Buoyancy

8 Key excerpts on "Buoyancy"

• 

  eBook - ePub

  Basic Engineering Mechanics Explained, Volume 1

  Principles and Static Forces

  • Gregory Pastoll, Gregory Pastoll(Authors)
  • 2019(Publication Date)
  • Gregory Pastoll

    (Publisher)

  Chapter 9 Buoyancy Definition and applications of Buoyancy The effect of the densities of the fluid and of the immersed object on flotation The fraction of a floating object that will be submersed Flotation of closed compartment and open vessels How Buoyancy affects submerged objects that are denser than the fluid

  Artist’s impression of the diving bell designed and built by Sir Edmond Halley (of comet fame) in 1790, for undersea work. The weighted barrel was filled with compressed air to replenish that used up by the divers. Illustration based upon contemporary engravings. The appearance of the diving suit and helmet are conjecture, based upon Halley’s partial description, as no detailed drawing of them could be found.

  Definition and applications of Buoyancy When an object is placed in a fluid (either a liquid or a gas) it experiences an upward force exerted on it by the fluid. This phenomenon is called Buoyancy.

  We have all had personal experience of Buoyancy. You have seen boats and balls float on water, and have experienced feeling ‘lighter’ when standing in water than when standing in air. You also know that it takes a great deal of effort to submerge a beach ball or a soccer ball fully in water.

  The first person to quantify the value of the Buoyancy force was Archimedes, whose famous principle states that the Buoyancy force on an object that is either immersed or floating, is equal to the weight of the fluid that has been displaced.

  Archimedes’ principle may be confirmed by a simple experiment.

  Suspend a heavy solid object, such as a stone, by a thin thread attached to a spring balance. Note the weight of the object, from the reading on the spring balance. While it is still attached to the thread, dip the object into a container of water that is full to the brim.

  Now observe the reading on the spring balance, which indicates the (apparently diminished) weight of the object. Collect the displaced water, and weigh it. You will find that the amount by which the weight of the object appears to be reduced is equal to the weight of the displaced water.

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  Introduction to Naval Architecture

  • E. C. Tupper(Author)
  • 2013(Publication Date)
  • Butterworth-Heinemann

    (Publisher)

  centre of Buoyancy .

  Since the Buoyancy force is equal to the weight of the body, m =ρ ∇.

  In other words the mass of a floating body equals the mass of the water displaced by that body.

  If the density of a body is greater than that of the water, the weight of water it can displace is less than its own weight and it will sink to the bottom. If held by a spring balance its apparent weight would be reduced by the weight of water it displaced – due to the water pressures acting upon it. This leads to Archimedes’ Principle which states that when a solid is immersed in a fluid it experiences an upthrust equal to the weight of the fluid displaced. This explains why divers find it easier to lift heavy items underwater.

  To illustrate the principle, consider a rowing boat in a swimming pool. If a large lump of iron in the boat is dropped over the side it will sink to the bottom of the pool. But will the depth of the water, as measured at the side of the pool, increase or decrease? It will in fact decrease. Initially the boat displaces water equal to the weight of itself and the lump of iron. After jettisoning the iron it displaces only its own weight. The iron is displacing its own volume, however, which is less than the volume of water equal to its weight. Hence the total volume of water displaced by the boat and iron decreases and the water level in the pool drops.

  Underwater Volume

  Once a ship form is defined the underwater volume can be calculated. If the immersed areas of a number of sections throughout the length of a ship are calculated, a sectional area curve can be drawn as in Figure 4.2 . The underwater volume, or volume of displacement

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  eBook - ePub

  Introduction to Engineering Mechanics

  A Continuum Approach, Second Edition

  • Jenn Stroud Rossmann, Clive L. Dym, Lori Bassman(Authors)
  • 2015(Publication Date)
  • CRC Press

    (Publisher)

  Eureka (I have found it),” so intoxicated by hydrostatics that he neglected to dry off or don a bathrobe. The next day, so the story goes, Archimedes dunked his friend’s crown, as well as a lump of gold equal to what he had provided to the goldsmith, and found that they did not displace equal amounts of water. The crown did, in fact, contain less gold than the King had specified. The goldsmith, unable to produce the remainder of the gold, was beheaded posthaste.

