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Unit IV
Testing of IC Engines&
Supercharging
Syllabus : IC Engines
UNIT-IV : TESTING OF IC ENGINESObjective of testing, various performance parameters for IC engines - indicated power, brake power, friction power, SFC, AF ratio etc, methods to determine various performance parameters, characteristic curves, heat balance sheet
SUPERCHARGINGSupercharging and turbo-charging methods and their limitations
Lecture No 23
Learning Objectives: • To understand objectives of engine testing• To learn about engine performance parameters
Objectives of Testing• Engine performance during development
• Engine performance after development/ sample testing after production by manufacturers
• Engine performance testing by Govt Testing Agencies for certification like ARAI
Objectives of Testing• Whether engine is performing as per design?
Indicated Power (IP) Brake Power (BP) Frictional Power (FP) Mechanical Efficiency Thermal Efficiency Brake Specific Fuel Consumption (bsfc) Heat Balance Sheet
• Whether engine is meeting emission norms? CO HC NOx PM Soot
Objectives of Testing• Leakage in engine?
• Smooth running of engine?
• Engine performance after overhaul/repairs?
Performance Parameters
Brake Power (BP)Power available at output shaft/ crank shaft
Mechanical Losses/Frictional Power (FP)Sum of frictional losses and pumping losses incl power required to operate engine accessories like water pump, dynamo etc
Indicated Power (IP)/ Theoretical PowerPower produced within eng cylinderIP = BP + FP
Tests Performed on IC Engines
1. Indicated Power (IP)2. Brake Power (BP)3. Frictional Power (FP)4. Mechanical Efficiency5. Air-Fuel Ratio6. Brake Specific Fuel Consumption (bsfc)7. Thermal Efficiency8. Working out Heat Balance Sheet
Engine/ Mechanical IndicatorIndicator PaperWrapped Drum
p-V diagramStylus
Weight
Coupling Nut
To Combustion Chamber
Piston
Rope connectedTo Piston Rod
Pulleys
Measurement of IP on Mech/Eng Indicator• To determine IP, p-V diagram is required, the area of which represents work developed by engine per cycle
• Apparatus used for drawing actual p-V diagram is called Mechanical/ Engine Indicator• Eng indicator consists of a cylinder, piston, piston rod coupling nut, straight line linkage with stylus, spring of required stiffness, indicator card wrapped drum, pulley, rope and weights . • Vertical movement of stylus and horizontal movement of the cord combines to produce a closed figure called Indicator diagram • Area of indicator diagram can be measured by Planimeter to a definite scale giving work developed
Mean Effective PressureIndicated Mean Effective Pressure (imep)• imep is the average pressure, which if acted over the entire stroke length, would produce the same work done by the piston as is actually produced by the engine during a cycle • Let ‘a’ be the net area of indicator diagram (cm2) ‘l’ be length of diagram (cm) and ‘k’ be spring stiffness N/cm2/cm
Hence, mean height of diagram = a/l
2/)( cmNxkl
aimeppm
Indicated Power (IP)Let pm= imep (N/cm2);
A= Piston top area (cm2)=Лd2/4;
L= Length of stroke (m)
n= Power stroke /min(=N/2 for 4 S eng as one power stroke per 2 rev & N for 2S eng)
N= RPM
Hence, Force on Piston=
WD per Cycle=
Pm x A (Newton)
pm x A x L (Nm)
Hence, IP = pm x A x L x n (Nm/min)/Cycle/Cylinder
)//(60
cylindersNmLAnpm
)(000,60
kWCylindersofNoxLAnp
IP m
Lecture No 24
Learning Objectives: • To learn working out/ measurement of Brake Power (BP) & Friction Power (FP)
Measurement of BP1. Rope Brake Friction Dynamometer
Spring Balance
Flywheel/Brake Drum
Rope
Weight
S
W
Rope Brake Friction Dynamometer (Contd.)
