Chapter 2 FLUID STATICS

About This Presentation

Title:

Chapter 2 FLUID STATICS

Description:

... heavier gage liquid fills the lower U tube up to 0-0; then the ... The volume of liquid displaced in each reservoir equals the displacement in the U tube ... – PowerPoint PPT presentation

Number of Views:1442

Avg rating:3.0/5.0

Slides: 116

Provided by: nata246

Category:

more less

Transcript and Presenter's Notes

Title: Chapter 2 FLUID STATICS

1
Chapter 2FLUID STATICS
2

The science of fluid statics
the study of pressure and its variation
throughout a fluid
the study of pressure forces on finite surfaces
Special cases of fluids moving as solids are
included in the treatment of statics because of
the similarity of forces involved.
Since there is no motion of a fluid layer
relative to an adjacent layer, there are no shear
stresses in the fluid
? all free bodies in fluid statics have only
normal pressure forces acting on their surfaces

3
2.1 PRESSURE AT A POINT

Average pressure dividing the normal force
pushing against a plane area by the area.
Pressure at a point the limit of the ratio of
normal force to area as the area approaches zero
size at the point.
At a point a fluid at rest has the same pressure
in all directions ? an element dA of very small
area, free to rotate about its center when
submerged in a fluid at rest, will have a force
of constant magnitude acting on either side of
it, regardless of its orientation.
To demonstrate this, a small wedge-shaped free
body of unit width is taken at the point (x, y)
in a fluid at rest (Fig.2.1)

4
Figure 2.1 Free-body diagram of wedge-shaped
particle
5

There can be no shear forces ? the only forces
are the normal surface forces and gravity ? the
equations of motion in the x and y directions
px, py, ps are the average pressures on the
three faces, ? is the unit gravity force of the
fluid, ? is its density, and ax, ay are the
accelerations
When the limit is taken as the free body is
reduced to zero size by allowing the inclined
face to approach (x, y) while maintaining the
same angle ?, and using
the equations simplify to
Last term of the second equation
infinitestimal of higher of smallness, may be
neglected

When divided by dy and dx, respectively, the
equations can be combined
? is any arbitrary angle ? this equation proves
that the pressure is the same in all directions
at a point in a static fluid
Although the proof was carried out for a
two-dimensional case, it may be demonstrated for
the three-dimensional case with the equilibrium
equations for a small tetrahedron of fluid with
three faces in the coordinate planes and the
fourth face inclined arbitrarily.
If the fluid is in motion (one layer moves
relative lo an adjacent layer), shear stresses
occur and the normal stresses are no longer the
same in all directions at a point ? the pressure
is defined as the average of any three mutually
perpendicular normal compressive stresses at a
point,
Fictitious fluid of zero viscosity (frictionless
fluid) no shear stresses can occur ? at a point
the pressure is the same in all directions

7
2.2 BASIC EQUATION OF FLUID STATICS

Pressure Variation in a Static Fluid
Force balance
The forces acting on an element of fluid at rest
(Fig. 2.2) surface forces and body forces.
With gravity the only body force acting, and by
taking the y axis vertically upward, it is -? dx
dy dz in the y direction
With pressure p at its center (x, y, z) the
approximate force exerted on the side normal to
the y axis closest to the origin and the opposite
e side are approximately
dy/2 the distance from center to a face normal
to y

8
Figure 2.2 Rectangular parallelepiped element of
fluid at rest
9

Summing the forces acting on the element in the y
direction
For the x and z directions, since no body forces
act,
The elemental force vector dF
If the element is reduced to zero size, alter
dividing through by dx dy dz dV, the expression
becomes exact.
This is the resultant force per unit volume at a
point, which must be equated to zero for a fluid
at rest.
The gradient ? is

-?p is the vector field f or the surface pressure
force per unit volume
The fluid static law or variation of pressure is
then
For an inviscid fluid in motion, or a fluid so
moving that the shear stress is everywhere zero,
Newton's second law takes the form
a is the acceleration of the fluid element, f -
j? is the resultant fluid force when gravity is
the only body force acting

In component form, Eq. (2.2.4) becomes
The partials, for variation in horizontal
directions, are one form of Pascal's law they
state that two points at the same elevation in
the same continuous mass or fluid at rest have
the same pressure.
Since p is a function of y only,
relates the change of pressure to unit gravity
force and change of elevation and holds for both
compressible and incompressible fluids
For fluids that may be considered homogeneous and
incompressible, ? is constant, and the above
equation, when integrated, becomes
in which c is the constant of integration. The
hydrostatic law of variation of pressure is
frequently written in the form
h -y, p is the increase in pressure from that
at the free surface

Example 2.1 An oceanographer is to design a sea
lab 5 m high to withstand submersion to 100 m,
measured from sea level to the top of the sea
lab. Find the pressure variation on a side of the
container and the pressure on the top if the
relative density of salt water is 1.020.
At the top, h 100 m, and
If y is measured from the top of the sea lab
downward, the pressure variation is

Pressure Variation in a Compressible Fluid
When the fluid is a perfect gas at rest at
constant temperature
When the value of ? in Eq. (2.2.7) is replaced by
?g and ? is eliminated between Eqs. (2.2.7) and
(2.2.9),
If P P0 when ? ?0, integration between limits
- the equation for variation of pressure with
elevation in an isothermal gas
- constant temperature gradient of atmosphere
?

