chapter 4 properties of working fluids - t...
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CHAPTER 4 PROPERTIES OF WORKING
FLUIDS
4.1 Introduction
4.2 Burned and Unburned Mixture Composition
4.3 Gas Property Relationships
4.4 A Simple Analytic Ideal Gas Model
4.5 Relation Between Unburned and Burned Mixture Charts
4.6 Exhaust Gas Composition
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4.1 Introduction
The study of engine operation through an
analysis of processes inside engine requires models
for compositions and working fluids inside the
engine, as well as, models for individual processes-
induction, compression, combustion, expansion,
and exhaust. This chapter deals with models for
working fluid composition, and thermodynamics
and transport properties.
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Process Spark-ignition engine Compression-ignition engine
Intake AirFuel1
Recycle exhaust2
Residual gas
Air Recycle exhaust2
Residual gas
Compression AirFuel vaporRecycle exhaustResidual gas
AirRecycle exhaustResidual gas
Expansion Combustion products (mixture of N2, H2O, CO2, CO, H2, O2, NO, OH, O, H, …)
Combustion products (mixture of N2, H2O, CO2, CO, H2, O2, NO, OH, O, H)
Exhaust Combustion products (mainly N2, CO2, H2O and either O2 if < 1 or CO, and H2, if > 1)
Combustion products (mainly N2, CO2, H2O and O2)
Table 4.1 Working fluid constituents
1 Liquid and vapor in the intake; mainly vapor within cylinder2 Sometimes used to control NOx emissions
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EXHAUST GAS RECIRCULATION (EGR)
SYSTEM
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Unburned mixture Burned mixture
1.Single ideal gas throughout operating cycle with cv and cp constant
2.Ideal gas; cv,u constant Ideal gas; cv,b constant3.Frozen mixture of ideal gases cv,u(T)
4.Frozen mixture of ideal gases cv,u(T)
Approximations fitted to equilibrium thermodynamic properties
5.Frozen mixture of ideal gases cv,u(T)
Mixture of reacting ideal gases in thermodynamic equilibrium
Note: Subscript i,u, and b denote species i in the gas mixture, the unburned mixture,
and burned mixture properties respectively.
Table 4.2 Categories of models for thermodynamics properties
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4.2 Burned and unburned mixture composition
The mass of charge trapped in the cylinder (mc) is the
inducted mass per cycle (mi) plus the residual mass (mr) left
over from the previous cycle.
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mc = mi + mr
Residual fraction xr
c
rr
m
mx
In some engines, a fraction of engine exhaust gases is
recycled to the intake to dilute the fresh mixture for control
of NOx
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Percent of exhaust gas recycled (% EGR)
100(%)
i
EGR
m
mEGR
EGRm
im
: the mass of Exhaust Gas Recycled
: inducted mass per cylinder
Burned gas fraction, xb
c
rEGRb
m
mmx
rr xxEGR )1)(100/( 7
mc = mi + mr
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► Burned mixture composition
Combustion of CHy with air
y = H/C ratio
Stoichiometric Combustion Equation:
22222 )4
1(773.32
)773.3)(4
1( Ny
OHy
CONOy
CH y
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Combustion equation (with equivalence ratio, )
written per mole O2:
222
2
222
222
222
)1(2NnOnHn
COnOHnCOnNOHc
NOH
COOHCO
(1)
where : N/O molar ratio (= 3.773 for air)
: 4/(4+y)ni : moles of species i per mole O2 reactant
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The ni are determined using these assumptions
1) CO and H2 can be neglected
2) O2 can be neglected3) water gas reaction can be assumed to be in equilibrium
OHCOHCO 222
with the equilibrium constant
22
2)(HCO
COOH
Pnn
nnTK
111
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or Kp can be determined from
3
9
2
63 102803.010611.110761.1743.2ln
TTTKP
Kp of 3.5 is often assumed which corresponds to the value at temperature T = 1740 K
The ni is obtained from element balance of equation (1), and
ni is shown in table 4.3
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Table 4.3 burned gas composition under 1700 K
species ni, moles/moles O2 reactant
CO2
H2OCOH2
O2
N2
Sum: (nb)
00
c
0
1 1
)1(2
1
1)1(
c
c )1(2
c )1(2
)2(
12
absent !
absent !
absent !
