Derive an expression for the velocity profile v x y

Derive an expression for the velocity profile v
x
(y) in the backflow region and find L
2
c) Determine the coating thickness h
∞
d) Determine the shear force per unit width (in the z-direction) that the liquid exerts on the sheet.
.
.

μ
d
2
v
x
dy 2
= −
Δ P
F
L
1
.
a) Integrating the governing equation twice yields
v
x
(y) = −
1
μ
Δ P
F
L
1
y 2
2
+ C
1
y + C
2
.
Since the velocity is U at y=0,
C
2
= U ,
and since the velocity is 0 at y=h,
0 = −
1
μ
Δ P
F
L
1
h
2
2
+ C
1
h + U
or
C
1
=
1
μ
Δ P
F
L
1
h
2
−
U
h
.
The velocity profile is therefore
v
x
(y) =
1
μ
Δ P
F
L
1
h 2
2
y
h
−
y 2
h 2
⎡
⎣
⎢
⎤
⎦
⎥
+ U 1 −
y
h
⎛
⎝
⎜
⎞
⎠
⎟
.
This expression gives the velocity profile in both regions (downstream and upstream of the vertical
slot). The only difference is the negative pressure gradient in the two regions. In the forward flow
region,
Δ P
F
L
1
=
P
1
− P
0
L
1
,
whereas in the backflow region
Δ P
B
L
2
=
P
0
− P
1
L
2
.
b) In the backflow region, the velocity profile is again
v
x
(y) =
1
μ
Δ P
B
L
2
h 2
2
y
h
−
y 2
h
2
⎡
⎣
⎢
⎤
⎦
⎥
+ U 1 −
y
h
⎛
⎝
⎜
⎞
⎠
⎟
.

To find the length of the backflow region, we use the fact that the volumetric flow across any cross-
section there must be zero. At steady state, if there is no flow at the left air/liquid boundary, then
there must be no flow at any x-position. We therefore have
Q = v
x
(y) = 0
0
h
∫
or
1
μ
Δ P
B
L
2
h 2
2
y
h
−
y 2
h
2
⎡
⎣
⎢
⎤
⎦
⎥
+ U 1 −
y
h
⎛
⎝
⎜
⎞
⎠
⎟
⎡
⎣
⎢
⎤
⎦
⎥
dy = 0
0
h
∫
.
Doing the integral yields
P
0
− P
1
L
2
= − μ U
6
h
2
or
L
2
=
h 2 P
1
− P
0
( )
6 μ U
.
c) The flow rate (per unit width) in the forward flow region is
Q =
1
μ
Δ P
F
L
1
h 2
2
y
h
−
y 2
h 2
⎡
⎣
⎢
⎤
⎦
⎥
+ U 1 −
y
h
⎛
⎝
⎜
⎞
⎠
⎟
⎡
⎣
⎢
⎤
⎦
⎥
dy
0
h
∫
with the negative pressure gradient as specified in (a). Then
Q =
1
μ
Δ P
F
L
1
h
3
12
+ U
h
2
⎡
⎣
⎢
⎤
⎦
⎥
= Uh
∞
,
where the last equality reflects the fact that, far downstream, the velocity of the film will equal U
everywhere. The final coating thickness is therefore
h
∞
=
h 3
12 μ U
Δ P
F
L
1
+
h
2
⎡
⎣
⎢
⎤
⎦
⎥
.

d) In both the forward flow and backflow regions, the force per unit width is given by
F
x
= τ
yx
y = 0
dx
− L
2
0
∫
+ τ
yx
y = 0
dx
0
L
1
∫
.
The shear stresses in each region have the same form, but the pressure gradient terms differ. They
take the form
τ
yx
y = 0
= μ
dv
x
dy
y = o
= −
Δ P
L
h
2
+ μ
U
h
⎛
⎝
⎜
⎞
⎠
⎟
.
The shear stresses are functions only of y, so they are constant at y=0, and the integrals are trivial
and yield
F
x
= − L
2
P
0
− P
1
L
2
h
2
+ μ
U
h
⎛
⎝
⎜
⎞
⎠
⎟
− L
1
P
1
− P
0
L
2
h
2
+ μ
U
h
⎛
⎝
⎜
⎞
⎠
⎟
or
F
x
= − μ
U
h
L
2
+ L
1
( )
.
The pressure gradient terms apparently cancel each other, so the net force is in the –x direction, and
is the same as it would be for a linear shear flow with no pressure gradient.