Arba-Minch University Institute Of Technology (AMiT)

Department of Hydraulic and Water Resources Eng’g

CHAPTER- THREE
3.1 HYDROSTATICS
INTRODUCTION

Hydrostatics deals with the study of fluids at rest or moving with uniform velocity as a
solid body, so that there is no relative motion between fluid elements (or layers). There is
no shear stress in a fluid at rest. Hence, only normal pressure forces are present in
hydrostatics.

Engineering applications of hydrostatics principles include the study of forces acting on

submerged bodies such as dam faces, gates & others and the analysis of stability of
floating bodies.
OR in other word
Fluid static’s is a branch of hydraulics that deals with fluids (water) at rest. The particles
of fluid are at rest, there is no tangential or shear stress between the fluid particles.
In hydrostatics, all forces act normally to the boundary surface and are independent of
viscosity. The analysis made on hydrostatics is based on straightforward application of
the mechanical principles of forces and moment and exact solution can be obtained
without experimental evidence.
As in solid mechanics we shall build our knowledge by first considering static’s followed
by the more difficult problem of dynamics. Considering Newton’s second law, that is, d
(mv)/dt = 0. This can be achieved either when the fluid velocity is constant or the very
special case where the acceleration is constant everywhere in the flow. The first case is
the case of fluid static’s (the branch of fluid mechanics, which is concerned with fluids at
rest), while the latter is the special case of solid body acceleration. The overriding
assumption necessary to achieve these two conditions is that there is no relative motion of
adjacent fluid layers, and consequently the shear stresses are zero. Therefore, only normal
or pressure forces are considered to be acting on the fluid surfaces.
Fluid Pressure
The pressure intensity or more simply the pressure on a surface is the pressure force per
unit area expressed by the relation P dFbut the force should be applied normal to the
dA
surface.
Pressure at a point
Consider a finite but small element (the small triangular prism) of liquid at rest, acted
upon by the fluid around it. The values of average unit pressures on the three surfaces are
P1, P2 and P3. In the Z direction the forces are equal and opposite and cancel each other.

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Y
Pdsdz (Surface force)
Z
ds

X
Pxdydz
dz dy
900-
dx
Pydxdz

1/2 dxdydz (body force)

Fig.3.1 Definition sketch for normal stress at a point.

Px and Py are the average pressure in the horizontal and vertical directions.

For equilibrium condition,

Fx= 0,
PxdydZ- PdSdZcos  = 0
But ds cos =dy
PxdydZ –PdZdy = 0
 Px=P
Fy =0
Pydxdz-Pdsdz sin -1/2dxdydz=0
ds sin=dx
Pydxdz-Pdxdz -1/2dxdydz=0
Py-P -1/2dy=0 as compared to others dy is small so, 1/2dy is ignored.

 Py=P
The pressure force can also consider and it will be the same with others.
 P=Pz=PX=Py

As the triangular prism approaches a point, dy approaches zero as a limit and the average
pressures become uniform or even “point pressures”. Then putting dy = 0 in equation, we
obtain p1=p3 and hence p1= p2 = p3. Therefore, the pressure is independent of its
orientation.

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3.2 Pressure Distribution PASCAL’s Law

The pressure variation throughout a fluid at rest can be obtained by again applying
Newton’s second law to a differential element such as shown in Fig.3.2. Note that the
pressures shown are all compressive. This, by convention, is defined as positive pressure,
since tensile stresses in fluids are relatively rare. The pressure on the left hand face is
taken as P. If the rate of change of pressure (or pressure gradient) in the x direction is
p/x, then the total change in pressure between the left face and the right face is the rate
of change of pressure times the distance between the two faces, or (p/x) dx.

P(dP)dz
dz
z
dz

P P(dP)dy
dy
y dxdydz dx

x dy

Fig.3.2 Definition sketch for pressure variation

For fluid element at rest FX=0, Fy=0, Fz=0, the pressure force in the opposite vertical
faces must be equal.
Fx  0  pdydz  p p.dxdydz0
 x 
p  0
x
 
Fy 0  pdxdz  p p.dydxdz0
 y 
p  0
y
The preceding two equations show, respectively, that the pressure does not change in the
x and y directions. Thus, the pressure is constant throughout a horizontal plane.

With reference to Fig.3.2 the vertical direction will now be examined.

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Similar to the foregoing procedure, if the pressure on the bottom face is taken as P, the
pressure on the top face becomes p + (P/z) dz.

Fz  pdxdy ppz dz dxdy dxdydz  0

p  
z
It has been shown that p is not a function of x or y. If it is further assumed that the
pressure does not change with time, the relationship may be replaced by the total
differential equation.

dp  
dz
From the above equation the pressure variation is not a function of x and y.
This equation can now be integrated to give the actual pressure variation in the vertical
direction. The negative sign indicates that as z gets higher up ward, the pressure gets
smaller. For incompressible fluids, (where  = constant) the above equation can be
directly used.

