Mechanics of Fluids

236 Laminar flow between solid boundaries
Similarly there is a net outflow (∂Qz/∂z)δz from the box in the z direction.
If, at the particular position considered, h does not vary with time (i.e. if
both the speed and the load are steady), the total net outflow of lubricant
from the box must be zero.
∴ ∂Qx
∂x
∂Qz
δx + δz = 0 (6.59)
∂z
To obtain ∂Qx/∂x we make appropriate substitutions in eqn 6.23:
� �
∂Qx ∂
= δz −
∂x ∂x
∂p h
∂x
3
��
�Rh
+
12µ 2
�
�
�R ∂h 1 ∂ h
= −
2 ∂x 12 ∂x
3
��
∂p
δz (6.60)
µ ∂x
where the negligible changes of elevation allow us to drop the asterisk from
p∗ and δz takes the place of b. The corresponding equation for the z (axial)
direction, in which there is no relative velocity between journal and bush, is
∂Qz
∂z =−1
�
∂ h
12 ∂z
3
�
∂p
δx (6.61)
µ ∂z
Substituting eqn 6.60 and 6.61 into eqn 6.59, we obtain
�
�R ∂h 1 ∂ h
δxδz −
2 ∂x 12 ∂x
3
�
∂
δxδz −
µ ∂x
1
�
∂ h
12 ∂z
3
�
∂p
δxδz = 0
µ ∂z
If the axes of journal bush are parallel, h will be independent of z, and
dependent only on x. With µ again assumed constant throughout, the
equation thus becomes
6µ�R dh
�
∂
− h
dx ∂x
3 ∂p
∂x
�
− h 3 ∂2p = 0 (6.62)
∂z2 The position z = 0 may be taken mid-way between the ends of the bearing;
for a given value of x, the pressure varies from, say, pc at z = 0 to atmospheric
at each end of the bearing (where z =±L/2). That is, p/pc varies
symmetrically along the length of the bearing from zero to 1 and back to zero.
It is reasonable to assume that the form of this variation is the same whatever
the value of x; that is, that f = p/pc is a function of z only. (We shall show
later that for a short bearing this assumption is justified.) Substituting pc for
p in eqn 6.62 gives
6µ�R dh d
−
dx dx
and division by h 3 pc then yields
�
h
�
f − h
dx
3 d
pc
2f = 0 (6.63)
dz2 3 dpc
a − b 2 f − d2f = 0 (6.64)
dz2