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Cone and Plate Viscometer Boundary Conditions |
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December 15, 2021, 14:06 |
Cone and Plate Viscometer Boundary Conditions
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New Member
Hans Foelsche
Join Date: Sep 2020
Posts: 6
Rep Power: 5 |
Hello,
I am trying to simulate a cone-and-plate viscometer (CPV) in Fluent, and I am having trouble with one of the boundary conditions. I am simulating the CPV from the CAD model attached. The petri dish will be seeded with cells on the bottom, and the cone will be driven by a motor. The circular edge of the cone is a set distance from the edge of the petri dish. When the fluid is placed in the petri dish, there will be some fluid that will ride up the side of the cone. I have been having trouble simulating the top portion of the fluid where the fluid will be exposed to the surrounding environmental conditions. I have tried various wall boundary conditions, and none of them are properly simulating a freely rotating fluid boundary condition. Whether I select stationary or moving wall, all of them result in this specific boundary as a stationary wall. I have tried both absolute and relative rotational velocity options, and neither does the trick. However, all of the other boundary conditions are working as expected. For some additional background, I implemented a periodic boundary condition so that I only have to simulate a small slice of the CPV, as it is axisymmetric. I also have been driving the fluid with a cone speed of 10 rad/s with water as the fluid to start. Also assumed steady and incompressible. The holes in the petri dish will have tubing to replace the fluid. Attached are the model of the cone and petri dish, as well as a velocity contour of the fluid. I only included the scale to show that the velocities are indeed correct according to v=w*r. As seen in the contour figure, the top part of the fluid is stationary. That is the issue I am trying to fix, as it should be free to rotate, not bounded as a stationary wall. The contour at the top wall should be a gradient from the highest speed to stationary, not all stationary. Thank you for any help. |
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