Chapter 2 FLUID STATICS - PowerPoint PPT Presentation

1 / 111
About This Presentation
Title:

Chapter 2 FLUID STATICS

Description:

Chapter 2 FLUID STATICS – PowerPoint PPT presentation

Number of Views:224
Avg rating:3.0/5.0
Slides: 112
Provided by: Natal214
Category:

less

Transcript and Presenter's Notes

Title: Chapter 2 FLUID STATICS


1
Chapter 2FLUID STATICS
2
  • The science of fluid statics will be treat in two
    parts
  • the study of pressure and its variation
    throughout a fluid
  • the study of pressure forces on finite surfaces.
  • Special cases of fluids moving as solids are
    included in the treatment of statics because of
    the similarity of forces involved.
  • Since there is no motion of a fluid layer
    relative to an adjacent layer, there are no shear
    stresses in the fluid.
  • Hence, all free bodies in fluid statics have only
    normal pressure forces acting on their surfaces.

3
2.1 PRESSURE AT A POINT
  • The average pressure is calculated by dividing
    the normal force pushing against a plane area by
    the area.
  • The pressure at a point is the limit of the ratio
    of normal force to area as the area approaches
    zero size at the point.
  • At a point a fluid at rest has the same pressure
    in all directions. This means that an element dA
    of very small area, free to rotate about its
    center when submerged in a fluid at rest, will
    have a force of constant magnitude acting on
    either side of it, regardless of its orientation.
  • To demonstrate this, a small wedge-shaped free
    body of unit width is taken at the point (x, y)
    in a fluid at rest (Fig. 2.1)

4
Figure 2.1 Free-body diagram of wedge-shaped
particle
5
  • There can be no shear forces, the only forces are
    the normal surface forces and gravity. The
    equations of motion in the x and y directions
  • px, py, ps are the average pressures on the
    three faces, ? is the unit gravity force of the
    fluid, ? is its density, and ax, ay are the
    accelerations.
  • When the limit is taken as the free body is
    reduced to zero size by allowing the inclined
    face to approach (x, y) while maintaining the
    same angle ?, and using the geometric relations
  • The equations simplify to
  • Last term of the second equation is an
    infinitestimal of higher of smallness, may be
    neglected.

6
  • When divided by dy and dx, respectively, the
    equations can be combined
  • (2.1.1)
  • Since ? is any arbitrary angle, this equation
    proves that the pressure is the same in all
    directions at a point in a static fluid.
  • Although the proof was carried out for a
    two-dimensional case, it may be demonstrated for
    the three-dimensional case with the equilibrium
    equations for a small tetrahedron of fluid with
    three faces in the coordinate planes and the
    fourth face inclined arbitrarily.
  • If the fluid is in motion (one layer moves
    relative to an adjacent layer), shear stresses
    occur and the normal stresses are no longer the
    same in all directions at a point. The pressure
    is defined as the average of any three mutually
    perpendicular normal compressive stresses at a
    point,

7
2.2 BASIC EQUATION OF FLUID STATICS
  • Pressure Variation in a Static Fluid
  • Force balance
  • The forces acting on an element of fluid at rest
    (Fig. 2.2) consist of surface forces and body
    forces.
  • With gravity the only body force acting, and by
    taking the y axis vertically upward, it is - ? dx
    dy dz in the y direction.
  • With pressure p at its center (x, y, z) the
    approximate force exerted on the side normal to
    the y axis closest to the origin and on the
    opposite side are approximately
  • where dy/2 is the distance from center to a face
    normal to y.

8
Figure 2.2 Rectangular parallelepiped element of
fluid at rest
9
  • Summing the forces acting on the element in the y
    direction gives
  • For the x and z directions, since no body forces
    act,
  • The elemental force vector dF is given by
  • If the element is reduced to zero size, alter
    dividing through by dx dy dz dV, the expression
    becomes exact.
  • (2.2.1)
  • This is the resultant force per unit volume at a
    point, which must be equated to zero for a fluid
    at rest.
  • The quantity in parentheses is the gradient,
    called ? (del)
  • (2.2.2)

10
  • The negative gradient of , -?p, is the vector
    field f of the surface pressure force per unit
    volume
  • (2.2.3)
  • The fluid static law of variation of pressure is
    then
  • (2.2.4)
  • For an inviscid fluid in motion, or a fluid so
    moving that the shear stress is everywhere zero,
    Newton's second law takes the form
  • (2.2.5)
  • where a is the acceleration of the fluid
    element, f - j? is the resultant fluid force when
    gravity is the only body force acting.

