M4A33: Flow in curved pipes: the Dean equations

Consider flow down a slowly curved pipe. In terms of cylindrical polar coordinates(r, , z) we shall model this as a portion of a torus, (r b)2 + z2 = a2 where b a, andseek solutions independent of , driven by a pressure gradient in the -direction.

The velocity u = (ur, u, uz) satisfies the Navier Stokes equations

1

r

r(rur) +

uz

z= 0

(Dur

Dtu2

r

)

= p

r+

(2ur

ur

r2

)

(Du

Dt+uur

r

)

= G(r, z, t) + (2u

u

r2

)

Duz

Dt=

p

z+ 2uz

(1)

where the material derivative D/Dt = /t+ ur/r + uz/z. Here G is the downpipepressure gradient, G = 1/r p/. We shall seek steady solutions to these equations.Let us first see if there is a unidirectional solution, as for the straight pipe. If we substituteur = 0 = uz, we find

p

z= 0 and

p

r=u2r

=u

z= 0 . (2)

So such a solution is only possible if u is constant on cylinders. Such a flow would beconsistent with a no-slip condition only for flows between concentric cylinders. Any curvedpipe-flow cannot be unidirectional.

However, if the pipe is almost straight, we might expect the flow to be almost unidi-rectional. Now r and z vary over the scale a and we assume

b a so that r = b+ ax ' b and

r

1

a

1

r(3)

We scale z = az and let U0 be a typical scale of u. Then we expect a suitable scale forthe pressure to be p U20 a/b and if we scale

urur

r uz

ur

zu2r

= ur uz U0(ab

) 12. (4)

We therefore write

u = U0u ur,z = U0

(ab

) 12ux,z p = U

20

(ab

)p (5)

1

and neglecting terms of order (a/b), equations (1) become

uxx

+uzz

= 0

U20b

(DuxDtu21

)

= U20b

p

x+U0

a2

(ab

)12 2ux

U20(ab)1/2

DuDt

= G+U0

a22u

U20b

(DuzDt

)

=U20b

p

z+U0

a2

(ab

) 12 2uz

(6)

We choose the scale U0 and define a parameter K such that

Ga2

U0= 1 and K =

U0a

(ab

)12. (7)

From now on we drop the from all the dimensionless variables to obtain the Deanequations.

ux

x+uz

z= 0

K

(Dux

Dt u2

)

= Kp

x+2ux

KDu

Dt= 1 +2u

KDuz

Dt= K

p

z+2uz

. (8)

These equations are essentially the two-dimensional Navier-Stokes equations with a bodyforce u2 acting towards the inside of the bend. If we write u = (u, v, w) in Cartesiancoordinates (x, y, z), and introduce a stream function, (x, z) where u ux = /zand w uz = /x, and v(x, z) u, then (8) reduce to

K(zvx xvz) = 1 +2v

K(zx xz) = 2 2Kvvz

}

(9)

where = 2 is the downpipe vorticity and suffices now denote partial derivatives.These equations are to be solved for v(x, z) and (x, z) subject to the no-slip conditions

= 0, v = 0 on the pipe boundary. (10)

There is one parameter in the problem, K, which is known as the Dean number and definedin (7). It is a Reynolds number modified by the pipe curvature, (a/b).

2