Hydrostatic Force on Curved Surface

12 Key excerpts on "Hydrostatic Force on Curved Surface"

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  eBook - PDF

  Mechanics of Fluids

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

    (Publisher)

  However, there is no escape from dealing separately with the surfaces in contact with each fluid. 2.5.2 Hydrostatic thrusts on curved surfaces On a curved surface the forces p δ A on individual elements differ in direc-tion, so a simple summation of them may not be made. Instead, the resultant thrusts in certain directions may be determined, and these forces may then be combined vectorially. It is simplest to calculate horizontal and vertical components of the total thrust. Any curved surface may be projected on to a vertical plane. Take, for Horizontal component example, the curved surface illustrated in Fig. 2.16 . Its projection on to the vertical plane shown is represented by the trace MN and the horizontal projection lines may be supposed in the x -direction. Let F x represent the component in this direction of the total thrust exerted by the fluid on the curved surface. By Newton’s Third Law the surface exerts a force -F x on the fluid. Consider the fluid enclosed by the curved surface, the projection lines and the vertical plane. For this fluid to be in equilibrium the force -F x must be equal in magnitude to the force F on the fluid at the vertical plane. Also the two forces must be in line, that is, -F x must act through the centre of pressure of the vertical projection. In any given direction, therefore, the horizontal force on any surface equals the force on the projection of that surface on a vertical plane perpendicular to the given direction. The line of action of the horizontal force on the curved surface is the same as that of the force on the vertical projection. Fig. 2.16 66 Fluid statics Fig. 2.17 The vertical component of the force on a curved surface may be determined Vertical component by considering the fluid enclosed by the curved surface and vertical projection lines extending to the free surface (see Fig. 2.17 ).

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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)

  Comment Note that the air pressure used in the calculation of the force was gage pressure. Atmospheric pressure does not affect the resultant force (magnitude or location), since it acts on both sides of the plate, thereby canceling its effect. V2.6 Pop bottle Video CG O C A B F H F V F 2 F 1 A C B ( b ( ) c (d) ) O B C √(F H ) 2 + (F V ) 2 F R = ( a) Photograph courtesy of Intex Marketing, Ltd. FIGURE 2.23 Hydrostatic force on a curved surface. Sukpaiboonwat/Shutterstock.com Feliks Gurevich/ Shutterstock.com 64 CHAPTER 2 Fluid Statics from the principles of statics, it is known that when a body is held in equilibrium by three nonparallel forces, they must be concurrent (their lines of action intersect at a common point) and coplanar. Thus, F H = F 2 F V = F 1 +  and the magnitude of the resultant is obtained from the equation F R = √ _______________ ( F H ) 2 + ( F V ) 2 The resultant F R passes through the point O, which can be located by summing moments about an appropriate axis. The resultant force of the fluid acting on the curved surface BC is equal and opposite in direction to that obtained from the free‐body diagram of Fig. 2.23c. The desired fluid force is shown in Fig. 2.23d. EXAMPLE 2.9 Hydrostatic Pressure Force on a Curved Surface Given A 6-ft-diameter drainage conduit of the type shown in Fig. E2.9a is half full of water at rest, as shown in Fig. E2.9b. Find Determine the magnitude and line of action of the resul- tant force that the water exerts on a 1-ft length of the curved sec- tion BC of the conduit wall. Solution We first isolate a volume of fluid bounded by the curved section BC, the horizontal surface AB, and the vertical surface AC, as shown in Fig.