11
  • In component form, Eq. (2.2.4) becomes
  • (2.2.6)
  • The partials, for variation in horizontal
    directions, are one form of Pascal's law they
    state that two points at the same elevation in
    the same continuous mass of fluid at rest have
    the same pressure.
  • Since p is a function of y only,
  • (2.2.7)
  • relates the change of pressure to unit gravity
    force and change of elevation and holds for both
    compressible and incompressible fluids.
  • For fluids that may be considered homogeneous and
    incompressible, ? is constant, and the above
    equation, when integrated, becomes
  • in which c is the constant of integration. The
    hydrostatic law of variation of pressure is
    frequently written in the form
  • (2.2.8)
  • h -y, p is the increase in pressure from that
    at the free surface.

12
  • Example 2.1
  • An oceanographer is to design a sea lab 5 m high
    to withstand submersion to 100 m, measured from
    sea level to the top of the sea lab. Find the
    pressure variation on a side of the container and
    the pressure on the top if the relative density
    of salt water is 1.020.
  • At the top, h 100 m, and
  • If y is measured from the top of the sea lab
    downward, the pressure variation is

13
  • Pressure Variation in a Compressible Fluid
  • When the fluid is a perfect gas at rest at
    constant temperature
  • (2.2.9)
  • When the value of ? in Eq. (2.2.7) is replaced by
    ?g and ? is eliminated between Eqs. (2.2.7) and
    (2.2.9),
  • (2.2.10)
  • If P P0 when ? ?0, integration between limits
  • (2.2.11)
  • (2.2.12)
  • which is the equation for variation of pressure
    with elevation in an isothermal gas.

14
  • The atmosphere frequently is assumed to have a
    constant temperature gradient expressed by
  • (2.2.13)
  • The density may be expressed in terms of pressure
    and elevation from the perfect-gas law
  • (2.2.14)

15
  • Example 2.2
  • Assuming isothermal conditions to prevail in the
    atmosphere, compute the pressure and density at
    2000 m elevation if P 105 Pa, ? 1.24 kg/m3 at
    sea level.
  • From Eq. (2.2.12)
  • Then, from Eq. (2.2.9)

16
2.3 UNITS AND SCALES OF PRESSURE MEASUREMENT
  • Pressure may be expressed with reference to any
    arbitrary datum. The usual data are
  • absolute zero
  • local atmospheric pressure.
  • Absolute pressure - difference between its value
    and a complete vacuum.
  • Gage pressure - difference between its value and
    the local atmospheric pressure.

17
  • The bourdon gage (Fig. 2.3) is typical of the
    devices used for measuring gage pressures.
  • The pressure element is a hollow, curved, flat
    metallic tube closed at one end the other end is
    connected to the pressure to be measured.
  • When the internal pressure is increased, the tube
    tends to straighten, pulling on a linkage to
    which is attached a pointer and causing the
    pointer to move.
  • The dial reads zero when the inside and outside
    of the tube are at the same pressure, regardless
    of its particular value.
  • The gage measures pressure relative to the
    pressure of the medium surrounding the tube,
    which is the local atmosphere.

18
Figure 2.3 Bourdon gage
19
  • Figure 2.4 illustrates the data and the relations
    of the common units of pressure measurement.
  • Standard atmospheric pressure is the mean
    pressure at sea level, 760 mm Hg.
  • A pressure expressed in terms of the length of a
    column of liquid is equivalent to the force per
    unit area at the base of the column. The 
    relation for variation of pressure with altitude
    in a liquid p ?h Eq. (2.2.8) shows the
    relation between head h, in length of a fluid
    column of unit gravity force ?, and the pressure
    p (p is in pascals, ? in newtons per cubic metre,
    and h in metres).
  • With the unit gravity force of any liquid
    expressed as its relative density S times the
    unit gravity force of water
  • (2.3.1)
  • For water ?w may be taken as 9806 N/m3.

