| Author | Message | |||
     Susanne Zibek (susanne) New member Username: susanne Post Number: 2 Registered: 12-2007 |
Dear Mr. Nelson, I have a big cylinder (region 'cylinder') with a hole in the centre of the bottom (region 'bottom_pore') for influx of concentration (u) with velocity v0. In timedependent simulation there will be a distribution of concentration in 'cylinder'. Now I'm interessted in volume integral of (u) in a special region- so I defined 'smallcylinder' (or is there another possibility? ) Before starting the timedependent simulation a steady state is necessary. Without 'smallcylinder' steady state and timedependent simulation are running well (attachment:with about 2500 nodes) but with 'smallcylinder' even steady state isn't running: error when 1,000,000 nodes achieved). How can I solve this problem: I just want to calculate the volume integral – the ‘smallcylinder’ should not have any influence in simulation. I want to run the simulation with the minimum possible number of nodes (example: timedependent simulation with 10 seconds and 20,000 nodes needs 24h CPUtime) Thank you very much for your suggestion and help Susan
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1050 Registered: 06-2003 |
FlexPDE must construct a tetrahedral mesh to perform the computation. This mesh must match the shape and size of features you put in the domain. You have specified tiny cylinder, running from top to bottom within the large cylinder. In order to resolve the shape of a cylinder, FlexPDE uses at least twelve cells around the circumference. Your domain is 700 times as long as this dimension, so there are about 700*12*2 cell faces on the surface of the cylinder and about 24 cell faces in the cross-section. This means about 60000 nodes to fill the cylinder. But the cells in the surrounding volume must conform to the size of the cells in the cylinder, so we end up with hundreds of thousands of cells to model the tiny cylinder. There is no mechanism to form a volume integral on a volume that is not bounded by mesh cells. I suggest you change the circular cross-section of your cylinder to a square. This can be modeled with far fewer cells, and the resulting time degradation will not be as severe. I used a square cylinder of side=2*RSphere, and the problem completed in under 10 minutes. | |||
| abbasali Member Username: abbasali Post Number: 4 Registered: 04-2010 |
Hi Dear Dr. Nelson How can I find following integrals 1- Assume the coordinate is cartesian2 and (x,y),(x1,y1) are two typical points. I want to find following integral VOL_INTEGRAL(sqrt((x-x1)^2+(y-y1)^2)*h(x1,y1) dx1dy1) in whole R^2, where h defined previously. As you know the result should be a function of (x,y) which I can apply it in next calculations. 2- With above assumptions, VOL_INTEGRAL(LOG10(1/sqrt((x-x1)^2+(y-y1)^2))*q(x1,y1) dx1dy1) where q is the unknown of our problem we should find it using next equations. | |||
| rgnelson Moderator Username: rgnelson Post Number: 1366 Registered: 06-2003 |
There is no mechanism for defining a scalar field where each point is the integral over a diminished range of the domain. You can, however, differentiate the VOL_INTEGRAL wrt x and y, resulting in a PDE for the integral value. Declare a VARIABLE to model the integral value at each point, and write the PDE in the EQUATIONS section. You will need to provide boundary conditions at each entry boundary. |