Summation of moles above
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The value of c is obtained by solving the quadratic
0)1(2)1(2)1(2)1( 2 KKccK
i
ib nn is given in the bottom line of Table 4.3
Hence the mole fractions are
ix~
b
i
n
n
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For alcohols, alcohol-hydrocarbon blends, the
))(24
1(1
22 NOzy
OCH zy
per mole O2:
222 )2
1()1(2 Nz
OHC
(2)
where )1(2
2
z
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*
21*
zIf we write and
The reactant expression (2) becomes:
222 *)1(2* NOHC
which is same form as the reactant expression for hydrocarbon fuel (1)
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Table 4.3 can still be used to give the burned gas composition, except
*Replacing with
*,, ,,
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Example 4.1 Calculate the low-temperature burned gas
composition resulting from the combustion of 7 g/s air
with 0.48 g/s ethane (C2H6)
Kp is assumed 3.5.
Stoichiometric combustion:
2222262 773.35.332)773.3(5.3 NOHCONOHC
16
sFA )/( 1.16
sAF )/(1.16
1
actualAF )/(g/sg/s
7
48.0
You’re supposed to
know how to
find (A/F)s!!given
(again )
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s
actual
AF
AF
)/(
)/( 11.1)1.16(
7
48.0
1We need c since based on Table 4.3
018.046.12 ccThus
134.0c (use smaller positive value)
mole fraction:
b
CO
COn
nx 2
2
~ 17
y = 6/3 from C3H6
Remember ??
0)1(2)1(2)1(2)1( 2 PPP KKccK
Where Kp = 3.5 and y
4
4
We can find c by solving the following equation
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bn )2(From Table 4.3,
345.5773.31.1)57.02(
2COn 495.0134.0)1.1(57.0)( c
2
~COx
0.4950.093
5.345 Ans
OHn2
c )1(2
88.0134.0)1.157.01(2
OHx2
~164.0
345.5
88.0 Ans
18
b
CO
COn
nx 2
2
~
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► Unburned mixture composition
Number of moles of fuel per mole O2 in mixture
depends on the molecular weight of fuel, Mf
Molecular formula of fuel )( yCH
)12( yM f
The fresh mixture per mole O2
22)( NOCH y
Substituting )12/( yM f from (3), then
the reactant expression becomes
22)()21(4
NOCHM
y
f
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The unburned mixture (fuel, air, and a burned gas fraction),
per mole O2 can be written:
)(22222 NOHCOOHCOb nnnnnnx
unburned reactants unburned residuals
The number of moles of each species in the unburned
mixture, per mole O2, is summarized in Table 4.4.
20
22)()21(
4)1( NOCH
Mx y
f
b
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Table 4.4 Unburned mixture composition
Species ni, moles/mole O2 reactant
fuelO2
N2
CO2
H2OCOH2
Sum1
00
1 1
fb Mx /)21)(1(4
bx1
bx
)1(2 bx
fb Mx /)21)(1(4
bx1
)( cxb
cxb )1(2
cxb
cxb )1(2
un un21
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The total number of moles of unburned mixture nu,
bb
f
bu nxM
xn
1
)21(4)1( , and nb is given in Table 4.3
Molecular weight of (burned and unburned) mixture
The mass of mixture (burned or unburned) per mole O2
16.28)21(432 RPm
Molecular weight
- burned mixture, Mb bRPb nmM /
- unburned mixture, Mu uRPu nmM / 22
Previous table for burned mixture !!!
