i.c. engine cycles - indian institute of technology...
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I.C. Engine Cycles
Thermodynamic Analysis
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AIR STANDARD CYCLES Air as a perfect gas
All processes ideal and
reversible
Mass same throughout
Constant Specific Heat
1. OTTO CYCLE
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OTTO CYCLE
Efficiency is given by
Efficiency increases with increase in
compression ratio and specific heat
ratio (γ) and is independent of load,
amount of heat added and initial
conditions.
1
11
r
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Efficiency of Otto cycle is given as
= 1- (1/r (-1))
Γ = 1.4
r
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 Pressure
It is that constant pressure which, if exerted
on the piston for the whole outward stroke,
would yield work equal to the work of the
cycle. It is given by
21
32
21
VV
Q
VV
Wmep
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OTTO CYCLE
Mean Effective Pressure
We have:
Eq. of state:
To give:
r
TMR
mpQ
mep1
1
10
132
rV
V
VVVV
11
1
1
1
2121
1
101
p
T
m
RMV
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OTTO CYCLE
Mean Effective Pressure
The quantity Q2-3/M is heat added/unit mass
equal to Q’, so
r
TR
mpQ
mep1
1
10
1
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OTTO CYCLE
Mean Effective Pressure
Non-dimensionalizing mep with p1 we get
Since:
101
11
1
TR
mQ
r
p
mep
10 vcm
R
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OTTO CYCLE
Mean Effective Pressure
We get
Mep/p1 is a function of heat added, initial
temperature, compression ratio and
properties of air, namely, cv and γ
111
1
11
r
Tc
Q
p
mep
v
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Choice of Q’
We have
For an actual engine:
F=fuel-air ratio, Mf/Ma
Ma=Mass of air,
Qc=fuel calorific value
M
QQ 32
cyclekJinQFM
QMQ
ca
cf
/
32
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Choice of Q’
We now get:
Thus:
M
QFMQ ca
rV
VVAnd
V
VV
M
MNow a
11
1
21
1
21
rFQQ c
11
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Choice of Q’
For isooctane, FQc at stoichiometric
conditions is equal to 2975 kJ/kg, thus
Q’ = 2975(r – 1)/r
At an ambient temperature, T1 of 300K and
cv for air is assumed to be 0.718 kJ/kgK,
we get a value of Q’/cvT1 = 13.8(r – 1)/r.
Under fuel rich conditions, φ = 1.2, Q’/ cvT1 =
16.6(r – 1)/r.
Under fuel lean conditions, φ = 0.8, Q’/ cvT1
= 11.1(r – 1)/r
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OTTO CYCLE
Mean Effective Pressure
We can get mep/p1 in terms of rp=p3/p2 thus:
We can obtain a value of rp in terms of Q’ as
follows:
11
11 1
1
r
rrr
p
mep p
11
1
rTc
Qr
v
p
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OTTO CYCLE
Mean Effective Pressure
Another parameter, which is of importance,
is the quantity mep/p3. This can be
obtained from the following expression:
1
11
1
1
13
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
11
1
rTc
Qr
p
c
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We can write the mep formula for the
diesel cycle like that for the Otto cycle in
terms of the η, Q’, γ, cv and T1:
111
1
11
r
Tc
Q
p
mep
v
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Diesel Cycle
We can write the mep in terms of γ, r and rc:
The expression for mep/p3 is:
11
11
1
r
rrrr
p
mep cc
rp
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.
We can obtain the mep in terms of the cut-
off and pressure ratios as before. This is
given in the next slide.
11
111
1
cpp
cp
rrr
rr
r
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Dual Cycle
For the dual cycle, the expression for mep/p3
is given in the next slide.
11
111
1
r
rrrrrrrr
p
mep cppcp
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Dual Cycle
For the dual cycle, the expression for mep/p3
is as follows:
3
1
13 p
p
p
mep
p
mep
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Dual Cycle
We can write an expression for rp the pressure ratio in terms of the peak pressure which is a known quantity:
We can obtain an expression for rc in terms of Q’ and rp and other known quantities as follows:
rp
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:
111
1
1
pv
crrTc
Qr
c
v
pr
rTc
Q
r1
11
1
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Air Standard Cycle
4. MILLER CYCLE
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AIR STANDARD ENGINE
EXHAUST PROCESS
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Exhaust Process
Begins at Point 4
Pressure drops Instantaneously to
atmospheric.
