1Parallel viscous flow
IB Fluid Dynamics
1.2 Steady parallel viscous flow
We are first going to consider a very simple type of flow, known as steady parallel
viscous flow, and derive the equations of motion of it. We first explain what this
name means. The word “viscous” simply means we do not assume the viscosity
vanishes, and is not something new.
Definition
(Steady flow)
.
A steady flow is a flow that does not change in time.
In other words, all forces balance, and there is no acceleration.
Definition
(Parallel flow)
.
A parallel flow is a flow where the fluid only flows
in one dimension (say the
x
direction), and only depends on the direction
perpendicular to a plane (say the
x −z
plane). So the velocity can be written as
u = (u(y), 0, 0).
These can be conveniently thought of as being two-dimensional, by forgetting
the z direction.
Note that our velocity does not depend on the
x
direction. This can be
justified by the assumption that the fluid is incompressible. If we had a velocity
gradient in the
x
-direction, then we will have fluid “piling up” at certain places,
violating incompressibility.
We will give a formal definition of incompressibility later. In general, though,
fluids are not compressible. For example, sound waves are exactly waves of com-
pression in air, and cannot exist if air is incompressible. So we can alternatively
state the assumption of incompressibility as “sound travels at infinite speed”.
Hence, compressibility matters mostly when we are travelling near the speed
of sound. If we are moving in low speeds, we can just pretend the fluid is indeed
incompressible. This is what we will do in this course, since the speed of sound
is in general very high.
For a general steady parallel viscous flow, We can draw a flow profile like
this:
y
The lengths of the arrow signify the magnitude of the velocity at that point.
To derive the equations of motion, we can consider a small box in the fluid.
y
x, y
x + δx
y + δy
We know that this block of fluid moves in the
x
direction without acceleration.
So the total forces of the surrounding environment on the box should vanish.
We first consider the
x
direction. There are normal stresses at the sides, and
tangential stresses at the top and bottom. The sum of forces in the
x
-direction
(per unit transverse width) gives
p(x)δy −p(x + δx)δy + τ
s
(y + δy)δx + τ
s
(y)δx = 0.
By the definition of τ
s
, we can write
τ
s
(y + δy) = µ
∂u
∂y
(y + δy), τ
s
(y) = −µ
∂u
∂y
(y),
where the different signs come from the different normals (for a normal pointing
downwards,
∂
∂n
=
∂
∂(−y)
).
Dividing by δxδy, we get
1
δx
(p(x) − p(x + δx)) + µ
1
δy
∂u
∂y
(y + δy) −
∂u
∂y
(y)
.
Taking the limit as δx, δy → 0, we end up with the equation of motion
−
∂p
∂x
+ µ
∂
2
u
∂y
2
= 0.
Performing similar calculations in the y direction, we obtain
−
∂p
∂y
= 0.
In the second equation, we keep the negative sign for consistency, but obviously
in this case it is not necessary.
This is the simplest possible case. We can extend this a bit by allowing,
non-steady flows and external forces on the fluid. Then the velocity is of the
form
u = (u(y, t), 0, 0).
Writing the external body force (per unit volume) as (
f
x
, f
y
,
0), we obtain the
equations
ρ
∂u
∂t
= −
∂p
∂x
+ µ
∂
2
u
∂y
2
+ f
x
0 = −
∂p
∂y
+ f
y
.
The derivation of these equations is straightforward, and is left as an exercise
for the reader on the example sheet. Here
ρ
is the density, i.e. the mass per unit
volume. For air, it is approximately
1 kg m
−3
, and for water, it is approximately
1000 kg m
−3
.
Along with equations, we also need boundary condition. In general, Newto-
nian fluids satisfy one of the following two boundary conditions:
(i)
No-slip condition: at the boundary, the tangential component of the fluid
velocity equals the tangential velocity of boundary. In particular, if the
boundary is stable, the tangential component of the fluid velocity is zero.
This means fluids stick to surfaces.
u
T
= 0. (1.5)
(ii)
Stress condition: alternatively, a tangential stress
τ
is imposed on the fluid.
In this case,
−µ
∂u
T
∂n
= τ.
The no-slip condition is common when we have a fluid-solid boundary, where
we think the fluid “sticks” to the solid boundary. The stress condition is more
common when we have a fluid-fluid boundary (usually liquid and gas), where we
require the tangential stresses to match up. In non-parallel flow, we will often
require the flow not to penetrate the solid boundary, but this is automatically
satisfied in parallel flow.
We are going to look at some examples.
Example
(Couette flow)
.
This is flow driven by the motion of a boundary, as
we have previously seen.
U
We assume that this is a steady flow, and there is no pressure gradient. So our
equations give
∂
2
u
∂y
2
= 0.
Moreover, the no-slip condition says
u
= 0 on
y
= 0;
u
=
U
on
y
=
h
. The
solution is thus
u =
Uy
h
.
Example
(Poiseuille flow)
.
This is a flow driven by a pressure gradient between
stationary boundaries. Again we have two boundaries
P
1
P
0
We have a high pressure
P
1
on the left, and a low pressure
P
0
< P
1
on the right.
We solve this problem again, but we will also include gravity. So the equations
of motion become
−
∂p
∂x
+ µ
∂
2
u
∂y
2
= 0
−
∂p
∂y
− gρ = 0
The boundary conditions are u = 0 at y = 0, h. The second equation implies
p = −gρy + f (x)
for some function f. Substituting into the first gives
µ
∂
2
u
∂y
2
= f
0
(x).
The left is a function of
y
only, while the right depends only on
x
. So both must
be constant, say G. Using the boundary conditions, we get
µ
∂
2
u
∂y
2
= f
0
(x) = G =
P
1
− P
0
L
,
where L is the length of the tube. Then we find
u =
G
2µ
y(h − y).
Here the velocity is the greatest at the middle, where y =
h
2
.
Since the equations of motion are linear, if we have both a moving boundary
and a pressure gradient, we can just add the two solutions up.