Design and Simulation of Two Stroke Engines

Design and Simulation of Two-Stroke Engines
also x3 = — = 0.825 x6 = 1.20 x 3.76 = 4.52
The actual combustion equation now becomes
CH165 Hj 65 + 4• 1.20(O2 1.20(O2_+ + 3.76? 3.76N2) 4
(4J
0.424CO + 0.576CO2 +"6.825H20 + 4.52N2
The total moles of combustion products are 6.342 which, upon division into the moles of
each gaseous component, gives the volumetric proportion of that particular gas. The exhaust
gas composition is therefore 6.7% CO, 9.1% C02, 13.0% H20, and the remainder nitrogen.
The emission of carbon monoxide is now significant, by comparison with the zero concentration
ideally to be found at stoichiometry, the 12.5% C02 concentration and the 14.1% H20
proportions by volume as seen in the equivalent analysis for octane in Sec. 2.1.6.
4.3.2.3 Lean mixture combustion
In lean mixture combustion, the value of the equivalence ratio is greater than unity:
X>\ (4.3.15)
In the first instance let dissociation effects be ignored, in which case Eqs. 4.3.5 to 4.3.8
are directly soluble, hence X5 is zero. There is excess oxygen to burn all of the fuel so there is
no free carbon monoxide and the value of xi is zero. The molecular balances become:
x2 = 1 x3 = - 2Xm = 2x2 + x3 + 2x4 x6 = Xmk
Hence, for the remaining unknown X4:
x 4=A, m-l-- (4.3.16)
Consider a practical example where the equivalence ratio is 50% lean of stoichiometry
for the combustion of diesel fuel, i.e., the value of n is 1.81 and the value of A, is 1.5.From Eq.
4.3.11 the value of A^ is found as:
Xm = x(l + -] = 1.5 x (1 + — j = 2.178
1 01
Hence x4 = 2.178 - 1 - — = 0.726
300