THE BOUNDARY LAYER OPPOSITE A ROTATING STIRRER
P.W. HEMKER
Abstract. How can we find the structure of the boundary layer that exists near an
(infinite) straight wall, opposite a rotating stirrer?
For a nonviscous, potential flow the flow pattern is easily calculated. For a mildly
viscous (high Re-number, but still laminar) flow a boundary layer appears near the
wall.
The flow is determined by the incompressible NS equations, by the non-slip boundary
conditions at the wall and by the potential flow solution away from the wall.
The wall being characterized by y=0, for the potential flow the stirrer is modelled by
a simple vortex at (x,y)=(0,a) and the straight wall by a mirrored vortex of opposite
strength at (x,y)=(0,-a).
I want to know the structure of the boundary layer. At the moment I am NOT
interested in a numercal (computational) solution. Later, the knowledge about the
BL might be used for the construction of a proper a-priori mesh.
More details are found at http://pieth.home.xs4all.nl/rotstir140417.pdf (this note).
1. Introduction
We study in the stationary flow over a flat silicon slide placed next to a rotating stirrer
in a cuvette, as shown in Figure 1. The purpose of this study is to support de modelling
of biochemical reactions at the silicone surface. The thickness of the layer of molecules
attached to the silicon depends on these reactions, and on the convection and diffusion
of the molecules in the solution. This motivates the study of the flow pattern in the
neighbourhood of the silicon slide.
Figure 1. The cuvette in which the ROTSTIR flow is modelled.
Date: April 17, 2014.
1
2
P.W. HEMKER
We consider the flow as two-dimensional, with a geometry as shown in Figure 2. In this
two-dimensional model we neglect the walls of the cuvette and we consider the silicon
slide as an infinite straight boundary, Thus, the flow occurs in the half-infinite plane in
which the stirrer is placed.
As a first approximation we neglect the no-slip boundary condition at the wall and
consider the flow as potential. Later we use the incompressible Navier-Stokes equations
to describe the situation near the wall. For these equations the no slip boundary
condition at the wall is taken into account and the remaining boundary conditions
require matching with the potential flow.
Figure 2. The geometry of the ROTSTIR problem
2. The potential flow
The flow field in the horizontal x-y-plane, induced by the rotating stirrer, is approximated by a single vortex with strength Γ, which is defined as the integral of the flow
velocity over a full circle. In a potential flow Γ is constant and so its value should be
equal to is value at the circumference of the stirrer: the velocity ωR times the distance
2πR, so that Γ = 2πωR2 , where R denotes the radius of the stirrer and ω its angular
velocity in rad.sec−1 .
The flow, induced by the stirrer placed at a distance d from the flat plate, is modelled
by two vortices, the second one obtained by mirroring the first one in the flat plate. We
choose the coordinate system such that the flat plate is situated at y = 0 and the centres
of the vortices are located at (x, y) = (0, a) and (x, y) = (0, −a). The flow is described
by its velocities u(x, y) and v(x, y), respectively in the x- and the y-direction.
For a single √
vortex at he origin, the flow can be described by the stream function
Γ
Ψ = − 2π log x2 + y 2 , for which u = ∂Ψ/∂y and v = −∂Ψ/∂x.
By substituting y → y + a and y → y − a, and adding the two vortices (with opposite signs because of the opposite directions) we obtain as velocities of the potential
flow
ωR2 (y − a)
ωR2 (a + y)
(2.1)
−
,
u0 (x, y) =
(a + y)2 + x2 (y − a)2 + x2
ωR2 x
ωR2 x
v0 (x, y) =
(2.2)
−
.
(y − a)2 + x2 (a + y)2 + x2
THE BOUNDARY LAYER OPPOSITE A ROTATING STIRRER
3
At the wall y = 0 we have v0 (x, 0) = 0 en u0 (x, 0) = 2ωR2 a/(a2 + x2 ).
Following the classical theory by Von Karman, for a non-slip boundary condition
we assume that, for a low viscosity, a viscous boundary layer appears near the flat
plate.
3. Incompressible Navier-Stokes
The incompressible Navier-Stokes equations. The incompressible Navier-Stokes
equations in two dimensions, without body forces, are
∂u ∂v
+
= 0,
∂x ∂y
(
)
( 2
)
∂u
∂p
∂ u ∂2u
∂u
∂u
(3.1)
= −
+ 2 ,
ρ
+u
+v
+µ
∂t
∂x
∂y
∂x
∂x2
∂y
(
)
( 2
)
∂v
∂v
∂v
∂p
∂ v
∂2v
ρ
+u
+v
= −
+µ
+
,
∂t
∂x
∂y
∂y
∂x2 ∂y 2
where p(x, y) denotes pressure, and ρ and µ are respectively mass density and viscosity.
We are interested in the stationary state, so that with ν = µ/ρ, the kinematic viscosity,
we have
(3.2)
u x + vy = 0 ,
uux + vuy = −px /ρ + ν(uxx + uyy ) ,
uvx + vvy = −py /ρ + ν(vxx + vyy ) .
