A Finite Element Thermo-Mechanical Stress Analysis of IC Engine Piston
Β
NOX PDF
1. Sam Cutlan
Malcolm McDonald
Powertrain and Sustainability
Combustion Emissions, Calculations and Mapping for
Efficiency
BEng (Hons) Automotive Engineering
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Table of Contents
1.0 Introduction ................................................................................................................................2
2.0 πππ₯ Concentration Levels..........................................................................................................2
2.1 Calculation of the Net Enthalpyβs for each Fuel Equivalence Ratio........................................2
2.2 Production of Temperature (K) vs βππ β βππ298 (kJ/kmol) Graph and Determination of
Initial Burning Temperatures..............................................................................................................3
2.3 Determining all Burning Temperatures with an increasing Pressure .....................................4
2.4 Determining πππ₯ concentration levels as a function of Fuel Equivalence Ratio...................6
3.0 The Effect of Adding E.G.R on NOx Concentration Levels ..........................................................8
3.1 Determining the Coefficients of the Products ........................................................................8
3.2 Determining the Initial Burning Temperatures and all Burning Temperatures......................8
3.3 Determining NOx Concentration Levels..................................................................................9
4.0 Analysis of NOx Model..............................................................................................................10
4.1 Assumptions Made ...............................................................................................................10
4.2 Effectiveness of Calculations.................................................................................................12
4.3 Mechanisms Responsible for NOx Formation.......................................................................12
5.0 References ......................................................................................................................................13
6.0 Appendices......................................................................................................................................14
6.1 Appendix 1 ..................................................................................................................................14
6.2 Appendix 2 ..................................................................................................................................14
6.3 Appendix 3 ..................................................................................................................................15
6.4 Appendix 4 ..................................................................................................................................16
6.5 Appendix 5 ..................................................................................................................................16
6.6 Appendix 6 ..................................................................................................................................17
6.7 Appendix 7 ..................................................................................................................................18
6.8 Appendix 8 ..................................................................................................................................18
6.9 Appendix 9 ..................................................................................................................................19
6.1.1 Appendix 10 .............................................................................................................................21
6.1.2 Appendix 11 .............................................................................................................................21
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1.0 Introduction
The aim of this report is to successfully develop a fully working πππ₯ model for Iso-octane running on
the lean side of stoichiometric. Prior to creating the πππ₯ model, there are a number of assumptions
that have to be considered; 1) the rise in pressure at the start of combustion (SOC) is linear and
increases in 2 bar/degree increments from 10 bar (SOC) to 50 bar, 2) the engine is a 1600cc four
cylinder naturally aspirated (N/A) and 3) the engine speed is considered a constant 1800 RPM.
From knowing these critical pieces of data, a model to predict the πππ₯ Parts per Million (PPM) VS
Fuel Equivalence Ratio and the effect of adding Exhaust Gas Recirculation (EGR) VS πππ₯
concentration levels will be created within Excel. Further to the initial assumptions made, there are a
number of assumptions made later on in the πππ₯ model which are later discussed.
2.0 πππ₯ Concentration Levels
2.1 Calculation of the Net Enthalpyβs for each Fuel Equivalence Ratio
To obtain the initial flame temperatures required to calculate the rising adiabatic flame
temperatures up to peak pressure, it is critical to calculate the net enthalpy released for each fuel
equivalence ratio. To do this, the standard enthalpy of formation and molecular weight of species
table was needed to calculate the net enthalpyβs released. Refer to Equation 1 below as a reference
for a fuel equivalence ratio of 0.6;
Equation 1
0.6 Γ πΆ8 π»18 + 12.5(π2 + 3.76π2) = 4.8πΆπ2 + 5.4π»2 π + 5π2 + 47π2
(McDonald, 2015)
As can be seen from the above equation, the enthalpy required to get the reactants to the zero level
will change due to the fuel equivalence ratio changing. As well as this, the coefficients of the
products will also change as a result of the fuel equivalence ratio changing and hence the enthalpy
released will change too. Once performed, the net enthalpy released is shared between the products
by forms of the following equation;
Equation 2
πππ‘ πΈππ‘βππππ¦ π πππππ ππ = 4.8 Γ πΈππ(πΆπ2) + 5.4 Γ πΈππ(π»2 π) + 5 Γ πΈππ(π2) + 47 Γ πΈππ(π2)
(McDonald, 2015)
Once the above equation has been set up with all net enthalpyβs and coefficients calculated for each
fuel equivalence ratio, it is required that the βEqnβ be determined. The βEqnβ is a quadratic equation
determined by producing a graph through the use of Excel of Temperature (K) VS βΜ π(π) β βΜ π π(298)
(kJ/kmol). From this graph, the equations of each line of πΆπ2, π»2 π, π2 πππ π2 can be found and
inserted into Equation 2 to find the initial temperatures (T) for each fuel equivalence ratio.
