Lecture note ice

8-Nov-13

1

Internal Combustion Engine

8-Nov-13 Internal Combustion Engine 1

Introduction


What is IC Engine?

An internal combustion engine is a thermal system (power plant) that

converts heat obtained from chemical energy sources (gasoline, natural

gas) into mechanical work.

Where are IC Engines Used?

IC engines are used as the propulsion systems for land transport

vehicles such as automobiles (cars, etc.), marine vehicles (boats, etc.)

and small airplanes.

IC engines are also used in portable electrical generators and as prime

mover in grass cutting machine, etc.

8-Nov-13

2

Introduction


Components of IC Engines

• Cylinder, piston, inlet valve and

exhaust valve.

• Piston moves from the top dead

center (TDC) to the bottom dead

center (BDC).

• Clearance volume, Vc is a spacing

between the top of the piston and

the valve’s heads when the piston is

at the end of the delivery stroke. piston

bore

stro

ke

TDC

BDC

Inlet valve (air)

Exhaust valve

(gas)

Engine Classification


Reciprocating internal combustion (IC) engines are

classified into two general categories:

a) Spark-ignition (SI) engines;

b) Compression-ignition (CI) engines.

Description of SI Engines

• Run on liquid fuel such as gasoline or petrol, which is

mixed with air.

• The air-fuel mixture enters the cylinder and is

compressed to a higher pressure and temperature.

8-Nov-13

3



Description of SI Engines

• A spark from a spark-plug ignites the combustible air-

fuel mixture.

• It burns and combustion gases is produced.

• The high pressure of the gases pushes the piston

downwards, producing a power stroke of the piston.

• The crankshaft transforms the reciprocating motion into

rotational motion (rpm), which is carried by gears and

drive shaft systems to the wheels, causing the vehicle to

move.



Description of CI Engines

• Run on diesel liquid fuel.

• The fresh atmospheric air enters the cylinder in which it

is compressed to about 1/22 of its original volume,

causing its temperature to raise to about 540oC (1000 oF)

or higher.

• Diesel fuel is then injected into the compressed air.

• The heat of compression of the air causes the diesel to

burn.

8-Nov-13

4



Description of CI Engines

• Thus producing high temperature combustion gases.

• The combustion gases pushes the piston downward during the

power stroke of the piston.

• As in the SI engines, the reciprocating motion is transformed into

rotational motion.

IN BOTH ENGINES, THE COMBUSTION GASES ARE EVENTUALLY

EXHAUSTED OUT OF THE CYLINDER SO THAT FRESH-AIR MIXTURE

CAN BE INDUCED INTO THE CYLINDER TO CONTINUE THE

THERMODYNAMICS CYCLES.

Compression Ratio

volumeClearance

volumeMaximum

c

scv

V

VVr

+=


Compression ratio =

ie.

Note: compression ratio is volume ratio and

it is not pressure ratio.

(1)

piston

bore

stro

ke

TDC

BDC

Inlet valve (air)

Exhaust valve

(gas)

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5

Mean Effective Pressure


• Mean effective pressure is a conceptual pressure or imagination.

• It is defined as the height of a rectangle on a pressure-volume (p-v)

diagram, having an area equals to that of the air-standard cycle drawn

on the same diagram.

∫= dvpW

p

vA D

p changes with

piston

movement

sm xVpdvpW == ∫

pm = mean effective pressure

pmC

B



• This pressure can give the same total net work as actual pressure.

• For the same engines size, pm can be used as a criteria or parameter to

compare the engines performance.

∫= dvpW

p

vA D

p changes with

piston

movement

sm xVpdvpW == ∫

pm = mean effective pressure

pmC

B

8-Nov-13

6


VxP

ALxP

LxPA

LxF

xWork

=

=

=

=

= distanceforce

smnet VvolumesweptxpW ,=

minmax VV

W

V

Wp net

s

netm −

==∴


ie.

(2)

Classification by Cycles


Reciprocating internal combustion engines operate either on two-

stroke or four-stroke cycle.

Four-stroke Cycle

• Most automotive engines operate on 4-stroke cycle.

• Every forth piston stroke is the power stroke.

• The crankshaft makes two revolutions to complete the cycle.