Let W=Dead Weight (mg) in Newton (N) S=Spring Balance Reading (N) Rb=Radius of Brake Drum (D+d)/2 (m) D=Brake Drum dia and ‘d’ rope dia N=Engine RPM
Hence, net Brake Load= (W – S)
Braking Torque = (W-S) x Rb
Hence, Brake Power (BP) =
)(000,60
2kW
NxxRSW b
Brake Mean Effective Pressure (bmep)
)(000,60
kWCylindersofxNoLAnp
BP bmep
2. Prony Brake Dynamometer
BrakeShoes
Flywheel/Brake Drum
Length, L
Weight W
LoadArm
Prony Brake Dynamometer (Contd.)Let W (=mg) be the weight (N)
Let L be the distance from centre of brake drum tohanger, called load arm (m)
Then, Torque=W x L (Nm)
)(
000,60
2
,
kWNLxW
BP
Hence
60
2&
N
xTBP
Frictional Power (FP)• Difference between IP and BP is called FP
• FP includes:- Pumping losses due to intake & exhaust processes
- Frictional losses in bearings, rotary/sliding parts- Power required to drive auxiliaries like governor, water, lub oil, fuel pumps, alternator/dynamo, valve operating mechanism etc
• FP increases as square of N but practically FP ∞ N1.6
• Higher FP results in: - Reduced power output - Decreased mech efficiency - Increased bsfc - Increased requirement of cooling
Methods of Measurement of FP
1. By measurement of IP and BP
2. Willan’s Line Method
3. Morse Test
4. Motoring Test
FP by Willan’s Line Method( Fuel Rate Extrapolation Method )
- 8 - 4 0 4 8 12 16 20
4
3
2
1
Fuel FlowRate (kg/h)
BP (kW)
A
At Constant Eng Speed, say 1500 RPM
FP by Willan’s Line Method• A graph between fuel consumption rate (kg/h) taken on y-axis and BP (kW) on x-axis is drawn, while engine is made to run at some constant speed, say 1500 RPM
• The graph is extrapolated back to zero fuel consumption, which cuts on –ve x-axis at point ‘A’
• The –ve intercept on x-axis represents FP at that speed of the engine
• Although when BP=0, some fuel consumption is there. This fuel is consumed to overcome engine friction
• Only for CI engine to be run at constant speed as Fuel consumption rate v/s BP plot is almost straight line in case of diesel engine, hence can be extrapolated
FP by Morse Test• Morse Test can be used for determining FP/IP of multi-cylinder IC engines, generally 3 cyl and more by cutting off each cylinder in turn
• In SI engines, each cylinder is rendered in-operative by short-circuiting the SP or cutting off fuel supply in MPFI systems. In CI engines, fuel supply is cut off
• Consider 4 stroke, 4 cylinder SI engine coupled with dynamometer
• Engine is run at constant speed N throughout one set of test parameters, as FP ∞ N2
• It is assumed that pumping & mech losses are same whether a cylinder is working or not• Throttle position is kept fixed, however, to attain same speed N, load is decreased by dynamometer
FP by Morse Test• Let B=BP of eng when all cylinders are working
B1=BP of eng when Cylinder No 1 is cut off
Similarly, B2=BP of eng when Cylinder No 2 is cut off
• Let I1, I2, I3 & I4 be the IPs developed by Cylinder Nos 1, 2, 3 & 4 respectively and their corresponding FPs be F1, F2, F3 & F4
B3=BP of eng when Cylinder No 3 is cut off
B4=BP of eng when Cylinder No 4 is cut off
• Total BP(B) = (I1+I2+I3+I4) - (F1+F2+F3+F4)
BP(B) = (I1+I2+I3+I4) - (F1+F2+F3+F4)
FP by Morse Test
Hence, B1=(I2+I3+I4) – (F1+F2+F3+F4)
On subtracting; B – B1 = I1Similarly, B – B2 = I2 B – B3 = I3 B – B4 = I4