  FIGURE 16.11Distributed force due to hydrostatic pressure on a submerged object. Left, distributed; right, net resultant upward vertical Buoyancy force.

  Archimedes’ principle states that the buoyant force on an object equals the weight of the volume of fluid the object displaces.

  The force on a submerged object due to the fluid’s hydrostatic pressure tends to be an upward vertical force, as the pressure in the fluid increases with depth and the resultant force is upward. Refer to Figure 16.11 to visualize this. If this Buoyancy force exactly balances the weight of the object, the object is said to be neutrally buoyant.

  The line of action of the Buoyancy force acts through the centroid of the displaced fluid volume. The stability of an object designed to float on or maneuver in a fluid depends on the moments due to the Buoyancy and weight forces on the object, and whether the resultant moment will tend to right or to capsize the craft. For submerged vessels that operate at a range of depths, mechanisms that allow active control of these forces are necessary. Tanks that can be flooded or filled with air to adjust the vessel’s weight mimic the swim bladder in fish to allow vessels to maintain the proper force balance.

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  eBook - ePub

  Pumps and Hydraulics, Part 1 (of 2)

  • N. (Nehemiah) Hawkins(Author)
  • 2018(Publication Date)
  • Perlego

    (Publisher)

  THE HYDROSTATIC BALANCE.

  Every body immersed in a liquid is submitted to the action of two forces: gravity which tends to lower it, and the Buoyancy of the liquid which tends to raise it with a force equal to the weight of the liquid displaced. The weight of the body is either totally or partially overcome by its Buoyancy, by which it is concluded that a body immersed in a liquid loses a part of its weight equal to the weight of the displaced liquid.

  This principle, which is the basis of the theory of immersed and floating bodies , is called the principle of Archimedes, after the discoverer. It may be shown experimentally by means of the hydrostatic balance (Fig. 92). This is an ordinary balance, each pan of which is provided with a hook; the beam being raised, a hollow brass cylinder is suspended from one of the pans, and below this a solid cylinder whose volume is exactly equal to the capacity of the first cylinder; lastly, an equipoise is placed in the other pan. If now the hollow cylinder be filled with water, the equilibrium is disturbed; but if at the same time the beam is lowered so that the solid cylinder becomes immersed in a vessel of water placed beneath it, the equilibrium will be restored. By being immersed in water the solid cylinder loses a portion of its weight equal to that of the water in the hollow cylinder. Now, as the capacity of the hollow cylinder is exactly equal to the volume of the solid cylinder the principle which has been before laid down is proved.

  Fig. 92.

  Minerals, if suspected of containing spaces, should be coarsely pulverized, and then the second method may be conveniently applied to determine their density—thus prepared, a higher result will be obtained, and even metals when pulverized were found to give a greater specific gravity than when this is determined from samples in their ordinary state. Very fine powders may also be examined by the method in use for ascertaining the specific gravity of fluids, viz.: by comparing the weight of a measured quantity with that of the same quantity of water.

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  Introduction to Fluid Mechanics, Sixth Edition

  • William S. Janna(Author)
  • 2020(Publication Date)
  • CRC Press

    (Publisher)

