sci.math.num-analysis SUNA39, Reduced quad for internal flow ====================================== For the problem at hand, incompressible and irrotational flow at the shell side of tube bundles (SUNA38), we will restrict ourselves to rectangular cylindrical symmetric elements. While considering the incompressibility condition, fluxes through cylinder surfaces are important. Accompanying surfaces are defined as follows: _________ < r2,z2 If the cylinder axis is on the left, then: | | O(3) | a | ^ | O(1) = pi.(r2.r2 - r1.r1) x O(4) | | O(2) O(2) = 2.pi.r2.(z2 - z1) i ^ | | ^ O(3) = O(1) s |_________| O(4) = 2.pi.r1.(z2 - z1) | r1,z1 > O(1) ^ Conservation of mass now gives rise to the following equation: - O(1).v1 + O(2).u2 + O(3).v3 - O(4).u4 = 0 Here the numbering of area's O and velocities (u,v) have been taken equal. Or, apart from the constants (pi): 2.(r2.u2 - r1.u4).(z2 - z1) + (r2.r2 - r1.r1).(v3 - v1) = 0 Divide the whole expression by 2.(r2 - r1).(z2 - z1) : r2.u2 - r1.u4 v3 - v1 ------------- + (r2 + r1)/2 . ------- = 0 r2 - r1 z2 - z1 In the limit for r2 - r1 -> 0 and z2 - z1 -> 0 this is consistent with the partial differential equation for incompressibility in cylinder coordinates: d(r.u)/dr + r.dv/dz = 0 Much easier to handle is the irrotational condition. In fact, it is completely equivalent with the condition in flat 2-D cartesian coordinates: (r2 - r1).u1 + (z2 - z1).v2 + (r1 - r2).u3 + (z1 - z2).v4 = 0 In the limit for r2 - r1 -> 0 and z2 - z1 -> 0 this is consistent with the partial differential equation for irrotationality (?) in cylinder coordinates: du/dz - dv/dr = 0 In the above, fluxes have been defined as if the velocities are "midside" ones. But in the first place we are stucked with the Reduced Quadrilateral, which has its velocities defined at the vertices. The relationship between the former and the latter kind of velocities is simply found by taking mean values. This means that, in the above discretizations, substitutions like the following have to be made: f1 := (f1 + f2)/2 midside vertex Such a procedure results in schemes like below: A(8) A(6) A(7) A(5) v v ____|____ ____|____ 4 ____|____ 3 A(7) | | A(5) A(8) | | A(6) u | | u _| |_ _| |_ _| |_ | | | | | | A(1) |____ ____| A(3) A(2) |____ ____| A(4) u |____ ____| u | | 1 | 2 A(2) A(4) A(1) A(3) v v Incompressible Irrotational Vertex velocities, each distributed A(2*k-1) = coefficient of Approximation for u(k) over two halves of A( 2*k ) = coefficient of Approximation for v(k) the quad's sides. Remember that the reduced quadrilateral is a "wrong" element. Despite of this, alas (:-), it frequently gives _exellent_ results. To be continued ... - * Han de Bruijn; Applications&Graphics | "A little bit of Physics * No * TUD Computing Centre; P.O. Box 354 | would be NO idleness in * Oil * 2600 AJ Delft; The Netherlands. | Mathematics" (HdB). * for * E-mail: Han.deBruijn@RC.TUDelft.NL --| Fax: +31 15 78 37 87 ----* Blood