Performance-Prediction Tools for Flat-Land Thrust Bearings

Dec. 9, 2010
How can an oil film between two parallel surfaces support a bearing load?

Authored by:
E.R. Booser
Engineering Consultant
Vero Beach, Fla.

M.M. Khonsari
Professor, Mechanical Engineering
Louisiana State University
Baton Rouge, La.

Edited by Jessica Shapiro
[email protected]

Key points:
• Flat-land thrust bearings are used for all sizes of lightly loaded applications.
• Empirically derived coefficients and equations can help engineers predict flat-land-bearing performance.
• Modifications that help generate and maintain oil-film pressure can boost flat-land bearings’ capabilities.

Louisiana State University,
“Know your thrust bearings,” Machine Design, April 12, 2007
“Cool ideas for thrust bearings,” Machine Design, June 16, 2009
Khonsari, M.M. and E.R. Booser, Applied Tribology – Bearing Design and Lubrication, 2nd Ed. John Wiley & Sons, Ltd., West Sussex, U. K. 2008.
Wilcock, D.F. and E.R. Booser, Bearing Design and Application, McGraw-Hill Book Co., 1957.

Millions of flat-land thrust bearings are used in everything from large turbines to gear sets, but engineers still don’t have good tools for predicting how the bearings will work in practice.

Flat-land thrust bearing surfaces are typically segmented, and an oil film separates them from the surfaces they support. The parallelism of the two surfaces theoretically prevents any oil pressure from developing in the film. That is, an oil film between two parallel bearing surfaces should not be able to support a load.

In practice, however, thermal expansion of the oil and bearing surfaces lets oil-film pressure develop to some extent. Flat-land thrust bearings will never hold more than 20 to 25% of the load supported by their tapered-land or stepped counterparts which have physical wedges to pressurize oil in the direction of sliding.

By understanding how flat-land thrust bearings work and what controls their performance characteristics such as load capacity, power loss, and temperature rise, engineers can get the most out of turbines, compressors, gear sets, and other rotating machines for light loads, simple axial positioning, and occasional reverse thrust.

Load limits
Flat-land thrust bearings depend on efficient oil delivery to the bearing surface for their thrust-load capabilities. Small flat-land bearings often lack surface features to help channel the oil. Bearings of this type, including small washers and thrust shoulders on sleeve bushings, are commonly used for loads of 20 to 35 psi.

Adding radial grooves to better distribute feed oil raises load capacity to 125 psi in larger bearings. Higher oil viscosities, faster speeds, greater oil-feed volumes, and self-lubricating bearing materials that don’t depend on oil films can all further boost load capacity.

Although these loads represent the suggested design envelope, in practice, the loads causing bearing failure commonly exceed these limits by a factor of two or more.

In addition to load capacity, engineers using flat-land thrust bearings need to predict operating characteristics like film thickness, oil-flow rate, power loss, and subsequent temperature rise.

Film thickness
The key characteristic in any thrust bearing is the thickness, or height, of the oil film between the bearing and rotating thrust runner. It is this film that actually supports the bearing load. At the oil inlet, the film is at its maximum thickness, h1. It thins out toward trailing edge to a minimum thickness, h2 that depends on surface velocity, U, and unit pressure, P.

Thrust-bearing surfaces are often tapered to enhance pumping action and pressurize the oil film for higher load capacity. But bearing surfaces without pumping wedge geometry generate no film pressure; the corresponding load capacity is nil.

In actual bearings, however, oil-film pressure is generated by two secondary effects: thermal expansion — of the passing oil and of the bearing surface as it is heated by the oil — and possible elastic deformation of the surfaces.

The following minimum-film-thickness relation describes behavior in many thrust bearings:
h2 = Kh (µ × U × B/P)0.5

where Kh = a dimensionless film thickness coefficient; μ = oil-film viscosity in reyns (lb-sec/in.2) at the mean bearing temperature; U = surface velocity at the mean diameter in in./sec (ips); B = bearing segment breadth (circumferential length at the mean diameter) in inches; and P = unit

projected load on the bearing surface in lb/in.2 P, in turn, is calculated from the load, W, in lb and the segment area, A, in in.2

P = W/A = W/(L × B)

where L = half the difference between the bearing’s outer and inner diameters in inches.