If the fluid can be assumed incompressible so that  = constant, this can be integrated to
give
P + z = constant

This expression defines what is often referred to as the hydrostatic pressure variation, in
which the pressure increases linearly with decreasing elevation. The constant of
integration can be absorbed by integrating between two elevations z1 and z2 with
corresponding pressure P1 and P2,

P2

Z
Z2

P1
Z1

Fig.3.3. Pressure relative to the surface of a liquid

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P2 Z2

P   Z
P1 Z1
P2- P1 = - (z2 - z1) Showing pressure decreases linearly with an increase in
elevation.

Since the pressure at the surface is atmospheric it can be taken to be zero gage pressure.
So, the above expression will be P1 = (z2 - z1) But z2-z1=z and substituting,

P1 = z

And the pressure is proportional to the depth below the free surface. In other words, the
pressure at a point in a stationary liquid is the product of the depth of the point and the
specific weight of the fluid. If a free surface does not exist, for example in a closed
container completely filled with liquid, The above equation can be applied in reverse to
determine the position of a line of zero pressure, provided that the actual pressure is
known at some point in the container.
When water fills a containing vessel, it automatically seeks a horizontal surface upon
which the pressure is constant everywhere. In practice, the free surface of water in vessel
is the surface that is not in contact with the cover of the vessel. Such a surface may be
subjected to the atmospheric pressure (open vessel) or any other pressure that is exerted
in the vessel (closed vessel).

N.B: The pressure in a homogeneous, incompressible fluid at rest depends on the depth
of the fluid relative to some reference plane, and it is not influenced by the size or shape
of the container in which the fluid is held.

3.2 Pressure measurement

3.2.1Absolute and gage pressures

The pressure at a point within a fluid mass can be designated as either an absolute pressure
or a gage pressure.
In a region such as outer space, which is virtually void of gases, the pressure is essentially
zero. Such a condition can be approached very nearly in a laboratory when a vacuum pump
is used to evacuate a bottle. The pressure in a vacuum is called absolute zero, and all
pressures referenced with respect to this zero pressure are termed absolute pressures.

Many pressure-measuring devices measure not absolute pressure but only difference in
pressure. For example, a Bourdon-tube gage indicates only the difference between the
pressure in the fluid to which it is tapped and the pressure in the atmosphere. In this case,
then, the reference pressure is actually the atmospheric pressure. This type of pressure
reading is called gage pressure. For example, if a pressure of 50 kPa is measured with a
gage referenced to the atmosphere and the atmospheric pressure is 100 kPa, then the
pressure can be expressed as either p = 50 kPa gage or p = 150 kPa absolute.

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Whenever atmospheric pressure is used as a reference, the possibility exists that the
pressure thus measured can be either positive or negative. Negative gage pressures are
also termed as vacuum or suction pressures. Hence, if a gage tapped into a tank indicates
a vacuum pressure of 31 kPa, this can also be stated as 70 kPa absolute, or -31 kPa gage,
assuming that the atmospheric pressure is 101 kPa absolute.

Water surface in contact with the earth’s atmosphere is subjected to the atmospheric
pressure, which is approximately equal to a 10.33-m- high column at sea level. In still
water, any element located below the water surface is subjected to a pressure greater than
the atmospheric pressure.

Fig.3.4. Graphical representation of gage and absolute pressure.

3.3 Measurement of pressure

Since pressure is a very important characteristic of a fluid field, it is not surprising that
numerous devices and techniques are used in its measurement
All the devices designed for measurement of the intensity of hydraulic pressure are based
on either of the two fundamental principles of measurement of pressure: firstly by
balancing the column of liquid (whose pressure is to be found) by the same or another
column of liquid and secondly by balancing the column of liquid by spring or dead
weight.
3.1. Mercury Barometer
The measurement of atmospheric pressure is usually accomplished with a mercury
barometer, which in its simplest form, consists of a glass tube closed at one end with the
open end immersed in a container of mercury as shown in Fig. The tube is initially filled
with mercury (inverted with its open end up) and then turned upside down (open end down)
with the open end in the container of mercury.

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The column of mercury will come to an equilibrium position where its weight plus the
force due to the vapor pressure (which develops in the space above the column) balances
the force due to the atmospheric pressure. Thus,
Patm = h + Pvapor
Where:  is the specific weight of mercury. For most practical purposes the contribution
of the vapor pressure can be neglected since it is extremely small at room temperatures
(e.g. 0.173 Pa at 20oC).

Pvapor

h
Patm

Fig. 3.5 a) Mercury barometer b) Piezometer tube

3.2. Manometry

A standard technique for measuring pressure involves the use of liquid columns in
vertical or inclined tubes containing one or more liquid of different specific gravities.
Pressure measuring devices based on this technique are called manometers. In using a
manometer, generally a known pressure (which may be atmospheric) is applied to one
end of the manometer tube and the unknown pressure to be determined is applied to the
other end. In some cases, however, the difference between pressures at ends of the
manometer tube is desired rather than the actual pressure at the either end. A manometer
to determine this differential pressure is known as differential pressure manometer.
The mercury barometer is an example of one type of manometer, but there are many
other configurations possible, depending on the particular application. The common types
of manometers include the piezometer tube, the U-tube manometer, micro- manometer
and the inclined - tube manometer.
3.2.1Piezometer Tube
The simplest type of manometer consists of a vertical tube, open at the top, and attached
to the container in which the pressure is desired, as illustrated in Fig.3.5. Since
manometers involve columns of fluids at rest, the fundamental equation describing their
use is the Eq.
P = h + P0

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Which gives the pressure at any elevation within a homogeneous fluid in terms of a
reference pressure p0 and the vertical distance h between p and p0? Remember that in
fluid at rest pressure will increase as we move downward, and will decrease as we move
upward. Application of this equation to the piezometer tube Fig.3.5 indicates that the
pressure PA can be determined by a measurement of h1 through the relationship.

