Ic Engine Cycles_1

8/8/2019 Ic Engine Cycles_1

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I.C. Engine Cycles

Thermodynamic Analysis


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AIR STANDARD CYCLES

1. OTTO CYCLE


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Efficiency of Otto cycle is given as

L= 1- (1/r(K-1))

= 1.4

rL

1 0

2 0.242

3 0.356

4 0.426

5 0.475

6 0.512

7 0.541

8 0.565

9 0.585

10 0.602

16 0.67

20 0.698

50 0.791

CR from 2 to 4, efficiency is 76%

CR from 4 to 8, efficiency is 32.6%

CR from 8 to 16, efficiency is 18.6%


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OTTO CYCLE

Mean Effective PressureIt is that constant pressure which, if exerted

on the piston forthe whole outward stroke,

would yield work equal to the work ofthe

cycle. It is given by

21

32

21

VV

Q

VV

Wmep

!

!


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OTTO CYCLE

Mean Effective PressureWe have:

Eq. of state:

To give:

r

TMR

p

ep1

1

10

1

32

!

L

!

!

rV

V

VVVV

11

1

1

1

2

121

1

10

1

pm

V !


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OTTO CYCLE

Mean Effective PressureThe quantity Q2-3/M is heatadded/unit mass

equal to Q, so

r

TRp

ep1

1

10

1

d!L


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OTTO CYCLE

Mean Effective PressureNon-dimensionalizing mep with p1 we get

Since:

- d

-

!

1011

1

1

TR

mQ

r

pmep L

10 ! Kvcm

R


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OTTO CYCLE

Mean Effective PressureWe get

Mep/p1 is a function ofheatadded, initial

temperature, compression ratio andproperties ofair, namely, cv and

? A1

1

1

1

11

-

d!

K

L

r

Tc

Q

p

mep

v


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Choice of Q

We have

Foran actual engine:

F=fuel-air ratio, Mf/Ma

Ma=Mass ofair,

Qc=fuel calorific value

M

QQ 32!d

cyclekJinQFM

QMQ

ca

cf

/

32

!

!


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Choice of Q

We now get:

Thus:

M

QFMQ ca!d

rV

VV

And

V

VV

M

MNow

a

11

1

21

1

21

!

}

!d

rFQQ c

11


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Choice of Q

For isooctane, FQc at stoichiometricconditions is equal to 2975 kJ/kg,thus

Q = 2975(r 1)/r

Atan ambienttemperature, T1 of 300K andcv forair is assumed to be 0.718 kJ/kgK,

we geta value of Q/cvT1 = 13.8(r 1)/r.

Under fuel rich conditions, = 1.2, Q/ cvT1 =16.6(r 1)/r.

Under fuel lean conditions, = 0.8, Q/ cvT1= 11.1(r 1)/r


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OTTO CYCLE

Mean Effective PressureWe can get mep/p1 in terms of rp=p3/p2 thus:

We can obtain a value of rp in terms of Q asfollows:

1111

1

1

!

K

K

rrrr

pmep p

11

1

d

!K

rTc

Qr

v

p


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OTTO CYCLE

Mean Effective PressureAnother parameter, which is of importance,

is the quantity mep/p3. This can be

obtained from the following expression:

1

11

1

1

13 d

!

K

K

rTc

Qrp

mep

p

mep

v


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Air Standard Cycles

2. DIESEL CYCLE


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Diesel Cycle

Thermal Efficiency of cycle is given by

rc is the cut-ff ratio, V3/V2

We can write rc in terms of Q:

-

!

1

111

1

c

c

r

r

r K

LK

K

11

1

d

!KrTc

Qr

p

c


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We can write the mep formula forthe

diesel cycle like that forthe Otto cycle interms ofthe , Q, , cv and T1:

? A11

1

1

11

-

d!

KL

r

TcQ

pmep

v


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Diesel Cycle

We can write the mep in terms of , rand rc:

The expression for mep/p3 is:

11

11

1

! K

K KK

r

rrrr

p

mep cc

!

Krp

mep

p

mep 1

13


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Air Standard Cycle

3. DUAL CYCLE


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Dual Cycle

The Efficiency is given by

We can use the same expression as before

to obtain the mep.

To obtain the mep in terms ofthe cut-offandpressure ratios we have the following

expression

-

!

11

111

1

cpp

cp

rrr

rr

r KL

K

K


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Dual Cycle

Forthe dual cycle,the expression for mep/p3is as follows:

11

111

1

!

K

K KKK

r

rrrrrrrr

p

mep cppcp


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Dual Cycle

Forthe dual cycle,the expression for mep/p3is as follows:

11

111

1

!