On adding; IP = I1 + I2 + I3 + I4
= 4B – (B1+B2+B3+B4)
B, B1, B2, B3 & B4 can be measured by Dynamometer,Hence IP can be calculated
Therefore, FP = IP - BP
Lecture No 25
Learning Objectives: • To understand working out of heat balance sheet• To learn measurement of air/fuel consumption
Theoretical/ Air Std Efficiencies
Otto Cycle: 1
11
r
Diesel Cycle:
1
111
1
r
Dual Cycle:
1.1
1.11
1
r
Some Definitions
Thermal Efficiencies:
i) Indicated Thermal EfficiencyCVxm
IP
fi
ii) Brake/Overall Thermal Efficiency CVxm
BP
fb
- Where mf is fuel consumed in kg/sec- CV is Calorific Value of fuel in kJ/kg- IP/BP is in kW
Some Definitions
Mechanical Efficiency:
100xIP
BPmech
Relative Efficiency : Defined as the ratio of Brake Thermal Efficiency to Air Standard Efficiency at sameCompression Ratio(CR)
100xa
br
Some DefinitionsVolumetric Efficiency:
TempAtmatvolswepttoingcorrespondechofMass
inductedechactualofMassv &Prarg
arg
• Reduced Volumetric Efficiency causes reduction in Power Output
VolumeSwept
conditionssuctionatinhaledeChofVolActualAlso v
arg,
• Ratio of actual mass of charge inducted during suction stroke to mass of charge corresponding to swept volume of the engine at atm pr & temp
• Volumetric Efficiency puts a limit on the amt of fuel that can be burnt and hence on its power, since output of eng depends on amt of air inducted
Some Definitions(Brake) Specific Fuel Consumption (bsfc/sfc):
)/(3600
kWhkgBP
mbsfc xf
Specific Output: • BP per unit of piston displacement
AxL
BPOutputSpecific
• bsfc is defined as the amount of fuel required to be supplied to eng to develop 1kW of power per hour at crankshaft
Heat Balance Sheet• Heat Balance Sheet is an account of heat released on combustion of fuel in the combustion chamber and its utilization in the engine
• To draw heat balance sheet, tests are carried out on engine, while it is run at some constant speed
Heat Supplied:
Heat Supplied = mf x CV (kJ/min)
Where mf = mass flow rate of fuel (kg/min)
CV = Calorific Value of fuel (kJ/kg)
Heat Balance SheetHeat Expenditure/Utilization:
1) Heat Equivalent to BP:
3) Heat carried away by Exhaust gases
Heat carried by exh gases=mgCpg(Tge – Tsa) kJ/min
Where mg=(ma+mf) =flue gases flow rate (kg/min)
Heat Equivalent to BP = BP x 60 (kJ/min)
2) Heat Rejected to Cooling Water:
Heat carried away by water =mwCpw(Two – Twi) kJ/min
Where mw=cooling water circulation kg/min& Cpw=4.187 kJ/kgK
4) Unaccounted Heat:
By difference
Heat Balance SheetHeat
SuppliedkJ/
min% Heat Utilization kJ/
min%
Heat Suppliedby comb ofFuel=mf x CV
100
Total 100
a) Heat to BP=BPx60
b) Heat to water =mwxCpw(Two – Twi)
c) Heat carried away by exhaust gases=mgxCpg(Tge – Tsa)
d) Heat Unaccounted(By difference)
100
100cc 200cc
3-Way Cock
3-Way Cock
Fuel to Eng
Fuel from Tank
Start
Stop
Start
Stop
Volumetric Fuel Flow Meter(Burette Type)
Fuel Measurement• Time required to supply given volume of fuel is noted
• Mass Flow Rate of Fuel Supply:
FuelofxDensityTime
Volumem f
• Density of Fuel = Sp Gravity of fuel x Density of water
• This method does not give very accurate mass flow rate due to variation in density with temp
WaterofDensity
materialofDensityGravitySpecific
Gravimetric Fuel Flow Meter
Weighing Machine
Flask
Fuel to Engine
Fuel Tank
Valves
A
B
Air Flow Meter
ΔH