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Figure 4.1 Molecular weight of unburned and burned
isooctane-air mixtures as a function of fuel/air
equivalence ratio and burned gas fraction
23
Mo
lec
ula
r w
eig
ht
Equivalence ratio
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Table 4.5 Factors for relating properties on molar and mass basis
Quantity, per mole O2 in the
mixtureGeneral equation1
Equation for C8H18-air mixtures
Moles of burned mixture (nb)
Moles of unburned mixture (nu)Mass of mixture2
Mass of air2 138.2
1,1)1( bn
1,)2( bn bn
773.436.0 bn
773.336.1
1
11
)21(4)1(
bb
f
b nxM
x
un
773.408.0
bx28.0
un
773.408.0
)128.1( bx
16.28)21(432 RPm 12.92.138
16.2832
* *
1For hydrocarbon fuel, for air = 3.773; for fuels containing oxygen, and
are replaced for and respectively
2Unit: kg/kmol
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4.3 Gas Property Relationships
Individual species in Unburned and Burned gas mixtures can
be modeled as ideal gases.
The most important relationship for property determination
for engine calculations are summarized below
Internal Energy u(T,v) and Enthalpy h(T,p)
Specific heat
TdT
duc
T
uc v
v
v
TdT
dhc
T
hc p
p
p
Constant volume
Constant pressure
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Specific heat ratioc
cv
p
0
0 v
Tc dT
Tu u
0
0 p
TdTc
Th h
The change from its value at the reference condition
T0, v0, p0 to the value at T, v, p
pv
o
dv dpccs s ds dT R dT R
T v v p
entropy
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Integrate
00 0
0
00ln ln
v
v
s vTdvc
ds dT RT v
s T v
Tc
dT R v vT
Ts s
Similarly,
0
0
0
lnp
Tp
dT RT p
T
cs s
The integral terms in the above equation are functions of
temperature only, we define
v
dvRdT
T
cds v
0
lnv
v
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0
0
( ) ( )
( ) ( )
v
p
TdT
T TT
T
TdT
T TT
T
c
c
then0
0
0
0
ln
ln
R
pR
p
vs s
v
s s
Thus the entropy change between state 1 and 2
22 12 1
1
lnp
Rp
s s
28
o
v
vRdT
T
TT
vcss ln
0
0
0
0
0
lnp
Tp
dT RT p
T
cs s
define Ψ(T) Ф(T)
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ss 12
Isentropic process:
2 12
1
lnp
Rp
Mixture properties are determined either on mass or molar basis
sx
hx
ux
ii
ii
ii
s
h
u
sx
hx
ux
ii
ii
ii
s
h
u
and
cxc
cxc
ipip
iviv
,
,
cxc
cxc
ipip
iviv
,
,
22 12 1
1
lnp
Rp
s s
Mass
basis
Molar
basis
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4.4 A SIMPLE ANALYTIC IDEAL GAS MODEL
- heat capacity cc vp, are constant but different for unburned
& burned gas mixture
hTcu
hTcu
bfbbvb
ufuuvu
,,
,,
hTch
hTch
bfbbpb
ufuupu
,,
,,
h uf , h bf ,& : enthalpies of formation of unburned
and burned gas mixture at 0 K
unburned
burned
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uu bu
hTchTc bfbbvufuuv .,,, (1)
Combustion process
• Constant volume adiabatic
We solve for Tb by using relations
MM
RR
b
u
u
b
/( 1)vC R
(2)
(3)
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From (1) hTchTc bfbbvufuuv .,,,
, , ,
,
1b uv u f u f b
v b
c h hT Tc
uv,u f
v,b
fv,u
u
v,b v,b
1+
= +
c hTc
hcT
c c
u fb
u
v,bb u
-1= +
-1
γR hT
cγR
vC R /( 1)
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u fb
u
v,bb u
-1= +
-1
γR hT
cγR
fb u
b bu uu
M= -1 +
-1M
hTγTγ R
Tb
b fb b
u
buu
b fb bb
uu uu
u
-1 -1M= +
-1M
-1 -1M M= +
M-1M
γ γhT
Rγ
γ γhT
Rγ
MM
RR
b
u
u
b
vC R /( 1)
MM
RR
b
u
u
b
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34
- Constant pressure adiabatic
hh bu
R
hT
MM
Tu
f
u
u
u
u
b
b
b
b
1
1
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35Figure 4.2 Internal energy versus temperature plot for
stoichiometric unburned and burned gas mixtures: iso-
octane fuel; unburned residual fraction 0.1.
