FINAL TEST

Questions based on PFI Engine:

Q1. What is the Compression ratio for the engine?

It is the ratio of total volume to the clearance volume of the cylinder.

Rc = Total volume/Clearance volume = (Vs + Vc) / Vc

= 5.73*10^-4 / 5.704*10^-5

= 10.04:1

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

We can change the compression ratio of the engine without changing geometrical parameters by using a turbocharger and supercharger.

The turbocharger provides more amount of intake air which increases the demand for fuel and more power is generated in power stroke increasing swept volume, If swept volume increases and clearance volume remains constant we get an increase in compression ratio.

But the demerit is it causes detonation, combustion of a substance that is initiated suddenly and propagates extremely rapidly giving rise to a shock wave.

The swept volume requirement for Cr of 10.3 and clearance volume constant is:

Cr = (Vs + Vc) / Vc

10.3 = (Vs + 5.704*10^-5) / 5.704*10^-5

Vs = 5.3047*10^-4 m³

The other way is to change the geometrical parameters:

• By changing the thickness of the cylinder head.
• By adding protruded bowl geometry on the piston surface.
• By increasing the stroke and bore of the engine.

We can calculate the value of swept volume by assuming the required compression ratio and clearance volume, using swept volume we can calculate the bore and stroke required.

Vs = (π/4) * B² * S

Where,

B = Bore(m)

S = Stroke(m)

• By increasing connecting rod and crankshaft length for higher reciprocation of piston.

Q3. Calculate Volumetric Efficiency for this engine?

It is the ratio of the actual volume of intake air V_a(m³) drawn into the cylinder/engine and the theoretical volume of the engine/cylinder V_d(m³) during the intake engine cycle.

η_v = (V_a / V_d)

The intake air volume can be calculated using the mass of air 'm_a' (kg) present in the cylinder to the density of air 'ρ_a' at 1bar and 293K temperature.

V_a = m_a/ρ_a

The mass of air inside the cylinder can be calculated by adding O2 and N2 masses.

The mass of air = 0.0004963 kg

ρ_a = (101325 pa / (293(K)*286.9(J/kg.K))

= 1.20 kg/ m³

V_a = (4.963*10^-4) / 1.20

= 4.1358*10^-4

η_v = (V_a / V_d)

= (4.1358*10^-4) / 5.1596*10^-4

= 80.1%

Volumetric efficiency is = 80.1%

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

The air mass flow rate can be calculated using:

ṁ = m_a *N / n_r

Where,

m_a = mass of air(kg)

N = speed in rpm

n_r = no of crankshaft rotation for a cycle.

ṁ= (0.0004963* 3000)/(2*60)

ṁ= 0.012 kg/s

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

Since we have provided adaptive mesh refinement in our setup if there is a change in parameter we included in AMR, and the value exceeds sub-grid-scale, then the cell is refined in that region. The AMR works based on the curvature(second derivative) change of a parameter.

When a cell gets refined the cell count increases, and this occurs throughout the engine geometry wherever there is a sudden change in temperature and velocity. This is the reason why cell count varies during simulation.

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

WHY?

In the case of the real engine to avoid the heat transfer and isolate the cylinder to preserve its maximum temperature we have valves in contact with the cylinder head, If there is any temperature loss it affects the ignition timing, slows combustion hence less power decreasing the overall efficiency of the engine.

While running simulation we should provide a gap because of intersection between the valve and cylinder will cause intersection error without resolving it we can't run our case. To resolve this we can translate the valve to a small distance which will avoid intersection.

To avoid any discharge of charged particles from this gap connecting triangles are created between two regions.

What happens if this gap is big?

If the gap between that valve and cylinder is big more amount of air is discharged into the cylinder before suction stroke to capture this extra mass, we should refine the mesh near the valve boundaries, increasing the computational time and effect the accuracy of the simulation. To avoid any issues the gap between the valve and cylinder must be small as possible such that connecting triangles are small too which doesn't interfere with the airflow inside the cylinder.

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

The degrees for which the combustion process occurred = 240.199˚

To convert it into seconds, we should first calculate the time taken for 1˚.

Time taken per cycle,

t = 2*60 / 3000

= 0.04s

For 1˚ = 0.04s / 720˚ (because 1 cycle for 4 stroke engine consist of 720˚)

= 5.555*10^-5 s

The time taken for 240.199˚ in sec = 240.199˚ * 5.555*10^-5

= 0.0133s

Questions based on DIESEL Engine:

Q1. What is an advantage of SECTOR Simulation?

Sector geometry can be used in case of closed-cycle analysis, i.e. In compression ignition engine.

In a CI engine fuel is injected by nozzles inside the chamber and if we use more nozzles we can divide the engine into parts equal to a number of nozzles because the fuel is distributed evenly in the whole cylinder and also each sector parts have the same simulation characteristics.

The sector simulation generally increases computational efficiency by reducing computational time and expense.

In sector simulation, we can only simulate the combustion process.

Q2. When will the sector approach not work?

In case if we want to simulate an open cycle simulation in which the intake valve and the exhaust valve opens and closes with suction of charge and exhaust of flue gases we cannot work with the sector approach, since it is restricted only for in-cylinder simulations, and also we can't simulate multi-cylinder engines since every cylinder has its own characteristics, valve timings, fuel injection timings, etc.

The sector approach can only work when the cylinder of the engine has an evenly distributed air-fuel mixture, such that every sector of the cylinder represents the whole cylinder itself.

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

In the sector approach since its closed-cycle analysis, we can only simulate combustion processes, hence the start of the crank angle is the start of the compression stroke, and the end of the crank angle is our end of the power/expansion stroke.

If we start our cycle at 0˚ the SCA is 180˚ and ECA is 540˚.

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

No, to simulate port fuel injection we should use open-cycle analysis since the air-fuel mixture is injected from the intake port and the distribution of this mixture is not equal throughout the cylinder, also we can't control the flow of charge, since in closed cycle analysis we can't control the translation of intake and exhaust port.

In case if we use direct and indirect fuel injection in which the fuel is injected inside the cylinder we can use a sector approach.