FINAL TEST

Q1. What is the Compression ratio for the engine?

From the Volume vs Crank Angle Plot

Here the maximum volume is = 0.0005738 and the minimum volume is = 5.7032 e-5 m^3

And the compression ratio is Maximum Volume / Minimum Volume = 0.0005738 / 5.7032e-5 = 10.06

Q2. Modify the compression ratio for this engine to 10.3 without changing geometrical parameters.

The engine compression ratio is purely a function of the engine geometrical parameters (bore, stroke, clearance volume, crank radius) because compression ratio is a ratio of stroke+clearance volume to the clearance volume.

compression ratio = (Stroke Volume + Clearance Volume)/Clearance Volume

where Stroke Volume = `pi*"Bore"^2/4*"Stroke Length"`

Storke Length = 2* Crank Radius

So, the compression ratio cannot be changed without changing the geometrical parameters like bore, stroke, clearance volume, crank radius

Geometrically there are two ways of changing the Compression ratio

1. By changing the clearance volume

Clearance volume can be changed by offsetting the crank pin

To get CR of 10.3

(Vc + Vs)/Vc = 10.3

We know that Vs = Max Volume - Min Volume = 0.0005738 - 5.7032e-5 = 5.1676e-4 m^3

Vc = Min Volume = 5.7032e-5 m^3

(Vc + Vs)/Vc = 10.3

Vc + Vs = 10.3Vc

Vc - 10.3 Vc = -Vs

Vs = 10.3Vc - Vc = ( 10.3 - 1 )* Vc = 9.3Vc

New Vc = Vs/9.3 = 5.1676e-4/9.3 = 5.556e-5 m^3

So, clearance volume needs to be changed from 5.7032e-5 m^3 to 5.556e-5 m^3 to increase the compression ratio from 10.06 to 10.3

2. By changing the Stroke Volume

(Vc + Vs)/Vc = 10.3

Vc + Vs = 10.3Vc

Vc - 10.3 Vc = -Vs

Vs = 10.3Vc - Vc = ( 10.3 - 1 )* Vc = 9.3Vc

New Vs = 9.3Vc = 9.3*5.7032e-5 m^3 = 5.3039 e-4 m^3

Now the Vs can be changed by changing either the bore, or the stroke or both.

`V_s = {pi*D^2}/d*L`

We can calculate the required(new) stroke length by simply substituting the new stroke volume in the above equation.

Q3. Calculate Volumetric Efficiency for this engine

Volumetric Efficiency = Density of air-fuel mixture drawn inside the cylinder / Density of air inside the intake manifold

Density of air-fuel mixture drawn inside the cylinder = mass of air after the intake stroke/max volume

Mass of air can be obtained from the mass plot in thermo_region0 mass plot (cylinder region)

mass = 0.0005005 kg

Volume can be obtained from the volume plot of thermo_region0(cylinder region)

Volume = 0.000574219 m^3

Density of air inside the cylinder = 0.0005005 kg/0.000574219 m^3 = 0.8716 kg/m^3

Density of air inside the intake manifold is a function of temperature and pressure

Pressure is atmospheric when the intake valve is closed

Temperature can be obtained from the mean temperature plot in thermo_region1(Intake port 1)

Temperaure varies between 360 to 373 K,

Say the mean temperature is around 367 K, density of air at 367K is

`PV =mRT`

`P = rho/{R*T}`

Where

P is the Atmospheric Pressure = 101325 Pa

R is the Gas Constant for air = 287 J/Kg K

T is the temperature of air inside the intake manifold.

`rho = P/{RT} = {101325 }/{287 *367} = 0.9619 {kg}/m^3`

So Volumetric Efficiency

`eta_{vol} = rho_"cylinder"/rho_"Intake manifold" = 0.8716/0.9619 = 0.9061`

or 90.61%

Q4. Measure the air mass flow rate for this engine in kg/s

Mass inside the cylinder is 0.0005005 kg when all the valves are closed as obtained from Q3.

