Lubrication - Example Calculation

1. Introduction
The standard assumptions are made in this calculation, see lubrication notes.
The design procedure used in this example makes use of dimensionless groups plotted against the Sommerfeld number on a series of charts (A A Raimondi and J Boyd, "A Solution for the Finite Journal Bearing and its Application to Analysis and Design", parts I, II and III, Trans. ASLE, vol 1, no. 1, 159-209) to determine other values for the bearing. These charts are reproduced in many texts on Mechanical Engineering Design. To follow this example you should have a copy of one of the texts that include these charts - marked * in the References.
A plain bearing with hydrodynamic lubrication is to be designed to carry a load of 2500 Newtons at a shaft speed of 30 revs per second = N.

Initial assumptions are that a SAE grade 40 oil will be used and the mean oil temperature in the bearing is 68.3oC (which corresponds to 155oF). At this temperature the oil has an absolute viscosity of 4 micro reyn (i.e. 27.5 mPa.s or 0.0275 Pa.s). The density is assumed to be 850 kg/m3 and the specific heat 1800 J/kg oC
As a starting point it will be assumed that the: length / diameter ratio = 1 and for a first calculation length = diameter = 2 radius = 40 mm will be used. l = d = 2 r = 40mm
It is also assumed that the bearing / shaft is between a 'free running fit' - H9/d9 and a 'close running fit' - H8/f7 with a radial clearance of 0.04 mm.

2. Calculation
First check that the 'unit load' (load / projected area of bearing) is acceptable:
The unit load = load / (length x diameter) = 2500 / (0.04 x 0.04) = 1.56 MPa. This is perhaps on the high side for machinery (but low for IC engines) but this value is accepted for this example.
Bearing designs are commonly based on charts using the 'Bearing Characteristic Number' or Sommerfeld Number, this is defined (in a non - dimensional way) as:

S = ( r / c )2(mu N/P)
where r is radius, c is the radial clearance, mu is the absolute viscosity, N is shaft speed in revs/s and P is the unit load.
As there is no information given about rotation of the bush or load vector, it is assumed that these are not rotating and that S = S'. So:

Substituting values: S = (0.02 / 0.00004)2(0.0275 x 30 / 1563000) = 0.132

The first chart in the series plots the minimum film thickness variable ho/c where ho is the minimum film thickness against the Sommerfeld number for a variety of l/d ratios. From this chart ho/c = 0.41 so the minimum film thickness

ho = 0.41 x 0.00004 = 0.0000164 m
It is important that the minimum film thickness be significantly greater than the maximum surface roughness. Keep in mind the fact that surface roughness is usually given as an average roughness amplitude (Ra) and the extreme peak to valley depths may well be 3 - 6 times greater than the Ra.
The minimum film thickness = radial clearance - eccentricity ho = r - e, divide both sides by c
ho / c = 1 - epsilon, where epsilon is the eccentricity ratio.

The eccentricity of this design is e = c - ho
e = 0.00004 - 0.0000164 = 0.0000236 m

Two lines indicating optima, maximum load and minimum power loss, are also marked on the first chart and the zone between these lines is normally considered to be a recommended operating region.

The second chart needed plots the coefficient of friction variable f r/c where f is the coefficient of friction, against the Sommerfeld number. From this chart the value of f r/c is 3.3 so

f = 3.3 x 0.00004 / 0.02 = 0.0066
The torque to overcome the friction = coefficient of friction x load x radius, substituting values T = 0.0066 x 2500 x 0.02 = 0.33 Nm.
The power lost is: torque x angular velocity, substituting values Power lost = 0.33 x 2 x 3.14159 x 30 = 62.2 watts.
The third chart used plots the flow variable, Q/r c N l, against S. From this chart Q/r c N l = 4.3, substituting values The total flow, Q = 4.3 x 0.02 x 0.00004 x 30 x 0.04 = 4.128 E-6 m3/s
The fourth chart used plots the (side) flow ratio (leakage ratio) against S and from this chart the (side) flow ratio: Qs/Q = 0.66, substituting values The side leakage, Qs = 0.66 x 4.128 E-6 = 2.724 E-6 m3/s

Rise in Oil Temperature. It is assumed that the oil that is retained in the bearing rises in temperature by deltaT and the oil that leaks out of the side of the bearing rises in temperature by deltaT/2. First the volume flow rates need to be converted to kg/s.

Total oil mass flow rate = 4.128 E-6 x 850 = 0.003509 kg/s
Side leakage mass flow rate = 0.66 x 0.003509 = 0.002316 kg/s
Mass flow rate remaining in bearing = 0.34 x 0.003509 = 0.001193 kg/s

Carrying out an energy balance assuming that all the energy put into the oil raises only the oil temperature:
62.2 = 1800 (0.002316 deltaT/2 + 0.001193 deltaT)
The rise in oil temperature: deltaT = 14.7o C, assume 15o C.

Oil cooling. As it was assumed at the outset that the mean oil temperature was 68.3o C, this means that the inlet oil temperature would be 68.3 - 7.5 = 60.8o C and the outlet temperature 75.8o C. Provided there is sufficient cooling available to ensure the oil from the bearing outlet is cooled to 60o C or less, then this design may well be acceptable. As the viscosity of most oils changes very rapidly (typically halving for a 10 - 15o C temperature rise) it is important to check operation will still be satisfactory outside these limits - for example when starting from cold and during short periods of overloading.

As well as the above charts there are also charts to provide additional information such as the maximum film pressure, the angular position where the oil film thickness is a minimum and where it terminates.

3. Interactive Java Applet
This problem can also be solved using the Java Applet to assist with journal bearing design given here
Comparison of Raimondi and Boyd chart results with Java applet (above) click here

David J Grieve. Revised: 23rd February 2010. Original: 24th June 2003.