MULTI DIMENSIONAL SIMULATION OF A PORT FUEL INJECTED ENGINE(PART 2)
INTAKE SYSTEM BOTTOM :
In this part of intake system, fuel is being injected via nozzles into an intake port and fuel and air mixes in this region and because of mixing , temperature of this part of intake port is slightly higher than top part of intake port.
Also some part fuel gets vapourize in this part .
Like in combustion chamber ,there is no chemical reaction occurring here and therefore stoichiometry is not applicable here .
Now to calculate the mass flow rate of fuel or total injected mass , we do the following,
To calculate total injected mass, we will run the simulation without spray and just with air only and will observe how much air is being trapped in the combustion chamber (i.e in cylinder) as that is what we need to burn it.
Calculations :
Mass of air trapped in the combustion chamber of an engine is
m(air) = 9.538*e-5 grams.
This value we got from CFD analysis as discussed above.
And we know , Molecular weight of air , (MWair) = 29 grams.
Therefore , moles of air , n(air) = (m(air)/MWair)
= ((9.538*e-5)/29)
n(air) = 3.289 * e-6 moles
Now , we know from stoichiometric combustion reaction,
For 1 mole of fuel (C8H18) , 12.5 moles of air is required.
Therefore , by unitary method,
For 3.289 * e-6 moles of air ,
Moles of fuel (C8H18) required = (3.289 * e-6 )/12.5
n(C8H18) = 2.6315 * e-7 moles
We know,
n(C8H18) = (m(C8H18)/MW(C8H18) )
Therefore, m(C8H18) = n(C8H18) * MW(C8H18)
m(C8H18) = 2.6315 * e-7 * 114
Therefore, m (C8H18) = 3 * e-5 grams .`
This is how we calculate the total injection mass and give it as an input.
Injection of fuel begins few crank angles before TDC i.e few crank angles before compression ends.
And here compression stroke is from -540⁰ to -360⁰.
Start of injection = -480⁰.
This is the reason we start our simulation at -520⁰ as spray phenomena occurs here.
Explaination of calculation of mass fraction of species in the Intake port bottom:
When the fuel (C8H18) is sprayed via nozzles , some part of the fuel get trapped in the intake port only due to fuel hitting the wall of the port .
Let say 5% of the fuel gets trapped in 1st cycle then in the 2nd cycle this 5% trapped fuel will come first into the combustion chamber .
Also we will be spraying fuel in the 2nd cycle and again some amount will be trapped in the port and this will go on.
After some cycles the amount of fuel getting trapped will reach the steady state and so we ran for the 10 cycles i.e crank angle = 720 *10 = 7200 and will get the values for mass fraction of each species in the intake port.
In the intake port , there is no reactions and so we cannot apply stoichiometry.
Source and sink modeling :
Modeling of source and sink means to model spark plug and spark plug electrode . there is a need to model this as without modeling , it will require a very fine mesh for spark plug electrode to ignite and running with such a small mesh size will take infinite time to simulate which is not feasible. Hence there is a need to model source and sink.
Spark plug and spark plug electrode
Spark plug ignition
Spray modeling :
Spray needs to be modeled as to capture the real picture of spray i.e how it is coming into the port , how it is breaking into ligaments and then into tiny tiny particles , how it is hitting the wall of the port , how and where it starts to evaporate.
Without modeling it , we will be unable to get the spray phenomena as it requires really very small mesh size and running with such a small mesh will take infinite time to simulate.
Spray phenomena
Nozzle orientation :
Nozzle needs to be oriented in a particular manner depends upon the system and how spray is desired. Nozzle orientation is essential as all above parameters depends upon nozzle orientation.
Results and Validation :
1) Cylinder pressure plot (CONVERGE CFD result)
Validation (JOHN HEYWOOD results)
2) Heat release rate (HRR) plot (CONVERGE CFD result)
Validation (JOHN HEYWOOD results) :
3) Integrated heat release (IHR) (CONVERGE CFD result) :
Validation (JOHN HEYWOOD results):
Minor differences are there in the results because of some difference in engine parameters of John Heywood and our engine parameters.
Engine parameters :
Crank speed = 3000 rpm (CONVERGE CFD)
Crank speed = 1500rpm (JOHN HEYWOOD)
Cylinder temperature plot :
Cylinder volume plot :
Calculations of INDICATED POWER and INDICATED TORQUE:
Converge provides ENGINE PERFORMANCE CALCULATOR which calculates the values for INDICATED WORK and IMEP as shown below,
Indicated Work is the work obtained during Combustion period. And the duration of Combustion is 240.18⁰
Now we will be converting this duration of combustion into seconds,
1 revolution --> 360⁰
x <-- 240.18⁰
x = `(240.18)/360`
x = 0.66721 revolutions
Now we also know the speed of an Engine
i.e N = 3000 RPM
i.e 1 min --> 3000 revolutions
y <-- 0.66721 revolutions
`∴`y = (0.66721)/3000
`y = 2.224*e^(-4) min
`y = 2.224*e^(-4)*60 seconds
y = 0.01334 seconds
Now we know,
Power = (Work (J))/(Time (s))
So,
Indicated Power (P_i) = (Indicated Work (J))/(Time for combustion(s))
= (467.302)/(0.01334)
P_i = 35130.884 Watts = Indicated Pressure
P_i = 35.130 KW
Also we know,
Power = (2*π*N*T)/60
P_i = (2*π*N*T_i)/60
T_i = (60*P_i)/(2*π*N)
= (60*35130.884)/(2*π*3000)
T_i = 111.8817 N-m = Indicated Torque
Calculating Combustion Efficiency:
Calorific value of an Iso-Octane = 44.8 MJ/kg
Total mass of fuel injected = 3*e^(-5) kg `
∴ Total input energy = (Total mass of fuel injected)* (Calorific value)
= 3*e^(-5) * 44.8 * e^6
∴ Total input energy = 1344 Joules
Total Output energy:
Total output energy is nothing but the Integrated heat release (IHR) which is given by Converge.
∴ Total Output energy = 1240.88 Joules = IHR
η_combustion = (Total Output energy)/(Total input energy)
= (1240.88)/1344
η_combustion = 92.327%`
LINKS for an Animations are given below,
1) Animation for COMBUSTION within the Cylinder,
2) Animation for SPRAY of fuel coming into the Cylinder from Intake Port,
Thank you.