Question

From a combustion process using a mixture of methane (1)/ethane (2) as fuel, the ORSAT analysis of the stack gases is given in the table below:

Component	Xi (molar basis)
CO2	0.0769
O2	0.0837
N2	0.8394

Assume that 100 kgmole of stack gases are produced.

DETERMINE:

1. Sketch the process. Label all streams.

2. Write the chemical reaction(s) that take place during the process

3. Find the composition of the fuel (mole fraction)

4. Find the % excess air.

Answer 1

Let x= moles of methane and y= moles of ethane

Given 100 kg moles of stack gases

Moles of N2= 83.94 kg moles. This has come from air used for combustion

Air contains 79% N2 and 21% O2. Moles of air supplied= 83.94/0.79= 106

Moles of oxygen supplied= 106*0.21 =22.3

Moles of oxygen remaining= 8.37 moles

Moles of Oxygen utilized= 22.3-8.37= 13.93

13.93*(1+x/100)= 22.3

Hence % excess air ( or oxygen)= 100*(8.37/22.3)=37.53%

The combustion reaction is

CH4+2O2----àCO2+ 2H2O and C2H6+ 3.5O2 ----à2CO2+3H2O

X moles of methane produces x moles of CO2 and y moles of C2H6 gives 2 moles of C2H6

Hence x+2y= 7.69 (1), x= 7.69-2y (1)

Moles of oxygen used for combustion of methane = 2x and mole of ethane= 3.5y

Hence 2x+3.5y=13.93

From Eq.1 , 2*(7.69-2y)+3.5y= 13.93

15.38-4y+3.5y= 13.93

0.5y= 15.38-13.93

Hence y= 2.9 kg moles and x= 7.69-2*2.9= 1.89 kg moles

Total moles of fuel = 2.9+1.89= 4.79

mole fraction= moles of given fuel/total moles

Mole fractions : Methane= 1.89/4.79= 0.39, ethane= 1-0.39=0.61