FINAL TEST

                                                                      FINAL TEST

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 = `(text(Maximum Volume of the Cylinder))/(text(Minimum Volume of the Cylinder))`

it is given by

`r_c=(V_max)/(V_min)=(V_d+V_c)/(V_c)`

Where,

`r_c` = Compression Ratio

`V_c` = Clearence Volume

`V_d` = 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 and 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 and 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 and keeping the Clearance volume constant sp as to achieve a 10.3 compression ratio.

We know that

For PFI Engine

`V_max=5.74xx10^-4m^3`

`V_min=5.70xx10^-5m^3= V_c`

Also,

`V_max= V_d+V_c`

`V_d=V_max-V_c`

`V_d=5.74xx10^-4-5.70xx10^-5`

`V_d= 5.17xx10^-4m^3`

Now, let us carry Clearance volume and keeping Displacement volume constant so as to obtain a 10.3 Compression Ratio.

`r_c=(V_max)/(V_min)=(V_d+V_c)/(V_c)=10.3`

Upon substituting the values to the above equation,

We get,

New clearance volume `V_c'=5.55xx10^-5`

(b) Lets us consider keeping constant Displacement volume and carrying the Clearance Volume constant so as to achieve a 10.3 compression ratio.

`r_c=(V_max)/(V_min)=(V_d+V_c)/(V_c)=10.3`

In the above equation substituting the value of `V_c= 5.70xx10^-5` 

we get,

New Displacement volume, `V_d' =5.336xx10^-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 and Old displacement volume in the Compression ratio equation.

OR 

By substituting Old clearance volume and New displacement volume in the Compression ratio equation we can obtain a 10.3 (desired) compression ratio.

Using case (a):

`r_c=(V_max)/(V_min)=(V_d+V_c')/(V_c')`

`r_c=(5.17xx10^-4+5.55xx10^-5)/(5.55xx10^-5)`

`r_c=10.3`

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 and sir into and 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 and 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,

`eta_v=(text(Volume of air sucked into the cylinder during intake stroke))/(text(Displacement Volume))`

`eta_v=(V_a)/(V_d)xx100`

Displacement volume was found previously and was equal to `5.17xx10^-4m^3`

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,

`P.V_a=m.R.T`

`:. V_a=(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

`= 0.000534 kg`

Substituting the values of P,m, R, and T in the ideal gas equation, we get,

`V_a=4.4317 xx 10^-4m^3`

Substituting the value in the Volumetric efficiency equation.

We get,

`eta_v=(4.4317xx10^-4)/(5.17xx10^-4)xx100=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 `3^(rd)` 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 ie, 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 and Fixed embedding we provide to simulate a CFD problem accurately. In the PFI engine simulation, we provided velocity and temperature AMR's to have accurate values of velocity and temperature at the end of the simulation. We provided a scale of 3 and subgrid scaling of 1 for velocity and a scale of 3 and 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 and 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 and also to dissipate the heat from cylinder walls to the cylinder head. In a real engine using a valve spring and retainer, the valves are kept in contact with the cylinder head.

In the simulation, the valve spring and the retainers were not designed, so a small gap was provided between the valve and cylinder head.

(b) What happens if this gap is big?

If the maintained gap becomes big, the heat transfer between the cylinder head and cylinder will not happen. These causes raise in temperature o the cylinder which will lead to detonation/pre-ignition/ induces knock and also engine might emit more unwanted gases like NOx and Hiroy soot which will tend to spoil the environment. This can also cause the life of the engine and 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`

`:.` The engine took 0.0133 seconds to cover 240 deg.

That is the combustion stroke lasted for about 0.0133 seconds.

DIESEL ENGINE:

Q1. What is an advantage of SECTOR Simulation?

The major advantage of the Sector Simulation is, for computational efficiency we take one sector of the domain and simulate. This will also help in a faster run time. As this is a closed cycle simulation, the number of parameters used and boundary conditions required is minimum. Since in a sector simulation only a part o the combustion chamber is simulated, the number of cells required to capture the effects is less. Thus, the simulation is done faster and the results are accurate.

Q2. When will the sector approach not work?

The sector approach does not hold good for open-cycle analysis. In the open cycle approach the engine valves ie, intake, and exhaust valves are kept open and 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) and End Crank Angle (ECA) for a sector simulation?

As the sector approach holds good for only closed-cycle analysis. The beginning o the simulation can be taken at the beginning o the compression stroke and the end of the simulation can be taken at the end of the expansion stroke.

Let us assume that the start o the fuel injection takes at 0 CAD, 180 CAD can be taken as start and 540 CAD can be taken as the end of the crank angle for the Diesel engine.

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

The PFI case is an open cycle analysis. As discussed before earlier sector simulation can not run an open cycle analysis. The sector simulation can only simulate a closed part of the 360 deg engine. In a PFI engine, the position of the fuel injector might not be symmetrical, and therefore, stimulating a part of the engine and then extrapolating to the complete PFI engine might not work. The sector simulation can not simulate the spark plug and capture its effect.

Considering all the above facts, it is highly impossible to convert a PFI case into a Sector Case.