Experience with industrial and marine steam turbines shows many flat-land bearings with radial oil grooves and sectors with L/B ~ 1 have h1/h2 = 1.3 to 1.5. With this film-thickness ratio and L/B, oil-film computer simulations show the dimensionless coefficient Kh to be 0.23.

Another common flat-land thrust bearing configuration with radially narrow, elongated sectors is seen in thrust shoulders on sleeve bearings such as those for industrial electric motors. Temperature-load capacity tests show this design to have a lower h1/h2 of 1.2.

The elongated arc causes oil loss from the outside diameter, so film pressure peaks near the leading edge and minimum film thickness is reached well before the trailing edge. An equivalent L/B of 0.75 is therefore appropriate for this geometry.

A corresponding Kh of 0.15 was found to describe the behavior of thrust faces with four, six, and 12 radial oil grooves.

Oil flow
Oil typically flows between a flat bearing segment and parallel thrust runner at approximately half the runner surface speed, U. Multiply by the cross-sectional area of the flow, the film height times the segment length, to get the flow rate:
Q = 0.5 × h × U × L.

In practice, film thickness can vary. The flow rate is also affected by pressure and side flow. Use the empirically derived dimensionless flow rate coefficient, Kq, to account for these phenomena:

Q = Kq × h2 × U × L.

Power loss
Engineers may look at viscous drag resistance, F, on a stationary flat-bearing segment as a measure of efficiency. It depends on fluid viscosity, μ; surface velocity, U; and film thickness, h; with:
F = μ × U × B × L/h.

But a more useful measure is power loss, E, which is F × U. Use the dimensionless flow-friction coefficient, Kf, to reflect secondary film flow and geometric effects:
E = Kf × μ × U2 × B × L/h2.

Temperature rise
Power loss causes oil temperature to increase by ∆T:
∆T = E/(Q × ρ × Cp)

where ρ = density (0.0307 lb/in.2 for mineral oils) and Cp = specific heat (4,400 in.-lbm/(°F-lbm) for mineral oils).

Substituting for E, Q, and h2 allows engineers to drop oil viscosity from the expression, because a thicker oil film compensates for the greater viscous drag of higher viscosity oil or higher speeds.
∆T = [Kf/(K q × Kh2 × ρ × Cp)] × P

This can be further simplified by using a dimensionless temperature coefficient, Kt, so that temperature rise is proportional to unit axial load, P.
∆T = Kt × P

Values of Kt drop slightly with increased loading. That’s because a more pronounced thermal wedge shape forms in the bearing with the higher temperature.

In addition, bearings where L/B deviates significantly from unity will have smaller Kt and lower load capacity. However research shows the number of oil-distributing grooves has no significant effect on ∆T.

Boosting load capacity
Flat-land thrust bearings have inherently low load capacity. However, some modifications can help engineers improve load capacity in these bearings.

Try increasing the base viscosity of the oil. This can create a thicker oil film without increasing bearing temperature. Adding radial grooves to distribute oil over the plain thrust face can also boost film thickness, especially when oil-feed pressure is maintained.

To get even more out of thrust bearings, create more bearing area by increasing the bearing’s OD. Or turn them into the higher-load tapered or step thrust bearings by incorporating those geometries into the thrust-face sector. If machining the thrust faces isn’t possible, mounting individual flat-land segments on rubber backings or springs will let them pivot to create an oil wedge.

Application example 1
Determine the performance characteristics for a 7-in.-ID, 10-in.-OD thrust bearing in a gear set. The bearing has 14 flat lands and operates at 1,800 rpm under 3,000-lb load. ISO 68 grade turbine oil is fed to the bearing at 120°F.

Assuming that oil grooves take up 20% of the area, the bearing area, A, is:
A = 0.8 × π × [(10 in.)2 – (7 in.)2]/4 = 32.04 in.2

Then the unit pressure, P, is:
P = 3,000 lb/32.04 in.2 = 94 psi

Radial length, L, and circumferential breadth, B, at the mean diameter for each of the 14 flat segments are:
L = (OD – ID)/2 = [10 in. – 7 in.]/2 = 1.5 in.
B = (A/14)/L = [32.04 in.2/14]/1.5 in.2 = 1.53 in.
L/B = 1.5 in./1.53 in. = 0.98, nearly square.

