dean
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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).
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