Problem 2.41
[Difficulty: 2]
Given:
Velocity distribution between parallel plates
Find:
Force on lower plate
Solution:
Basic equations
F
τ
yx
A
⋅
=
τ
yx
μ
du
dy
⋅
=
du
dy
d
dy
u
max
1
2 y
⋅
h
⎛
⎜
⎝
⎞
⎠
2
−
⎡
⎢
⎣
⎤
⎥
⎦
⋅
=
u
max
4
h
2
−
⎛
⎜
⎝
⎞
⎠
⋅
2
⋅
y
⋅
=
8 u
max
⋅
y
⋅
h
2
−
=
so
τ
yx
8
μ
⋅
u
max
⋅
y
⋅
h
2
−
=
and
F
8 A
⋅
μ
⋅
u
max
⋅
y
⋅
h
2
−
=
At the lower surface
y
h
2
−
=
and
h
0.1 mm
⋅
=
A
1 m
2
⋅
=
u
max
0.05
m
s
⋅
=
μ
1.14
10
3
−
×
N s
⋅
m
2
⋅
=
(Table
A.8)
Hence
F
8
−
1
×
m
2
⋅
1.14
×
10
3
−
×
N s
⋅
m
2
⋅
0.05
×
m
s
⋅
0.1
−
2
×
mm
⋅
1 m
⋅
1000 mm
⋅
×
1
0.1
1
mm
⋅
1000 mm
⋅
1 m
⋅
×
⎛
⎜
⎝
⎞
⎠
2
×
=
F
2.28 N
⋅
=
(to the right)

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Problem 2.57
[Difficulty: 2]
Given:
Flow of three fluids between two plates
Find:
Upper plate velocity; Interface velocities; plot velocity distribution
Solution:
The shear stress is the same throughout (the velocity gradients are linear, and the stresses in the fluids at the interfaces must be
equal and opposite).
Given data
F
100 N
⋅
=
h
1
0.5 mm
⋅
=
h
2
0.25 mm
⋅
=
h
3
0.2 mm
⋅
=
A
1 m
2
⋅
=
μ
1
0.15
N s
⋅
m
2
⋅
=
μ
2
0.5
N s
⋅
m
2
⋅
=
μ
3
0.2
N s
⋅
m
2
⋅
=
The (constant) stress is
τ
F
A
=
τ
100 Pa
=
For each fluid
τ
μ
∆
V
∆
y
⋅
=
or
∆
V
τ ∆
y
⋅
μ
=
where
∆
V is the overall change in velocity over distance
∆
y
Hence
V
12
τ
h
1
⋅
μ
1
=
V
12
0.333
m
s
=
where V
12
is the velocity at the 1 - 2 interface
Hence
V
23
τ
h
2
⋅
μ
2
V
12
+
=
V
23
0.383
m
s
=
where V
23
is the velocity at the 2 - 3 interface
Hence
V
τ
h
3
⋅
μ
3
V
23
+
=
V
0.483
m
s
=
where V is the velocity at the upper plate
0
0.1
0.2
0.3
0.4
0.5
0.25
0.5
0.75
1
Velocity (m/s)
Position (mm)