PA = 1h1
Where, 1 is the specific weight of the liquid in the container. Note that since the tube is
open at the top, the pressure Po can be set equal to zero (we are now using gage pressure),
with the height h1 measured from the meniscus at the upper surface to point (1). Since
point (1) and point A within the container are at the same elevation, PA =P1.

Although the piezometer tube is a very simple and accurate pressure-measuring device, it
has several disadvantages. It is only suitable if the pressure in the container is greater than
atmospheric pressure (otherwise air would be sucked into the system), and the pressure to
be measured must be relatively small so that required height of the column is reasonable.
Also, the fluid in the container in which the pressure is to be measured must be a liquid
rather than a gas.

3.3.2.2U- Tube Manometer

To overcome the difficulties noted previously, another type of manometer, which is
widely used, consists of a tube formed into the shape of U as is shown in Fig.3.5. The
fluid in the manometer is called the gage fluid. To measure larger pressure differences
we can choose a manometer with higher density, and to measure smaller pressure
differences with accuracy we can choose a manometer fluid which is having a density
closer to the fluid density.
To find the pressure pa in terms of the various column heights, we can use one of the two
ways of manometer reading techniques:
I) Surface of equal pressure(SEP)
II) Step by step procedure(SS)
a) Start at one end and write the pressure there
b) Add the change in pressure there
+ If next meniscus is lower.
- If next meniscus is higher
c) Continue until the other end of the gage and equate the pressure
at that point
Thus, for the U- tube manometer shown in Fig.3.5, using SS method we will start at point
A and work around to the open end. The pressure at points A and (1) are the same, and as
we move from point (1) to (2) the pressure will increase by 1h1. The pressure at point (2)
is equal to the pressure at point (3), since the pressures at equal elevation in a continuous
mass of fluid at rest must be the same. Note that we could not simply “jump across” from
point (1) to a point at the same elevation in the right – hand tube since these would not be
points within the same continuous mass of fluid. With the pressure at point (3) specified
we now move to the open end where the pressure is zero. As we move vertically upward

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the pressure decreases by an amount 2h2. In equation form these various steps can be
expressed as
PA + 1h1 - 2h2 = 0
And therefore, the pressure PA can be written in terms of the column heights as
PA = 2h2 - 1h1
A major advantage of the U- tube manometer lies in the fact that the gage fluid can be
different from the fluid in the container in which the pressure is to be determined. For
example, the fluid in A in Fig. 3.5b can be either a liquid or a gas. If A does contain a
gas, the contribution of the gas column, 1h1, is almost always negligible so that PA  p2
and in this instance the above Eq. becomes.
PA = 2h2
Thus, for a given pressure the height, h2 is governed by the specific weight, 2, of the
gage fluid used in the manometer. If the pressure PA is large, then a heavy gage fluid,
such as mercury, can be used and a reasonable column height (not too long) can still be
maintained. Alternatively, if the pressure PA is small, a lighter gage fluid, such as water,
can be used so that a relatively large column height (which is easily read) can be
achieved.

Fig.3.5 Simple U-tube and Differential U-tube manometer

The U- tube manometer is also widely used to measure the difference in pressure
between two containers or two points in a given system. Consider a manometer
connected between container A and B as is shown in Fig.3.5. The difference in pressure
between A and B can be found by again starting at one end of the system and working
around to the other end. For example, at A the pressure is PA, which is equal to p1, and as
we move to point (2) pressure increases by 1h1. The pressure at p2 is equal to p3, and as
we move upward to from point (4) to (5) the pressure decreases by 3h3. Finally, P5 = PB,
since they are at equal elevation. Thus,

PA + 1h1 - 2h2 - 3h3 = PB

And the pressure difference is

PA - PB = 2h2 + 3h3 - 1h1
When substituting in numbers, be sure to use a consistent system of units!

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3.3.2.3) Differential U-tube

Inverted U-tube manometer is used for measuring pressure differences in liquids. The
space above the liquid in the manometer is filled with air which can be admitted or
expelled through the tap on the top, in order to adjust the level of the liquid in the
manometer.

Capillarity due to surface tension at the various fluid interfaces in the manometer is
usually not considered, since for a simple U –tube with a meniscus in each leg, the
capillary effects cancel (assuming the surface tension and tube diameters are the same at
each meniscus), or we can make the capillary rise negligible by using relatively large
bore tubes (with diameters of about 0.5 in, or larger). Two common gage fluids are water
and mercury. Both give a well –defined meniscus, a very important characteristic for a
gage fluid, and their properties are well known. Of course, the gage fluid must be
immiscible with respect to the other fluids in contact with it. For highly accurate
measurements, special attention should be given to temperature since the various specific
weights of the fluids in the manometer well vary with temperature.

