FLUID STATICS

1
Chapter 3 FLUID STATICS
Fluid Mechanics, Spring Term 2007
2
Shear Forces
Normal Forces (pressure)
where F is a force normal to area A
3
Flow of an unconfined viscous fluid down an
incline.

• Flowing viscous fluid exert shear forces.
• Static fluids only exert normal forces.
• Moving fluids (dynamics) will be covered later.

4
Pressure is a scalar quantity
Figure 3.1 (p. 31)
Force balance in the x-direction
5
Force balance in the z-direction
Vertical force on lower boundary
Vertical force on DA
Total weight of wedge element
specific weight
6
From last slide
Divide through by
to get
Now shrink the element to a point
This can be done for any orientation a, so
7
For your Culture (i.e., not required for this
course)

It is possible to have different normal
stresses. Consider a small cubic fluid element
that is part of a larger fluid mass Different
normal forces in one (coordinate) orientation are
equal to shear forces in another orientation.
8

• (contd)
• So what is the pressure then?
• Pressure is the average of the normal forces
  acting at a point.
• Differences between normal forces are due to
  fluid motion.

In this case, if the force vectors are equal in
magnitude, then p 0
9
Pressure Transmission
Hydraulic Lift Figure 3.2 (p. 32)
In a closed system, pressure changes from one
point are transmitted throughout the entire
system (Pascals Law).
10
Absolute Pressure, Gage Pressure, and Vacuum
Figure 3.3 (p. 34) Example of pressure relations

• Pressure in a vacuum is p 0.
• Absolute pressure is referenced to absolute
  pressure.
• Gage pressure is referenced to another pressure,
  typically atmospheric pressure (most gages
  measure relative pressures).

11
Pressure Variation with Elevation
Static fluid All forces must balance as there
are no accelerations. Look at force balance in
direction of D l
Figure 3.4 (p. 35)
12
From figure, note that
Shrink cylinder to zero length
(from previous slide)
or
13
Pressure Variation for a Uniform-Density Fluid
The pressure-elevation relation derived on the
previous slide,
is perfectly general (applies also to variable
g). But if g is constant, the above equation is
easy to integrate
The quantity is known as the
piezometric pressure and is
called the piezometric head.
14
For an incompressible fluid, g is
constant. Pressure and elevation at one point can
thus be related to pressure and elevation at
another point
for
or
15
Example 3.3 What is the water pressure at a
depth of 35 ft?
With the information given, all we can calculate
is the pressure difference between points 1 and
2.
(Do yourself a favor and work in SI-units!)
16
Example 3.4 What is the gage pressure at point 3
?

• Two step solution
• Calculate
• Calculate

(relative to atmospheric pressure at point 1)
17
Pressure Measurements
Figure 3.7 (p. 42) U-tube manometer Better for
higher pressures. Possible to measure pressure in
gases.
Figure 3.6 (p. 41) Piezometer or simple manometer
18
Can start either at open end or inside
pipe. Here we start at open end
p at open end
Change in p from 1 to 2
Change in p from 3 to 4
p in pipe
19
The complete path from point 1 to point 2 may
include several U-tubes. In general
From example 3.9 (p. 44)
20
Differential Manometer
Figure 3.8 (p. 44)
Used for measuring pressure differences between
points along a pipe.
21
Example 3.10 Find the change in piezometric
pressure and in piezometric head between points 1
and 2.
( from
)
22
The cancel out to give
(piezometric pressure)
(piezometric head)
23
Hydrostatic Forces on Plane Surfaces
The white area AB in the figure is a plane of
irregular shape. Line A-B is an edge view of that
area. What is the net force due to pressure
acting on the sloping plane AB?
First, note that hydrostatic pressure increases
along y as
(since y is not vertical)
24
This figure is absolutely awful Line AB
represents the true location of the surface. The
white surface is not drawn in its actual location.
Line 0-0 is horizontal the white area has been
rotated about axis A-B from its proper location.
In other words, the apparent depth of the white
area within the fluid is not as it appears.
25
From the definition of pressure
or
so that the total force on a plane area A is
or
or, since g and sin a are constants
26
But the first moment of the area is defined as
so that the total force can be written as
where is the pressure at the centroid of the
area. The boxed equation is known as the
hydrostatic force equation.
We have thus replaced an integral involving a
variable pressure by a constant resultant
pressure
27
Vertical Location of Line of Action of Resultant
Hydrostatic Force
In English We wish to represent the distributed
pressure force by an equivalent point force.
Where (in the vertical) does that force act?
2 weights on a beam supported at ycp
28
So, for the moment about a point at ycp we have
But with and
we get
The integral on the right-hand side is the second
moment of the area (about point y0)
29
The book just refers to the parallel axis
theorem to write
Mathematically, I think it is easy to see that
using
where is the
2nd moment about y0
Is the 2nd moment about
(thus yielding eq. on top)
and
30
As an aside, you may recall that
The moment of inertia of an object about an axis
through its center of mass Icm is the minimum
moment about any axis in that direction. The
moment about any other parallel axis is equal to
Icm plus the moment of inertia about distance d
of the entire object treated as a point mass
located at the center of mass.
Our system of pressures has nothing to do with
rotations, but the equations are of the same form
31
Back to the problem at hand
Recall from a few slides ago that
so that
or
or
Note that at great depth , the
difference between the centroid and the center of
mass gets very small.
32
Example 3.12
Find the normal force required to open the
elliptical gate if it is hinged at the
top. First find Ftotal, the total hydrostatic
force acting on the plate
With (Appendix p. A-5) we
get
33
Now calculate the slant distance between and
The slant distance to the hinge is 8m x 5m/4m
10m, and the slant distance from the hinge to the
center of mass is 2.5m. Hence,
The two moments about the hinge must add to zero
34
(No Transcript)
35
Hydrostatic Forces on Curved Surfaces
We could integrate the vector forces along
segment AB, but it is often easier to find
equivalent forces on a free body as illustrated
above. FAC acts at the center of pressure as from
previous section, FCB acts at centroid of area
CB, and W acts at the center of mass of the free
body ABC.
36
Example 3.14
Find magnitude and line of action of equivalent
force F. Force balance in x and y
37
The line of action of the horizontal force is
Where we just read directly off
the figure.
The line of action for the vertical force can be
found by summing the moments about C (or any
other point)
(notice that we could add a constant to every
x-coordinate since )
38
From Appendix p. A-5 (Figure A.1)
Distance from C to centroid is
So that xcp is found to be
39
The comlete result is summarized below
40
(No Transcript)
41
(No Transcript)
42
(No Transcript)
43
(No Transcript)
44
(No Transcript)