• The sequence of events in this cycle is as follows:

� Intake stroke: The intake valve opened. The piston moving

downward, allowing the air fuel mixture to enter the cylinder.

� Compression stroke: The intake valve closed. The piston is

moving upward, compressing the mixture.

8-Nov-13

7



Four-stroke Cycle

� Power stroke: The ignition system delivered a spark

to the spark plug the ignites the compressed mixture.

As the mixture burns, it creates high pressure that

pushes the piston down.

� Exhaust stroke: Exhaust stroke: The exhaust valve

opened. The piston moves upward as the burned

gases escape from the cylinder.



Four-stroke Cycle

• The ignition occurs before

the compression process

end, ie. at point x.

• p > patm during exhaust

stroke.

• p < patm during intake stroke.TDC BDC

1

2

4

3

p

V

8-Nov-13

8

Classification by Cycles–4-stroke cycle

The cylinders are

arranged in a line

in a single bank

The cylinders are

arranged in 2

banks set at an

angle to one

another The cylinders are

arranged in 2 banks on

opposite sides of the

engine

Classification by Cycles–4-stroke cycle

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9



Two-stroke Cycle

BDC

TDC

Exhaust port

Intake port

PistonPower stroke

Piston

PistonCompression stroke



Two-stroke Cycle

• Only two strokes – power stroke

and compression stroke.

• One revolution per cycle.

• The exhaust gases exits from the

cylinder during the end of the

power stroke, while the mixture

of fuel/air enters the cylinder.

• This cycle is simple and cheap –

suitable for low power

consumption machine such as

motorbike, etc.

PistonPower stroke

Piston

PistonCompression stroke

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10

The Air Standard Cycles


Actual Cycle

• The working fluid is gas.

• Energy obtained from the combustion gases.

• This process is complicated since chemical analysis is involved.

As an engineer we need to simplify the analysis for design purposes.

Therefore several standard cycles have been developed. Those

standards are based on several assumptions and are known as ‘Air

Standard Assumption’.



The Air Standard Assumption

• The working fluid is air. It follows the ideal gas laws and no chemical

reaction.

• Each process is reversible and isentropic.

• The combustion process is replaced with heat supply process.

• The exhaust process is replaced with heat rejection process.

• Air specific heats (cp and cv) are constant.

The cycle uses those assumptions is called ‘The Air Standard Cycles’.

Models developed from those cycles are simple and able to study the

effects of major parameters towards actual engines performance.

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11



The Otto Cycle

• Otto cycle is an ideal air-standard

cycle for spark-ignition (SI) or

petrol engine, the gas engine, and

high-speed oil engines.

• The cycle is shown on pressure -

volume (p-v) diagram.

= Vmin = Vmax



The Otto Cycle

The Process Sequence

The processes making up the air-

standard Otto cycle are,

1-2 Isentropic compression

2-3 Reversible constant volume

heating

3-4 Isentropic expansion


cooling

= Vmin = Vmax

8-Nov-13

12


v

p

c

c=γ


The Otto Cycle Analysis

= Vmin = Vmax

Compression / expansion index

(3)


(4) ie.

volumeClearance

meSwept volu volumeClearance

volumeMinimum

volumeMaximum ration Compressio

2

1

v

vrv =

+=

=



8-Nov-13

13


( )( )23

14

23

41

23

4123

supply

1

1

TTcm

TTcm

Q

Q

Q

QQ

Q

W

v

v

netotto

−−

−=

−=

−==η

( )( )23

141TT

TTotto −

−−=∴ η

8-Nov-13Internal Combustion Engine25


= Vmin = Vmax

Thermal efficiency of the cycle

(5)


1

3

4

4

3

−

=

γ

v

v

T

T

1

2

1

1

2

−

=

γ

v

v

T

T



= Vmin = Vmax

Since processes 1-2 and 3-4 are

both isentropic, then,

and

8-Nov-13

14


1

2

1

2

1

4

3

T

T

v

v

T

T=

=

−γ

2314 and vvvv ==



= Vmin = Vmax

but


( )( ) 3

4

23

14

4

14

3

23

4

1

3

2

4

1

3

2

1

2

4

3

or

or

11or

rearrange ie.,

T

T

TT

TT

T

TT

T

TT

T

T

T

T

T

T

T

T

T

T

T

T

=−−

−=

−

−=−

==



( )( )