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  Advanced Transport Phenomena

  Analysis, Modeling, and Computations

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

    (Publisher)

  The calculation on a plane surface (surface with no curvature) is shown here as an illustrative example. See Fig. 4.9. 171 4.2 The equation of hydrostatics Let η be the coordinate measured from the surface of the liquid in the direction of the plate. Let us consider the force on the plate between η = L 1 and η = L 2 . The pressure at any point is ρ gd , where d is the height measured from the top. Since d = η cos α the pressure locally is equal to ρ g η cos α d η . Hence the force on the plate between η = L 1 and η = L 2 is given by Pressure force = L 2 L 1 ρ g η cos α d η Upon integrating and after some minor algebra: Pressure force = L 2 L 1 ρ g [ ( L 1 + L 2 )/ 2 ] cos α( L 2 − L 1 ) This can be interpreted as the pressure at the center ( ( L 1 + L 2 )/ 2) multiplied by the area ( L 2 − L 1 ) of the section of the plate. The moment of the force is equal to the force multiplied by the lever arm and is given as M = L 2 L 1 ρ g η 2 cos α d η Upon expressing this as M = F η c.p. we find that the center of pressure η c.p. is located at η c.p. = 2 3 L 2 1 + L 1 L 2 + L 2 2 L 1 + L 2 This is a rather interesting result: the value of the pressure at the centroid is used to calculate the (net) pressure force, but the center of pressure (the point at which the net force can be presumed to act) is not at the centroid and is somewhat below the centroid. 4.2.3 Force on a curved surface Consider a submerged surface immersed in contact with a fluid as shown in Fig. 4.10 . The calculation of this force requires a vector integration. This is because the pressure forces at each location act in different directions. In other words, the vector n is not along a constant direction along the surface. However, the calculations can be simplified by using what is called the free-body diagram in statics. This is illustrated in Figs. 4.10 and 4.11. The simpler “rules” for cal-culation of the horizontal and vertical components of the pressure force can be stated as follows.

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  Practical Hydraulics and Water Resources Engineering

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

    (Publisher)

  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 2

  )

  =

  force 

  ( kN )

  area 

  (

  m 2

  )

  .

  Force is in kilo-Newtons (kN), area is in square metres (m2 ) and so pressure is measured in kN/m2 . Sometimes pressure is measured in Pascals (Pa) in recognition of Blaise Pascal (1620–1662) who clarified much of modern day thinking about pressure and barometers for measuring atmospheric pressure.

  1 Pa = 1 N/m2 .

  One Pascal is a very small quantity and so kilo-Pascals are often used so that

  1 kPa = 1 kN/m2 .

  Although it is in order to use Pascals, kN/m2 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. The difference between them is best illustrated by an example:

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  Engineering Fluid Mechanics

  • William Graebel(Author)
  • 2018(Publication Date)
  • CRC Press

    (Publisher)

  Figure 2.11b. Pressure prism for surface of Figure 2.11a. The height is pressure, the base is area. 4. Forces on Surfaces 66 Hydrostatics and Rigid-Body Motions Sought: Force exerted by water on one side of a flat plate and the center of pressure. Given: The specific weight is 62.4 lb/ft 3 . Assumptions: The liquid is of constant specific weight. The fluid is at rest. Solution: Using s as the distance measured along the plate from the top edge, we have as the pressure distribution p=γ(h+s cos θ) and dA=b ds. Putting these together, we have s c =M/F=w(h/2+w cos θ/3)/(h+w cos θ/2). This result coincides with that of the previous example, as of course it should. Example 2.4.4. Hydraulic brakes An important application of hydrostatics is the transmission of power in hydraulic devices. Figure 2.12 shows a typical automobile braking system. The driver applies a force F D to the brake pedal. This force is transferred to the piston in the master brake cylinder. What force is transmitted to the brakes? Sought: Force transmitted to brakes. Given: The areas of the various cylinders and the lever arms. Assumptions: The brake fluid is of constant specific weight. The fluid is at rest. Figure 2.12. Vehicle hydraulic brake system. 67 Figures: See Figure 2.12. Solution: By elementary statics, the force applied to the master cylinder is (b/a)F D . If A m is the area of the master cylinder piston, the resulting pressure in the hydraulic fluid is p=F D b/aA m . This pressure is transmitted equally to all wheel cylinders; hence F B =pA w =F D bA w /aA m , where A w is the area of the wheel piston. (Often this area is nearly the same as the master cylinder area.) The brake shoe is then pressed against the brake disk or drum with a force F B . Typically in both disk and drum brakes the wheel cylinders are double-acting, so that each end produces a force F B .

