24. Laminar flow between parallel boundaries
We shall now consider laminar flow between two parallel flat walls (Fig. 48). Placing the origin of our coordinate system halfway between the two walls, the x axis is pointed in the direction of flow and the у axis is normal to the walls.
Passing two cross-sections normal to the flow the
distance between which is l,
consider a rectangular volume of thickness 2y and unit
width
perpendicular to the page (Fig. 48). The condition for uniform
motion
of the control volume along the x
axis
is:
where pf = рг — p2 is the difference between the pressure intensities at the two cross-sections. The minus is due to the negative value of the derivative du/dy.
From the foregoing, determine the velocity increment dv corresponding to the coordinate increment dy:
Integrating,
(6.11)
As,
at у
=
,
и
=
0, then
,
whence finally
То compute the rate of discharge per unit width, first take two elementary areas of size I X dy located symmetrically relative to the z axis and express the elementary discharge:
whence
(6.12)
From
this, expressing the pressure loss in terms of the mean velocity
(6.12)
The relations obtained can be used to investigate flow through a space between two nested cylinders parallel to their centre lines if the space a is small in relation to the cylinder diameters: otherwise the law is more complicated, and we shall not examine it here.
When one of the walls is moving parallel to the other with velocity U, the pressure intensity in the space being constant along its length, the moving wall carries the fluid along. The velocity distribution is linear (Fig. 49) and therefore expressed in the form
(6.14)
The
rate of discharge per unit width of the spacing perpendicular to
the page is determined from the mean velocity, which is
,
i.
e.,
(6.15)
I
f
there is a pressure drop in the fluid filling the spacing, the
velocity
distribution law across the space is
found as the sum (or difference, depending
on the direction of motion
of the wall) of expressions (6.11)
and (6.14):
The resultant velocity distribution across the space is shown in Fig. 50 when (a) the wall moves in the direction of flow due to the pressure drop, and (b) the wall moves in the opposite direction.
The rate of discharge is determined as the sum of the discharges given by Eqs (6.12) and (6.15):
This kind of flow through narrow spaces is found in pumps and other hydraulic machines.
Example.
Determine whether an aircraft lubricating system with the
characteristics
given below will function in horizontal flight at an altitude of
16,000m (
mm
Hg) (i. e., determine absolute pressure at pimp intake
in mm Hg). The length of the intake pipeline is / = 2 m, the diameter d -18 mm; elevation of the oil surface in the tank above the pump z = 0.7 m; tin pressure in the oil tank is atmospheric (Fig. 51). The required flow rate to ensure the necessary heat transfer into the oil at maximum performance of the engiiu is Q = 16 lit/min; viscosity of MK-8 oil v = 0.11 cm2/sec, yoil=900 kg/m. Neglect local losses.
Solution,
(i)
Velocity of oil in pipeline:
(ii) Reynolds number:
(iii) Friction loss in intake pipe:
(iv) Pressure at pump intake from Bernoulli's equation between sections 0-0 and 1-1:
whence
or
Hg