In a combustion process a feed stream containing mainly propane, with impurities

Question

In a combustion process a feed stream containing mainly propane, with impurities being ethane and butane, is burned at the rate of 100 kmol per day with 65% excess air to produce heat. Assuming that the fuel gas mixture is completely burned and no carbon monoxide is formed, calculate: The molar composition of the fuel feed stream. The molar flow rate of air. The composition of the flue gas. Composition of air may be taken as 21% O_2 and 79% N_2 by volume. Relative Molar Masses: C=12, H=1, O=16 and N=14 The composition of the fuel gas is given below: Component Mass% Ethane 2 propane 92 Butane 6

Explanation / Answer

(a) Molar composition of the fuel feed stream

with 100 kmol per day of fuel combustion

we would have,

For a 100 mass % fuel

mol of ethane in mixture = 2/30.07 = 0.066 mols

mol of Propane in mixture = 92/44.1 = 2.086 mols

mol of butane in mixture = 6/58.12 = 0.103 mols

Total mols = 0.066 + 2.086 + 0.103 = 2.255 mols

mole fraction of ethane = 0.066/2.255 = 0.029

mole fraction of propane = 2.086/2.255 = 0.925

mole fraction of butane = 0.103/2.255 = 0.103/2.255 = 0.046

Thus, molar composition of fuel = ethane : propane : butane = 2.9 kmol : 92.5 kmol : 4.6 kmol

(b) molar flow rate of air

2 C2H6 + 7 O2 + 25.97 N2 = 4 CO2 + 6 H2O + 25.97 N2

air requirement = 32.97 + 21.43 = 54.40

For 2.9 kmol = 2.9 x 54.40/2 = 78.88 kmol

C3H8 + 5 O2 + 18.55 N2 = 3 CO2 + 4 H2O + 18.55 N2

air requirement = 23.55 + 15.31 = 38.86

For 92.5 kmol = 92.5 x 23.55 = 2178.375 kmol

2 C4H10 + 13 O2 + 48.23 N2 = 8 CO2 + 10 H2O + 48.23 N2

air requirement = 61.23 + 39.80 = 101.03

For 4.6 kmol = 4.6 x 101.03/2 = 232.37 kmol

Total molar flow rate of air required = 78.88 + 2178.375 + 232.37 = 2489.625 kmol per day

(c)The composition of flue gas

2 C2H6 + 7 O2 + 25.97 N2 = 4 CO2 + 6 H2O + 25.97 N2

air requirement = 32.97 + 21.43 = 54.40

For 2.9 kmol = 2.9 x 54.40/2 = 78.88 kmol

C3H8 + 5 O2 + 18.55 N2 = 3 CO2 + 4 H2O + 18.55 N2

air requirement = 23.55 + 15.31 = 38.86

For 92.5 kmol = 92.5 x 23.55 = 2178.375 kmol

2 C4H10 + 13 O2 + 48.23 N2 = 8 CO2 + 10 H2O + 48.23 N2

air requirement = 61.23 + 39.80 = 101.03

For 4.6 kmol = 4.6 x 101.03/2 = 232.37 kmol

CO2 = 5.8 + 277.5 + 18.4 = 301.7 kmol

H2O = 8.7 + 370 + 23 = 401.7 kmol

O2 = 16.07 kmol

N2 = 60.465 (from excess air) + 37.656 + 857.94 + 69.93 = 1026 kmol

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