( )( )6

11 ie.,

1111

1otto

1

2

13

4

23

14

−

−

−=

−=−=

−−

−=∴

γ

γ

η

η

v

otto

r

v

vT

T

TT

TT

8-Nov-13

15


γη

η

η

γ index and ration compressio offunction 1

1

res temperatuoffunction 1

basic

1

23

14

⋯

⋯

⋯

−−=

−−

−=

=

v

otto

otto

in

netotto

r

TT

TT

Q

W



Summary

Example 1


An Otto cycle has an inlet pressure and temperature of 100

kN/m2 and 17 oC respectively. The compression ratio is

8/1. If 800 kJ/kg heat is supplied to the system at constant

volume calculate,

a) The maximum cycle temperature;

b) The maximum cycle pressure;

c) The net work;

d) The engine thermal efficiency;

e) The mean effective pressure.

For air, cv = 0.718 kJ/kgK and γ = 1.4.

8-Nov-13

16

Example 2


When working on the Otto with air as the working fluid, an

engine has an air standard efficiency of 54.5% and rejects

heat at the rate of 520 kJ/kg of air used. The engine has a

single cylinder with a bore of 72 mm and a stroke of 85 mm.

The pressure and temperature at the beginning of

compression are 0.98 bar and 66oC respectively. Determine,

a) The compression ratio of the engine;

b) The net work done per kg of air;

c) The pressure and temperature at the end of compression;

d) The maximum pressure and temperature in the cycle;

e) The mean effective pressure.

For air, cv = 0.718 kJ/kgK and γ = 1.4.



The Diesel Cycle

Diesel cycle is an ideal air

standard cycle for compression-

ignition (CI) engines.

p

vv1v2

p3 = p2

1

4

32

pvγ = const

8-Nov-13

17



The Diesel Cycle



standard Diesel cycle are,


2-3 Reversible constant pressure

heating



cooling

p

vv1v2

p3 = p2

1

4

32

pvγ = const


v

p

c

c=γ


The Diesel Cycle Analysis

Compression / expansion index

(3)

p

vv1v2

p3 = p2

1

4

32

pvγ = const

8-Nov-13

18


( )1441 . TTcmQ v −=

( )2323 . TTcmQ p −=



Heat added to the engine

(7)

p

vv1v2

p3 = p2

1

4

32

pvγ = const

Heat rejected from the engine

(8)


( )( )

( )( )

( )( )

( ) res temperatuoffunction 9 1

ie.,

1.

.11

or

basic or

23

14

23

14

23

14

23

41

23

4123

⋯

⋯

TT

TT

TT

TT

TTcm

TTcm

Q

Q

Q

QQ

Q

W

Diesel

p

vDiesel

Diesel

in

netDiesel

−−

−=

−−

−=−/

−/−=−=

−==

γη

γη

ηη



Thermal efficiency

8-Nov-13

19


β

β

γγ

γ

23

2

3

2

3

1

211

2

1

21

1

2

1

1

2

ie

3 2 process isobaricfor ratio off-cut

TT

v

v

T

T

r

TT

v

v

TT

v

v

T

T

v

=

→==

=→

=→

= −−

−

⋯



Thermal efficiency in terms of compression ratio rv

and cut-off

ratio, b p

vv1v2

p3 = p2

1

4

32

pvγ = const


[ ]( )[ ]

( )101

11

ie

Also,

1

12

1

2

1

34

11

4

2

2

3

1

4

3

3

4

−

−−=∴

=

=

=

=

=

=

−

−

−−

−−−

βγβ

η

βββ

β

β

γ

γ

γ

γγγ

γγγ

v

Diesel

vvv

v

r

rT

rT

rTT

rv

vx

v

v

v

v

T

T



Thermal efficiency in terms of compression ratio rv

and cut-off

ratio, b

p

vv1v2

p3 = p2

1

4

32

pvγ = const

8-Nov-13

20

Example 3


A Diesel cycle has an inlet pressure

and temperature of 0.1 MPa and 300

K respectively. The compression ratio

is 18 and the cut-off ratio is 2.

