1Parallel viscous flow
IB Fluid Dynamics
1.4 More examples
Since we are probably bored by the Couette and Poiseuille flows, we do another
more interesting example.
Example
(Gravity-driven flow down a slope)
.
Suppose we have some fluid
flowing along a slope.
y
x
g
u
p
0
atmospheric
pressure
α
Here there is just atmosphere above the fluid, and we assume the fluid flow is
steady, i.e.
u
is merely a function of
y
. We further assume that the atmospheric
pressure does not vary over the vertical extent of the flow. This is a very good
approximation because ρ
air
ρ
liq
.
Similarly, we assume
µ
air
µ
liq
. So the air exerts no significant tangential
stress. This is known as a free surface.
We first solve the
y
momentum equation. The force in the
y
direction is
−gρ cos α. Hence the equation is
∂p
∂y
= −gp cos α.
Using the fact that p = p
0
at the top boundary, we get
p = p
0
− gρ cos α(y −h).
In particular, p is independent of x. In the x component, we get
µ
∂
2
u
∂y
2
= −gρ sin α.
The no slip condition gives
u
= 0 when
y
= 0. The other condition is that there
is no stress at y = h. So we get
∂u
∂y
= 0 when y = h.
The solution is thus
u =
gρ sin α
2µ
y(2h − y).
This is a bit like the Poiseuille flow, with
gp sin α
2µ
as the pressure gradient. But
instead of going to zero at
y
=
h
, we get to zero at
y
= 2
h
instead. So this is
half a Poiseuille flow.
It is an exercise for the reader to calculate the volume flux q.
For a change, we do a case where we have unsteady flow.
Example. Consider fluid initially at rest in y > 0, resting on a flat surface.
y
x
At time
t
= 0, the boundary
y
= 0 starts to move at constant speed
U
. There is
no force and no pressure gradient.
We use the x-momentum equation to get
∂u
∂t
= ν
∂
2
u
∂y
2
,
where
ν
=
µ
ρ
. This is clearly the diffusion equation, with the diffusivity
ν
. We
can view this as the diffusion coefficient for motion/momentum/vorticity.
Definition (Kinematic viscosity). The kinematic viscosity is
ν =
µ
ρ
.
The boundary conditions are
u
= 0 for
t
= 0 and
u →
0 as
y → ∞
for all
t
.
The other boundary condition is obviously u = U when y = 0, for t > 0.
You should have learnt how to solve this in, say, IB Methods. We will
approach this differently here.
Before we start, we try to do some dimensional analysis, and try to figure
out how far we have to go away from the boundary before we don’t feel any
significant motion.
We first list all the terms involved. We are already provided with a velocity
U
. We let
T
be our time scale. We would like to know how fast the movement
of fluid propagates up the
y
axis. We note that in this case, we don’t really
have an extrinsic length scale — in the case where we have two boundaries, the
distance between them is a natural length scale to compare with, but here the
fluid is infinitely thick. So we need to come up with a characteristic intrinsic
length scale somewhat arbitrarily. For example, at any time
T
, we can let
δ
be
the amount of fluid that has reached a speed of at least
U
10
(that was completely
arbitrary).
Instead of trying to figure out the dimensions of, say,
ν
and trying to match
them, we just replace terms in the differential equation with these quantities
of the right dimension, since we know our differential equation is dimensionally
correct. So we obtain
U
T
∼ ν
U
δ
2
.
Since the U cancels out, we get
δ ∼
√
νT .
We can figure out approximately how big this is, using the table below:
µ(kg m
−1
s
−1
) ρ(kg m
−3
) ν(m
2
s
−1
)
water 10
−3
10
3
10
−6
air 10
−5
1 10
−5
These are, in general, very tiny. So we expect to feel nothing when we move
slightly away from the boundary.
We now solve the problem properly. In an infinite domain with no extrinsic
length scale, the diffusion equation admits a similarity solution. We write
u(y, t) = Uf(η),
where
f
(
η
) is a dimensionless function of the dimensionless variable
η
. To turn
y into a dimensionless variable, we have to define
η =
y
δ
=
y
√
νt
.
We substitute this form of the solution into the differential equation. Then we
get
−
1
2
ηf
0
(η) = f
00
(η),
with boundary condition f = 1 on η = 0. Then the solution is by definition
f = erfc
η
2
.
Hence
u = U erfc
y
2
√
ηt
.
U
δ ∼
√
ηt
We can compute the tangential stress of the fluid in the above case to be
τ
s
= µ
∂u
∂y
= µ
U
√
νt
2
√
π
e
−y
2
y=0
=
µU
√
πνt
.
Using our previous calculations of the kinematic viscosities, we find that
µ
√
ν
is 1
for water and 10
−3
for air.
This is significant in, say, the motion of ocean current. When the wind blows,
it causes the water in the ocean to move along with it. This is done in a way
such that the surface tension between the two fluids match at the boundary.
Hence we see that even if the air blows really quickly, the resultant ocean current
is much smaller, say a hundredth to a thousandth of it.