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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
1
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 processesinduction, compression, combustion, expansion,
and exhaust. This chapter deals with models for
working fluid composition, and thermodynamics
and transport properties.
2
Table 4.1 Working fluid constituents
Process
Spark-ignition engine
Air
Fuel1
 Recycle exhaust2
Residual gas
Compression Air
Fuel vapor
Recycle exhaust
Residual gas
Expansion Combustion products (mixture
of N2, H2O, CO2, CO, H2, O2,
NO, OH, O, H, …)
Intake
Exhaust
1
Compression-ignition engine
Air
 Recycle exhaust2
Residual gas
Air
Recycle exhaust
Residual gas
Combustion products (mixture of N2, H2O, CO2,
CO, H2, O2, NO, OH, O, H)
Combustion products (mainly Combustion products (mainly N2, CO2, H2O and O2)
N2, CO2, H2O and either O2 if
 < 1 or CO, and H2, if  > 1)
3
Liquid and vapor in the intake; mainly vapor within cylinder
2 Sometimes used to control NO emissions
x
EXHAUST GAS RECIRCULATION (EGR)
SYSTEM
4
Table 4.2 Categories of models for thermodynamics properties
Unburned mixture
Burned mixture
1.Single ideal gas throughout operating cycle with cv and cp constant
Ideal gas; cv,b constant
2.Ideal gas; cv,u constant
3.Frozen mixture of ideal gases cv,u(T)
4.Frozen mixture of ideal gases
Approximations fitted to equilibrium
thermodynamic properties
cv,u(T)
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,
5
and burned mixture properties respectively.
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.
mc = mi + mr
Residual fraction xr
mr
xr 
mc
In some engines, a fraction of engine exhaust gases is
recycled to the intake to dilute the fresh mixture for control
of NOx
6
Percent of exhaust gas recycled (% EGR)
 mEGR
EGR(%)  
 mi
mEGR
mi

 100

: the mass of Exhaust Gas Recycled
: inducted mass per cylinder
Burned gas fraction, xb
mc = mi + mr
mEGR  mr
xb 
mc
 ( EGR / 100)(1  xr )  xr
7
► Burned mixture composition
Combustion of CHy with air
y = H/C ratio
Stoichiometric Combustion Equation:
y
y
y
CH y  (1  )(O2  3.773N 2 )  CO2  H 2 O  3.773(1  ) N 2
4
2
4
8
Combustion equation (with equivalence ratio,  )
written per mole O2:
 c  2(1   )H 2  O2  N 2
where
nCO2 CO 2  n H 2O H 2 O  nCOCO
(1)

 n H 2 H 2  nO2 O2  n N 2 N 2
 : N/O molar ratio (= 3.773 for air)

ni
: 4/(4+y)
: moles of species i per mole O2 reactant
9
The ni are determined using these assumptions
1)   1 CO and H2 can be neglected
2)   1 O2 can be neglected
3)   1 water gas reaction can be assumed to be in equilibrium
CO2  H 2  CO  H 2 O
with the equilibrium constant
K P (T ) 
n H 2O nCO
nCO2 n H 2
10
or Kp can be determined from
1.761  10 1.611  10
0.2803  10
ln K P  2.743 


2
T
T
T3
3
6
9
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
11
Table 4.3 burned gas composition under 1700 K
species
CO2
H2O
CO
H2
O2
N2
Sum: (nb)
ni, moles/moles O2 reactant
 1
 1

2(1   )
absent !
absent !
  c
2(1   )  c
0
0
1

(1   )  1 
c
2(  1)  c
absent !
0

(2   )  
12
Summation of moles above
The value of c is obtained by solving the quadratic
( K  1)c 2  cK 2(  1)     2(1   )  2K (  1)  0
nb   ni
is given in the bottom line of Table 4.3
i
Hence the mole fractions are
ni
~
xi 
nb
13
For alcohols, alcohol-hydrocarbon blends, the
1
y z
CH y Oz  (1   )(O2  N 2 ) 

