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Modern Engineering Thermodynamics

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15.4 Fuel Modeling 601<br />

equation, since it results from the oxidation of the hydrogen in the fuel. For convenience, we write the combustion<br />

reaction for 100. moles of dry product formed by burning 1.00 mole of the fuel model C n H m . Using the given<br />

combustion analysis, we have<br />

C n H m + a½O 2 + 3:76ðN 2 ÞŠ ! 7:10ðCO 2 Þ + 0:800ðCOÞ + 9:90ðO 2 Þ + bðH 2 OÞ + 82:2ðN 2 Þ<br />

The element balances are<br />

Carbon (C) balance: n = 7.10 + 0.800 = 7.90<br />

Hydrogen (H) balance: m = 2b<br />

Nitrogen (N 2 ) balance: 3:76a = 82:2, or a = 82:2/3:76 = 21:9<br />

Oxygen (O 2 ) balance: a = 21:9 = 7:10 + 0:800/2 + 9:90 + b/2, or b = 9:00:<br />

Then, from the preceding hydrogen balance, m = 2b = 18:0: Consequently, the fuel model is C 7.90 H 18.0 , which is<br />

approximately octane, C 8 H 18 . The final reaction equation is<br />

C 7:90 H 18:0 + 21:9½O 2 + 3:76ðN 2 ÞŠ ! 7:10ðCO 2 Þ + 0:800ðCOÞ + 9:90ðO 2 Þ + 9:00ðH 2 OÞ + 82:2ðN 2 Þ<br />

b. On a molar basis, 1.00 mole of fuel contains 7.90 moles of C and 18.0 moles of H, and on a molar percentage basis,<br />

this becomes [7.90/(7.90 + 18.0)](100.) = 31.0% C and [18.0/(7.90 + 18.0)](100) = 69.0% H. The molecular mass of<br />

the fuel in this model is<br />

and so the fuel’s composition on a mass basis is<br />

and<br />

M fuel = 7:90ð12Þ + 18:0ð1Þ = 113 kg/kgmole = 113 lbm/lbmole<br />

ð7:90 kgmole C/kgmole fuelÞð12:0 kg C/kgmole CÞ/ ð113 kg fuel/kgmole fuelÞ<br />

= 0:840 kg C/kg fuel = 0:840 lbm C/lbm fuel<br />

ð9:00Þð2:016Þ/113 = 0:161 kg H/kg fuel = 0:161 lbm H/lbm fuel<br />

Therefore, the fuel can be said to consist of 31% carbon and 69% hydrogen on a molar basis or 84% carbon and 16%<br />

hydrogen on a mass basis.<br />

c. Referring to the final combustion equation determined in part a, the air/fuel ratio on a molar basis is<br />

ðA/FÞ molar = n air 21:9 × ð1 + 3:76Þ moles of air<br />

= = 104 moles air/mole fuel<br />

n fuel 1 mole of fuel<br />

and on a mass basis it is<br />

<br />

<br />

28:97 kg air/kgmole air<br />

ðA/FÞ mass<br />

= ð104 kgmole air/kgmole fuelÞ× 113 kg fuel/kgmole fuel<br />

= 26:7 kg air/kg fuel = 26:7 lbm air/lbm fuel<br />

d. To determine the percent of theoretical air used, we must first determine the minimum air required for complete<br />

combustion. The reaction for 100.% theoretical air has the form<br />

C 7:90 H 18:0 + a½O 2 + 3:76ðN 2 ÞŠ ! bðCO 2 Þ + cðH 2 OÞ + dðN 2 Þ<br />

The element balances are<br />

Carbon (C) balance: 7.90 = b<br />

Hydrogen (H) balance: 18.0 = 2c, orc = 11.0<br />

Oxygen (O 2 ) balance: a = b + c/2 = 7:90 + 9:00/2 = 12:4<br />

Nitrogen (N 2 )balance:3:76a = d = 3:76ð12:4Þ = 46:6<br />

Then the theoretical molar air/fuel ratio (for 100% theoretical air) is<br />

12:4ð1 + 3:76Þ<br />

ðA/FÞ molar<br />

= = 59:0 mole air/mole fuel<br />

1<br />

theoretical<br />

finally, the percent of theoretical air used in the actual combustion process is<br />

" ,<br />

#<br />

% of theoretical air = ðA/FÞ molar<br />

actual<br />

ðA/FÞ molar<br />

theoretical<br />

or 77.0% excess air.<br />

= 104/59:0 ð100Þ = 177%<br />

× 100<br />

(Continued )

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