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Design and Simulation of Two Stroke Engines

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The other equations reveal for air at 293 K:<br />

Chapter 2 - Gas Flow through <strong>Two</strong>-<strong>Stroke</strong> <strong>Engines</strong><br />

CP= 1022 J/kgK Cv = 734J/kgK y = 1.393<br />

It will be seen that the value <strong>of</strong> the ratio <strong>of</strong> specific heats, y, is not precisely 1.4 at st<strong>and</strong>ard<br />

atmospheric conditions as stated earlier in Sec. 2.1.3. The reason is mostly due to the fact that<br />

air contains argon, which is not included in the above analysis <strong>and</strong>, as argon has a value <strong>of</strong> y<br />

<strong>of</strong> 1.667, the value deduced above is weighted downward arithmetically.<br />

The most important point to make is that these properties <strong>of</strong> air are a function <strong>of</strong> temperature,<br />

so if the above analysis is repeated at 500 <strong>and</strong> 1000 K the following answers are found:<br />

for air:<br />

T = 500K CP = 1061 J/kgK Cv = 773 J/kgK y = 1.373<br />

T=1000K CP= 1143 J/kgK Cv = 855 J/kgK y= 1.337<br />

As air can be found within an engine at these state conditions it is vital that any simulation<br />

takes these changes <strong>of</strong> property into account as they have a pr<strong>of</strong>ound influence on the<br />

characteristics <strong>of</strong> unsteady gas flow.<br />

Exhaust gas<br />

Clearly exhaust gas has a quite different composition as a mixture <strong>of</strong> gases by comparison<br />

with air. Although this matter is discussed in much greater detail in Chapter 4, consider the<br />

simple <strong>and</strong> ideal case <strong>of</strong> stoichiometric combustion <strong>of</strong> octane with air. The chemical equation,<br />

which has a mass-based air-fuel ratio, AFR, <strong>of</strong> 15, is as follows:<br />

2CoHig + 25<br />

79<br />

02 + — No<br />

21<br />

= 16C02 + 18H20 + 94.05N2<br />

The volumetric concentrations <strong>of</strong> the exhaust gas can be found by noting that if the total<br />

moles are 128.05, then:<br />

16 9 94 05<br />

X>co = = 0.125 uH 0 = = 0.141 t>N = —•— = 0.734<br />

2 128.05 2 128.05 2 128.05<br />

This is precisely the same starting point as for the above analysis for air so the procedure<br />

is the same for the determination <strong>of</strong> all <strong>of</strong> the properties <strong>of</strong> exhaust gas which ensue from an<br />

ideal stoichiometric combustion. A full discussion <strong>of</strong> the composition <strong>of</strong> exhaust gas as a<br />

function <strong>of</strong> air-to-fuel ratio is in Chapter 4, Sec. 4.3.2, <strong>and</strong> an even more detailed debate is in<br />

the Appendices A4.1 <strong>and</strong> A4.2, on the changes to that composition, at any fueling level, as a<br />

function <strong>of</strong> temperature <strong>and</strong> pressure.<br />

In reality, even at stoichiometric combustion there would be some carbon monoxide in<br />

existence <strong>and</strong> minor traces <strong>of</strong> oxygen <strong>and</strong> hydrogen. If the mixture were progressively richer<br />

than stoichiometric, the exhaust gas would contain greater amounts <strong>of</strong> CO <strong>and</strong> a trace <strong>of</strong> H2<br />

67

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