For safety reasons, the maximum allowable temperature for a reactor is set to be

Question

For safety reasons, the maximum allowable temperature for a reactor is set to be 20000C. A fuel mixture comprising 80 mol% CH4 and 20 mol% C2H6 is fed into the reactor at 250C. Oxygen (O2 ) is also fed into the reactor in the form of air (21 % O2 and 79% N2 ) at 250C. The following reactions occur: CH4 (g)+2O2 (g)CO2 (g)+2H2O(g): H0 r (250C)=802.3 kJ/mol C2H6 (g)+7 2 O2 (g)2CO2 (g)+3H2O(g): H0 r (250C)=1,427.9 kJ/mol Determine the percentage of excess air so that the maximum outlet temperature does not exceed 2,0000C. Assume that the reactor is adiabatic and that CH4 and C2H6 are completely burned, while N2 is inert. Given enthalpy of each species at 2,0000C (with reference to that at 250C) is as follows: O2 69.6 kJ/mol; N2 66.0 kJ/mol; CO2 101.9 kJ/mol; H2O 87.0 kJ/mol. 5.84. % excess air

Explanation / Answer

For adabatic reactor

Enthalpy of products + Heat of reaction – enthalpy of reactants= 0

Enthalpy of products= -heat of reaction (1)

Basis : 1mol of mixture

Moles of CH4 =0.8moles and C2H6= 0.2 moles

Since the reactants are entering at 25 deg.c, their enthalpy =0

The reactions are CH4+2O2------à CO2 +2H2O

Moles of O2 required =2*0.80=1.6

And C2H6+ 3.5O2------à 2CO2 + 3H2O

Moles of oxygen required= 3.5*0.2=0.7

Total moles of oxygen = 1.60+0.7= 2.30 moles

Moles of air =2.30/0.21=10.95 moles

Let x= %excess air supplied

Total air supplied =(1+x/100)*10.95

N2 in air = (1+x/100)*10.95*0.79

Moles of Oxygen in air = (1+x)*10.95*0.21

Products contains

CO2 ( from CH4)= 0.8 moles and CO2 from C2H6= 2*0.2= 0.4 moles

Total of CO2= 1.2 moles

H2O= 2*0.8+3*0.2= 2.2 moles

N2= (1+x)*10.95*0.79

Since complete combustion takes place, oxygen = x*10.95/100*0.21= 2.3x

So enthalpy of products = 1.2* 101.9+ 2.2*87 + (1+x/100)*10.95*0.79*66 +2.3x/100* 69.6

122.8+ 191.4 +562.6*(1+x/100)+1.6x

Heat of reaction at 25 deg.c = 0.8* 802.3 +0.2*1427.9 =-927.42 Kj

Hence, from Eq.1

927.42 =122.8+191.4 +570.3*(1+0.01x)+ 1.6x

x= excess air = 5.86%

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