Buoyancy
In our common experience we know that wooden objects float on water and small dust particles also float in air. 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 the hydrostatic principle. With reference to figure consider a fluid element of area dAH . The net upward force on the fluid element is
dFB= P2 dA H
=W( h2 - h 1 ) dAH
The upward buoyant force becomes
FB= W( h2 - h 1 ) dA H = W (body volume)
This result shows that the buoyant acting on the object is equal to the weight of the fluid it displaces.
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 00 can be taken and equated to the moment of the resultant forces. The equation gives the distance to the centroid to the object volume.
The centroid 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 Archimidies 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 P1and P 2 . The buoyant force FB is
FB=蝌 dF B =r1 gdV 1 + r 2 gdV 2
= r g displaced volume i ( )i where dV1and dV 2 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) (90-white) the buoyant force of the body can be computed using equation (…). Since the specific weight of the air (11.8 N/ m2 ) is very 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. So, equation (….) becomes
FB = pg (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 coincidies 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 on the body which may overturn it. If so the floating body is unstable otherwise after imposing the 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 of major
Stability of a submerged body :-
Consider a body fully submerged in a fluid in the case given in figure (….), 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 floating body:- A body floating in equilibrium (i.e. FB = W ) is displaced through an angular displacement q . 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 moves relative to body from Bto B1 . Since the buoyant force FB and the weight W are not in the same straight line, a turning movement proportional to ‘WXq ’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 .q provided q is small; sinq= q (in radians).
The distance GM is the metacentric height. We can observe in figure that (a) Stable equilibrium: when M lies above G, a restoring moment is produced. Metacentric height GM is positive. (b) Unstable equilibrium:When M lies below G an overturning moment is produced and the metacentric height GM is negative. (c) Natural equilibrium: If M coincides with G neither restoring nor overturning moment is produced and GM is zero.
Determination of Metacentric Height:-
(1) Experimental method:
The metacentric height of a floating body can be determined in an experimental set up figure with a movable load arrangement. Because of the movement of the load, the floating object is tilted with angle q for its new equilibrium position. The measurement of q 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 ‘
= P. X The restoring moment is = W. GM .q
For equilibrium in the tilted position the restoring moment must equal to overturning So, equating
P. X= W . GM .q Metacentric weight, P. X GM = W.q and the true metacentric height is the value of GM as q 0 . This may be determined by plotting a graph between the calculated value of GM for various q values and the angle q .
(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 q the centre of buoyancy will move to new position B as a result the change in shape of displaced fluid. In the new position Aⅱ 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. Therefore 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= xq da
Integrating over the whole wedge and multiplying by the specific weight w of the liquid,
Weight of wedge 0AA= Wq xda OAA
Similarly, Weight of wedge OCC= Wq xda OCC
Equating Equations ( ) and ( ),
Wq蝌 xda= W q xda OAA OCC xda = 0
Where 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 waterlineplane.
Computation of metacentric height GM :
From the figure the distance BM is BM= BB /q
The distance BB is calculated by taking moment about the centroidal axis YY .
BB wv= xwdv + xwdv - xwdv Aⅱ ECC O 蝌 AAⅱ ECO OCC OAA
The integral nwdv equal to zero, because YY axis symmetrically divides the AA ECO submerged portion AA ECO .
At a distance x, dv= Lxtanq . dx
Substituting it in the above equation.
BB VAECC O =0 +蝌 xLx tanq dx - xL( - x tan q ) dx OCCⅱ OAA = tanq x2 dAwaterline waterline
= tanq IO
where Io is the second moment of area of water line plane about OO . Thus,
Distance BM= BB /q I tanq = o q.VAECC O I = o VAECC O Since, BM= GM + BG I GM=o - BG Vsubmerged
Periodic time of transverse oscillation.
When an overturning moment which results an angular displacement q to a floating body is suddenly removed, the floating body may be set in a state of oscillation. This oscillation behave as in the same manner of a simple pendulum suspended at metacentre M. Only the restoring moment (W. GMq ) sets it in a state of oscillation. So, it is equal to the rate of change of angular momentum.
d 2q Where, KG is the radius of gyration about its axis of rotation and is 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, periodic time
Displacement T = 2p Acceleration q = 2p GMq g/ K 创 G2 = 2p K / GM . g G2
From the above equation it can be observed that a large metacentric height gives greatest stability to a floating object. However it reduces the time period of oscillation which results in discomfort for the passenger in a passenger ship.
Some typical metacentric heights of various floating vessels are-
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 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 fluid particles the shear stress is zero throughout the liquid. At this equilibrium it is said to be fluid 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 fluid particles assumes acceleration ‘a’ as a solid body.
