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BDA 3043

APPLIED THERMODYNAMIC

CHAPTER 5INTERNAL COMBUSTION

ENGINES

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5.1 INTRODUCTIONOne of the most significant inventions of the 20th century is the internal combustion (IC) engineDefinition

An engine in which the chemical energy of the fuel is released inside the engine and used directly for mechanical work E : Exhaust cam shaft

I : Intake Cam shaft

S : Spark plug

V : Inlet and exhaust valve

W : Water Jacket for cooling flow

P : Piston

R : Connecting Rod

C : Crank shaft

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IC engines use reciprocating piston in a cylinder (block)The piston operates between the “top dead center” (TDC) and the “bottom dead center” (BDC)Valves are used to control the flow of gas into and out of engineStroke is the largest distance the piston travelsBore is the diameter of the pistonOther components are piston, block, crankshaft, connecting rod etc.

tdc

bdc

bore

stroke

l

aθ

l = connecting rod

a = crank shaft

θ = crank angle

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5.2 TYPES & CLASSIFICATIONS OF IC ENGINES

IC engine can be classified according to:

applicationsAutomobile, truck, locomotive, light aircraft, marine, portable, power system etc

basic engine designReciprocating engine, rotary engine

no of cylinders1, 2, 3, 4, 5, 6, 8, 10, 12 etc.

arrangement of cylinderIn-line, V-type, opposed, radial

working cycle4-stroke, 2-stroke

fuelGasoline, diesel, nitro methane, alcohol, natural gas, hydrogen etc

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ENGINE DESIGN & CYLINDER ARRANGEMENT

V-type, 6 cylinder

(V6)

Inline, 4-cylinder

(Straight 4)

6

Opposed, 4-cylinder

(Flat 4)

ENGINE DESIGN & CYLINDER ARRANGEMENT

Rotary egine

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4-STROKE ENGINE4-Stroke

1. Requires 4 stroke of piston to complete a cycle

1-2 Induction strokeInlet valve open. Exhaust valve is closed. BDC to TDC. Air + fuel is induced.

2-3 Compression strokeAir + fuel is compressed to TDC. Spark occurred at S and combustion occurs mainly at constant volume. Large increase in pressure and temperature.

3-4 Working strokeHot gas expand pushing the piston down to BDC. Exhaust valve open at E to assist exhaustion. Inlet valve is still closed.

4-1 Exhaust stroke

The gas is force to exit the cylinder. Piston moved to TDC. Inlet valve is still closed.

2. 2 revolution of crank shaft per cycle

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START INTAKE COMPRESSION SPARK

POWEREXHAUST

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2-Stroke1. Requires 2 stroke of piston to complete a cycle

First stroke : BDC – TDC (Both compression and induction stroke)

As piston ascends on the compression stroke, the next charge is drawn into crankcase C as the spring loaded valve, S open automatically. Ignition occur before TDC. Both transfer and exhaust port is uncovered.

Second stroke: TDC – BDC ( Both working and exhaust stroke)

At TDC working stroke begin. As the piston descend through about 80%, the exhaust port is uncovered and exhaust begin. The transfer port is uncovered later due to the shape of the piston and the position of the port. The descending piston push the air to enter the cylinder through the transfer port.

2. 1 revolution of crank shaft per cycle

3. Less efficient compared to 4 stroke

4. High power-to-weight ratio

5. Suitable for small applications

2-STROKE ENGINE

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5.3 THE AIR STANDARD CYCLES

Before we could discuss in depth about IC engines, let us look at several types air standard cycles.We will discuss three standard cycles :

Otto cycleDiesel cycleDual combustion cycle

The air standards cycles are ideal cycles used as a yardstick for the actual cycles.There are few assumptions applied to the cycles:

Working fluid is air behaving as ideal gasAll process in the cycle are internally reversibleCombustion process is replaced by a heat addition process from an external sourceExhaust process is replaced by a heat rejection processNo chemical reaction

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OTTO CYCLE

The Otto cycle is the ideal cycle for the petrol engine, the gas engine and the high speed oil engine.

1-2: Isentropic compression2-3: Reversible constant volume

heating3-4: Isentropic expansion4-1: Reversible constant volume

cooling

2

1

4

3P

V

CPV =γ

CPV =γ

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OTTO CYCLE – ENERGY BALANCE

( )

( )

( ) 112

1

1

2

1

2

1

1

2

12

2

1

2

1

1

2

1212

Also,

ratio ncompressio

process, isentropicFor

Law1st Fromncompressio Isentropic :21

−

−−

Γ=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

Γ=

=Γ=+

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−=−=

−

γ

γγγ

γ

γ

v

v

vc

cs

V

TT

pP

VV

TT

PP

VVV

VV

VV

PP

TTmCWδWδQΔU

0

2

1

4

3

P

VVc Vs

CPV =γ

CPV =γ

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OTTO CYCLE – ENERGY BALANCE

( )

