Mathematics
Hydrostatic Pressure
Hydrostatic pressure refers to the pressure exerted by a fluid at rest due to the force of gravity. It is directly proportional to the depth of the fluid and the density of the liquid. This concept is important in mathematics for understanding fluid dynamics and is often used in calculations related to fluid pressure and buoyancy.
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12 Key excerpts on "Hydrostatic Pressure"
eBook - PDF- Donald F. Elger, Barbara A. LeBret, Clayton T. Crowe, John A. Robertson(Authors)
- 2016(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 = −ρ g • If density is constant, the hydrostatic differential equa- tion 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 con- stant everywhere in a static body of fluid. p γ + z = constant 3.8 Summarizing Key Knowledge Problems 91 Pressure Distributions and Forces Due to Pressure • A fluid in contact with a surface produces a pressure dis- tribution, 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 distribu- tions due to gases are typically idealized as uniform pres- sure 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 distribu- tion, 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.
eBook - PDF- 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.
eBook - PDF- 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.
eBook - PDF- Donald F. Elger, Barbara A. LeBret, Clayton T. Crowe, John A. Roberson(Authors)
- 2019(Publication Date)
- Wiley(Publisher)
56 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. A traditional application of fluid statics is described in Fig. 3.1. 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 57 3.1 Describing Pressure Because engineers use pressure in the solution of nearly all fluid mechanics problems, this section introduces fundamental ideas about pressure.
- Melvyn Kay(Author)
- 2017(Publication Date)
- CRC Press(Publisher)
Chapter 2Water standing still
Hydrostatics2.1 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 =Introducing the units of measurementforce area.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 that1 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:
- Philip M. Gerhart, Andrew L. Gerhart, John I. Hochstein(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
The assumption of zero shearing stresses will still be valid as long as the fluid mass moves as a rigid body; that is, there is no relative motion between adjacent elements. After completing this chapter, you should be able to: ■ determine the pressure at various locations in a fluid at rest. ■ explain the concept of manometers and apply appropriate equations to determine pressures. ■ 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. Learning Objectives Fluid Statics 2.2 Basic Equation for Pressure Field 41 The equations of motion (Newton’s second law, F = m a ) in the y and z directions are, respectively, ∑ F y = p y δx δz − p s δx δs sin θ = ρ δx δy δz 2 a y ∑ F z = p z δx δy − p s δx δs cos θ − γ δx δy δz 2 = ρ δx δy δz 2 a z where p s , p y , and p z are the average pressures on the faces, γ and ρ are the fluid specific weight and density, respectively, and a y , a z the accelerations. Note that a pressure must be multiplied by an appropriate area to obtain the force due to the pressure. It follows from the geometry that δy = δs cos θ δz = δs sin θ so that the equations of motion can be rewritten as p y − p s = ρa y δy 2 p z − p s = ( ρa z + γ ) δz 2 Since we are really interested in what is happening at a point, we take the limit as δx , δy , and δz approach zero (while maintaining the angle θ ), and it follows that p y = p s p z = p s or p s = p y = p z . The angle θ was arbitrarily chosen so we can conclude that the pressure at a point in a fluid at rest, or in motion, is independent of direction as long as there are no shearing stresses present . This important result is known as Pascal’s law, named in honor of Blaise Pascal (1623– 1662), a French mathematician who made important contributions in the field of hydrostatics.
