Design and Simulation of Two Stroke Engines

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Design and Simulation of Two-Stroke Engines Considering air as the example gas at a temperature of 20°C, or 293 K, the molal specific heats of oxygen and nitrogen are found using Eqs. 2.1.33 and 34 as: Gas o2 N2 CO C02 H20 H2 Oxygen, 02: CP = 31,192 J/kgmol Cv = 22,877 J/kgmol Nitrogen, N2: CP = 29,043 J/kgmol Cv = 20,729 J/kgmol Table 2.1.1 Properties of some common gases found in engines M 31.999 28.013 28.011 44.01 18.015 2.016 Ko -9.3039E6 -8.503.3E6 -8.3141 E6 -1.3624E7 -8.9503E6 -7.8613E6 K1 2.9672E4 2.7280E4 2.7460E4 4.1018E4 2.0781 E4 2.6210E4 K2 2.6865 3.1543 3.1722 7.2782 7.9577 2.3541 From a mass standpoint, these values are determined as follows: Hence the mass related values are: M Cv _ = w M Oxygen, 02: CP = 975 J/kgK Cv = 715 J/kgK Nitrogen, N2: CP = 1037 J/kgK Cv = 740 J/kgK K3 -2.1194E-4 -3.3052E-4 -3.3416E-4 -8.0848E-4 -7.2719E-4 -1.2113E-4 (2.1.35) For the mixture of oxygen and nitrogen which is air, the properties of air are given generally as: R air ~ 2/( e gas R gas) C Vair " X( E gas C Vgas) Yair " S C Pair - X( e gas C PgasJ 'gas gas 'gas; (2.1.36) Taking just one as a numeric example, the gas constant, R, which it will be noted is not temperature dependent, is found by: Rair = I Mgas) = ^ { ^ ) + 0 J 6 { ^ ) = 288 J / kgK U 1.999 V 28.011. 66
The other equations reveal for air at 293 K: Chapter 2 - Gas Flow through Two-Stroke Engines CP= 1022 J/kgK Cv = 734J/kgK y = 1.393 It will be seen that the value of the ratio of specific heats, y, is not precisely 1.4 at standard atmospheric conditions as stated earlier in Sec. 2.1.3. The reason is mostly due to the fact that air contains argon, which is not included in the above analysis and, as argon has a value of y of 1.667, the value deduced above is weighted downward arithmetically. The most important point to make is that these properties of air are a function of temperature, so if the above analysis is repeated at 500 and 1000 K the following answers are found: for air: T = 500K CP = 1061 J/kgK Cv = 773 J/kgK y = 1.373 T=1000K CP= 1143 J/kgK Cv = 855 J/kgK y= 1.337 As air can be found within an engine at these state conditions it is vital that any simulation takes these changes of property into account as they have a profound influence on the characteristics of unsteady gas flow. Exhaust gas Clearly exhaust gas has a quite different composition as a mixture of gases by comparison with air. Although this matter is discussed in much greater detail in Chapter 4, consider the simple and ideal case of stoichiometric combustion of octane with air. The chemical equation, which has a mass-based air-fuel ratio, AFR, of 15, is as follows: 2CoHig + 25 79 02 + — No 21 = 16C02 + 18H20 + 94.05N2 The volumetric concentrations of the exhaust gas can be found by noting that if the total moles are 128.05, then: 16 9 94 05 X>co = = 0.125 uH 0 = = 0.141 t>N = —•— = 0.734 2 128.05 2 128.05 2 128.05 This is precisely the same starting point as for the above analysis for air so the procedure is the same for the determination of all of the properties of exhaust gas which ensue from an ideal stoichiometric combustion. A full discussion of the composition of exhaust gas as a function of air-to-fuel ratio is in Chapter 4, Sec. 4.3.2, and an even more detailed debate is in the Appendices A4.1 and A4.2, on the changes to that composition, at any fueling level, as a function of temperature and pressure. In reality, even at stoichiometric combustion there would be some carbon monoxide in existence and minor traces of oxygen and hydrogen. If the mixture were progressively richer than stoichiometric, the exhaust gas would contain greater amounts of CO and a trace of H2 67
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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
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Chapter 4 Combustion in Two-Stroke
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Chapter 4 - Combustion in Two-Strok
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a stratilpl o tions to ivior. If m
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CD 30 -, W 20 - LU CO iS _J LU OC 1
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181 also x3 = — = 0.905 x6 = 2.17
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to be cur .3.20) .3.21) .3.22) for
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3.23) com- :on is hhas has a /eng-
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stroke .3.29) mean me to 3land ssio
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a LU z: en 3 CO a LU z EC D m O ho
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CD 1.2 -i 1.0 - di „„ LU 0.8 cc
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CO —> 111" cc LU < LU _J LU CC £
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Combustion in compression-ignition
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stroke ari more often lseful lental
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lat the ssi ;tribu- ELEVATION VIEWS
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CYLINDER HEAD TYPE IS CENTRAL BORE,
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CO E 50 40 - O O _i Hi > I CO 30 Z)
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CURRENT INPUT DATA FOR 2-STROKE 'SP
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Cylin- PP £ Pa- tics of Paper Chap
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vfitro- ;titi >tants 1970. i(In- Ch
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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 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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0.35 • 0.34 LU ° 0.33 - c 03 DC
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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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• • / . • 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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Simulation, engine See Computer mod
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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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