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) The process flow diagram is shown below

Basis

1 mole ethylene

10% of ethylene and 90% CO2

Feed = 1/0.1 = 10 mol

The reaction

The dry bulb temperature = 30°C

R. H = 30%

From perry handbook

Saturation pressure at T = 30°C = 4.2470 KPa

R. H = PA/Pas

PA = 0.30(4.2470) = 1.2741 KPa

P = 101.325 KPa

y = 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) = 0.780

From stiochiometry

Amount of oxygen required for 1 mole ethylene = (5/2) = 2.5 moles

Excess oxygen = 50%

Oxygen supplied = 1.5(2.5) = 3.75 moles

Air supplied = (3.75/0.2073) = 18.0897 moles

Amount of water vapor in air that condensed = 18.0897(0.01257) = 0.22738 moles

Exit gas analysis

Component	moles
N2	18.0897(0.78) =14.109
O2	3.75-2.5 = 1.25
CO2	2(1) +9 = 11
H2O	2
Total	28.359

All water in exit gas is condensed

Water condensed = 2 moles

Water stream leaving the system

2+0.22738 = 2.22738 moles

Dry gas analysis

Component	moles	mol%
N2	14.109	53.526
O2	1.25	4.74
CO2	11	41.73
Total	26.359	100

P2 = 150 atm = 150(1.013×105) = 1.5195×107 Pa

T2 = 20°C = 293 K

Van der waals equation

For nitrogen

Tc = -147°C = 126 K

Pc = 33.5 atm = 33.5(1.013×105) = 3.393×106 Pa

a = 27(8.314) 2(126) 2/(64×3.393×106)

a = 0.1364 Nm4/mol2

b = (8.314×126) /(8×3.393×106)

b = 3.8592×10-5

Subsituting in van der waal equation

Solving we get

V 1= 1.5555×10-4 m3/mol

For oxygen

Tc = -118. 6°C = 154.4 K

Pc = 49.8 atm

Solving equations for a and b from previous part we get

a = 0.1378 Nm4/mol2

b = 3.1807×10-5 m3/mol

Solving we get

V2 = 1.4274×10-4 m3/mol

For CO2

Tc = 31.2°C = 304.2 K

Pc = 72.8 atm

Solving for a and b we get

a = 0.3659 Nm4/mol2

b = 4.2868×10-5 m3/mol

Solving we get V3 = 6.992×10-5 m3/mol

Total volume of combustion gas without water

V = (0.53526) (V1) +(0.0474) (V2) +(0.4173) (V3)

V = (0.53526) (1.555×10-4) +(0.0474) (1.4274×10-4) +(0.4173) (6.992×10-5)

V = 1.1917×10-4 m3/mol

V = 0.1191 L/mol

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