unburned
burned
TdT
duc
T
uc v
v
v
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36
4.5 THERMODYNAMIC CHARTS (แผนภมิู)
1. Chart that relates the mixture temperature, pressure and
volume in “compression process”2. Chart that presents internal energy and enthalpy as
functions of temperature
ASSUMPTIONS :
•Compression process is reversible and adiabatic
•Fuel is in vapor phase
•Mixture composition is homogeneous and frozen
•Each species can be modeled as an ideal gas
•The burned gas fraction is zero
4.5.1 unburned mixture charts
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37
22 1
1
22 1
1
( ) ( ) ln
( ) ln
u
u
vRnT T
v
pRnT T
p
un : Moles of unburned gas per kilogram of fuel
3
uair air
Jmp Pa v n R T Kkg kg .K
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38Figure 4.3 Sensible enthalpy and internal energy of unburned
isooctane-air mixtures as function of temperature.
Units:kJ/kg air in mixture.
hs
us
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39Figure 4.4 Isentropic compression functions, and , as
function of temperature for unburned isooctane-air mixtures. Units: J/kg airK.
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40
Table 4.6 Unburned mixture composition for charts
22 1
1
22 1
1
( ) ( ) ln
( ) ln
u
u
vRnT T
v
pRnT T
p
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Example 4.2 The compression process in an internal combustion
engine can be modeled approximately as adiabatic and reversible
(i.e., as an isentropic process). A spark-ignition engine with a
compression ratio of 8 operates with a stochiometric fuel vapor-air
mixture which is at 350 K and 1 atm at the start if the compression
stroke. Find the temperature, pressure, and volume per unit mass
of air at the end of the compression stroke. Calculate the
compression stroke work.
Give = 350 K at the start of compression, find at the end
= 350 K,
= 150 J/kg air K. From Eq. (4.25a),
of compression using the isentropic chart, Fig. 4-4 For
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42
Figure 4-4 then gives T2 = 682 K
The ideal gas law gives
and 2 12 1
1 2
6828 15.5 atm
350
T vp p
T v
1150 292ln 757 J / kg air K
8
From table 4.6
Figure 4-4
This is wrong in
your handout. Please correct it !!
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43Figure 4.4 Isentropic compression functions, and , as
function of temperature for unburned isooctane-air mixtures. Units: J/kg airK.
350 K
150
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44Figure 4.4 Isentropic compression functions, and , as
function of temperature for unburned isooctane-air mixtures. Units: J/kg airK.
757
682 K350 K
150
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45
Note that p2 can also be obtained from Fig. 4-4:
The compression stroke work, assuming the process
is adiabatic and using the data in Fig. 4-3, is
2 2 1
1
2
980 180ln 2.74
292
15.5 atm
u
p
p n R
p
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46Figure 4.3 Sensible enthalpy and internal energy of unburned
isooctane-air mixtures as function of temperature.
Units:kJ/kg air in mixture.
T1=350 K
u1=40
T2=682 K
u2=350
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47
Problem 2 (20 points)
A 3-liter, six cylinder SI engine is running at 3600 rpm. The engine has an equal size
of bore and stroke. Its compression ratio is 9.5. The connecting rod length is 16.6 cm.
If combustion ends at 20o after top center (ATC), calculate
•Bore
•Mean piston speed
•Clearance volume
•Piston velocity as soon as the combustion ends• Distant between the crank axis and the piston pin axis as soon as the combustion
ends
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48
Problem 3 (22 points)
In an analysis of combustion of propane (C3H8)with air, it is found that the exhaust
gas composes of CO2, O2, H2O and N2 in which CO2 is found to be 9 % by mole.
