Derivation and solution of the equations for fluid flow … · Derivation and solution of the equations for ﬂuid ... the continuity equation ... Derivation and solution of the equations

Derivation and solution of the equations for fluid

flow in a helical channel

Hayden TronnoloneSupervisor: Yvonne Stokes

The University of Adelaide

School of Mathematical Sciences

November 2, 2011

Overview Equations Thin-film approximation Solution Existence

Gravity separation spirals

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Derivation and solution of the equations for fluid flow in a helical channel University of Adelaide


Secondary flow



Water slides

www.kulin.wa.gov.au

commons.wikimedia.org www.copacabanaresort.com



Flow in a river

Deposited

material

Outer bank

eroded

Surface

Rotating

secondary

flow

Inner bank

Outer bank



◮ This can result in the development of a meander.

◮ Cauto River (Rıo Cauto), Cuba (from Wikimedia Commons)



Mathematical modelling

◮ It is very difficult to measure thin flows without interferingwith them.

◮ Aim to develop a mathematical model of the fluid flow; asystem of equations that describes the real-world process.

◮ The equations need to capture the important aspects of thesystem while remaining tractable.



◮ The governing equations are◮ the Navier–Stokes equations for an incompressible fluid

(Newton’s second law); and◮ the continuity equation (conservation of mass).





◮ We seek steady-state solutions; we assume the velocity is notchanging with time.





◮ We seek steady-state solutions; we assume the velocity is notchanging with time.

◮ We also seek helically-symmetric solutions; we assume thevelocity (and hence geometry) in any cross section along thechannel.



◮ The helix can be describedby its curvature and torsion.

◮ We assume both are small.



Boundary conditions

◮ To complete our description, we need to say what happens atthe boundaries.

◮ Friction causes the fluid to “stick” to the channel wall, so thevelocity is zero there.



Boundary conditions

◮ To complete our description, we need to say what happens atthe boundaries.

◮ Friction causes the fluid to “stick” to the channel wall, so thevelocity is zero there.

◮ On the free surface, we have two boundary conditions.◮ Fluid particles on the free surface stay on the free surface.◮ There is no stress on the free surface (we ignore the effects of

surface tension).



Governing equations

Continuity equation:∂v

∂y+

∂w

∂z= 0

Navier–Stokes equations:

v∂u

∂y+ w

∂u

∂z= ∇2u +

R sinα

F2

v∂v

∂y+ w

∂v

∂z−

1

2Ku2 = −

∂p

∂y+∇2v

v∂w

∂y+ w

∂w

∂z= −

∂p

∂z+∇2w −

R2 cosα

F2

R = Uaν, F = U

√ga, K = 2ǫR2



Thin-film approximation



Thin-film approximation

◮ Rescale the vertical co-ordinate. Let

z =z

δ,

where 0 < δ ≪ 1 is some small aspect ratio.

◮ We substitute into the previous system and determinecorresponding scales on v ,w and p. We keep terms of leadingorder in δ (the largest terms).



Thin-film equations

Continuity equation:∂v

∂y+

∂w

∂z= 0

Navier–Stokes equations:

∂2u

∂z2+ sinα = 0

−∂p

∂y+

∂2v

∂z2+ χu2 = 0

−∂p

∂z− cosα = 0

χ = δK2R



Thin-film solution

◮ The thin-film solution is (dropping checks on variables)

p(y , z) = cosα(H + h − z)

u(y , z) =sinα

2(z − H)(H + 2h − z)

v(y , z) =−χ sin2 α

120(z − H){(z − H)3[(H + 2h − z)

× (H + 4h − z) + 2h2]− 16h5}

−cosα

2(z − H)(H + 2h − z)

∂

∂y(H + h)



◮ To complete the solution we need to find the free surfaceshape, which is given by solving

d

dy(H + h) =

6η

35h4

where η = χ sin2 αcosα

.



◮ How “good” is the thin-film model? How well does itcompare to the more complicated model?

◮ To answer this, we must try to solve the more complicatedequations using a numerical method (aided by ComsolMultiphysics).

◮ We compare the thin-film and numerical solutions to assessthe accuracy of the thin-film model.

◮ Compare solutions in a channel of rectangular cross section aswe can find analytical solutions for this case.



−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

y

z

Navier–Stokes model (numerical solution)Thin-film approximation ( δ = 0.08 )



−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

y

z




−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

y

z




Flux down the channel

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0

0.1

0.2

0.3

0.4

0.5

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0

0.1

0.2

0.3

0.4

0.5

Thin-film (left) and Navier–Stokes (right).



Existence

◮ We are interested in the existence of the thin-film solution asδ increases.

◮ The thin-film solution depends upon determining the shape ofthe free-surface, h, which is given by solving

dh

dy=

6η

35h4, η =

χ sin2 α

cosα, χ =

δK

2R.

◮ Whether we can solve this or not depends upon the quantity η.



◮ Integrating the differential equation for the free surface shapeyields

h(y) =

(

h−3ℓ −

18η

35(1 + y)

)

−13

,

where hℓ is a constant.

◮ Solving for this constant reduces to finding the roots of apolynomial.



0.5 1 1.5 2 2.5 3 3.5

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

η

Roots

We find no root, and hence no solutions, for η > 3.28129.



Conclusions

◮ We have developed a mathematical model of a fluid flow in ahelical channel and derived a simplified model for shallowflows.

◮ For a rectangular channel cross-section, the thin-film solutionagrees with the numerical solution, especially along the freesurface.

◮ In this geometry, the ability to find a solution depends on theparameter η.

◮ It is possible to extend this analysis to different channelgeometries.


Derivation and solution of the equations for fluid flow … · Derivation and solution of the equations for ﬂuid ... the continuity equation ... Derivation and solution of the equations

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