Example 2.2 Assuming isothermal conditions to
prevail in the atmosphere, compute the pressure
and density at 2000 m elevation if P 105Pa, ?
1.24 kg/m3 at sea level.
From Eq. (2.2.12)
Then, from Eq. (2.2.9)

15
2.3 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any
arbitrary datum
absolute zero
local atmospheric pressure
Absolute pressure difference between its value
and a complete vacuum
Gage pressure difference between its value and
the local atmospheric pressure

16
Figure 2.3 Bourdon gage.
17

The bourdon gage (Fig. 2.3) typical of the
devices used for measuring gage pressures
pressure element is a hollow, curved, flat
metallic tube closed at one end the other end is
connected to the pressure to be measured
when the internal pressure is increased, the tube
tends to straighten, pulling on a linkage to
which is attached a pointer and causing the
pointer to move
the dial reads zero when the inside and outside
of the tube are at the same pressure, regardless
of its particular value
the gage measures pressure relative to the
pressure of the medium surrounding the tube,
which is the local atmosphere

18
Figure 2.4 Units and scales for pressure
measurement
19

Figure 2.4 the data and the relations of the
common units of pressure measurement
Standard atmospheric pressure is the mean
pressure at sea level, 760 mm Hg.
A pressure expressed in terms of the length of a
column of liquid is equivalent to the force per
unit area at the base of the column. The
relation for variation of pressure with altitude
in a liquid p ?h Eq. (2.2.8) (p is in
pascals, ? in newtons per cubic metre, and h in
metres)
With the unit gravity force of any liquid
expressed as its relative density S times the
unit gravity force of water
Water ? may be taken as 9806 N/m3.

Local atmospheric pressure is measured by
mercury barometer
aneroid barometer (measures the difference in
pressure between the atmosphere and an evacuated
box or tube in a manner analogous to the bourdon
gage except that the tube is evacuated and
sealed)
Mercury barometer glass tube closed at one end,
filled with mercury, and inverted so that the
open end is submerged in mercury.
It has a scale the height of column R can be
determined
The space above the mercury contains mercury
vapor. If the pressure of the mercury vapor hv is
given in millimetres of mercury and R is measured
in the same units, the pressure at A may be
expressed as (mm Hg)

Figure 2.5 Mercury barometer
21

Figure 2.4 a pressure may be located vertically
on the chart, which indicates its relation to
absolute zero and to local atmospheric pressure.
If the point is below the local-atmospheric-pressu
re line and is referred to gage datum, it is
called negative, suction, or vacuum.
Example the pressure 460 mm Hg abs, as at 1,
with barometer reading 720 mm, may be expressed
as -260 mm Hg, 260 mm Hg suction, or 260 mm Hg
vacuum.
Note
Pabs pbar pgage
Absolute pressures P, gage pressures p.

Example 2.3 The rate of temperature change in
the atmosphere with change in elevation is called
its lapse rate. The motion of a parcel of air
depends on the density of the parcel relative to
the density of the surrounding (ambient) air.
However, as the parcel ascends through the
atmosphere, the air pressure decreases, the
parcel expands, and its temperature decreases at
a rate known as the dry adiabatic lapse rate. A
firm wants lo burn a large quantity of refuse. It
is estimated that the temperature of the smoke
plume at 10 m above the ground will be 11oC
greater than that of the ambient air. For the
following conditions determine what will happen
to the smoke.
(a) At standard atmospheric lapse rate ß
-0.00651oC per meter and t0 20oC.
(b) At an inverted lapse rate ß 0.00365oC per
meter.

By combining Eqs. (2.2.7) and (2.2.14),
The relation between pressure and temperature
for a mass of gas expanding without heat transfer
(isentropic relation, Sec. 6.1) is
in which T1 is the initial smoke absolute
temperature and P0 the initial absolute pressure
k is the specific heat ratio, 1.4 for air and
other diatomic gases.
Eliminating P/P0 in the last two equations
Since the gas will rise until its temperature is
equal to the ambient temperature,
the last two equations may be solved for y. Let
Then
For ß -0.00651oC per metre, R 287
m?N/(kg?K), a 2.002, and y 3201 m. For the
atmospheric temperature inversion ß -0.00365oC
per metre, a -0.2721, and y 809.2 m.

24
2.4 MANOMETERS

Manometers are devices that employ liquid columns
for determining differences in pressure.
Figure 2.6a the most elementary manometer
piezometer
It measures the pressure in a liquid when it is
above zero gage
Glass tube is mounted vertically so that it is
connected to the space within the container
Liquid rises in the tube until equilibrium is
reached
The pressure is then given by the vertical
distance h from the meniscus (liquid surface) to
the point where the pressure is to be measured,
expressed in units of length of the liquid in the
container.
Piezometer would not work for negative gage
pressures, because air would flow into the
container through the tube

25
Figure 2.6 Simple manometers.
26

Figure 2.6b for small negative or positive gage
pressures in a liquid
With this arrangement the meniscus may come to
rest below A, as shown. Since the pressure at the
meniscus is zero gage and since pressure
decreases with elevation,
units of length H2O
Figure 2.6c for greater negative or positive
gage pressures (a second liquid of greater
relative density employed)
It must be immiscible in the first fluid, which
may now be a gas
If the relative density of the fluid at A is S1
(based on water) and the relative density of the
manometer liquid is S2, the equation for pressure
at A
hA - the unknown pressure, expressed in length
units of water,
h1, h2 - in length units