4 2
per mole O2:
 C  2 (1   ) H 2  O2  (1 
where
z
2
) N 2
(2)
2

2  z (1   )
14
If we write
*  
 z 
and  *  1  
2

The reactant expression (2) becomes:
 * C  2 (1   ) H 2  O2   * N2
which is same form as the reactant expression for
hydrocarbon fuel (1)
Table 4.3 can still be used to give the burned gas
composition, except
Replacing
,,


with
*
,,
*
15
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: (again )
C2 H 6  3.5(O2  3.773N2 )  2CO2  3H 2O  3.5  3.773N 2
You’re supposed to ( A / F ) s  16.1
know how to
1
find (A/F)s!!
( F / A) s 
16.1
( F / A) actual  0.48 g/s
7 g/s
given
16
( F / A) actual
 0.48 


(16.1)  1.1  1
( F / A) s
 7 
We need c since   1 based on Table 4.3
We can find c by solving the following equation
( K P  1)c 2  cK P 2(  1)     2(1   )  2K P  (  1)  0
4
Where Kp = 3.5 and  
4 y
y = 6/3 from C3H6
Remember ??
Thus c 2  1.46c  0.18  0
c  0.134
nCO2
~
mole fraction: xCO2 
nb
(use smaller positive value)
17
From Table 4.3,
nb  (2   )  
 (2  0.57)1.1  3.773  5.345
nCO2
~
xCO2 
nb
nCO2  (  c)  0.57(1.1)  0.134  0.495
0.495
~
 0.093
xCO2 
5.345
Ans
n H 2O  2(1   )  c
 2(1  0.57  1.1)  0.134  0.88
~x
H 2O
0.88

 0.164
5.345
Ans
18
► 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 (CH y )
M f   (12  y)
The fresh mixture per mole O2

(CH y )   O2  N 2

Substituting   M f /(12  y) from (3), then
the reactant expression becomes
4
(1  2 ) (CH y )  O2  N 2
Mf
19
The unburned mixture (fuel, air, and a burned gas fraction),
per mole O2 can be written:
 4

(1  xb ) 
(1  2 ) (CH y )  O2  N 2 
M
 f

unburned reactants
xb (nCO2  nH 2O  nCO  nH 2  nO2  n N2 )
unburned residuals
The number of moles of each species in the unburned
mixture, per mole O2, is summarized in Table 4.4.
20
Table 4.4 Unburned mixture composition
Species
fuel
O2
N2
CO2
H2O
CO
H2
Sum1
ni, moles/mole O2 reactant
 1
 1
4(1  xb )(1  2 ) / M f
4(1  xb )(1  2 ) / M f
1  x b
1  xb


xb 
xb (  c)
2 xb (1   )
0
0
nu
xb 2(1   )  c
xb c
xb 2(  1)  c
nu
21
The total number of moles of unburned mixture nu,
 4(1  2 )

nu  (1  xb ) 
 1     xb nb , and nb is given in Table 4.3
 M f

Previous table for burned mixture !!!
Molecular weight of (burned and unburned) mixture
The mass of mixture (burned or unburned) per mole O2
mRP  32  4 (1  2 )  28.16
Molecular weight
- burned mixture, Mb
M b  mRP / nb
- unburned mixture, Mu
M u  mRP / nu
22
Molecular weight
Equivalence ratio
Figure 4.1 Molecular weight of unburned and burned
23
isooctane-air mixtures as a function of fuel/air
equivalence ratio and burned gas fraction
Table 4.5 Factors for relating properties on molar and mass basis
Quantity, per
mole O2 in the
mixture
Moles of burned
mixture (nb)
Moles of
unburned
mixture (nu)
General equation1
nb  (1   )  1   ,   1
nb  (2   )  ,   1
 4(1  2 )