Compute the fuel/air equivalence ratio and the corresponding A/F.
(Molecular weight of air: 28.96, atomic weight H: 1, O: 16, C: 12, N: 14)
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49
Problem 4 (15 points)
Gaseous mixture in a closed tank composes of CO2, CO and O2. These gases
are in chemical equilibrium based on
2
Kp
2 O2CO2CO
Pressure of the mixturein the tank is 1.4 atm. If there are 6 moles of CO,
calculate the equilibrium constant Kp of this reaction.
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50
4.5.2 Burned mixture charts
1. For products of combustion at high temperature2. During expansion process
ASSUMPTIONS :
•Expansion process is reversible and adiabatic.
•Each species in mixture can be modeled as an ideal gas.
•Mixture is in thermodynamic equilibrium at temperature
above 1700 K. Mixture composition is frozen below 1700 K.
•At the datum state of 298.15 K and 1 atm chemical
elements in their naturally occurring form (N2, O2, H2 as
diatomic gases and C as solid graphite) are assigned zero
enthalpy and entropy.
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51Figure 4.5 Internal energy versus entropy chart for
equilibrium burned gas mixture, isooctane fuel, equivalence
ratio 1.0
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52
Example 4.3 the expansion process in an internal combustion
engine, following completion of combustion, can be modeledapproximately as an adiabatic and reversible process (i.e.,
isentropic). Under full-center immediately following
combustion is 7100 kPa. Find the gas state at the end of the
expansion stroke and the expansion stroke work. The
compression ratio is 8, the mixture is stoichimetric, and the
volume per unit mass of air at the start of expansion is 0.125
m3/kg air
Locate 7100 kPa and 0.125 m3/kg air on the
= 1.0 burned gass chart (Fig. 4-8). This gives 2825 K,
and s1 = 9.33 kJ/kg airK. The gas expands at
m3/kg air. Following a constantconstant entropy to
entropy process from state 1 on Fig. 4-8 gives
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53Figure 4.5 Internal energy versus entropy chart for
equilibrium burned gas mixture, isooctane fuel, equivalence
ratio 1.0
1
2
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54
= 1840 K, = 570 kPa, and = -1540 kJ/kg air
The expansion stroke work,
assuming the process is adiabatic, is
kJ/kg air
Looked up from chart at point 2,
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55
4.5 Relation between unburned and burned mixture charts (ความสัมพนัธ์ระหว่างแผนภูมสิารผสมเผาไหม้และไม่เผาไหม้)
1T 1p1vGiven unburned mixture at , , , the combustion could
be considered as
(1) constant-volume adiabatic combustion
(2) constant-pressure adiabatic combustion
Charts
Unburned Burned
- Expansion process
- Enthalpy and entropy of the
chemical elements in their naturally occurring form form ( , and and C (solid)) are
assigned zero2O 2N
2H
- Compression process
- Enthalpy and internal
energy of the mixture = 0 at T = 298 K
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GOING FROM UNBURNED TO BURNED
56
Combustion
End of compression
(unburned)
Start of expansion
(burned)
Adiabatic
Constant volume Constant pressure
ub = uu hb = hu
Relations between unburned and burned
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57
ub = uu hb = hu
Constant volume Constant pressure
ufusu uuu ,,
ufusu hhh ,,
Fig. 4-3 Fig. 4-3
4.0buf xh 11819.51,
buf xu 11833.47, , ,
6.0buf xh 17718.77,
buf xu 1779.70, , ,
8.0buf xh 23618.103,
buf xu 23656.94, , ,
0.1buf xh 29517.129,
buf xu 29562.118,
, ,
2.1buf xh 27596.155,
buf xu 27699.141, , ,
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58
The datum for internal energy and enthalpy for the
unburned mixture in unburned chart is different from
that used in the burned chart. In order to locate the
state of the unburned mixture in the burned chart, the
reference datum has to be changed. These data can be
related through the enthalpies of formation.
