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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 AC 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 DDxda 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 . BBwv AECCO xwdv AAECO The integral OCC xwdv xwdv OAA xwdv equals to zero, because YY axis symmetrically divides the AAECO submerged portion AAECO . At a distance x, dv Lx tan.dx Substituting it into the above equation gives BBVAECC 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 cp 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.