Mechanics of Fluids

Fundamentals of the theory of hydrodynamic lubrication 223
Over a small length δx the clearance h may be assumed unchanged, and
from eqn 6.34 we have
�
dp V q
= 12µ −
dx 2h2 h3 �
(6.35)
Equation 6.35 involves two noteworthy assumptions: first, that the variation
of velocity with y is the same as if the boundaries were parallel; second,
that the acceleration of the fluid as h decreases requires inertia forces that are
negligible compared with the viscous forces. For the usual very small inclination
of the slipper, these assumptions are justified. Furthermore, components
of velocity in the y direction are negligible.
If µ is assumed constant throughout, and (a − x)δ is substituted for h,
eqn 6.35 may be integrated:
�
V
q
p = 12µ
−
2(a − x)δ2 2(a − x) 2δ3 �
+ A
The integration constant A may be determined from the condition that
p = p0, the ambient pressure, when x = 0.
The discharge divided by width, q, still remains to be determined. However,
p again equals p0 when x = l, and inserting this condition into the
equation enables q to be calculated. After simplification the final result is
p = p0 +
6µVx(l − x)
δ2 (a − x) 2 (2a − l)
(6.36)
This equation expresses the relation between p, the pressure in the lubricant,
and x. Since, under the slipper, x is always less than l and l is less
than 2a, the last term of eqn 6.36 is positive; p is thus greater than p0 and
the bearing can sustain a load. If the distance a approaches infinity, that is,
if the surfaces become parallel, then p = p0 for all values of x − as we found
in Section 6.7. The inclination of the slipper must be such that the fluid is
forced into a passage of decreasing cross-section: if V, and therefore the fluid
velocity, were reversed the last term of eqn 6.36 would be made negative,
p would become less than p0 and the bearing would collapse.
Thus the fluid is, as it were, used as a wedge. This wedge principle is, in
fact, one of the fundamental features of any hydrodynamically lubricated
bearing, whether plane or not.
The load that such a bearing can support is determined by the total net
thrust which the fluid exerts on either bearing surface.
The net thrust divided by the width of the bearing is obtained from the
integral
� l
Substituting from eqn 6.36 we have
Net thrust divided by width = T =
0
(p − p0)dx (6.37)
6µV
δ2 � l x(l − x)
dx
(2a − l) 0 (a − x) 2