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 in the cylinder of an internal combustion engine. 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.

Here maximum volume = 0.00057 meter cube and minimum volume is 5.75e-05 meter cube and hence the compression ratio of the engine is 10:1.

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

Just need to show the calculation

To modify the compression ratio of the engine without changing its geometrical parameters, either the clearance volume (or) swept volume should be changed.

METHOD 1: swept volume is constant and clearance volume is varied.

`Compression ratio =(Vs + Vc)/(Vc)`

Vc = 5.75e-05

Vs = 5.125e-04

For CR to be 10.3 by above formula Vc needs to be 5.51e-05 meter cube

METHOD 2: clearance volume is constant and swept volume is varied.

`Compression ratio =(Vs + Vc)/(Vc)`

Vc = 5.75e-05

Vs = 5.125e-04

For CR to be 10.3 by above formula Vs needs to be 5.34e-04 meter cube.

Q3. Calculate Volumetric Efficiency for this engine

Volumetric efficiency (VE) in internal combustion engine engineering is defined as the ratio of the mass density of the air-fuel mixture drawn into the cylinder at atmospheric pressure (during the intake stroke) to the mass density of the same volume of air in the intake manifold.

We know that swept volume, Vs = 5.125e-04

P * Va = m * R * T

Va = (m * R * T) / P

where,
P – Pressure of air at atmospheric condition (101325 N/m2)

m – Mass of air

R – Universal gas constant (287 J/Kg-K)

T – Temperature of air at atmospheric condition (293 K)

Mass is taken from the plot and

m = 0.0005 kg

Va = [(0.000534 * 287 * 293)] / 101325

Va = 4.4317 * 10-4 m3

Volumetric Efficiency = (4.4317 * 10-4) / (5.125*10-4) * 100

Volumetric Efficiency = 86.47 %

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

The mass of air (m) is taken as 0.0005 Kg.

This mass is the mass flow rate of air into the cylinder per 360 degree.

Speed of the engine (N) = 3000 rpm = 3000 / 60 = 50 rps.

Angle covered in one second = 50 * 360 = 18000 degrees.

Mass of air entering for 18000 degrees = 0.0005 * 50 = 0.025 Kg/s.

ṁa = 0.025 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 have 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

• WHY?
• What happens if this gap is big?

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.

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.

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

Duration of combustion is 240.199 degrees.

Engine speed (N) = 3000 rpm => 50 rps

Revolution in degrees = 360*50 => 18000degrees/sec

Hence for 240.199 degrees, t = 240.199/18000

t = 0.0133 seconds

So, combustion lasts for 0.0133 seconds.

Questions based on 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 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 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 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.

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

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

Let us assume that the start of the fuel injection takes 0 CAD,180 CAD can be taken at the start and 540 CAD can be taken at 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 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.