EXAMPLE 12-5. EXCESS AIR CALCULATIONS. gas that contains 7 percent oxygen by vol

Question

EXAMPLE 12-5. EXCESS AIR CALCULATIONS. gas that contains 7 percent oxygen by volume on a wet basis at standard conditions. A liquid injection incinerator has a stack incinerator is burning toluene at a rate of 184 lb/h with air. (a) What percent excess air is required? (b) If the stack gas had been on a dry basis, what would the excess air be? c) What is the combustion efficiency of the system if the CO content of the flue gas is 500 ppmv? olution )There is no other source of oxygen in the system except the combustion air. Toluene, CsHsCH3 has a molecular weight of 92 and a heating value of 18.252 Btu/lb. CA,CH, + 902 ? 7CO2 + 4H3O

Explanation / Answer

First calculate amount of product-

Molecular weight of toluene is 92,

Heating value is 18,252 BTU/lb

Therefore,184 lb/h / 92 = 2 lb mole/h

C6H5CH3+9O2=7CO2+4H2O

2lb mol/h. 18. 14 . 8

18/0.21=85.71 lb mole of air/h

Of which 79/21*18=67.7 lb mol/h of N2 remain in the gas after burning.

Amount of gas at theoretical air = 14+8+67.7=89.7lb mol/h

Let x be oxygen, Total gas= 89.7+X+X(79/21)

Since O2 is 7% given

We have, x / 89.7+x+x(79/21)=0.07

x = 9.42 lb mol/ h

Therefore 9.42/0.21=44.86lb mol/h

%Excess air = excess/ theoretical*100 = 44.86/85.71*100=52.34%

b) on dry basis theoretical moles of gas is 89.7-8(H2O)=81.7 lb mol/h

Let x be the O2

Total gas=81.7 + x +x(79/21)

x/81.7+x+x(79/21)=0.07

x = 8.59 lb mol/h

Excess air required= 8.59/0.21=40.91 lb mol/h

c) 500 ppmv CO = 500/10^6*122.6=0.061 lb mol/h

Whereas 122.6 is total amount of gas on dry basis

CO2 flow rate= 2lb mol/h * 7

= 14 lb mol/h

Combustion efficiency=CO2-CO/CO2*100

= 14-0.061/14*100

= 99.56%

%Excess air = excess/ theoretical*100 = 44.86/85.71*100=52.34%

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