Fluid Flow in Open Helical Channels
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1 Hayden Tronnolone Supervisor: Yvonne Stokes School of Mathematical Sciences September 24, 2010
2 Much research has been conducted into flows in closed, helical pipes. Such flows have application to modelling blood flow in arteries.
3 Much research has been conducted into flows in closed, helical pipes. Such flows have application to modelling blood flow in arteries. We investigate flows in open helical channels. Focus on the application to spiral particle separators used in the mineral processing industry.
4 Problem Background Mathematical Formulation Thin-film Approximation Numerical Solution Comparison of Solution Methods C Water slides commons.wikimedia.org
5 Spiral Particle Separators 2004 Tiomin Resources INC. (TIO) All rights reserved.
6 Secondary Flow
7 Mathematical Formulation Radius A, pitch 2πP. Curvature κ = A A 2 +P 2, torsion τ = P A 2 +P 2.
8 Helical co-ordinate system - from Germano (1982). Tangent T Normal N Binormal B For an orthogonal co-ordinate system we must use N and B in place of N and B. These rotate with s.
9 The governing equations are the Navier Stokes equations for an incompressible fluid, along with the continuity equation. These equations are expressed in our helical co-ordinate system.
10 The governing equations are the Navier Stokes equations for an incompressible fluid, along with the continuity equation. These equations are expressed in our helical co-ordinate system. We seek steady-state solutions so drop time derivatives.
11 Definition A fluid flow in a helical domain is helically symmetric if it is invariant along the length of the helix.
12 Definition A fluid flow in a helical domain is helically symmetric if it is invariant along the length of the helix. We will seek helically symmetric solutions so set derivatives with respect to s to zero. This is justified by Holtham (1992).
13 Definition A fluid flow in a helical domain is helically symmetric if it is invariant along the length of the helix. We will seek helically symmetric solutions so set derivatives with respect to s to zero. This is justified by Holtham (1992). We can now consider our domain to be a cross section through the channel. This reduces the problem from three spacial dimensions to two, although we are still solving for the velocity perpendicular to the plane.
14 In addition, we non-dimensionalise the problem. This introduces the new dimensionless parameters: ɛ : the non-dimensional curvature λ : the ratio of the torsion to the curvature
15 In addition, we non-dimensionalise the problem. This introduces the new dimensionless parameters: ɛ : the non-dimensional curvature λ : the ratio of the torsion to the curvature We have the standard Reynolds and Froude numbers, R and F. We also introduce the Dean number, K = 2ɛR 2
16 We assume that ɛ and λ are sufficiently small so that 0 < ɛ, λ R 1 so we can drop terms of this order.
17 We assume that ɛ and λ are sufficiently small so that 0 < ɛ, λ R 1 so we can drop terms of this order. Finally, we transform to a Cartesian co-ordinate system in the cross section.
18 Cartesian co-ordinates
19 Equation System Continuity equation: Navier Stokes equations: v y + w z = 0 tan α = λ v u y + w u z = 2 u + R sin α F 2 v v y + w v z 1 2 Ku2 = p y + 2 v v w y + w w z = p z + 2 w R2 cos α F 2
20 Equation System Continuity equation: Navier Stokes equations: v y + w z = 0 tan α = λ v u y + w u z = 2 u + R sin α F 2 v v y + w v z 1 2 Ku2 = p y + 2 v v w y + w w z = p z + 2 w R2 cos α F 2
21 Equation System Continuity equation: Navier Stokes equations: v y + w z = 0 tan α = λ v u y + w u z = 2 u + R sin α F 2 v v y + w v z 1 2 Ku2 = p y + 2 v v w y + w w z = p z + 2 w R2 cos α F 2
22 Boundary Conditions We apply the no-slip condition along the channel walls.
23 Boundary Conditions We apply the no-slip condition along the channel walls. On the free surface, we have two boundary conditions. The kinematic boundary condition states that fluid particles on the free surface stay on the free surface. The dynamic boundary condition states that there is no stress on the free surface (we ignore the effects of surface tension).
24 Thin-film Approximation This approximation arises from the thin fluid depth compared to its width. We can reduce this to a system linear differential equations using a thin-film approximation.
25 Thin-film Approximation This approximation arises from the thin fluid depth compared to its width. We can reduce this to a system linear differential equations using a thin-film approximation. Rescale the vertical co-ordinate ž = z δ where 0 < δ 1 is some small aspect ratio.