  Substituting, we have d R b = ρ g z 2 − ρ g z 1 d A = ρ g z 2 − z 1 d A But the volume of the element is (z 2 − z 1) dA, and our equation becomes d R b = ρ g d V Integrating over the entire volume gives the total vertical force: R b = ρ g ∭ d V (2.32) Thus, the buoyant force equals the weight of the volume of fluid displaced. This concept is known as Archimedes’ principle. (Recall that pressure does not vary with horizontal distance, so there are no unbalanced forces in the x- or y -direction.) With reference to Figure 2.30, we can now evaluate the moment of the buoyant force about the origin: R b x r = ρ g ∭ x d V Combining with Equation 2.32, we obtain an expression for the line of action of R b : x r = ∭ x d V V (2.33) Thus, the buoyant force acts through the centroid of the submerged volume: the center of Buoyancy. When the buoyant force exceeds the object’s weight while submerged in a liquid, the object will float in the free surface. A portion of its volume will extend above the liquid surface, as illustrated in Figure 2.31. In this case, FIGURE 2.31 A floating body. d R b = ρ g z 2 − ρ a g z 1 d A where ρ a is the air density or density of the fluid above the liquid. For most liquids, ρ a ≪ ρ and thus we can write d R b = ρ g z 2 d A = ρ g d V s where d V S is the submerged volume. Integration gives R b = ρ g ∭ d V s (2.34) Thus, the buoyant force exerted on a floating body equals the weight of the displaced volume of liquid. It can be shown that this force acts at the center of Buoyancy of only the submerged volume. Example 2.17 Figure 2.32 shows a 4 cm diameter cylinder floating in a basin of water, with 7 cm extending above the surface. If the water density is 1 000 kg/m 3, determine the density of the cylinder. FIGURE 2.32 A cylinder floating in the surface of a liquid. Solution A free-body diagram of the cylinder includes gravity and Buoyancy forces

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  eBook - ePub

  Practical Civil Engineering

  • P.K. Jayasree, K Balan, V Rani(Authors)
  • 2021(Publication Date)
  • CRC Press

    (Publisher)

  ABC is equilibrium (as the fluid is at rest),

  FIGURE 12.10    Resultant force on a curved submerged surface.

  R =

  R H 2

  +

  R Y 2

  (12.10)

  and acts through O at an angle of θ .

  The angle the resultant force makes to the horizontal is

  θ =

  tan

  − 1

  (

  R Y

  R H

  )

  (12.11)

  12.3   Buoyancy and Flotation

  When a body is immersed in a fluid, an upward force is exerted by the fluid on the body. This upward force is equal to the weight of the fluid displaced by the body and is called the force of Buoyancy or simply Buoyancy.

  Centre of Buoyancy:

  It is defined as the point, through which the force of Buoyancy is supposed to act. As the force of Buoyancy is a vertical force and is equal to the weight of the fluid displaced by the body, the center of Buoyancy will be the center of gravity of the fluid displaced.

  Metacenter:

  It is defined as the point about which a body starts oscillating when the body is tilted by a small angle. The metacenter may also be defined as the point at which the line of action of the force of Buoyancy will meet the normal axis of the body when the body is given a small angular displacement. Consider a body floating in a liquid as shown in Figure 12.11a . Let the body is in equilibrium and G is the center of gravity and B is the center of Buoyancy. For equilibrium, both the points lie on the normal axis, which is vertical.

  FIGURE 12.11    Buoyancy and metacentric height (a and b).

  Metacentric Height:

  The distance MG , i.e., the distance between the metacenter of a floating body and the center of gravity of the body, is called metacentric height.

  Analytical Method for Metacenter Height:

  Figure 12.11a shows the position of a floating body in equilibrium. The location of center of gravity and center of Buoyancy in this position is at G and B . The floating body is given a small angular displacement in the clockwise direction. This is shown in Figure 12.11b , the vertical line through B 1 cuts the normal axis at M . Hence M is the metacenter and GM

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  eBook - ePub

  Fluid Mechanics, Hydraulics, Hydrology and Water Resources for Civil Engineers

  • Amithirigala Widhanelage Jayawardena(Author)
  • 2021(Publication Date)
  • CRC Press

    (Publisher)

  Archimedes’ principle ’ (220 BC). The location of the line of action of the Buoyancy force determines its stability. The Centre of Buoyancy is the centroid of the displaced fluid.

