Page 194 - Tribology in Machine Design

Sliding-element bearings 179

Substitution for dp/dr, obtained by the differentiation of eqn (5.11), gives

the latter term is obtained by substitution for p 0 from eqn (5.12). The total
flow through a cylindrical section of total height h, radius r and length 2nr,
is

This is the minimum oil delivery required from the pump for a desired film
of thickness h. Let V be the average velocity of the flow in the line, A its
cross-sectional area and Y\ the mechanical efficiency of the pump. Then the
power required from the pump is

If the circular pad, shown in Fig. 5.4, is rotated with speed n about its axis,
the tangential fluid velocity vv t may be found from eqn (5.2b) by substituting
Wi =0, W 2 =2nrn and for dp/dz the quantity dp/d(r®) = (l/r)dp/d&. But
since h= const, dp/d&=0. Thus

The torque required for rotation is

therefore

If, over a portion of the pad, the flow path in one direction X is short
compared with that in the other direction Z, as shown in Fig. 5.5, the flow
velocity w and the pressure gradient dp/dz will be relatively small and eqn
2
2
(5.9) may be approximated by d p/dx = 0, i.e. parallel flow is assumed for a
distance b through each slot of approximate width /. Integration of the
differential equation, together with the use of the limits p =p 0 at x =0 and
f*
p = 0 at x — b gives the pressure distribution. Integration q x from eqn
J-i
(5.3) gives the flow Q across one area bl. The slot equations for one area are

The force or torque required to move a hydrostatic bearing at slow speed is
extremely small, less than in ball- or roller-bearings. Also, there is no