A general procedure in working all manometer
problems
Start at one end (or any meniscus if the circuit
is continuous) and write the pressure there in an
appropriate unit (say pascals) or in an
appropriate symbol if it is unknown.
Add to this the change in pressure, in the same
unit, from one meniscus to the next (plus if the
next meniscus is lower, minus if higher). (For
pascals this is the product of the difference in
elevation in metres and the unit gravity force of
the fluid in newtons per cubic metre.)
Continue until the other end of the gage (or the
starting meniscus) is reached and equate the
expression to the pressure at that point, known
or unknown.
The expression will contain one unknown for a
simple manometer or will give a difference in
pressures for the differential manometer. In
equation form,

28
Figure 2.7 Differential manometers
29

A differential manometer (Fig. 2.7) determines
the difference in pressures at two points A and B
when the actual pressure at any point in the
system cannot be determined
Application of the procedure outlined above to
Fig. 2.7a produces
For Fig. 2.7b
If the pressures at A and B are expressed in
length of the water column, the above results can
be written, for Fig. 2.7a,
For Fig 2.7b

Example 2.4 In Fig. 2.7a the liquids at A and B
are water and the manometer liquid is oil. S
0.80 h1 300 mm h2 200 mm and h3 600 mm.
(a) Determine pA - pB, in pacals.
(b) If pB 50kPa and the barometer reading is
730 mm Hg, find the pressure at A, in meters of
water absolute.
(a)
(b)
(a)?

Micromanometers
For determining very small differences in
pressure or determining large pressure
differences precisely several types of
manometers
One type very accurately measures the differences
in elevation of two menisci of a manometer.
By means of small telescopes with horizontal
cross hairs mounted along the tubes on a rack
which is raised and lowered by a pinion and slow
motion screw so that the cross hairs can be set
accurately, the difference in elevation of
menisci (the gage difference) can be read with
verniers.

32
Figure 2.8 Micromanometer using two gage liquids
33

Fig. 2.8 two gage liquids, immiscible in each
other and in the fluid to be measured ? a large
gage difference R can be produced for a small
pressure difference.
The heavier gage liquid fills the lower U tube up
to 0-0 then the lighter gage liquid is added to
both sides, filling the larger reservoirs up to
1-1.
The gas or liquid in the system fills the space
above 1-1. When the pressure at C is slightly
greater than at D, the menisci move as indicated
in Fig. 2.8.
The volume of liquid displaced in each reservoir
equals the displacement in the U tube ?
Manometer equation
?1, ?2 and ?3 are the unit gravity force

Example 2.5 In the micromanometer of Fig 2.8
the pressure difference is wanted, in pascals,
when air is in the system, S2 1.0, S3 1.10,
a/A 0.01, R 5 mm, t 20oC, and the barometer
reads 760 mm Hg.
The term ?1(a/A) may be neglected. Substituting
into Eq. (2.4.1) gives

35
Figure 2.9 Inclined manometer

The inclined manometer frequently used for
measuring small differences in gas pressures.
Adjusted to read zero, by moving the inclined
scale, when A and B are open. Since the inclined
tube requires a greater displacement of the
meniscus for given pressure difference than a
vertical tube, it affords greater accuracy in
reading the scale.
Surface tension causes a capillary rise in small
tubes. If a U tube is used with a meniscus in
each leg, the surface-tension effects cancel.

36
2.5 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure
throughout a fluid have been considered.
The distributed forces resulting from the action
of fluid on a finite area may be conveniently
replaced by a resultant force, insofar as
external reactions to the force system are
concerned.
In this section the magnitude of resultant force
and its line of action (pressure center) are
determined by integration, by formula, and by use
of the concept of the pressure prism.

37
Horizontal Surfaces

A plane surface in a horizontal position in a
fluid at rest is subjected to a constant
pressure.
The magnitude of the force acting on one side of
the surface is
The elemental forces pdA acting on A are all
parallel and in the same sense ? a scalar
summation of all such elements yields the
magnitude of the resultant force. Its direction
is normal to the surface and toward the surface
if p is positive.
Fig. 2.10 arbitrary xy axes - to find the line
of action of the resultant, i.e., the point in
the area where the moment of the distributed
force about any axis through the point is zero,
Then, since the moment of the resultant must
equal the moment of the distributed force system
about any axis, say the y axis,
x the distance from the y axis to the
resultant

38
Figure 2.10 Notation for determining the line
of action of a force
39
Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis, for example, x
k, the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

40
Figure A.1 Notation for first and second moments
41
(2) Second moment

The second moment of an area A (the moment of
inertia of the area)

The moment about a parallel axis, for example, x
k, the moment

Figure A.2 Moments of inertia of simple areas
about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes
parallel to the xy axes.
43
Inclined Surfaces

Fig. 2.11 a plane surface is indicated by its
trace A'Bit is inclined ?o from the horizontal.
x axis intersection of the plane of the area and
the free surface. y axis taken in the plane of
the area, with origin O in the free surface. The
xy plane portrays the arbitrary inclined area.
The magnitude, direction, and line of action of
the resultant force due to the liquid, acting on
one side of the area, are sought.
For dA
Since all such elemental forces are parallel, the
integral over the area yields the magnitude of
force F, acting on one side of the area,
Magnitude of force exerted on one side of a plane
area submerged in a liquid is the product of the
area and the pressure at its centroid
The presence of a free surface is unnecessary

