Hi, everyone!!

Well, I am having a problem with CFX 5.I am using 5.4.1, and I am modelling a siphon tube with some guidvanes and a shaft on it.

My problem is I had some results for two different designs, and in Fluid models, in the Domain Form,I set the Heat Transfer model to NONE, and then, I could not choose if I want bouyant forces like gravity.Why???

And,beyond that,then I changed the HTransfer model to Total Energy, and then, you have the option of setting gravitational forces.Ok, fine, but my big surprise was to see that at the end of the day, I am having EXACTLY the same results and plots I was having in the non-buoyant problem.

Then, I set a problem (slightly different design), and I ran it with the Heat Transfer model set to Total Energy, and non-buoyant,and then same HT model but buoyant, and the results were the same!!!!Does it not suppose to make a difference, if you put gravity or not???This does not make sense.

Well, hope you can help me with this.Thank you for your pacience for reading such a long message

Hi Gilberto,

Is there a temperature difference in your model to drive buoyant forces?

Robin

Thanks for your response.No,there is not a temperature difference (I already thought about it),but gravity must act regardless of the temperature difference.How can I set it so I can see the efect of gravity???

Hello,

Is there a density difference in your problem?

If not, there is no need to include gravity. If so, let use know what it is.

No, there is not any density difference,I am running water through a siphon tube.

So you say that gravity is included already??If it is so,something does not make sense,because I should be having a difference in pressure of 19.6 kPa between the water level upstream, and the tailwater level, because I have a 2m height difference.And I am having differences of around 5 kPa, or something like that.

Thanks, mate, I hope you can give me some hints about it.

If there is no density differences, either due to temp difference, componant or phase differences, gravity does not effect the flow.

If your inlet is at a higer level than your outlet then you need to set both boundaries as pressure boundaries with the relative pressure rho*g*h higher at the top than at the bottom

Jon,

Thanks,mate,for answering me so quickly!!Well, mate, I thought that the computer could take in acount that thing, about having different levels, because, at the end of the day,that is why you draw your drawing,to give the computer your exact dimensions.

I am currently fixing mass flow rate at the inlet and pressure at the outlet.Are my results right and reliable??Why am I not getting the difference in pressure that you normally expect?Should not the program be aware of that?Is the program including the term ro*g*h in its calculations.

Sorry about my big array of questions,Jon.I hope you can answer some of them.

Cheers,mat

Set it as as my last suggestion and see what happens

I am running it now, and first thing, it is telling me that a Wall has been placed to prevent the fluid coming out of the domain.That always happens when I try to fix pressure b.c.!!!The reason is my drawing is a siphon, so the first bit,the fluid has to go upwards, at an angle of 45, and apparently the computer does not like that!!!!

Well, when it finishes, I'll tell you.

Use an opening instead of an outlet

Are, you are using CFX5.4.1.

Dont set the pressure how I told you. Set it as an expression = rho (or perhaps density, check on the variables list) * g * z (or which ever you vertical direction).

Also, use an opening instead of an outlet.

Hi Gilberto,

The following excerpt from the help doc may help:

Note that the Results File does not contain the hydrostatic contribution to pressure, and this should be added to obtain actual values of pressure relative to the Reference Pressure in these cases.

Basically, when you set your gravitational acceleration, you also set a reference density and temperature. The reference temperature is used for the Bousinesque model (if you did not choose an ideal gas) and the reference density is used to calculate a background pressure (ie RHO*g*h).

If what you really want is to see the guage pressure at the second height, set your refernce density to that of air, not water.

Regards, Robin

Hi, Robin,thanks for your answer.

Well, I don't understand it.Why should I change the density to the one of air, if I have water??My problem is I want to see the pressure everywhere,with the term ro*g*z included.

Is there any way I can set my problem to see my pressure,including hidrostatic contribution?Because if I set the pressure to be only ro*g*z,will I not miss the other contributions to the change in pressure?

The extract from the manual is all right, but it is so in the way that they tell you the stuff in the manual "this hould be added", but HOW???The manual help is really useful to spot your problems, but they never ever tell you a precise solution, to sort it out.

Regards,

Gilbert

Robin, he is not two phase or density variable.

Gilberto, have you set it up as I last suggested?

I have fixed pressure at inlet and outlet,with a difference of 19.6 kPa.

Well, first, I had this problem with the wall.Does this mean that the results from that are not reliable,because of the wall(the maximum was 20% of the are)???

And second, I find it pretty stupid having to tell the computer what the result is beforehand,instead of it giving me the appropriate results.Shouldn't it be the other way around???

How can I fix the pressure as you told me,Jon?In the expression editor??Is it not going to affect my results??I will have the pressure equal to ro*g*h?And the other contributions to pressure change??????What happens to them?My god, this looks like easy,but it is getting complicated.

Thanks a lot for your time and your help.I hope you can answer some of my questions,although some of them are more on the whole philosophy of the program,hahaha

Regards,

Gilberto

You dont have to 'tell the computer the results beforehand'

You have to set up the boundary conditions as they are in physical reality. Set them as rho*g*h NOT constant and I said previously, when I realised you were using cfx5.

Yes, you set this up in the expression editor, just for the two boundaries.

The code will then calculate the hydrodynamics from the bondary conditions you will have set up. It is necessary to set up the boundary conditions correctly, otherwise the results cannot be correct.

Which problem with the wall are you talking about? I told you to change to openings not outlets. Have you done this?

Ok, I go to Boundary Conditions, and in VALUES, I go to Expression Editor, I created Pinlet and Poutlet, as density*g*z, it creates the expression, but it won't let me put it as Static Pressure, in the Values window.How should I set that???

I have set the pressures at inlet and outlet with values 19.6kPa and 0.I also changed the b.c. type to opening in the inlet.

How do you set the pressure in the b.c. through an expression??As I told you, I created the expression, but when I write it in the Static Pressure box in the VALUES window, and apply, it does not recognise my expression

The problem with the opening type was almost exactly the same as before with the Inlet type of b.c.

If I fix these two pressures, I am getting a mass flow rate of 2312kg/s. But, what happens if I want to fix the mass flow rate, and I want hydrostatic effects to show?????

I would really appreciate your answers.

Cheers,

Gilberto

Just to make it clear, and I know that I am probably bothering you all, but I really need an answer.

The question is:

As the program does not consider hydrostatic contribution to pressure, is it possible to set the mass flow rate and to add this effect???

If it is possible,HOW can it be done????

Regards,

Gilberto