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Transient simulation of flow through a valve

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Old   March 28, 2018, 11:19
Default Transient simulation of flow through a valve
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Aadit Shroff
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Hello all!

I am simulating a transient flow (water) through a choke valve (in SimScale) using the PIMPLE solver.

I have only the following conditions known to me:
1) Inlet Pressure - 8.27 MPa (gauge)
2) Inlet Mass Flow Rate - 5.52 kg/s
3) Inlet Temperature - 65.5 C

My objective is to find the pressure drop across the valve.

For this I have calculated the flow velocity at the inlet through the mass flow rate (m*=roh.A.V) which I supply as the inlet velocity BC. Outlet pressure BC is set to 0 Pa gauge (hence, exhausting to the atmosphere). Is my approach correct? In this case, when I get the pressure results, does it mean that the highest pressure value is the total drop in pressure? Does specifying the inlet velocity automatically take the pressure into account?

When I set the inlet BC as the inlet pressure, I get insanely high valves of pressure and velocity in the valve. Is there any way I can use the pressure inlet BC and still run a simulation? I mean, could I use the inlet pressure and outlet mass flow as my BC combination? (inlet and outlet mass flow are the same as the valve has no leakage.)

Looking for general advise regarding this as I am very new to CFD. Thank you!

Regards,
Aadit
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Old   March 28, 2018, 13:10
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Quote:
Originally Posted by aadit.shroff View Post
Hello all!

I am simulating a transient flow (water) through a choke valve (in SimScale) using the PIMPLE solver.

I have only the following conditions known to me:
1) Inlet Pressure - 8.27 MPa (gauge)
2) Inlet Mass Flow Rate - 5.52 kg/s
3) Inlet Temperature - 65.5 C

My objective is to find the pressure drop across the valve.

For this I have calculated the flow velocity at the inlet through the mass flow rate (m*=roh.A.V) which I supply as the inlet velocity BC. Outlet pressure BC is set to 0 Pa gauge (hence, exhausting to the atmosphere). Is my approach correct? In this case, when I get the pressure results, does it mean that the highest pressure value is the total drop in pressure? Does specifying the inlet velocity automatically take the pressure into account?

When I set the inlet BC as the inlet pressure, I get insanely high valves of pressure and velocity in the valve. Is there any way I can use the pressure inlet BC and still run a simulation? I mean, could I use the inlet pressure and outlet mass flow as my BC combination? (inlet and outlet mass flow are the same as the valve has no leakage.)

Looking for general advise regarding this as I am very new to CFD. Thank you!

Regards,
Aadit

Being the fluid incompressible, I suppose you are using a divergence-free approach, right? Temperature does not enter in this model and you solve only momentum equation with the divergence-free constraint.
You can set the velocity inflow but I don't think you can set the outflow corresponding to the environmental pressure. Where is located the section outlet? If is far from the valve you could set a condition on the velocity
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Old   March 28, 2018, 13:17
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Aadit Shroff
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Hi @FMDenaro,

You are right about the incompressible nature of the fluid. The outlet is of the valve itself. In a nutshell the design consists of a short inlet pipe, the main valve body and a short outlet pipe.

As I set the inlet as the calculated velocity, I get peak pressure at the inlet itself, about 1.5 MPa, which is way less than my actual inlet pressure. My reasoning behind this value was that the result is of the pressure loss itself, and not of the pressure created at the inlet due to the mass flow through it.

The outlet pressure set to the atmosphere is because I do not know the exact environment in which the valve would be operating. At the moment, I have set inlet pressure and exit mass flow rate (as I know the entry mass flow rate and the valve is assumed to be leak free). I was under the impression that I should set at least one condition in term of pressure, which I have done but Im not so sure if this would work. :/

Regards,
Aadit
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Old   March 28, 2018, 13:48
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Originally Posted by aadit.shroff View Post
Hi @FMDenaro,

You are right about the incompressible nature of the fluid. The outlet is of the valve itself. In a nutshell the design consists of a short inlet pipe, the main valve body and a short outlet pipe.

As I set the inlet as the calculated velocity, I get peak pressure at the inlet itself, about 1.5 MPa, which is way less than my actual inlet pressure. My reasoning behind this value was that the result is of the pressure loss itself, and not of the pressure created at the inlet due to the mass flow through it.

