Question

Air at 30.0% relative humidity is cooled isobarically at 1 atm absolute from 75.0°C to 35.0°C. a. Estimate the dew point and the degrees of superheat of the air at 75.0°C. Dew point: °C Degrees of superheat: °C b. How much water condenses (mol) per cubic meter of feed gas? (See Example 6.3–2) mol c. Suppose a sample of the 75.0°C air is put in a closed variable-volume chamber containing a mirror and the pressure is raised at constant temperature until a mist forms on the mirror. At what pressure (atm) would the mist form? (Assume ideal gas behavior.) atm

Answer 1

Answer:

a. Dew point: 48.7°C. Degrees of superheat 26.3°C

b. $3.33mol/m^3$

c. $P=3.34atm$

Explanation:

a. Based on the psychometric chart of air, the specific volume of air at the given conditions is:

$v =\frac{RT}{PM} =\frac{0.082\frac{atm*L}{mol*K}*348.15K}{1atm*28.97g/mol} *\frac{1m^3}{1000L}*\frac{1000g}{1kg} =0.98544m^3/kg$

The dew point at the specific volume and the 30%-humidity has a value of 48.7°C, it means that there are 75°C-48.7°C=26.3°C of superheat.

b. At 75°C the molar fraction of water is 11580Pa/101625Pa=0.114 moles per cubic meter of feed gas are:

$\frac{n}{V}=y_{H_2O}\frac{P}{RT}=0.114\frac{1atm}{0.082\frac{atm*L}{mol*K}*348.15K} }*\frac{1000L}{1m^3} \\\frac{n}{V}=4mol/m^3$

Once the 35°C are reached, the mole fraction of water is 1688Pa/101325Pa=0.017 and remaining moles per cubic meter of feed gas are:

$\frac{n}{V}=y_{H_2O}\frac{P}{RT}=0.017\frac{1atm}{0.082\frac{atm*L}{mol*K}*308.15K} }*\frac{1000L}{1m^3} \\\frac{n}{V}=0.673mol/m^3$

So the condensed moles per cubic meter of feed gas are:

$4mol/m^3-0.673mol/m^3=3.33mol/m^3$

c. Considering the Raoult's law, one computes the pressure as follows:

$P=\frac{P^{sat}_{H_2O}}{y_{H_2O}}$

At 75°C and 30%-humidity, the saturation water vapor pressure has a value of 38599Pa, thus:

$P=\frac{38599Pa}{0.114}*\frac{1atm}{101325Pa} \\P=3.34atm$

Best regards.