Challenge

Week 5 - Rayleigh Taylor Instability

Objectives:

  1. What are some practical CFD models that have been based on the mathematical analysis of Rayleigh Taylor waves? In your own words, explain how these mathematical models have been adapted for CFD calculations.
  2. Perform the Rayleigh Taylor instability simulation for 2 different mesh sizes with the base mesh being 0.5 mm. Compare the results by showing the animations( Attach them from your google drive). Observe the differences in the animation and explain the effect of mesh size on simulation. Also, explain why a steady-state approach might not be suitable for these types of simulation.[ Choose Air and Water for this simulation]
  3. Run one more simulation with water and user-defined material(density = 400 kg/m3, viscosity = 0.001 kg/m-s) for refined mesh.
  4. Define the Atwood Number. Find out the Atwood number for both the cases and explain how the variation in Atwood number in the above two cases affects the behavior of the instability.

Procedure:

Introduction:

Rayleigh Taylor Instability :

The Rayleigh–Taylor instability is an instability of an interface between two fluids of different densities which occurs when the lighter fluid is pushing the heavier fluid. Examples include the behaviour of water suspended above oil in the gravity of Earth,mushroom clouds like those from volcanic eruptions and atmospheric nuclear explosion, supernova explosions in which expanding core gas is accelerated into denser shell gas, instabilities in plasma fusion reactors and inertial confinement fusion.

Some practical CFD models that have been based on the mathematical analysis of Rayleigh Taylor waves are:

  1. Kelvin Helmholtz instability Model:

This instability can occur when there is velocity shear in a continuous fluid, or where there is a velocity difference across the interface between two fluids. An example is wind blowing over the water. other examples showing this instability are clouds,oceans,saturns bands.

2. Richtmeyer meshkow instability :

This instability occurs when a shock wave interacts with an interface separating two different fluids or in other words when two fluids of different densities are impulsively accelerated.The development of instability begins with small amplitude pertubations which initially grow linear with time.This instability can be considered the impulsive acceleration limit of the Rayleigh Taylor Instability.

3. Plateau Rayleigh instability :

This instability explains why and how a falling stream of fluid breaks up into smaller packets with the same volume but different or small surface area . It is often called as Rayleigh instability.The driving force of this instability is that the liquids by virtue of their surface tensions ,tends to minimize their surface area .

4.Rayleigh Taylor Instability :

The Rayleigh–Taylor instability is an instability of an interface between two fluids of different densities which occurs when the lighter fluid is pushing the heavier fluid.

Case 1) Rayleigh Taylor instability simulation for base mesh 0.5 mm :

Step 1) Geometry creation :

BASELINE MODEL :

Now here the RT instability simulation for a 2D model is performed using the air and water as the two fluids.

We have used alaminar model with volume of Fluid method to simulate the calculation for the two different fluid phases.

for the geometry we have used two square plates of length 20 mm each sharing a common edge between them as shown in

the following image. Also we have turned on the share topology to have a conformal mesh.

Step 2) Mesh :

For the Meshing , element size of 0.5 mm was used for the baseline case. The elements we by default quadratic which is good

and we can use this mesh for our fluent setup.

Step 3) Setup and solution:

We will setup the solver type to pressure based. where we will set the time to transient and apply gravity in the negatve Y

direction .

here we have two materials forst is the air and another is water , both having all the default properties set by fluent

we will set the viscous model to lanimar and for the calculation in mutliple phases we will be using volume of fluid model,

with implicit formulation for the volume of fluid parameters.

Phases we have set our primary phase to air and the secondary phase to water

to initilize the simulation we will be hitting the initilize button and then patch the solution . while patching we select the phase to water and the variable as volume fraction and use value '0' for the air surface zone and use value '1' for teh water sueface zone as shown below.

now a countour will be set to view the initial patching of the phase for the volume fraction of water on all the surfaces.

the contour shown 1 for the water in red and 0 for the air in blue colour so we can see that water is above the air initally.

Now we will create a solution animation for the above contour so that we can see the distribution of phase over the time until is reaches a steady state.

To run the calculation we will use a time step value of 0.005 s and run the calculation for about 300 time steps and then increase the time step to 0.05 s and run for another 50 time steps so that our solution reaches steady state faster and also the solution wont blow up.

Then we have our final solution where the calculation are complete the hase contour looks like as shown below

Time step size (sec)= 0.005

No of Time step : 1500

Result:

Residual plot:

the transient simulation animation is given in the link

Animation Link :Mesh size : 0.5mm

https://youtu.be/6c1znpmk6O4

Case 2) Mesh size: 0.3 mm

Time step size (sec)= 0.005

No of Time steps : 1500

Result:

Residual plot:

Animation Link: Mesh size : 0.3mm

https://youtu.be/uxIzKTwzH_8

Case 3) Mesh size: 0.2 mm

Time step size (sec)= 0.005

No of Time steps : 1500

Result:

Residual plot:

Animation Link :Mesh size : 0.2mm

https://youtu.be/qriKrtRCYkU

CHANGING THE MATERIALS :

Running one more simulation with water and user-defined material(density = 400 kg/m3, viscosity = 0.001 kg/m-s) for

refined mesh.

Time step size (sec)= 0.005

No of Time steps : 1500

Result :

Residual plot:

Animation Link: Mesh size : 0.2mm (water-UD material)

https://youtu.be/0VsS0z3GwKU

Observations :

  • The finer mesh captures the instability better than the coarse mesh.but it takes more time for the heavier fluid to settle down.
  • The computational time is also increased for fine mesh but the result obtained having better accuracy.
  • The steady state simulation is performed if we are concerned more about the final state results or the equilibrium state,
  • In RT instability CFD model we are more concerned to learn about the transition of the irregularities that start developing when we pour the high densiy fluid upon the low density fluid under the effect of gravity and we want to know how the interface collapses as a function of time so In this case Transient state is more suitable over the steady state.
  • For the water and user defined material the bubbles formed are smaller due to lesser difference between the densities of both the fluids

Atwood Number:

It is a dimensionless number that is used to study hydrodynamic instabilities in density stratified flows.

It is denoted by letter A

`A = (rho_1 − rho_ 2)/ (rho_ 1 +rho_ 2 ) → 1`

where,

`rho1`=Density of heavier fluid `

`rho_2`=Density of lighter fluid

Calculate the Atwood Number for Air-water RT instability

`rho_1 =998.2 (kg)/m^3` (water density)

`rho_2 =1.225 (kg)/m^3` (Air density)

The above values are substituting in 'equation 1' we get A=0.9975

Calculate the Atwood Number for Water-UD material RT instability

`rho_1 =998.2 (kg)/m^3` (water density)

`rho_2 =400 (kg)/m^3`(User defined material density)

substituting the abovevalues in 'equation 1' we get A=0.4278

we can clearly see that for second case the artwood number is small so the bubbe formed in this case are small as compared

to the bubbles formed for the first case and it take less time to seperate after the instability.

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