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DESCRIPTION
performance parametersTRANSCRIPT
Mechanical efficiency a factor called Indicated Power (ip) is considered. It is defined as the power developed by combustion of fuel in the combustion chamber. It is always more than brake power.The power developed by an engine and measured at the output shaft is called the Brake Power (BP)
BP(BRAKE POWER) =kw
Indicated power =Brake power +friction power
Mechanical efficiency =
So the mechanical efficiency is defined as ratio of brake power to the indicated power.
Volumetric Efficiency It is the ratio of the actual volume of the charge drawn in during the suction stroke to the swept volume of the piston. The amount of air taken inside the cylinder is dependent on thevolumetric efficiencyof an engine and hence puts a limit on the amount of fuel which can be efficiently burned and the power output. The value of volumetric efficiency of a normal engine lies between 70 to 80 percent, but for engines with forced induction it may be more than 100 percent.The engine output is limited by the maximum amount of air that can be taken in during the suction stroke, because only a certain amount of fuel can be burned effectively with a given quantity of air
Mean effective pressure Mean effective pressure is an important parameter for comparing the performance of different engines. It is defined as the average pressure acting overpiston throughout a power stroke.If mean effective pressure is based on brake power(bp) then it is referred to as brake mean effective pressure(bmep). If it is based on indicated power(ip) it is called indicated mean effective pressure(imep).imep is a fictitious constant pressure that would produce the same work per cycle if it acted on the piston during the power stroke.Frictionmean effective power is the difference of imep and bmep,
Indicated mean effective pressure =
Brake mean effective pressure =
Volumetric efficieny
In a four-stroke naturally aspirated engine, the theoretical maximumvolumeof air that each cylinder can ingest during the intake cycle is equal to the swept volume of that cylinder .Since each cylinder has one intake stroke every two revolutions of the crankshaft, then the theoretical maximum volume of air it can ingest during each rotation of the crankshaft is equal to one-half its displacement. The actual amount of air the engine ingests compared to the theoretical maximum is called volumetric efficiency (VE). An engine operating at 100% VE is ingesting its total displacement every two crankshaft revolutions.
Volumetric Efficiency =
Brake thermal efficiency =
Indicated thermal efficiency =
THERMAL ANALYSIS
FUEL CONSUMPTION AND AIR CONSUMPTION
AT 2500 RPM S.noSpeed(rpm)Load(kg)Fuel consumption(time in sec)Air consumption
H1H2H2-H1
12500023.3516215
22500511.9817247
32500109.21142410
42500156.76132613
52500205.6503030
62500254.9553530
AT 4000 RPM
S.noSpeed(rpm)Load(kg)Fuel consumption (time in sec)Air consumption
H1H2H2-H1
14000011.80266438
2400057.05347339
34000105.06367640
44000153.64408040
54000203.56418241
64000252.9489042
MODEL CALCULATIONS
AT 2500 RPM:At load =15kg and speed =2500rpm BP(BRAKE POWER) =kw = =15.41kwTotal fuel consumption (TFC) = TFC= =4.42 kg/hr Indicated power =Brake power +friction power =15.41+6.5 =21.91kw Indicated mean effective pressure = IMEP = =3.37MPa Brake mean effective pressure = BMEP = =1.48 MPa
Specific fuel consumption (SFC) = = =0.287 kg/Kwhr Mechanical efficiency = = =0.703 =70.3%Brake thermal efficiency = = =29.86%Indicated thermal efficiency = = =42.4% Volumetric Efficiency = where n = = = 0 .911=91.1%
For air fuel ratio: = =1000 kg/ = 13mm at 15 kg = = =1300 mm =13mts
Velocity = = =15.97Discharge Q= AVA= Area of orifice = = =1.809Mass of air =QQ= AV =1.809 =0.02889 Q= 0.02889 =0.02889 =0.02889 =103.68
=23.45
At load =15kg and Speed =4000rpm
BP =kw = =24.655kwTotal fuel consumption (TFC) = TFC= =8.208 kg/hr Indicated power =Brake power +friction power =24.655+15 =39.655kw Indicated mean effective pressure =IMEP = =3.813MPaBrake mean effective pressure =BMEP = = 2.37 MPa
