Cooling Moist Air at Constant Volume
An air–water vapor mixture is contained in a rigid, closed vessel with a volume of 35 m ^{3} at 1.5 bar, 120°C, and \phi=10 \%. The mixture is cooled at constant volume until its temperature is reduced to 22°C. Determine (a) the dew point temperature corresponding to the initial state, in °C, (b) the temperature at which condensation actually begins, in °C, and (c) the amount of water condensed, in kg.
Known A rigid, closed tank with a volume of 35 m ^{3} containing moist air initially at 1.5 bar, 120°C, and \phi=10 \% is cooled to 22°C.
Find Determine the dew point temperature at the initial state, in °C, the temperature at which condensation actually begins, in °C, and the amount of water condensed, in kg.
Schematic and Given Data:
Engineering Model
1. The contents of the tank are taken as a closed system. The system volume remains constant.
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 is a saturated liquid at the system temperature.
Analysis
a. The dew point temperature at the initial state is the saturation temperature corresponding to the partial pressure p_{ v 1}.
With the given relative humidity and the saturation pressure at 120°C from Table A-2
p_{ v 1}=\phi_{1} p_{ g 1}=(0.10)(1.985)=0.1985 bar
Interpolating in Table A-2 gives the dew point temperature as 60°C, which is the temperature condensation would begin if the moist air were cooled at constant pressure.
b. Whether the water exists as a vapor only, or as liquid and vapor, it occupies the full volume, which remains constant. Accordingly, since the total mass of the water present is also constant, the water undergoes the constant specific volume process illustrated on the accompanying T–υ diagram. In the process from state 1 to state 1′, the water exists as a vapor only. For the process from state 1′ to state 2, the water exists as a two-phase liquid–vapor mixture. Note that pressure does not remain constant during the cooling process from state 1 to state 2.
State 1′ on the T–υ diagram denotes the state where the water vapor first becomes saturated. The saturation temperature at this state is denoted as T′. Cooling to a temperature less than T′ results in condensation of some of the water vapor present. Since state 1′ is a saturated vapor state, the temperature T′ can be found by interpolating in Table A-2 with the specific volume of the water at this state. The specific volume of the vapor at state 1′ equals the specific volume of the vapor at state 1, which can be evaluated from the ideal gas equation
v_{ v 1}=\frac{\left(\bar{R} / M_{ v }\right) T_{1}}{p_{ v 1}}=\left(\frac{8314}{18} \frac{ N \cdot m }{ kg \cdot K }\right)\left(\frac{393 K }{0.1985 \times 10^{5} N / m ^{2}}\right)
=9.145 \frac{ m ^{3}}{ kg }
1 Interpolation in Table A-2 with v_{ v 1}=v_{ g } \text { gives } T^{\prime}=56^{\circ} C. This is the temperature at which condensation begins.
c. The amount of condensate equals the difference between the initial and final amounts of water vapor present. The mass of the water vapor present initially is
m_{ v 1}=\frac{V}{v_{ v 1}}=\frac{35 m ^{3}}{9.145 m ^{3} / kg }=3.827 kg
The mass of water vapor present finally can be determined from the quality. At the final state, the water forms a two-phase liquid–vapor mixture having a specific volume of 9.145 m ^{3} / kg. Using this specific volume value, the quality x_{2} of the liquid–vapor mixture can be found as
x_{2}=\frac{v_{ v 2}-v_{ f 2}}{v_{ g 2}-v_{ f 2}}=\frac{9.145-1.0022 \times 10^{-3}}{51.447-1.0022 \times 10^{-3}}=0.178
where v_{ f 2} \text { and } v_{ g 2} are the saturated liquid and saturated vapor specific volumes at T_{2}=22^{\circ} C, respectively.
Using the quality together with the known total amount of water present, 3.827 kg, the mass of the water vapor contained in the system at the final state is
m_{ v 2}=(0.178)(3.827)=0.681 kg
The mass of the condensate, m_{ W 2}, is then
m_{ w 2}=m_{ v 1}-m_{ v 2}=3.827-0.681=3.146 kg
1 When a moist air mixture is cooled at constant mixture volume, the temperature at which condensation begins is not the dew point temperature corresponding to the initial state. In this case, condensation begins at 56°C, but the dew point temperature at the initial state, determined in part (a), is 60°C.
Skills Developed
Ability to…
• apply psychrometric terminology and principles.
• demonstrate understanding of the onset of condensation when cooling moist air at constant volume.
• retrieve property data for water.
Quick Quiz
Determine the humidity ratio at the initial state and the amount of dry air present, in kg. Ans. 0.0949, 40.389 kg.
