Hydrostatic Force

11 Key excerpts on "Hydrostatic Force"

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  Practical Hydraulics and Water Resources Engineering

  • Melvyn Kay(Author)
  • 2017(Publication Date)
  • CRC Press

    (Publisher)

  27 Chapter 2 Water standing still Hydrostatics 2.1 INTRODUCTION Hydrostatics is the study of water when it is not moving; it is standing still. It is important to civil engineers who are designing water storage tanks and dams. They want to work out the forces that water creates in order to build reservoirs and dams that can resist them. Naval architects designing submarines want to understand and resist the pressures created when they go deep under the sea. The answers come from understanding hydrostatics. The science is simple both in concept and in practice. Indeed, the theory is well established and little has changed since Archimedes (287–212 bc) worked it out over 2000 years ago. 2.2 PRESSURE The term pressure is used to describe the force that water exerts on each square metre of some object submerged in water, that is, force per unit area. It may be the bottom of a tank, the side of a dam, a ship or a submerged submarine. It is calculated as follows: pressure force area = . Introducing the units of measurement . pressure (kN/m ) force (kN) area (m ) 2 2 = Force is in kilo-Newtons (kN), area is in square metres (m 2 ) and so pres- sure is measured in kN/m 2 . Sometimes pressure is measured in Pascals (Pa) in recognition of Blaise Pascal (1620–1662) who clarified much of modern 28 Practical Hydraulics and Water Resources Engineering day thinking about pressure and barometers for measuring atmospheric pressure. 1 Pa = 1 N/m 2 . One Pascal is a very small quantity and so kilo-Pascals are often used so that 1 kPa = 1 kN/m 2 . Although it is in order to use Pascals, kN/m 2 is the measure of pressure that engineers tend to use and so this is used throughout this text (see example of calculating pressure in Box 2.1). 2.3 FORCE AND PRESSURE ARE DIFFERENT Force and pressure are terms that are often confused.

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  Introduction to Fluid Mechanics, Sixth Edition

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

    (Publisher)

  2   Fluid Statics

  In this chapter, we study the forces present in fluids at rest. Knowledge of force variations—or, more appropriately, pressure variations—in a static fluid is important to the engineer. Specific examples include water retained by a dam or bounded by a levee, gasoline in a tank truck, and accelerating fluid containers. In addition, fluid statics deals with the stability of floating bodies and submerged bodies and has applications in ship hull design and in determining load distributions for flat-bottomed barges. Thus, fluid statics concerns the forces that are present in fluids at rest, with applications to various practical problems.

  After completing this chapter, you should be able to:

  • Discuss pressure and pressure measurement;
  • Apply the hydrostatic equation to manometers;
  • Develop equations for calculating forces on submerged surfaces;
  • Examine problems involving stability of partially or wholly submerged objects.

  2.1 PRESSURE AND PRESSURE MEASUREMENT

  Because our interest is in fluids at rest, let us determine the pressure at a point in a fluid at rest. Consider a wedge-shaped particle exposed on all sides to a fluid as illustrated in Figure 2.1a . Figure 2.1b is a free-body diagram of the particle cross section. The dimensions Δx , Δy , and Δz are small and tend to zero as the particle shrinks to a point. The only forces considered to be acting on the particle are due to pressure and gravity. On either of the three surfaces, the pressure force is F = pA . By applying Newton’s second law in the x - and z -directions, we get, respectively,

  FIGURE 2.1 A wedge-shaped particle.

  ∑ F x = p x Δ z Δ y − p s Δ s Δ y sin θ = ρ 2 Δ x Δ y Δ z a x = 0

  ∑ F z = p z Δ x Δ y − p s Δ s Δ y cos θ − ρ g Δ x Δ y Δ z 2         = ρ 2 Δ x Δ y Δ z a z = 0

  where

  p x , p z , and p s are average pressures acting on the three corresponding faces

  a x and a z are the accelerations

  ρ is the particle density

  The net force equals zero in a static fluid. After simplification, with a x = az

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  Casing and Liners for Drilling and Completion

  • Ted G. Byrom(Author)
  • 2013(Publication Date)
  • Gulf Publishing Company

    (Publisher)

  normal meaning perpendicular to the surface.