3.3.2.4 Inclined – tube Manometer

To measure small pressure changes, a manometer of the type shown in Fig. 3.6 is
frequently used. One leg of the manometer is inclined at an angle, and the differential
reading 2 is measured along the inclined tube. The difference in pressure PA – PB can be
expressed as

PA 1h12 22 sin  3h3 PB

Or pA  pB  22 sin 3h3 1h1
Where it is to be noted that the pressure difference between points (1) and (2) is due to
the vertical distance between the points, which can be expressed as 2 sin. Thus, for
relatively small angles the differential reading along the inclined tube can be made large
even for small pressure differences. The inclined- tube manometer is often used to
measure small differences in gas pressures so that if pipes A and B contain a gas then

pA pB  22 sin

Or
2  pA  pB
2 sin
Where the contributions of the gas columns h1 and h3 have been neglected. The above
Equation shows that the differential reading 2 (for a given pressure difference) of the
inclined –tube manometer can be increased over that obtained with a conventional U-tube
manometer by the factor 1/sin. Recall that sin  0 as   0.

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Fig.3.6 Inclined Tube manometer

4. Mechanical and Electronic pressure measuring devices

Although manometers are widely used, they are not well suited for measuring very high
pressures, or pressures that are changing rapidly with time. In addition, they require the
measurement of one or more column heights, which although not particularly difficult,
can be time consuming. To overcome some of these problems numerous other types of
pressure –measuring instruments have been developed. Most of these make use of the
idea that when a pressure acts on an elastic structure the structure will deform, and this
deformation can be related to the magnitude of the pressure. Probably the most familiar
device of this kind is the Bourdon pressure gage, which is shown in Fig.3.7.

The essential mechanical element in this gage is the hollow, elastic curved tube (Bourdon
tube) which is connected to the pressure source as shown in Fig. As the pressure within
the tube increases the tube tends to straighten, and although the deformation is small, it
can be translated into the motion of a pointer on a dial as illustrated. Since it is the
difference in pressure between the outside of the tube (atmospheric pressure) and the
inside of the tube that causes the movement of the tube, the indicated pressure is gage
pressure. The Bourdon gage must be calibrated so that the dial reading can directly
indicate the pressure in suitable units. A zero reading on the gage indicates that the
measured pressure is equal to the local atmospheric pressure. This type of gage can be
used to measure a negative gage pressure (vacuum) as well positive pressure.

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Figure 3.7 Bourdon Gauge

Manometers-advantages and limitations

The manometer in its various forms is an extremely useful type of pressure measuring
instrument, but suffers from a number of limitations.
While it can be adapted to measure very small pressure differences, it cannot be used
conveniently for large pressure differences - although it is possible to connect a number
of manometers in series and to use mercury as the manometric fluid to improve the range.
(Limitation)
A manometer does not have to be calibrated against any standard; the pressure difference
can be calculated from first principles. (Advantage)
Some liquids are unsuitable for use because they do not form well-defined menisci.
Surface tension can also cause errors due to capillary rise; this can be avoided if the
diameters of the tubes are sufficiently large - preferably not less than 15 mm diameter.
(Limitation)
A major disadvantage of the manometer is its slow response, which makes it unsuitable
for measuring fluctuating pressures. (Limitation)
It is essential that the pipes connecting the manometer to the pipe or vessel containing the
liquid under pressure should be filled with this liquid and there should be no air bubbles
in the liquid. (Important point to be kept in mind)

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3.5 Hydrostatic pressure on plane and curved surfaces

When a surface is submerged in a fluid, forces develop on the surface due to the fluid.
The determination of these forces is important in the design of storage tanks, ships,
dams, and other hydraulic structures. For fluid at rest we know that the force must be
perpendicular to the surface since there are no shearing stresses present. We also know
that the pressure will vary linearly with depth if the fluid is incompressible.

1. Forces on plane surface

The distributed forces resulting from the action of fluid on a finite area can be
conveniently replaced by resultant force. The magnitude of resultant force and its line
of action (pressure center) are determined by integration, by formula and by using the
concept of the pressure prism.

i. 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

x
A A PdA=PdA=PA

The elemental forces PdA acting on A are all parallel. The summation of all elements
yields the magnitude of the resultant force. Its direction is normal to the surface.

 To find line of action of the resultant, the moment of resultant is equated to the
moment of the distributed system about any axis (y-axis).

i.e. PAx1=A x PdA

x1 is the distance from the y axis to the resultant.

AA
x1=
1 x dA x p-is constant.

x is the distance to the centroid of the area.

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Hence, for a horizontal area subjected to static fluid pressure, the resultant passes
through the centroid of the area.
ii. Inclined surfaces
A plane surface which is inclined to the water surface may be subjected to hydrostatic
pressure. For a plane inclined  0 from the horizontal, the intersection of the plane of area
and the free surface is taken as the x-axis. The y-axis is taken in the plane of the area with
origin 0 at the free surface. Thus, the x-y plane portrays the arbitrary inclined area. We
wish to determine the magnitude, direction and line of the action of the resultant force
acting on one side of this area due to the liquid in contact with the area.
For an element area A at y distance from the origin, the magnitude of the force F
acting on it is

F = P A = hA = y sin  A
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.