K

K KKK

r

rrrrrrrr

p

mep cppcp

!

3

1

13pp

pmep

pmep


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Dual Cycle

We can write an expression for rp thepressure ratio in terms ofthe peakpressure which is a known quantity:

We can obtain an expression for rc in termsof Q and rp and other known quantities asfollows:

!Krp

prp

1

1

3


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Dual Cycle

We can also obtain an expression for rp in

terms of Q and rc and other known

quantities

as follows:

-

d

!

111

1

1

KK K pv

crrTc

Qr

KK

K

-

d

!

c

v

p

r

rTc

Q

r1

11

1


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AIR STANDARD ENGINE

EXHAUST PROCESS


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Exhaust Process

Begins at Point 4

Pressure drops Instantaneously to

atmosp

heric.

Process is called Blow Down

Ideal Process consists of 2 processes:

1. Release Process2. Exhaust Process


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Release Process

Piston is assumed to be stationary at end of

Expansion stroke at bottom center

Charge is

assumed

to be divided in

to 2 p

arts

One part escapes from cylinder, undergoes

free (irreversible) expansion when leaving

Other part remains in cylinder, undergoes

reversible expansion

Both expand to atmospheric pressure


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Release Process

State ofthe charge that remains in the

cylinder is marked by path 4-4, which in

ideal case will be isentropic and extension

of path 3-4.

Expansion ofthis charge will force the

second portion from cylinder which will

escape into the exhaust system.


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Release Process

Considerthe portion that escapes fromcylinder:

Will expand into the exhaust pipe and

acquire high velocityKinetic energy acquired by first element will

be dissipated by fluid friction and

turbulence into internal energy and flowwork. Assuming no heattransfer, it willreheatthe charge to final state 4


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Release Process

Succeeding elements will startto leave at statesbetween 4 and 4, expand to atmosphericpressure and acquire velocity which will beprogressively less. This will again be dissipated

in friction.End state will be along line 4-4, with first element

at 4 and lastat 4

Process 4-4 is an irreversible throttling processand

temper

ature

atpoin

t4 will be

higher

thanat

4 thus

v4 > v4


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Expansion of Cylinder Charge

The portion that remains is assumed to

expand, in the ideal case, isentropically to

atmospheric.

Suchan ideal cycle drawn on the pressure

versus specific volume diagram will

resemble an Atkinson cycle orthe

Complete Expansion Cycle


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COMPLETE EXPANSION

If V is the total volume and v the specific

volume,then mass m is given by

And if m1 is the TOTAL MASS OF CHARGE:

vVm !


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COMPLETE EXPANSION

Let f be the residual gas

fraction, given by

!

!

!

!

!

'

'

4

1

4

1

1

2

6

1

1

6

1

1

6

6

1

1

v

v

r

v

vx

V

V

v

vx

V

V

v

Vv

V

m

mf e


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65vv !

65

65

mm

VV

@

m6 = me or mass of charge remaining in cylinderat

end of exhaust stroke or residual gas


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Residual Gas Fraction

f = (1/r)(v1/v4)

- d

!

K

K

K

1

1

1

1

1

1

rTcQ

pp

rf

v

i

e


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Temperature of residual gas T6 can be obtained

from the following relation:

K

K

K

K 1

1

1

1

1

61

d

!

rTc

Q

p

p

T

T

vi

e


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INTAKE PROCESS


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Intake Process

Intake process is assumed to commence whenthe inlet valve opens and piston is at TDC.

Clearance volume is filled withhot burnt charge

with mass me and internal energy ue attime t1. Fresh charge of mass ma and enthalpy ha enters

and mixes with residual charge. Piston moveddownwards to the BDCattime t2.

This is anon-steady flow process. It can beanalyzed by applying the energy equation to theexpanding system defined in the figure. Since


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Intake Process

Q W = [Eflow out Eflow in + (Esystem]t1 to t2.. (1)

and, since the flow is inward, Eflow out is

zero. Process is assumed to be adiabatic

therefore Q is zero. Thus

- W = Eflow in + (Esystem . (2)


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Intake Process

Assume flow is quasi-steady. Neglect

kinetic energy. Energy crossing a-aand

entering into the cylinder consists of

internal energy ua and the flow energy pavaso that

Eflow in,t1 to t2 = ma (ua + pava) . (3)


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Intake Process

Change in energy ofthe system,(Esystem,between times t1 and t2 is entirely a change ininternal energy and since

m1 = ma + me (4)@(Esystem = m1u1 - meue (5)

The mass ofthe charge in the intake manifold

can be ignored or made zero by proper choice ofthe boundary a-a. The work done by the air onthe piston is given by


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Intake Process

This is Eq. 6

Integrated from tdc to bdc

! pdVW


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Intake Process

This integration is carried out from TDCto

BDC. Substituting from Eq. 3, 5 and 6 in

Eq. 2 to give

This is the basic equation of the

Intake Process.