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2.2 Production of Temperature (K) vs βΜ π( π) β βΜ π π(298) (kJ/kmol) Graph and
Determination of Initial Burning Temperatures
From directly inserting the Temperature (K) and βΜ π(π) β βΜ π π(298) (kJ/kmol) results from
πΆπ2, π»2 π, π2 πππ π2 obtained from the Thermodynamic Properties of Ideal Fuels Tables; a graph
could be produced to find the equations of each line required to calculate the initial temperatures;
Figure 1
Figure 1 - Enthalpy required from 298K vs Temperature.
The equations representing each line can be seen below;
Table 1
Equations of each line
CO2 0.0033408353x2
+ 46.4931007649x - 15,563.7522326659
H20 0.0044468568x2
+ 32.6521535886x - 10,643.8864074281
O2 0.0016322369x2
+ 31.1507970936x - 9,827.6586302809
N2 0.0015490125x2
+ 29.4904374322x - 9,267.6068460288
Table 1 - Equations of each line for CO2, H20, O2 and N2 respectively.
As can be seen from Table 1, each equation is in the form ππ2
+ ππ2
+ π, where π₯ in the above
equations represents the initial temperature (T). From inserting these equations into Equation 2, the
entire equation was summated in Excel such that an online quadratic equation calculator could be
used to find (T) (Math.com, 2005). The initial temperatures could then be found for each fuel
equivalence ratio, see Table 2.
y = 0.003x2 + 46.493x - 15,563.752
y = 0.004x2 + 32.652x - 10,643.886
y = 0.002x2 + 31.151x - 9,827.659
y = 0.002x2 + 29.490x - 9,267.607
-20000
0
20000
40000
60000
80000
100000
120000
140000
160000
180000
0 500 1000 1500 2000 2500 3000 3500
HT-H298(KJ/Kmol)
Temperature (K)
Enthalpy Required from 298K VS Temperature
CO2 H2O O2 N2
Poly. (CO2) Poly. (H2O) Poly. (O2) Poly. (N2)
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Table 2
Fuel Equivalence Ratio (ΙΈ) Initial Temperatures ππ΅1
(K)
0.6 1711.482252
0.65 1807.515705
0.7 1900.877098
0.75 1991.664636
0.8 2079.972673
0.85 2165.891804
0.9 2249.508968
0.95 2330.907571
0.96 2346.928159
0.97 2362.863842
0.98 2378.715234
0.99 2394.482948
Table 2 - Fuel Equivalence Ratio against Initial Temperature.
2.3 Determining all Burning Temperatures with an increasing Pressure
The initial flame temperatures at each fuel equivalence ratio are the burning temperatures produced
at SOC where the crank angle is 0Β° and the pressure is 10 bar. It is then essential that the flame
temperatures be calculated for an increasing crank angle (from 0Β° to 20Β°) and an increasing pressure
(from 10 bar to 50 bar). To do this, if;
ππ π
= ππππ π‘πππ‘
(1) π1 π1
π
= π2 π2
π
From the Ideal Gas Law;
π1 π1 = ππ π1
And;
π2 π2 = ππ π2
Rearrange to make π1 and π2 the subject of the equation;
π1 =
ππ π1
π1
And;
π2 =
ππ π2
π2
Therefore equation (1) becomes;
π1 Γ (
ππ π1
π1
)
π
= π2 Γ (
ππ π2
π2
)
π
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m x R will cancel out, therefore the equation becomes;
π1 Γ (
π1
π1
)
π
= π2 Γ (
π2
π2
)
π
By multiplying out the brackets the equation becomes;
π1 Γ π1
π
π1
π =
π2 Γ π2
π
π2
π
And by applying the following rule;
π
π π
=
1
π πβ1
The equation will become;
π1
π
π1
πβ1 =
π2
π
π2
πβ1
Then from cross multiplying the brackets the equation becomes;
π2
π
π1
π =
π2
πβ1
π1
πβ1 = (
π2
π1
)
πβ1
Hence;
(
π2
π1
)
π
= (
π2
π1
)
πβ1
From taking n over the equals sign the equation will become;
π2
π1
= (
π2
π1
)
πβ1
π
And finally the equation will be;
Equation 3
π2 = π1 Γ (
π2
π1
)
πβ1
π
Since the initial flame temperature is represented by (π1 = ππ΅1
)then (π2 = ππ΅2
) can be calculated
by using Equation 3 above (See appendix 1).