20
Figure 2.4 Units and scales for pressure
measurement
21
  • Local atmospheric pressure is measured by
  • mercury barometer (Fig. 2.5) or
  • aneroid barometer, which measures the difference
    in pressure between the atmosphere and an
    evacuated box or tube in a manner analogous to
    the bourdon gage except that the tube is
    evacuated and sealed.
  • Mercury barometer glass tube closed at one end,
    filled with mercury, and inverted so that the
    open end is submerged in mercury.
  • It has a scale so arranged that the height of
    column R can be determined.
  • The space above the mercury contains mercury
    vapor. If the pressure of the mercury vapor hv is
    given in millimetres of mercury and R is measured
    in the same units, the pressure at A may be
    expressed as

Figure 2.5 Mercury barometer
mm Hg
22
  • In Figure 2.4 a pressure may be located
    vertically on the chart, which indicates its
    relation to absolute zero and to local
    atmospheric pressure.
  • If the point is below the local-atmospheric-pressu
    re line and is referred to gage datum, it is
    called negative, suction, or vacuum.
  • For example, the pressure 460 mm Hg abs, as at 1,
    with barometer reading 720 mm, may be expressed
    as -260 mm Hg, 260 mm Hg suction, or 260 mm Hg
    vacuum.
  • It should be noted that
  • Pabs pbar pgage
  • Absolute pressures are symbolized by P, gage
    pressures - p.

23
  • Example 2.3
  • The rate of temperature change in the atmosphere
    with change in elevation is called its lapse
    rate. The motion of a parcel of air depends on
    the density of the parcel relative to the density
    of the surrounding (ambient) air.
  • However, as the parcel ascends through the
    atmosphere, the air pressure decreases, the
    parcel expands, and its temperature decreases at
    a rate known as the dry adiabatic lapse rate. A
    firm wants to burn a large quantity of refuse. It
    is estimated that the temperature of the smoke
    plume at 10 m above the ground will be 11oC
    greater than that of the ambient air. For the
    following conditions determine what will happen
    to the smoke.
  • (a) At standard atmospheric lapse rate ß
    -0.00651oC per meter and t0 20oC.
  • (b) At an inverted lapse rate ß 0.00365oC per
    meter.

24
  • By combining Eqs. (2.2.7) and (2.2.14),
  • The relation between pressure and temperature
    for a mass of gas expanding without heat transfer
    is
  • in which T1 is the initial smoke absolute
    temperature and P0 the initial absolute pressure
    k is the specific heat ratio, 1.4 for air and
    other diatomic gases.
  • Eliminating P/P0 in the last two equations
  • Since the gas will rise until its temperature is
    equal to the ambient temperature,
  • the last two equations may be solved for y. Let
  • Then
  • For ß -0.00651oC per metre, R 287
    m?N/(kg?K), a 2.002, and y 3201 m. For the
    atmospheric temperature inversion ß -0.00365oC
    per metre, a -0.2721, and y 809.2 m.

25
2.4 MANOMETERS
  • Manometers are devices that employ liquid columns
    for determining differences in pressure.
  • The most elementary manometer piezometer
    (Figure 2.6a).
  • It measures the pressure in a liquid when it is
    above zero gage.
  • Glass tube is mounted vertically so that it is
    connected to the space within the container.
  • Liquid rises in the tube until equilibrium is
    reached.
  • The pressure is then given by the vertical
    distance h from the meniscus (liquid surface) to
    the point where the pressure is to be measured,
    expressed in units of length of the liquid in the
    container.
  • Piezometer would not work for negative gage
    pressures, because air would flow into the
    container through the tube.

