Question

A natural gas (85 mol% CH4, 15% C2H6) at 20C and 80atm is burned completely with 30% excess air. Natural gas enters the furnace at a rate of 285 L/s. The stack gas emerges at 280C and 1 atm d) Calculate the molar composition (mole fractions) of the stack gas e) Calculate the volumetric flow rate (L/s) of the stack gas

Answer 1

a)

V = 285 L/s

PV = nRT

n = PV/(RT) = (80*285)/(0.082*(20+273)) = 948.9

mol of CH4 = 85/100*948.9= 806.565

mol of C2H6 = 15/100*948.9= 142.33

CH4 + 2O2 = CO2 + 2H2O

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

mol of O2 for CH4 = 2*CH4 = 2*806.565= 1613.13

mol of O2 for CH4 = 7/2*C2H6 = 2*142.33= 284.66

Total O2 = 1613.13+284.66=1897.79

total O2 at 30% excess = 1897.79*1.3 = 2467.127 mol of O2

mol of N2 = (0.79/0.21) * 2467.127= 9281.09 mol of N2

after rection:

mol of CO2 formed = 2*1613.13+2*284.66 = 3795.58

mol of H2O = 5*948.9 = 4744.5

then..

mol of O2 left = 2467.127 -1897.79 = 569.337

Total mol at the end:

mol of O2 = 569.337

mol of N2 = 9281.09

mol of CO2 = 3795.58

mol of H2O = 4744.5

mol = 569.337+9281.09 +3795.58+4744.5 = 18390.507

mol frac:

x frac. of O2 = 569.337/18390.507= 0.0309

x frac. of N2 = 9281.09 /18390.507= 0.5046

x frac. of CO2 = 3795.58/18390.507= 0.2063

x frac. of H2O = 4744.5/18390.507= 0.2579

b)

volumetric flow

PV = nRT

V = nRT/P

V = (18390.507)(0.082)(280+273)/1

V = 833935.9 L/s