( )

( )1441

3

4

11

2

1

3

4

2323

involved is work Nocooling Isometric :14

1

expansion Isentropic :4-3

involved is work Noheeting Isometric :3-2

TTmCQQ

PP

VV

TT

TTmCQQ

Vout

vv

v

Vin

−=−=

−

Γ=Γ

=

Γ=⎟⎟⎠

⎞⎜⎜⎝

⎛=

−==

−

−−

γγ

γγ

2

1

4

3P

V

CPV =γ

CPV =γ

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OTTO CYCLE – THERMAL EFFICIENCY

Thermal efficiency can be found usingtherefore

in

out

in

netth Q

QQ

W−== 1η

( )( )23

141TTmCTTmC

v

vth −

−−=η

Since 1-2 and 3-4 are isentropic

Then

Hence substituting

( ) 11

3

4

4

3

1

2

1

1

2 and −−−

Γ=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= γ

γγ

vVV

TT

VV

TT

( ) ( ) 143

112 and −− Γ=Γ= γγ

vv TTTT

( )( ) 11

14

14 111 −− Γ−=

Γ−−

−= γγηvv

th TTTT

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OTTO CYCLE – MEAN EFFECTIVE PRESSURE

To understand MEP, let us look at the PV diagram given.The diagram can be translated into a constant pressure pidiagram which reflect the same amount of work between V1 and V2

That pressure, pi is called the mean effective pressure.

( )21

21

VVpW

orVV

Wimepp

inet

neti

−×=

−==

1

2

3

4

Pb

P

V

V2 V1

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EXAMPLE 5.1

Calculate the ideal air standard cycle efficiency based on Otto cycle for petrol engine with cylinder bore of 50 mm, a stroke of 75mm and a clearance volume of 21.3 cm3

SOLUTION

tdc

bdc

50mm

75mm

VcWe know

333 1011

75and 50

cmmmmmLmm

−×=

==φ

Cycle efficiency can be determine using formula

1

11 −Γ−= γη

vth

Find cΓ

c

csc V

VV +=Γ

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( )75

450

4

22

××

=×=ππ LdVs

33 26.147or 15.262,147 cmmm=

3.213.2126.147 +

=+

=Γc

csv V

VV9.7=

( ) 14.11 9.71

11

1 −− −=Γ

−= γηv

th %56atau56.0=

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EXAMPLE 5.2

An ideal Otto cycle has a compression ratio of 8:1. At the beginning of the compression process, air is at 100 kPa and 17ºC and 800 kJ/kg of heat is transferred to air during the constant-volume heat –addition process. Accounting for the variation of specific heats of air with temperature, Determine:

i. The maximum temperature ,

ii. The maximum pressure,

iii. The net work out-put

iv. The thermal efficiency

v. The mean effective pressure (MEP),Pe

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DIESEL CYCLE

Invented by Rudolph Diesel in 1892.It works on the idea of spontaneous ignition which blasted into the cylinder by compressed air.Also known as a modified constant pressure cycle

1-2: Isentropic compression2-3: Reversible constant pressure

heating3-4: Isentropic expansion4-1: Reversible constant volume

cooling

2

1

4

3P

V

CPV =γ

CPV =γ

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DIESEL CYCLE – ENERGY BALANCE

( )

( )

( ) 112

1

1

2

1

2

1

1

2

12

2

1

2

1

1

2

1212

Also,

ration compressio

process, isentropicFor

Law1st Fromncompressio Isentropic :21

−

−−

Γ=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

Γ=

=Γ=+

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−=−=

−

γ

γγγ

γ

γ

v

v

vc

cs

V

TT

pP

VV

TT

PP

VVV

VV

VV

PP

TTmCWδWδQΔU

0 2

1

4

3P

V

CPV =γ

CPV =γ

sVcV

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( )( )

( )

( )1441

3

4

11

2

1

3

4

2323

2323

0cooling Isometric :14

1

expansion Isentropic :4-3

WheatingIsobar :3-2

TTmCQQW

PP

VV

TT

TTmCQQVVp

Vout

vv

v

pin

−=−==

−

Γ=Γ

=

Γ=⎟⎟⎠

⎞⎜⎜⎝

⎛=

−==−=

−

−−

γγ

γγ

DIESEL CYCLE – ENERGY BALANCE

2

1

4

3P

V

CPV =γ

CPV =γ

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DIESEL CYCLE – THERMAL EFFICIENCY

Thermal efficiency can be found usingtherefore

in

out

in

netth Q

QQ

W−== 1η

( )( )23

141TTmCTTmC

p

vth −

−−=η

Besides using the above formula, thermal efficiency can also be determine using

( ) γββη γ

γ

111

1 −Γ−−

−=v

th

Where ratio off-cut2

3 ==VVβ

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EXAMPLE 5.3

An ideal Diesel cycle with air as the working fluid has a compression ratio 18:1 and a cutoff ratio of 2:1. At the beginning of the compression process, the working fluid is 0.1 MPa and 300K.