Thermometer
Orifice Plate (A, Cd)
Manometer
Air SurgeTank
Air Intake to Eng
Measurement of Air Consumption by Air Flowmeter
• Surge tank is connected to intake side of the engine
• Manometer measures the pressure difference
• Vol Flow Rate wd HgxAxC .2
-Cd – Coeff of discharge for given orifice-A – Orifice cross sectional area-ΔHw-Head of water to be converted to air head
Lecture No 26
Learning Objectives: • To learn about engine characteristic curves
SI Engine Characteristic Curves
POWER(kW)
Speed
IP
BP
FP
SI Engine Characteristic Curves• Lab tests carried out to determine eng performance
• During tests, throttle is kept full (full /rated load, max fuel consumption) and speed is varied by adjusting the brake load
• IP, BP, FP, bsfc, mechanical & volumetric efficiencies etc are worked out
• Same tests can be repeated at half load
• At rated output, max p-V diagram area, hence max imep; For given torque; power ∞ N
kWNTx
hp000,60
2
SI Engine Characteristic Curves• IP increases when imep or speed or both increase
• IP initially increases faster with speed, if inlet conditions are kept constant
• However, after certain limit, rate of increase of IP reduces with speed due to reduction in vol efficiency as air/charge velocity increase results in inlet pr drop
• Mech losses increase with increase in speed(FP∞ N2) due to which increase in IP is off-set by steep increase in FP
SI Engine Characteristic Curves
IPBPbsfcMech Eff
Speed
IP
BP
bsfc
Mech Efficiency
x
SI Engine Characteristic Curves• As FP ∞ N2, mech efficiency reduces due to steep increase in FP
• At lower speeds, due to lower charge velocity because of low piston speed, bsfc reduces since volumetric efficiency increases and mech efficiency also increases
• After certain speed, bsfc increases due to reduction in volumetric efficiency and increase in mech losses
x• Point x represents economical speed of eng for min fuel consumption
SI Engine Characteristic Curve
Vol Efficiency
Speed
SI Engine Characteristic Curve• Volumetric Efficiency reduces with increase in speed due to increase in intake velocity resulting in drop of suction pressure
• Higher the speed, lesser the time available for induction of charge
• Suction valve fully opens only when pressure inside cylinder slightly below the surrounding pressure, thus reducing effective suction stroke
CI Engine Characteristic Curves
Power
bsfc
Speed
BP
bsfc
CI Engine Characteristic Curves• IP and BP increase with speed but due to steep increase in FP, IP and BP start coming down
• For bsfc curve, same reasons as in SI engine
Engine Characteristic Curves
bsfc
BP
SI
CI
CI Engine Characteristic Curve
Brake/OverallEfficiency
A/F Ratio
Stoichiometric Mixture
LeanMixture
RichMixture
Lecture No 27
Learning Objectives: • To understand working out of engine parameters through numerical problems
Q1. Obtain cylinder dimensions of a twin-cylinder, 2-SIC engine from the following data:Engine speed=4000RPM; Volumetric efficiency=77%;Mech Efficiency=75%; Fuel consumption=10 lit/hr;Sp. Gr. Of fuel=0.73; A/F ratio=18; Piston speed= 600m/min; imep=5 bar.Also, determine power output at STP conditions(p=101325 N/m2; Ta=25˚C; R for air=0.287 kJ/kgK)
Solution:
Cylinder Dimensions=? D & L
Piston speed=2LN Since speed & N are given, L=?