- John I. Hochstein, Andrew L. Gerhart(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
This type of pressure distribution is commonly called a hydrostatic distribution, and Eq. 2.7 shows that in an incompressible fluid at rest the pressure varies linearly with depth. The pressure must increase with depth to “hold up” the fluid above it. It can also be observed from Eq. 2.6 that the pressure difference between two points can be specified by the distance h, since h = p 1 − p 2 _ γ In this case, h is called the pressure head and is interpreted as the height of a column of fluid of specific weight γ required to give a pressure difference p 1 − p 2 . For example, a pressure dif- ference of 10 psi can be specified in terms of pressure head as 23.1 ft of water ( γ = 62.4 lb/ft 3 ), or 518 mm of Hg ( γ = 133 kN/m 3 ). As illustrated by the adjacent figure, a 23.1-ft-tall column of water with a cross-sectional area of 1 in. 2 weighs 10 lb. V2.1 Pressure on a car Video V2.2 Demonstration of atmospheric pressure. Video 23.1 ft Water = 10 lb pA = 0 pA = 10 lb A = 1 in. 2 z x y z 1 z 2 p 1 p 2 h = z 2 – z 1 Free surface (pressure = p 0 ) FIGURE 2.3 Notation for pressure variation in a fluid at rest with a free surface. 40 CHAPTER 2 Fluid Statics When one works with liquids, there is often a liquid–vapor interface we call a free surface, as is illustrated in Fig. 2.3, and it is convenient to use this surface as a reference plane. The ref- erence pressure p 0 would correspond to the pressure acting on the free surface (which would frequently be atmospheric pressure). If we let p 2 = p 0 in Eq. 2.7, it follows that the pressure p at any depth h below the free surface is given by the equation: p = γ h + p 0 (2.8) As is demonstrated by Eq. 2.7 or 2.8, the pressure in a homogeneous, incompressible fluid at rest depends only on the depth of the fluid relative to some reference plane. It is not influ- enced by the size or shape of container in which the fluid is held.
eBook - PDF- Joseph Katz(Author)
- 2010(Publication Date)
- Cambridge University Press(Publisher)
In this section we try to calculate the forces that are due to pressure in a stationary fluid (called hydrostatic) acting on a submerged surface or on a closed body. Consider a body with surface S , as shown in Fig. 3.11, submerged in an incompressible fluid. The force acting at a point on this body was defined by Eq. (1.4) such that F = − p ndS , (1.4) and dS is an infinitesimal surface element, as shown. So if we need to find the total force F acting on a specific area S 1 because of a variable-pressure field, then we need to integrate this equation over a desirable surface S 1 : F = − S 1 p ndS . (3.29) Assuming that only gravitational forces act on the fluid, then the pressure will vary with the depth h , as given by Eq. (3.9b) p ( h ) = p a + ρ gh . (3.9b) The meaning of this relation is explained visually in Fig 3.12, where the pressure (and resulting force) will increase on the sides of a container with the depth h . At the bottom of the container, at a depth of H , the pressure is constant, as shown in the figure. So basically all the weight of the liquid is supported by the bottom sur-face, but in addition, the sides will experience horizontally acting hydrostatic force (because pressure at a point acts equally to all directions). In comparison, if the p a p = p a + ρ gh p = p a + ρ gh H h Figure 3.12. Variation of the Hydrostatic Pressure in a stationary container. 78 Fluid Statics dS p x dF x Figure 3.13. The force components acting on a surface are split by the unit normal vector n = ( n x , n y , n z ) . same weight of bricks were piled in the container, then only the loads on the bottom surface will be present. To continue the determination of the force acting on the submerged surface we substitute the pressure from Eq. (3.9b) into force equation (3.29) to get F = − S 1 ( p a + ρ gh ) ndS . (3.30) Note that Eq. (3.30) is a vector expression and the force components must be eval-uated in a well-defined coordinate system.
- 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.
- Robert W. Fox, Alan T. McDonald, John W. Mitchell(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
In a static, homogeneous fluid, or in a fluid undergoing rigid-body motion, a fluid particle retains its identity for all time, and fluid elements do not deform. We may apply Newton’s second law of motion to evaluate the forces acting on the particle. 3.1 The Basic Equation of Fluid Statics The first objective of this chapter is to obtain an equation for computing the pressure field in a static fluid. We will deduce what we already know from everyday experience, that the pressure increases with depth. To do this, we apply Newton’s second law to a differential fluid element of mass dm = ρ dV ---, with sides dx, dy, and dz, as shown in Fig. 3.1. The fluid element is stationary relative to the stationary rectangular coordinate system shown. From our previous discussion, recall that two general types of forces may be applied to a fluid: body forces and surface forces. The only body force that must be considered in most engineering problems is due to gravity. We will not consider body forces caused by electric or magnetic fields. For a differential fluid element, the body force is dF B = g dm = g ρ dV --- where g is the local gravity vector, ρ is the density, and dV --- is the volume of the element. In Cartesian coordinates dV --- = dx dy dz , so dF B = ρ g dx dy dz In a static fluid there are no shear stresses, so the only surface force is the pressure force. Pressure is a scalar field, p = p x, y, z and in general we expect the pressure to vary with position within the fluid. The net pressure force that results from this variation can be found by summing the forces that act on the six faces of the fluid element. Let the pressure be p at the center, O, of the element. To determine the pressure at each of the six faces of the element, we use a Taylor series expansion of the pressure about point O. The pressure at the left face of the differential element is O Pressure, p y dx dz dy z p p dy (dx dz) ( j ) 2 + – y p p dy (dx dz) (–j ) 2 y x ^ ^ Fig.
