# Aula Teórica 6 ```Aula Te&oacute;rica 6
Pressure distribution in “Rigid Body”
type flows.
What is a rigid body flow?
• A flow is of the “Rigid body” if the adjacent
elementary volumes maintain their relative
position, i.e. if there is no shear.
• In this case forces are exclusively resulting
from:
– Pressure or
– Gravity
Force resultant

du i
dt
Weight   g  dxdydz


p
xi
 g i
Examples
• Uniform linear acceleration:

x3

du i

dt
xi
du 3
p

dt
x1
0


p
p
x3
du 1
x3
 g 3
 g 3

p
dt
 x1
du 1
p
dt
 g i

 x1
 g1
If there is a free surface
• The pressure is hydrostatic and consequently there is
inclination….
• If acceleration is constant, the horizontal pressure
gradient is constant and the free surface is a plane:

x3
p
 x1
x1  0    0
  a x1
x1  0    0  C 1  0
p  g
p
x1
 x1

 
g
 x1
a x1
g
 
   a x1
x1  C 1
 
a x
g
7
1
x1
x
g
   0 . 7 x   z  0 . 021 m
Rotating flow

• Rotation implies acceleration.
du i
dt

p
xi
 g i
• Cylindrical coordinates
are the most adequate
• i.e. the parabola vertices is at z=0 and r=0.
• The line along which p=p1 is:
• At free surface p=p0 and thus:
r 
2
z surf 
2
2g
• i.e. the free surface is a parabola.
Free surface formation
• The volume above the resting
level is equal to the volume below
that level.
• The free surface can cut the
bottom.
Curvilinear stream lines
• Can we have any curvature?
• We are limited by the pressure gradient.
Summary
• If there is acceleration in a free surface flow,
free surface inclination will appear.
• If acceleration is linear and constant, the free
surface is plan
• If there is rotation at constant angular velocity,
the free surface is a parabola.
```