Question

Q3. (7.5 marks) A gas mixture containing 88.0 mol% N2 and 12.0 mol% ethanol vapor, flows through a pipe at a rate of 100 m/h. The pressure is 760 mmHg absolute and the temperature is 50°C. a) What is the partial pressure of ethanol vapor in the gas mixture? b) What is the molar flowrate of the gas mixture in kmol/h c) Is the gas mixture saturated? if not, to what temperature (in °C) would it be cooled at constant pressure in order to begin condensing ethanol d) To what temperature (in°C) would the gas be cooled at constant pressure in order to condense 60% of the ethanol. What is the mole fraction of the ethanol vapor in the gas mixture after condensation? mmHg m3 R = 62.366 kgmol K

Answer 1

A)

Gas mixture contains 88 mol% N₂ and 12 mol% ethanol vapor

Volumetric flowrate (V) = 100 m³/h

P = 760 mmHg

T = 50°C = 323 K

Mole fraction of ethanol = 0.12

Partial pressure of ethanol vapor (Pa) = yP = (0.12) (760) = 91.2 mmHg

B)

n = PV/(RT)

R = 62.366 mmHg m3/(kgmol K)

n = 760(100) /(62.366×323)

n = 3.7727 Kgmol/h

Molar flowrate of gas mixture = 3.7727 kmol/h

C)

Antonie equation

ln(P*) = A -(B/(T+C) )

T - K

P* - mmHg

For ethanol

A - 18.5242

B- 3578.91

C- (-50.5)

P* = e^{(A-(B/(T+C) )}

At T = 50°C = 323 K

P* = e^{(18.5242-(3578.91/(323-50.5) )}

P* = 219.3317 mmHg

Partial pressure = 91.2 mmHg

Partial pressure < saturation pressure

The mixture is not saturated

At dew point saturation pressure = partial pressure

At P* = 91.2 mmHg

We get

91.2 = e^{(18.5242-(3578.91/(T-50.5) )}

T = 305.933 K

In order to condense at constant pressure the temperature to which mixture is cooled = 305.933 K = 32.783°C

D)

Our aim is to condense 60% of ethanol in feed

n = 3.7727 Kmol/h

Nitrogen in feed = 3.7727(0.88) = 3.31997 Kmol/h

Ethanol in feed = 3.7727(0.12) = 0.452724 kmol/h

Ethanol to be condensed (L) = 0.452724(0.60) = 0.271634 Kmol/h

Outlet gas product (P) =

3.31997 +( 0.452724 -0.271634)=

3.50106 kmol/h

mole fraction of ethanol in outlet gas stream = (0.452724-0.271634) /(3.50106) = 0.05172

Partial pressure of ethanol in outlet gas = (0.05172) (760) = 39.31049 mmHg

The liquid and vapor will be in equilibrium so saturated pressure= partial pressure

Saturation pressure = 39.31049 mmHg

39.31049 = e^{(18.5242-(3578.91/(T-50.5) )}

T = 291.460 K = 18.310°C

To condense 60% of ethanol the temperature to which mixture to be cooled = 18.310°C

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