Now, to find out D=? CylinderstrokeoneforLDVs /4
2
engineSforNSforNStrokesPower
strokespowerofxNocylofxNoVsRateFlowVolTotal
42/&2
min/
Solution(Contd):
vv
VaVs
Vs
Va
TaRmaVapaVaRateFlowVolactualFor ...
)(18/ givenRatioFAm
mmaFor
f
a
ff xh
mx
h
litm
331010
wxfuelofgrSpxh
mx .1010
33
hkg
hkgxxx
/3.7
/100073.01010 3
Solution(Contd):
hmVa
Vs
Vs
VaSince
v
v
/14477.0
9.110 3
hm
xxVaRateFlowHence
TaRmaVapa
/9.110
101325
273252874.131,
...
3
hkghkgxmm
ma
f
a /4.131/3.71818
Solution(Contd):
22
600
4min60
144 23
xNxN
xDm
CylindersxNxLDVsRateFlowVol 24
2
232 10094.5 mxD
mmmD 4.710714.0
mm
mx
LAnd
75
075.040002
600
Q2. A 6 cylinder gasoline engine operates on 4 strokecycle. Bore of cylinder is 80mm and stroke 100mm. Clearance volume per cylinder is 70CC. At 4000 RPM,Fuel consumption is 20kg/hr and the torque developed is 150Nm. Calculate:-(a)BP (b) Brake mean effective pressure (c) Brakethermal efficiencyIf CV of the fuel is 43000kJ/kg, find relative efficiencyon brake power basis, assuming engine works on Constant volume cycle and gamma for air =1.4.
Solution:
kWNTx
BP000,60
2 kW
xx83.62
60000
40002150
?bmep
cylidersofnoxAxLxnxbmep
BP60000
barmNx
xxxxbmep
25.6/1025.6
660000
24000
1.008.0483.62
25
2
CVxm
BP
fb
;a
br
%3.26263.043000
360020
83.62
x
1
11
ra
;c
cs
V
VVr
CCxVs 6.502108.0
42
;18.870
706.502
r 5686.0
18.8
11
14.1 a
%25.464625.05686.0
263.0r
Lecture No 28
Learning Objectives: • To understand working out of engine parameters and heat balance sheet through numerical problems
Q3. During trial of a single cylinder, 4 stroke oil enginethe following results were obtained:Cyl bore=200mm, Stroke=400mm, mep=6 bar,Torque=407Nm, speed=250 RPM, Oil consumption=4kg/hr, CV of fuel=43MJ/kg, Cooling water rate=4.5kg/min, Air used per kg of fuel=30kg, Rise in cooling water temp=45°C, Temp of Exhaust gases=420°C, Room temp=20°C, mean sp. heat of exhaust gases=1kJ/kgK, Sp. Heat of water=4.18kJ/kgK, Barometric pressure=1.01325 barFind IP, BP and draw up heat balance sheet in kJ/hr.
Solution:
)(000,60
kWcylofnoxAxLxnximep
IP
60000
2 NxTBP
kWx
xxxxIP 7.151
000,602250
4.02.04
106 25
kWxx
65.1060000
2502407
Heat Balance Sheet
1. Heat supplied by fuel to eng =mfxCV
=4x43000kJ/hr=172,000kJ/h
2. Heat utilized
(ii) Heat to cooling water=mw x Cpw x ∆Tw
=4.5x60x4.18x45 = 50,787 kJ/h
(iii) Heat to exhaust gases=mg x Cpg x (Te-Ta )
To find mg : ma=30kg/kg of fuel; hence mg=31kg
Since fuel consumption is 4kg/h; mg=31x4kg/h
Hence, Heat to exhaust gases=31x4x1(420-20)=49,600kJ/h
(iv) Unaccounted Heat=33,255kJ/h (by difference)
(i) Heat to Power Output=BPx3600kJ/h
=10.65x3600=38,358kJ/h
Heat Supplied
kJ/hr % Heat Utilized kJ/hr %
Heat supplied by fuel= mfxCV
172,000 100 To BP= BPx3600
38,358 22.3
To Cooling Water =mw.Cpw.∆Tw
50,787 29.5
To exhaust gases =mg.Cpg.∆Tg
49,600 28.8
Unaccounted Heat
33,255 19.3
Total 172,000 100 Total 172,000 100
Heat Balance Sheet
Q4. During a test on a 4 stroke oil engine, the following data were obtained:Mean height of indicator diagram = 21mmIndicator spring number/stiffness=27kN/m2/mmSwept volume=14 lit, effective brake load=77kg, Effective brake radius= 0.7m, speed=6.6 rev/s, fuel consumption=0.002kg/s, CV of fuel=44MJ/kg, Cooling water rate=0.15kg/s, water inlet temp=38°C,cooling water outlet temp=71°C, Sp. heat of water=4.18kJ/kgK, energy carried by exhaust gases=33.6kJ/sDetermine IP, BP and mech efficiency and draw up heat balance sheet in kJ/s and %.