  2.3.1 Hydrostatic Pressure

  Hydrostatic pressure is more or less intuitive. It might be described as a force per unit area exerted on a static fluid or by a static fluid on some material body. The hydrostatic pressure is a result of the body force plus any additional pressure that may be applied to the fluid. We also can write the fundamental equation of fluid statics in terms of pressure instead of body force. But, before we do that, it is necessary to introduce a coordinate system to make things a little easier. In almost all texts on mechanics, the coordinate systems used for illustration show the vertical coordinate to be positive in the upward direction, however since almost everything we measure in regard to a well in the oil field is measured from the surface downward, we start out with such a coordinate system and stay with it throughout this text. Figure 2-2 shows the coordinate system we most often use.

  Figure 2-2 An earth-oriented Cartesian coordinate system.

  Note that this is still a conventional right-hand coordinate system. It may look a bit awkward at first, because the positive z -axis points downward and the positive y -axis is to the right of the positive x -axis when viewed from above, but if viewed from below it is exactly what we are more accustomed to. The main advantage to us is that the z -axis is positive downward, so that our depth measurements in a well correspond in sign and direction to the z -axis. Another advantage to this coordinate system is that, if we assume the x -axis is positive in the North direction, then the y -axis is positive in the East direction, and directional azimuth and trigonometric functions are compatible since all angles are measured in the same direction. Directional azimuth in a well is measured from North in a clockwise direction, but trigonometric functions are such that the angle is measured counterclockwise from the x -axis towards the y

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  Mechanical Engineering Systems

  • Richard Gentle, Peter Edwards, William Bolton(Authors)
  • 2001(Publication Date)
  • Newnes

    (Publisher)

  In the previous section we looked at the subject of hydrostatic pressure variation with depth in a liquid, and used the findings to explore the possibilities for measuring pressure with columns of liquids (manometry). In this section we are going to extend this study of hydrostatic pressure to the point where we can calculate the total force due to liquid pressure acting over a specified area. One of the main reasons that we study fluid mechanics in mechanical engineering is so that we can calculate the size of forces acting in a situation where liquids are employed. Knowledge of these forces is essential so that we can safely design a range of devices such as valves, pumps, fuel tanks and submersible housings. Although we shall not look at all the many different situations that can arise, it is vital that you understand the principles involved by studying a few of the most common applications.

  Hydraulics

  The first thing to understand is the way that pressure is transmitted in fluids. Most of this is common sense but it is worthwhile spelling it out so that the most important features are made clear.

  Look at Figures 3.1.14 and 3.1.15 . In Figure 3.1.14 a short solid rod is being pushed down onto a solid block with a force F . The cross-sectional area of the rod is A and so a localized pressure of

  Figure 3.1.14 Force applied to a solid block

  Figure 3.1.15 Force applied to a liquid

  is felt underneath the base of the rod. This pressure will be experienced in the block mainly directly under the rod, but also to a much lesser extent in the small surrounding region. It will not be experienced in the rest of the solid block towards the sides.

  Now look at Figure 3.1.15 where the same rod is used as a piston to push down with the same force on a sealed container of liquid. The result is very different because the pressure of

  will now be experienced by all the liquid equally. If the container were to spring a leak we know that the liquid would spurt out normal to the surface that had the hole in it. If the leak were in the top surface of the container then the liquid would spurt upwards, in completely the opposite direction to the force that is being applied to the piston. This means that the liquid was being pushed in that upwards direction and that can only happen if the pressure acts normally to the inside surface.

