Cooling Moist Air at Constant Pressure
A 1-lb sample of moist air initially at 70°F, 14.7 lbf / in.^{2}, and 70% relative humidity is cooled to 40°F while keeping the pressure constant. Determine (a) the initial humidity ratio, (b) the dew point temperature, in °F, and (c) the amount of water vapor that condenses, in lb.
Known A 1-lb sample of moist air is cooled at a constant mixture pressure of 14.7 lbf / in.^{2} from 70 to 40°F. The initial relative humidity is 70%.
Find Determine the initial humidity ratio, the dew point temperature, in °F, and the amount of water vapor that condenses, in lb.
Schematic and Given Data:
Engineering Model
1. The 1-lb sample of moist air is taken as the closed system. The system pressure remains constant at 14.7 lbf / in.^{2}
2. The gas phase can be treated as an ideal gas mixture. The Dalton model applies: Each mixture component acts as an ideal gas existing alone in the volume occupied by the gas phase at the mixture temperature.
3. When a liquid water phase is present, the water vapor exists as a saturated vapor at the system temperature. The liquid present is a saturated liquid at the system temperature.
Analysis
a. The initial humidity ratio can be evaluated from Eq. 12.43. This requires the partial pressure of the water vapor, p_{ v 1}, which can be found from the given relative humidity and p_{ g } from Table A-2E at 70°F as follows
\omega=0.622 \frac{p_{ v }}{p-p_{ v }} (12.43)
p_{ v 1}=\phi p_{ g }=(0.7)\left(0.3632 \frac{ lbf }{ in .^{2}}\right)=0.2542 \frac{ lbf }{ in .^{2}}
Inserting values in Eq. 12.43
\omega_{1}=0.622\left(\frac{0.2542}{14.7-0.2542}\right)=0.011 \frac{ lb (\text { vapor })}{ lb ( dry \text { air })}
b. The dew point temperature is the saturation temperature corresponding to the partial pressure, p_{ v 1}. Interpolation in Table A-2E gives T = 60°F. The dew point temperature is labeled on the accompanying property diagram.
c. The amount of condensate, m_{ w }, equals the difference between the initial amount of water vapor in the sample, m_{ v 1}, and the final amount of water vapor, m_{ v2}. That is,
m_{ w }=m_{ v 1}-m_{ v 2}
To evaluate m_{ v 1}, note that the system initially consists of 1 lb of dry air and water vapor, so 1 lb =m_{ a }+m_{ v 1}, where m_{ a } is the mass of dry air present in the sample. Since \omega_{1}=m_{ v 1} / m_{ a }, m_{ a }=m_{ v 1} / \omega_{1}. With this we get
1 lb =\frac{m_{ v 1}}{\omega_{1}}+m_{ v 1}=m_{ v 1}\left(\frac{1}{\omega_{1}}+1\right)
Solving for m_{ v 1}
m_{ v 1}=\frac{1 lb }{\left(1 / \omega_{1}\right)+1}
Inserting the value of \omega_{1} determined in part (a)
m_{ v 1}=\frac{1 lb }{(1 / 0.011)+1}=0.0109 lb (\text { vapor })
1 The mass of dry air present is then m_{ a }=1-0.0109=0.9891 lb (dry air).
Next, let us evaluate m_{ v 2}. With assumption 3, the partial pressure of the water vapor remaining in the system at the final state is the saturation pressure corresponding to 40°F: p_{ g }=0.1217 lbf / in .^{2} Accordingly, the humidity ratio after cooling is found from Eq. 12.43 as
\omega_{2}=0.622\left(\frac{0.1217}{14.7-0.1217}\right)=0.0052 \frac{ lb (\text { vapor })}{ lb (\text { dry air })}
The mass of the water vapor present at the final state is then
m_{ v 2}=\omega_{2} m_{ a }=(0.0052)(0.9891)=0.0051 lb (\text { vapor })
Finally, the amount of water vapor that condenses is
m_{ w }=m_{ v 1}-m_{ v 2}=0.0109-0.0051=0.0058 lb \text { (condensate) }
1 The amount of water vapor present in a typical moist air mixture is considerably less than the amount of dry air present.
Skills Developed
Ability to…
• apply psychrometric terminology and principles.
• demonstrate understanding of the dew point temperature and the formation of liquid condensate when pressure is constant.
• retrieve property data for water.
Quick Quiz
Determine the quality of the two-phase liquid–vapor mixture and the relative humidity of the gas phase at the final state. Ans. 47%, 100%.