44
Figure 2.11 Notation for force of liquid on one
side of a plane inclined area.
45
Center of Pressure

Fig. 2.11 the line of action of the resultant
force has its piercing point in the surface at a
point called the pressure center, with
coordinates (xp, yp). Center of pressure of an
inclined surface is not at the centroid. To find
the pressure center, the moments of the resultant
xpF, ypF are equated to the moment of the
distributed forces about the y axis and x axis,
respectively ?
- may be evaluated conveniently through
graphical integration, for simple areas they may
be transformed into general formulas

When either of the centroidal axes is an axis of
symmetry for the surface, vanishes and
the pressure center lies on x x- . Since
may be either positive or negative, the
pressure center may lie on either side of the
line x x-. To determine yp by formula, with
Eqs. (2.5.2) and (2.5.6)
In the parallel-axis theorem for moments of
inertia
in which IG is the second moment or the area
about its horizontal centroidal axis. If IG is
eliminated from Eq. (2.5.9)

47
Example 2.6 The triangular gate CDE (Fig. 2.12)
is hinged along CD and is opened by a normal
force P applied at E. It holds oil, relative
density 0.80, above it and is open to the
atmosphere on its lower side Neglecting the
weight of the gate, find (a) the magnitude of
force exerted on the gate by integration and by
Eq. (2.5.2) (b) the location of pressure center
(c) the force P needed to open the gate.
Figure 2.12 Triangular gate
48

(a) By integration with reference to Fig. 2.12,
When y 4, x 0, and when y 6.5, x 3, with
x varying linearly with y thus
in which the coordinates have been substituted
to find x in terms of y. Solving for a and b
gives
Similarly, y 6.5, x 3 y 9, x 0 and x
6/5(9 - y). Hence,
Integrating and substituting for ?sin? leads to
By Eq. (2.5.2)

(b) With the axes as shown,
In Eq. (2.5.8)
I-xyis zero owing to symmetry about the
centroidal axis parallel to the x axis hence
In Eq. (2.5.11),
i.e., the pressure center is 0.16 m below the
centroid, measured in the plane of the area.
(c) When moments about CD are taken and the
action of the oil is replaced by the resultant,

50
The Pressure Prism

Pressure prism another approach to determine the
resultant force and line of action of the force
on a plane surface - prismatic volume with its
base the given surface area and with altitude at
any point of the base given by p ?h. h is the
vertical distance to the free surface, Fig. 2.13.
(An imaginary free surface may be used to define
h if no real free surface exists.) (in the
figure, ?h may be laid off to any convenient
scale such that its trace is OM)
The force acting on an elemental area dA is
(2.5.12)
- an element of volume of the pressure prism.
After integrating, F ?
From Eqs. (2.5.5) and (2.5.6),
(2.5.13)
? xp, yp are distances to the centroid of the
pressure prism ? the line of action of the
resultant passes through the centroid of the
pressure prism

51
Figure 2.13 Pressure prism
52
Effects of Atmospheric Pressure on Forces on
Plane Areas

In the discussion of pressure forces the pressure
datum was not mentioned p ?h ? the datum
taken was gage pressure zero, or the local
atmospheric pressure
When the opposite side of the surface is open to
the atmosphere, a force is exerted on it by the
atmosphere equal to the product of the
atmospheric pressure P0 and the area, or P0A ,
based on absolute zero as datum. On the liquid
side the force is
The effect P0A of the atmosphere acts equally on
both sides and in no way contributes to the
resultant force or its location
So long as the same pressure datum is selected
for all sides of a free body, the resultant and
moment can be determined by constructing a free
surface at pressure zero on this datum and using
the above methods

Example 2.8 An application of pressure forces on
plane areas is given in the design of a gravity
dam. The maximum and minimum compressive stresses
in the base of the dam are computed from the
forces which act on the dam. Figure 2.15 shows a
cross section through a concrete dam where the
unit gravity force of concrete has been taken as
2.5? and ? is the unit gravity force of water. A
1 m section of dam is considered as a free body
the forces are due to the concrete, the water,
the foundation pressure, and the hydrostatic
uplift. Determining amount of hydrostatic uplift
is beyond the scope of this treatment. but it
will be assumed to be one-half the hydrostatic
head at the upstream edge, decreasing linearly to
zero at the downstream edge of the dam. Enough
friction or shear stress must be developed at the
base of the dam to balance the thrust due to the
water that is Rx 5000?. The resultant upward
force on the base equals the gravity force of the
dam less the hydrostatic uplift Ry 6750?
2625? - 1750? 7625? N. The position of Ry is
such that the free body is in equilibrium. For
moments around O,

54
Figure 2.15 Concrete gravity dam
55

It is customary to assume that the foundation
pressure varies linearly over the base of the
dam, i.e., that the pressure prism is a trapezoid
with a volume equal to Ry thus
in which smax, smin are the maximum and minimum
compressive stresses in pascals. The centroid of
the pressure prism is at the point where x 44.8
m. By taking moments about 0 to express the
position of the centroid in terms of smax and
smin,
Simplifying gives
When the resultant falls within the middle third
of the base of the dam, smin will always be a
compressive stress. Owing to the poor tensile
properties of concrete, good design requires the
resultant to fall within the middle third of the
base.