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  Engineering Fluid Mechanics

  • Donald F. Elger, Barbara A. LeBret, Clayton T. Crowe, John A. Roberson(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  Hydrostatic Equilibrium • A hydrostatic condition means that the weight of each fluid particle is balanced by the net pressure force. • The weight of a fluid causes pressure to increase with increasing depth, giving the hydrostatic differential equation. The equations that are used in hydrostatics are derived from this equation. The hydrostatic differential equation is dp _ dz = −γ = −ρg • If density is constant, the hydrostatic differential equation can be integrated to give the hydrostatic equation. The meaning (i.e., physics) of the hydrostatic equation is that piezometric head (or piezometric pressure) is constant everywhere in a static body of fluid. p _ γ + z = constant Pressure Distributions and Forces Due to Pressure • A fluid in contact with a surface produces a pressure distribution, which is a mathematical or visual description of how the pressure varies along the surface. • A pressure distribution is often represented as a statically equivalent force F p acting at the center of pressure (CP). • A uniform pressure distribution means that the pressure is the same at every point on a surface. Pressure distributions due to gases are typically idealized as uniform pressure distributions. • A hydrostatic pressure distribution means that the pressure varies according to dp/dz = −γ. Force on a Flat Surface • For a panel subjected to a hydrostatic pressure distribution, the Hydrostatic Force is F p = _ p A • This Hydrostatic Force • Acts at the centroid of area for a uniform pressure distribution. • Acts below the centroid of area for a hydrostatic pressure distribution. The slant distance between the center of pressure and the centroid of area is given by y cp − _ y = _ I _ _ y A Hydrostatic Forces on a Curved Surface • When a surface is curved, one can find the pressure force by applying force equilibrium to a free body comprised of the fluid in contact with the surface.

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  Advanced Transport Phenomena

  Analysis, Modeling, and Computations

  • P. A. Ramachandran(Author)
  • 2014(Publication Date)
  • Cambridge University Press

    (Publisher)

  169 4.2 The equation of hydrostatics The equation of hydrostatics ............................................................................... 4.2 In a stationary fluid, there are no viscous forces. Hence we need to consider only the pressure force and gravity. Hence Pressure force + Gravity forces = zero. From the representation of the forces, this leads to ∇ p = ρ g (4.11) which is the equation of hydrostatics. This equation gives the local variation of the pressure at a given point. For an incompressible fluid, the density is treated as a constant. Then the following equation holds for the pressure variation in a column of, for example, a liquid: p = p 0 + ρ gd (4.12) where d is the depth into the liquid (taken to be positive in the downward direction) and p 0 is the pressure at the surface of the liquid. The above equation can be used for gases where the pressure changes are moderate so that an average density can be used. However, it is easy to combine this with the ideal-gas law to provide the following equation for the pressure variation in a column of gas: p p 0 = exp − M w z R G T (4.13) The derivation is left out, but it involves using the density as a function of p in Eq. (4.11) followed by integration. Here z is the elevation or height measured from the surface or the datum plane with a base pressure p 0 . In deriving this expression the variation of temperature with height is ignored. The effect of temperature variation is minor and is also left as an exercise for the reader. An important application of the equation of hydrostatics leads to the famous Archimedes principle discussed in the following section. 4.2.1 Archimedes’ principle Consider a solid body immersed in a liquid as shown in Fig. 4.8 . The pressure force acts on the surface of the body and is given by Pressure force = A ( − p ) n dA = V −∇ p dV We now use the equation of hydrostatics as ∇ p = ρ l g , where ρ l is the liquid density.

