Question

A natural gas mixture having a molar analysis 60% CH4,
30% C2H6, 10% N2 is supplied to a furnace like the one
shown in Fig. P13.8, where it burns completely with 20%
excess air. Determine
(a) the balanced reaction equation.
(b) the air–fuel ratio, both on a molar and a mass basis

Answer 1

a)

assume a basis of 100 moles

60 mol of CH4, 30 mol of C2H6, 10 mol of N2,

the reaction:

CH4 + 2O2 = CO2 + 2H2O

C2H6 + 7/2O2 = 2CO2 + 3H2O

N2 =intert, no reaction

b)

moles of O2 for CH4 = 2*60 = 120 , at 20% excess --> 120*1.2 = 144 mol of O2

mol of O2 for C2H6 = 3.5*30 = 105, at 20% --> 105*1.2 = 126 mol of O2

total mol of O2 = 126+144 = 270

mol of air = mol of O2/0.21 = 270/0.21 = 1285.7

ratio --> mol of air / mol of fuel = 1285.7/(144 +126 ) = 4.76

c)

for mass:

moles of O2 for CH4 = 2*60 = 120 , at 20% excess --> 120*1.2 = 144 mol of O2

mol of O2 for C2H6 = 3.5*30 = 105, at 20% --> 105*1.2 = 126 mol of O2

total mol of O2 = 126+144 = 270

mol of air = mol of O2/0.21 = 270/0.21 = 1285.7

mass of air = mol*MW = 1285.7*29 = 37285.3 g of air

Fuel = 60mol of CH4 * 16 g/mol + 30 mol of C2H6 * 30 g/mol = 1860

ratio --> g of air / g of fuel = 37285.3 /(1860) = 20.05