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Modern Engineering Thermodynamics

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15.8 Adiabatic Flame Temperature 617<br />

For example, the reaction for the combustion of liquid octane with Y % theoretical air is<br />

C 8 H 18 + ðY/100Þ12:5 O 2 + 3:76ðN 2 Þ ! 8CO ð 2 Þ+ 9H ð 2 OÞ+ ðY/100 − 1ÞO 2 + 47ðY/100ÞðN 2 Þ<br />

To simplify matters, assume that, before the combustion, the reactants are at the SRS. Then, hðT R Þ = hðT°Þ and<br />

hðT R Þ − hðT°Þ = 0 for all the reactants. The heat produced by this reaction when the combustion products are at<br />

temperature T P is<br />

q r<br />

= ∑<br />

R<br />

ðn i /n fuel<br />

Þðh f °Þ i<br />

−∑<br />

P<br />

ðn i /n fuel Þ½h° f + hðT P Þ − hðT°ÞŠ i<br />

= ðh f °Þ C8H 18<br />

− 8½h f ° + hðT P Þ − hðT°ÞŠ CO2<br />

− 9½h f ° + hðT P Þ − hðT°ÞŠ H2O<br />

− ðY/100 − 1Þ½h f ° + hðT P Þ − hðT°ÞŠ O2<br />

− 47ðY/100 − 1Þ½h f ° + hðT P Þ − hðT°ÞŠ N2<br />

The molar specific heat equations in kJ/(kgmole·K), accurate to at least 0.43% over the range of 300 to 3500 K,<br />

can be found in Table C.14b in Thermodynamic Tables to accompany <strong>Modern</strong> <strong>Engineering</strong> <strong>Thermodynamics</strong> as<br />

Carbon dioxide: ðc p Þ CO2<br />

= −3:7357 + 30:529θ 0:5 − 4:1034θ + 0:024198θ 2<br />

Water: ðc p Þ H2 O = 143:05 − 183:54θ0:25 + 82:751θ 0:5 − 3:6989θ<br />

Oxygen: ðc p Þ O2<br />

= 37:432 + 0:020102θ 1:5 − 178:57θ −1:5 + 236:88θ −2<br />

Nitrogen: ðc p Þ N2<br />

= 39:060 − 512:79θ −1:5 + 1072:7θ −2 − 820:40θ −3<br />

where θ° =T°/100 = 298/100 = 2:98, and θ P = T p /100. Integrating these equations from the SRS (θ°) tothe<br />

temperature of the combustion products (θ P ) gives<br />

½hðT P Þ − hðT°ÞŠ CO2<br />

= 100 ×<br />

Z θP<br />

θ°<br />

ðc p Þ CO2<br />

dθ<br />

= −373:57ðθ P − θ° Þ+ 2035:3 ðθ P Þ 1:5 − ðθ°<br />

Þ 1:5<br />

− 205:17 ðθ P Þ 2 − ðθ°Þ 2 + 0:8066 ðθ P Þ 3 − ðθ°Þ 3<br />

½hðT P Þ − hðT°ÞŠ H2 O<br />

Z θP<br />

= 100 × ðc p Þ H2 O dθ<br />

θ°<br />

= 14,305: ðθ P − θ° Þ− 14,683:2 ðθ P Þ 1:25 − ðθ°<br />

Þ 1:25<br />

+ 5516:7 ðθ P Þ 1:5 − ðθ°Þ 1:5 − 184:95 ðθ P Þ 2 − ðθ°Þ 2<br />

½hðT P Þ − hðT°ÞŠ O2<br />

= 100 ×<br />

Z θP<br />

θ°<br />

ðc p Þ O2<br />

dθ<br />

= 3743:2ðθ P − θ° Þ+ 0:80408 ðθ P Þ 2:5 − ðθ°<br />

Þ 2:5<br />

+ 35:714: ðθ P Þ −0:5 − ðθ°Þ −0:5 − 23,688 ðθ P Þ −1 − ðθ°Þ −1<br />

½hðT P Þ − hðT°ÞŠ N2<br />

= 100 ×<br />

Z θP<br />

θ°<br />

ðc p Þ N2<br />

dθ<br />

To simplify the algebra, we define the following terms:<br />

= 3906:0ðθ P − θ° Þ+ 102,558 ðθ P Þ −1/2 − ðθ°<br />

− 107,270: ðθ P Þ −1 − ðθ°Þ −1 + 41,020 ðθ P Þ −2 − ðθ°Þ −2<br />

Þ −1/2<br />

A = θ P − θ° B = ðθ P Þ 1:25 − ðθ°Þ 1:25 C = ðθ P Þ 1:5 − ðθ°Þ 1:5<br />

D = ðθ P Þ 2 − ðθ°Þ 2 E = ðθ P Þ 2:5 − ðθ°Þ 2:5 F = ðθ P Þ 3 − ðθ°Þ 3<br />

G = ðθ P Þ −1/2 − ðθ°Þ −1/2 H = ðθ P Þ −1 − ðθ°Þ −1 I = ðθ P Þ −2 − ðθ°Þ −2

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