Setting the pressure derivative and the stress

Setting the pressure derivative and the stress derivatives in the x- and z-directions to zero, one finds
d τ
yz
dy
= ρ g
or
τ
yz
= ρ g(y − H) ,
where it has been used that
τ
yz
= 0 at y = H .
The fluid begins to flow when the magnitude of the stress (regardless of the sign) exceeds the yield
stress, which will happen first at the wall where the shear stress is greatest. The paint begins to flow
when
τ
yx
> τ
0
or when H >
τ
0
ρ g
.
The same conclusion can be reached by performing a force balance on a section of fluid as shown in the
diagram below. The shear stress acting on an area LW must balance the weight ρ gLHW,
τ
yz
LW + ρ gLWH = 0 .
The paint begins to flow when the shear stress becomes equal to the yield stress, or when

H =
τ
0
ρ g
.
4. Whitaker 5-15.
Use coordinate axes such that x points down, y points from the solid/water surface toward the solid/air
surface, so that y=0 is the water/solid surface and y=h
I
is the air/liquid interface. Neglect the density of
the air, and require that the velocity v
x
and shear stress τ
yx
be continuous at the air/water interface.
a) Derive the velocity profiles in both the air and the water.
b) Identify a criterion that must be satisfied for the velocity profile in the water to yield a shear stress
of zero at the air/water interface.
Solution
y
z
ρ g
τ
yz
L
H

Also, for the y- and z-directions,
∂ p
∂ y
= 0 and
∂ p
∂ z
= 0 .
The equations from the y- and z-directions imply the pressure only depends on x. But pressure does
not depend on x at the air/water interface, and pressure is independent of y and z, so the pressure is
constant everywhere in this problem. The flow is driven purely by gravity. We have
d
2
v
x
I
dy
2
= −
ρ
w
g
μ
w
so integrating twice gives
v
x
I
y
( )
= −
ρ
w
g
μ
w
y
2
2
+ C
1
y + C
2
.
The same analysis applies in the air, but the density in the air is negligible (this assumption is
consistent with the problem statement, in which pressure variations in the air are neglected). Flow in
the air is driven by friction (i.e., the no-slip surface) between the water and air. In the air we have
v
x
II
y
( )
= C
3
y + C
4
.
The no-slip condition at y=0 yields C
2
=0, so in the water
v
x
I
y
( )
= −
ρ
w
g
μ
w
y
2
2
+ C
1
y 0 ≤ y ≤ h
I
.
At the air/wall interface the no-slip condition is
v
x
II
= 0 at y=h
I
+ h
II
so
0 = C
3
h
I
+ h
II
( )
+ C
4
or
v
x
II
y
( )
= C
4
1 −
y
h
I
+ h
II
⎛
⎝
⎜
⎞
⎠
⎟
for h
I
≤ y ≤ h
I
+ h
II
.
Two constants remain and for those we use the conditions at the air/water interface. We require that the
shear stress be continuous, so

μ
w
dv
x
I
dy
= μ
A
dv
x
II
dy
at y=h
I
and the velocity is continuous so
v
x
I
= v
x
II
at y=h
I
.
These two conditions yield the equations
−
ρ
w
g
μ
w
h
I
+ C
1
⎛
⎝
⎜
⎞
⎠
⎟
= −
μ
A
μ
w
C
4
h
I
+ h
II
and
−
ρ
w
g
μ
w
h
I
2
+ C
1
=
C
4
h
I
1 −
h
I
h
I
+ h
II
⎛
⎝
⎜
⎞
⎠
⎟
.