  3.7.1 Metacentre

  When the floating body is in equilibrium condition (Figures 3.10 and 3.11), the CG and the centre of Buoyancy (B) are on the same vertical line. The centre of Buoyancy changes when the body is displaced. In the new position, it is at B’ (Figure 3.10 and 3.11). Where a vertical line through B’ intersects the line through B and CG is called the Metacentre, M. When M is above the CG, the body is stable. When M is below CG, the body is unstable. When they coincide, the body is in neutral equilibrium.

  Figure 3.10 Center of Buoyancy and metacentre. (a) Equilibrium condition and (b) disturbed condition.

  Figure 3.11 Conditions of equilibrium. (a) Stable, (b) unstable, and (c) neutral.

  The distance between CG and M is called the Metacentric Height, GM . It should be positive for stability and negative for instability. The relative position of M and CG determines the stability of the body. For stability, the CG should be as low as possible. The greater the distance between CG and M, the greater will be the stability.

  3.7.1.1 Calculation of metacentric height GM

  Referring to Figure 3.12 , considering an elemental area dA at a distance x from the centreline of the floating object in the equilibrium position, the elemental volume displaced is given by

  Figure 3.12 Calculation of metacentric height: (a) Equilibrium position; (b) Displaced position.

  d V = z d A

  This when integrated should be equal to

  V

  x ¯

  0

  =

  ∫

  (

  z d A

  )

  x

  (3.47)

  where

  x ¯

  0

  is the x coordinate of the centre of Buoyancy and V is the immersed volume. If the submerged part is symmetrical about the yz plane,

  x ¯

  0

  = 0

  . After a small tilt θ , the new

  x ¯

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  eBook - ePub

  Engineering Science

  • W. Bolton(Author)
  • 2015(Publication Date)
  • Routledge

    (Publisher)

  AY , then:

  Figure 29.6 Example

  The second moment of area of a rectangular plate about an axis which is its upper edge is bd3 /3, where b is its breadth and d its depth.

  Example

  Determine the position of the centre of pressure of a vertical rectangular plate of width b and depth d immersed in a fluid with one edge at the free surface of the fluid (Figure 29.6 ).

  The second moment of area of a rectangular plate about an axis through its upper edge is bd 3 /3. Hence:

  29.5 Archimedes' principle

  When an object is immersed in a fluid of density ρ , then the pressure p on its lower 2 surface must be greater than the pressure p at its upper surface since it is at a 1 greater depth in the fluid, for example the cube shown in Figure 29.7 .

  If the height difference between the upper and lower faces is h then the pressure difference is:

  p 2 - p 1 = hρg

  Figure 29.7 Cube immersed in a fluid

  For the cube shown in Figure 29.7 , we have p1 = F1 /A and p2 = F2 /A, thus we can write the above equation as:

  F 2 - F 1 = Ahρg

  But Ah is the volume of the cube and so Ahρg is the weight of the fluid displaced by the cube. Thus there is an upthrust acting on an immersed object equal to the weight of fluid it displaces. This is known as Archimedes’ Principle and applies to all objects immersed in fluids, regardless of their shape.

  Example

  What will be the upthrust acting on an object of volume 100 cm3 when immersed in a liquid of density 950 kg/m3 ?

  The upthrust is the weight of fluid displaced by the object and is thus, for a volume V , given by:

  Upthrust = Vρg = 100 x 10−6 x 950 x 9.8 = 0.93 N

  29.5.1 Floating

  When an object floats in a fluid then the weight of the object is just balanced by the upthrust. Thus if an object of volume V floats in a liquid with a fraction f of its volume below the surface of the liquid, then the upthrust is fVρg , where ρ is the density of the liquid. If the object has a density σ then its weight is Vσg and so fVρ = Vσg . Hence the fraction immersed is f = σ /ρ

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