The unburned mixture enthalpy hu with the same datum as
the burned mixture is
ufusu hhh ,,
where hs,u is the sensible enthalpy
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59
is the enthalpy of formation of unburned mixture per
kilogram of air, defined as
ifh ,
~
ufh ,
~
if
i
i hn ,
~ ,
where ni = number of kilomoles of species i per kilogram of air
Similarly, the internal energy uu is given by
ufusu uuu ,,
where if
i
iuf unu ,,~
Alternatively,
ufu , can be obtained from
TRnnhu RPufuf
~)(,,
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60
ufh ,
ufu , Expressions of and (kJ/kgair) at several :
4.0buf xh 11819.51,
buf xu 11833.47, , ,
6.0buf xh 17718.77,
buf xu 1779.70, , ,
8.0buf xh 23618.103,
buf xu 23656.94, , ,
0.1buf xh 29517.129,
buf xu 29562.118,
, ,
2.1buf xh 27596.155,
buf xu 27699.141, , ,
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61
bx
uT
usu , upuv
Example 4.4 Calculate the temperature and pressure after constant pressure adiabatic combustion of unburned mixture (with = 1.0 and = 0.08) at the state corresponding
= 682 K,
= 350 kJ/kgair, = 1.57 MPa and = 0.125 m3/kgair
to the end of compression process:
Analysis
b uh h
, ,b s u f uh h h (1)
For = 1.0
ufh , bx29517.129 (2)
2367.129 airkJ/kg366
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From figure 4.3 at T = 682 K, = 1.0,
ush , airkJ/kg465we obtain (3)
Replacing (2) and (3) into (1), resulting in
bh = 465 - 366 = 99 kJ/kgair
ANS
bvbu
kPa1570p
airkJ/kg665bu 2440KbT 3
air0.485 m /kgbv
Using a trial and error procedure to find and at
, from the burned chart (figure 4.5) for = 1.0
, we finally get
,
,
62
For constant pressure combustion, Pb = Pu = 1.57 MPa
b b b bu h p v airkJ/kgbv31057.199
, ,b s u f uh h h
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63Figure 4.5 Internal energy versus entropy chart for
equilibrium burned gas mixture, isooctane fuel, equivalence
ratio 1.0
1570
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TRIAL AND ERROR PROCEDURE
64
1. Make a reasonable guess of vb
2. Compute ub
3. For ub and pb = 1570 kPa locate point on the
burned chart
4. Find vb from the chart and check if
this vb is equal to vb in (1.)
5. Use vb in (4.) for computing ub as steps (2.) to
(4.) repeat.
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65
4.6 Intro. to Exhaust Gas Composition
While the formulas for the products of combustion used
in chapter 3 are useful for determining unburned mixture
stoichiometry, they do not correspond closely to the actual
burned gas composition. In the actual case, not all the fuel
which enters the engine is fully burned inside the chamber; the
combustion inefficiency even when excess air is present in a
few percent. Also, the contents of each cylinder are not
necessarily uniform in composition, and the amounts of fuel
and air fed to each cylinder of a multi-cylinder engine are not
exactly the same. For all these reasons, the composition of the
engine exhaust gases cannot be easily calculated.
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66
It is now routine to measure the composition of engine
exhaust gases (e.g., CO2, NOx and unburned hydrocarbonsand particulates). This is done to determine engine emissions.
It is also done to determine the relative proportions of fuel and
air which enter the engine so that its operating equivalence
ratio can be computed.
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67Figure 4.6 Spark-ignition engine exhaust gas composition data in
mole fraction as a function of fuel/air equivalence ratio.
Fuel-air equivalence ratio
Mole
fra
ctions o
f
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68Figure 4.7 Hydrogen concentration in spark-ignition engine
exhaust as a function of carbon monoxide concentration. Units:
percent by volume.
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69Figure 4.8 Exhaust gas composition from several diesel engines
in mole fractions on a dry basis as a function of fuel/airequivalence ratio.