| Author | Message | |||
     Alex Shulgin (alex2000) New member Username: alex2000 Post Number: 1 Registered: 10-2007 |
I have been trying to solve a heat equation Div(-k*grad(Temp)) + dt(Temp) = 0. The region consist of inner circle (where k = 1, material No.1) and outer ring (where k = 0.23, material No.2). On outer boundary I have VALUE(Temp) = Tw or NATURAL(Temp) = alfa*(Temp – Tw), where alfa and Tw are known. There isn’t problem. But at the interface (on the boundary between of two materials) BC is lamda*(n<dot>grad(Temp))2 = (n<dot>grad(Temp))1 + f(Temp)2, where lamda is known constant, f is known function of Temp. There where (…)1 is the value at the left hand side of interface, (…)2 is the value at the right hand side of interface How FlexPDE can deal with this case? I think that that attached script isn’t deal this BC correct.
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 980 Registered: 06-2003 |
The default condition for your PDE on internal boundaries is normal(k*grad(Temp))1 = normal(k*grad(Temp))2. That is, energy is conserved. So if your lambda by chance equals k2/k1, then that part of your requirement is satisfied by default. In this case, an internal source such as f(Temp) can be applied simply as Natural(Temp)=f(Temp). Assuming that Temp is continuous at the boundary, f(Temp)2=f(Temp)1. The Natural on an internal boundary means the difference between the boundary flux as seen on the left and the boundary flux as seen on the right. So Natural(Temp)=f means (k*grad(Temp))2 = (k*grad(Temp))1 + f. See documentation of Natural BC. If lambda is not k2/k1, then you will have by implication some kind of magic energy dump at the boundary, and I don't know how this should be modeled without more information about the physics. If Temp is not continuous at the boundary, then you need a CONTACT boundary condition to support two separate values. See the documentation of the CONTACT boundary condition. | |||
| muneer Ismael (muneer) New member Username: muneer Post Number: 1 Registered: 12-2007 |
My Dear Dr. Nelson I am solving two variable say V1 and V2 using flexpdfe in 3-dimensional extruded domain. For each variable there exists a separate pde. The problem is that the nuemann boundary condition of variable V2 on the extruded cylindrical surface equals to the normal component of a vector composed of the other variable i.e. Natural (V2)=normal(grad(V1)). Now my question is how I can verify this boundary condition on the extruded curved surface. My very best Sincere regards | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1021 Registered: 06-2003 |
The Natural BC defines the outward normal component of some vector characteristic of how you have written your PDE. Let us say your equation is Div(K*grad(u))+S=0 The natural BC therefore means the outward normal component of K*grad(u). If you have set Natural(u)=Normal(v), then you can plot Normal(K*grad(u)) on an extrusion surface and compare it with a similar plot of Normal(v). Or you can plot the difference of the two normals. For example, if you have a surface named "Top", you can write CONTOUR(Normal(K*grad(u))) ON SURFACE "Top" See "Help->Problem Descriptor Reference->The Sections of a Descriptor->Monitors and Plots->Controlling the Plot Domain", or see "ON SURFACE" in the Help Index. | |||
| muneer Ismael (muneer) New member Username: muneer Post Number: 2 Registered: 12-2007 |
My Dear Dr Nelson Thank you very much indeed your replaying and efforts. By the virtue of you I obviated the deffinition of Bcs. But I have another problem if you permit this is as follow: I need to calculate the surface integral of normal(V1)on a portion of the sidewall. where this portion laies in say layer '2'and occupies only 15 degrees of the sidewall(i.e. on a portion of the cylinderical extruded sidewall) Thank you very much for your efforts | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1024 Registered: 06-2003 |
Give a name to the layer. Overlay a named FEATURE on the segment of the wall where you want to form the integral. Then use the form result = SURF_INTEGRAL(<integrand>,<boundary_name>,<layer_name>) See "Integrals in three dimensions", "Features" and "Boundary Paths and Path names" in the Help Index. |
SampleP1