[dreamchaser]

Hello,

I am doing a grid independence study where I am doubling the number of elements each time. I would like to know what the smallest length of an element at the wall is in the y-direction (Picture attached). Is there anyway to see this in CFX or CFD-Post?

The other question I had was on integrating in CFD-Post. I got a pressure profile as expected with some negative pressures (pic attached). I would like to get the force acting in the y-direction (force pushing the wall) for only positive pressures. So far reading up on this, I have came up with the expression to get the force on the wall. I wanted to make sure I use lengthint and and not area integral? And I am not sure how to change this expression so it integrates over the domain for positive pressure.

lengthint(pressure*x)@wall

I thank you in advance!

[ghorrocks]

You will want to do an area integral of pressure to get force, but then divide by the extrusion length to get force per unit length.

I am puzzled why you would want to only integrate the positive pressures. Does not sound realistic to me.

Try defining a new variable with a CEL expression like if(p>0[Pa],p,0[Pa])

[dreamchaser]

Originally Posted by ghorrocks

You will want to do an area integral of pressure to get force, but then divide by the extrusion length to get force per unit length.

I am puzzled why you would want to only integrate the positive pressures. Does not sound realistic to me.

Try defining a new variable with a CEL expression like if(p>0[Pa],p,0[Pa])

Hi Glenn,

Thanks for the advice. My last question was how do I do the same thing with the x values? So how do i define a new variable of x-values that corresponds with the positive pressures?

To clarify, I made a new variable called NewP which was equal to:
if(p>0[Pa],p,0[Pa])

My expression to get the force acting on the wall is:
areaInt(NewP *X )@wall

However, I would like to integrate X only over the positive pressure values.

Any advice would be appreciated.
Thanks

[ghorrocks]

Sounds trivial. Define a variable with:

if(p>0[Pa],x,0[m])

[dreamchaser]

Originally Posted by ghorrocks

Sounds trivial. Define a variable with:

if(p>0[Pa],x,0[m])

Thanks Glenn!!

[dreamchaser]

Originally Posted by dreamchaser

Thanks Glenn!!

Hi Glenn,

I had a question regarding the physical meaning of my results. I am running the same case (pic attached). I am trying to get the force on the wall as a function of the rotating wall. I realized that as I increase the rpm of the rotating wall, the force on the top wall starts to decrease at a point and eventually becomes negative everywhere. (There is no convergent film present anymore)

I attached a plot of the pressure on the top wall for two different RPM's. Red=1833 RPM and Blue is 2855 RPM.
For example, for 96 RPM, I get a pressure developed on the top wall whereas the pressures are all negative at the wall when I increase the RPM to 2855RPM.

My explanation is that since I modeling a shear thinning fluid (viscosity decreases with increasing shear rate) using a Non Newtonian model (Bird Carreau), the increase in RPM of the rotating wall (velocity) increases the shear rate which reduces the viscosity. As a result, a higher RPM of 2855 produces higher shear rates, making the simulation run at the "low shear viscosity" thereby producing a smaller pressure ( or negative) at the wall.

Does this make sense? Do you have any alternate explanations for this?
Its just funny how there is no positive pressure at all in the domain and every value is negative.

Thanks in advance.

[ghorrocks]

Your explanation sounds feasible.

If this is an incompressible simulation then the fact that all the pressures are negative is not important. The pressure level could be anything and it makes no difference. It is only for flows where something is a function of absolute pressure (such as an ideal gas or cavitation) where the absolute pressure becomes important.

[dreamchaser]

Originally Posted by ghorrocks

Your explanation sounds feasible.

If this is an incompressible simulation then the fact that all the pressures are negative is not important. The pressure level could be anything and it makes no difference. It is only for flows where something is a function of absolute pressure (such as an ideal gas or cavitation) where the absolute pressure becomes important.

Hi Glenn,

I realized that the pressures started to go negative because my domain is not closed I believe (like a real journal bearing). When the domain is closed, mass flow rate becomes a function of RPM and thus there is a delta P created between the gap. I ran a simulation where I put a box around the sphere (attached). I realized that the pressure did not go negative as I increased the RPM and thus mass flow rate was a function of RPM.


However, I realized that for this new domain the positive pressures did not equal the negative pressure. (Picture attached). I believe this has to do with my boundary conditions. With the initial domain, where I was just modeling the gap, I would set the pressure to be zero at the inlet and outlet which allowed a converging and diverging pressure of equal magnitude. However this setup did not work for high RPM's because of mass continuity due to the domain being open.