56
2.6 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p dA vary in direction,
as in the case of a curved surface, they must be
added as vector quantities
their components in three mutually perpendicular
directions are added as scalars, and then the
three components are added vectorially.
With two horizontal components at right angles
and with the vertical component - all easily
computed for a curved surface - the resultant can
be determined.
The lines of action of the components also are
readily determined.

57
Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a
curved surface is equal to the pressure force
exerted on a projection of the curved surface.
The vertical plane of projection is normal to the
direction of the component.
Fig. 2.16 the surface represents any
three-dimensional surface, and dA an element of
its area, its normal making the angle ? with the
negative x direction. Then
Projecting each element on a plane perpendicular
to x is equivalent to projecting the curved
surface as a whole onto the vertical plane
?force acting on this projection of the curved
surface is the horizontal component of force
exerted on the curved surface in the direction
normal to the plane of projection.
To find the horizontal component at right angles
to the x direction, the curved surface is
projected onto a vertical plane parallel to x and
the force on the projection is determined.

58
Figure 2.16 Horizontal component of force on a
curved surface
Figure 2.17 Projections of area elements on
opposite sides of a body
59

When looking for the horizontal component of
pressure force on a closed body, the projection
of the curved surface on a vertical plane is
always zero, since on opposite sides of the body
the area-element projections have opposite signs
(Fig. 2.17).
Let a small cylinder of cross section dA with
axis parallel to x intersect the closed body at B
and C. If the element of area of the body cut by
the prism at B is dAB and at C is dAC, then
and similarly for all other area elements
To find the line of action of a horizontal
component of force on a curved surface, the
resultant of the parallel force system composed
of the force components from each area element is
required. This is exactly the resultant of the
force on the projected area, since the two force
systems have an identical distribution of
elemental horizontal force components. Hence, the
pressure center is located on the projected area
by the methods of Sec. 2.5.

Example 2.9 The equation of an ellipsoid of
revolution submerged in water is x2/4 y2/4
z2/9 1. The center of the body is located 2 m
below the free surface. Find the horizontal force
components acting on the curved surface that is
located in the first octant. Consider the xz
plane to be horizontal and y to be positive
upward.
Projection of the surface on the yz plane has an
area of
Its centroid is located m below the free
surface ?

61
Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a
curved surface is equal to the weight surface and
extending up to the free surface
Can be determined by summing up the vertical
components of pressure force on elemental areas
dA of the surface
In Fig.2.18 an area element is shown with the
force p dA acting normal to it. Let ? be the
angle the normal to the area element makes with
the vertical. Then the vertical component of
force acting on the area element is p cos ? dA,
and the vertical component of force on the curved
surface is given by
(2.6.2)
p replaced by its equivalent ?h cos ? dA is the
projection of dA on a horizontal plane ? Eq.
(2.6.2)
(2.5.3-4)
in which d? is the volume of the prism of height
h and base cos ? dA, or the volume of liquid
vertically above the area element

62
Figure 2.18 Vertical component of force on a
curved surface
Figure 2.19 Liquid with equivalent free surface
63

Fig. 2.19 the liquid is below the curved surface
and the pressure magnitude is known at some point
(e.g., O), an imaginary or equivalent free
surface s-s can be constructed p/? above O, so
that the product of unit gravity force and
vertical distance to any point in the tank is the
pressure at the point.
The weight of the imaginary volume of liquid
vertically above the curved surface is then the
vertical component of pressure force on the
curved surface.
In constructing an imaginary free surface, the
imaginary liquid must be of the same unit gravity
force as the liquid in contact with the curved
surface otherwise, the pressure distribution
over the surface will not be correctly
represented.
With an imaginary liquid above a surface, the
pressure at a point on the curved surface is
equal on both sides, but the elemental force
components in the vertical direction are opposite
in sign ? the direction of the vertical force
component is reversed when an imaginary fluid is
above the surface.
In some cases a confined liquid may be above the
curved surface, and an imaginary liquid must be
added (or subtracted) to determine the free
surface.

The line of action of the vertical component is
determined by equating moments of the elemental
vertical components about a convenient axis with
the moment of the resultant force. With the axis
at O (Fig.2.18),
in which is the distance from 0 to the
line of action
Since Fv ??
the distance to the centroid of the volume
? the line of action of the vertical force passes
through the centroid of the volume, real or
imaginary, that extends above the curved surface
up to the real or imaginary free surface

Example 2.10 A cylindrical barrier (Fig. 2.20)
holds water as shown. The contact between
cylinder and wall is smooth. Considering a 1-m
length of cylinder, determine (a) its gravity
force and (b) the force exerted against the wall.
(a) For equilibrium the weight of the cylinder
must equal the vertical component of force
exerted on it by the water. (The imaginary free
surface for CD is at elevation A.) The vertical
force on BCD is
The vertical force on AB is
Hence, the gravity force per metre of length is
(b) The force exerted against the wall is the
horizontal force on ABC minus the horizontal
force on CD. The horizontal components of force
on BC and CD cancel the projection of BCD on a
vertical plane is zero ?,
since the projected area is 2 m2 and the
pressure at the centroid of the projected area is
9806 Pa.