Determine:

i. Thermal efficiency

ii. The mean effective pressure

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5.4 THE DUAL- COMBUSTION CYCLEKnown as the mixed cycle or semi-diesel cycleThe working cycle of modern diesel engine invented by Ackroyd-stuart in

1888, where it is a combination of the otto and diesel cycle.

1-2: Isentropic compression (adiabatic and reversible)2-3: Heat addition at constant volume3-4: Heat addition at constant Pressure4-5: Isentropic expansion5-1: Heat rejection at constant volume

1

2

34

5

Qin

Qin

Qout

P(bar)

V(m3)

Isentropic

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5.4 THE DUAL-COMBUSTION CYCLE

The air standard efficiency of the dual combution cycle can defined;ηthDual=net work output/heat supplied=Wnet/QinHeat supplied,Q32+Q34Q32=mCv(T3-T2)Q34=mCp(T4-T3)Heat rejected, Q51Q51=mCv(T5-T1)Apply the first thermodynamics to the cycle , the net work is, Wnet=Net heat energy transferred

=(Q32+Q34)-(Q51)

Simplication gives )()()(

1

)()()()()(

)()()(

3423

15

3423

153423

153423

TTTTTT

TTmCTTmCTTmCTTmCTTmC

Hence

TTmCTTmCTTmCW

pv

vpv

vpvnet

−+−−

−=

−+−

−−−+−=

−−−+−=

γη

η

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5.4 THE DUAL- COMBUSTION CYCLE

Or

Where ; ε=V1/V2 is Compression ratioβ=V4/V3 is cut-off ratiok=P3/P4 is pressure ratio

[ ] 11()1(11 −−+−

−−= γ

γ

εβγβη

kkk

thdual

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EXAMPLE 5.4A diesel engine works on the dual combustion cycle and has a compression ratio of 18/1. At the start of compression the air is at the temperature of 22ºC. In the cycle, heat is added at constant volume until the pressure has increased by 50% and then at constant pressure for 7% of the stroke.Calculate the air standard efficiency of the cycle. For air assume γ=1.4;Cp=1.005 kJ/kgK;Cv=0.718 kJ/kgK.

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5.5 PERFORMANCE CRITERIA OF IC ENGINESPerformance Criteria I C Engine(a) Indicated Power (iP)(b) Brake Power (bP)(c) Thermal efficiency (ηTh)(d) Volumetric efficiency (ηV)

(e) Specific fuel Consumption (sfc)GEOMETRICAL PARAMETERS

Referring to the diagramDisplacement;Compression Ratio,ε or Γv=

Note:“Indicated” refers to the values obtained by analysis on the cycle (i.e. Indicated Power, Indicated MEP)“Brake” refers to the values obtained through experimental methods (i.e. Brake Power, BMEP)Vs is multiplied by no of cylinder for multi-cylinder engines

tdc

bdc

b

L

l

a

θ

VC

Vs

C

Cs

s

V+VV

Lb= V 2

4π

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INDICATED POWERDefinition: The rate of indicated cycle work.How it’s being measured: Using indicator diagram obtained

from the engine.

Net work done per cycle = area of power loop – area of pumping loopTherefore indicated mean effective pressure, Pi

Pi=(a/l)load x s

POWER LOOP

PUMPING LOOP

P

V

springindicator theof diagram oflength

diagramindicator of area stiffnesspi ×=

Note: the constant will depends on the scales of the recorder. Normally the units of the constant are either bar/mm or kPa/mm

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INDICATED POWERFor one cylinder engine,

Indicates mean effective pressure (MEP);where L = stroke

A = area of piston

For N rpm and n cylinder, indicated power can be written as:

LAW

VW

VVWp net

s

netneti ×

==−

=21

engine stroke- 2for

engine stroke- 4for 2NnLApip

NnLApip

i

i

××××=

××××=

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In the test on four-stroke four-cylinder automobile engine an indicator diagram is taken and found to have an area of 670 mm2 and a length of 82 mm. The spring in the indicator has a stiffness of 0.9 bar/mm.

Determine

i. the indicated power, iP of the engine at a crankshaft speed of 3200 rev/min

ii. If the cylinders have a bore of 80 mm and the piston stroke is 105 mm, What is the capacity of the engine.

