Question

A closed container of air and water vapor is at 95 degrees Celsius and 5000 mm...

A closed container of air and water vapor is at 95 degrees Celsius and 5000 mm Hg and contains 12% water by volume. Antoine constants for water are A=7.97, B=1668.2, C=228.0. a) Calculate the dew point temperature and degrees of superheat of the air-water mixture. b) If the air-water mixture is cooled isobarically to a point where the mole fraction of air is .915, what is the temperature and percentage of the water vapor that condenses?

Homework Answers

Answer #1


a) Calculate the dew point temperature and degrees of superheat of the air-water mixture.
A=7.97, B=1668.2, C=228.0
Log(pv) = A-B/(T+C), T temperature in degC and pv = vapor pressure of pure substance in mmHg.
mole fraction of water vapor, yH2O = 0.12
Total pressure, P = 5000 mmHg
partial pressure of water vapor, pH2O = yH2O*P = 0.12*5000 = 600 mmHg

At temperature, T=Tdp(dew point) pv = pH2O = 600mmHg

Log(pH2O) =Log(600) = 2.78 = A-B/(Tdp+C) = 7.97-1668.2/(Tdp+228)
2.78 = 7.97-1668.2/(Tdp+228)
7.97-2.78 = 5.19 = 1668.2/(Tdp+228)
(Tdp+228) = 321.4
Dew point temperature, Tdp = 93.4 degC
Temperature, T = 95 degC
Degree of superheat = T - Tdp = 95-93.4 = 1.6 degC


b) If the air-water mixture is cooled isobarically to a point where the mole fraction of air is .915,
what is the temperature and percentage of the water vapor that condenses?

mole fraction of water vapor, yH2O = 1-0.915 = 0.085
partial pressure of water vapor, pH2O = yH2O*P = 0.085*5000 = 425 mmHg

Log(pH2O) =Log(425) = 2.6 = A-B/(Tdp+C) = 7.97-1668.2/(Tdp+228)
2.6 = 7.97-1668.2/(Tdp+228)
7.97-2.6 = 5.37 = 1668.2/(Tdp+228)
(Tdp+228) = 310.65
Dew point temperature, Tdp = 82.65 degC
There is equilibrium between water vapor and condensed liquid.
So, temperature, T = 82.65 degC

Basis: Let total 100 moles of molecules initially present in the container

Balance dry air:

Initially present:
mole fraction of water vapor, yH2O = 0.12
moles of water vapor=100*0.12= 12 mol
Moles of air, n1 = 100 - 12 = 88 mol

Finally present after isobaric compression:
Total moles of molecules = n2
mole fraction of air, yair = 0.915
Moles of air, n1 = n2*0.915
88 = n2*0.915
Total moles of molecules present after compression, n2 = 96.2
Moles of water vapor =n2*(1-0.915)=96.2*0.085=8.2
Moles of molecules condensed = 12-8.2 = 3.8 mol

percentage condensed = (moles H2O condensed)/moles H2O present initially*100 = 3.8/12*100 = 31.67%

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