Determine the velocity, U, at the 8.5-in. mean diameter, D:
U = π × D × N = π × 8.5 in. × 1,800 rpm/60 sec/min
= 801 ips

Next, calculate the bearing’s temperature rise (∆T), minimum film thickness (h2), power loss (E), and oil flow (Q).
T = Kt × P = 0.17 × 94 psi = 16°F
T2 = 120°F + 16°F = 136°F where viscosity μ
= 3.7 × 10-6 reyns
h2 = Kh (µ × U × B/P)0.5
h2 = 0.23 [(3.7 × 10-6 reyns) × (801 ips)
× (1.53 in.) / (94 psi)]0.5 = 0.0016 in.
E = Kf × μ × U2 × B × L/h2
E = 0.83 × (3.7 × 10-6 reyns) × (801 ips)2 × (1.53 in.)
× (1.5 in.)/(0.0016 in.) =2,826 in.-lb/sec
= 0.43 hp/pad or 6 hp/bearing.
Q = Kq × h2 × U × L
Q = 0.68 × (0.0016 in.) × (801 ips) × (1.5 in.) = 1.31 in.3/sec

Multiplying by 14 sectors gives a total oil flow of 18.3 in.3/sec.

Application example 2
Calculate the performance characteristics of a 5-in.-OD × 4-in.-ID flat-land thrust shoulder on an 3,600-rpm electric-motor sleeve bearing. The bearing supports 400 lb and has six 1/8-in.-wide radial oil grooves that carry 130°F ISO 32 grade turbine oil.

With six radial oil grooves, each 1/8-in. wide, the bearing area, A, is:
A = π × [(5 in.)2 – (4 in.)2)]/4 – 6 × (1/8 in.) × [(5 in. – 4 in.)/2] = 7.07 in.2 – 0.38 in.2 = 6.69 in.2
Then the unit pressure, P, is:
P = 400 lb/6.69 in.2 = 59.8 psi

Radial length, L, and mean circumferential breadth, B, for each of the six flat pads are:
L = (OD – ID)/2 = [5 in. -– 4 in.)/2 = 0.5 in.
B = (A/L) / 6 = (6.69 in.2/0.5 in.)/6 = 2.23 in.
L/B = 0.5 in./2.23 in. = 0.22 indicating radially narrow, circumferentially elongated flat segments.
The curvature shortens the equivalent circumferential breadth, B, so it’s appropriate to use an equivalent L/B of 0.75 for all thrust faces. This effective L/B is reflected in the dimensionless coefficients we’ll use going forward.

Determine the velocity, U, at the 4.5-in. mean diameter, D, of the flat pads at 3,600 rpm:
U = π × D × N = π × 4.5 in. × 3,600 rpm/60 sec/min = 848 ips.

Next, calculate the bearing’s performance characteristics: temperature rise (∆T), minimum film thickness (h2), power loss (E), and oil flow (Q).
∆T = Kt × P = 0.52 × 59.8 psi = 31°F
T2 = 130°F + 31°F = 161°F where viscosity μ = 1.25 × 10-6 lb-sec/in.2 (reyns)

h2 = Kt × (µ × U × B/P)0.5
h2 = 0.23 [(1.25 × 10-6 reyns) × (848 ips) × (2.23 in.)/(59.8 psi)]0.5 = 0.0014 in.

E = Kf × μ × U2 × B × L/h2
E = 0.91 (1.25 × 10-6 reyns) × (848 ips)2 × (1.53 in.) × (0.5 in.)/(0.0014 in.) = 447 in.-lb/sec
This translates to 0.068 hp/pad or 0.41 hp for all six pads.

Q = Kq × h2 × U × L = 0.58 × (0.0014 in.) × (848 ips) × (0.5 in.) = 0.34 in.3/sec
This becomes 2.0 in.3/sec for the entire six-sector bearing.

© 2010 Penton Media, Inc.

About the Author

Jessica Shapiro

Jessica serves as Associate Editor - 3 years service, M.S. Mechanical Engineering, Drexel University.

Work experience: Materials engineer, The Boeing Company; Primary editor for mechanical and fastening & joining.

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