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High order FINITE VOLUME "upwind" scheme for Convective terms in Momentum equation |
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January 5, 2016, 13:46 |
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#21 |
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Filippo Maria Denaro
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I also suggest to see a chapter in a old book where the multidimensional implementation (while taking into account the presence of the surface integral) is illustrated for the QUICK.
QUICKEST - Quadratic Upstream Interpolation for Convective Kinematics with Estimated Streaming Terms B. P. Leonard. Elliptic systems: Finite-difference method IV. In W. J. Minkowycz, E. M. Sparrow, G. E. Schneider, and R. H. Pletcher, editors, Handbook of Numerical Heat Transfer, pages 347–378. Wiley, New York, 1988. |
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January 5, 2016, 13:49 |
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#22 | |
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Mihir Makwana
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based on the AVERAGE value of velocity "u" ( first ) i.e whether +ve or -ve at the face of u-CV, the upwinding for "u" (second) can be applied. I will look at the paper ( section 4 ) |
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January 5, 2016, 13:51 |
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#23 | |
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Mihir Makwana
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I will take a look at the book |
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January 5, 2016, 13:53 |
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#24 | |
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Filippo Maria Denaro
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but in a FV method, you do not work on d (uu)/dx ...just think about the term Div (vv) in the differential form that must be integrated to produce the integral form. Therefore, you have to discretize (1/|V|) Int [S] n.(vv) dS. You see that no derivative appears and the upwind must be applied on the flux function reconstruction. |
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January 5, 2016, 14:01 |
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#25 |
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Mihir Makwana
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if say its a 1D flow,
after integrating duu/dx w.r.t x i get uu at the faceS of the u-C.V Now, based on the AVERAGE value of velocity "u" ( first ) i.e whether +ve or -ve at the face of u-CV, the upwinding for "u" (second) can be applied. |
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January 5, 2016, 14:10 |
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#26 | |
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Filippo Maria Denaro
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If you consider ONLY the 1D case, you can work exactly as described in Section 4 in the paper I linked. The integral will simply produce: [(uu)est - (uu)west ]/h Note that, whatever high order reconstruction you use for the flux, the time derivative acts on the volume averaged velocity, therefore you must always expect a second order accuracy in space in the error analys. A proper higher order scaling in the error curve requires special reconstruction bewteen averaged and point-wise velocity as described in Sec.4 |
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January 5, 2016, 14:12 |
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#27 | |
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Mihir Makwana
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This will reduce the overall order of the scheme. Is there a way ( you did mentioned building lopsided stencils that retain 5th order of which i have no idea ) where overall 5th order is maintained. |
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January 5, 2016, 14:16 |
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#28 | |
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Mihir Makwana
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January 5, 2016, 14:20 |
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#29 | |
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Filippo Maria Denaro
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to mantain the same order on a boundary you have to use one-sided interpolation...however, let me say that some types of boundary can be treated. For example, on a wall you set just that the normal component of the convective flux integrated on the face of the FV is zero. |
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January 5, 2016, 14:34 |
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#30 | |
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Michael Prinkey
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January 5, 2016, 14:43 |
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#31 | |
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Filippo Maria Denaro
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As I wrote above, the solution is in the one-sided interpolation at high degree that does not depend on the upwinding. However, near a wall the normal velocity component will be very small and the diffusive flux will be predominant over the convective (in turbulence, we have at least 3-4 nodes within y+=1, i.e. the viscous sub-region). That justifies the fact one forces the stencil according to the wall location |
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January 5, 2016, 14:51 |
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#32 |
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Filippo Maria Denaro
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just consider that on a wall du/dx=dw/dz=0 (x streamwise, z spanwise), therefore from the continuity equation dv/dy=0 with v=0 at the wall...
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January 5, 2016, 15:02 |
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#33 | |
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Michael Prinkey
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Chasing uniform high-order accuracy is a difficult task, as I'm sure you well know. |
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January 7, 2016, 08:31 |
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#34 |
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Mihir Makwana
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The paper
THREE-DIMENSIONAL INCOMPRESSIBLE FLOW CALCULATIONS WITH ALTERNATIVE DISCRETIZATION SCHEMES by Panos Tamamidis & Dennis N. Assanis to which i referred for the 5th order upwind scheme given by eqn. 17 is http://i.imgur.com/GlFRysZ.jpg?1 The paper to which they refer i.e Rai[13] is http://i.imgur.com/Z8cEyLf.jpg?1 to which i don't have access Now when i fit a 4th order polynomial with 3 upstream points and 2 downstream points then i get different co-efficients http://i.imgur.com/EgNDdw9.jpg?1 the way i fitted the polynomial is phi(x=2.5dx) = L0*phi_d2 + L1*phi_d1 + L2*phi_u1 + L3*phi_u2 + L4*phi_u3 where Li are lagrange polynomial at i ( = 0 to 4) with x_i = i*dx I used the same analogy for quick scheme and it gave me the correct co-efficients. So, why a discrepancy with the formula ( eqn. 17 ) ?? |
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January 7, 2016, 09:35 |
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#35 | |
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Michael Prinkey
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January 12, 2016, 15:19 |
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#36 |
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Biswajit Ghosh
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Did you tried MUSCL?
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