66
Figure 2.20 Semifloating body
67
Tensile Stress in a Pipe and Spherical Shell

Fig. 2.21 a circular pipe under the action of an
internal pressure is in tension around its
periphery assuming that no longitudinal stress
occurs, the walls are in tension
Consider a section of pipe of unit length (the
ring between two planes normal to the axis and
unit length apart). If one-half of this ring is
taken as a free body, the tensions per unit
length at top and bottom are respectively T1 and
T2
The horizontal component of force acts through
the pressure center of the projected area and is
2pr, in which p is the pressure at the centerline
and r is the internal pipe radius.
For high pressures the pressure center may be
taken at the pipe center then T1 T2, and
T is the tensile force per unit length. For wall
thickness e, the tensile stress in the pipe wall
is

68
Figure 2.21 Tensile stress in pipe
69

Example 2.11 A 100 mm-ID steel pipe has a 6 mm
wall thickness. For an allowable tensile stress
of 70 MPa, what is the maximum pressure?
From Eq. (2.6.6)

70
2.7 BUOYANT FORCE

Buoyant force the resultant force exerted on a
body by a static fluid in which it is submerged
or floating
Always acts vertically upward (there can be no
horizontal component of the resultant because the
projection of the submerged body or submerged
portion of the floating body on a vertical plane
is always zero)
The buoyant force on a submerged body is the
difference between the vertical component of
pressure force on its underside and the vertical
component of pressure force on its upper side
Figure 2.22

71
Figure 2.22 Buoyant force on floating and
submerged bodies
72

Fig. 2.22 the upward force on the bottom is
equal to the gravity force of liquid, real or
imaginary, which is vertically above the surface
ABC, indicated by the gravity force of liquid
within ABCEFA. The downward force on the upper
surface equals the gravity force of liquid
ADCEFA. The difference between the two forces is
a force, vertically upward, due to the gravity
force of fluid ABCD that is displaced by the
solid. In equation form
FB is buoyant force, V is the volume of fluid
displaced, and ? is the unit gravity force of
fluid
The same formula holds for floating bodies when V
is taken as the volume of liquid displaced

Fig.2.23 the vertical force exerted on an
element of the body in the form of a vertical
prism of cross section dA is
dV is the volume of the prism. Integrating over
the complete body gives
? is considered constant throughout the volume
To find the line of action of the buoyant force,
moments are taken about a convenient axis O and
are equated to the moment of the resultant, thus,
i is the distance from the axis to the line of
action.
This equation yields the distance to the centroid
of the volume ? the buoyant force acts through
the centroid of the displaced volume of fluid
this holds for both submerged and floating
bodies.
The centroid of the displaced volume of fluid is
called the center of buoyancy.

74
Figure 2.23 Vertical force components on element
of body
75

Determining gravity force on an odd-shaped object
suspended in two different fluids yields
sufficient data to determine its gravity force,
volume, unit gravity force, and relative density.
Figure 2.24 two free-body diagrams for the same
object suspended and gravity force determined in
two fluids, F1 and F2 ?1 and ?2 are the unit
gravity forces of the fluids. W and V, the
gravity force and volume of the object, are to be
found.
The equations of equilibrium are written and
solved

76
Figure 2.24 Free body diagrams for body
suspended in a fluid
77

A hydrometer uses the principle of buoyant force
to determine relative densities of liquids
Figure 2.25 a hydrometer in two liquids with a
stem of prismatic cross section a
Considering the liquid on the left to be
distilled water (unit relative density S 1.00),
the hydrometer floats in equilibrium when
V0 is the volume submerged, ? is the unit
gravity force of water, and W is the gravity
force of hydrometer
The position of the liquid surface is marked 1.00
on the stem to indicate unit relative density S.
When the hydrometer is floated in another 1iquid,
the equation of equilibrium becomes
where ?V a?h. Solving for ?h with Eqs. (2.7.2)
and (2.7.3)

78
Figure 2.25 Hydrometer in water and in liquid of
relative density
79

Example 2.12 A piece of ore having a gravity
force of 1.5 N in air is found to have a gravity
force 1.1 N when submerged in water. What is its
volume, in cubic centimetres, and what is its
relative density?
The buoyant force due to air may be neglected.
From Fig. 2.24

80
2.8 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical
stability.
A small upward displacement decreases the volume
of liquid displaced ? an unbalanced downward
force which tends to return the body to its
original position.
Similarly, a small downward displacement results
in a greater buoyant force, which causes an
unbalanced upward force.
A body has linear stability when a small linear
displacement in any direction sets up restoring
forces tending to return it to its original
position.
A body has rotational stability when a restoring
couple is set up by any small angular
displacement.

Methods for determining rotational stability are
developed in the following discussion
A body may float in
stable equilibrium
unstable equilibrium (any small angular
displacement sets up a couple that tends to
increase the angular displacement)
neutral equilibrium (any small angular
displacement sets up no couple whatever)
Figure 2.26 three cases of equilibrium
a light piece of wood with a metal mass at its
bottom is stable
when the metal mass is at the top, the body is in
equilibrium but any slight angular displacement
causes it to assume the position in a
a homogeneous sphere or right-circular cylinder
is in equilibrium for any angular rotation i.e.,
no couple results from an angular displacement

82
Figure 2.26 Examples of stable, unstable, and
neutral equilibrium
83

A completely submerged object is rotationally
stable only when its center of gravity is below
the center of buoyancy (Fig. 2.27a)
When the object is rotated counterclockwise, the
buoyant force and gravity force produce a couple
in the clockwise direction (Fig. 2.27b)

Figure 2.27 Rotationally stable submerged body
84

Normally, when a body is too heavy to float, it
submerges and goes down until it rests on the
bottom.
Although the unit gravity force of a liquid
increases slightly with depth, the higher
pressure tends to cause the liquid to compress
the body or to penetrate into pores of solid
substances and thus decrease the buoyancy of the
body
Example a ship is sure to go to the bottom once
it is completely submerged, owing to compression
of air trapped in its various parts

85
Determination of Rotational Stability of Floating
Objects

Any floating object with center of gravity below
its center of buoyancy (centroid of displaced
volume) floats in stable equilibrium (Fig.
2.26a). Certain floating objects, however, are in
stable equilibrium when their center of gravity
is above the center of buoyancy.
Figure 2.28a a cross section of a body with all
other parallel cross sections identical. The
center of buoyancy is always at the centroid of
the displaced volume, which is at the centroid of
the cross-sectional area below liquid surface in
this case.

