A stream of Fuel Gas containing Ethane (C_2 H_6), Propane (C_3 H_8) and Butane (

Question

A stream of Fuel Gas containing Ethane (C_2 H_6), Propane (C_3 H_8) and Butane (C_4 H_10) is fed into a combustion chamber where it is totally combusted with the stoichiometric amount of atmospheric air consisting of Oxygen, Nitrogen and I mol %% of Carbon Dioxide. (a) Sketch the process and clearly indicating all the known and relevant unknown parameters. (b) Confirm that the molar flow rate of Air entering the combustion chamber is 3628.03 mol s^-1. (c) Determine the mole percentages of Oxygen and Nitrogen in the combustion air. (d) The Butane flow in the Fuel Gas entering the combustion chamber is 3.524 kg s^-1. Determine i) The molar flow rale of Fuel Gas being fed to the combustion chamber ii) The composition of the Fuel Gas expressing your answer as mole percentages. Data: Relative molar masses(g mol^-1) H = 1 C = 12 N = 14 O = 16

Explanation / Answer

a)Given

Mole fraction of Co2= 0.1213

Mole fraction of N2= 0.7321

Mole fraction of H2O=0.1466

Number moles of product flue gas = 3.915 k mol

Mole of co2= 0.1213*3.915=0.4748

Mole of N2= 0.7321*3.915=2.866

Mole of H2O=0.1466*3.915=0.564

Reaction be

XC2H6+YC3H8+ZC4H10+ A*(O2+3.76N2)-------->0.4748CO2+

2.866N2

+0.574H2O

From reaction

3.76*A =2.866

Therefore A=0.76228

b)

Number of moles of O2 used= A=0.76228

Therefore moles air used = (1+3.76)*0.76228

=3.62845k mol

c)

Mole % of O2 in air= 21%

Mole % of N2 in air=79%

d)

Given mass flow rate of butane =3.524 kg/ s

Mol Wt of butane = 58 gm/mole

Molar flow rate of Butane =3.524/58. =0.060kmol

Implies Z= 0.060

Atomic carbon balance from reaction

2X+3Y+4Z=0.474---------I

Hydrogen atomic balance

6X+8Y+10Z=1.148-------------II

Given

Z= 0.060 k mol------------III

From I ,II ,III; we get

X= 0.114 k mol

y=0.154 k mol

Z=0.060 k mol

Total molar flow rate of fuel =X+Y+Z=0.328 kmol/s

Mol% of C2H6=0.114*100/0.328 =34.75%

Mol% of C3H8=46.9512%

Mol % of C4H10= 18.29%

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