Dalton model.
Analysis
a. At steady state, the mass flow rates of the dry air entering and exiting are equal. The common mass flow rate of the dry air can be determined from the volumetric flow rate at the inlet
\dot{m}_{ a }=\frac{( AV )_{1}}{v_{ a 1}}
The specific volume of the dry air at inlet 1, v _{ a 1}, can be evaluated using the ideal gas equation of state, so
\dot{m}_{ a }=\frac{( AV )_{1}}{\left(\bar{R} / M_{ a }\right)\left(T_{1} / p_{ a 1}\right)}
The partial pressure of the dry air p_{ a 1} can be determined from p_{ a 1}=p_{1}-p_{ v 1}. Using the relative humidity at the inlet \phi_{1} and the saturation pressure at 30°C from Table A-2
p_{ v 1}=\phi_{1} p_{ g 1}=(05)(0.04246)=0.02123 bar
Thus, p_{ a 1}=1.013-0.02123=0.99177 bar. Inserting values into the expression for \dot{m}_{ a } gives
\dot{m}_{ a }=\frac{\left(280 m ^{3} / min \right)\left(0.99177 \times 10^{5} N / m ^{2}\right)}{(8314 / 28.97 N \cdot m / kg \cdot K )(303 K )}=319.35 kg / min
b. Conservation of mass for the water requires \dot{m}_{ v 1}=\dot{m}_{ v 2}+\dot{m}_{ w }. With \dot{m}_{ v 1}=\omega_{1} \dot{m}_{ a } \text { and } \dot{m}_{ v 2}=\omega_{2} \dot{m}_{ a }, the rate at which water is condensed per unit mass of dry air is
\frac{\dot{m}_{ w }}{\dot{m}_{ a }}=\omega_{1}-\omega_{2}
The humidity ratios \omega_{1} \text { and } \omega_{2} can be evaluated using Eq. 12.43. Thus, \omega_{1} is
\omega=0.622 \frac{p_{ v }}{p-p_{ v }} (12.43)
\omega_{1}=0.662\left(\frac{p_{ v 1}}{p_{1}-p_{ v 1}}\right)=0.622\left(\frac{0.02123}{0.99177}\right)=0.0133 \frac{ kg (\text { vapor })}{ kg (\text { dry air })}
Since the moist air is saturated at 10°C, p_{ v 2} equals the saturation pressure at 10°C: p_{ g }=0.01228 bar from Table A-2. Equation 12.43 then gives \omega_{2}=0.0076 kg(vapor)/kg(dry air). With these values for \omega_{1} \text { and } \omega_{2}
\frac{\dot{m}_{ w }}{\dot{m}_{ a }}=0.0133-0.0076=0.0057 \frac{ kg (\text { condensate })}{ kg (\text { dry air })}
c. The rate of heat transfer \dot{Q}_{ cv } between the moist air stream and the refrigerant coil can be determined using an energy rate balance. With assumptions 1 and 2, the steady-state form of the energy rate balance reduces to
0=\dot{Q}_{ cv }+\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) (a)
With \dot{m}_{ v 1}=\omega_{1} \dot{m}_{ a }, \dot{m}_{ v 2}=\omega_{2} \dot{m}_{ a }, \text { and } \dot{m}_{ w }=\left(\omega_{1}-\omega_{2}\right) \dot{m}_{ a }, this becomes
\dot{Q}_{ cv }=\dot{m}_{ a }\left[\left(h_{ a 2}-h_{ a 1}\right)-\omega_{1} h_{ g 1}+\omega_{2} h_{ g 2}+\left(\omega_{1}-\omega_{2}\right) h_{ f 2}\right] (b)
which agrees with Eq. 12.55. In Eq. (b), the specific enthalpies of the water vapor at 1 and 2 are evaluated at the saturated vapor values corresponding to T_{1} \text { and } T_{2}, respectively, and the specific enthalpy of the exiting condensate is evaluated as h_{ f } \text { at } T_{2}. Selecting enthalpies from Tables A-2 and A-22, as appropriate, Eq. (b) reads
0=\dot{Q}_{ cv }+\dot{m}_{ a }\left[\underline{\left(h_{ a 1}-h_{ a 2}\right)}+\underline{\left.\omega_{1} h_{ g 1}+\left(\omega_{2}-\omega_{1}\right) h_{ w }-\omega_{2} h_{ g 2}\right.}\right] (12.55)
\begin{aligned}\dot{Q}_{ cv }=&(319.35)[(283.1-303.2)-0.0133(2556.3)\\&+0.0076(2519.8)+0.0057(42.01)]\end{aligned}
= -11,084 kJ/min
Since 1 ton of refrigeration equals a heat transfer rate of 211 kJ/min (Sec. 10.2.1), the required refrigerating capacity is 52.5 tons.
Alternative Psychrometric Chart Solution Let us consider an alternative solution using the psychrometric chart. As shown on the sketch of the psychrometric chart, Fig. E12.11b, the state of the moist air at the inlet 1 is defined by \phi=50 \% and a dry-bulb temperature of 30°C. At 2, the moist air is saturated at 10°C. Rearranging Eq. (a), we get
\dot{Q}_{ cv }=\dot{m}_{ a }\left[\underline{\left(h_{ a }+\omega h _{ v }\right)_{2}}-\underline{\left(h_{ a }+\omega h_{ v }\right)_{1}}+\left(\omega_{1}-\omega_{2}\right) h_{ w }\right]
The underlined terms and humidity ratios, \omega_{1} \text { and } \omega_{2}, can be read directly from the chart. The mass flow rate of the dry air can be determined using the volumetric flow rate at the inlet and v_{ a 1} read from the chart. The specific enthalpy h_{ w } is obtained (as above) from Table A-2: h_{ f } \text { at } T_{2}. The details are left as an exercise.
Skills Developed
Ability to…
• apply psychrometric terminology and principles.
• apply mass and energy balances for a dehumidification process in a control volume at steady state.
• retrieve property data for dry air and water.
• apply the psychrometric chart.
Quick Quiz
Using the psychrometric chart, determine the wet-bulb temperature of the moist air entering the dehumidifier, in °C. Ans. ≈22°C.
