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