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Modern Engineering Thermodynamics

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618 CHAPTER 15: Chemical <strong>Thermodynamics</strong><br />

Then, the heat of the combustion reaction becomes<br />

q r<br />

= ∑<br />

R<br />

ðn i /n fuel Þðh°Þ f i<br />

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

P<br />

= ðh f °Þ C8H 18<br />

− 8½h f ° − 373:57A + 2035:3C − 205:17D + 0:8066FŠ CO2<br />

− 9½h f °+14,305:A − 14,683:2B + 5516:7C − 184:95DŠ H2O<br />

− ðY/100 − 1Þ½3743:2A + 0:80408E + 35,714:G − 23,688HŠ O2<br />

− 47ðY/100 − 1Þ½3906:0A + 102,558G − 107270:H + 41,020IŠ N2<br />

From Table 15.1, we find<br />

h° f C = −249:952 MJ/kgmole, h° 8H 18ðlÞ f CO 2<br />

= −393:522 MJ/kgmole,<br />

h f ° H 2 OðgÞ = −241:827 MJ/kgmole, and h f ° N 2<br />

= h f ° O 2<br />

= 0 because they are elements:<br />

While these equations are difficult to handle using a hand calculator, they are easily solved using an equation<br />

solver or a spreadsheet. The spreadsheet in Figure 15.7 illustrates the process for the combustion of liquid<br />

octane with 200% theoretical air. Typing control + tilde (Ctrl and Shift + ~) reveals the details, shown in<br />

Figure 15.8.<br />

FIGURE 15.7<br />

Solving the equations with a spreadsheet.

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