3Dynamics

IB Fluid Dynamics


3.4 A case study: stagnation point flow (u = 0)
Example
(Stagnation point flow)
.
We attempt to find a solution to the Navier-
Stokes equation in the half-plane
y ≥
0, subject to the boundary condition
u → (Ex, −Ey, 0) as y → ∞, where E > 0 is a constant, and u = 0 on y = 0.
Intuitively, we would expect the flow to look like this:
x
y
The boundary conditions as
y → ∞
gives us a picture of what is going on at
large
y
, as shown in the diagram, but near the boundary, the velocity has to
start to vanish. So we would want to solve the equations to see what happens
near the boundary.
Again, this problem does not have any extrinsic length scale. So we can seek
a similarity solution in terms of
η =
y
δ
, δ =
r
ν
E
.
First of all, we make sure this is the right similarity solution. We show that
δ
has dimensions of length:
– The dimension of E is [E] = T
−1
– The dimension of ν is [ν] = L
2
T
−1
.
Hence
E
ν
= L
2
,
and δ has dimension L. Therefore η =
y
δ
is dimensionless.
Note that we make
η
a function of
y
instead of a function of
x
so that we
can impose the boundary condition as y → ∞.
What would the similarity solution look like? Applying the incompressibility
conditions ∇ · u = 0, it must be of the form
u = (u, v, 0) = (Exg
0
(ν), −Eδg(η), 0).
To check this is really incompressible, we can find the streamfunction. It turns
out it is given by
ψ =
√
νExg(η).
We can compute
u =
∂ψ
∂t
=
√
νExg
0
(η)
1
δ
= Exg
0
(η),
and similarly for the
y
component. So this is really the streamfunction. Therefore
we must have
∇ · u
= 0. Alternatively, we can simply compute
∇ · u
and find it
to be zero.
Finally, we look at the Navier-Stokes equations. The
x
and
y
components are
u
∂u
∂x
+ v
∂u
∂y
= −
1
ρ
∂p
∂x
+ ν
∂
2
u
∂x
2
+
∂
2
u
∂y
2
u
∂v
∂x
+ v
∂v
∂y
= −
1
ρ
∂p
∂y
+ ν
∂
2
v
∂x
2
+
∂
2
v
∂y
2
.
Substituting our expression for u and v, we get
Exg
0
Eg
0
− EδgExg
00
1
δ
= −
1
ρ
∂p
∂x
− νExg
000
1
δ
2
.
Some tidying up gives
E
2
x(g
02
− gg
00
) = −
1
ρ
∂p
∂x
− E
2
xg
000
.
We can do the same thing for the y-component, and get
E
√
νEgg
0
= −
1
ρ
∂p
∂y
− E
√
νEg
00
.
So we’ve got two equations for
g
and
p
. The trick is to take the
y
derivative of
the first, and
x
derivative of the second, and we can use that to eliminate the
p
terms. Then we have
g
0
g
00
− gg
000
= g
(4)
.
So we have a single equation for g that we shall solve.
We now look at our boundary conditions. The no-slip condition gives
u
=
0
on y = 0. So g
0
(0) = g(0) = 0 when η = 0.
As as
y → ∞
, the boundary conditions for
u
gives
g
0
(
η
)
→
1,
g
(
η
)
→ η
as
η → ∞.
All dimensional variables are absorbed into the scaled variables
g
and
η
. So
we only have to solve the ODE once. The far field velocity
u
= (
Ex, −Ey,
0) is
reached to a very good approximation when
η & 1, y & δ =
η
E
.
η
g
0
(η)
δ = O(1)
We get the horizontal velocity profile as follows:
x
y
Ex
δ
At the scale δ, we get a Reynolds number of
Re
δ
=
Uδ
ν
∼ O(1).
This is the boundary layer. For a larger extrinsic scale L δ, we get
Re
L
=
UL
2
1.
When interested in flow on scales much larger than
δ
, we ignore the region
y < δ
(since it is small), and we imagine a rigid boundary at
y
=
δ
at which the no-slip
condition does not apply.
When Re
L
1, we solve the Euler equations, namely
ρ
Du
Dt
= −∇p + f
∇ · u = 0.
We still have a boundary condition — we don’t allow fluid to flow through the
boundary. So we require
u · n = 0 at a stationary rigid boundary.
The no-slip condition is no longer satisfied.
It is an exercise for the reader to show that
u
= (
Ex, −Ey,
0) satisfies the
Euler equations in y > 0 with a rigid boundary of y = 0, with
p = p
0
−
1
2
ρE
2
(x
2
+ y
2
).
We can plot the curves of constant pressure, as well as the streamlines:
x
y
p
0
As a flow enters from the top, the pressure keeps increases, and this slows down
the flow. We say the
y
-pressure gradient is adverse. As it moves down and flows
sideways, the pressure pushes the flow. So the x-pressure gradient is favorable.
At the origin, the velocity is zero, and this is a stagnation point. This is also
the point of highest pressure. In general, velocity is high at low pressures and
low at high pressures.