86
Figure 2.28 Stability of a prismatic body
87

? when the body is tipped (Fig. 2.28b), the
center of buoyancy is at the centroid B' of the
trapezoid ABCD, the buoyant force acts upward
through B', and the gravity force acts downward
through G, the center of gravity of the body
When the vertical through B' intersects the
original centerline above C, as at M, a restoring
couple is produced and the body is in stable
equilibrium
The intersection of the buoyant force and the
centerline is called the metacenter (M)
When M is above G, the body is stable when below
G, it is unstable and when at G, it is in
neutral equilibrium
The distance MG is called the metacentric height
and is a direct measure of the stability of the
body. The restoring couple is
in which ? is the angular displacement and W the
gravity force of the body

Example 2.13 In Fig. 2.28 a scow 6 m wide and 20
m long has a gross mass of 200 Mg. Its center of
gravity is 30 cm above the water surface. Find
the metacentric height and restoring couple when
?y 30 cm.
The depth of submergence h in the water is
The centroid in the tipped position is located
with moments about AB and BC,
By similar triangles AEO and B'PM,

G is 1.97 m from the bottom hence
The scow is stable, since is positive
the righting moment is

90
Nonprismatic Cross Sections

For a floating object of variable cross section
(e.g., a ship) (Fig. 2.39a), a convenient formula
can be developed for determination of metacentric
height for very small angles of rotation ?
The horizontal shift in center of buoyancy r
(Fig. 2.29b) is determined by the change in
buoyant forces due to the wedge being submerged,
which causes an upward force on the left, and by
the other wedge decreasing the buoyant force by
an equal amount ?FB on the right.
The force system, consisting of the original
buoyant force at B and the couple ?FB x s due to
the wedges, must have as resultant the equal
buoyant force at B'. With moments about B to
determine the shirt r,

91
Figure 2.29 Stability relations in a body of
variable cross section
92

The amount of the couple can be determined with
moments about O, the centerline of the body at
the liquid surface
For an element of area dA on the horizontal
section through the body at the liquid surface,
an element of volume of the wedge is x? dA. The
buoyant force due to this element is ?x? dA, and
its moment about O is ?x2? dA, in which ? is the
small angle of tip in radians.
By integrating over the complete original
horizontal area at the liquid surface, the couple
is determined to be
I is the moment of inertia of the area about the
axis y-y (Fig.2.29a) Substitution into the above
equation produces
V is the total volume of liquid displaced
Since ? is very small
and

Example 2.14 A barge displacing 1 Gg has the
horizontal cross section at the waterline shown
in Fig. 2.30. Its center of buoyancy is 2.0 m
below the water surface, and its center of
gravity is 0.5 m below the water surface.
Determine its metacentric height for rolling
(about y-y axis) and for pitching (about x-x
axis).
GB 2 0.5 1.5 m
For rolling
For pitching

94
Figure 2.30 Horizontal cross section of a ship
at the waterline
95
2.9 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses ? the variation
of pressure is simple to compute
For fluid motion such that no layer moves
relative to an adjacent layer, the shear stress
is also zero throughout the fluid
A fluid with a translation at uniform velocity
still follows the laws of static variation of
pressure.
When a fluid is being accelerated so that no
layer moves relative to an adjacent one (when the
fluid moves as if it were a solid), no shear
stresses occur and variation in pressure can be
determined by writing the equation of motion for
an appropriate free body
Two cases are of interest, a uniform linear
acceleration and a uniform rotation about a
vertical axis
When moving thus, the fluid is said to be in
relative equilibrium

96
Uniform Linear Acceleration

Fig. 2.31 a liquid in an open vessel is given a
uniform linear acceleration a
After some time the liquid adjusts to the
acceleration so that it moves as a solid, i.e.,
the distance between any two fluid particles
remains fixed ? no shear stresses occur
By selecting a cartesian coordinate system with y
vertical and x such that the acceleration vector
a is in the xy plane (Fig. 2.31a), the z axis is
normal to a and there is no acceleration
component in that direction
Fig. 2.31b the pressure gradient ?p is then the
vector sum of -?a and -j?
Since ?p is in the direction of maximum change in
p (the gradient), at right angles to ?p there is
no change in p. Surfaces of constant pressure,
including the free surface, must therefore be
normal to ?p

97
Figure 2.31 Acceleration with free surface
98

To obtain a convenient algebraic expression for
variation of p with x, y, and z, that is, p
p(x, y, z), Eq. (2.2.5) is written in component
form
Since p is a function of position (x, y, z), its
total differential is
Substituting for the partial differentials gives
which can be integrated for an incompressible
fluid,