Calculate,

a) The cycle thermal efficiency;

b) The mean effective pressure, MPa.

For air, cp = 1.005 kJ/kgK , cv = 0.718

kJ/kgK and γ = 1.4.

p

vv1v2

p3 = p2

1

4

32

pvγ = const



The Dual-Combustion Cycle

The air-standard dual-combustion

cycle is a basis for comparing the

performance of modern oil engines

(vegetables oil).

p

vv1v2 = v3

p3 = p4

1

43

2

pvγ = const

5

8-Nov-13

21






standard Dual-combustion cycle are,



heating

3-4 Reversible constant pressure

heating



cooling

p

vv1v2 = v3

p3 = p4

1

43

2

pvγ = const

5




Note:

The heat is supplied to the air in

two parts; the first part at constant

volume and the remainder at

constant pressure, hence the name ‘

dual-combustion’.

p

vv1v2 = v3

p3 = p4

1

43

2

pvγ = const

5

8-Nov-13

22



The Dual-combustion Cycle Analysis

Thermal efficiency

To fix the thermal efficiency completely, three factors are necessary, and

these are:

3

4

v

v=β

2

1

v

vrv =Compression ratio,

Ratio of pressures,

Ratio of volumes,

(11)

(12)

2

3

p

p=κ

(13)


( ) ( )[ ] 111

11 −−+−

−−= γ

γ

βγκκκβ

ηvr



Thermal efficiency

Then it can be shown that the thermal efficiency of the dual-combustion

cycle is,

(14)

8-Nov-13

23


in

outinth

Q

QQ −=η



Thermal efficiency

Equation (14) is too cumbersome to use. The best method to evaluate

the thermal efficiency of dual-combustion cycle is to use the relation,

(15)


( ) ( )3423 TTcTTcQ pvin −⋅+−⋅=

( )15 TTcQ vout −⋅=



Thermal efficiency

where Qin is the total amount of heat added to the air and Qout is the heat

rejected from the air, which are given by

(16)

(17)

8-Nov-13

24

Example 4


A dual-combustion cycle has an

inlet pressure and temperature

of 0.1 MPa and 300 K respectively. The

compression ratio is 18.5, ratio of

pressures is 1.5/1 and the cut-off ratio is

1.2. Calculate,

a) The cycle thermal efficiency;

b) The mean effective pressure.

For air, cp = 1.005 kJ/kgK , cv = 0.718

kJ/kgK, g = 1.4 and R = 0.287 kJ/kgK.

p

vv1v2 = v3

p3 = p4

1

43

2

pvγ = const

5

Criteria of Performance for ICE


In order that different types of engines or different

engines of the same type may be compared, certain

performance criteria must be defined.

They are obtained by measurement of the quantities

concerned during bench tests, and calculation is by

standard procedures.

8-Nov-13

25



Indicated Power (ip)

Indicated power (ip) is

defined as the rate of work

done by the combustion gases

on the piston.

It is evaluated based on the

indicator (p-V) diagram

obtained during engine

testing, typically shown as in

the figure. V

p

3


nNLApip i ⋅⋅⋅⋅=

2

nNLApip i ⋅⋅⋅⋅=



a) Two-stroke engines

For two-stroke engines, the

indicated power is given by

(18)

b) Four-stroke engines

For the 4-stroke engines, the indicated

power is given by

(19)

V

p

3

l

8-Nov-13

26


pressure. effectivemean indicated

cylinders; ofnumber

(rpm); engine theof speed rotational

piston; theof stroke oflenght

piston; theof area where,

=

=

=

=

=

ip

n

N

L

A






-Indicated mean effective pressure

When testing an engine to determine its

performance, it is necessary to measure

the power which is actually produced

inside the cylinder. This is done using a

device known as an indicator.

The purpose of any indicator is to

reproduce the relationship between the

pressure and the volume of the working

fluid as the piston moves through a

complete cycle in the cylinder. V

p

3

l

8-Nov-13

27


Mechanical engine indicator


sxl

api

=



The indicator produces a pressure-

volume diagram for an actual cycle.