We want to solve these equations in a thin layer along the wall: (x, y) ∈ [−∞, +∞] ×
[−L, 0]. So, we want to find the solution of (3.2) with boundary conditions u(x, 0) =
v(x, 0) = 0 at the solid wall at y = 0, and the potential flow solution u(x, y) = u0 (x, y)
and v(x, y) = v0 (x, y) at y = −L and at x = ±∞. For the pressure in the thin area we
use the pressure from the potential flow. We know that p0 + 21 ρu20 + 21 ρv02 is constant,
and hence
(
)
8a2 R4 xω 2 x2 + y 2 + a2
p0x /ρ = (
)2 ,
a4 + 2a2 (x − y)(x + y) + (x2 + y 2 )2
(
)
8a2 R4 yω 2 x2 + y 2 − a2
p0y /ρ = (
)2 .
a4 + 2a2 (x − y)(x + y) + (x2 + y 2 )2
4
The outer solution
in BL coordinates
P.W. HEMKER
4. The boundary layer equations
In order to handle the thin infinite strip along the flat plate we introduce a coordinate
transformation. By taking ξ = arctan(x/a) and η = −y/(εa) for some ε > 0, the strip
Ω is characterized by −π/2 < ξ < π/2 and 0 < η < L/(aε). In these coordinates the
potential solution reads
(4.1)
ū0 (ξ, η) =
(
)
2R2 ω sec2 (ξ) − ε2 η 2
aU ((εη − 1)2 + tan2 (ξ)) ((εη + 1)2 + tan2 (ξ))
(4.2)
v̄0 (ξ, η) =
−4εηR2 ω tan(ξ)
aU ((εη − 1)2 + tan2 (ξ)) ((εη + 1)2 + tan2 (ξ))
(4.3)
p̄0 (ξ, η) =
Expansion in ε of
the outer solution
in BL coordinates
The NS equations
in BL coordinates
a2 U 2 ((εη
−
1)2
−2R4 ω 2
+ tan2 (ξ)) ((εη + 1)2 + tan2 (ξ))
where the scaled functions ū0 , v̄0 and p̄0 , are defined by ū0 (ξ, η) = u0 (x, y)/U ,
v̄0 (ξ, η) = v0 (x, y)/U , and
p̄0 (ξ, η) = (u0 (x, y)2 + v0 (x, y)/(2U 2 ρ), with x = a tan(ξ) and y = −aεη. In order to
simplify the formulas we scale with U = 1/a. For small values of εη the equations
(4.1)-(4.3) are expanded as
(4.4)
(
)
ū0 (ξ, η) = 2R2 ω cos2 (ξ) + 2R2 ω cos4 (ξ)(2 cos(2ξ) − 1)η 2 ε2 + O (ηε)4
(4.5)
(
)
v̄0 (ξ, η) = −4R2 ω sin(ξ) cos3 (ξ)ηε + O (ηε)3
(4.6)
(
)
p̄0 (ξ, η) = −2R4 ω 2 cos4 (ξ) − 4R4 ω 2 cos6 (ξ) cos(2ξ)η 2 ε2 + O (ηε)4
By introducing ū(ξ, η) = u(a tan(ξ), −aεη), v̄(ξ, η) = v(a tan(ξ), −aεη) and p̄(ξ, η) =
Figure 3. Pressure and potential flow field
THE BOUNDARY LAYER OPPOSITE A ROTATING STIRRER
5
p(a tan(ξ), −aεη), the Navier-Stokes equations (3.2) become
(4.7)
ε cos2 (ξ)ūξ (ξ, η) − v̄η (ξ, η) = 0
(4.8)
ε cos2 (ξ)p̄ξ (ξ, η) + ε cos2 (ξ)ū(ξ, η)ūξ (ξ, η) − ūη (ξ, η)v̄(ξ, η)
)
ν( 2
−ε cos4 (ξ)ūξξ (ξ, η) + ε2 sin(2ξ) cos2 (ξ)ūξ (ξ, η) − ūηη (ξ, η) = 0
ε
(4.9)
−p̄η (ξ, η) + ε cos2 (ξ)ū(ξ, η)v̄ξ (ξ, η) − v̄(ξ, η)v̄η (ξ, η) +
)
ν( 2
−ε cos4 (ξ)v̄ξξ (ξ, η) + ε2 sin(2ξ) cos2 (ξ)v̄ξ (ξ, η) − v̄ηη (ξ, η) = 0
ε
Now, by setting ε = ν for small values of the viscosity, ε becomes a small parameter
and the equations loose their singular perturbation character.
As usual we assume that the boundary layer has little effect on the pressure and so we
substitute p̄ = p̄0 . The resulting Navier-Stokes equations are expanded in powers of ε
to obtain
v̄η (ξ, η) − ε cos2 (ξ)ūξ (ξ, η) = 0 ,
(4.10)
(4.11)
ε
(4.12)
(
ūη (ξ, η)v̄(ξ, η) + ūηη (ξ, η) −
cos2 (ξ)ū(ξ, η)ū
ξ (ξ, η)
)
( )
+ 8R4 ω 2 sin(ξ) cos5 (ξ) + O ε2 = 0 ,
( )
v̄(ξ, η)v̄η (ξ, η) + v̄ηη (ξ, η) − ε cos2 (ξ)ū(ξ, η)v̄ξ (ξ, η) + O ε2 = 0 .
For the boundary layer domain [−π/2, +π/2] × [0, L/(εa)], where L = O (εa) is small
enough, the boundary conditions are:
(4.13)
(4.14)
(4.15)
(4.16)
ū(ξ, 0) = v̄(ξ, 0) = 0 ,
ū(−π/2, η) = v̄(−π/2, η) = ū(+π/2, η) = v̄(+π/2, η) = 0 ,
ū(ξ, L/(εa)) = ū0 (ξ, L/(εa)) ,
v̄(ξ, L/(εa)) = v̄0 (ξ, L/(εa)) .
Expansion in ε of
the NS equations
in BL coordinates