26
Figure 2.6 Simple manometers
27
  • Figure 2.6b shows that for small negative or
    positive gage pressures in a liquid the tube may
    take the form.
  • With this arrangement the meniscus may come to
    rest below A, as shown. Since the pressure at the
    meniscus is zero gage and since pressure
    decreases with elevation,
  •   
     units of length H2O
  • Figure 2.6c - for greater negative or positive
    gage pressures a second liquid of greater
    relative density is employed.
  • It must be immiscible in the first fluid, which
    may now be a gas.
  • If the relative density of the fluid at A is S1
    (based on water) and the relative density of the
    manometer liquid is S2, the equation for pressure
    at A may be written thus, starting at either A or
    the upper meniscus and proceeding through the
    manometer,
  • hA - the unknown pressure, expressed in length
    units of water,
  • h1, h2 - in length units.

28
  • A general procedure should be followed in working
    all manometer problems
  • Start at one end (or any meniscus if the circuit
    is continuous) and write the pressure there in an
    appropriate unit (say pascals) or in an
    appropriate symbol if it is unknown.
  • Add to this the change in pressure, in the same
    unit, from one meniscus to the next (plus if the
    next meniscus is lower, minus if higher). For
    pascals this is the product of the difference in
    elevation in metres and the unit gravity force of
    the fluid in newtons per cubic metre.
  • Continue until the other end of the gage (or the
    starting meniscus) is reached and equate the
    expression to the pressure at that point, known
    or unknown.
  • The expression will contain one unknown for a
    simple manometer or will give a difference in
    pressures for the differential manometer. In
    equation form,

29
  • A differential manometer (Fig. 2.7) determines
    the difference in pressures at two points A and B
    when the actual pressure at any point in the
    system cannot be determined.
  • Application of the procedure outlined above to
    Fig. 2.7a produces
  • Similarly, for Fig. 2.7b
  • If the pressures at A and B are expressed in
    length of the water column, the above results can
    be written, for Fig. 2.7a,
  • Similarly, for Fig 2.7b
  • in which S1, S2, and S3 are the applicable
    relative densities of the liquids in the system.

30
Figure 2.7 Differential manometers
31
  • Example 2.4  
  • In Fig. 2.7a the liquids at A and B are water
    and the manometer liquid is oil. S 0.80 h1
    300 mm h2 200 mm and h3 600 mm.
  • (a) Determine pA - pB, in pascals.
  • (b) If pB 50 kPa and the barometer reading is
    730 mm Hg, find the pressure at A, in meters of
    water absolute.
  • (a)
  • (b)
  • From (a)

32
  • Micromanometers
  • Use for determining very small differences in
    pressure or determining large pressure
    differences precisely.
  • One type very accurately measures the differences
    in elevation of two menisci of a manometer.
  • By means of small telescopes with horizontal
    cross hairs mounted along the tubes on a rack
    which is raised and lowered by a pinion and slow
    motion screw so that the cross hairs can be set
    accurately, the difference in elevation of
    menisci (the gage difference) can be read with
    verniers.

33
Figure 2.8 Micromanometer using two gage liquids
34
  • With two gage liquids, immiscible in each other
    and in the fluid to be measured, a large gage
    difference R (Fig. 2.8) can be produced for a
    small pressure difference.
  • The heavier gage liquid fills the lower U tube up
    to 0-0 then the lighter gage liquid is added to
    both sides, filling the larger reservoirs up to
    1-1. The gas or liquid in the system fills the
    space above 1-1. When the pressure at C is
    slightly greater than at D, the menisci move as
    indicated in Fig. 2.8.
  • The volume of liquid displaced in each reservoir
    equals the displacement in the U tube ?
  • Manometer equation
  • in which ?1, ?2 and ?3 are the unit gravity
    force. Simplifying for ?y
  • (2.4.1)

35
  • Example 2.5
  • In the micromanometer of Fig 2.8 the pressure
    difference is wanted, in pascals, when air is in
    the system, S2 1.0, S3 1.10, a/A 0.01, R
    5 mm, t 20oC, and the barometer reads 760 mm
    Hg.
  • The term ?1(a/A) may be neglected. Substituting
    into Eq. (2.4.1) gives

36
Figure 2.9 Inclined manometer
  • The inclined manometer (Fig. 2.9) is frequently
    used for measuring small differences in gas
    pressures.
  • It is adjusted to read zero, by moving the
    inclined scale, when A and B are open. Since the
    inclined tube requires a greater displacement of
    the meniscus for given pressure difference than a
    vertical tube, it affords greater accuracy in
    reading the scale.
  • Surface tension causes a capillary rise in small
    tubes. If a U tube is used with a meniscus in
    each leg, the surface-tension effects cancel.