From the diagram, the indicated mean

effective pressure, pi, can be obtained.

(20)

V

p

3

l

8-Nov-13

28




where, a = net area of the indicator

diagram;

l = length of the indicator

diagram;

s = a recorder constant.

V

p

3

l

Example 5


In a test on a 4-stroke, 4-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 the

indicated power of the engine at a crankshaft speed of

3200 rpm if the cylinders have a bore of 80 mm and the

piston stroke is 105 mm. What is the capacity of the

engine?

8-Nov-13

29

Example 5

kW 4.41

4x60

3200x0.105x10x03.5x10x35.7x

2

1

2

1

m 10x03.508.0x4

x4

piston, of Area

bar 35782

670x90x

m 0.105 mm 105 m; 0.08 mm 80

bar/mm 0.9 mm; 82 ;mm 670

4 engine; stroke-4 :Data

3-2

2322

2

=

=

⋅⋅⋅⋅⋅=∴

===

===

====

===

=

−

nNLApip

πd

π A

. .

l

as p

Ld

sla

n

i

i


Example 5

3

63-

3

33-

3-

cm 2113

10 x 10 x 2.113 capacity

hence ;cmin capacity engine express topracticecommon isIt

m 10 x 2.113

0.105 x 10 x 5.03 x 4

xx

cylinders all of meswept volu Total capacity Engine

=

=

=

=

=

=

LAn


8-Nov-13

30



Brake Power (bp)

The power measured at the output shaft of the engine is known as the

brake power of the engine.

The indicated power developed by the engine is the power available at

the piston.

This mechanical power, in the form of linear motion of the piston, is

transmitted through the connecting rod and the crankshaft, to be

transformed into rotary power at the output shaft.

However, during this process some of the power is used to overcome

the frictional resistance between the moving parts.



Brake Power (bp)

Hence, the power available at the

output shaft is less than the

indicated power ……..that is the

brake power.

The brake power is measured using

a dynamometer which provides

resistance to engine torque by

opposing the rotation of the shaft.

Water-cooled brake

drum coupled to the

engine shaft

W = force due to weights, N

S = spring balance load, N

Resisting torque, T = (W- S)R Nm

8-Nov-13

31


TNbp ⋅⋅⋅= π2

( )

length. arm torque arm, brakebetween forces andwith

x

x i.e.

r;dynamomete by the measured torque

(rpm); speed rotational engine where,

==

=

−=

=

=

RSW

RF

RSWT

T

N


Brake Power (bp)

(21)


bpipfp −=


Friction Power (fp)

Friction power (fp) is the amount of power needed to

overcome friction resistance in many moving parts of the

engine. We have,

(22)

8-Nov-13

32


ip

bpm =η


Mechanical Efficiency (ηηηηm)

The mechanical efficiency of an IC engine is defined as

(23)

Typical IC engines have mechanical efficiency of 80 to 90 %.

Example 6


The specification for a 4-stroke, single cylinder internal combustion engine is

as follows:

Bore = 146 mm

Stroke = 280 mm

Speed at full load = 475 rpm

Not brake load = 433 N

Torque arm length = 0.45 m

Area of indicator diagram = 578 mm2

Length of indicator diagram = 70 mm

Recorder spring rating = 0.815 bar/mm

Determine for the engine:

a) Indicated power

b) Brake power

c) Mechanical efficiency

8-Nov-13

33

Example 6

kW 2.461

1x60

475x280.0x0167.0x10x73.6x

2

1

2

1

m 0167.0146.0x4

x4

piston, of Area

bar 73.670

578x8150x


2

222

=

=

⋅⋅⋅⋅⋅=∴

===

===

=

nNLApip

πd

π A

.

l

as p

n

i

i


Example 6

kW .699

85.194x60

475xx2

2

85.19445.0x433x Torque,

=

=

⋅⋅⋅=∴

===

π

π TNbp

Nm R FT

% 77.8

778.046.12

69.9

=

=== ip

bp mη


8-Nov-13

34

Example 7


The engine in Example 5 is connected to a rope brake in

order to measure the brake power. The brake drum

diameter is 0.9 meter and the rope 20 mm in diameter. At

3200 rpm the dead load is 250 N and the spring balance

reads 18 N. Neglecting the weight of the rope, what is the

brake power of the engine?

