1. 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 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 = Pa.s
1 Pa = 1N/m2 B U X

2. 1+k = h1/h2 = 10/5 = 2, k = 1
Xo/B= (1+k)/(2+k) = 2/3, xo = 24x2/3 = 16 cm
ho=h2[2(1+k)/2+k] = 5(2)(2)/3 = 20/3 = 6.66 mm
= cm
We need to find load W from the equation

3. Substituting the values we get
Therefore W = (0.42 x 0.076/0.021)2
~ 2.3 N

4. Temperature 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

6. Tilting 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

7. Ball 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

8. Tapered roller bearings (takes on radial as well as thrust/axial loads)
Radial load
Outer race
Axial load
Inner race
Ref:

9. Determination of viscosity

10. Viscosity 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

11. Force 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

12. Forces 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)ltr+Dr
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
r+Dr
Pressure p2
Momentum out by flow

13. Equilibrium 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,

14. Shear 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

15. Velocity in terms of radius
From Newton’s 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

16. Max. 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

17. Expression 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