FINAL TEST

PFI Engine:

1. What is the Compression ratio for the engine?

Compression ratio is defined as the ratio of the maximum to minimum volume in the cylinder of an internal combustion

engine.

Compression ratio of an engine = $\frac{\text{Maximum Volume of the Cylinder}}{\text{Minimum Volume of the Cylinder}}$

 Ie, it is given by

$\underset{c}{r}=\frac{\underset{max}{V}}{\underset{min}{V}}=\frac{\underset{d}{V}+\underset{c}{V}}{\underset{c}{V}}$

 

Where,

 $\underset{c}{V}$

= Clearance Volume and 

$\underset{d}{V}$

= Displacement Volume/ Swept 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 heat

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.

 

2. 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/swept volume.

 

(a) Let us consider varrying Displacement volume and keeping the Clearance volume constant so as to achieve 10.3

compression ratio.

 wkt, for the PFI engine

$\underset{max}{V}$= $5.74\times {10}^{-4}$

   and

$\underset{min}{V}=5.70\times {10}^{-5}=\underset{c}{V}$

 

 Also,

$\underset{max}{V}=\underset{d}{V}+\underset{c}{V}$

 $\underset{d}{V}=\underset{max}{V}-\underset{c}{V}$

  $\underset{d}{V}$ = $5.74\times {10}^{-4}-5.70\times {10}^{-5}$

  $\underset{d}{V}$ = $5.17\times {10}^{-4}{m}^3$

 

 Now, let us varry Clearance volume and keeping Displacement volume as constant so as to obtain 10.3 Compression ratio.

$\underset{c}{r}=\frac{\underset{max}{V}}{\underset{min}{V}}=\frac{\underset{d}{V}+\underset{c}{V}}{\underset{c}{V}}=10.3$

 

Upon substituting the values to the above equation,

we get,

New clearance volume $\underset{c}{V}\text{'}$= $5.55\times {10}^{-5}$

(b) Let us consider keeping constant Displacement volume and carrying the Clearance volume constant so as to achieve 10.3

compression ratio.

 $\underset{c}{r}=\frac{\underset{max}{V}}{\underset{min}{V}}=\frac{\underset{d}{V}+\underset{c}{V}}{\underset{c}{V}}=10.3$

 In the above euqation substituting the value of $\underset{c}{V}=5.70\times {10}^{-5}$

 we get,

New Displacement volume, $\underset{d}{V}\text{'}=5.336\times {10}^{-3}$

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 susbstituting Old Clearance volume and New displacement volume in the Compression ratio equation we can obtain 10.3

(desired) compression ratio.

Using case (a):

$\underset{c}{r}=\frac{\underset{max}{V}}{\underset{min}{V}}=\frac{\underset{d}{V}+\underset{c}{V}\text{'}}{\underset{c}{V}\text{'}}$

$\underset{c}{r}=\frac{5.14\times {10}^{-4}+10.55\times {10}^{-5}}{5.55\times {10}^{-5}}$

$\underset{c}{r}=10.3$

 

3. 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 air 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 if the air density in the cylinder were

equal to the ambient air density.

The Volumetric effieciency of an Engine is given by,

$\underset{v}{\eta }=\frac{\text{Volume of air sucked into the cylinder during intake stroke}}{Displacementvolume}$

$\underset{v}{\eta }=\frac{\underset{a}{V}}{\underset{d}{V}}\times 100$

Displacement volume was found previously and was equal to $5.17\times {10}^{-4}{m}^3$

To find the amount of air sucked into the cylinder during intake stroke, assuming the air used is ideal in nature.

Now by using ideal gas equation,

$P.\underset{a}{V}=m.R.T$

 $\therefore$

$\underset{a}{V}=\frac{m.R.T}{P}$

Where,

P = Pressure of air at atmospheric conditions = 101325$N{m}^{-2}$

m = Mass of the air 

R = Universal gas constant = 287 J/Kg-k

T = Temperature of air at atomospheric conditions = 293k

Plotting Mass V/S Crank angle to find out the mass of the air inducted into the engine,

 

m = 0.000534 Kg

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

$\underset{a}{V}=4.4317\times {10}^{-4}{m}^3$

Substituting the value in the volumetric efficiency equation.

We get,

$\underset{v}{\eta }=\frac{4.4317\times {10}^{-4}}{5.17\times {10}^{-4}}\times 100=85.719\%$

The volumetric efficiency of the engine, $\underset{v}{\eta }$=  85.719 %

 

4. Measure the air mass flow rate for this engine.

To find the air mass flow rate for the engine we need to know the mass of air. The mass of air can be found by the Mass V/S

Crank plot which is plotted in the 3rd question. 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 = $\frac{3000}{60}=50RPS$

 

Thus, the actual angle covered in one second is $50\times 360=18000deg$

 

To find the mass of air entering into the engine at 18000 deg,

wkt, Mass of air entering at 360 deg = 0.000534 Kg

$\therefore$

For Mass of air entering at 18000 deg = $0.000534\times 50=0.0267$

Kg/s

Air mass flow rate for this engine = 0.267 Kg/s

 

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

Total cell count variation is directly dependent on the Adoptive 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 to come into action which thereby varies the total cell count.

 

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

(a) why?

In 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 cylinder head. In

real engine using a valve spring and retainer, the valves are kept in contact with the cylinder head. 

In simulation the valve spring and retainer 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 trasnfer between the cylinder head and cylinder will not happen. This causes

raise in temperature of the cylinder which will lead to detonation/pre-igniton/induces knock and also engine might emitt 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 decrease by a great margin.

 

7. 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 50 RPS. In the previous questions we saw that engine covered 18000

deg in one second ie, 360x60 = 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.

$t=\frac{240}{18000}=0.0133\sec$

 

$\therefore$

the engine took 0.0133 seconds to cover 240 deg.

That is the combustion stork lasted for about 0.0133 seconds.

 

 

DIESEL Engine:

1. 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 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 of the combustion chamber is simulated, the

number of cells required to capture the effect are less. Thus, the simulation is done faster and the results are accurate.

 

2. When will the sector approach not work?

The sector approach does not holds good for open cycle analysis. In open cycle approach the engines vales ie, intake and

exhaust vavles are kept open and they required to be modelled. As the position of valves varry 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 centre axis

of the engine.

 

3.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 of the simulation can be taken at the

beginning of 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 of the fule injection takes at 0 CAD, 180 CAD can be taken as start and 540 CAD can be taken

as end of the crank angle for the Diesel engine.

 

4. 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 a open cycle analysis. The

sector simulation can only simulate a closed part of a 360 deg engine. In PFI engine the position of the fuel injector might

not be symmetrical and therfore, simulating 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 impossiblt to convert PFI case into a Sector case.