Total Mass entering the cylinder (air+fuel) in 720 degrees => 0.0005005 Kg

Mass of Fuel sprayed in 720 degrees => 3e-5 Kg

So, Mass of air entering the cylinder in 720 degrees = Total Mass entering the cylinder (air+fuel) in 720 degrees - Mass of fuel injected in 720 degrees

Mass of air entering the cylinder in 720 degrees = (0.0005005 - 3e-5) Kg

Mass of air entering the cylinder in 720 degrees = 4.705e-4 Kg

Mass of air entering the cylinder in 360 degrees (1 Rotation) = (4.705e-4)/2 Kg = 2.3525e-4 Kg

Engine RPM = 3000

Engine RPS = RPM/60 = 3000/60 = 50 Rotations per second

Mass Flow Rate of Air = RPS* Mass of air entering the cylinder in 1 Rotation = 50*2.3525e-4 = 0.0117625 Kg/s

Q5.Why is the cell count varying during the simulation?

The cell count varies because of Adaptive mesh refinement(AMR) and fixed embedding. AMR is provided to refine the mesh size based on the Velocity and Temperature gradients, it refines the mesh when the gradients are steep. Fixed Embedding is provided to refine the mesh in the critical regions and boundaries where the flow needs to be captured more accurately. Giving Permenant mode of Fixed Embedding will increase the cell count but it will not cause variation in cell count during the simulation. Whereas a Cyclic type of fixed embedding will cause the cell count to vary during the simulation.

Q6. In a real engine, valves are in contact with the cylinder head but while running simulation, there is a small gap

• WHY?

During the simulation there is a small gap is kept to prevent the formation of intersecting triangles in converge. To control the flow of fluid through the valves we use Events in converge

• What happens if this gap is big?

Smaller the gap the solution will be more accurate but it will consume more time. Bigger the gap, the solution will be less accurate but faster. So there is a tradeoff between speed and accuracy.

Q7. How many seconds does combustion last in this case?

From the Plot of mass of IC8H18 vs crank angle we can see that the mass of the fuel starts to drops from -5 degrees till 45 degrees which indicates combustion.

So combution duration in terms of crank angles is 5+45 =50 degrees

Engine RPM = 3000

Engine RPS = 3000/60 = 50

Degrees per second = 50*360 = 18000 degrees/s

Duration of Combustion = combution duration in terms of crank angle/Degrees per second = 50/18000 = 0.0027777 seconds = 2.78 milliseconds

Questions based on DIESEL Engine:

Q1. What is an advantage of SECTOR Simulation?

Sector Simulation is computationally cheaper. It generally is used for initial screening and to simulate the effect of the piston profile on the engine performance and emissions. In sector simulation only one sector is simulated and hence the cell count required is lesser than complete simulation. And it is a closed cycle simulation so only compression and power strokes are simulated.

Q2. When will the sector approach not work?

For Closed cycle simulation, sector approach will only work if the piston profile is axisymmetric (i.e. symmetrical around the axis).

It cannot be used for open cycle simulation where we need to model the intake and exhaust processes.

Q3. How to choose Start Crank Angle (SCA) and End Crank Angle (ECA) for a sector simulation?

In a closed cycle simulation we only simulate the compression and power stroke. We can choose the Start Crank Angle (SCA) as the start of compression and End Crank Angle (ECA) as the end of power stroke respectively. If we assume that intake stroke starts at 0 degrees, then compression will start at 180 degrees (SCA) and power stroke will end at 540 degrees (ECA).

Q4. Can we convert the PFI case into a SECTOR case? Explain your answer?

No. PFI case cannot be converted into a sector case. PFI stands for Port Fuel Injection which means fuel is injected inside the intake port. So it is very necessory to simulate the fuel droplet vaporization in the intake port in case of a PFI simulation. Sector simulation is a closed cycle simulation where intake and exhaust processes are not simulated, only the compression and power strokes are simulated. In case of Sector simulation we cannot divide the intake and exhause valve for one sector. Hence it is not possible to convert a PFI case into a sector case.