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2.4 Determining πππ₯ concentration levels as a function of Fuel Equivalence
Ratio
Throughout the following calculations the fuel equivalence ratio is taken as 0.6 to show workings
out. The calculations consider that an alternate fuel equivalence ratio will change the value of the
coefficients of the reactants; therefore changing the PPM of NOx. It is assumed that the engine has a
volumetric efficiency of 40% and since each cylinder has 400cc of displacement, the volume sucked
in is; 0.4*400=160ccs, hence each cylinder has 160*10^-6 m^3 of displacement. As well as this, it is
also assumed that the injection mixture comes in at 293k = 20Β°C.
From using the Ideal gas law;
ππ = ππ π’π
The number of kmols of mixture in the cylinder is (n) = PV/RuT
Since the pressure (bar) of a N/A engine mixture comes in at 1 atmosphere = 10^5 N/m^2, the
equation can be solved for n.
n = 6.57 Γ 10β6
Therefore, since;
π = βΓ (π Γ 1 + 12.5 Γ (1 + 3.76)) = 6.57 Γ 10β6
Where Ο=0.6, the equation can be solve to find the β multiplier;
βΓ 60.1 = 6.57 Γ 10β6
β=
6.57 Γ 10β6
60.1
= 1.09287 Γ 10β7
β was solved for all fuel equivalence ratios by performing the same procedure above. To find the
Kmols of N2 and O2 in each cylinder, β would then be multiplied by the coefficients of the products
of N2 and O2 for that specific fuel equivalence ratio. At a fuel equivalence ratio of 0.6;
kmols of N2 in each cylinder= βΓ 47 = 5.14 Γ 10β6
kmols of O2 in each cylinder= βΓ 0.625 = 5.46 Γ 10β8
Therefore the kmols/m^3 of N2 and O2 in each cylinder is;
N2 = 0.128412037
O2 = 0.013660855
Once this is performed the Exponential^(-67837/TB), d(NO)/dt and [NOx]/degree can be calculated.
By firstly knowing the burnt temperatures, the Exponential^(-67837)/TB) can be calculated; where
TB represents the burnt temperatures. The Zeldovich Model which is donated by dβNOβ/dt is
calculated by means of the following equation;
Equation 4
πβππβ
ππ‘
= 4.7 Γ 1013
Γ βπ2β Γ βπ2β0.5
Γ πΈππ(β67837
ππ΅β )
(McDonald, 2015)
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And the βNOxβ/degree can be calculated by using the following equation;
Equation 5
βππβ
ππππππ
=
πβππβ
ππ‘
Γ π‘πππ πππ 1Β° =
πβππβ
ππ‘
Γ
1
6 Γ π ππ
(McDonald, 2015)
These calculations for the fuel equivalence ratio = 0.6 can be seen in appendix 2.
From calculating the βNOxβ/degree, the total NOx during combustion can be found;
The total NOx during combustion/m^3 for fuel equivalence ratio 0.6 = 2.07191Γ 10β5
To find the kmols in the exhaust, βΓ (4.8 + 5.4 + 5 + 47) is calculated and is = 6.79764Γ 10β6
The kmols of NOx made is = βΓ 40 Γ 10β6
= 1.13532Γ 10β6
Finally, the Parts Per Million (PPM) of NOx produced =
πππππ ππ πππ₯ ππππ
πππππ ππ ππ₯βππ’π π‘
Γ 106
= 121.9194269
The NOx concentration levels were calculated for all fuel equivalence ratios and a table of values, as
well as a graphical representation of how the NOx PPM changes can be seen below respectively:
Table 3
Fuel Equivalence Ratio NO PPM
0.6 121.9194
0.65 590.8381
0.7 2311.211
0.75 7533.467
0.8 20904.44
0.85 49883.94
0.9 101449.5
0.95 163840
0.96 171257.1
0.97 172822.1
0.98 163962.7
0.99 134346.4
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Figure 2
Figure 2 - NOx concentration level as a function of fuel equivalence ratio.