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  Young, Munson and Okiishi's A Brief Introduction to Fluid Mechanics

  • John I. Hochstein, Andrew L. Gerhart(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  35 35 • calculate the hydrostatic pressure force on a plane or curved submerged surface. • calculate the buoyant force and determine the stability of floating or submerged objects. • determine the pressure at various locations in a fluid at rest. • explain the concept of manometers and apply appropriate equations to determine pressures. LEARNING OBJECTIVES After completing this chapter, you should be able to: In this chapter, we will consider an important class of problems in which the fluid is either at rest or moving in such a manner that there is no relative motion between adjacent particles. In both instances there will be no shearing stresses in the fluid, and the only forces that develop on the surfaces of the particles will be due to the pressure. Thus, our principal concerns are to investigate pressure and its variation throughout a fluid and the force on submerged surfaces due to that pressure variation. The absence of shearing stresses greatly simplifies the analysis and, as we will see, allows us to obtain relatively simple solutions to many important practical problems. 2.1 Pressure at a Point As we briefly discussed in Chapter 1, the term pressure is used to indicate the normal force per unit area at a given point acting on a given plane within the fluid mass of interest. A question that immediately arises is how the pressure at a point varies with the orientation of the plane passing through the point. To answer this question, consider the free-body diagram, illustrated in Fig. 2.1, that represents a small triangular wedge of fluid from some arbitrary location within a fluid mass. Since we are considering the situation in which there are no shearing stresses, the only external forces acting on the wedge are due to the pressure and gravity. For simplicity the forces in the x direction are not shown, and the z axis is taken as the vertical axis so gravity acts in the negative z direction.

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  Munson, Young and Okiishi's Fundamentals of Fluid Mechanics

  • Andrew L. Gerhart, John I. Hochstein, Philip M. Gerhart(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  35 35 • calculate the hydrostatic pressure force on a plane or curved submerged surface. • calculate the buoyant force and determine the stability of floating or submerged objects. • determine the pressure at various locations in a fluid at rest. • explain the concept of manometers and apply appropriate equations to determine pressures. LEARNING OBJECTIVES After completing this chapter, you should be able to: In this chapter we will consider an important class of problems in which the fluid is either at rest or moving in such a manner that there is no relative motion between adjacent particles. In both instances there will be no shearing stresses in the fluid, and the only forces that develop on the surfaces of the particles will be due to the pressure. Thus, our principal concerns are to investigate pressure and its variation throughout a fluid and the force on submerged surfaces due to that pressure variation. The absence of shearing stresses greatly simplifies the analysis and, as we will see, allows us to obtain relatively simple solutions to many important practical problems. 2.1 Pressure at a Point As we briefly discussed in Chapter 1, the term pressure is used to indicate the normal force per unit area at a given point acting on a given plane within the fluid mass of interest. A question that immediately arises is how the pressure at a point varies with the orientation of the plane passing through the point. To answer this question, consider the free-body diagram, illustrated in Fig. 2.1, that represents a small triangular wedge of fluid from some arbitrary location within a fluid mass. Since we are considering the situation in which there are no shearing stresses, the only external forces acting on the wedge are due to the pressure and gravity. For simplicity the forces in the x direction are not shown, and the z axis is taken as the vertical axis so gravity acts in the negative z direction.

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  Essentials of Hydraulics

  • Pierre Y. Julien(Author)
  • 2022(Publication Date)
  • Cambridge University Press

    (Publisher)

  1.1 Units and Water Properties 5 1.2 Hydrostatic Pressure Pressure is the force per unit area perpendicular to a surface. In a fluid, the pressure is the same in all directions. Atmospheric pressure is discussed in Section 1.2.1, followed by hydrostatic pressure in Section 1.2.2. The difference between relative and absolute pressure is explained in Section 1.2.3. Finally, the hydraulic grade line is covered in Section 1.2.4 with vapor pressure (absolute in Section 1.2.5 and relative in Section 1.2.6). 1.2.1 Atmospheric Pressure The atmospheric pressure was first measured by the French scientist Blaise Pascal, in the seventeenth century. The atmospheric pressure decreases with altitude, as shown in Table 1.5. At sea level, the atmospheric pressure is p atm ¼ 101:3 kPa ¼ 14:7 psi ¼ 2,116 psf . Note that 1 psi is the pressure generated by a column of 2.31 ft, or 70.4 cm of water. 1.2.2 Hydrostatic Pressure Hydrostatics refers to fluids at rest. The free surface of a fluid at rest is horizontal, which defines a surface of equal pressure (SEP). A datum is a reference SEP. Commonly used datum elevations are the mean sea level, the floor of a building and a benchmark elevation. Starting from the atmospheric pressure p 0 at the free surface, the hydrostatic pressure distribution can be obtained from the analysis of forces on an infinitesimal element of fluid of volume 8 and weight W shown in Figure 1.1: W ¼ ρg8 ¼ γ8 ¼ γ Δx Δy Δz: Table 1.5.

