Download Buoyancy

Buoyancy
In our common experience we know that wooden objects float on water, but a small
needle of iron sinks into water. This means that a fluid exerts an upward force on a body
which is immersed fully or partially in it. The upward force that tends to lift the body is
called the buoyant force, Fb .
The buoyant force acting on floating and submerged objects can be estimated by
employing hydrostatic principle.
With reference to figure(#), consider a fluid element of area dAH . The net upward force
acting on the fluid element is
dFB  ( P2  P1 )dAH
 w  h2  h1  dAH
The total upward buoyant force becomes
FB   W  h2  h1 dAH  W  volume of the body 
This result shows that the buoyant force acting on the object is equal to the weight of the
fluid it displaces.
Center of Buoyancy
The line of action of the buoyant force on the object is called the center of buoyancy. To
find the centre of buoyancy, moments about an axis OO can be taken and equated to the
moment of the resultant forces. The equation gives the distance to the centeroid to the
object volume.
The centeroid of the displaced volume of fluid is the centre of buoyancy, which, is
applicable for both submerged and floating objects. This principle is known as the
Archimedes principle which states:
“A body immersed in a fluid experiences a vertical buoyant force which is equal to the
weight of the fluid displaced by the body and the buoyant force acts upward through the
centroid of the displaced volume”.
Buoyant force in a layered fluid
As shown in figure an object floats at an interface between two immiscible fluids of
density 1 and 2 .
The buoyant force FB is
FB   dFB   1 gdV1    2 gdV2
n
  i g  displaced volume 
1
i
where dV1 and dV2 are the volumes of fluid element submerged in fluid 1 and 2
respectively. The centre of buoyancy can be estimated by summing moments of the
buoyant forces in each fluid volume displaced.
Buoyant force on a floating body
When a body is partially submerged in a liquid, with the remainder in contact with air (as
shown in figure), the buoyant force of the body can also be computed using equation
(…). Since the specific weight of the air (11.8 N / m3 ) is negligible as compared with the
specific weight of the liquid (for example specific weight of water is 9800 kN / m3 ),we
can neglect the weight of displaced air. Hence, equation (….) becomes
FB   g (Displaced volume of the submerged liquid)
= The weight of the liquid displaced by the body.
The buoyant force acts at the centre of the buoyancy which coincides with the centeroid
of the volume of liquid displaced.
Stability
Floating or submerged bodies such as boats, ships etc. are sometime acted upon by
certain external forces. Some of the common external forces are wind and wave action,
pressure due to river current, pressure due to maneuvering a floating object in a curved
path, etc. These external forces cause a small displacement to the body which may
overturn it. If a floating or submerged body, under action of small displacement due to
any external force, is overturn and then capsized, the body is said to be in unstable.
Otherwise, after imposing such a displacement the body restores its original position and
this body is said to be in stable equilibrium. Therefore, in the design of the
floating/submerged bodies the stability analysis is one of major criteria.
Stability of a Submerged body
Consider a body fully submerged in a fluid in the case shown in figure (….) of which the
center of gravity (CG) of the body is below the centre of buoyancy. When a small angular
displacement is applied a moment will generate and restore the body to its original
position; the body is stable.
However if the CG is above the centre of buoyancy an overturning moment rotates the
body away from its original position and thus the body is unstable. Note that as the body
is fully submerged, the shape of the displaced fluid remains the same when the body is
tilted. Therefore the centre of buoyancy in a submerged body remains unchanged.
Stability of a floating body
A body floating in equilibrium ( FB  W ) is displaced through an angular displacement  .
The weight of the fluid W continues to act through G. But the shape of immersed volume
of liquid changes and the centre of buoyancy relative to body moves from B to B1. Since
the buoyant force FB and the weight W are not in the same straight line, a turning
movement proportional to ‘ Wx ’is produced.
In figure (…) the moment is a restoring moment and makes the body stable. In figure
(…) an overturning moment is produced. The point ‘M ‘at which the line of action of the
new buoyant force intersects the original vertical through the CG of the body, is called
the metacentre. The restoring moment
 W .x  W .GM.
Provided  is small; sin   (in radians).
The distance GM is called the metacentric height. We can observe in figure that
Stable equilibrium: when M lies above G, a restoring moment is produced. Metacentric
height GM is positive.
Unstable equilibrium: When M lies below G an overturning moment is produced and the
metacentric height GM is negative.
Natural equilibrium: If M coincides with G neither restoring nor overturning moment is
produced and GM is zero.
Determination of Metacentric Height
Experimental method
The metacentric height of a floating body can be determined in an experimental set up
with a movable load arrangement. Because of the movement of the load, the floating
object is tilted with angle  for its new equilibrium position. The measurement of  is
used to compute the metacentric height by equating the overturning moment and
restoring moment at the new tilted position.
The overturning moment due to the movement of load P for a known distance, x, is
 P.x
The restoring moment is
 W .GM.
For equilibrium in the tilted position, the restoring moment must equal to the overturning
moment. Equating the same yields
P.x  W .GM.
The metacentric height becomes
P .x
GM 
W .
And the true metacentric height is the value of GM as   0 . This may be determined
by plotting a graph between the calculated value of GM for various  values and the
angle  .
(ii) Theoretical method:
For a floating object of known shape such as a ship or boat determination of metacentric
height can be calculated as follows.
The initial equilibrium position of the object has its centre of Buoyancy, B, and the
original water line is AC. When the object is tilted through a small angle  the center of
buoyancy will move to new position B  . As a result, there will be change in the shape of
displaced fluid. In the new position AC is the waterline. The small wedge OCC is
submerged and the wedge OAA is uncovered. Since the vertical equilibrium is not
disturbed, the total weight of fluid displaced remains unchanged.
Weight of wedge OAA = Weight of wedge OCC .
In the waterline plan a small area, da at a distance x from the axis of rotation OO uncover
the volume of the fluid is equal to
DDxda  x da
Integrating over the whole wedge and multiplying by the specific weight w of the liquid,
Weight of wedge
0 AA 