BDC

a a r rTDC

pdV m h mu m u !


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Intake Process

There are THREE cases of operation ofan engine. These are as follows:

1. For the spark ignition engine operating at

full throttle. This is also similarto theconventional (naturally aspirated)compression ignition engine. Atthisoperating condition, exhaust pressure,

pex, is equal to inlet pressure, pin,that is

pex/pin = 1


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Intake Process

2. For the spark ignition engine operating at idleand partthrottle. Atthis operating condition,exhaust pressure is greaterthan inletpressure,that is

pex/pin > 1

There are two possibilities in this case:

(i) Early inlet valve opening. Inlet valve opens

before piston reaches TDC

.(ii) Late inlet valve opening. Inlet valve opens

when piston reaches near orat TDC.


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Intake Process

3. For the spark and compression ignition

engine operating witha supercharger. At

this operating condition,the inlet

pressure is greaterthan the exhaust

pressure,that is

pex/pin < 1

C 1 Wid O Th ttl SI


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Case 1: Wide Open Throttle SI or

Conventional CI Engine.

Fig.1


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WOT SIand Conventional CI

Since intake process is at manifold pressure(assumed constant) and equal to pa

Thus p1 = pa = p6 hence

By definition, m = V/v so thatW = m1p1v1 - mep6v6

= m1pava - mepeve

!!1

6

611VVppdVW


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WOT SIand Conventional CI

Substituting in the basic equation forthe intakeprocess, forW,and simplifying

m1hm = maha + mehe

Dividing through by m1 and remembering thattheratio me/m1 is the residual gas fraction, f, we get

h1 = (1 f) ha + fheThis gives the equation ofthe ideal intake process

at wide open throttle foran Otto cycle engineand can be applied to the dual cycle engine aswell.

C 2( ) P t th ttl SI i


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Case 2(a): Part throttle SI engine.

Early inlet valve opening.

Fig.2


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Part Throttle: Early IVO

Ifthe inlet valve opens before the pistonreaches TDC,the residual charge will firstexpand into the intake manifold and mix

withthe fresh charge and then reenterthecylinderalong withthe fresh charge.

Now

= p1v1m1 p1v7me

!!

1

7

711VVppdVW


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Part Throttle: Early IVO

Hence:

-(p1v1m1 p1v7me) = -maha + m1u1 - meue

Upon simplification,this becomesm1h1 = maha + meu7 + p1v7me

Thus we get

h1 = (1- f) ha + f (u7 + p7v7)= (1 f) ha + fh7


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Part Throttle Late IVO

The residual at the end of the exhaust stroke is atpoint 6. In this case, the valve opens when the pistonreaches the TDC. The piston starts on its intakestroke when the fresh charge begins to enter.

However, since the fresh charge is at a lowerpressure, mixing will not take place until pressureequalization occurs. Thus before the charge enters,the residual charge expands and does work on thepiston in the expansion process, 7-7. This process,in the ideal case, can be assumed to be isentropic.Once pressure equalization occurs, the mixture ofthe residual and fresh charge will press against thepiston during the rest of the work process, 7-1.


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Now:

During the adiabatic expansion,the work

done by the residuals is given by

-U = me(u7 u7)Hence,W = me(u7 u7) + p1(V1 V7)

d

d

!!BDC

TDC

pdVpdVpdVW

1

7

7

7


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And since m = V/v,

W = me(u7 u7) + m1p1v1 mep7v7Thus, m1h1 = maha + meh7

Which reduces to hm = (1 f) ha + fh7This gives the equation forthe case where the inlet

valve opens late,that is,afterthe piston reachesthe top dead center ofthe exhaust stroke.

Althoughthe throttle may drop the pressureradically,this has little effect on eithertheenthalpy ofthe liquid orthe gases, being zerofor gases behaving ideally.


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Case 3: Supercharged Engine

Fig. 4


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Supercharged Engine

Here,the intake pressure is higherthan theexhaust pressure. Pressure p6 or p1represents the supercharged pressure and

p5 or p6 the exhaust pressure. Intake startsfrom point 6

As before

= p1v1m1 p1v6me

dd

!!

1

6611

VVppdVW


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Supercharged Engine

Hence

- (p1v1m1 p1v6me) = -maha + m1u1 - meue

Upon simplificat

ion,th

is becomesm1h1 = maha + meu6 + p1v6me

Thus we get

h1 = (1- f) ha + f (u6 + p6v6)= (1 f) ha + fh6

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