6.14. Air at 50% relative humidity is cooled isobarically at 1 atm absolute from

Question

6.14. Air at 50% relative humidity is cooled isobarically at 1 atm absolute from 90°C to 25°C. (a) Estimate the dew point and degrees of superheat of the air at 90°c. (b) How much water condenses (mol) per cubic meter of feed gas? (See Example 6.3-2.) (c) Suppose a sample of the 90°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.)

Explanation / Answer

Given

Relative humidity

Hr = 50%

Part a

Hr = partial pressure of H2O x 100 / vapor pressure of H2O

vapor pressure of H2O at 90 C = 525.80 mmHg

Partial pressure of H2O = 50 x 525.80 / 100 = 262.90 mmHg

Dew point temperture = 72.70 °C

Degrees of superheat = 90 - 72.70 = 17.30 °C

Part b

At 90 °C

Mol fraction of H2O = Partial pressure of H2O / total pressure

= 262.90 mmHg / (1 atm x 760 mmHg /atm)

= 0.3459

At 25 °C (saturated conditions)

Partial pressure of H2O = vapor pressure of H2O = 23.7560 mmHg

Mol fraction of H2O = Partial pressure of H2O / total pressure

= 23.7560 mmHg / (1 atm x 760 mmHg /atm)

= 0.03126

Total Moles at 90 °C = PV/RT

= [1 atm x 101325 Pa/atm x 1 m3 ]/[ 8.314 J/mol·K x (90+273)]

= 33.574 mol

Moles of H2O at 90 C = 0.3459 x 33.574 = 11.613 mol

Moles of H2O at 25 C = 0.03126 x 33.574 = 1.0495 mol

Moles of H2O condensed = 11.613 - 1.0495 = 10.563 mol

Part C

From Raoult’s law at 90 C

Mol fraction of H2O x total pressure = partial pressure of H2O

total pressure = vapor pressure of H2O / Mol fraction of H2O

= 525.80 mmHg / 0.3459

= 1520.09 mmHg x 1atm/760mmHg

= 2 atm

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