Solution:
)(000,60
kWcylofnoxAxLxnximep
IP
)( kWcylofnoxnxLxAximepIP
mmmkNxmmkxl
aimep ./27)(21 2
22 /567./27)(21 mkNmmmkNxmm
kWmxlmkNIP 2.262
6.6).(10)(14)./(567 332
000,1
2 NxRSWBP b
kW
xxxx
92.21
000,1
6.627.081.977
%65.831002.26
92.21 xmech
Heat Supplied
kJ/s % Heat Utilized kJ/s %
Heat supplied by fuel=mf x CV =0.002x44000(=88kJ/s)
88 100 To BP 21.92 24.9
To Cooling Water =mw.Cpw.∆Tw
0.15x4.18x (71-38)
20.69 23.5
To exhaust gases (given)
33.6 38.2
Unaccounted Heat
11.79 13.4
Total 88 100 Total 88 100
Heat Balance Sheet
Q5. A 4 cylinder 4 stroke SI engine has a bore of5.7cm and stroke 9cm. Its rated speed is 2800 RPM and it is tested at this speed against a brake which has a torque arm of 356mm. The net brake load is 155N and fuel consumption is 6.74 lit/hr. Sp gravity ofpetrol is 0.735 and CV is 44200kJ/kg. A Morse test iscarried out and cylinders are cut off in order of 1, 2, 3 & 4 with corresponding brake loads of 111, 106.5, 104.2 and 111N. Determine engine torque, bmep,brake thermal efficiency, sfc, mech efficiency and imep.Solution:
Engine torque= (W-S).Rb=WxL
= 155x0.356=55.18Nm
)(60000
2
000,60kW
NTxcylofnox
AxLxnxbmepBP
kWxxxxbmep
17.164000,60
22800
09.0)057.0(4
2
kWx
BP 17.1660000
2800.218.55
xCVm
BP
fb
barbmep 55.7
skgx
xxmxhlitm f
/10376.1
3600
1000735.0)(10)/(74.6
3
33
kWhkg
xx
BP
mbsfc f
/306.017.16
360010376.1 3
%58.262658.0
4420010376.1
17.163
or
xxxCVm
BP
fb
We know that IP = 4BP - (BP1+BP2+BP3+BP4)
60000
2 NxWxRBP b
60000
24 4321
NxRWWWWWIP b
60000
28002356.01112.1045.1061111554
xxxIP
kW54.19
%75.8210054.19
17.16 x
IP
BPm
8275.055.7
.