3.0 The Effect of Adding E.G.R on NOx Concentration Levels
3.1 Determining the Coefficients of the Products
When adding Exhaust Gas Recirculation (EGR) back into the cylinder, it as a result increases the mass
burned. Therefore, the coefficients of the products burned will increase (more mass will need to be
burned). Hence, each product was multiplied by the percentage of EGR added, and added on to the
original amount of product. An example of how this was performed is seen below;
Effect on Exhaust Mixtures by adding 1% EGR:
CO2 7.6
H2O 8.55
O2 0.625
N2 47
CO2 + (CO2 X 0.01) = 7.676
H20 + (H2O X 0.01) = 8.6355
O2 + (O2 X 0.01) = 0.63125
N2 + (N2 X 0.01) = 47.47,
As stated above the amount burned will increase, this is proven by the coefficients of the products
increasing with an increasing EGR. All coefficients of the products burned were calculated (see
Appendix 3).
3.2 Determining the Initial Burning Temperatures and all Burning
Temperatures
As there is now a greater number of kmols of products being burned in each cylinder as a result of
adding EGR, this means that the flame temperatures will subsequently decrease. To find the initial
0
20000
40000
60000
80000
100000
120000
140000
160000
180000
200000
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
NOxPPM
Fuel Equivalence Ratio
NOx Concentration Level as a Function of Fuel
Equivalence Ratio
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flame temperatures, the same process as section 2.2 was undertaken. By a process of summating
the quadratic equation through the use of Excel from all the equations of each line (Figure 1), the
quadratic equation could be inputted into the online quadratic equation calculator to find the initial
temperature. It is important to note that the enthalpy released stays the same, as adding EGR does
the exact same on each side of the equation. For example, at ΙΈ=0.9 adding EGR=20%;
0.9πΆ8 π»18 + 12.5(π2 + 3.76π2) + 0.2(7.2πΆπ2 + 8.1π»2 π + 1.25π2 + 47π2)
β 1.2((7.2πΆπ2 + 8.1π»2 π + 1.25π2 + 47π2)
From using the equations in Table 1 and taking into consideration that the coefficients of products
are increasing, each initial flame temperature was found for each EGR percentage from 0% to 20% in
1% intervals. The initial flame temperatures can be seen in Appendix 4.
To find all of the burning temperatures, the same process was undertaken by using Equation 3. The
flame temperatures can be seen Appendix 5.
3.3 Determining NOx Concentration Levels
It is important to note that adding EGR back into the cylinder will change the alpha multiplier
required to determine the number of kmols/m^3 of N2 and O2. To find alpha, the coefficients of the
reactants for a fuel equivalence ratio of 0.95 will increase as the EGR increases. For example;
Adding 1% EGR to a fuel equivalence ratio of 0.95 is:
0.95πΆ8 π»18 + 12.5(π2 + 3.76π2) + 0.01(7.6πΆπ2 + 8.55π»2 π + 0.625π2 + 47π2)
0.95 Γ 1 + 12.5(4.76) + (0.01 Γ 63.775)
60.45 + 0.63775 = 61.08775
As can be seen with an increasing EGR, the air/fuel mass will increase. This was completed for all
EGR percentages; see Appendix 6.
To then find the alpha multiplier to find the correct trapped number of kmols/cycle, the number of
kmols of mixture in the cylinder is divided by 60.45 + [πΈπΊπ (%) Γ 63.775]. The alpha multiplier in
relation to an increasing EGR(%) can be seen in Appendix 7.
To find the kmols of N2 and O2, the alpha multiplier is multiplied by the coefficients of the products
of N2 and O2 for a fuel equivalence ratio of 0.95 (see appendix 8). Once found, the kmols/m^3 of N2
and O2 that are burned during combustion in the head volume of 40*10^-6m^3 can be found (also
seen in appendix 8).
Once this is calculated the Exponential^(-67837/TB) was found and then d(NO)/dt and [NOx]/degree
can be calculated by using the same process as before (in section 2.4) but taking into account that
the kmols/m^3 of N2 and O2 will be changing with an increasing EGR (%) (see Appendix 9).
Finally, the last procedure is to find the NOx PPM with a changing EGR (%). To perform this, the total
NOx during combustion/m^3 is summed up from the NOx (kmol/m^3) for each degree. Then it is
essential to multiply this summed value by 40*10^-6 to obtain the total NOx during combustion in
kmols. Then from there, the kmols in the exhaust is equal to the kmols in the exhaust multiplied by
(1+EGR(%)) from 0% to 20% EGR. To finish, the NOx PPM was calculated for all EGR percentages by
using the following equation;
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πππ₯ πππ =
πππππ ππ πππ₯ ππππ
πππππ ππ ππ₯βππ’π π‘
Γ 106
The values of the calculations can be seen in Appendix 10.