  1
(1  xb ) 
 1     xb nb 
 M f

  1
Mass of
mixture2
mRP  32  4 (1  2 )  28.16
Mass of air2
1For hydrocarbon fuel,
kg/kmol
nb  0.36  4.773
nb  1.36  3.773
nu  0.08  4.773
 0.28 xb
nu  0.08  4.773
 xb (1.28  1)
138.2  9.12
138.2
32  28.16
 for air = 3.773; for fuels containing oxygen,
are replaced for  and 
2Unit:
Equation for C8H18-air
mixtures
respectively
 * and
*
24
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
Constant volume
u  du

cv     dT  cv T 
 T v
Constant pressure
h  dh

c p     dT  c p T 
 T  p
25
Specific heat ratio
  cp
cv
The change from its value at the reference condition
T0, v0, p0 to the value at T, v, p
u u
0

T
 c dT
v
T0
h h
0

T
 c dT
p
T0
entropy
dv c p
dp
cv
s  so  ds  dT  R  dT  R
T
v
v
p
26
dv
cv
ds  dT  R
T
v
Integrate
s

ds 
s0
s
T

T0
 s0 
cv
dT 
T
T

T0
Similarly,
s
v

R
v0
dv
v
cv
dT  R  ln v  ln v 0 
T
 s0 
T

T0
c
T
p
dT  R ln
v
ln  
 v0 
p
p0
The integral terms in the above equation are functions of
temperature only, we define
27
 v
c
v
 s0   dT  R ln
 vo
T

T0
T
s




define Ψ(T)
then
s
 s0 
T

T0
c
T
p
dT  R ln
p
p0
Ф(T)
s
v 
 s 0    R ln 


 v0 
 (T ) 
s
 p 
 s 0    R ln 

p
 0
 (T ) 
T

c v(T )
dT
T

c p (T )
dT
T
T0
T
T0
Thus the entropy change between state 1 and 2
 p2 




R
ln


s 2 s1  2 1
p
 1
28
Isentropic process:
s2  s1
 p2 




R
ln


s 2 s1  2 1
p
 1
p 

ln  2    2 1
R
 p1 
Mixture properties are determined either on mass or molar basis
Mass
basis
u   xi u i
h   xi hi
s   xi s i
u   xiu i
Molar
basis
h   xi hi
s   xi s i
and
c  x c
c  x c
v
p
i
i
v ,i
p ,i
c  xc
c  xc
v
p
i
i
v ,i
29
p ,i
4.4 A SIMPLE ANALYTIC IDEAL GAS MODEL
- heat capacity
c ,c
p
v
are constant but different for unburned
& burned gas mixture
u c T h
u c T h
u
v ,u
u
f ,u
b
v ,b
b
f ,b
unburned
h c T h
h c T h
u
p ,u
u
f ,u
b
p ,b
b
f ,b
burned
h
f ,u &
h
f ,b
: enthalpies of formation of unburned
and burned gas mixture at 0 K
30
Combustion process
• Constant volume adiabatic
u u
u
b
c T h
v ,u
u
f ,u
 cv,b T b  h f .b
(1)
We solve for Tb by using relations
Rb
Ru
 Mu
(2)
Mb
Cv  R /(  1)
(3)
31
From (1)
c T h
v ,u
u
T
b
f ,u

 cv,b T b  h f .b
1
c
c T  h
 h f ,b 
u
v
,
u
f
,
u


v ,b
1 

+ h f 
c
T
u
v,u


c
c
=
T
c
v,b
v,u
Cv  R /(  1)
+
u
v,b
h f
c
v,b
γ -1

R
=
T
R  γ -1
u
b
b
u
+
u
h f
c
v,b
32
Tb
γ -1

R
=
T
R  γ -1
u
b
Rb
Ru
 Mu
Mb
=
Mb
Mu
b
+
u
u
h f
c
Cv  R /(  1)
v,b
 γ -1 + h  γ -1
T
-1
R M
R
γ 
R
 γ -1 + M h  γ -1
T
 γ -1 M R
f
b
b
u
b
b
u
u
=
Mb
Mu
b
b
f
Mb
b
u
u
u
 M b   T u h f
T b= γ b-1  M u   γ -1 +
Ru
 u
 
