Transcript Coaxial cylinders method

Coaxial cylinders method
Consider laminar flow of an incompressible viscous
fluid between two vertical coaxial cylinders
The outer one is rotating with an angular velocity W
r
It is assumed that
Inner
cylinder
R1
•There are no end effects
•No-slip condition prevails in the cylinder-fluid
contact
If tr is the shear stress on a liquid layer at a distance r
from the axis of rotation, then the torque T on the
liquid shell by the outer layer of the liquid is
T = (2prl).tr.r where l is the length of the inner
cylinder
l
R2
Outer
cylinder
Shear stress at radius r
Therefore the shear stress at radius r is
T
tr 
2pr 2 l
From Newton’s law of viscous flow the shear stress at radius r is
u
w
t r  h  hr.
y
r
Where w is the angular velocity, h is the dynamic coefficient of
viscosity. The distance of separation y = r and the change in
velocity du = rdw
Constant of integration
• Now
• Therefore
T
 w  2pr 3lh . r
T
w
C
2
2plh.2r
• Where C is a constant of integration
• At r = R1, w = 0, at r = R2, w = W
• Therefore
T
C
2
4plhR1
Expression for coefficient of viscosity
Substituting we get:
4plhR1 R 2 W
T
2
2
R 2  R1
2
T(R 2  R 1 )
h
...(20)
2
2
4plR 1 R 2 W
2
Or
2
2
Knowing values of the other terms, the coefficient of
viscosity h can be calculated
Solid sphere method
• When a solid body is allowed to fall from rest in a
homogenous fluid of infinite extent, it will initially accelerate
till the gravitational force is balanced by buoyant and viscous
forces.
• Consider a sphere of radius r, moving in a fluid with viscosity
h and attaining a uniform velocity V, the viscous resistance is
given by Stoke’s law as 6prhV
• If the densities of the material of the solid sphere and the
liquid are rs and rl respectively, then the gravity and buoyant
forces are respectively
4 3
pr rs g
3
and
4 3
pr r l g
3
Solid sphere in liquid- expression for viscosity
Therefore balancing forces we get:
Weight of
sphere
Sphere
4 3
4 3
pr rs g  pr rl g  6phrV
3
3
Or
2gr 2 (rs  rl )
h
...(21)
3V
Force of
buoyancy
Lubricant
Therefore the viscosity can be determined if we know values for
the other terms
Journal bearing- process at startup
Shaft/journal
e = eccentricity
Bearing
Stationary
journal
Instant of starting (tends to While running (slips due to loss
climb up the bearing)
of traction and settles eccentric
to bearing)
Because of the eccentricity, the wedge is maintained
(lack of concentricity)
Journal bearing- geometry
Bearing: center O and radius R1
Bearing
Shaft: center C and radius R2
OC is the eccentricity measured as e
Shaft
G
All angular distances are measured
from the position of maximum film
thickness (where the extension of line
CO cuts the bearing surface at G)
O
Consider a point B on the bearing
surface such that the angle GOB = q
OB is the radius R1 of the bearing.
The line OB cuts the shaft at point A and
AB is the film thickness h
Draw a line from the shaft center C
parallel to OB cutting the shaft at E and
bearing at F
q=0
q
C
q
Direction of rotation
Increase in q
Journal bearing- film thickness
Distances AB and EF are very small compared to the
radii
OB and CF are so close together (e being very small
compared to radii)
Therefore ABFE is considered a rectangle
G
From O, drop a perpendicular to CE cutting CE at D
The oil film thickness h = EF = OB - DE = OB-(CE-CD)
O
CD = eCosq
OB – CE = R1 - R2 = c, where c is the radial clearance
of the bearing
Hence h = c + ecosq = c{1+(e/c)cosq}
The ratio e/c is called the eccentricity ratio of the
shaft and is written as e, so
h  c(1  e cos q)...( 22)
q
C
q