F= PdA sin ydA sin YA hA P .A G

=hc.A

Fig.
hc=yc.sin
Y sin  h andpG  h ; The pressure at the centroid of the area.

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Hence, the 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 point on the plane surface where this resultant force acts is known as the center of
pressure. Considering the plane surface as free body we see that the distributed forces
can be replaced by a single resultant force at the pressure center with out altering any
reactions or moments in the system.

F  dF  PdA
A
Let xp and yp be distances measured from the y-axis and x-axis to the pressure
center respectively, then

F.yp   y. dF, yP  1  ydF

F
But F= sin A Y and dFy sin dA

yp  1 y2 sin dA 1  y2 dA Io

 sinAY AY AY
2
But, Io = AY +Ig

Ig
yp Y 
AY
yp Y 0,b/ c Ig is positive
.
This shows that center of pressure is below the center of gravity (or centroid).

Where Ig-is the moment of inertia of the plane with respect to its own centroid.

xp.F   x.dF  x.ysin.dA

I
Xp  1  xysin.dA 1  x.y.dA xy
YAsin A AY A AY

Ixy X.Y.A Ixyg

I
Xp  xyg  X (Product of inertia at (x, y)).
AY

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3.6 The pressure prism

The pressure prism is an approach, which is developed for determining the resultant
hydrostatic force and line of action of the force on a plane surface. It is a prismatic
volume with its base the given surface area and with altitude at any point of the base
given by p=h. Where h is the vertical distance to the free surface.

Fig. Pressure prism

Force acting on the element area A is:

 F=h A= , which is an element of volume of the pressure prism.
After integrating, F=, the volume of the pressure prism equals the magnitude of the
resultant force acting on one side of the surface. The center of pressure is given by

 
xp =
1 x.d and y  1 y.d.
p

This shows that the resultant force passes through the centroid of the pressure prism.

Therefore; the pressure force is the volume of the prism in magnitude acting at the
centroid of the prism normal to the surface.
2. Forces on curved surfaces
When the elemental forces pA vary in direction, as in the case of a curved surface, they
must be added as vector quantities that is, their components in three mutually
perpendicular directions are added as scalars and then the three components are added
vector ally.

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With two horizontal components at right angle and with 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.
3.7 Horizontal and Vertical component of Forces on a curved surface
The horizontal component of pressure force on a curved surface is equal to the pressure
force exerted on a vertical projection of the cured surface. The vertical plane of the
projection is normal to the direction of the component.
Thus, the magnitude and the line of action of the horizontal component of force on a
curved surface can be determined by using the relations developed for plane surface.
Vertical component of force on a curved surface
The vertical component of pressure force on a curved surface is equal to the weight of
liquid vertically above the curved surface and extending up to the free surface and acts
through the center of gravity of the fluid mass within the volume.

Fig. Forces on curved surface

3.8 Tensile stress in a pipe and spherical shell
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, as shown in Fig.
below.

1 unit
T1 T1
T1 T2
FH
y
T2 T2

Fig. Internal forces on walls of a pipe. A section of pipe of unit length is considered

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The bursting of a pipe can be thought of as a tendency for the top half to separate from
the bottom half. The only force acting against this tendency is the hoop tension (T) of the
pipe walls then the bursting pressure force must exactly equal the hoop tension.

Total bursting pressure = P* 2r * 1

P = pressure at the centre line
r = the internal pipe radius

For high pressures the pressure centre can be taken at the pipe centre; then T1 = T2

 T = Pr
For wall thickness t, the tensile stress in the pipe wall, =
T  pr
t t
For larger variations in pressure b/n top and bottom of pipe, the location of pressure
centre y is computed.

[FH = 0}  T1+ T2 =FH =2pr

[ M @T2 ] 2rT1 -2pry = 0 (Neglecting the vertical component)

 T1 = py T2 = p (2r-y)

Thin spherical shell subjected to an internal pressure

Fluid force FH = Pr2 (considering half of the sphere)
Resisting force = stress in the wall * cut wall area =  * 2 r *t
Neglecting the weight
  = Pr/2t

3.9 Relative Equilibrium

Translation and Rotation of fluid masses
A general class of problems involving fluid motion in which there are no shearing
stresses occur when a mass of fluid undergoes rigid body motion. For example, if a
container of fluid accelerates along a straight path, the fluid will move as a rigid mass
(after the initial sloshing motion has died out) with each particle having the same
acceleration. Since there is no deformation, there will be no shearing stresses and
similarly if a fluid is contained in a tank that rotates about a fixed axis, the fluid will
simply rotate with the tank as a rigid body. In both cases there is no relative motion
between particles; hence no shear stress occurs in the fluid. This condition of fluid is
called relative equilibrium. Generally there is no motion between the fluid and the
containing vessel, however, there is an additional force acting to cause the acceleration.
Specific results for these two cases (rigid body uniform motion and rigid body rotation)
are developed in the following two sections.