u
u




33
- Constant pressure adiabatic
h h
u
b
hf 
 b  1  M b   u


Tb   
T
   1 u

M
R
u 
u 
b 
u
34
u  du

cv     dT  cv T 
 T v
unburned
burned
Figure 4.2 Internal energy versus temperature plot for
stoichiometric unburned and burned gas mixtures: isooctane fuel; unburned residual fraction 0.1.
35
4.5 THERMODYNAMIC CHARTS (แผนภูมิ)
4.5.1 unburned mixture charts
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
36
v 
 (T 2)   (T 1)  n u R ln  2 
 v1 
 p2 
 (T 2)   T 1  n u R ln 

 p1 
nu
: Moles of unburned gas per kilogram of fuel
3

  n R  J
T K
m
p  Pa  v 

 kg .K   
u
kg
air 
air



37
hs
us
Figure 4.3 Sensible enthalpy and internal energy of unburned
38
isooctane-air mixtures as function of temperature.
Units:kJ/kg air in mixture.
Figure 4.4 Isentropic compression functions,  and , as39
function of temperature for unburned isooctane-air mixtures.
Units: J/kg airK.
Table 4.6 Unburned mixture composition for charts
v 
 (T 2)   (T 1)  n u R ln  2 
 v1 
 p2 
 (T 2)   T 1  n u R ln 

p
 1
40
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
of compression using the isentropic chart, Fig. 4-4 For
= 150 J/kg air K. From Eq. (4.25a),
at the end
= 350 K,
1
 150  292ln    757 J / kg air  K
8
From table 4.6
Figure 4-4
Figure 4-4 then gives T2 = 682 K
The ideal gas law gives
This is wrong in
your handout.
Please correct it !!
 T2  v1  682
and p2  p1    
 8  15.5 atm
 T1  v2  350
42
150
350 K
Figure 4.4 Isentropic compression functions,  and , as43
function of temperature for unburned isooctane-air mixtures.
Units: J/kg airK.
757
150
350 K
682 K
Figure 4.4 Isentropic compression functions,  and , as44
function of temperature for unburned isooctane-air mixtures.
Units: J/kg airK.
Note that p2 can also be obtained from Fig. 4-4:
 p2   2  1 980  180
ln   

 2.74

nu R
292
 p1 
p2  15.5 atm
The compression stroke work, assuming the process
is adiabatic and using the data in Fig. 4-3, is
45
u2=350
u1=40
T1=350 K
T2=682 K
Figure 4.3 Sensible enthalpy and internal energy of unburned
46
isooctane-air mixtures as function of temperature.
Units:kJ/kg air in mixture.
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
47
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)
48
Problem 4 (15 points)
Gaseous mixture in a closed tank composes of CO2, CO and O2. These gases
are in chemical equilibrium based on
Kp
2CO 2 
2CO  O 2
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.
49
4.5.2 Burned mixture charts
1. For products of combustion at high temperature
2. 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.
50
Figure 4.5 Internal energy versus entropy chart 51 for
equilibrium burned gas mixture, isooctane fuel, equivalence
ratio 1.0
Example 4.3 the expansion process in an internal combustion
engine, following completion of combustion, can be modeled
approximately 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 airK. The gas expands at
constant entropy to
m3/kg air. Following a constant
entropy process from state 1 on Fig. 4-8 gives
52
1
2
Figure 4.5 Internal energy versus entropy chart 53 for
equilibrium burned gas mixture, isooctane fuel, equivalence
ratio 1.0
Looked up from chart at point 2,
= 1840 K,
= 570 kPa, and
= -1540 kJ/kg air
The expansion stroke work,
assuming the process is adiabatic, is
kJ/kg air
54
4.5 Relation between unburned and burned mixture charts
(ความสั มพันธ์ ระหว่ างแผนภูมสิ ารผสมเผาไหม้ และไม่ เผาไหม้ )
Given unburned mixture at T1 , p1 , v1 , the combustion could
be considered as
(1) constant-volume adiabatic combustion
(2) constant-pressure adiabatic combustion
Charts
Unburned
- Compression process
- Enthalpy and internal
energy of the mixture = 0
at T = 298 K
Burned
- Expansion process
- Enthalpy and entropy of the
chemical elements in their
naturally occurring form form
(O2 , N 2 and H and C (solid))55are
2
assigned zero
GOING FROM UNBURNED TO BURNED
Combustion
End of compression
(unburned)
Start of expansion
(burned)
Adiabatic
Constant volume
ub = u u
Constant pressure
hb = hu
Relations between unburned and burned
56
Constant pressure
Constant volume
ub = u u
hb = h u

