Title: In the analysis of a tilting pad thrust bearing, the following dimensions were measured: h1 = 10 mm, h2 = 5mm, L = 10 cm, B = 24 cm The shaft rotates at a speed of 120 RPM. Find the position on the x axis where the pressure differential would be
1In the analysis of a tilting pad thrust bearing,
the following dimensions were measured h1 10
mm, h2 5mm, L 10 cm, B 24 cmThe shaft
rotates at a speed of 120 RPM. Find the position
on the x axis where the pressure differential
would be maximum. Calculate the height of
separation when the pressure differential is max.
Find the position on x axis corresponding to the
center of pressure. Also calculate the load on
the bearing. Coefficient of dynamic viscosity
40 cp.
Z
Approximate U as the speed of the periphery of
the collar with dia. 23 cm
h1
h
h2
NOTE 1 cp 0.001 Pa.s 1 Pa 1N/m2
B
X
U
2- 1k h1/h2 10/5 2, k 1
- Xo/B (1k)/(2k) 2/3, xo 24x2/3 16 cm
- hoh22(1k)/2k 5(2)(2)/3 20/3 6.66 mm
-
13.71 cm - We need to find load W from the equation
3Substituting the values we get
Therefore W (0.42 x 0.076/0.021)2 2.3 N
4Temperature rise in tilting pad bearing
- The rate of work done on the lubricant is the
frictional force F multiplied by the velocity U - It can be approximated (especially at high
speeds) that all the heat is removed by the oil
and no heat is removed by the pad or runner - Hence if the temperature rise is Dt, the density
of oil r and its specific heat capacity is c, the
energy balance, with J as the mechanical
equivalent of heat, we get
5- Now where ho is the
distance of separation when the pressure is
maximum - And
- Therefore
- Friction is given by
6Tilting pad bearing- Temperature rise
- In non-dimensional form, the friction would be
- From the load equation,
- Where W is the non-dimensional load
- Substituting these values into the energy balance
equation we get temperature rise
7Ball thrust bearings
Housing collar (attaches to housing and is
stationary)
Shaft collar (attaches to shaft and rotates)
Radial load
Axial load
Inner race
Outer race
Radial ball/roller bearing- takes on radial
loads Shaft is attached to inner race while outer
race is attached to machinery housing Inner race
rotates with shaft while outer race remains
stationary
Thrust ball bearing - takes on axial loads
8Tapered roller bearings(takes on radial as well
as thrust/axial loads)
Radial load
Outer race
Axial load
Inner race
Ref http//www.jtekt.co.jp/e/company/news/images/
20060404_3e.jpg
9Determination of viscosity
10Viscosity determination- capillary tube method
- Consider steady laminar flow of an incompressible
fluid of density r through a long vertical tube
of radius R - The flow may be assumed to be similar to sliding
of numerous cylindrical shells
Force of flow
Flow takes place due to pressure drop
Viscous drag
11Force balance on a fluid shell element
Momentum in by flow
Pressure p1
- Flow takes place from a region of higher pressure
to a region of lower pressure - The forces acting on the shell element are
- Pressure force
- Gravitational force
- Shear stress
r
R
Tube wall
l
Momentum out by flow
Pressure p2
12Forces on a cylindrical fluid element
Momentum in by flow
Pressure p1
Shear force at surface at radius r 2prltr Shear
force at outer surface of element -2p(r
Dr)ltrDr Pressure force at top (z 0)
p12prDr Pressure force at bottom (z l)
-p22prDr Gravity force on shell element
2prDrlrg Where tr and tr Dr are the shear
stresses at r and r Dr respectively Upward
forces have been indicated with a negative sign
r
R
l
rDr
Pressure p2
Momentum out by flow
13Equilibrium of the forces
From the equilibrium of dynamic forces we have
Which reduces to
or
p1 (p2 rgl) is the pressure difference Dp
causing flow Hence,
14Shear stress on element at radius r
From previous slide
Integrating we get
where A is a constant of
integration From theory on shear flow in laminar
region, the shear stress is maximum at the center
line (where r 0). Therefore
When r 0, dtr/dr 0, which gives A 0.
Therefore
15Velocity in terms of radius
From Newtons law of viscosity
The negative sign is incorporated because the
velocity decreases with increase in radius.
Velocity is zero at surface of tube and maximum
at center
Hence
On integrating we get
At r R, u 0. Therefore
16Max. velocity and avg. velocity
Hence we get the expression for velocity in terms
of radius as
From the above expression it is clear that the
velocity profile is parabolic and is maximum at r
0. The average velocity is half the maximum
velocity.
Therefore
17Expression for viscosity
The volume rate of flow Q Average velocity x
Area
This is known as Hagen-Poiseuille law and can be
used to determine the viscosity. Therefore
Where V is the volume of flow in time t