Question

Chemical Process Calculations (Chem Eng Question)

Assume the combustion of Butane (C4H10) in air to be complete.

a) Use a degree-of-freedom analysis to prove that if the precentage excess air and the percentage conversion of butane are specified, the molar composition of the product gas can be determined.

(b) Calculate the molar composition of the product gas for each of the following three cases:

i) theorectical aire supplied, 100% conversion of butane;

ii) 20% excess air, 100% conversion of butane; and

iii) 20% excess air, 90% conversion of butane.

Answer 1

b)

The combustion reaction of butane is given as:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
First we calculate the theoretical air supply. For the combustion of 1 mol butane 6.5 mol O₂ are needed and 4 mol CO_2, 5 mol H₂O are formed respectively. We know that 1 mol of air consist of 0.21 mol O₂ and 0.79 mol N₂

n(N₂) = (0.79 x 6.5)/0.21 = 3.76 x 6.5 = 24.44 mol

number moles of theoretical air supply = 4.76 x 6.5 = 30.94 mol.
1) number of moles of theoretical air (n^th), 100 % conversion
Number of mol of exhaust gas = n(But)+n(O₂)+n(N₂)+n(CO₂)+n(H₂O)
= 0 + 0 + 30.94 mol + 4 mol + 5 mol = 33.4 mol
Exhaust gas composition:
O₂: 0 % ; N₂: 73.2 % (24.44/33.4); CO₂: 11.9 % (4/33.4); H₂O: 14.9 % (5/33.4)

2) n(air) = 1.2x number of moles of theoretical air supply ; 100% conversion
n(air) = 37.12 mol containing 7.79 mol O₂ and 29.32 mol N₂.
After combustion 7.79 mol - 6.5 mol O₂= 1.29 mol O₂ remain in the exhaust gas so
number moles of exhaust gas = n(O₂) + n(N₂) +n(CO₂) +n(H₂O) // n(But)= 0 because of 100 % conversion
number moles of exhaust gas = 1.29 + 29.32 +4 + 5 mol = 39.61 mol
Composition of exhaust gas:
O₂: 3.3 % ; N₂: 74 % ; CO₂: 10.1 % ; H₂O: 12.6 %

3) n(air) = 1.2x number of moles of theoretical air supply; 90 % conversion
n(air) = 37.12 mol containing 29.32 mol N₂ and 7.79 mol O₂
For the combustion of 0.9 mol Butane 5.85 mol O₂ are needed and 3.6 mol CO₂ and 4.5 mol H₂O are produced. In the exhaust gas there is 0.1 mol butane and 1.94 mol O₂. The number of mol of the exhaust gas = n(But)+n(O₂) + n(CO₂)+n(H₂O)
number moles of exhaust gas = 0.1 + 29.32 + 1.94 +3.6 +4.5= 39.46 mol
Composition: But: 0.25 % ;O₂: 4.91 %; N₂: 74.3 % ; CO₂: 9.12 % ; H₂O: 11.4 %