How to check size scale effect in a butterfly valve
 

Hello.
I want to compare a 12 inch butterfly valve to a 36 in butterfly valve and find what differences occurs only by changing size specially the discharge coefficient and cavitation index.

I use total pressure inlet and static pressure outlet boundaries.
I use static outlet pressure = 101325 pa ( 1 atm ) since my valve open to atmosphere and changing inlet pressure from 110000 pa to 300000 pa .
Now my question:

1-If i use some pressure for 36 inch valve what pressure should i use for 12 inch valve ??? The same pressure ?? Or change it by some none-dimensional number like Reynolds or something else ??

2-Is my pressure inlet suitable for this purpose ??? Or should I use another type boundary ??

3- I do simulation by constant pressure in two size and after getting result i saw that discharge coefficient and sigma don't change at all !! :(

What is the problem ??
Thanks for your helps. :)

No one has any idea ??? :(

I'm not sure I understand what you're trying to do. Are you comparing two geometries? Do you know the flow conditions for them? Would they be operating in the same place?

To find out what boundary condition you need to set, take a look at how your equipment operates in real life and work from that. Since it is a valve, it's very likely that it operates under a given upstream pressure. So setting pressures at inlet and outlet are ok.

No bruno , I am not comparing two geometries. I am comparing one geometry by two different size.
Everything should be same in two size and i only want to find size changing effect on cavitation index.

If your boundary conditions have been obtained from experiments and the locations of these boundaries are far enough from the location of your valve then you will not have issues using the same boundary conditions while swapping out the geometry.

Ok Sarang.
So what is problem that i get similar result in same geometry in different size ?
-------
Case 1 :
Inlet Pressure : 200000 Pa
Outlet pressure : 101325 pa
Valve size : 36 inch
Valve Opening : 30 Degree
Cavitation index : 1.67
Discharge coefficient : 0.5586
-------
Case 2 :
Inlet Pressure : 200000 Pa
Outlet pressure : 101325 pa
Valve size : 12 inch
Valve Opening : 30 Degree
Cavitation index : 1.73
Discharge coefficient : 0.5602
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Two different size of valves but approximately same result in cavitation index and Discharge coefficient . :(
What is my problem ?

1. I am assuming you are using a steady state solver. Did you check if the runs converged ?
2. Can you post a plot of your solution domain ? (Looking into the pipe and Looking from side)
3. How are you calculating discharge co-efficient ?

Yes , I am using steady state solver and my runs converged to 1e-6 without no problem.

I attached my model picture for you.
I calculate Cd by this formula :

ave(Fluid 1.Velocity)@inlet/(2*g*(ave(Pressure)@Up-ave(Pressure)@Down )*0.0001[m Pa^-1]+ave(Fluid 1.Velocity)@inlet ^2)^.5

and "Up" plane and "Down" plane is 5 and 10 valve diameter from valve.

Ok, so your upstream section seems long enough. Let us try the following

1. As a sanity check make sure that the outflow mass flow rate + Inflow mass flow rate is reaching a steady state.

2. How many cells do you have between the valve and the wall at the location of the smallest gap ?

3. What is your boundary condition at the valve boundary ?

massFlow()@inlet = 5477.53 [kg s^-1]
massFlow()@outlet = -5477.13 [kg s^-1]
So it correct i think.
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I attached my mesh on valve position.
I think it is about 20 cells between valve and wall at the location of the smallest gap.
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The wall boundary condition of valve and pipe is "no slip wall" --> "smooth wall"

1.That looks ok. What turb model and wall treatment options are you using ? Also, what is the RE#
2. I typically look at the profile of the mass flow rate and not at a single number.

I would expect that you would have more pressure loss in the case with a larger valve. Instead of going after discharge co-efficient let us try the following

1. Measure deltaP values for both cases
2. Compare trend against existing data. If you google for pressure drop curves for check valves, you should get something that you can compare your solutions against.

Thank you for your answer Sarang.
1. I am using SST turbulence model with automatic wall function.
If you mean Reynolds number for Re , It is different for one size to another.
For example for 36 inch Butter Fly valve and in 30 degree opening on the 160000 pa inlet pressure it is 5807008.

2- How can i draw profile of the mass flow rate ? I don't know how to do it ?

3- I measured delta P in all my case. I attached them for 36 and 12 inch valve for you.

4- My biggest problem is that there is no trend in data !!!
You can see result files and see that all thing in 36 valve and 12 inch valve are similar . :(

What are you comparing your results with? Do you have a reference model/ manual calculations/ benchmark results..?

@fresty:
I have a reference experimental result for a 120 inch butterfly valve.
My 120 inch CFD valve in OK with that experiment.
But in other size no significant change in result were observed.

My bad, misread part of your expression... ignore my earlier comment regarding the error... However:

you may need to use a mass flowrate inlet and pressure outlet boundary condition to see how for equal mass flow rates the pressure drop changes in different sizes of flow domains.. that would probably be the way to go...

Ok fresty .
I saw my reference and it has this formula for calculating Cd:
Cd = V / sqrt (2*g*H + v^2)
and my ansys formula is this .
About using "areaAve" instead of "ave" i will check it. I think as you said it should give a better result than using only "ave" .
---
About second part of your writing i should say that i know if i get an equal massflow rate in all size every thing will change by size since massflow in related to pipe diameter. But i should get velocity or pressure constant. that they are not related to pipe diameter.
I should find the effect of valve size in cavitation in constant pressure drop.

Are you using compressible flow model..? If your flow is incompressible then change in area at constant pressure drop would simply proportionally increase or decrease the velocity.. if i am right then the fraction would always generate similar value in both cases isn't it?
I am assuming the "V" & "v^2" in your formula both represent inlet velocity??

I am using 2 phase model with two fluid :
1 - Water at 25 Celsius
2- Water vapor at 25 Celsius
I defined both of them incompressible.
You said :
"change in area at constant pressure drop would simply proportionally increase or decrease the velocity"
How ?? with which equation ?? assume that the flow rate in not constant in different valve size.
--
Yes , V is inlet velocity .

Alright.. well to think over and correct myself, as you reminded me that the flow rate in both the valves would not be constant so yes the velocity is not proportional to area...
However as deduction from your results it shows incompressible flow relation/ mass conservation (v=m'/rho*A) i.e. velocity would remain constant as change in area of valve only changes the mass flow, hence your equation of Cd would just render similar results as it is a function of two constants..
i might be wrong at all this as not really an expert in valve simulations but to me it seems that the solution lies in compressible flow model... thoughts?

You mean i should use compressible flow condition ?
For both water and water vapor ?
I found from some article that bigger valve has a bigger cavitation.
This cause from bigger space for bubble to travel and collapse.
Have you any idea of how to see this phenomena in ansys ??