Mechanics of Fluids

part of the bearing then gives
whence
Vh1
2 − h3 1 ps
=
12µ l1
Vh2
2 − h3 �
2
−
12µ
ps
�
l2
ps
6µV =
�
h 3 1 /l1
Fundamentals of the theory of hydrodynamic lubrication 229
h1 − h2
�
+
�
h3 2 /l2
� (6.46)
Since ps must be positive, V(h1 − h2) must be positive; that is, the lubricant
must flow from the larger clearance to the smaller. The load carried
divided by the width of the bearing corresponds to the area of the pressure
graph (Fig. 6.21)
� l
0
p dx = 1
2 psl
where l = l1 + l2. The maximum load is determined by the maximum value
of the expression 6.46 and this occurs when h1/h2 = 1+ √ 3/2 = 1.866 and
l1/l2 = (5 + 3 √ 3)/4 = 2.549. The maximum load divided by width is then
4
9 (2√ 3 − 3)µVl 2 /h 2 2 = 0.2063µVl2 /h 2 2
an appreciable increase above the maximum value, 0.1602µVl 2 /h 2 2 ,
obtained with the inclined-plane type of bearing. Unfortunately, for bearings
that are not infinite in width, sideways leakage is greater with a Rayleigh step,
and thus in practice its superiority in load-carrying is lessened. Refinements
have been devised, however, to improve the performance.
Example 6.7 Consider a step bearing of breadth b, with the step
centrally positioned.
(a) Show that the volumetric flow rate Q through the bearing is
given by
Q = V(1 + H2 )
2(1 + H3 ) bh1
where H = h1/h2.
(b) A bearing has the following dimensions: h1 = 0.5 mm, h2 =
0.25 mm, l = 100 mm, b = 100 mm. The bearing is used in
conjunction with an oil of relative density 0.87 and kinematic viscosity
2 × 10−4 m2 · s−1 . The relative velocity between the bearing
surfaces is 10 m · s−1 . Determine the volumetric flow rate of oil.
(c) Determine the load supported by the bearing.