uu  u s ,u  u f ,u
hu  hs ,u  h f ,u
Fig. 4-3
Fig. 4-3

  0.4, hf ,u  51.9  1181xb ,u f ,u  47.3  1183xb
  0.6 , hf ,u  77.8  1771xb , u f ,u  70.9  177 xb
  0.8 , hf ,u  103.8  2361xb ,u f ,u  94.6  2365xb
  1.0 ,h f ,u



u
 129.7  2951xb , f ,u  118.2  2956 xb
  1.2 , hf ,u  155.6  2759 xb , u f ,u
 141.9  2769 xb
57
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

hu  hs ,u  h f ,u
where hs,u is the sensible enthalpy
58
~
h f ,i is the enthalpy of formation of unburned mixture per
kilogram of air, defined as
~
~

n

h
h f ,u  i f ,i ,
i
where ni = number of kilomoles of species i per kilogram of air
Similarly, the internal energy uu is given by
uu  u s ,u  u f ,u
where
u f ,u   ni u~ f ,i
i


u
Alternatively,
f ,u can be obtained from
~


u f ,u  h f ,u  (nP  nR ) RT
59

Expressions of h f ,uand
u f ,u (kJ/kgair) at several :
  0.4, hf ,u  51.9  1181xb ,u f ,u  47.3  1183xb
  0.6 , hf ,u  77.8  1771xb , u f ,u  70.9  177 xb
  0.8 , hf ,u  103.8  2361xb ,u f ,u  94.6  2365xb

  1.0 ,hf ,u  129.7  2951xb ,u f ,u  118.2  2956 xb
  1.2 , hf ,u  155.6  2759 xb , u f ,u
 141.9  2769 xb
60
Example 4.4 Calculate the temperature and pressure after
constant pressure adiabatic combustion of unburned
mixture (with  = 1.0 and x b= 0.08) at the state corresponding
to the end of compression process: Tu = 682 K,
u s ,u = 350 kJ/kgair, pu = 1.57 MPa and vu = 0.125 m3/kgair
Analysis
hb  hu
hb  hs ,u  hf ,u
For

(1)
= 1.0
h

f ,u
 129.7  2951xb
 129.7  236  366 kJ/kgair
(2)
61
From figure 4.3 at T = 682 K,  = 1.0,
we obtain hs ,u  465 kJ/kgair
(3)
Replacing (2) and (3) into (1), resulting in
hb  hs ,u  hf ,u
hb = 465 - 366 = 99
kJ/kgair
For constant pressure combustion, Pb = Pu = 1.57 MPa
ub  hb  pbvb  99  1.57  103 vb kJ/kgair
Using a trial and error procedure to find v b and u b at
p  1570 kPa , from the burned chart (figure 4.5) for  = 1.0 ,
we finally get ub  665 kJ/kgair , Tb  2440K, vb  0.485 m3/kg air
62ANS
1570
Figure 4.5 Internal energy versus entropy chart 63 for
equilibrium burned gas mixture, isooctane fuel, equivalence
ratio 1.0
TRIAL AND ERROR PROCEDURE
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.
64
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.
65
It is now routine to measure the composition of engine
exhaust gases (e.g., CO2, NOx and unburned hydrocarbons
and 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.
66
Mole fractions of
Fuel-air equivalence ratio
Figure 4.6 Spark-ignition engine exhaust gas composition data in
mole fraction as a function of fuel/air equivalence ratio.
67
Figure 4.7 Hydrogen concentration in spark-ignition engine
68
exhaust as a function of carbon monoxide concentration. Units:
percent by volume.
Figure 4.8 Exhaust gas composition from several diesel engines
69
in mole fractions on a dry basis as a function of fuel/air
equivalence ratio.