A mixture of fuels, ethane (C2H6) and methane (CH4), is burned in excess air. Th

Question

A mixture of fuels, ethane (C2H6) and methane (CH4), is burned in excess air. The fuel stream enters the furnace at 25°C and 1 atm, while a stream of air enters the furnace at 200°C and 1 atm. At steady state, the flue gas stream leaves the furnace at 800°C and 1 atm and 257 mol/s. The composition of the flue gas is 7.39 mol% O2, 73.2 mol % N2, 12.3 mol% H2O, 5.43 mol% CO2, and 1.67 mol% CO.

Use the information above and the PFD to answer the following questions:

1. Write balanced reactions for the complete and incomplete combustion of methane and ethane (4 reactions total).

Explanation / Answer

Nitrogen comes from air supplied and does not participate in the reaction and is known as tie substance.

Molar flow rate of products= 257 mol/s, flow rate of nitrogen= 0.732*257 =188 mole/s

Air contains 79% N2 and 21%O2, molar flow rate of air supplied= 188/0.79= 238 mole/s

Balance equations for complete and incomplete combustion of methane(CH4): CH4 on complete combustion gives CO2 while on incomplete combustion gives CO. This is the case with ethane as well.

CH4+ 2O2----->CO2+ 2H2O (1), CH4+1.5O2--->CO+2H2O (2)

Balance equations for complete and incomplete combustion of ethane

C2H6+ 3.5O2---->2CO2+3H2O (3) , C2H6+2.5 O2--->2CO+ 3H2O (4)

Nitrogen comes from air supplied and does not participate in the reaction and is known as tie substance.

Molar flow rate of products= 257 mol/s, flow rate of nitrogen= 0.732*257 =188 mole/s

Air contains 79% N2 and 21%O2, since N2 comes from air and does not participate, molar flow rate of air supplied= 188/0.79= 238 mole/s

Molar flow rate of air = 238*0.21= 50 mol/s, molar flow rate of oxygen remaining in the products= 257*7.39/100 mol/s =19 mol/sec

Oxygen consumed there fore will be = oxygen supplied-Oxygen in the product= 50-19=31 mol/sec

Molar flow rate of CO2= 257*5.43/100 = 14 mol/s

Molar flow rate of CO= 257* 1.67/100 = 4.3 mol/s

Molar flow rate of H2O= 257*12.3/100 = 31.6 mol/s

Let x1= flow rate of methane combusted to produce CO2, x2= flow rf ate methane combusted to produce CO.

Moles of CO2 formed= same as moles of methane consumed through reaction-1= x1

Moles of CO formed= x2 ( from reaction-2)

Molar flow rate of water= 2x1+2x2

Let y1= flow rate of ethane combusted to produce CO2, y2= flow rate of ethane combusted to produce CO.

Molar flow rate of CO2= 2y1 (from reaction-3), molar flow rate of CO2= 2y2

Molar flow rate of water= 3y1+3y2

Molar flow rate of oxygen used= 2x1+1.5x2+3.5y1+2.5y2 ( from reactions-1,2,3 and 4)

Total molar flow rate of water= 2x1+2x2+3y1+3y2

Total molar flow rate of CO2= x1+2y1

Molar flow rate of CO= x2+2y2

Hence 2x1+1.5x2+3.5y1+2.5y2= 31   (1),

x1+2y1=14 (2) , x1= 14-2y1 (2A)

x2+2y2=4.3 (3), x2= 4.3-2y2 (3A)

2x1+2x2+3y1+3y2=31.6 (4)

Substituting the values of x1 and x2 from Eq.2A and 3A into Eq.1

2*(14-2y1)+5*(4.3-2y2)+3.5y1+2.5y2 =31  

28-2y1+21.5-10y2+3.5y1+2.5y2=31

1.5y1+7.5y2= 28+21.5-31= 18.5 (5)

Substituting the values of x1 and x2 in Eq.4

2*(14-2y1)+ 2*(4.3-2y2)+3y1+3y2= 31.6

28-4y1+8.6-4y2+3y1+3y2= 31.6

Hence y1+y2= 28+8.6-31.6= 5 (6)

Or y1= 5-y2 , substituting the value of y1 in Eq.5 gives 1.5*(5-y2)+7.5y2=18.5

Or 7.5-1.5y2+7.5y2= 18.5

6y2=11, y2= 11/6= 1.83

Hence y1= 5-1.83= 3.17, x1=14-2*3.17= 7.66 and x2=4.3-2*1.83= 0.64

Hence x1+x2=7.66+0.64=8.3 mol/s of methane

Ethane is y1+y2= 5 mol/s

Oxygen is supplied for complete combustion =8.3 ( as if all the methane becomes CO2)+5*3.5 ( as if all the CH4 becomes CO2)= 25.8 mol/s

Oxygen supplied= 50 moles/s

Excess oxygen= 50-25.8= 24.2 mol/s

Excess % of oxygen= 100*24.2/25.8= 94%

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