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  Introduction to Engineering Mechanics

  A Continuum Approach, Second Edition

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

    (Publisher)

  Figure 16.8 . This general formulation can then be applied to a wide range of problems.

  FIGURE 16.7Pressure prisms for (a) distributed beam loading and (b) hydrostatic pressure.

  FIGURE 16.8Hydrostatic Force on an inclined plane surface of arbitrary shape.

  Essentially, we are once again considering the effect of a distributed force, and in order to deal with the equivalent concentrated load, we must find the “center of pressure” at which this equivalent load acts. We choose coordinates, as shown in Figure 16.8 , that are convenient for the surface in question, and we must find the point (xR , yR )—the center of pressure, at which the resultant force acts.

  The total force exerted on the plane surface by the fluid is simply the integral of the fluid pressure over the surface’s entire area:

  F R = ∫ A d F = ∫ A p   d A ,

  (16.20)

  where p is the gage pressure. For a fluid at rest, the pressure distribution is hydrostatic, and dF = ρgh = ρgy sin θ. For constant ρg and θ, we thus have

  F R = ρ g   sin θ ∫ A y   d A ,

  (16.21)

  and we recognize that the integral ∫ y dA = yc A, where yc is the position of the centroid of the entire submerged surface. So, the resultant force is simply

  F R = ρ g A y c sin  θ=ρ g h c A ,

  (16.22)

  where hc is the vertical distance from the fluid surface to the centroid of the area. We notice that this force’s magnitude is independent of the angle θ and depends only on the fluid’s specific weight, the total area, and the depth of the centroid.

  Although we might suspect that this resultant force passes through the centroid of the surface, if we remember that pressure is increasing with increasing depth, we realize that the center of pressure must actually be below the centroid. We can find the coordinates of the center of pressure by summing moments around the x

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  Mechanics of Fluids

  • John Ward-Smith(Author)
  • 2018(Publication Date)
  • CRC Press

    (Publisher)

  This is because, if the imaginary fluid were in fact present, pressures at the two sides of the surface would be identical and the net force reduced to zero. If a free surface does not actually exist, an imaginary free surface may be considered at a height p /% g above any point at which the pressure p is known. The density of the imaginary fluid must, of course, be supposed the same as that of the actual fluid so that the variation of pressure over the sur-face is correctly represented. The vertical component of the total force is then equal to the weight of the imaginary fluid vertically above the curved surface. In general the components of the total force must be considered in three Resultant thrust mutually perpendicular directions, two horizontal and one vertical. These three components need not meet at a single point, so there is, in general, no single resultant force. In many instances, however, two forces lie in the same Hydrostatic thrusts on submerged surfaces 67 plane and may then be combined into a single resultant by the parallelogram of forces. If there is a vertical plane on which the surface has no projec-tion (e.g. the plane perpendicular to the horizontal axis of a cylindrical surface) there is no component of Hydrostatic Force perpendicular to that plane. The only horizontal component then needing consideration is the one parallel to that plane. When the two sides of a surface are wholly in contact with a single fluid of uniform density but the level of the free (atmospheric) surface on one side is different from that on the other, the net effective pressure at any point depends only on the difference in free surface levels. The effective pressure is therefore uniform over the area and so the components of the resultant force then pass through the centroids of the vertical and horizontal projections respectively. Example 2.3 A sector gate, of radius 4 m and length 5 m, controls the flow of water in a horizontal channel.

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