OAA
Similarly,
w xda
OCC 
Weight of wedge

w xda
OCC 
Equating Equations ( ) and ( ),

W

xda  W 
OAA
xda
OCC
 xda  0
in which, this integral represents the first moment of the area of the waterline plane about
OO, therefore the axis OO must pass through the centeroid of the waterline plane.
Computation of the Metacentric Height
Refer to Figure(#), the distance BM is
BM  BB '

The distance BB is calculated by taking moment about the centroidal axis YY  .
BBwv AECCO 

xwdv 
AAECO
The integral


OCC 
xwdv 

xwdv
OAA
xwdv equals to zero, because YY  axis symmetrically divides the
AAECO
submerged portion AAECO .
At a distance x, dv  Lx tan.dx
Substituting it into the above equation gives
BBVAECC O  0 

xLx tan  dx 
OCC 
 tan

xL   x tan  dx
OAA

x 2dA waterline
waterline
 tan IO
Where I0 is the second moment of area of water line plane about OO . Thus,
Distance
BM  BB ' 


Io tan
 .VAECC O
Io
VAECC O
Since,
BM  GM  BG
GM 
Io
Vsubmerged
 BG
Periodic Time of Transverse Oscillation
When an overturning moment which results an angular displacement  to a floating body
is suddenly removed, the floating body may be set in a state of oscillation. This
oscillation behaves as in the same manner as a simple pendulum suspended at metacentre
M.
Only the restoring moment W .GM  sets it in a state of oscillation. So, it is equal to the
rate of change of angular momentum.
W
d 2
W .GM.   KG 2 2
g
dt
Where, K G is the radius of gyration about its axis of rotation, and
d 2
the angular
dt 2
acceleration.
The negative sign indicates the acceleration is in the opposite direction to the
displacement. As it corresponds to simple harmonic motion, the periodic time is
T  2
 2
 2
Displacement
Acceleration