Consider a small fluid element of dx, dy, dz dimensions as shown in figure The hydrostatic equation (…) is applied with the acceleration component as -�P = r g r a 轾Net surface 轾 Body forceper 轾 Mass 轾 Acceleration 犏 犏 犏 犏 犏Per unit volume+ 犏 unit volumeof = 犏 per unit 犏 of fluid 臌犏of a fluidparticle 臌犏 afluidparticle 臌犏 volume 臌犏 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
dp d p d p iˆ+ ˆj + k ˆ +r gk ˆ = - r a i ˆ + a jˆ + a k ˆ dx d y d z ( x y z )
where ax, a y and a z are the acceleration components in the x ,y ,z directions respectively.
In scalar form equation (…) becomes
dp d p d p = -ra, = - r a , = - r ( a + g ) dxx d y y d z 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
dp d p d p = -ra, = - r a , = - r ( a + g ) dxx d y y d z z
Therefore, Pressure is a function of position (x, z) and the total differential becomes 抖r r dr = dx + dz 抖x z
Substituting for the partial differentials yields
dP= -r ax dx - r ( g + a z ) dz
For an incompressible fluid (r isconstant) . Pressure variations in the liquid can be computed by integration.
P= -r ax. X - r ( g + a z ) . Z + e where e is the constant of integration.
Let, at origin,
the pressure P= P°
then, e= P° and finally the above equation becomes
Pressure variation, P= P° -r ax. X - r ( g + a z ) . Z
If the accelerated liquid has a free surface, vertical rise between two points located on the free surface is computed as follows
骣 ax Dz12 = Z 1 - z 2 =琪 -( x 1 - x 2 ) 桫 g+ az
Note that the pressures at both points is the atmospheric pressure.
The slope of the free surface is
Z- z a tanq =1 2 = - 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 size 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 kˆ in the vertical upward direction, is selected.
A fluid particle ‘p’ rotating with a constant angular velocity ‘w’ has a centrepetal acceleration w2 r direct radially toward the axis of rotation (-ve directon). By substituting the acceleration component the pressure equation (…) for the fluid particle becomes
�P- -r gkˆ - r ( w2 r) i ˆ
Expanding equation ( )
抖p p p iˆ+ jˆ + k ˆ = -r gk ˆ + r w2 ri ˆ 抖r y z
The scalar components are
抖p p p =rw2 r, = 0, = - r g 抖r y z
Since P= P( r, z) , the total differential is
抖P P dP= dr + dz 抖r z
抖p p Substituting for and then for an incompressible fluid gives and integrating 抖r z
r 2 p=r w2 - r gz + c 2 where c is the constant of integration.
The equation for the surface of constant pressure (for example free surface) is
w 2 z= r2 + e 2g 1 c- P where e = 1 rg and this equation indicates that the isobar are paraboloids of revolutions.
Special case : Cylinder container.
Let, the point (1) on the axis of rotation at height he from the origin.
The pressure at point (1) is zero. (Atmospheric pressure).
Substituting pressure and position of (1) the equation (…) gives e= r ghe
The equation of the free surface becomes
w 2 z= r2 + he 2g
Consider a cylinder element of radius r, free surface height z and thickness dr. The volume of the element is dv= 2p rdrz
The volume of paraboloid generated by the free surface is
R V= 2p zr dr r =0 R 骣w 2 =2p 琪 r2 + he rdr 0 桫2g 骣w2 R 2 =pR2 琪 + he 桫 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= p R2 hi
Equating these two volumes we get w2 R 2 he= hi - 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 analyse the flow behaviour, a function of space and time, we follow one of the alternative approaches
(1) Lagarangian. (2) Eularian.
In the lagarangian approaches a fluid particles of fixed mass is selected and follows during the course of motion with time (fig *)
The fluid particles may change their shape, size and state as they move. As fluid particles of definite mass are selected, 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 re-entry of a spaceship into the earth’s atmosphere and flow measurement system based on particle imagery.
In the Eularian method a finite volume of through which fluid flows in and out is used. Here it does not keep track position and velocity of fluid particles of definite mass within the finite volume field variables which are continuous functions of space (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. Sometimes this finite volume 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) Velocity field , a vector field variable becomes v= v( x, y , z , t ) Similarly shear stress r is a tensor field variable and 轾rxx r xy r xz 犏 r= 犏 ryx r yy r yz 犏r r r 臌zx zy zz
Note that we have defined the fluid flow as a three dimensional flow in a Cartesian co- ordinates system.
Types of fluid flow:-
(1) 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 describe as uniform. Mathematically the velocity field is defined as v= v( t ) , independent to space coordinates.
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 flow:-
The flow in which the field variables don’t change with time is said to be steady flow. For steady flow,
dv = 0 or v= u( x, y , z) dt Many engineering devices and systems are generally designed to operate in a steady flow condition.
For example pipe flow in a pumping system becomes steady flow condition. So, we use often steady flow analysis in designing the devices.
If the field variables in a fluid region vary with time the flow is said to be unsteady flow. dv 0 v= v( x, y , z , t ) dt
One , two and three dimensional flow.
Although fluid flows generally occurs in three dimensions in which the velocity field vary with three space co-ordinates and time, 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 q 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 (4.3 pouglas), we can use two co-ordinate 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