Question

Butane (at 75°F and 1 atm) is fed to a boiler at 100 mol/s with 50% excess air. 70% of the butane is consumed, and the product gas contains 10 moles CO2 per mol CO.

a) Calculate the molar composition of the stack gas on i) wet basis, and ii) dry basis.

b) Compute the volume of stack gas at 170°F and 1 atm pressure per ft^3 of Butane.

Answer 1

Mole/s of butane supplied = 100mol/s

The reaction of combustion of butane is C4H10+6.5 O2----> 4CO2 +5 H2O and C4H10+4.5O2---> 4CO+5 H2O

Oxygen required for complete combustion= 100*6.5 =650 mole/s

Air required ( sicne air contains 79% N2 and 21% Oxygen) mole of air = 650/0.21=3095.24 mol/s

air is supplied 50% excess . hence air supplied= 1.5*3095.24=4643mol/s

Nitrogen in air supplied= 3668 mol/s and O2= 4643-3668=975 mol/s

since 70 mole of butane is consumed= butane consumed = 70 moles/sec

let x = moles/s of butane used for formation of CO2

moles of butane consumed= 70% and hecne moles of butane consumed= 70 mol/s

moles of butane consumed for CO2 formation=x and moles of Butane consumed = 70-x

1 mole of C4H10 gives 4 moles of CO2 and 1 mole of C4H10 gives 4 mole of CO

x moles of C4H10 gives 4x moles of CO2 and 4*(70-x) moles of CO

given CO2/CO in the exhaust gas = 4x/(4*(70-x)= 10/1

x/(70-x)= 10 and hence x =70*10-10x and hence 11x= 70*10 and x= 700/11=63.63 mole of C4H1O used for formation of CO2 and rest for formation of CO which is 70-63.63=6.37 moles

Products

CO2= 4*63.63= 254.52moles/s CO= 4*(70-63.63) =4*6.37=25.48 mole/s

N2 ( un reacted )= 3668 mol/s

The reaction of combustion of butane is C4H10+6.5 O2----> 4CO2 +5 H2O and C4H10+4.5O2---> 4CO+5 H2O

Oxygen remaining = oxygen supplied- oxygen consumed for formation of CO2- oxygen consumed for formation of CO

Moles of Oxygen consumed for producing CO2= 63.63*6.5 (1 mole of C4H10 requires 6.5 mole of oxygen and 6.37 mole of C4H1O requires 4.5*6.37 moles of Oxygen)

=413.595+28.665 =442.26 mole/s of oxygen

Oxygen remaining= 975-442.26=532.74 mole/s

H2O formed= Water formed from formation of CO2 + water formed from formation of CO

= 5*63.63+5*6.37 =350 mol/s of water

Products (dry basis )

CO2= 254.42 mole/s CO= 25.48 mol/s N2= 3668 mol/s O2= 532.74 mol/s

Total moles= 254.42+25.48+3668+532.74=4480.64 mols/

Percentages of these products ( dry basis ); CO2= 100*(254.42/4480.64)= 5.67% CO= 100*25.48/4480.64=0.568% N2= 100*(3668/4480.64)=81.86% Oxygen= 100*532.74/4468.64=11.89%

Products ( wet basis ) : CO2= 254.42 mol/s. CO= 25.48 mol/s N2= 3668 mol/s O2= 532.74 mol/s and H2O= 350 mol/S

Percentages : CO2= 5.27% CO= 0.53%, N2= 75.93% O2=11.03% and H2O= 7.25%

Volume of butane (V)= nRT/P, = 0.7302ft3.atm/Role P=1 atm , T =75+460=535 R , n=100 mol/s, V= 100*0.7302/(1(535) V= 39065.7 ft3/s Volume of exhaust gas = 4830.64*0.7302*635/1= 2239857 ft3/s Volume of stack gas per ft3 of butane = 57.33