Transcript CE319F Elementary Fluid Mechanics

CE 319 F
Daene McKinney
Elementary Mechanics of Fluids
Viscosity
Some Simple Flows
• Flow between a fixed and a moving plate
Fluid in contact with the plate has the same
velocity as the plate
u = x-direction component of velocity
y
Moving plate
u=V
V
B
u( y) 
V
y
B
Fluid
Fixed plate
x
u=0
Some Simple Flows
• Flow through a long, straight pipe
Fluid in contact with the pipe wall has the same
velocity as the wall
u = x-direction component of velocity
R
r
x
  r 2 
u (r )  V 1    
  R  
V
Fluid
Fluid Deformation
• Flow between a fixed and a moving plate
• Force causes plate to move with velocity V
and the fluid deforms continuously.
y
Moving plate
t0
u=V
t1 t2
Fluid
Fixed plate
x
u=0
Fluid Deformation
Shear stress on the plate is proportional
to deformation rate of the fluid
da 
dL
dy
dL
dt 
dV
da dV

dt dy

da

dt
dV
dy
y
dL
t
da
dy
Moving plate
u=V+dV
t+dt
dx
Fluid
Fixed plate
x
u=V
Shear in Different Fluids
• Shear-stress relations for
different types
of fluids
• Newtonian fluids: linear
relationship
• Slope of line (coefficient of
proportionality) is “viscosity”
dV
dy
dV
 
dy

Viscosity
•
Newton’s Law of Viscosity   

•
Viscosity
•
Units
•
Water (@ 20oC)
–
•

dV / dy
N / m2 N  s

m / s / m m2
V+dv
 = 1x10-3 N-s/m2
Air (@ 20oC)
–
•
dV
dy
 = 1.8x10-5 N-s/m2
Kinematic viscosity



V
Flow between 2 plates
Force is same on top
and bottom
F1  1 A1   2 A2  F2
A1  A2
1   2
y
1  
du
du

2
dy 1
dy 2
Thus, slope of velocity
profile is constant and
velocity profile is a st. line
Moving plate
u=V
V
B
u( y) 
V
y
B
Fluid
Fixed plate
Force acting
ON the plate
x
u=0
Flow between 2 plates
Shear stress anywhere
between plates
 
du
V

dy
B
  0.1 N  s / m 2 ( SAE 30 @ 38o C )   (0.1 N  s / m 2 )( 3 m / s )
0.02 m
V  3 m/s
B  0.02 m
y
 15 N / m 2
Moving plate
V
B
u( y) 
V
y
B
u=V


Fixed plate
Shear
on fluid
x
u=0
Flow between 2 plates
• 2 different coordinate systems
B
r
x
  r 2 
u (r )  V 1    
  B  
V
u ( y )  C  y  B  y 
y
x
Example: Journal Bearing
• Given
–
–
–
–
–
–
Rotation rate, w = 1500 rpm
d = 6 cm
l = 40 cm
D = 6.02 cm
SGoil = 0.88
oil = 0.003 m2/s
• Find: Torque and Power
required to turn the bearing at
the indicated speed.
Example: cont.
• Assume: Linear velocity profile in oil film
Shear Stress   
dV
w (d / 2)

dy
(D  d ) / 2
 2

*
1500

(0.06 / 2)
60

 (0.88 * 998 * 0.003) 
 124 kN / m 2
(0.0002) / 2
d d
Torque M  (2 l )
2 2
0.06
0.06
 (2 *124,000 *
* 0.4)
 281 N  m
2
2
Power P  Mw  281*157.1  44,100 N  m / s  44.1kW
Example: Rotating Disk
• Assume linear velocity profile: dV/dy=V/y=wr/y
• Find shear stress