GM    g
KG 2
GM.g
KG 2
From the above equation it can be observed that a large metacentric height gives higher
stability to a floating object. However it reduces the time period of oscillation which may
cause discomfort for passengers in a passenger ship.
Some typical metacentric heights of various floating vessels are given below
Ocean going vessels: 0.3m to 1.2m.
War ship: 1m to 1.5m.
River crafts: > 3.6m.
Liquids in Rigid Body Motion
Many liquids such as water, milk and oil are transported in tankers. When a tanker is
being accelerated at constant rate, the liquid within the tanker starts splashing. After that
a new free surface is formed, each liquid particle moves with same acceleration. At this
equilibrium stage the liquid moves as if it were a solid. Since there is no relative motion
between liquid particles the shear stress is zero throughout the liquid. At this equilibrium
it is said to be liquid in rigid body motion.
Uniform linear acceleration
A liquid in a vessel is subjected to a uniform linear acceleration, a as discussed in
previous section after sometime the liquid particles assumes acceleration a as a solid
body.
Consider a small fluid element of dx, dy and dz dimensions as shown in figure. The
hydrostatic equation (…) is applied with the acceleration component as
P  g  a
Net surface
 Body forceper  Mass   Acceleration
Per unit volume   unit volumeof    per unit   of fluid


 
 
 

of afluidparticle afluidparticle   volume  particle

Note that each term of equation (…) represents respective force per unit volume.
If g  kˆ , the relation can be resolved into their vectorical components as
p ˆ p ˆ p ˆ
i
j
k   gkˆ    ax iˆ  ay jˆ  az kˆ
x
y
z


where ax , ay and az are the acceleration components in the x ,y ,z directions respectively.
In scalar form equation (…) becomes
p
p
p
   ax ,
   ay ,
    az  g 
x
y
z
Special case I :- Uniform acceleration of a liquid container on a straight path.
Consider a container partly filled with a liquid, moving on a straight path with a uniform
acceleration ‘a’. In order to simplify the analysis the projection of the path of motion on
the horizontal plane is assured to be the x-axis, and the projection on the vertical plane to
be the z-axis. Note that there is no acceleration component in the y direction. i.e.
ay  0
The equations (…) of motion for acceleration fluid becomes
p
p
p
   ax ,
   ay ,
    az  g 
x
y
z
Therefore,
Pressure is a function of position  x, z  and the total differential becomes
dp 
p
p
dx 
dz
x
z
Substituting for the partial differentials yields
dP  ax dx    g  az  dz
For an incompressible fluid
computed by integration.
  isconstant  .
Pressure variations in the liquid can be
P  ax . X    g  az .Z  c
where c is the constant of integration.
Let, at origin,
the pressure
then,
P P
c P
and finally the above equation becomes
Pressure variation,
P  P  ax .X    g  az .Z
If the accelerated liquid has a free surface, vertical rise between two points located on the
free surface is computed as follows

a 
z12  Z1  z2    x   x1  x2 
 g  az 
Note that the pressure at both points is the atmospheric pressure.
The slope of the free surface is
tan 
Z1  z2
a
 x
X1  X 2
g  az
The line of constant pressure isobars are parallel to the free surface (shown in figure).
The conservation of mass of an incompressible fluid implies that the volume of the liquid
remains constant before and during acceleration. The rise of the liquid level on one side
must be balanced by liquid level drop on the other side.
Uniform rotation about a vertical axis
When a liquid in a container is rotated about its vertical axis at constant angular velocity,
after sometime the liquid will move like a solid together with the container. Since every
liquid particle moves with the same angular velocity: no shear stresses exit in the liquid.
This type of motion is also known as forced vortex motion.
As shown in figure a cylindrical coordinate system with the unit vector iˆ in the radial
direction and k̂ in the vertical upward direction, is selected.
A fluid particle ‘p’ rotating with a constant angular velocity ‘  ’ has a centripetal
acceleration  2r direct radially toward the axis of rotation (-ve direction). By
substituting the acceleration component the pressure equation (…) for the fluid particle
becomes


P    gkˆ    2r iˆ
Expanding equation ( )
p
iˆ

r
p ˆ p
jˆ
k
   gkˆ   2riˆ
y
z
The scalar components are
p
p
p
  2r ,
 0,
 g
r
y
z
Since P  P  r , z  , the total differential is
dp 
p
p
dr 
dz
r
z
Substituting for
p   2
p
p
and
r
z
then for an incompressible fluid gives and integrating
r2
  gz  c
2
where c is the constant of integration.
The equation for the surface of constant pressure (for example free surface) is
z
2
2g
r 2  c1
c1 
where
cp
and this equation indicates that the isobars are paraboloids of
g
revolutions.
Special case: Cylinder liquid-filled container
Let, the point (1) on the axis of rotation is at height hc from the origin. Since the pressure
at point (1) is at atmospheric pressure, we can neglect the effect of the pressure.
Substituting pressure and position of (1) the equation (…) gives
c   ghc
The equation of the free surface becomes
w2 2
z
r  hc
2g
Consider a cylinder element of radius r, free surface height z and thickness dr.
The volume of the element is
dv  2 .rdr .z
The volume of paraboloid generated by the free surface is
R
V 
 2 zr dr
r 0
w2 2