EXAMPLE 5.5

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BRAKE POWER

It is the measured output of the engine (actual power).The engine is connected to a brake dynamometer which can be loaded in such a way that the torque exerted by the engine can be measured.The torque is obtained by reading off a net load, W at a known radius, r from the axis rotation and hence the torque is given by

The brake power is given by

Nowadays torque can be measured directly and bp is obtained directly using above equation.

rW ×=τ

( )Nbp πτωτ 2=×=τπNbp 2=

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ENGINEDYNAMOMETER

35

36

37

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FRICTION POWER & MECHANICAL EFFICIENCY

The difference between the ip and bp is the friction power, fp. It is defined as the power required to overcome the frictional resistance of the engine parts.

The mechanical efficiency is defined as

The value lies between 80 to 90%.

bpipfp −=

ipbp

m =η

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Example 5.6A basic test engine 4-stroke and 4 cylinder are got the reading as a following: Brake Load, 320.8 N, Torque Arm, r=60 cm, engine speed=1500 rev/min. Taken Pi and iP from example 5.5Calculate:i. Brake PowerIi. Friction PowerIii. Mechanical efficiencyIv. Brake mean effective pressure. V. if brake power of a engine is 50 kW, specific fuel consumption, sfc=3.2 liter of 10 minute, and fuel viscoscity,0.8 kg/liter. Take LCV as a 40200 kJ/kg. Determine thermal break efficiency

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BRAKE MEAN EFFECTIVE PRESSURE

The brake power of an engine can be obtained accurately using dynamometer. From equation andwe will get

Since ηm and ip are difficult to obtain, they may be combined and replace by a brake mean effective pressure pb.

ipbp m ×= η2

or NALnpipALnNpip ii ==

2or

NALnpbpALnNpbp imim ×=×= ηη

2or

NALnpbpALnNpbp bb ==

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BRAKE THERMAL EFFICIENCY

Generally we define efficiency as

For IC engine, the actual output is brake power and the input is the chemical energy of the fuel supplied

The value of NKR/LCV is a standard value. For diesel, NKR = 45.5 MJ/kg and for petrol, NKR = 44.2 MJ/kg

inputoutput

=η

valuecalorific low fuel of rate flow mass

powerbrake

=

=

×=

NKRmwhere

NKRm

f

fbt

&

&η

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INDICATED THERMAL EFFICIENCY

Indicated thermal efficiency is defined as

If we divide ηbt with ηit

NKRmfit ×=

&powerindicatedη

itmbt

mti

bt

it

bt

IPBP

NKRmfIP

NKRmfBP

ηηη

ηηη

ηη

×=

==

×

×=

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SPECIFIC FUEL CONSUMPTION

It is defined as mass flow rate of fuel per unit power output.It is a criterion of economical power production

).

(3600jamkW

kgbpmsfc f&

×=

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5.5 OTHER TYPE OF ENGINE TESTING-MORSE TEST

Since it is very difficult to obtain indicated power, Morse Test is introduce to give alternative solution.The test is suitable for multi cylinder engine.Testing procedures:

1. Engine is set to run at certain speed, N and torque is measured2. One cylinder is cut out by shorting the plug.3. When the speed falls, load is reduced to restore the engine speed.4. The torque is measured again.5. The plug is placed back and another cylinder is cut out by shorting its

plug.6. Speed is again restored and torque is again measured.7. The procedures is repeated until all cylinder is simultaneously cut out.

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MORSE TEST

If it is a 4 cylinder engine:

( ) ( ) ( ) ( )totalSSSS

SSSSSSSS

SSSS

FPIPIPIPIPBPFPIPFPIPFPIPFPIPBP

BPBPBPBPBP

−+++=

−+−+−+−=

+++=

4321

44332211

4321

When cylinders are cut off

totalSSSoff

totalSSSoff

totalSSSoff

totalSSSoff

FPIPIPIPBPFPIPIPIPBPFPIPIPIPBPFPIPIPIPBP

−+++=

−+++=

−+++=

−+++=

−

−

−

−

0

0

0

0

3214

4213

4312

4321

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MORSE TEST

Subtracting the equations, for cylinder 1

So for each cylinder,

Then for the engine

( )( )

11

432

43211

0

Soff

totalSSS

totalSSSSoff

IPBPBPFPIPIPIP

FPIPIPIPIPBPBP

=−

−+++

−−+++=−

−

−

offSoffS

offSoffS

BPBPIPBPBPIPBPBPIPBPBPIP

−−

−−

−=−=

−=−=

4433

2211

4321 SSSS IPIPIPIPIP +++=

( ) ( ) ( ) ( )[ ]offoffoffoff WWWWWWWWNRIP −−−− −+−+−+−= 43212π

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A Morse test is carried out to a 4 cylinder, 4 stroke petrol engine. Based on the data given, determine the mechanical efficiency of the engine:

W = 120N

W1-off = 86.8N

W2-off = 81.4N

W3-off = 88.6N

W4-off = 82.1N

EXAMPLE 5.5

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END