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  Dynamics of Offshore Structures

  • Minoo H Patel(Author)
  • 2013(Publication Date)
  • Butterworth-Heinemann

    (Publisher)

  Chapter 7 Hydrostatics of floating bodies Hydrostatic pressures within a fluid at rest can exert very large forces on the submerged parts of offshore structures, particularly at large water depths. Hydrostatic pressures also impart rather more subtle properties which affect the stability of floating bodies. Both of these features have a profound influence on the design of floating and fixed offshore structures and are considered here in more detail. 7.1 Basic properties of a fluid at rest A fluid may be defined as a substance that deforms due to the effect of a shear stress, however small. It follows from this statement that no shearing stresses can exist in a fluid at rest with the reactions between adjacent layers of fluid being confined to normal stresses only. These normal stresses are called pressures and are defined as the normal force per unit area on an infinitesimal plane surface at any orientation in the fluid. By resolving pressure induced forces on an infinitesimal tet-rahedral fluid element with three mutually perpendicular faces, it can readily be shown that the pressure at a point in a fluid in equilibrium is the same in all directions. This is also true for a fluid in bulk motion in which there are no shearing stresses. The pressure at a point in a fluid undergoing generalized motion is considered further in Section 3.5 Although the above properties apply both to liquids and gases, the hydrostatics of offshore structures is principally concerned with liquids of large density (compared to gases) which induce substantial pressure gradients in the Earth's gravitational field - see, for example, Figure 2.6(b) for such gradients in the oceans and the atmosphere. A qualitative characteristic that distinguishes hquids from gases is the fact that a gas will expand in volume indefinitely to fill its containing volume fully, whereas a hquid will be of an essentially constant volume with no definite shape.

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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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  A Concise Handbook of Mathematics, Physics, and Engineering Sciences

  • Andrei D. Polyanin, Alexei Chernoutsan(Authors)
  • 2010(Publication Date)
  • CRC Press

    (Publisher)

  In the case of walls of constant curvature (cylindrical and spherical walls), the total pressure force passes trough the wall center of curvature. In the case of liquid excess pressures on the side of the wetted side of the wall, all components and the total force are directed from the liquid to the wall (outside from the inside). In the case of two-sided action of liquids on the wall, one first determines the horizontal and vertical components on each side of the wall under the assumption of one-sided action E4.1. H YDROSTATICS 777 of the liquid, and then the total horizontal and vertical components of action of both liquids are determined. If the curvilinear surface under study is intersected by the vertical axis at more than one point, it is necessary to divide this surface into simple parts (a simple part of a surface is its part intersected by the vertical axis only at a single point) and calculate the vertical component for each of the parts. Then the obtained vertical components are summed algebraically. The same process is used if the surface under study is intersected by the horizontal axis at more than one point. E4.1.3. Archimedes Principle. Stability of Floating Bodies ◮ Archimedes principle. The body (completely or partly) immersed in a liquid is under the action of the buoyancy force F A numerically equal to the weight of the liquid in the volume displaced by the body ( Archimedes principle ). Thus, we have F A = ρgV b , where V b is the volume of the liquid displaced by the body. The force F A passes through the center of gravity of the volume displaced by the body, which is called the center of buoyancy . ◮ Stability of floating bodies. If a floating body is in equilibrium, then its center of gravity and the center of buoyancy lie on the common vertical line called the axis of buoyancy (for a body symmetric with respect to a plane, the axis of buoyancy lies in this plane).

