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Problem 2.73
[Difficulty: 4]
ds
AA
U=
ω
r
dz
z
r
a
Section
AA
Given:
Conical bearing geometry
Find:
Expression for shear stress; Viscous torque on shaft
Solution:
Basic equation
τμ
du
dy
⋅
=
dT
r
τ
⋅
dA
⋅
=
Infinitesimal shear torque
Assumptions: Newtonian fluid, linear velocity profile (in narrow clearance gap), no slip condition
tan
θ
()
r
z
=
so
r
z tan
θ
⋅
=
Then
du
dy
⋅
=
μ
∆
u
∆
y
⋅
=
μ
ω
r
⋅
0
−
a0
−
⋅
=
μω
⋅
z
⋅
tan
θ
⋅
a
=
As we move up the device, shear stress increases linearly (because rate of shear strain does)
But from the sketch
dz
ds cos
θ
⋅
=
dA
2
π
⋅
r
⋅
ds
⋅
=
2
π
⋅
r
⋅
dz
cos
θ
⋅
=
The viscous torque on the element of area is
dT
r
τ
⋅
dA
⋅
=
r
⋅
z
⋅
tan
θ
⋅
a
⋅
2
⋅
π
⋅
r
⋅
dz
cos
θ
⋅

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