26 Thin-film Approximation This approximation arises from the thin fluid depth compared to its width. We can reduce this to a system linear differential equations using a thin-film approximation. Rescale the vertical co-ordinate ž = z δ where 0 < δ 1 is some small aspect ratio. We substitute into the previous system and determine corresponding scales on v, w and p. We keep only terms of order O(1).
27 Thin-film Equations Continuity equation: Navier Stokes equations: ˇv y + ˇw ž = 0 2 u ž 2 + sin α = 0 ˇp y + 2ˇv ž 2 + χu2 = 0, ˇp ž cos α = 0 χ = δk 2R
28 Thin-film Equations Continuity equation: Navier Stokes equations: ˇv y + ˇw ž = 0 2 u ž 2 + sin α = 0 ˇp y + 2ˇv ž 2 + χu2 = 0, ˇp ž cos α = 0 χ = δk 2R
29 Thin-film Equations Continuity equation: Navier Stokes equations: ˇv y + ˇw ž = 0 2 u ž 2 + sin α = 0 ˇp y + 2ˇv ž 2 + χu2 = 0, ˇp ž cos α = 0 χ = δk 2R
30 Thin-film Equations Continuity equation: Navier Stokes equations: ˇv y + ˇw ž = 0 2 u ž 2 + sin α = 0 ˇp y + 2ˇv ž 2 + χu2 = 0, ˇp ž cos α = 0 χ = δk 2R
31 Let the channel shape be H(y) and the free surface be H(y) + h(y).
32 Let the channel shape be H(y) and the free surface be H(y) + h(y). The thin-film solution is then (dropping checks on variables) p(y, z) = cos α(h + h z) u(y, z) = sin α (z H)(H + 2h z) 2 v(y, z) = χ sin2 α 120 (z H){(z H)3 [(H + 2h z) (H + 4h z) + 2h 2 ] 16h 5 } cos α (z H)(H + 2h z) (H + h) 2 y
33 Numerical Solution To solve the full equations we must use a numerical solution method. We use the finite element package, Comsol Multiphysics.
34 Numerical Solution To solve the full equations we must use a numerical solution method. We use the finite element package, Comsol Multiphysics. Initially, we do not know the shape of the free surface. We use the Arbitrary Lagrangian Eulerian method. This method allows us to adjust the location of the free surface during the solution process.
35 ALE Method
36 1. Make an initial guess for the free surface shape.
37 1. Make an initial guess for the free surface shape. 2. Solve for the velocity and pressure, subject to the no-slip condition on the channel wall and dynamic condition on the free surface.
38 1. Make an initial guess for the free surface shape. 2. Solve for the velocity and pressure, subject to the no-slip condition on the channel wall and dynamic condition on the free surface. 3. Move the free surface in the direction of the normal component of velocity. If the flux is positive, the surface shifts up. If the flux is negative, the surface shifts down.
39 1. Make an initial guess for the free surface shape. 2. Solve for the velocity and pressure, subject to the no-slip condition on the channel wall and dynamic condition on the free surface. 3. Move the free surface in the direction of the normal component of velocity. If the flux is positive, the surface shifts up. If the flux is negative, the surface shifts down. 4. Repeat steps 2 and 3 until the kinematic boundary condition is satisfied to some tolerance ( free surface u n ds tolerance).
40 We can use several different measures to assess the accuracy of the solution. The flux through the free surface. The area of the flow domain.
41 We can use several different measures to assess the accuracy of the solution. The flux through the free surface. The area of the flow domain. We solve for a rectangular channel shape.
42 Streamlines of Numerical Solution z (thin film) y R = 20, F = 0.2, K = 375
43 We have error in area The absolute value of the flux through the free surface is approximately The flux down the channel is 0.051, which agrees with previous calculations (Stokes, Wilson & Duffy 2004).
44 Comparison of Solution Methods z (thin film) y Navier Stokes model (numerical solution) Thin-film approximation ( δ = 0.1 )
45 The Navier Stokes model gives the flux down the channel as 0.051, while the thin-film approximation gives
46 Conclusions Fluid flows in open, helical channels have been investigated. Full numerical solutions have been found which agree with the literature.
47 Conclusions Fluid flows in open, helical channels have been investigated. Full numerical solutions have been found which agree with the literature. A thin-film approximation has also been found which matches the numerical solution closely, except at the channel boundaries.
48 Conclusions Fluid flows in open, helical channels have been investigated. Full numerical solutions have been found which agree with the literature. A thin-film approximation has also been found which matches the numerical solution closely, except at the channel boundaries. Future work will consider investigating this agreement under different parameters. Solutions under different channel geometries could also be explored.
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