 2  
r  hc rdr
2g

0
R
 w 2R 2

  R2 
 hc 
 4g

Since the liquid mass is conserved and incompressible this volume must be equal to the
initial volume of the liquid before rotation.
The initial volume of fluid in the container is
V   R 2hi
Equating these two volumes we get
hc  hi 
w 2R 2
4g
In the case of a closed container with no free surface or with a partly exposed free surface
rotated about the vertical axis an imaginary free surface based on equation(##) can be
constructed.
FLUID KINEMATICS
The fluid kinematics deals with description of the motion of the fluids without reference
to the force causing the motion.
Thus it is emphasized to know how fluid flows and how to describe fluid motion. This
concept helps us to simplify the complex nature of a real fluid flow.
When a fluid is in motion, individual particles in the fluid move at different velocities.
Moreover at different instants fluid particles change their positions. In order to analyze
the flow behavior, a function of space and time, we follow one of the following
approaches
(1) Lagarangian approach
(2) Eularian approach
In the Lagarangian approach a fluid particle of fixed mass is selected. We follow the fluid
particle during the course of motion with time (fig *)
The fluid particles may change their shape, size and state as they move. As mass of fluid
particles remains constant throughout the motion, the basic laws of mechanics can be
applied to them at all times. The task of following large number of fluid particles is quite
difficult. Therefore this approach is limited to some special applications for example reentry of a spaceship into the earth’s atmosphere and flow measurement system based on
particle imagery.
In the Eularian method a finite region through which fluid flows in and out is used. Here
we do not keep track position and velocity of fluid particles of definite mass. But, within
the region, the field variables which are continuous functions of space dimensions (x, y,
z) and time (t), are defined to describe the flow. These field variables may be scalar field
variables, vector field variables and tensor quantities. For example, pressure is one of the
scalar fields. Sometimes this finite region is referred as control volume or flow domain.
For example the pressure field ‘P’ is a scalar field variable and defined as
P  P  x, y, z, t 
Velocity field, a vector field, is defined as v  v  x, y, z,t 
Similarly shear stress  is a tensor field variable and defined as
 xx  xy  xz 


   yx  yy  yz 
 zx  zy  zz 


Note that we have defined the fluid flow as a three dimensional flow in a Cartesian coordinates system.
Types of Fluid Flow
Uniform and Non-uniform flow: If the velocity at given instant is the same in both
magnitude and direction throughout the flow domain, the flow is described as uniform.
Mathematically the velocity field is defined as v  v  t  , independent to space
dimensions (x, y, z).
When the velocity changes from point to point it is said to be non-uniform flow. Fig.()
shows uniform flow in test section of a well designed wind tunnel and ( ) describing non
uniform velocity region at the entrance.
Steady and unsteady flows
The flow in which the field variables don’t vary with time is said to be steady flow. For
steady flow,
v
 0 Or v  v  x, y, z 
t
It means that the field variables are independent of time. This assumption simplifies the
fluid problem to a great extent. Generally, many engineering flow devices and systems
are designed to operate them during a peak steady flow condition.
If the field variables in a fluid region vary with time the flow is said to be unsteady flow.
v
0
t
v  v  x, y, z,t 
One, two and three dimensional flows
Although fluid flow generally occurs in three dimensions in which the velocity field vary
with three space co-ordinates and time. But, in some problem we may use one or two
space components to describe the velocity field. For example consider a steady flow
through a long straight pipe of constant cross-section. The velocity distributions shown in
figure are independent of co-ordinate x and  and a function of r only. Thus the flow
field is one dimensional.
But in the case of flow over a weir of constant cross-section (), we can use two coordinate system x and z in defining the velocity field. So, this flow is a case of two
dimensional flow. The reduction of independent space variable in a fluid flow problem
makes it simpler to solve.