Question

A gas mixture consisting of 10% ethylene gas (C2H4) and 90% CO2 by mol is combusted with 50% excess air via the following unbalanced reaction: C2H4 (g) + O2 (g) à CO2 (g) + H2O (g) Air for the reaction is brought in from outdoors and cooled through a chiller to remove any humidity. The combustion product gas is pressurized so that all water vapor is condensed into liquid, which then combines with the water from the air chiller and exits through a separate H2O stream. a) Draw the process flow diagram. b) If the outdoor air has a dry bulb temperature of 30°C and 30% relative humidity (RH), how many mol of water leave the system (in the combined H2O exit stream) per mol of ethylene combusted? c) What is the molar volume (in L/mol) of the water-free combustion gas if it is pressurized to 150 atm at 20°C? CO2: Tc = 31.2°C, Pc = 72.8atm O2: Tc = -118.6°C, Pc = 49.8atm N2: Tc = -147°C, Pc = 33.5atm

Answer 1

A)

Basis :

1 mole of ethylene

Feed is 10% ethylene and 90% CO2

Total feed = (1/0.10 )= 10 moles

The air enters at 30% R. H and DBT of 30°C

From Perry handbook

At T = 30°C

Saturation pressure = 4.2470 = Pas

R. H = PA/Pas

PA = 0.30(4.2470) = 1.2741 KPa

P = 101.325 KPa

y = PA/P = (1.2741/101.325 )= 0.01257

Mole fraction of water vapor = 0.01257

Mole fraction of oxygen = (1-0.01257) (0.21) = 0.2073

Mole fraction of nitrogen = (1-0.01257) (0.79) = 0.78

The reaction occuring is

According to stiochiometry

Oxygen needed for 1 mole ethylene = 5/2 = 2.5 moles

Oxygen is 50% in excess

Oxygen supplied = 1.5(2.5) = 3.75 moles

Air supplied = 3.75/(0.2073) = 18.089 moles

Amount of water vapor in air = 18.089(0.01257) = 0.2273 moles

All water vapor is condensed.

Component	moles
N2	18.089(0.78) = 14.109
O2	3.75-2.5= 1.25
CO2	1(2) + 9 = 11
H2O	1(2)= 2
Total	28.359

Alm water vapor is condensed in condenser

Water condensed = 2 moles

Total water leaving the system = 2+0.2273= 2.2273 moles/ mole Etylene burned

Dry gas composition

Component	moles	mol%
N2	14.109	53.526
O2	1.25	4.742
CO2	11	41.73
Total	26.359	100

C) To find volume

P = 150 atm = 150×1.013×105 = 1.519×107 Pa

T = 20°C

For CO2

Tc = 31.2°C = 304.35 K

Pc = 72.8 atm= 72.8×1.013×105 Pa

Van der waals equation is given by

For 1 mole the above equation is valid

a = 27R2T2c/(64Pc)

b = RTc/8Pc

For CO2 a = 27(8.314) 2(304.35) 2/(64×7.374×106)

a = 0.3662 Nm4/mol2

b = 8.314(304.35) /(8×7.374×106) =

4.289×10-5 m3/mol

We get

V1 = 5.851×10-4 m3/mol

For oxygen

Tc = -118. 6°C = 154.4 K

Pc = 49.8 atm = 49.8×1.013×105 atm

From above formulae

We get

a = 0.1378 Nm4/mol2

b = 3.1807×10-5 m3/mol

we get

V2 = 1.5090×10-4 m3/mol

For Nitrogen

Tc = -147°C= 126 K

Pc = 33.5 atm = 33.5×1.013×105 Pa

Subsituting all values for a and b we get

a = 0.1364 Nm4/mol2

b = 3.858×10-5 m3/mol

we get

V3 = 1.6308×10-4 m3/mol

Total Volume of combustion gas without water

V = V1x1 + V2x2 + V3x3

V = (5.851×10-4) (0.4173) +(1.5090×10-4) (0.04742) +(1.6308×10-4) (0.53526) =

V = 3.3860×10-4 m3/mol

V = 0.33860 L/mol

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