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Although problems relating to fluids having rigid body motion are not strictly speaking,
“fluid static” problems, they are induced in this chapter because as we will see the
analysis and resulting pressure relationships are similar to those for fluid at rest. (Laws of
fluid static’s can still be applied by modifying to allow for effects of acceleration.)

i. Uniform linear acceleration

Consider a small rectangular element of fluid of size x, y and z as shown in the figure
below, z being measured perpendicular to the paper.
Constant pressure
planes
ay
y y

ax x P ax

W=xyz
x

Fig. Linear acceleration of a liquid with a free surface

The pressure on the left face of the elemental fluid

P   p 1x
 x  2
 p
and on the right face P 1/ 2x
 x 
For equilibrium in the x –direction

P   p1/ 2xP p1/ 2xyz  xyz a

  x   
 x  
 
x

 pxyz  xyz a
 x  x

p   a (*)
x x

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Similarly in the y direction

 p
 y  xyz  xyz   x y z ay
p   pa  
y y

p     ay 1 (**)
y  g 

dp  p.dxp dy  p dz (az  0)
x y z
a 
dp ax dx   y 1dy
g 
 
P   axx ay1 Y C
g 
If x = y = 0

P = C = P0

 y 1Y  P0
a
P = -  ax x -
g 
If the origin is taken at a point on the free surface, P0 = 0 (= Patm)

  y 1Y
a
Thus, p = -ax x –
g 
 ay1 Y   P  axx
g 

Y  P  axx
  y 1  y 1
a a
g  g 
Y   ax x  p (***)
ay  g
 y 1
a
g 
Y=mx+b

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 The slope of the free surface will be m = -

ax
ay  g
and the y – intercept b P
  y 1
a
g 
Along a free surface the pressure is constant, so that for the accelerating mass shown in
the figure the free surface will be inclined if ax  0. In addition, all lines of constant
pressure will be parallel to the free surface.

For the special circumstance in which ax = 0, ay  0, which corresponds to the mass of

fluid accelerating in the vertical direction, equation (***) indicates that the fluid surface
will be horizontal. However, from Eq (**) we see that the pressure distribution is not
p    ay 1   (a  g)
hydrostatics, but is given by the equation
y  g  y

For fluids of constant density this equation shows that the pressure will vary linearly with
depth but the variation is due to the combined effects of gravity and the externally
induced acceleration,  (g +ay) rather than simply the specific weight g.

Note:
Pressure along the bottom of a liquid filled tank which is resting i.e accelerating
upward will be increased over that which exists when the tank is at rest (or moving
with a constant velocity).
For a freely falling fluid mass (ay = -g) the pressure gradients in all the three
coordinate directions are zero which means that if the pressure surrounding the
mass is zero the pressure throught will be zero.

ii. Vortex flow

The flow of a fluid along a curved path or the flow of a rotating mass of fluid is known as
vortex flow. The vortex flow is of two types namely: forced vortex flow and free vortex
flow.

Forced vortex flow:

It is defined as that type of vortex flow, in which some external torque is required to
rotate the fluid mass. The fluid mass in this type of flow rotates at constant angular
velocity, . The tangential velocity of any fluid particle is given by
=  r
Where r = radius of fluid particle from the axis of rotation.

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Central axis

Liquid

Cylinder stationary Cylinder is rotating

Fig. Forced vortex flow

Hence angular velocity  is given by

    constant
r
Examples of forced vortex are:
1. A vertical cylinder containing liquid which is rotated about its central axis with a
constant angular velocity, as shown in figure above.
2. Flow of liquid inside the impeller of a centrifugal pump.
3. Flow of water through the runner of a turbine.

Free vortex flow

Type of flow when no external torque is required to rotate the fluid mass is called free
vortex flow. Thus the liquid incase of free vortex is rotating due to the rotation which is
imparted to the fluid previously. Examples of the free vortex flow are:
1. Flow of a liquid through a hole provided at the bottom of a container.
2. Flow of liquid around a circular bend in a pipe.
3. Flow of fluid in a centrifugal pump casing.
Free vortex flow will be treated in the coming chapter.

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3.9.1 Equation of forced vortex flow

Uniform rotation about a vertical axis

Consider a small element of fluid to move in a circular path about an axis with radius r,
and angular velocity.

Axis of rotation Axis of rotation

r2
r r1 z2
z1

 

Fig. Rigid body rotation

For constant angular velocity, the particle will have an acceleration of 2r directed
radically inward.

p  2r   wr  F ma
r g r r

P  0

p     F 
y y
These results show that for this type of rigid body rotation, the pressure is a function of
two variables r and y, and therefore the differential pressure is
dp dpdy dpdr
dy dr
 2
dp dy  dr
g
Thisequation givesthevariationof pressure
of arotatingfluid.
p  y  2 r C
2

g 2
For r=0, y=0, P=C=P0

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 r
2 2
P=P0-y +
2g
Consider two points 1 and 2in the fluid as shown above. Integrating the above equation
for point 1 and 2 we get,

2
 2 2
dp  rdr dy
2

1 g
1 1

p2  p1  r2  y121

 2 
2

 2g 1
p2  p1  r22 r12 y2 y1
2

2g

  r22 r12 y2 y1
2g

If the points 1 and 2 lie on the free surface of the liquid, then p 1=p2 and hence the above
equation becomes

2g
 1 2 1 
 r 2 r2 y y =0
2

y2-y1=
2g 2

1 r 2 r2
1 
If the point 1 lies on the axis of rotation, then the above equation becomes
2r22
y2 y1 
2g
Let y2-y1 = y, then
2r22
y=
2g
Thus y varies with the square of r. Hence the equation is an equation of parabola. This
means the free surface of the liquid is a paraboloid.
Note: Volume of paraboloid of revolution is half of the volume of the circumscribing
cylinder.