Do you have any ideas on why the positive pressures are not equal in magnitude to the diverging part for this new geometry? I also put a picture of another domain I tried. I am not sure with this new domain if I can just add a 'line' and impose a new boundary condition of P=0 to create an equal pressure distribution.

Do you have any advice on how I can model this sphere spinning against a flat wall and make a closed domain?

Thanks in advance

[ghorrocks]

Why not simply model the entire domain - so include the domain around the space around the cylinder, like you did for your 4-way touching simulation?

[dreamchaser]

Originally Posted by ghorrocks

Why not simply model the entire domain - so include the domain around the space around the cylinder, like you did for your 4-way touching simulation?

Hi Glenn,

Okay. I thought about trying that also. I put up a pic of the new domain I will try to simulate. I had a question about it. I put up comments of my boundary conditions I intend to use.

First I want to examine the effect of one wall which I have put. I would like to make everything an 'opening' boundary condition and change the rpm of the sphere. Does this setup make physical sense?

More importantly, in my real project, this sphere will be covered in grease. The whole domain will not have a lubricant. Thus, I wanted to initialize the domain with a thin circular 'grease' around the rotating sphere. Everywhere else, is a vacuum. Can you please give me some advice about how to do this initialization of grease?

Thanks

[dreamchaser]

Originally Posted by dreamchaser

Hi Glenn,

Okay. I thought about trying that also. I put up a pic of the new domain I will try to simulate. I had a question about it. I put up comments of my boundary conditions I intend to use.

First I want to examine the effect of one wall which I have put. I would like to make everything an 'opening' boundary condition and change the rpm of the sphere. Does this setup make physical sense?

More importantly, in my real project, this sphere will be covered in grease. The whole domain will not have a lubricant. Thus, I wanted to initialize the domain with a thin circular 'grease' around the rotating sphere. Everywhere else, is a vacuum. Can you please give me some advice about how to do this initialization of grease?

Thanks

and does it make physical sense to model this thin layer of grease for a steady state simulation?

[ghorrocks]

It depends on what the grease does. If the grease does not move then just model it as a solid. If it affects something subtle like wall shear then you need a model for it. If it moves then you need to have a model for that.

Surrounding your domain with an opening is probably a bad idea. I would make it a wall, and possibly use a small opening at the top to keep the absolute pressure level correct (if that is important).

[dreamchaser]

Originally Posted by ghorrocks

It depends on what the grease does. If the grease does not move then just model it as a solid. If it affects something subtle like wall shear then you need a model for it. If it moves then you need to have a model for that.

Surrounding your domain with an opening is probably a bad idea. I would make it a wall, and possibly use a small opening at the top to keep the absolute pressure level correct (if that is important).

Okay. Thanks.

I wanted to clarify a question with you regarding getting force from the pressure. I plotted the pressure acting along the wall. I then exported the data to matlab and integrated the pressure data along the length of the wall (x).

So this answer I get is force per length?? If I want to know the force (in newtons) acting at the wall, I need to multiply this number by the depth of the wall to get newtons?

And I assume multiplying by a depth, one is assuming the length of the wall in the z-direction (into the board)??

I just wanted to clarify this because I want to know to display force data in a report, then I need to assume a depth for the wall.

Thanks

[ghorrocks]

CFX is a 3D modelling package so integrating the pressure gives you force. When you do 2D simulations in CFX you use a small thickness in the z direction to make it a 3D simulation. So the force reported in 2D simulations is over the z thickness you used to generate the mesh.

[dreamchaser]

Originally Posted by ghorrocks

CFX is a 3D modelling package so integrating the pressure gives you force. When you do 2D simulations in CFX you use a small thickness in the z direction to make it a 3D simulation. So the force reported in 2D simulations is over the z thickness you used to generate the mesh.

Hi Glenn,

I forgot about that. Just to make sure I understand, so if I graph the pressure along the wall and integrate with trapezoidal rule, I get 1600 which (according to CFX) is really F/z(depth)=1600? So if I want to know the force, I just multiply this by the extrusion distance in the z-direction?

I did mesh statistics in CFX on my mesh and saw the domain extents for z is -0.00152476 [m], 0 [m]. So I just multiply the result from the trap rule by .00154 to know the force acting on wall with some depth z, correct?

