Question

Q11 The composition of the fuel gas to the cracker on the ethylene plant is:

Hydrogen, H2: 20 vol%
Methane, CH4: 50 vol%
Carbon monoxide, CO: 15 vol%
Carbon dioxide, CO2: 10 vol%
Ethane, C2H6: 5 vol%

a) Calculate the molecular weight of the fuel gas. (5 marks)
b) Calculate the composition of the fuel gas in mass%. (10 marks) (15 marks)

Answer 1

a) If we assume fuel gas is treated as ifeal gas then vol% = mol%

1- Hydrogen, H2 = 20 mol%

2- Methane, CH4 = 50 mol%

3 - Carbon monoxide, CO = 15mol%

4 - Carbon dioxide CO2 = 10mol%

5 - Ethane C2H6 = 5mol%

If we take mole fraction of given composition

Components    Mole fraction   Molecular weight (gm/mol)

H2,    y1 = 0.2 M1= 2

CH4, y2 = 0.5 M2 = 16

CO, y3 = 0.15 M3 = 28

CO2, y4 = 0.10 M4 = 44

C2H6, y5 = 0.05. M5 = 30

Average molecular weight :

M = y1M1 + y2M2 + y3M3 + y4M4 + y5M5

M = 0.2*2 + 0.5*16 + 0.15*28+ 0.1*44 + 0.05* 30

M = 18.5 gm/mol

b) basis of fuel gas = 1mol

Then moles of components

H2 = 0.2 mole

CH4 = 0.5 mole

CO = 0.15 mole

CO2 = 0.1 mole

C2H6 = 0.05 mole

We know ,

mass m = mole * molecular weight

For H2, m1 = 0.2*2 = 0.4 gm

For Ch4, m2 = 0.5*16 = 8 gm

For CO, m3 = 0.15*28 = 4.2 gm

For CO2, m4 = 0.10*44 = 4.4 gm

For C2H6, m5 = 0.05*30 = 1.5 gm

mt = m1 + m2 +m3 + m4 + m5 = 18.5 gm

Mass fraction of components = mass of components /total mass of gas fuel

Mass fraction of hydrogen x1 = m1/ mt = 0.4/18.5 = 0.0216

Mass fraction of CH4, x2 = m2/mt = 8/18.5 = 0.432

Mass fraction of CO x3 = m3/mt = 4.2/18.5 = 0.227

Mass fraction of CO2, x4 = m4/mt = 4.4/18.5 = 0.237

Mass fraction of C2H6, x5 = m5/mt = 1.5/18.5 = 0.081

Fuel composition by mass %

H2 = 0.0216*100 = 2.16%

CH4 = 0.432*100 = 43.2 %

CO = 0.227*100 = 22.7 %

CO2 = 0.237*100= 23.7%

C2H6 = 0.081*100 = 8.1%