Specific fuel consumption (SFC) = = =0.333 kg/Kwhr Mechanical efficiency = = =0.622 =62.2%Brake thermal efficiency = = =25.74%Indicated thermal efficiency = = =38.27% Volumetric Efficiency =
where n = = = 0.8391 = 83.91%
For air fuel ratio: = =1000 kg/ = 40mm at 15 kg = = =4000mm =40mts
Velocity = = =Discharge Q= AVA= Area of orifice = = =1.809Mass of air =QQ= AV =1.809 = Q= 0.0507 =0.0507 =0.0507 =182.52
=22.236
TABLES
CALCULATED PARAMETERS
AT 2500 RPM
Load(kg)Indicated power(kw)Brake power(kw)Total fuel consumption(kg/hr)Specific fuel consumption(kg/kwhr)Indicate mean effective pressure(Mpa)Brake mean effective pressure(Mpa)Air fuel ratio(a/f)
06.501.2791050.38
511.6365.1362.4940.4851.790.7930.6
1016.77310.2733.2440.3162.581.5828.07
1521.9115.414.420.2873.372.3723.45
2027.0420.5465.2880.2574.163.1629.8
2532.1825.6826.030.2744.953.9526.14
AT 4000 RPM
Load(kg)Indicated power(kw)Brake power(kw)Total fuel consumption(kg/hr)Specific fuel consumption(kg/kwhr)Indicate mean effective pressure(Mpa)Brake mean effective pressure(Mpa)Air fuel ratio(a/f)
01502.5321.44070.237
523.2188.2184.2380.5152.2330.7942.557
1031.43616.4365.9050.3593.0221.5830.848
1539.65524.6558.2080.333.8132.3722.19
2047.87532.8768.3930.2554.6033.1622.001
2556.0941.0910.8000.2525.3933.9517.3
EFFICIENCIES
Load(kg)Mechanical efficiencyBrake thermal efficiencyIndicated thermal efficiencyVolumetric efficiency
00043.556.52
At 2500 rpm
544.117.6539.966.94
1061.227.1444.379.89
1570.329.8642.491.11
2075.933.3343.8138.3
2579.836.545.74138.3
00050.7481.92
At 4000 rpm
535.416.6246.9583.084
1052.223.8545.6383.91
1562.225.7438.27783.91
2068.633.548.8985.07
2573.232.644.5286.06
GRAPHS
THERMAL DESIGN
INTRODUCTIONCritical TemperatureGases can be converted to liquids by compressing the gas at a suitable temperature.Gases become more difficult to liquefy as the temperature increases because the kinetic energies of the particles that make up the gas also increase.Thecritical temperatureof a substance is the temperature at and above which vapor of the substance cannot be liquefied, no matter how much pressure is applied.Every substance has a critical temperatureCritical PressureThecritical pressureof a substance is the pressure required to liquefy a gas at its critical temperature.
For a diesel cycle we have
(
At 2500 rpm , w =15 kgFor a diesel cycle we have = 300 =944.77k Also
( = 1 ( = 55.42 bar =55.42 bar THE ABOVE PRESSURE IS THE CRITICAL PRESSURE
= =51.6 kw= = 0.6192 KJ==1.47Kg= = =1374.86 K
= 944.77 + 1374.86 =2319.639 K
At 4000 rpm , w =15 kgFor a diesel cycle we have = 300 =944.77k Also
( = 1 ( = 55.42 bar =55.42 bar THE ABOVE PRESSURE IS THE CRITICAL PRESSUREQ =
Q = =95.784 KW= = 0.7183 KJ==1.47Kg
= = =1374.86 K
=2319.639K
CRITICAL TEMPERATURESS.noLoad(kg)Total heat supplied(Q)(Kw)Mass of the fuel (kg)Critical temperature (T3 K)
1549.447.41700.155
21068.911.2310-51987.28
31595.7841.7110-52393.68
42098.1281.810-52405.33
525126.0432. 2510-52802.975
S.noLoad(kg)Total heat supplied(Q)(Kw)Mass of the fuel (kg)Critical temperature (T3 K)
1529.0998.311995.86
21037.851.0810-52091.5
31551.61.4710-52319.639
42061.641.7610-52024.25
52570.362. 0110-52174.71
MECHANICAL DESIGN
CYLINDER:MATERIAL: High Grade Cast IronD=Bore Diameter L=Stroke Length = 1.4D D= = =72.9 mmL=102.06 mm
CYLINDER HEADMATERIAL:HIGH GRADE CAST IRON
= Thickness of Cylinder Head = 0.5 x 72.9 x = 14.5 mm = 1.45 cm
PISTONMATERIAL: ALUMINIUM ALLOY
= 70 For aluminium alloy = = 8.88 mm
PISTON RINGSMATERIAL: FINE GRAINED ALLOY CAST IRONTHICKNESS OF RIB = (0.3 0.5) = 0.4 D = 0.4 = 3.552 mm
Axial thickness of piston rings = where
= = 2.43 mm
Radial Thickness of Piston Ring = 0.045D = 0.045 = 3.28 mm
GUDGEON PINMATERILAL: CASE HARDENED STEEL
Length = 0.45D = 0.45 = 32.8 mmMaximum Load ( F )= AF = =23148.603 N = = = 210941.65 N-mmFor diameter of Gudgeon pinWe know
= = 16.48 mm
CONNECTING RODMATERIAL : ALLOY STEEL MAX LOAD(F) = 23148.603 NAREA OF CROSS SECTION = = ALLOWABLE ULTIMATE STRESS OF ALLOY STEEL = 1425 MPa
= t =1.215=flange thicknessWidth of flange=4t = 4.86 mmDepth of Web =5t = 6.075 mm