The volume flux per unit width (in the
z
-direction) of fluid along the channel is thus
Q
=
integraldisplay
d
0
v
x
d
y
=
G
(
x
)
d
3
(
x
)
12
μ
+
1
2
U d
(
x
)
.
(9.68)
Of course, in a steady state, this flux must be independent of
x
. Hence,
dp
dx
=
−
G
(
x
)
=
6
μ
bracketleftBigg
U
d
2
(
x
)
−
2
Q
d
3
(
x
)
bracketrightBigg
,
(9.69)
where
d
(
x
)
=
d
1
−
α
x
. Integration of the above equation yields
p
(
x
)
−
p
0
=
6
μ
α
U
parenleftBigg
1
d
−
1
d
1
parenrightBigg
−
Q
1
d
2
−
1
d
2
1
,
(9.70)
where
p
0
=
p
(0). Now, assuming that the sliding block is completely immersed in fluid of uniform ambient pressure
p
0
, we would expect the pressures at the two ends of the lubricating layer to both equal
p
0
, which implies that
p
(
l
)
=
p
0
.
It follows from the above equation that
Q
=
U
parenleftBigg
d
1
d
2
d
1
+
d
2
parenrightBigg
,
(9.71)

Incompressible Viscous Flow
179
and
p
(
x
)
−
p
0
=
6
μ
U
α
[
d
1
−
d
(
x
)] [
d
(
x
)
−
d
2
]
d
2
(
x
) (
d
1
+
d
2
)
.
(9.72)
Note that if
d
1
>
d
2
then the pressure increment
p
(
x
)
−
p
0
is positive throughout the layer, and vice versa. In other
words, a lubricating layer sandwiched between two solid bodies in relative motion only generates a positive pressure,
that is capable of supporting a normal load, when the motion is such as to drag (by means of viscous stresses) fluid
from the wider to the narrower end of the layer. The pressure increment has a single maximum in the layer, and its
value at this point is of order
μ
l U
/
d
2
, assuming that (
d
1
−
d
2
)
/
d
1
is of order unity. This suggests that very large
pressures can be set up inside a thin lubricating layer.
The net normal force (per unit width in the
z
-direction) acting on the lower plane is
f
y
=
−
integraldisplay
l
0
[
p
(
x
)
−
p
1
]
dx
=
−
6
μ
U
α
2
bracketleftBigg
ln
parenleftBigg
d
1
d
2
parenrightBigg
−
2
parenleftBigg
d
1
−
d
2
d
1
+
d
2
parenrightBiggbracketrightBigg
.
(9.73)
Moreover, the net tangential force (per unit width) acting on the lower plane is
f
x
=
integraldisplay
l
0
μ
parenleftBigg
∂v
x
∂y
parenrightBigg
y
=
0
dx
=
−
2
μ
U
α
bracketleftBigg
2 ln
parenleftBigg
d
1
d
2
parenrightBigg
−
3
parenleftBigg
d
1
−
d
2
d
1
+
d
2
parenrightBiggbracketrightBigg
.
(9.74)
Of course, equal and opposite forces,
f
′
x
=
−
f
x
=
μ
U l
2
d
2
0
3
2
k
2
bracketleftBigg
ln
parenleftBigg
1
+
k
1
−
k
parenrightBigg
−
2
k
bracketrightBigg
,
(9.75)
f
′
y
=
−
f
y
=
μ
U l
d
0
1
k
bracketleftBigg
2 ln
parenleftBigg
1
+
k
1
−
k
parenrightBigg
−
3
k
bracketrightBigg
,
(9.76)
act on the upper plane. Here,
d
0
=
(
d
1
+
d
2
)
/
2 is the mean channel width, and
k
=
(
d
1
−
d
2
)
/
(
d
1
+
d
2
). Note that if
0
<
d
2
<
d
1
then 0
<
k
<
1. The e
ff
ective coe
ffi
cient of friction,
C
f
, between the two sliding bodies is conventionally
defined as the ratio of the tangential to the normal force that they exert on one another. Hence,
C
f
=
f
x
f
y
=
f
′
x
f
′
y
=
4
3
d
0
l
H
(
k
)
,
(9.77)
where
H
(
k
)
=
k
bracketleftBigg
ln
parenleftBigg
1
+
k
1
−
k
parenrightBigg
−
3
k
2
bracketrightBiggslashBiggbracketleftBigg
ln
parenleftBigg
1
+
k
1
−
k
parenrightBigg
−
2
k
bracketrightBigg
.
(9.78)
The function
H
(
k

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- Fluid Dynamics, Fluid Mechanics, stress tensor, Fluid Motion