Example 7

kW 6.543

04.109x60

3200xx2

2

04.10947.0x232x motion, opposing torqueHence,

47.002.02

9.0

2

radius,mean aat applied is force This

23218250

rpm 3200 N; 18 N; 250

0.02m mm 20 d m; 0.9 D


=

=

⋅⋅⋅=∴

===

=+=+=

=−=−=

===

===

=

π

π TNbp

NmR FT

mdD

R

N S WF

NSW

n


8-Nov-13

35



( )24 2

1

where,

2

1x

engine, stroke-4for Therefore

x

nNALpbp

pp

nNALpbp

ipbp

b

imb

im

m

⋅⋅⋅⋅⋅=∴

⋅=

⋅⋅⋅⋅⋅=

=

η

η

η

Brake Mean Effective Pressure (bmep)

From the definition of mechanical efficiency, we can write the

expression for the engine’s brake power (bp) as


pressure effectivemean brake

pressure effectivemean indicated

pressure effectivemean standard

=

=

=

b

i

m

p

p

p


Note

8-Nov-13

36


•==E

bpbt

suppliedenergy of Rate

outputpower brakeη


Brake Thermal Efficiency (ηηηηbt)

The brake thermal efficiency, ηbt is the measure of the

overall efficiency of the engine. It is defined as

(25)




The rate of thermal energy supplied to the engine, is given

by

•

E

( )

fuel theof valuecalorificnet

fuel of rate flow mass where,

26.

=

=

=

•

••

C.V

m

VCxmE

f

f

8-Nov-13

37




( )

( ) ( )28 kg/kW.hr ,

where,

27

.x

1

.x

Therefore,

bp

msfcptionuel consumspecific f

VCbp

mVCm

bp

f

ff

bt

•

••

=

==η




VCsfc .x

1

Therefore,

bt =η

8-Nov-13

38



Indicated Thermal Efficiency (ηηηηit)

Based on the definition of the brake thermal efficiency, the

indicated thermal efficiency is defined as

VCm

ipη

f

it

.x•= (29)



Indicated Thermal Efficiency (ηηηηit)

Dividing equation (27) by equation (29), we obtain

itmbtη ηη x= (30)

8-Nov-13

39

Example 8


An engine produces 50 kW of brake power. A test shows

that the time for the engine to consume 4 liter of fuel was

recorded as 12 minutes. If the fuel density is 1.1 kg/liter,

determine,

a) the sfc of the engine when the C.V is 40400 kJ/kg;

b) the brake thermal efficiency.

Example 8

203.040400x44.0

3600

.x

1 ,efficiency thermalBrake (b)

kg/kW.hr44.060x50

3667.0 (a)

Therefore,

kg/min 3667.012

4x1.1x

, fuel, of rate flow Mass

kg/ 1.1

minutes 12;liter 4

kJ/kg 40400 kW; 50 :Data

===

===

===

=

==

==

•

•

•

VCsfc η

bp

msfc

t

V m

m

l

tV

C.Vbp

bt

f

fff

f

f

f

ρ

ρ


8-Nov-13

40



Volumetric Efficiency (ηηηηv)

Is defined as the ratio of the actual volume of air drawn in

during the suction stroke to the swept volume of the cylinder at

atmospheric pressure and temperature or

( )31

s

o

s

ov

V

V

V

Vη •

•

==


f

a

m

m

fuelofrateflowmass

airofrateflowmassRatioFuelAir •

•

==/


Air/Fuel Ratio

(32)

8-Nov-13

41

Example 9


A 4-stroke, 4 cylinder engine has a bore of 57 mm and a

stroke of 90 mm. When tested at 2800 rpm, the engine fuel

consumption is 0.001376 kg/s and the air fuel/ratio is

14.5/1. If atmospheric conditions in the test room is 1.013

bar and 15oC, determine the volumetric efficiency.


