Design and Simulation of Two Stroke Engines

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Chapter 4 - Combustion in Two-Stroke Engines The concept of equivalence ratio, A, is useful in the solution of these equations and in engine technology generally. It is defined as: The combination of Eqs. 4.3.4, 4.3.9 and 4.3.10 reveals: Obviously, the equivalence ratio is unity at stoichiometry. Equivalence Ratio, A = ,A a t n\ AFRS AFR Ar A = _ ,v m AFRS 1 + £ (4.3.11) 4.3.2.2 Rich mixture combustion In rich mixture combustion, the value of the equivalence ratio is less than unity: A
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
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Design and Simulation of Two-Stroke
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A Second Mulled Toast When as a stu
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Acknowledgments As explained in the
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Table of Contents Nomenclature xv C
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Table of Contents 2.10.1 Flow at pi
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Table of Contents 3.5.3.1 The use o
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Table of Contents 6.1.3 The effect
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Nomenclature NAME SYMBOL UNIT (SI)
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speed of rotation speed of rotation
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attenuation or transmission loss wa
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Design and Simulation ofTwo-Stroke
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Design and Simulation ofTwo'Stroke
Page 270 and 271: Design and Simulation of Two-Stroke
Page 302 and 303: Chapter 4 Combustion in Two-Stroke
Page 304 and 305: Chapter 4 - Combustion in Two-Strok
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Page 342 and 343: Combustion in compression-ignition
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£2 = Pi P2 I Pi J Chapter 4 - Comb
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CHAINSAW AT 9600 rpm. Chapter 4- Co
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LLI •z. O N -z. 0.21 -, 0.20 DC 0
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Chapter 4 - Combustion in Two-Strok
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Chapter 7 Reduction of Fuel Consump
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Chapter 7 • Reduction of Fuel Con
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Chapter 7 - Reduction of Fuel Consu
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^ s C£ Q 2^ E Q. Q_ z' o ISSI HCEM
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§ 2000
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y, ,6 Chapter 7 - Reduction of Fuel
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Chapter 7- Reduction of Fuel Consum
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INLET FOR r AIR AND FUEL Chapter 7
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Chapter 8 Reduction of Noise Emissi
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Chapter 8 • Reduction of Noise Em
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Chapter 8 - Reduction of Noise Emis
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• • / . • Appendix Listing of
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Active radical (AR) combustion and
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initial conditions, assumptions reg
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in two-stroke engine (schematic dia
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exhaust timing control valve, effec
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I Exhaust emissions/exhaust gas ana
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Exhaust systems (continued) untuned
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gas constant/universal gas constant
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in crankcase heat transfer analysis
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Mercury Marine air-assisted fuel in
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I PISTON POSITION (computer program
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Pressure wave reflection (in pipes)
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local acoustic velocity, 60 local M
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temperature vs. crankshaft angle (c
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work per thermodynamic (Otto) cycle
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Design and Simulation of Two Stroke Engines

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