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February 22, 2017, 09:34 |
Setting Gravity for a VERTICAL flow
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#1 |
New Member
Leo
Join Date: Jul 2014
Posts: 13
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Hello everyone,
Keeping it short and simple: How do I consider gravity in a CFX simulation? I have an one-phase-flow but it is strong affect by quote differences, it means, the flow is vertical (from the bottom to the top - which in my flow means the gravity would be -9.81 in the Y direction). Going further into details: I had of course searched for an answer last week and I read about activating "buoyancy effects" and so I have been doing this whole week. Now I was checking for plausibility and my flow is not plausible at all. I mean, changing the value of gravity acceleration in the buoyancy conditions makes no difference in my flow. Even considering gravity in + ou - Y or X-Axis or changing its value to 2, 10, 100 m/s² makes absolute no difference. So Buoyancy is not the point here. Thank you in advance! |
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February 22, 2017, 10:42 |
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#2 |
New Member
Leo
Join Date: Jul 2014
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Quick update: I've just created a Subdomain and there I have defined a "Momentum Source" equal to rho*g (in my case 997 kg m-3 * 9.81 m s-1 = 9780 kg m-2 s-2).
Now my results make no sense, streamlines and velocity on plane seem very weird. Was it a wrong guess? |
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February 22, 2017, 17:38 |
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#3 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
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What does gravity do in your flow? Do you have pressure dependent material properties (such as density a function of pressure)? You have stated it is a single phase flow so you do not have free surface effects.
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February 22, 2017, 21:37 |
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#4 |
New Member
Leo
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Hello Glenn, thanks for the attention.
My flow hast a strong elevation change (Delta H). If we consider the Bernoulli's equation, I wonder if and how CFX considers the term "rho*g*Delta h" from Bernoulli when solving my case. Giving möge details: I have one inlet (in the bottom), which I define with a known mass flow, and a single Outlet (in the top), which a define with a Static Pressure = 0 (Opening Pressure and Dirn). |
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February 22, 2017, 22:28 |
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#5 |
Super Moderator
Glenn Horrocks
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Unless the pressure affects the fluid then you should not model this with buoyancy. This is why the results you have been getting so far are weird. You should model this with no gravity and in post-processing add a hydrostatic pressure to the results (which will just be a linear function against height). You will need to account for the hydrostatic head in any pressure boundaries you apply.
So in your case do the simulation with no gravity and use the outlet at the top with pressure =0. In post-processing define a variable which is just the hydrostatic pressure and add it to the pressure from the simulation result. |
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February 24, 2017, 05:54 |
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#6 |
New Member
Leo
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Hello,
Thank you. I did what you recommended in post-processing and it worked pretty well. What seems a little bit "not intuitive" for me is that the gravity will have no effect at all in the internal flow, but only in the total pressure of the system (which is added just in the end, after the simulation). I attached a figure to this reply. There you see a rough sketch of my system. It is like a cuboid, inlet in the bottom, outlet at the top. So, it means, doesn't matter how we stand this cuboid in the ground, it will have absolutely no effect in the internal flow??? (for example, what if the gravity in this picture was not in -y but instead in + or -z ?). |
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February 24, 2017, 06:50 |
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#7 |
Super Moderator
Glenn Horrocks
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Be careful with the word "total pressure". In high speed flows this is defined as the static pressure plus the dynamic pressure. A better description for your case is absolute pressure.
The gravity direction will affect the pressure difference between the inlet and outlet as the height difference changes. Then the internal flow is just driven by the pressure difference. |
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February 24, 2017, 07:45 |
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#8 | |
New Member
Leo
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Quote:
But what if am I modeling my boundaries with mass flow rate? Let's imagine two situations (for the picture above): A: horizontal flow (gravity z axis) Inlet: 0.01 L/s Outlet: Opening Pressure 0 Pa B: vertical flow (gravity -y axis) Inlet (bottom): also 0.01 L/s Outlet: Also Opening Pressure 0 Pa (because this outlet is set on the top) So, is it right that both flows look exactly the same, or am I modeling my boundaries wrong? |
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February 25, 2017, 06:48 |
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#9 |
Super Moderator
Glenn Horrocks
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There will be a small difference between your two cases due to the hydrostatic pressure difference across the pressure boundary. But it will be a small difference, overall the two simulations will be very similar.
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February 28, 2017, 03:55 |
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#10 |
New Member
Leo
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Thank you.
So, just correct me if I am wrong: I can model my boundaries the way I am doing, with mass flow at the inlet and opening 0 Pa at the outlet, and since two simulations are very similar, I don't need to worry about anything. Thank you for your help. |
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February 28, 2017, 05:26 |
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#11 |
Super Moderator
Glenn Horrocks
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I would not say "I don't need to worry about anything". There is always something to worry about
You should check that this assumption is valid. Do a sensitivity check on it and see if it makes any difference. If it makes no difference then you can stop worrying. |
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March 6, 2017, 04:55 |
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#12 |
New Member
Leo
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Alright!
So, I did a sensitivity check and it's true, it makes apparently no difference. Thank you so much Ghorrocks! |
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May 10, 2019, 19:08 |
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#13 | |
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Newfoundland and Labrador
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Quote:
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May 11, 2019, 06:35 |
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#14 |
Super Moderator
Glenn Horrocks
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hydrostatic pressure is simply density*g*height, so you can easily add that in post processing if you like. But you can only do this if the free surface is level.
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