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Create pressure drop across components in a system

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Old   June 20, 2019, 15:25
Default Create pressure drop across components in a system
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Hello All,

I'm new to StarCCM+. I'm trying to study flow through a piping system. Aim is to study the flow distribution in the system. The piping system contains multiple components attached to it. These components have complicated geometries.

I'm not particularly interested in the flow distribution within the components. I know the pressure drop that each of these added components create. Is it possible to replace them with simple geometries and add the pressure drop they create. I did some search and I found that periodic boundary condition could be a way to do it. Any advise on defining the geometry and the corresponding boundary conditions would be really helpful. Thank you.
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Old   June 21, 2019, 07:34
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Quote:
Originally Posted by bram94 View Post
Hello All,

I'm new to StarCCM+. I'm trying to study flow through a piping system. Aim is to study the flow distribution in the system. The piping system contains multiple components attached to it. These components have complicated geometries.

I'm not particularly interested in the flow distribution within the components. I know the pressure drop that each of these added components create. Is it possible to replace them with simple geometries and add the pressure drop they create. I did some search and I found that periodic boundary condition could be a way to do it. Any advise on defining the geometry and the corresponding boundary conditions would be really helpful. Thank you.

You can simply create a small region as a porous region. This is common practice for such cases. Get the pressure drop across components to be replaced. From pressure drop you can calculate porous coefficients. Refer user guide for this. And then insert a porous region.


Hope this helps.
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Old   June 21, 2019, 10:57
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You could also use a porous baffle interface if you are looking for an instantaneous change, rather than one that happens over some length along the flow path..
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Old   June 21, 2019, 11:16
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Thanks for guiding me. Really appreciate it. Will get back after I setup the case .
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Old   July 8, 2019, 15:15
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Hey Guys,

I used your advice and was successfully able to get the setup running. However, I have doubt which came up now. I know it's really fundamental, your advise would be really helpful.

1) I used dP/L vs velocity plot to obtain the inertial and viscous coefficients. I’ve used second degree polynomial curve fitting (ax^2+bx). What is length that should be considered to calculate dp/L. Is it the dimension of the actual component or the geometry of the porous media that I'm using for my setup. dP – Pressure drop, L – length.

Thank you.
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Old   July 8, 2019, 15:17
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L will be the streamwise width of your porous region, assuming you have 3D porous regions and not 2D baffles.

So if you have a rectangular porous region of w x h x t where flow moves along t, you would set L = t.

Make sure to use fundamental units too. Pa, m and m/s so that your units work out to a = kg/m^4 and b = kg/m^3-s
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Old   July 8, 2019, 15:21
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Quote:
Originally Posted by fluid23 View Post
L will be the streamwise width of your porous region, assuming you have 3D porous regions and not 2D baffles.

So if you have a rectangular porous region of w x h x t where flow moves along t, you would set L = t.

Make sure to use fundamental units too. Pa, m and m/s so that your units work out to a = kg/m^4 and b = kg/m^3-s
Thanks for the quick inputs. I have one final query. Is it required to have the porous region as a straight section. In my porous geometry the streamwise direction has 2 bends (i.e., U loop connecting input and output pipes).
Sincerely appreciate your help .
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Old   July 9, 2019, 08:14
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I am not very sure. But you have to give resistance coefficients along coordinate axis directions, so I think bends might change your desired flow.

Instead you could create a straight porous region and the bend you can model it as slip wall condition.
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Old   July 9, 2019, 16:11
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I am not very sure. But you have to give resistance coefficients along coordinate axis directions, so I think bends might change your desired flow.

Instead you could create a straight porous region and the bend you can model it as slip wall condition.
Thanks for your inputs nalamyogesh. Giving a slip boundary condition would take care of the frictional loses in the bends, will it also reduce the losses due to bend (change in direction of flow)?, so that the pressure drop due to the bends is minimum.

Thanks again!
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Old   July 9, 2019, 16:19
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You can assign an axisymmetric tensor and define on axis and cross-axis values. Then use interpolateDirection() to get a field function that assigns a vector field that is parallell to the closest wall. Set axisymmetric tensor axis to your calculated values and set off-axis properties to be anywhere from 10x to 100x your on-axis properties. The flow will follow the bends and should result in (close to) the expected DP.

The value L will be the integrated path length along your bent section.
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Old   July 9, 2019, 16:30
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Quote:
Originally Posted by fluid23 View Post
You can assign an axisymmetric tensor and define on axis and cross-axis values. Then use interpolateDirection() to get a field function that assigns a vector field that is parallell to the closest wall. Set axisymmetric tensor axis to your calculated values and set off-axis properties to be anywhere from 10x to 100x your on-axis properties. The flow will follow the bends and should result in (close to) the expected DP.

The value L will be the integrated path length along your bent section.
Thanks for your input will try to implement this method and get back .
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Old   July 11, 2019, 12:37
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Hello All,

I tried the asymmetric tensor suggest by @fluid23. But the pressure drop values are really high. One thing I noted is that my inertial resistance is one order of magnitude higher than my viscous resistance. Is this trend normal ? As mentioned earlier I used Ax^2+Bx for curve fitting, where A is my inertial resistance and B is my viscous resistance. Thank you.
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