- Andrew L. Gerhart, John I. Hochstein, Philip M. Gerhart(Authors)
- 2023(Publication Date)
- Wiley(Publisher)
The next-generation LMTs may have movable secondary mirrors to allow a larger portion of the sky to be viewed. 80 CHAPTER 2 | Fluid Statics CHAPTER SUMMARY In this chapter, the pressure variation in a fluid at rest is considered, along with some important consequences of this type of pressure var- iation. It is shown that for incompressible fluids at rest the pressure varies linearly with depth. This type of variation is commonly referred to as a Hydrostatic Pressure distribution. For compressible fluids at rest, the pressure distribution will not generally be hydrostatic, but Eq. 2.4 remains valid and can be used to determine the pressure dis- tribution if additional information about the variation of the specific weight is specified. The distinction between absolute and gage pres- sure is discussed along with a consideration of barometers for the measurement of atmospheric pressure. Pressure-measuring devices called manometers, which utilize static liquid columns, are analyzed in detail. A brief discussion of mechanical and electronic pressure gages is also presented. Equa- tions for determining the magnitude and location of the resultant fluid force acting on a plane surface in contact with a static fluid are developed. A general approach for determining the magnitude and location of the resultant fluid force acting on a curved surface in contact with a static fluid is developed. For submerged or floating bodies, the concept of the buoyant force and the use of Archimedes’ principle are reviewed. The following checklist provides a study guide for this chap- ter. When your study of the entire chapter has been completed, you should be able to • write out meanings of the terms listed and understand each of the related concepts. These terms are particularly important and are set in bold black type in the text. • calculate the pressure at various locations within an incom- pressible fluid at rest.
eBook - PDF- John Ward-Smith(Author)
- 2018(Publication Date)
- CRC Press(Publisher)
Thus atmospheric pressure is usually effective, even if indirectly, on all surfaces, and over the differences of height normally encountered it is sensibly constant. Consequently it is often simpler to regard atmospheric pressure as the zero of the pressure scale. A pressure expressed relative to atmospheric pressure is known as a gauge pressure . Equation 2.4 then reduces to p = % gh . As we shall see in Section 2.3 , this relation forms the basis of a number of methods of measuring pressure. The direct proportionality between gauge pressure and h for a fluid of constant density enables the pressure to be simply visualized in terms of the vertical distance h = p /% g . The quotient p /% g is termed the pressure head corresponding to p . So useful is the concept of pressure head that it is employed whether or not an actual free surface exists above the point in question. For a liquid without a free surface, as for example in a closed pipe, p /% g corresponds to the height above the pipe to which a free surface would rise if a small vertical tube of sufficient length and open to atmo-sphere – known as a piezometer tube – were connected to the pipe ( Fig. 2.2 ). 46 Fluid statics Fig. 2.3 Provided that % is constant, all pressures may be expressed as heads. Thus pressures are sometimes quoted in terms of millimetres of mercury or metres of water. Equation 2.3 may be divided by % g to give ( p /% g ) + z = constant. That is, Piezometric pressure the sum of the pressure head and the elevation above the chosen horizontal datum plane is constant. This constant is known as the piezometric head and corresponds to the height of the free surface above the datum plane. The quantity p + % gz is termed the piezometric pressure . The fact that an increase of pressure in any part of a confined fluid is transmitted uniformly throughout the fluid is utilized in such devices as the hydraulic press and the hydraulic jack.
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