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  Fluid Mechanics for Civil and Environmental Engineers

  • Ahlam I. Shalaby(Author)
  • 2018(Publication Date)
  • CRC Press

    (Publisher)

  g , assuming the gage pressure scale. (a) Horizontal plane. (b) Vertical plane. (c) Sloping plane. (d) Curved surface. (e) Curved surface.

  2.4.2Magnitude and Location of the Hydrostatic Force for Plane Surfaces

  Many submerged surfaces of interest are planar, such as a gate valve in a dam or the wall of a liquid storage tank. The distributed hydrostatic force of the pressure prism can be replaced by a point force, F in the analysis of static pressure forces. Furthermore, determination of both the magnitude and location of the point force that is equivalent to the distributed force is required. The shape of the pressure prism and thus the computation of the magnitude and the location of the resultant force will vary as a function of the type of fluid, the type of surface (in this case, plane), the direction of slope of the plane, an open or an enclosed fluid, and the assumed pressure scale. If the shape of the pressure prism is uniformly distributed over the submerged surface area, A of a plane (see Figure 2.19b , for instance), the resultant force, F is equal to the pressure, p times the area, A and is given as follows:

  F = p A	(2.111)

  and the location of the resultant force, F is called the “center of pressure” and is located at the center of gravity of the pressure prism, which is at the centroid of the area, A . However, if the shape of the pressure prism in not uniformly distributed over the submerged area (see Figure 2.20b , for instance), in general, the magnitude of the resultant force, F acting on a plane surface area, A , which is submerged in a fluid, is equal to the product of the pressure acting at the centroid (center of area) of the surface area,

  pca

  and the surface area, A and is given as follows:

  F = p c a A	(2.112)

  One may note that the pressures (p or

  pca

  ) are assumed to be gage pressures unless specifically indicated as absolute pressure. Thus, in the case where the absolute pressure is assumed, or the plane is subjected to an air pressure,

  pg

  (enclosing the plane in a pressurized tank), the magnitude of the resultant force, F acting on a plane surface area, A

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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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  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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  Engineering Fluid Mechanics

  • Donald F. Elger, Barbara A. LeBret, Clayton T. Crowe, John A. Robertson(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  60 Fluid Statics CHAPTER ROAD MAP This chapter introduces concepts related to pressure and describes how to calcu- late forces associated with distributions of pressure. The emphasis is on fluids in hydrostatic equilibrium. CHAPTERTHREE FIGURE 3.1 The first man-made structure to exceed the masonry mass of the Great Pyramid of Giza was Hoover Dam. The design of dams involves calculations of hydrostatic forces. (U.S. Bureau of Reclamation) LEARNING OUTCOMES PRESSURE (§3.1). ● Define pressure and convert pressure units. ● Describe atmospheric pressure and select an appropriate value. ● Define and apply gage, absolute, vacuum, and differential pressure. ● Know the main ideas about hydraulic machines and solve relevant problems. THE HYDROSTATIC EQUATIONS (§3.2). ● Define hydrostatic equilibrium. ● Know the main ideas about the hydrostatic differential equation. ● Know the main ideas about the hydrostatic algebraic equation and solve relevant problems. PRESSURE MEASUREMENT (§3.3). ● Explain how common scientific instruments work and do relevant calculations (this LO applies to the mercury barometer, piezometer, manometer, and Bourdon tube gage). THE PRESSURE FORCE (§3.4). ● Define the center of pressure. ● Sketch a pressure distribution. ● Explain or apply the gage pressure rule. ● Calculate the force due to a uniform pressure distribution. ● Know the main ideas about the panel equations and be able to apply these equations. CURVED SURFACES (§3.5). ● Solve problems that involve curved surfaces that are acted on by uniform or hydrostatic pressure distributions. BUOYANCY (§3.6). ● Know the main ideas about buoyancy and be able to apply these ideas to solve problems. Describing Pressure 61 3.1 Describing Pressure Because engineers use pressure in the solution of nearly all fluid mechanics problems, this section introduces fundamental ideas about pressure.

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