The outlet pressure set to the atmosphere is because I do not know the exact environment in which the valve would be operating. At the moment, I have set inlet pressure and exit mass flow rate (as I know the entry mass flow rate and the valve is assumed to be leak free). I was under the impression that I should set at least one condition in term of pressure, which I have done but Im not so sure if this would work. :/

Regards,
Aadit

What you could done is to set only velocity BC.s in inlet and outlet. The problem is that if the computational outlet is close to the valve, the developed flow assumption is not very resonable. If you are solving only div v = 0 and the momentum equation you have to remember that the computed pressure field is determined up to an arbitrary constant. You can fix a value of the pressure inlet in a cell corresponding to you known value.
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Old   March 29, 2018, 01:38
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As far as I understand this I detect a physical ambiguity.

You have the velocity and the (difference) pressure at the inlet. You need only one of this and you should only set one.

If you set a velocity you get a pressure drop form the features of your geometry. Or you set the pressure drop and get the velocity field which is developed in your geometry.

Of course, this may be deviate form the measuring values to some degree. There remains room to think about the reasons.
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Die der Hauptbewegung überlagerte Schwankungsbewegung ist in ihren Einzelheiten so hoffnungslos kompliziert, daß ihre theoretische Berechnung aussichtslos erscheint. (Hermann Schlichting, 1950)
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Old   April 11, 2018, 03:04
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Originally Posted by FMDenaro View Post
What you could done is to set only velocity BC.s in inlet and outlet. The problem is that if the computational outlet is close to the valve, the developed flow assumption is not very resonable. If you are solving only div v = 0 and the momentum equation you have to remember that the computed pressure field is determined up to an arbitrary constant. You can fix a value of the pressure inlet in a cell corresponding to you known value.

Hello FMDenaro,

Wouldn't it be sufficient to set the vel. BC at the inlet and a gage pressure=0 BC at the outlet? I'm just thinking that a choke valve cannot have a uniform flow distribution at the outlet. Does that make sense?
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Old   April 11, 2018, 03:09
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Originally Posted by piu58 View Post
As far as I understand this I detect a physical ambiguity.

You have the velocity and the (difference) pressure at the inlet. You need only one of this and you should only set one.

If you set a velocity you get a pressure drop form the features of your geometry. Or you set the pressure drop and get the velocity field which is developed in your geometry.

Of course, this may be deviate form the measuring values to some degree. There remains room to think about the reasons.
So, I too, like the OP am confused about obtaining the pressure drop. If the BC at the outlet is p=0, do we then integrate the total pressure (or, density-normalized pressure x density) and then divide it by the area of the patch? Will that give us the pressure drop in the due to the geometry?
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Old   April 11, 2018, 04:29
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You can set mathematically correct BC.s for your problem using pressure but that does not mean they match correctly the physics you want to describe.
Are you sure you can set a constant pressure outlet BC.s in your problem?

Again, depending on the formulation you are using, if you are solving a divergence-free velocity field the computed pressure has no thermodinamic meaning
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Old   April 11, 2018, 04:36
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Quote:
Originally Posted by FMDenaro View Post
You can set mathematically correct BC.s for your problem using pressure but that does not mean they match correctly the physics you want to describe.
Are you sure you can set a constant pressure outlet BC.s in your problem?

Again, depending on the formulation you are using, if you are solving a divergence-free velocity field the computed pressure has no thermodinamic meaning
That is true.

Pardon my ignorance, but what do you mean by the phrase, divergence-free?
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Old   April 11, 2018, 04:41
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That is true.

Pardon my ignorance, but what do you mean by the phrase, divergence-free?
for incompressible assumption, the continuity equation is simply div v = 0
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Old   April 11, 2018, 05:28
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for incompressible assumption, the continuity equation is simply div v = 0
Sure. Yeah, that's fair. I thought that you might've meant the non-time-varying component of the velocity. Thanks for clearing that up.

So, with the original question I asked, if one were to put a uniform velocity BC at the inlet and outlet, how would the velocity distribution due to losses at the outlet be accurate?
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Old   April 11, 2018, 05:34
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Sure. Yeah, that's fair. I thought that you might've meant the non-time-varying component of the velocity. Thanks for clearing that up.

So, with the original question I asked, if one were to put a uniform velocity BC at the inlet and outlet, how would the velocity distribution due to losses at the outlet be accurate?

No, at the outlet you cannot use a uniform velocity...you could set some conditions like a vanishing normal derivative
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Old   April 12, 2018, 05:20
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Being the fluid incompressible, I suppose you are using a divergence-free approach, right? Temperature does not enter in this model and you solve only momentum equation with the divergence-free constraint.
You can set the velocity inflow but I don't think you can set the outflow corresponding to the environmental pressure. Where is located the section outlet? If is far from the valve you could set a condition on the velocity
Question, the change in area will cause a change in pressure. Do we need to account for that when we try to calculate the pressure drop in the circuit?
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