.
imepLAnimep
LAnbmepm
barimep 12.9
Q6. During trial of a 4 cylinder 4 stroke SI enginerunning at 50 rev/s, the brake load was 267N when all cylinders were working. When each cyl was cut off in turn and speed returned to same 50 rev/s, brake readings were 178N, 187N, 182N and 182N. Determine BP, IP and mech efficiency of the engine.For brake, BP=F.N/455(kW), where F is brake load in Newtons and N rev/s. The following results were obtained: Fuel consumption=0.568lit in 30 seconds, SG of fuel=0.72, CV=43000kJ/kg, A/F ratio=14:1,Exh temp=760°C, Cpg=1.015kJ/kg, Water inlet temp=18°C and outlet temp=56°C, water flow rate=0.28kg/s,Ambient temp=21°C. Draw heat balance sheet in kJ/s.Solution:
kWxNF
BP 34.29455
50267
455
.
We know that IP = 4BP - (BP1+BP2+BP3+BP4)
kWxNF
BP 56.19455
50178
455
.1
kWxNF
BP 55.20455
50187
455
.2
43 20455
50182
455
.BPkW
xNFBP
Therefore IP= 4x29.34 - (19.56+20.55+20+20) =37.25kW
%76.7810025.37
34.29 x
IP
BPmech
Heat utilized
(ii) Heat to cooling water=mw x Cpw x ∆Tw
=0.28x4.187x(56-18) =44.55 kJ/s (7.6%)(iii) Heat to exhaust gases=mg x Cpg x (Te-Ta )
To find mg :(ma +1)xmf=(14+1)x0.01363=0.204kg/s
Hence, Heat to exhaust gases=0.204x1.015(760-21)=153kJ/s (26.1%)
(iv) Unaccounted Heat=356kJ/s (61%) (by difference)
(i) Heat to BP=BP= 29.34kJ/s (5%)
Heat Balance SheetHeat supplied =mfxCV
skJx
xxxx
/2.5864300001363.0
4300030
100072.010568.0 3
Supercharging&
Turbocharging
Syllabus : IC Engines
SUPERCHARGINGSupercharging and turbo-charging methods and their limitations
Lecture No 29
Learning Objectives: • To learn effects of supercharging /turbo-charging and limitations
• Increasing Eng speed (BP= T x 2πN) (FP ∞ N2 & Volumetric η ↓ )
How can engine power be increased?
• Higher CR (Peak Pr increases; Thermal Load increases; Weight to Power ratio increases) (HUCR limited due to knocking/detonation in SI engines and heat load in CI engines)
• Utilization of exh energy in gas turbine, thus ↑ BP
• Use of 2-stroke cycle; but cooling, emission problems, lower volumetric & thermal efficiency
• Increasing charge density by - Lowering charge temp (Cooling) and / or - Increasing induction pressure
Objectives of Supercharging • To increase power output
• To increase power to weight ratio
• To compensate loss of power at high altitude
• To reduce bulk size of engine
Supercharging• Supplying air /Air-Fuel mixture at higher pressure than the pressure, at which the engine naturally aspirates, by a boosting device is called supercharging
• The device which boosts the pressure is called supercharger.
