Question

Twenty normal cubic meters per minute of butane (20 Nm^3 C4H10/min) is combusted in air. Calculate the combustion discharge gas composition (volume %, wet basis). There is not any excess oxygen requirement

Answer 1

Assuming butane behaves ideally,

normal T = 25 C, P= 101325 Pa

By ideal gas Law,

PV = nRT

.: n = 101325*20 / [8.31451*(25+273.15)]

.: n = 817.47 mol/min

molecular weight of butane = 4*12+10 = 58 g/mol

mass flow rate of butane = 817.47 mol/min x 58 g/mol = 47413.53 g/min

combustion reaction is as follows

C4H10 + (13/2)O2 --------> 5H2O + 4CO2

as O2 is not in excess, moles of O2 in product = 0

moles of O2 supplied = 13/2 x 817.47 mol/min = 5313.55 mol/min

moles of Air supplied = 5313.55 mol/min / 0.21.............(as air contains 21% O2 by volume)

= 25302.64 mol/min

N2 supplied = N2 in product stream = 25302.64 mol/min x 0.79.........(as air contains 79% N2 by volume)

= 32028.66 mol/min

Assuming 100 % conversion of butane

H2O formed = 5 x 817.47 mol/min = 4087.35 mol/min

CO2 formed = 4 x 817.47 mol/min = 3269.88 mol/min

Total moles of product = 32028.66 mol/min + 4087.35 mol/min + 3269.88 mol/min

= 39385.9 mol/min

For gases volume % = mole %

Volume % of N2 = [ 32028.66 / 39385.9] x 100 = 81.32

Similarly,

Volume % of H2O = 10.38

Volume % of CO2 = 100 - 10.38 - 81.32 =8.3