Process is called Blow Down
Ideal Process consists of 2 processes:
1. Release Process
2. 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 into 2 parts
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 of the 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 of this charge will force the
second portion from cylinder which will
escape into the exhaust system.
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Release Process
Consider the portion that escapes from cylinder:
Will expand into the exhaust pipe and acquire high velocity
Kinetic energy acquired by first element will be dissipated by fluid friction and turbulence into internal energy and flow work. Assuming no heat transfer, it will reheat the charge to final state 4”
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Release Process
Succeeding elements will start to leave at states between 4 and 4’, expand to atmospheric pressure and acquire velocity which will be progressively less. This will again be dissipated in friction.
End state will be along line 4’-4”, with first element at 4” and last at 4’
Process 4-4” is an irreversible throttling process and temperature at point 4” will be higher than at 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.
Such an ideal cycle drawn on the pressure
versus specific volume diagram will
resemble an Atkinson cycle or the
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:
v
Vm
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COMPLETE EXPANSION
1
11
v
Vm
Let me be the RESIDUAL CHARGE
MASS, then
6
6
v
Vme
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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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6
66
5
55
v
Vm
v
Vm
Mass of charge remaining in cylinder after blow
down but before start of exhaust stroke is:
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65 vv
65
65
mm
VV
m6 = me or mass of charge remaining in cylinder at
end of exhaust stroke or residual gas
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Residual Gas Fraction
f = (1/r)(v1/v4’)
1
1
1
1
1
1
rTc
Q
p
p
rf
v
i
e
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Temperature of residual gas T6 can be obtained
from the following relation:
1
1
1
1
1
6 1
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 when the inlet valve opens and piston is at TDC.
• Clearance volume is filled with hot burnt charge with mass me and internal energy ue at time t1.
• Fresh charge of mass ma and enthalpy ha enters and mixes with residual charge. Piston moved downwards to the BDC at time t2.
• This is a non-steady flow process. It can be analyzed by applying the energy equation to the expanding 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-a and
entering into the cylinder consists of
internal energy ua and the flow energy pava
so that
• Eflow in, t1 to t2 = ma (ua + pava) …. (3)
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Intake Process
• Change in energy of the system, Esystem, between times t1 and t2 is entirely a change in internal energy and since
m1 = ma + me … (4)
Esystem = m1u1 - meue … (5)
• The mass of the charge in the intake manifold can be ignored or made zero by proper choice of the boundary a-a. The work done by the air on the 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 TDC to
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 of an engine. These are as follows:
1. For the spark ignition engine operating at full throttle. This is also similar to the conventional (naturally aspirated) compression ignition engine. At this operating 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 idle and part throttle. At this operating condition, exhaust pressure is greater than inlet pressure, 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 or at TDC.
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Intake Process
3. For the spark and compression ignition
engine operating with a supercharger. At
this operating condition, the inlet
pressure is greater than the exhaust
pressure, that is
pex/pin < 1
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Case 1: Wide Open Throttle SI or
Conventional CI Engine.
• Fig.1
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WOT SI and 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 that
W = m1p1v1 - mep6v6
= m1pava - mepeve
1
6
611 VVppdVW
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WOT SI and Conventional CI
Substituting in the basic equation for the intake process, for W, and simplifying
m1hm = maha + mehe
Dividing through by m1 and remembering that the ratio me/m1 is the residual gas fraction, f, we get
h1 = (1 – f) ha + fhe
This gives the equation of the ideal intake process at wide open throttle for an Otto cycle engine and can be applied to the dual cycle engine as well.
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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
If the inlet valve opens before the piston reaches TDC, the residual charge will first expand into the intake manifold and mix with the fresh charge and then reenter the cylinder along with the fresh charge.
Now
= p1v1m1 – p1v7me
1
7
711 VVppdVW
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Part Throttle: Early IVO
Hence:
-(p1v1m1 – p1v7me) = -maha + m1u1 - meue
Upon simplification, this becomes
m1h1 = maha + meu7 + p1v7me
Thus we get
h1 = (1- f) ha + f (u7 + p7v7)
= (1 – f) ha + fh7
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Case 2(b): Part throttle SI engine.