## Energy DistributionEnergy Distribution kJ/skJ/s %%

1.1. Fuel energyFuel energy XX 100100

2.2. Brake powerBrake power XX11 XX11/X/X * 100* 100

3.3. Heat to cooling water Heat to cooling water XX22 XX22/X/X * 100* 100

4. 4. Heat to exhaustHeat to exhaust XX33 XX33/X/X * 100* 100

5. 5. Other reductionsOther reductions XX44 = X = X -- ((XX11 + X+ X22 + X+ X33)) XX44/X /X * 100* 100


Energy Balance of an Engine

8-Nov-13

42



The Morse Test

Morse test is an experimental procedure for determining

the engine’s indicated power without having to compute

the mean effective pressure, pi of the engine.

This indicated power is evaluated outside the cylinders.

This test is only applicable to multi-cylinder engines, is

carried out at constant speed (rpm).



The Morse Test

The steps in conducting the test is described as follows:

Assume that n = 4, engine speed = N rpm, torque = T, force = F

and torque arm length = R.

1. The engine is run at desired constant speed (rpm),

combustion occurred in all cylinders, and the brake power

(bp) is measured using a dynamometer.

2. The first cylinder is being cut-off by disconnecting the cable

to the spark plug (in a SI engine) or fuel-injector line (in a CI

engine) for that cylinder. The engine is still run at constant

speed (rpm). The new value of the brake power is measured.

8-Nov-13

43



The Morse Test

3. Step number 2 is repeated but now cylinder number 2 is

being cut-off. While cylinder number 1, 3 and 4 are firing.

Another new value of the brake power is measured.

4. Step number 2 is repeated but cylinder number 3 is being

cut-off. While cylinder number 1, 2 and 4 are firing. The new

value of the engine’s brake power is measured.

5. Finally, cylinder number 4 is being cut-off. Cylinder number

1, 2 and 3 are firing. Another new value of the brake power

is measured.


123443211234

123460

2fpipipipip

TNbp −+++=

⋅⋅⋅=

π


The Morse Test

During the 1st step, that all the cylinders are firing, the brake

power can be expressed as follows,

8-Nov-13

44


1234432234

234 060

2fpipipip

TNbp −+++=

⋅⋅⋅=

π


The Morse TestThe Morse TestThe Morse TestThe Morse Test

During the 2nd step, cylinder number 1 is not functioning, the

brake power is,

Since the engine is still run at constant speed, N rpm, the

frictional power fp is still the same.


12341234 ipbpbp =−


The Morse Test

When we subtract both equations, we will get,

8-Nov-13

45


41231234

31241234

21341234

ipbpbp

ipbpbp

ipbpbp

=−

=−

=−


The Morse Test

By continuing this procedure, we will get,


( ) ( )( ) ( )12312341241234

134123423412344321

bpbpbpbp

bpbpbpbpipipipip

−+−+

−+−=+++


The Morse Test

Therefore, the sum of all the indicated powers for 4 cylinders

is,

8-Nov-13

46


( )F

RNRFNTNbp

bpbpbpbpbpip

x60

2

60

x2

60

2

that,know also We

4 12312413423412341234

⋅⋅⋅=

⋅⋅⋅=

⋅⋅⋅=

−−−−⋅=

πππ


The Morse Test

i.e.,


[ ]12312413423412341234 460

2FFFFF

RNip −−−−⋅

⋅⋅⋅=

π


The Morse Test

Therefore,

(34)

8-Nov-13

47

Example 10


A 4-cylinder petrol engine has a bore and a stroke of 57 mm and 90 mm

respectively. At 2800 rpm the net load on the friction brake is 155 N and

the torque arm is 0.356 m. The engine consumes 6.74 liter of fuel/hour.

The relative density of the fuel is 0.735 and the lower calorific value of the

fuel is 44200 kJ/kg. A Morse test is carried out on the engine. The engine

cylinder is cut out one after another following the order of 1, 2, 3 and 4

and producing the brake loads of 111, 106.5, 104.2 and 111 N

respectively. Calculate,

i. the engine torque;

ii. the brake mean effective pressure;

iii. the brake thermal efficiency;

iv. the specific fuel consumption;

v. the mechanical efficiency;

vi. the indicated mean effective pressure.

Lecture note ice

Education

brake thermal

air standard

cycle thermal

air standard

engines brake

thermal efficiency

constant speed

mechanical