A stream of vapor - free air is saturated with acetone vapor at a total flow rat

Question

A stream of vapor - free air is saturated with acetone vapor at a total flow rate of 1 m 3/s and a pressure of 10 psig. The temperature of air (plus solvent) is 60 C and the air has 45% relative saturation. Assume that the air stream follows ideal gas law and answer to the following problems, knowing that acetone is the only condensable component:

a) calculate the molar flow rate of dry air, oxygen and acetone in the feed

b) calculate molal saturation, absolute saturation and percent saturation in the air

c) calculate dew temperature

d) if the air stream is cooled to room temperature (25 C), what fraction of theacetone in the feed is condensed

Explanation / Answer

a) Assuming that the air streams follow ideal gas law

i. Molar flow rate of acetone = n = Pv/RT

P-10 psi, v -1 m3/s, R - 0.00120591 psi-m3/gmol K, T -60C

n - 2.4 gmol/s

mass of acetone - 139.2 g

ii. molar flow rate of air

relative sat = 45% = (vap pr / sat vap pr ) * 100

sat vap pr at 60 C -2.69 psig

vap pr - 1.21 psig

n = Pv/RT

= 3.01 gmol/s

iii. o2 is 21% of air

n o2 - .21*3.01 = 0.6327 gmol/s

B) molar sat = n of acetone / n of dry air = 2.4/3.1 = 0.77

absolute sat = mass of acetone/ mass of air = 1.54

c) dew point = (241.88 × ln(vp/610.78))/(17.558 - ln(vp/610.78)

= 44 0C

d) 129 gmol of acetone present in excess in air will condense at 25C

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