Once calculated, a graph the effect of adding EGR on NOx concentration levels can be plotted (see
Figure 3).
Figure 3
Figure 3 - Effect of Adding EGR on NOx Concentration Levels.
4.0 Analysis of NOx Model
4.1 Assumptions Made
When creating the NOx model there were a number of assumptions that were made, these included;
1) Rise in pressure after SOC is linear (in 2 bar/degree intervals)
2) Assuming the same volumetric efficiency of 40%
3) Pressure in the cylinder after inlet valve is closing is 1 bar
4) Temperature of air and fuel coming in is 20Β°C
5) Assume a Universal Gas Constant of 8316JKMOL^-1K^-1
6) Assume a Polytropic Index of 1.2
7) Assume the combustion volume in the head = 40ccβs
8) N2 and O2 concentrations are constant during combustion
When considering bullet point 1), the pressure rise after SOC is considered constant in 2 bar intervals
up to a pressure of 50bar. This is untrue to assume a constant rise in pressure; as firstly the NOx
0.000
20000.000
40000.000
60000.000
80000.000
100000.000
120000.000
140000.000
160000.000
180000.000
0 0.05 0.1 0.15 0.2 0.25
NOxPPM
EGR (%)
NOx Concentration Levels VS EGR Percentage
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model uses an increasing fuel equivalence ratio (from 0.6 to 0.99), therefore meaning that there will
be more kmols of mixture burned inside the cylinder. As more mass is burned it is expected that the
pressure rise accordingly, and the model does not take this into account.
Bullet point 2) considers that the volumetric efficiency stays constant at 40% throughout
combustion. However, this does not take into account that some constants in the NOx model change
the percentage of volumetric efficiency. These include;
ο· Mixture temperature (which is influenced by heat transfer)
ο· Ratio of exhaust to inlet manifold pressures
ο· Compression ratio
ο· Intake and manifold port design
ο· Intake and exhaust valve geometry, size, lift and timings.
(Heywood, 1988).
Bullet point 3) considers that when the inlet valve closes the pressure will be 1 bar (atmospheric
pressure). This however does not concern that there will be reverse flow of gas when drawn into the
cylinder. Before the intake valve closes there is a pressure drop inside the cylinder as the piston
reaches BDC (drawing in air). The air will be drawn into the cylinder even when the piston passes
BDC; this inward movement of air combined with the upward movement of the piston on the
following stroke causes a pressure rise (so will not be exactly atmospheric); also stopping the inward
flow of gas. This is the point at which the intake valve should close (Hillier & Coombes, 2004).
Plenum size will also alter the intake pressure. The NOx model does not take these factors into
account.
Concerning bullet point 4) the temperature of the fuel and air mixture coming into the cylinder is
considered to be 20Β°C. However the NOx model doesnβt take into account the radiated heat (heat
transfer) from the surroundings. It is therefore likely that the temperature of the air and fuel coming
in be higher than 20Β°C. It is suggested that an increase in fuel temperature coming into the cylinder
can reduce NOx levels (Chen, 2008).
Bullet point 5): The Universal Gas Constant (UGC) comes from the Ideal Gas Law and has been
discovered through past physical experiments undertaken. The UGC is a constant of proportionality
that relates between the energy scale in physics and the temperature scale. This value of the UGC
will not change due to it being a set value, however it is suggested that there is a slight uncertainty
in the value of 9.1*10^-7Jmol^-1K^-1 (Winterbone, 2015). This can therefore alter the NOx PPM if
there is uncertainty!
Bullet Point 6): A polytropic Index can vary from 1 β 1.4 (Engineering Toolbox, 2015), which as tested
on the NOx model at a fuel equivalence ratio of 0.95 can vary the temperature by 34 kelvin. Hence it
was calculated that this change in the polytropic index could vary the NOx PPM produced by a total
of 51,671.52 (between the values of 1 β 1.4). This is a 31.53% change in NOx PPM produced when
compared to the NOx PPM at a fuel equivalence ratio of 0.95.