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3.5 Buoyancy and stability of floating bodies

1. Buoyant force (Resultant fluid force in a body)

The buoyant force on a submerged body is the difference between the vertical
components of pressure force on its underside and the vertical component of pressure
force on its upper side. The buoyant force 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.

PBdA
hB

PDdA
PBdA
hC
O x

PCdA

Fig 3.13 Buoyant force on a submerged body

Assume a vertical cylindrical element of cross- sectional area dA. As dA is small, the
pressure on the exposed ends of the cylinder may be taken as p1 and p2.
Since p2> p1, there will be an upward force (p2 –p1) dA acting on the cylindrical element.

 dFB = (p2 – p1 ) dA = (h2-h1) dA = dv

Where dv = volume of the prism

The entire body may be considered to be made up of small cylindrical elements, then
integrating over the complete body gives

v v
FB  dFB  dv   dv  V
 is assumed constant through out the volume.

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V= Volume of the bod

The basic principle of buoyancy and flotation was fist discovered and stated by
Archimedes over 2200 years ago. Archimedes principle states that the up thrust or the
buoyancy on a body immersed in a fluid is equal to the weight, of the fluid displaced. The
up thrust will act through the center of gravity of the displaced fluid, which is called the
center of buoyancy.

By applying Archimedes’s principle, volumes of irregular solids can be found by

determining the apparent loss of weight when a body is wholly immerse in a liquid of
known specific gravity. Specific gravities of liquids can be determined by observing the
depth of flotation of a hydrometer. Further applications include problems of general
flotation and of naval architectural design.

To find the line of action of the buoyant force, moments are taken about a convenient
axis 0.

V x  x. dv x  The distance from the axis to the line of action.
x  1 x.dv
v
(Centroid of the displaced volume of fluid) i.e. B.
v
A body immersed in two different fluids
Up thrust on body = weight of fluid displaced by the body (Archimedes principle.)

If the body is immersed so that part of its volume V1 is immersed in a fluid of density 1
and the rest of its volume V2 in another immiscible fluid of mass density 2,
Up thrust on upper part, R1 = 1gV1
acting through G1, the centroid of V1,
Up thrust on lower part,R2 = 2gV2
acting through G2, the centroid of V2,
Total up thrust = 1gV1 + 2gV2.
The positions of G1 and G2 are not necessarily on the same vertical line, and the centre of
buoyancy of the whole body is, therefore, not bound to pass through the centroid of the
whole body.

Hydrometers
Precise measurement of the specific weight of a liquid is done by utilising the principle of
buoyancy. The device used for this, the hydrometer, is a glass bulb that is weighted on
one end to make the hydrometer float in a vertical position and has a stem of constant
diameter extending from the other end. The hydrometer is so designed that only the stem
end extends above the liquid surface. Therefore, appreciable vertical movement of the
hydrometer is required to change the buoyant force or displaced volume of the device.
Because the buoyant force (equal to the weight of the hydrometer) must be constant, the
hydrometer will float deeper or shallower depending on the specific weight of the liquid.
Consequently graduation on the stem, corresponding to different depths of submergence

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of the hydrometer, can be made to indicate directly the specific weight or specific gravity
of the liquid being measured.

Consider the following figure

1.0
1.0
h

(Vo-V)S
Distilled Water
Vo

W
W

Fig.3.14 Hydrometer in water and in liquid of specific gravity S

In the distilled water, the hydrometer floats in equilibrium when V0 = W

In which V0 is the volume submerged,  is the specific weight of water, and W is the
weight of the hydrometer. The position of the liquid surface is marked as 1.0 on the stem
to indicate unit specific gravity S. When the hydrometer is floated in another liquid, the
equation of equilibrium becomes
(Vo-V)S = W in which V = ah. Solving for h with the above equations gives

h  Vo S 1
a S
One common use of hydrometers is in checking the state of charge of a car battery. When
a battery is fully charged the specific gravity of the acid in it is about 1.28, and during
discharge this specific gravity falls. The instrument used to check the state of charge is
called a battery tester and it consists of a small hydrometer inside a glass container.
Floating in salt water and in fresh water

Exercise: People find that it is easier to float in salt water than in fresh water. Explain
If an egg is placed in a tall vessel and water is added, the egg remains on the
bottom, but if salt is added and the water is stirred, the egg rises and floats.
Why?

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39.2. Stability of submerged and floating bodies.

3 – Possible conditions of equilibrium of solid body.

1. Stable equilibrium – A small displacement from the equilibrium produces a

righting moment tending to restore the body to the equilibrium position.
2. Unstable equilibrium – A small displacement produces an over turning moment
tending to displace the body further from its equilibrium position
3. Neutral equilibrium - The body remains at rest in any position to which it may be
displaced. No couple.

