FINAL TEST
Questions based on PFI Engine:
Q1: What is the Compression ratio for the engine?
Compression ratio is defined as the ratio of the maximum to minimum volume ratio in the cylinder of an internal combustion engine.
The compression ratio of an engine =
Max. Vol. of the cylinder/Min. Vol. of the cylinder
it is given by
rc = Vmax/Vmin = (Vd+Vc)/Vc
Where,
rc = Compression Ratio
Vc = Clearence Volume
Vd = Displacement Volume / Swepy Volume
A high compression ratio is desirable because it allows an engine to extract more mechanical energy from a given mass of air-fuel mixture due to its higher thermal efficiency. This occurs because internal combustion engines are basically heating engines & higher efficiency is created because higher compression ratios permit the same combustion temperature to be reached with less fuel, while giving a longer expansion cycle, creating more mechanical power output & lowering the exhaust temperature.
Q2. Modify the compression ratio for this engine to 10.3 without changing geometrical parameters.
To modify the compression ratio for an engine without changing its geometrical parameters is possible only by changing its clearance volume or its displacement volume swept volume.
(a) Let us consider varying Displacement volume & keeping the Clearance volume constant sp as to achieve a 10.3 compression ratio.
We know that
For PFI Engine
Vmax = 5.74*10^-4 m^3
Vmin = 5.70*10^-5 m^3 = Vc
Vmax = Vd + Vc
Vd = 5.17*10^-4 m^3
Now, let us carry Clearance volume & keeping Displacement volume constant so as to obtain a 10.3 Compression Ratio.
rc = Vmax/Vmin = 10.3
Upon substituting the values to the above equation,
We get,
New clearance volume Vc' = 5.55*10^-5 m^3
(b) Lets us consider keeping constant Displacement volume & carrying the Clearance Volume constant so as to achieve a 10.3 compression ratio.
rc = Vmax/Vmin = 10.3
In the above equation substituting the value of
Vc = 5.70*10^-5 m^3
we get,
New Displacement volume, Vd' = 5.336*10^-3 m^3
Now by using any of the two methods, we can get a compression ratio of 10.3
That is by substituting New clearance volume & Old displacement volume in the Compression ratio equation.
Q3. Calculate Volumetric Efficiency for this engine.
Volumetric efficiency in an internal combustion engine design refers to the efficiency with which the engine can move the Charge of fuel & sir into & out of the cylinders. It also denotes the ratio of air volume drawn into the cylinder to the cylinder's swept volume. More specifically, volumetric efficiency is a ratio of the mass of air & fuel that is trapped by the cylinder during induction divided by the mass that would occupy the displaced volume is the air density in the cylinder was equal to the ambient air density.
The volumetric efficiency of an engine is given by,
ηv=Volume of air sucked into the cylinder during intake strokeDisplacement Volumeηv=Volume of air sucked into the cylinder during intake strokeDisplacement Volume
ηv=VaVd×100ηv=VaVd×100
Displacement volume was found previously & was equal to 5.17×10−4m35.17×10-4m3
To find the amount of air sucked into the cylinder during the intake stroke, assuming the air used is ideal in nature.
Now by using the ideal gas equation,
Va = m*R*T/P
where,
P= Pressure of air at atmospheric conditions = 101325 Nm^-2
m = Mass of air
R = Universal gas constant = 287 J/Kg-k
T = temperature of air at atmospheric conditions = 293k
Hence,
Va = 4.4317*10^-4 m^3
Volumetric Efficiency = 85.719 %
Q4. Measure the air mass flow rate for this engine in Kg/s.
To find the air mass flow rate for the engine we need to know the mass of the air. The mass of the air can be found by the Mass vs Crank plot in the 3rd3rd equation. According to the plot
Mass of the air, m = 0.000534 kg
The mass of the air measured is the mass flow rate of air into the cylinder per one cycle that is for 360 deg.
The speed of the engine is 3000 RPM. Converting it into RPS,
Speed of the engine in RPS =3000/60 = 50 RPS
Thus, the actual angle covered in one second is 50 x 360 = 18000 deg
To find the mass of air entering into the engine at 18000 deg,
we know that,
Mass of air entering at 360 deg = 0.000534 kg
For Mass of air entering at 18000 deg = 0.000534 x 50 = 0.0267 kg/s
Air mass flow rate for this engine = 0.267 kg/s
Q5. Why is the cell count varying during the simulation?