Bullet point 7) assumes that the combustion volume in the head is 40ccβs. However, this may not be
the exact value for the engine in question. A change in this volume will also change the volumetric
efficiency of the engine as a result of the compression ratio changing. This will therefore result in the
amount of NOx produced changing; a higher volume will decrease the amount of NOx PPM
produced whereas a lower volume will increase the amount of NOx PPM (Hill, 2006).
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Bullet point 8) assumes that N2 and O2 concentrations are constant during combustion. This is due
to the fact that the quantity produced during combustion is very small and change would not be
great enough to effect the NOx PPM much (McDonald, 2015).
4.2 Effectiveness of Calculations
Taking into consideration the assumptions made, the calculations have been very effective as they
have shown that NOx concentration levels as a result of fuel equivalence ratio and the introduction
of EGR on NOx concentration levels can be predicted. At the choice of the user, Excel can calculate
results to a large number of decimal places for each calculation performed. This then makes sure
that the end result (as the model performs many calculations) is accurate. A large part of the NOx
model was calculating the initial flame temperatures for each fuel equivalence ratio and for each
EGR percentage. To obtain the flame temperatures, a quadratic formula for each line of
πΆπ2, π»2 π, π2 πππ π2 was gathered from Figure 1. The number of decimal places outputted by the
formula trend line label could also be chosen by the user (see Figure 4), this would ensure that the
initial temperatures were very accurate. It was essential that this initial stage be accurate as it
greatly affects the NOx PPM at the end of the NOx model.
Figure 4
Figure 4 - Trend line decimal places.
4.3 Mechanisms Responsible for NOx Formation
There are a number of factors responsible for the NOx PPM being so high. These factors are;
1) There are a lot of free O2βs
2) Increasing Fuel Equivalence Ratio
3) The burning temperatures are too high because the polytropic index is too high
4) The pressure is too high
(McDonald, 2015).
Considering bullet point 2), the increasing fuel equivalence ratio increases the amount of NOx PPM
produced as there is more Iso-octane being burned in the cylinder. As discussed before, the
polytropic index in bullet point 3) can range between 1 β 1.4, it is a direct cause of calculating the
temperatures after finding the initial temperature so will increase the NOx PPM produced
considerably (as previously mentioned in section 4.1). Since the EXP^(67837/TB) is very sensitive to
alterations in TB, this will alter the NOx PPM greatly. When concerning bullet point 4), the pressure
maybe too high for lower fuel equivalence ratioβs but more suited to higher equivalence ratioβs. This
is due to the fact that there is more mass in the cylinder being burned so the pressure in the cylinder
will increase. A factor that has to also be taken into account is that the combustion process will last
for greater than 20Β° of crank angle. It would be interesting to be able to produce a NOx model from
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SOC to when the flame eventually dissipates (sometimes up to a maximum of 90Β° crank angle)
(Heywood, 1988).
5.0 References
Chen, G., 2008. Gas Turbines Power. Study of Fuel Temperature Effects on Fuel Injection,
Combustion, and Emissions of Direct-Injection Diesel Engines, 131(2), p. 8.
Engineering Toolbox, 2015. Compression and Expansion of Gases. [Online]
Available at: http://www.engineeringtoolbox.com/compression-expansion-gases-d_605.html
[Accessed 15 December 2015].
Heywood, J., 1988. Combustion in Spark-Ignition Engines. In: J. Holman, ed. Internal Combustion
Engine Fundamentals. New York: McGraw-Hill, p. 374.
Heywood, J., 1988. Volumetric Efficiency. In: J. Holman, ed. Internal Combustion Engine
Fundamentals . New York: McGraw-Hill Book Company, p. 209.
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COMBUSTION EMISSIONS, CALCULATIONS AND MAPPING FOR EFFICIENCY
15 | P a g e
6.3 Appendix 3EGR(%)00.010.020.030.040.050.060.070.080.090.10.110.120.130.140.150.160.170.180.190.2
CO27.67.6767.7527.8287.9047.988.0568.1328.2088.2848.368.4368.5128.5888.6648.748.8168.8928.9689.0449.12
H2O8.558.63558.7218.80658.8928.97759.0639.14859.2349.31959.4059.49059.5769.66159.7479.83259.91810.003510.08910.174510.26
O20.6250.631250.63750.643750.650.656250.66250.668750.6750.681250.68750.693750.70.706250.71250.718750.7250.731250.73750.743750.75
N24747.4747.9448.4148.8849.3549.8250.2950.7651.2351.752.1752.6453.1153.5854.0554.5254.9955.4655.9356.4