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May 24, 2020, 06:03 |
Convergence problem - 2D flow around sphere
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#1 |
New Member
elad
Join Date: May 2020
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Hi guys,
I'm trying to simulate 2D flow around a sphere (free fall of sphere with air resistance). Some data before I'll explain the problem: Mesh properties: Element Type: triangles Nodes#: 180,417 Elements#:356543 Element Quality:0.96, Aspect Ratio: 1.21:1, Skewness:0.052, Orthogonal Quality: 0.97 Boundary layer: First layer Thickness: 5.7e-05[m], maximum layers: 5, growth rate: 1.1. Air properties: density (kg/m3): 1.06, dynamic viscosity=1.999e-05 (Re will be approx. 8000) The problem is transient as the sphere's velocity and angular velocity changing with time. Boundary Condition will be: inlet: vx=23.914tanh(0.409t), Turbulent Intensity: 0.052, Turbulent length scale: 0.002. outlet (pressure outlet): Gauge Pressure (pascal)=0, Backflow Pressure Specification= Total Pressure, Turbulent parameters same as inlet. sphere wall: wall motion: moving wall: rotational - speed (rad/s)=1594.3tanh(0.409t) Turbulent model: k-w sst or transition sst gravity: x(m/s2)=9.81 (fall with the positive direction of x axis) first i've tried to run steady solution for 200 time steps (intial guess)(SIMPLE FIRST ORDER, PRESSURE - STANDARD)then i use trasient solution for 668 time steps where time step size is 6.8e-04 (total of 0.454sec). max iterations/time step will be 30.(SIMPLE SECOND ORDER, PRESSURE - SECOND ORDER, SECOND ORDER IMPLICIT) well i know that the flow regime in this case is transition. there will be von karman vortex shedding. Residuals: 1e-12 The problem: The solution is not converging! i've tried to use FSM, coupled, PISO and simple with no luck. the drag coeff is pretty much constant and the vortex average is increasing due to the increse of the angular velocity. the mass balance is also good (power of 1e-07). don't know what im doing wrong? please help me |
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May 25, 2020, 12:15 |
Rotating Sphere
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#2 |
Senior Member
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You cannot solve a rotating sphere with 2D axisymmetric approach since then the problem is not axisymmetric. If you are using 2D planar, then it is not a sphere but a cylinder with a length of 1 m.
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May 25, 2020, 16:43 |
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#3 |
New Member
elad
Join Date: May 2020
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ok right and in case im solving for sphere that is not rotating it is possible to solve it that way?
i want to simulate flow around sphere in 2D. |
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May 26, 2020, 09:59 |
Sphere in 2D
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#4 |
Senior Member
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Yes, if the sphere is not rotating, then you can solve it using 2D axisymmetric. If it is rotating, then use 3D.
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May 26, 2020, 10:18 |
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#5 |
New Member
elad
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ok thanks.
still not converging for the other solution (where the sphere is not rotating), any idea? |
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May 26, 2020, 10:56 |
Dynamic Mesh
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#6 |
Senior Member
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What are the case settings? Is the solver set to 2D axisymmetric? Do you have boundary of domain and the sphere aligned with x-axis? How large is the domain upstream and downstream of the sphere? And could you share a snapshot of the domain?
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Regards, Vinerm PM to be used if and only if you do not want something to be shared publicly. PM is considered to be of the least priority. |
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May 26, 2020, 11:23 |
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#7 |
New Member
elad
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snapshot of the domain: https://pasteboard.co/JaaOAAH.png
I didn't set the solver to axisymmetric. flow is along x-axis (drag direction = positive x) gravity on = -9.81 (negative x direction) boundary conidtions: 1. left side - velocity - inlet 2. right side - pressure outlet 3. up and down walls = symmetry 4. sphere wall = stationary wall, no slip I've tried 2 viscous models: sst k-omega and transition sst solution method: PISO(trasient flow) --> spatial discretization all second order, trasient formulation--->second order implicit Residuals = 1e-012 standard initialization = compute from inlet st number=0.2 so number of time steps = 668 + time step size = 6.8e-04 (to get 0.454 sec) 30 iterations per time step |
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May 26, 2020, 11:31 |
Domain
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#8 |
Senior Member
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Then, it is not a sphere but a cylinder. What's the Reynolds number based on sphere diameter?
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Regards, Vinerm PM to be used if and only if you do not want something to be shared publicly. PM is considered to be of the least priority. |
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May 26, 2020, 11:32 |
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#9 |
New Member
elad
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Reynolds number = 8000
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May 26, 2020, 11:38 |
Turbulence
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#10 |
Senior Member
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Then, you don't need transition model. Gravity is not required since the density is constant. Top and bottom should be wall with specified shear of 0 and not symmetry.
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May 26, 2020, 11:42 |
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#11 |
New Member
elad
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i dont need gravity althoug im trying to simulate free fall?
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May 26, 2020, 12:01 |
Free Fall
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#12 |
Senior Member
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What you are modeling does not represent free fall. In a free fall, the velocity of the object always increases until it reaches terminal velocity. You are using a fixed velocity, so, either it can represent a terminal velocity or a fixed cylinder, but not a free fall. For a free fall, you have to use either Dynamic mesh or a UDF to change inlet velocity in accordance with Newton's law.
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May 26, 2020, 12:06 |
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#13 |
New Member
elad
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actually i've already mentioned that vx equles to 23.914*tanh(0.409*t).
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May 26, 2020, 12:27 |
Velocity
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#14 |
Senior Member
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I apologize, I didn't see that earlier. That's good then. In any case, gravity is not needed since the effect of gravity is to cause velocity increase, which you are adding via the equation. As far as convergence is concerned, is it diverging or not converging enough? Setting a lower value for residual is just a way to tell Fluent not to assume convergence until the condition is reached but it does not affect the solution. 1e-12 is a very small number and will not be achievable for each time-step, until the time-step is very very small, which may not be needed. If you are getting convergence below 1e-3 for each time step, then that is more than enough. Even 2e-3 is good enough. More important criteria are monitors, such as, drag and lift. If those are constant or statistically constant, then you have convergence.
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Regards, Vinerm PM to be used if and only if you do not want something to be shared publicly. PM is considered to be of the least priority. |
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May 26, 2020, 12:31 |
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#15 |
New Member
elad
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it is not converging enough.
ok thx a lot! |
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