Question

In a combustion chamber 14 standard m³ per hour of butane is burnt with 616 kg per hour of air. What is the % O2 in the flue gas assuming 100% conversion? (Round your answer to 2 decimals and enter only numerical digits of your answer)

Answer 1

1 kg mole of any gas at standard conditions occupies 22.4 m3

molar mass of butane ( C4H10)= 4*12+10*1= 58 g/mole

density of Butane at standard conditions = 58/22.4 kg/m3

mass flow rate of butane = volumetric flow rate of butane*density = 14 m3/hr* 58/22.4 kg/m3= 36.2 kg/hr, molar flow rate of butane = mass of butane/ molar mass= 36.2/58 = 0.62 moles/hr

moles of air used= mass of air/ molar mass of air = 616/32 =19.25 kg/moles

the combustion reaction is C4H10 + 6.5O2 ------->4CO2+ 5H2O (1)

Theoretical moles ratio of C4H10: O2= 1:6.5

Actual moles ratio of C4H10 : O2= 0.62: 19.25 = 0.62/0.62: 19.25/0.62= 1: 31

so excess is O2. All the butane gets combusted. As per the reaction-1, 1 mole of butane undergoes combustion with 6.5 moles of oxygen to produce 4 moles of CO2 and 5 moles of water.

moles of O2 consumed = 6.5* moles of butane = 6.5*0.62 = 4.03 kg moles/hr

oxygen remaining = oxygen suplied- oxygen consumed through reactiion-1= 19.25-4.03 = 15.22 kg moles/hr

moles of CO2 produced =4*0.62= 2.48 kg moles/hr and that of H2O ( assuming H2O is in gasoue form) = 5*0.62= 3.1 kg moles/hr

Products constains ( assuming H2O is in gaseous form)

CO2= 2.48 kg moles/hr, O2=15.22 kg moles/hr and H2O= 3.1 kg moles/hr

% of O2 in products = 100* molar flow rate of O2/ total moles of products= 100*15.22/ (15.22+2.48+3.1)=100*15.22/20.8 = 73.2%

% of O2 in products ( assuming water is in liquid form)= 100*15.22/ ( 15.22+2.48)= 86%