• Purpose of supercharging to have small displacement engines but developing more power and to meet emission legislation on fuel consumption for emission control
• More power is achieved by raising density of charge, thus more mass of air making available more oxygen for combustion
Supercharging & Turbo Charging Systems
Effects of Supercharging • Increased Eng Output (p-V diagrams)
• Turbulence Effect (Higher BP)
• Power required to drive supercharger, thus ↓ BP
• Mech Efficiency increases
• SI Engines→ Knocking tendency as ign delay ↓
• bsfc ↑ for SI (due to reduced delay) but ↓ for CI engines due to better combustion & higher mechanical efficiency
• Better scavenging; Increase in power output
• For CI Engines→ Smoother running, low F/A ratio, ↑ durability & reliability and lower bsfc
Effects of Supercharging • Better atomization
• Better mixing of air and fuel
• Reduced exhaust smoke
• Better torque characteristic over whole speed range
• Increased gas load
• Better and smoother combustion
• Increased thermal stresses
• Increased valve overlap period of 60° to 160° of crank angle
• Increased cooling requirements of piston and valves
-(b)
p
patm
V
+(a)
71
5
62
4
3
+(d)
+(c)
patm
p
V
4
3
71
5
62
Naturally Aspirated Engine Supercharged Engine
P-V Diagrams of Naturally Aspirated &Supercharged Engines
Limitations of Supercharging• Power o/p limited by knock, thermal & mech loads
• For SI engines, knocking reached earlier
• In CI Engs, thermal & mech loads reached earlier
• Increase in peak pr, increases bearing loads
• Increase in intake pr increases peak pr leading to increase in weight of cylinder (limitation on peak pr)
• Increase in peak pr→ ↑ tendency to detonate (SI)
• Increase in peak pr increases friction losses
• ↑ peak pr → ↑ peak T →Reqmt of better cooling sys
• ↑T → ↑ exh gas temp →overheating of exh valves
Due to the above reasons, supercharging generally limited to 2.5 bar
Limitations of Supercharging In SI Engs.• Detonation is the limitation as it increases with ↑ pr, ↑ T, ↑ CR, ↑ density of charge
• Strongest detonation at stoichiometric A/F ratio
• CR limited due to detonation for given Octane Rating of fuel used
• Detonation can be reduced by reducing CR but BP & thermal efficiency decreases & bsfc increases
For the above reasons, SI Engines are normally NOT supercharged except for aircraft, high altitude compensation or higher power of aero engines required at the time of take off of aircraft at the expense of higher fuel consumption
Limitations of Supercharging In CI Engs.
• Increased induction pr helps in suppressing knocking tendency, improve combustion, higher power output & thermal efficiency and hence can use lower Cetane fuels
• Supercharging is limited by: - peak pressure →mechanical loading - peak temp →thermal loading - thermal stresses developed - mean temp of cylinder walls - loads on bearings
Lecture No 30
Learning Objectives: • To understand methods of supercharging• To learn various methods of turbocharging
Types of Superchargers• Centrifugal Type Supercharger
• Root’s Type Supercharger
• Vane Type Supercharger
Centrifugal Type Supercharger
Root’s Type Supercharger
Vane Type Supercharger
Vane Type Supercharger
Arrangements of Supercharging
Engine Load
Compressor
Air inlet toCompressor
Gears
Exhaust from Engine
Inlet to EngineAir outlet fromCompressor
After Cooler
Arrangements of Turbocharging
Engine Load
Compressor
Air inlet toCompressor
Exhaust from Engine
Inlet to Engine
Air outlet fromCompressor
After CoolerTurbine
Exhaust from Turbine
Method of Super/Turbocharging
EngineLoad
Compressor
Air inlet toCompressor
Exhaust from Engine
Inlet to Engine
Air outlet fromCompressor
After Cooler
Turbine
Exhaust from Turbine
Method of Super/Turbocharging
Engine Load
Compressor
Air inlet toCompressor
Gears
Exhaust from Engine
Inlet to Engine
Air outlet fromCompressor
After Cooler
Exhaust from Turbine
Turbine Load
Turbochargers
• Exhaust gases carry about 1/3 of the total energy generated in the eng cylinder
• In order to utilize this energy, hot gases can be allowed to expand further in a gas turbine and its work output can be utilized to drive a supercharger. This system of supercharger coupled to Turbine is called Turbocharger
• Due to cyclic fluctuations of the pressure in exhaust pipe, turbo charging is not employed in single cylinder eng, however, system is suitable for engines having 4 or more cylinders
Turbo Charger
• Turbocharger does not consume eng power• No gearing required between turbine and compressor as both are connected by single shaft
• Gain in power at nominal cost
• Exhaust energy, which is 1/3 of total energy generated in the engine, is gainfully utilized
• Exhaust noise level reduces
• Suitable for high speed engines
Advantages
Disadvantages• Increase in fuel consumption at low power output
• Total cost of unit increases
Blowers.rar
End of Unit - IV