Question

A natural gas containing 82 mol% CH4 and the balance C2H6 is burned with 25% excess air in a boiler furnace. The fuel gas enters the furnace at 298 K, and the air is preheated to 423 K. The heat capacities of the stack gas components may be assumed to have the following constant values:

CO2: Cp=50.0 J/(mol.K)

H2O(v): Cp=38.5 J/(mol.K)

O2: Cp=33.1 J/(mol.K)

N2: Cp=31.3 J/(mol.K)

Assuming complete combustion of the fuel, calculate the adiabatic flame temperature in K.

Answer 1

Let 1 mole of feed is entering the reactor.

ie., 0.82 mole methane and 0.18 mole ethane.

0.82 mole methane requires 2*0.82=1.64 mole Oxygen and 0.18 mole ethane requires 3.5*0.18=0.63 mole oxygen.

Thus theoretically 1.64+.63=2.27 mole oxygen is required. 25% excess supplied means, 1.25*2.27=2.8375 mole oxygen is supplied.

Air is having 79 mole% nitrogen and 21 mol% oxygen. Thus for 2.8375 mole oxygen, mole nitrogen is also present in the air supplied.

standard enthalpy of combustion is when reaction is at 298K.

standard enthalpy of combustion of methane = -882 KJ/mol

standard enthalpy of combustion of ethane = -1560 KJ/mol

Air is entering at 423K. we have enthalpy of combustion at 298K. So we have to assume like air is cooled to 298 and products is taken to adiabatic flame temperature.

Assuming complete combistion, total heat released by combustion= 0.82*-882+0.18*1560= -1004 KJ

heat released by converting air ( oxygen and nitrogen) from 423 to 298 = (298-423)(2.8375*33.1+10.6744*31.3)= -53503.765 J

Total = -1004 KJ + -53.503765 KJ = -1057.503765 KJ

This heat is used to attain adiabatic flame temperature.

Product stream contains remaining 25% oxygen, nitrogen fed, Carbon dioxide and water formed.

Oxygen remaining = 2.8375-2.27 = 0.5675 mole

Nitrogen = 10.6744 mole

Carbon dioxide formed = 0.82 (from methane) + 2*0.18 (from ethane) = 1.18 mole

water formed= 2*0.82+3*0.18=2.18 mole

Let T be the adiabatic flame temperature.

Thus

(0.5675*33.1+10.6744*31.3+1.18*50+2.18*38.5)(T-298)-1057.503765*10³=0

Thus T=2430K

Hope this is clear for you.

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