Fig. 3.14 Conditions of equilibrium

1. Submerged body

*B *G
* B, G
*G *B
B>G G>B G=B

Stable equilibrium (+ve stability) Unstable equilibrium (-ve stability) Neutral

equilibrium (0 stability)

For a submerged body, the centre of buoyancy remains constant. If an object is

fully submerged, whether it is a balloon in air or a submarine in water, it must be
designed that the centre of buoyancy lies some distance above the centre of gravity.

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Exercise
Explain with example why the centre of buoyancy and the centre of gravity are
located at different points for a fully submerged object.

2. Floating body

The following figure shows a solid body floating in equilibrium (weight acts through G &
the buoyancy through B). Both act in the same straight line. When the body is displaced
from its equilibrium, weight continues to act at G. The volume of liquid displaced
remains constant but the shape of this volume will change and the position of its G and B
will move relative to the body.

The point at which the line of action of the buoyant force for the displaced position cuts
the original vertical through the center of gravity of the body G is called metacenter,
designated M stable equilibrium. Metacentric height is the distance GM.

Fig.3.16 Stable equilibrium a) b)

The displaced fluid is rectangular in section (fig. a) but it is triangular in fig.b and the
center of buoyancy moves to B1. As a result F8 and W are not in the same straight line
producing a turning moment WX that is a righting moment.

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Fig.3.17 Unstable equilibrium a b)

Comparing the above figures, it can be seen that:

1. If M lies above G a righting moment is produced, GM is regarded as positive, and

equilibrium is stable.
2. If M lies below G an overturning moment is produced, GM is regarded as
negative, and equilibrium is unstable.
3. If M and G coincide the body is in neutral equilibrium.

Evaluation of Metacentric height

l
c
a c 
FB
a b O
a’ b’
G a
a G
B O
x
B B’ FB b

d
d
Fig.3.18 x –section and plan in upright position

Consider a non –prismatic floating object, such as a ship. Assume an outside force is
applied causing the body to tilt through a small angle . The relative position of the G
remains unchanged but B shifts from B to B’. The volume of fluid displaced is of course
unchanged and in effect a wedge shaped volume of water represented by aoa’ has shifted
across the central axis to bob’. These wedges represent a gain in the buoyant force on the
right side and a corresponding loss in buoyancy on the left side of c-d.

The buoyant force FB acting through B’ may be considered as the resultant of the original
buoyant force through B and the gain & loss of buoyant force.

Taking moment about B, we have

F * BM sin  F *
B B (Moment of resultant =  moment of
components.)

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Consider an element of area dA in plan at a distance x, from O. The buoyant force acting
on this element is  x  dA.  << Small tan   sin   

FB =  * volume =  x  dA

Then FB = xdA (integrated half of the water line)

Moment of this force about O

FB */ 2   x2dA

FB * 2  x2 dA
If the integration is performed over the entire area
A A
FB *  x dA   x2 dA
2

A
x dA Moment of inertia I of a horizontal section of the body taken at the surface of
2

the fluid

FB * I
FB BM sin  FB *
WBM   I
BM  I  I  I
W v V
V = volume of water displaced by the vessel.

GM  BM BG
 I  BG
Metacentric height
V
I
If the G is below B, then GM   BG
V
GM  I  BG
V
Attend the laboratory session for experimental determination of the metacentric height.

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Time of oscillation
Consider a floating body, which is tilted through an angle by an overturning couple as
shown below. Let the overturning couple is suddenly removed. The body will start
oscillating. Thus, the body will be in a state of oscillation as if suspended at the meta-
center M. This is similar to a case of a pendulum. The only force acting on the body is
due to the restoring couple due to the weight w of the body force of buoyancy FB.

Y M


W
W FB
FB

Fig. Tilted floating body

Restoring couple = W GM sin

Angular acceleration of the body, =

d2
dt2
Negative sign has been introduced as the restoring couple tries to decrease the angle .
Torque due to inertia = IY-Y (
d2 )
dt2
But IY-Y = (W/g) K2
Where W=weight of body, K=radius of gyration about Y-Y

Inertia torque = - (W/g) K2 (

d2 )
dt2
Equating the above equations
d2
 2 d2

W GM sin = - (W/g) K2 ( 2 ) or GM sin = - (K /) ( 2 )
dt dt
For small angle , sin  =

GM  = - (K2/g) (
d2 ) or (K2/g) ( d2 ) + GM  = 0
dt2 dt2


d2 + Gg  0
dt2 K2
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This is second-degree differential equation, the solution is

  C1sin GM2g *t  C2 Cos GM2g *t

K K
Where C1 and C2 are constants of integration.
The values of C1 and C2 are obtained from boundary conditions, which are
i) at t=0, =0
ii) at t=(T/2), =0
Where T=time of one complete oscillation
Substituting the first boundary condition, C2=0

Substituting the second boundary condition, we get 0= C1sin

Mg * T
K2 2
But C1 cannot be equal to zero and so the other alternative is

sin GM2g *T  0  sin

K 2
GMg *T  or T  2 K2
K2 2 GMg
This gives the time period of oscillation or rolling of a floating body.

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Hydraulics-1 Lecture Note 2014/2015 Page 33

 Arba-Minch University Institute Of Technology (AMiT)
Department of Hydraulic and Water Resources Eng’g

Hydraulics-1 Lecture Note 2014/2015 Page 34