Late inlet valve opening.
Fig. 3
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Part Throttle Late IVO
The residual at the end of the exhaust stroke is at point 6. In this case, the valve opens when the piston reaches the TDC. The piston starts on its intake stroke when the fresh charge begins to enter. However, since the fresh charge is at a lower pressure, mixing will not take place until pressure equalization occurs. Thus before the charge enters, the residual charge expands and does work on the piston in the expansion process, 7-7’. This process, in the ideal case, can be assumed to be isentropic. Once pressure equalization occurs, the mixture of the residual and fresh charge will press against the piston during the rest of the work process, 7’-1.
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Part Throttle Late IVO
Now:
During the adiabatic expansion, the work
done by the residuals is given by
-ΔU = me(u7 – u7’)
Hence, W = me(u7 – u7’) + p1(V1 – V7’)
BDC
TDC
pdVpdVpdVW
1
7
7
7
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Part Throttle Late IVO
And since m = V/v,
W = me(u7 – u7’) + m1p1v1 – mep7’v7’
Thus, m1h1 = maha + meh7’
Which reduces to hm = (1 – f) ha + fh7’
This gives the equation for the case where the inlet valve opens late, that is, after the piston reaches the top dead center of the exhaust stroke.
Although the throttle may drop the pressure radically, this has little effect on either the enthalpy of the liquid or the gases, being zero for gases behaving ideally.
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Case 3: Supercharged Engine
Fig. 4
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Supercharged Engine
Here, the intake pressure is higher than the exhaust pressure. Pressure p6’ or p1 represents the supercharged pressure and p5 or p6 the exhaust pressure. Intake starts from point 6’
As before
= p1v1m1 – p1v6’me
1
6
611 VVppdVW
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Supercharged Engine
Hence
- (p1v1m1 – p1v6’me) = -maha + m1u1 - meue
Upon simplification, this becomes
m1h1 = maha + meu6’ + p1v6’me
Thus we get
h1 = (1- f) ha + f (u6’ + p6’v6’)
= (1 – f) ha + fh6’
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Effect of Variation in
Specific Heat of Air
A more realistic solution for Air
Cycles
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Effect of Variation in specific
heat and gamma
Because of variation in specific heat and gamma, the cycle analysis will be different.
We cannot use the standard formulas for determining the air standard efficiencies.
We must determine the temperatures and pressures taking into account the variation in cp and γ and determine the net work and heat supplied or heat supplied and heat rejected to determine the efficiencies.
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Correlations for air
A number of correlations are available for
determining the specific heat of air (at
constant pressure or at constant volume)
as function of temperature. Some
correlations for gamma are also available.
For example:
γ = 1.4 – 7.18 x 10-5T
A few of the correlations are given below:
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Correlations for air
cp = 0.9211 + 0.0002306 T kJ/kg-K
T is in Kelvin
Other properties can be obtained.
A third order equation was proposed by Partha Pratim Saha, 89085,ex student of this course
cp = 26.430213692 + 8.4435671*10-3T
– 2.1567692496*10-6T2
+ 1.9461954*10 -10T3 kJ /kmole K
• T is in Kelvin
• Molecular weight of air is 29
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More correlations for air
According to Lucas, the cp of any gas is
given as follows:
cp= aij (T/1000)i-1
where i = 1 to 7 and j represents the
particular species, isooctane, oxygen or
nitrogen. The units are kJ kmole-1 K-1
Values of aij are given elsewhere
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kgKkJT
CT
CT
CCcp /100010001000
3
3
2
210
Gas C0 C1 C2 C3
Air 1.05 -0.365 0.85 -0.39
Methane 1.2 3.25 0.75 -0.71
CO2 0.45 1.67 -1.27 0.39
Steam 1.79 0.107 0.586 -0.20
O2 0.88 -0.0001 0.54 -0.33
N2 1.11 -0.48 0.96 -0.42
Another correlation for different gases and air is given below:
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The next slide gives tabulated
data of air as function of
temperature.
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RTc
TcT
p
p
kgKkJisairforR /287.0
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