Total cell count variation is directly dependent on the Adaptive Mesh Refinement & Fixed embedding we provide to simulate a CFD problem accurately. In the PFI engine simulation, we provided velocity & temperature AMR's to have accurate values of velocity & temperature at the end of the simulation. We provided a scale of 3 & subgrid scaling of 1 for velocity & a scale of 3 & subgrid scaling of 2.5 for temperature.
We had provided two levels of fixed embedding to the cylinders whose effect had the cell sizes decreased inside the engine. Also, separate fixed embedding was provided for both injector & spark plug. During simulation, the subgrid criterion changes making the AMR come into action which thereby varies the total cell count.
Q6. In a real engine, valves are in contact with the cylinder head but while running simulation, there is a small gap
(a) Why?
In the real engine, the valves are in contact with the cylinder head because to ensure that the heat inside the cylinder is sealed so that the engine does not lose its thermal efficiency & also to dissipate the heat from cylinder walls to the cylinder head. In a real engine using a valve spring & retainer, the valves are kept in contact with the cylinder head.
In the simulation, the valve spring & the retainers were not designed, so a small gap was provided between the valve & cylinder head.
(b) What happens if this gap is big?
If the maintained gap becomes big, the heat transfer between the cylinder head & cylinder will not happen. These causes raise in temperature o the cylinder which will lead to detonation/pre-ignition/ induces knock & also engine might emit more unwanted gases like NOx & Hiroy soot which will tend to spoil the environment. This can also cause the life of the engine & its efficiency to decreases by a great margin.
Q7. How many seconds does combustion last in this case?
From the engine performance calculator available in CONVERGE we had found that the engine is run through 240 deg in the combustion stroke. The engine speed was 3000 RPM or 50RPS. In the previous questions, we saw that engine covered 18000 deg in one second ie,
360 x 60 = 18000 deg
Now we need to find how many seconds will engine take to cover 240 deg if the engine took 1 second to cover 18000 deg.
t = 240/18000 = 0.0133 sec
Thus, Combustion stroke lasted for 0.0133 sec.
Questions based on DIESEL Engine:
Q 1. What is an advantage of SECTOR Simulation?
The primary advantage of a SECTOR simulation is the reduction of computation time. In a sector simulation, only a portion(sector) of the combustion chamber is simulated & hence the number of cells required will be less. this is a closed cycle simulation, the number of parameters used & boundary conditions required is minimum.
Q 2. When will the sector approach not work?
The SECTOR approach will not work in the following conditions;
The sector approach does not hold good for open-cycle analysis. In the open cycle approach the engine valves ie, intake, & exhaust valves are kept open & are required to be modeled. As the position of valves varies according to time, a sector approach will fail to capture the effects of the valves with respect to time.
If a multi-injector engine is considered, the sector approach will not work. This is because in the sector approach the injector is supposed to be symmetric about the center axis of the engine.
Q3. How to choose Start Crank Angle (SCA) & End Crank Angle (ECA) for a sector simulation ?
In a sector simulation the strokes which are simulated are compression & expansion stroke. So, the Start Crank Angle (SCA) can be choosen as the start of the compression stroke & the End Crank Angle (ECA) can be choosen as the end of the expansion stroke. If the start of the suction stroke is taken as 0 degree crank angle, then the start of the compression stroke becomes 180 degree crank angle & the end of the expansion stroke becomes 540 degree crank angles. So, the the Start Crank Angle (SCA) is 180 degree crank angle & the End Crank Angle (ECA) is 540 degree crank angle.
Q4. Can we convert the PFI case into a SECTOR case? Explain your answer ?
No, the PFI case cannot be convereted into a sector case because it has intake & exhaust ports & the injectfion point is at the intake port which cannot be divided into a sector element which can represent the entire PFI geometry. The PFI case has an inlet & outlet which cannot allow a sector simulation as the sector simulation is suited for closed & symmentric geometry. In this case, all the four strokes are to be performed which requires the PFI case to remain open with valves & ports.