Thanks

[ghorrocks]

Don't do the integration yourself. Use the force() function in CFX to calculate the force as that uses the integration points of the mesh which are more accurate than the method you describe. This will give you a force over the whole face, so to convert this to force per unit length then divide by the z extrusion length.

[dreamchaser]

Originally Posted by ghorrocks

Don't do the integration yourself. Use the force() function in CFX to calculate the force as that uses the integration points of the mesh which are more accurate than the method you describe. This will give you a force over the whole face, so to convert this to force per unit length then divide by the z extrusion length.

Hi Glenn,

Thanks for the insight. You have been very helpful and I thank you.

Regarding the force, I just did your earlier suggestion and wanted to make sure I did it right.

First I defined the positive pressures acting at the wall since negative pressures in journal bearings rarely occur and did:

Positive Pressure:
if (p>0[Pa],p,0[Pa])

After, I defined an expression doing:

areaInt(Positive Pressure)@wall

which gave me 2.46N.

So this method is correct, right?

I did this method because I believe the force function is taking the negative pressures into account. Acting on the face, I assume there is negative pressure also. There was true when I just plotted the pressure along a line of the wall.

And can you please explain why we have to divide this by the extrusion length (z) and not the length of the wall (x)?

Not sure if it makes sense more sense to report force/length data or just the force? I assume,if someone was to build this thing, we do know really know the length of the wall in the z-direction (extrusion length) to resist the force of the spinning sphere. So reporting force/extrusion length would be better from a design perspective?

Thanks in advance Glenn.

[ghorrocks]

Why do you take the positive pressures? You say "First I defined the positive pressures acting at the wall since negative pressures in journal bearings rarely occur" - but when they do occur you should take it into account! So just integrate the pressure variable.

Even better, do as I previously suggested and use the force() function which automatically includes both the pressure and viscous forces.

Force per length: I have no idea what is the important parameter for you, only you know that. So you divide by which ever length gives you a meaningful value.

[dreamchaser]

Originally Posted by ghorrocks

Why do you take the positive pressures? You say "First I defined the positive pressures acting at the wall since negative pressures in journal bearings rarely occur" - but when they do occur you should take it into account! So just integrate the pressure variable.

Even better, do as I previously suggested and use the force() function which automatically includes both the pressure and viscous forces.

Force per length: I have no idea what is the important parameter for you, only you know that. So you divide by which ever length gives you a meaningful value.

Hi Glenn,

Thanks for the information. I am now using the force() function.

I am trying to do a 3D simulation of this problem to essentially validate in my report that 2D simulations are okay. I wanted to make sure my reasoning is clear:

I know CFX is a 3D solver and in my case I specify the front and back face as a symmetry plane to do a 2D simulation. The extrusion distance is very small (.0015). I get the force/length. Regarding the 2D case the force per length was found to be 96.15 N/m. Predicting for an extrusion of 1 inch is then (96.15 N/m)*.0254m =2.44 N

I then run a 3D case (specifying the front and back faces as free slip walls) where my extrusion distance is 1 inch. Free slip wall most accurately describes the problem set up.*I use the force function on the wall and get 2.34N. Comparing with the previous predicted 2D value of 2.44 N, I can thus justify that 2D simulations are adequate enough. Does this justification and method sound right?

Also I am only integrating the positive pressures because as a paper states:

"The tendency to form sub-ambient pressures will be arrested by cavitation, or rupture, of the oil film and the influence of subambient pressures must be very small in low conformity contacts such as we are concerned here. Accordingly, in the integration of the pressure to obtain the load capacity, we only consider the contribution of the regions of positive pressures".

Does this explanation make sense to you?

Thanks for the help!

[ghorrocks]

Accuracy - If you are happy with 2.34N versus 2.44N then yes, it looks good. It depends on how accurate you need to be.

I do not like the justification your paper states. If you are not modelling cavitation it is true that the regions with pressure less than absolute zero are likely to have cavitated. However, the cavitation will then not behave as a simple single phase incompressible fluid - specifically the cavitation region is likely to be much larger than the negative pressure region in a single phase simulation.

So I strongly doubt the accuracy of the assumption that cavitation can be accounted for by ignoring negative absolute pressures. I would recommend you just work out the force over the whole thing, and if you have significant cavitation regions then you do a cavitation model to do it properly.