Vorticity Stream function Solver in Cylindrical Coordinates

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1 Vticity Stream function Solver in Cylindrical Codinates L. Brieda May 27, 2016 This document summarizes equations used to solve flow in a cylindrical pipe using the stream function approach as seen in 1 Governing Equations 1.1 Velocity Components Velocity components of an incompressible axisymmetric flow can be described using Stokes stream function ψ as [1] u z = u = 1 r u r = v = 1 r z The azimuthal component u θ does not depend on the stream function and can be defined independently. In this writeup, u θ = 0. Also, the volumetric flow bounded by streamtube ψ is Q = 2πψ. This fmulation automatically satisfies continuity u = 0, since 1.2 Vticity 1 (ru r ) + u z r z = 1 r z + 1 r z = 0 Vticity is defined as ω = v. F axisymmetric flow with / θ = 0 and u θ = 0 only the w θ component survives and we have ω = ω θ = v z u (2) 1.3 Stream function governing equation By substituting definitions of velocity in terms of Stokes stream function, Eq. 1, into Eq. 2 f vticity, we obtain ω = ( ) 1 ( ) 1 z r z r = 1 r z 2 1 r r 2 z ψ 2 1 = ωr (3) r The left side of this equation is not 2 ψ due to the negative sign on the last term on the left side. (1) 1

2 1.4 Vticity Transpt Equation The tempal evolution of vticity is given by the vticity transpt equation. This equation is nmally derived by taking curl of the momentum equation, f instance see [2] f details, and is given by F the axisymmetric flow we have t + v ω = ν 2 ω t + u w z + v w = ν [ 2 ω z ω r ] (4) 2 Finite Difference Fm 2.1 Stream function Equation 3 is discretized using standard central difference as 1 2 z (ψ i 1,j 2ψ i,j + ψ i+1,j ) r (ψ 1 i,j 1 2ψ i,j + ψ i,j+1 ) 2r i,j (ψ i,j+1 ψ i,j 1 ) = ω i,j r i,j (5) This equation is solved using an SOR-accelerated Jacobi solver, with convergence check ψ k+1 ψ k < ɛ tol. 2.2 Vticity The vticity transpt equation, Equation 4, is advanced using the Runge-Kutta fourth-der (RK4) method. Letting vticity transpt equation be given by [ 2 t = ν ω z ω ] u w r z v w t = R(ω) we have [3] w (1) = w k + t 2 Rk (6) w (2) = w k + t 2 R(1) (7) w (3) = w k + tr (2) (8) w k+1 = w k + t 6 3 Boundary Conditions ( R k + 2R (1) + 2R (2) + R (3)) (9) In der to solve this equation (using finite difference method) we need to specify boundary conditions. There are five types of boundaries to consider f the tube problem: 1) wall, 2) axis of revolution, 3) inlet, 4) outlet on zmax, and 5) outlet on rmax. These are sketched in Figure Inlet We assume the flow entering through the inlet is parallel to the cylinder axis. Thus at the inlet u r = v = 0 = 0 (10) z inlet 2

3 Figure 1: Boundary types f pipe problem Substituting v = 0 into the vticity equation gives us ω inlet = u (11) These equations are discretized using first der scheme as ψ 0,j = ψ 1,j and ω 0,j = (u 0,j 1 u 0,j+1 )/(2). 3.2 Axis of Revolution We require v = 0 as there can be no flow across the axis of revolution. Therefe / z = 0, and value of ψ is constant along the axis. We set this value to zero, giving us ψ axis = 0 (12) Zero radial velocity also implies that along the axis ω = u/. Axial symmetry implies ()/ = 0 at r = 0, and ω axis = 0 (13) 3.3 Wall Q = 2π(ψ 2 ψ 1 ) is the volumetric flow rate between two stream tubes. With ψ 1 = 0 being the axis of revolution, we have Dirichlet ψ wall = 1 2 u 0r 2 inlet (14) along the outer wall. Vticity boundaries along the wall are derived using similar approach to [2]. Since the stream function is constant along a wall, derivatives of ψ in Equation 3 vanish in the wall direction. Along a left wall we have z 2 wall = ωr Assuming the wall is at i = L, we can write the following Tayl series expansion ψ L+1,j = ψ L,j + z + 2 ψ L,j 2 L,j z 2 2 Since / z = vr and finally ψ L+1,j = ψ L,j v L,j r + 2 ψ L,j 2 z 2 2 L,j z 2 = 2(ψ L+1,j ψ L,j ) 2 + 2v L,jr ω L,j = 2(ψ L,j ψ L+1,j ) r 2 2v L,j (15) 3

4 Using similar approach, the boundary condition f a right wall at i = R is found to be ω R,j = 2(ψ R,j ψ R 1,j ) r 2 Along the top wall, / z = 0 and Equation 3 reduces to [ 2 1 ] r Again we start by expanding the second derivative, Using / = ur, the above reduces to ψ i,t 1 = ψ i,t wall + 2v R,j = ωr + 2 ψ i,t 2 i,t 2 2 (16) Substituting into the iginal equation, i,t 2 = 2(ψ i,t 1 ψ i,t 2 2(ψ i,t 1 ψ i,t ) 2 + 2u i,t r + 2u T,jr u i,t = ωr ω i,t = 2(ψ i,t ψ i,t 1 ) r 2 2u i,t + u i,t (17) r There is no bottom wall in this problem, but f generality, the matching boundary condition can be found to be 3.4 Zmax outlet ω i,b = 2(ψ i,b ψ i,b+1 ) r 2 + 2u i,b + u i,b r In general, the flow will be aligned with the z-axis, however, there may be some non-zero v component due to jet expansion. As such, simply setting / z zmax = 0 may not be valid. Following approach in [2], on zmax we let = vr (19) z zmax which is differenced as ψ ni 1,j = ψ ni 2,j v ni 1,j r j. Vticity boundary condition on the outlet is set as = 0 (20) z zmax ω ni 1,j = ω ni 2,j 3.5 Rmax outlet This is the trickiest of all boundaries and I am not particularly sure what boundary condition is most applicable. Generally, we expect there to be very little / no flow here. Setting no-flow boundary is analogous to making this boundary a wall, with ψ = ψ wall and ω set from Equation 17. However, I think me appropriate boundary may be requiring that any flow there may be is perpendicular to the wall, hence u = 0 and rmax = 0 (21) which is differenced as ψ i,nj 1 = ψ i,nj 2. Vticity boundary condition is set similarly to the zmax outlet = 0 (22) rmax ω ni 1,j = ω ni 2,j (18) 4

5 References [1] Wikipedia, Stokes Stream Function, Accessed May 10th, 2016, function [2] Salih, A., Streamfunction-Vticity Fmulation, Department of Aerospace Engineering Indian Institute of Space Science and Technology, March 2013, [3] Tannehill, J., Anderson, D., Pletcher, R., Computational Fluid Mechanics and Heat Transfer, Tayl & Francis, 2nd ed.,

F11AE1 1. C = ρν r r. r u z r

F11AE1 1. C = ρν r r. r u z r F11AE1 1 Question 1 20 Marks) Consider an infinite horizontal pipe with circular cross-section of radius a, whose centre line is aligned along the z-axis; see Figure 1. Assume no-slip boundary conditions