9-Practice Problems.pdf

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ssm_ch9.pdf
Chapter 9
Surface Resistance
Problem 9.1
An aluminum cube of density 2700 kg/m 3 slides with a constant speed of 20 cm/s
down a plate that is at an angle of 30 ! with respect to the horizontal. The plate is
covered with a stationary layer of 0.1-mm-thick oil of viscosity ! = 0"008 N · s/m 2 "
The cube has dimensions of #×#×#" Find #"
Solution
The weight of the cube is
$ = # 3 % "#
A free-body diagram is
Balancing forces in the & -direction gives
' shear = $ SIN(30 ! )
=
¡
# 3 % "#
¢
SIN(30 ! )
(1)
75
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76
CHAPTER 9. SURFACE RESISTANCE
The shear force is
' shear = (# 2
(2)
Assuming Couette ß ow, the shear stress is
( = ! )
*
(3)
where ) is the speed of the block and * is the thickness of the oil layer. Combining
Eqs. (2) and (3) gives
Μ
! )
*
' shear =
# 2
(4)
Combining Eqs. (1) and (4) gives
Μ
! )
*
¡
# 3 % "#
¢
# 2 =
SIN(30 ! )
or
# = ! )
*(% "# SIN30 ! )
= 0"008 0"2
0"0001(2700×SIN30)
= 11.9 mm
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77
Problem 9.2
A 1.000-in.-diameter shaft of length 2 inches rotates at an angular speed of + = 800
rpm within a stationary cylindrical housing. The gap between the stationary hous-
ing and the shaft has a dimension of 0.001 in. The gap is Þ lled with oil of viscosity
0"0003 lbf · s/ft 2 " Find the torque and power required to rotate the shaft. Assume
the oil motion in the gap can be described by planar Couette ß ow.
!
Solution
As the shaft rotates, a clockwise applied torque is required to balance the moment
caused by forces associated with ß uid friction (viscosity).
Figure 1 Sketch of the shaft
The force ,' in Fig. 1 is the viscous force on the top surface of the pie-shaped
region. Since shear stress is force per area
,' = (,-
= ( (.,/)#
(1)
The shear stress is
( = ! ,)
,0
which simpli Þ es, for Couette ß ow, to
( = ! )
*
where ) = .+ is the speed at the outer surface of the rotating shaft. Thus
( = ! .+
*
(2)
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78
CHAPTER 9. SURFACE RESISTANCE
Combining Eqs. (1) and (2) gives
,' = ! .+
* (.,/)#
The force ,' will cause a frictional torque of ,1 .
,1 = .,'
= ! . 3 #+
*
,/
The net frictional torque balances the applied torque.
Applied torque = frictional torque
2$
! . 3 #+
*
=
,/
0
= 22! . 3 #+
*
Μ
¶Μ
¶Μ
¶Μ
0"0003 lbf · s
ft 2
0"5 3 ×2
0"001 in. 3
ft 3
12 3 in. 3
22 ×800
60
1
s
= 22
= 0"0228 ft-lbf
Power 3 is
3 = ( applied torque ) (angular speed )
Μ
¶Μ
22 ×800
60
1
s
hp-s
550 ft-lbf
= (0"0228 ft-lbf )
= 0"00347 hp
Z
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79
Problem 9.3
Oil with viscosity 0.0014 lbf-s/ft 2 and density 1.71 slug/ft 3 ß ows between two par-
allel plates that are spaced 0.125 in. apart and inclined at a 45 ! angle. Pressure
gages at locations - and 4 indicate that 5 " !5 % = 5 psi. The distance between
the pressure gages is 2 ft. Each plate has a dimension (i.e., depth into the paper)
of 1.5 ft. Determine the rate of volume ß ow of oil.
Solution
Laminar ß ow between parallel plates (planar Poiseuille ß ow) is described by:.
Μ
12!
4 3
6 = ,5
,7 +% ,8
!
,7
(1)
The left side of Eq. (1) is
Μ
12!
4 3
Ã
12×0"0014 lbf-s/ft 2
(0"125912) 3 ft 3
!
!
6 = !
6
³
´
!14:864 lbf-s/ft 5
=
6
The pressure gradient is
,7 = !5
!7
! Μ
! 5 lbf/in. 2
2 ft
144 in. 2
ft 2
=
= !360 lbf/ft 3
The slope term is
% ,8
,7 = %SIN(45 ! )
= 1"71×32"2×SIN(45 ! )
= 38"93 lbf/ft 3
,5
Ã
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