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Part 30

The document describes problems related to torque required to turn bearings and shafts inside cylinders. It provides analysis and equations to calculate torque based on bearing area, viscosity, angular velocity, radii of cylinders. It then uses the developed equations to calculate torque and power required to rotate a shaft at a given velocity.

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Kerlos Saeed
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91 views4 pages

Part 30

The document describes problems related to torque required to turn bearings and shafts inside cylinders. It provides analysis and equations to calculate torque based on bearing area, viscosity, angular velocity, radii of cylinders. It then uses the developed equations to calculate torque and power required to rotate a shaft at a given velocity.

Uploaded by

Kerlos Saeed
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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PROBLEM 9.

14
Situation: A bearing is described in the problem statement.
Find: Torque required to turn bearing.

ANALYSIS

τ = µV /δ
T = τ Ar

where T = torque, A = bearing area = 2πrb

T = τ 2πrbr = τ 2πr2 b
= (µV /δ)(2πr2 b)

where V=rω. Then

= (µ/δ)(rω)(2πr2 b)
= (µ/δ)(2πω)r3 b
= (0.1/0.001)(2π)(200)(0.009)3 (0.1)
T = 9.16 × 10−4 N · m

741
PROBLEM 9.15
Situation: A shaft turning inside a stationary cylinder is described in the problem
statement.
Find: Show that the torque per unit length acting on the inner cylinder is given by
T = 4πµωrs2 /(1 − (rs2 /ro2 ).

ANALYSIS
Subscript s refers to inner cylinder. Subscript o refers to outer cylinder. The cylinder
is unit length into page.

∆r
r

τ s

το

Ts = τ (2πr)(r)
To = τ (2πr)(r) + d/dr(τ 2πr · r)∆r
Ts − To = 0
d/dr(τ 2πr2 )∆r = 0; d/dr(τ r2 ) = 0

Since there is no angular acceleration, the sum of the torques must be zero. Therefore

Ts − To = 0
d/dr(τ 2πr2 )∆r = 0
d/dr(τ r2 ) = 0

Then

τ r2 = C1
τ = µr(d/dr)(V /r)

So

µr3 (d/dr(V /r)) = C1


µ(d/dr(V /r)) = C1 r−3

Integrating,
µv/r = (−1/2)C1 r−2 + C2

742
At r = ro , v = 0 and at r = rs , v = rs ω so

C1 = 2C2 r02
µω = C2 (1 − r02 /rs2 )
C2 = µω/(1 − r02 /rs2 )

Then

τ s = C1 rs−2 = 2C2 (r0 /rs )2 = 2µωr02 /(rs2 − r02 ) = 2µω/((rs2 /r02 ) − 1)

So
Ts = τ 2πrs2 = 4πµωrs2 /((rs2 /r02 ) − 1)
which is the torque on the fluid. Torque on shaft per unit length

T = 4πµωrs2 /(1 − (rs2 /r02 )

743
PROBLEM 9.16
Situation: A shaft turning inside a stationary cylinder is described in the problem
statement.
Find: Power necessary to rotate shaft.

APPROACH
Apply the equation developed in Problem 9.15.

ANALYSIS

T = 4πµω rs2 /(1 − (rs2 /r02 ))


= 4π × 0.1 × (50)(0.01)2 0.03/(1 − (1/1.1)2 )
= 0.00109 N · m
P = Tω
= (0.00109 N · m) (50 s−1 )
P = 0.0543 W

744

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