To evaluate the constant of integration c let x
0, y 0, p p0 then c p0 and
When the accelerated incompressible fluid has a
free surface, its equation is given by setting p
0 in the above eq. Solving it for y gives
The lines of constant pressure, p const, have
the slope
and are parallel to the free surface. The y
intercept of the free surface is

100

Example 2.15 The tank in Fig. 2.32 is filled
with oil, relative density 0.8, and accelerated
as shown. There is a small opening in the rank at
A. Determine the pressure at B and C and the
acceleration ax required to make the pressure at
B zero.
By selecting point A as origin and by applying
Eq. (2.9.2) for ay 0
At B, x 1.8 m, y - 1.2 m, and p 2.35 kPa.
Ft C, x -0.15 m, y -1.35 m, and p 11.18
kPa. For zero pressure at B, from Eq. (2.9.2)
with origin at A,

101
Figure 2.32 Tank completely filled with liquid
102

Example 2.16 A closed box with horizontal base 6
by 6 units and a height of 2 units is half-filled
with liquid (Fig. 2.33). It is given a constant
linear acceleration ax g/2, ay -g/4. Develop
an equation for variation of pressure along its
base.
The free surface has the slope
hence, the free surface is located s shown in
the figure. When the origin is taken at 0, Eq.
(2.9.2) becomes
Then, for y 0, along the bottom,

103
Figure 2.33 Uniform linear acceleration of
container
104
Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid, moving
as a solid, about an axis
Every particle of fluid has the same angular
velocity
This motion is to be distinguished from
free-vortex motion, in which each particle moves
in a circular path with a speed varying inversely
as the distance from the center
A liquid in a container, when rotated about a
vertical axis at constant angular velocity, moves
like a solid alter some time interval.
No shear stresses exist in the liquid, and the
only acceleration that occurs is directed
radially inward toward the axis of rotation.
By selecting a coordinate system (Fig. 2.34a)
with the unit vector i in the r direction and j
in the vertical upward direction with y the axis
of rotation, the following equation may be
applied to determine pressure variation
throughout the fluid
(2.2.5)

105
Figure 2.34 Rotation of a fluid about a vertical
axis
106

For constant angular velocity w, any particle of
fluid P has an acceleration w2r directed radially
inward (a -iw2r)
Vector addition of -j? and -?a (Fig. 2.34b)
yields ?p, the pressure gradient. The pressure
does not vary normal to this line at a point ? if
P is taken at the surface, the free surface is
normal to ?p
Expanding Eq. (2.2.5)
k is the unit vector along the z axis (or
tangential direction). Then
p is a function of y and r only
For a liquid (? const) integration yields
c is the constant of integration

107

If the value of pressure at the origin (r 0, y
0) is p0, then c p0 and
When the particular horizontal plane (y 0) for
which p0 0 is selected and the above eq. is
divided by ?,
the head, or vertical depth, varies as the
square of the radius. The surfaces of equal
pressure are paraboloids of revolution.

108

When a free surface occurs in a container that is
being rotated, the fluid volume underneath the
paraboloid of revolution is the original fluid
volume
The shape of the paraboloid depends only upon the
angular velocity with respect to the axis (Fig.
2.35). The rise of liquid from its vertex to the
wall of the cylinder is w2r02/rg (Eq. (2.9.6)),
for a circular cylinder rotating about its axis.
Since a paraboloid of revolution has a volume
equal to one-half its circumscribing cylinder,
the volume of the liquid above the horizontal
plane through the vertex is
When the liquid is at rest, this liquid is also
above the plane through the vertex to a uniform
depth of
Hence, the liquid rises along the walls the same
amount as the center drops, thereby permitting
the vertex to be located when w, r0, and depth
before rotation are given

109
Figure 2.35 Rotation of circular cylinder about
its axis
110

Example 2.17 A liquid, relative density 1.2, is
rotated at 200 rpm about a vertical axis. At one
point A in the fluid 1 m from the axis, the
pressure is 70 kPa. What is the pressure at a
point B which is 2 m higher than A and 1.5 m from
the axis?
When Eq. (2.9.5) is written for the two points,
Then w 200 x 2p/60 20.95 rad/s, ? 1.2 x
9806 11.767 N/m3, rA 1 m, and rB 1.5 m.
When the second equation is subtracted from the
first and the values are substituted,
Hence

111

Example 2.18 A straight tube 2 m long, closed at
the bottom and filled with water, is inclined 30o
with the vertical and rotated about a vortical
axis through its midpoint 6.73 rad/s. Draw the
paraboloid of zero pressure, and determine the
pressure at the bottom and midpoint of the tube.
In Fig. 2.36, the zero-pressure paraboloid passes
through point A. If the origin is taken at the
vertex, that is, p0 0, Eq. (2.9.6) becomes
which locates the vertex at O, 0.577 m below A.
The pressure at the bottom of the tube is or
At the midpoint, .289 m and

112
Figure 2.36 Rotation of inclined tube of liquid
about a vertical axis
113
Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area
in contact with a liquid accelerating as a rigid
body can be obtained by integrating over the
surface
The nature of the acceleration and orientation of
the surface governs the particular variation of p
over the surface
When the pressure varies linearly over the plane
surface (linear acceleration), the magnitude of
force is given by the product of pressure at the
centroid and area, since the volume of the
pressure prism is given by pGA
For nonlinear distributions the magnitude and
line of action can be found by integration.

114
???