Chemical Process Calculations (Chem Eng Question) Assume the combustion of Butan

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.

Explanation / Answer

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 CO2, 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 (nth), 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 %

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