Question

A fuel gas containing 77% methane, 7% ethane and rest propane by volume. The fuel flows to a combustion chamber at the rate of 1321 m^3/h at 15C and 150 kPA (gauge) where it is burned with 8% excess air entering at 20C. ROund your answer to the nearest whole number.

a) WHat is the propane percent in the fuel gas?

b) What is the actual number of moles of fuel flowing into the combustion chamber?

c) What is the actual number of moles of oxygen flow into the combustion chamber?

d) What is the actual number of moles of air flowing into the combustion chamber?

e) What is the ACMH?

Answer 1

(a): The percent of propane in the fuel = 100% - 77% - 7% = 16 % propane by volume (answer)

(b): Given the volume of fuel flowing per hour, V = 1321 m3/h

T = 15 DegC = 15+273 = 288 K

P = 150 KPa = 150000 Pa

Now applying ideal gas equation, PV = nRT

=> n = PV / RT = 150000 Pa x 1321 m3 / (8.314 JK-1mol-1 x 288 K) = 82755 mol (answer)

(c): Moles of CH4 in the fuel gas = 82754 mol x (77/100) = 63721 mol

Moles of C2H6 in the fuel gas = 82754 mol x (7/100) = 5793 mol

Moles of C3H8 in the fuel gas = 82754 mol x (16/100) = 13241 mol

Balanced chemical reaction for the combustion of CH4 is

CH4 + 2 O2 ---- > CO2 + 2 H2O

1 mol, 2 mol, ----- 1 mol, 2 mol

1 mole of CH4 reacts with 2 mol of O2.

Hence  63721 mol CH4 that will react with the stoichiometric moles of O2

=  63721 mol CH4 x (2 mol O2 / 1 mol CH4) = 127442 mol O2

Balanced chemical reaction for the combustion of C2H6 is

C2H6 + 7/2 O2 ---- > 2 CO2 + 3 H2O

1 mol, 7/2(=3.5) mol, 2 mol, 3 mol

1 mole of C2H6 reacts with 7/2(=3.5) mol of O2.

Hence 5793 mol C2H6 that will react with the stoichiometric moles of O2

= 5793 mol C2H6 x (3.5 mol O2 / 1 mol C2H6) = 20275 mol O2

Balanced chemical reaction for the combustion of C3H8 is

C3H8 + 5 O2 ---- > 3 CO2 + 4 H2O

1 mol, 5 mol, ----- 3 mol, 4 mol

1 mole of C3H8 reacts with 5 mol of O2.

Hence 13241 mol C3H8 that will react with the stoichiometric moles of O2

= 13241 mol C3H8 x (5 mol O2 / 1 mol CH4) = 66205 mol O2

Hence total stoichiometric moles of O2 required for the complete combustion

= 127442 mol O2 + 20275 mol O2 + 66205 mol O2

= 213922 mol O2

However we have taken 8% excess air

Hence actual number of moles of oxygen flow into the combustion chamber

= 213922 mol O2 x 108 / 100 = 231036 mol O2 (answer)

(d) Air contains around 21 % O2

Hence actual number of moles of air flowing into the combustion chamber

= 231036 mol O2 x (100 % air / 21 % O2) = 1100171 mol air (answer)

(e) Actual moles of air entering = 1100171 mol air

T = 20 DegC = 293 K

P = 1 atm = 1.013 x 10⁵ Pa

Applying ideal gas equation, PV = nRT

=> V = nRT / P = 1100171 mol x 8.314 Jmol-1K-1 x 293 K / 1.013 x 10⁵ Pa = 26456 m³ / hr (answer)