37
2.5 FORCES ON PLANE AREAS
  • In the preceding sections variations of pressure
    throughout a fluid have been considered.
  • The distributed forces resulting from the action
    of fluid on a finite area may be conveniently
    replaced by a resultant force, insofar as
    external reactions to the force system are
    concerned.
  • In this section the magnitude of resultant force
    and its line of action (pressure center) are
    determined by integration, by formula, and by use
    of the concept of the pressure prism.

38
Horizontal Surfaces
  • A plane surface in a horizontal position in a
    fluid at rest is subjected to a constant
    pressure.
  • The magnitude of the force acting on one side of
    the surface is
  • The elemental forces p dA acting on A are all
    parallel and in the same sense therefore a
    scalar summation of all such elements yields the
    magnitude of the resultant force. Its direction
    is normal to the surface and toward the surface
    if p is positive.
  • To find the line of action of the resultant,
    i.e., the point in the area where the moment of
    the distributed force about any axis through the
    point is zero, arbitrary xy axes may be selected,
    as in Fig. 2.10.
  • Then, since the moment of the resultant must
    equal the moment of the distributed force system
    about any axis, say the y axis,
  • x is the distance from the y axis to the
    resultant.

39
Figure 2.10 Notation for determining the line
of action of a force
40
Inclined Surfaces
  • In Fig. 2.11 a plane surface is indicated by its
    trace A'B' it is inclined ?o from the
    horizontal. The intersection of the plane of the
    area and the free surface is taken as x axis. The
    y axis is taken in the plane of the area, with
    origin O in the free surface. The xy plane
    portrays the arbitrary inclined area.
  • The magnitude, direction, and line of action of
    the resultant force due to the liquid, acting on
    one side of the area, are sought.
  • For an element with area dA (2.5.1)
  • Since all such elemental forces are parallel, the
    integral over the area yields the magnitude of
    force F, acting on one side of the area,
  • (2.5.2)
  • The magnitude of force exerted on one side of a
    plane area submerged in a liquid is the product
    of the area and the pressure at its centroid. In
    this form the presence of a free surface is
    unnecessary.

41
Figure 2.11 Notation for force of liquid on one
side of a plane inclined area
42
Center of Pressure
  • The line of action of the resultant force has its
    piercing point in the surface at a point called
    the pressure center, with coordinates (xp, yp)
    (Fig. 2.11). The center of pressure of an
    inclined surface is not at the centroid. To find
    the pressure center, the moments of the resultant
    xpF, ypF are equated to the moment of the
    distributed forces about the y axis and x axis,
    respectively thus
  • (2.5.3, 2.5.4)
  • (2.5.5, 2.5.6)
  • Eqns may be evaluated conveniently through
    graphical integration for simple areas they may
    be transformed into general formulas
  • (2.5.7)
  • (2.5.8)

43
  • When either of the centroidal axes is an axis of
    symmetry for the surface, vanishes and
    the pressure center lies on x x-. Since
    may be either positive or negative, the
    pressure center may lie on either side of the
    line x x-. To determine yp by formula, with
    Eqs. (2.5.2) and (2.5.6),
  • (2.5.9)
  • In the parallel-axis theorem for moments of
    inertia
  • in which IG is the second moment of the area
    about its horizontal centroidal axis.

44
If Ix is eliminated from Eq. (2.5.9)
(2.5.10) or (2.5.11) IG is always positive
hence, yp y is always positive and the
pressure center is always below the centroid of
the surface.
45
Example 2.6 The triangular gate CDE (Fig. 2.12)
is hinged along CD and is opened by a normal
force P applied at E. It holds oil, relative
density 0.80, above it and is open to the
atmosphere on its lower side Neglecting the
weight of the gate, find (a) the magnitude of
force exerted on the gate by integration and by
Eq. (2.5.2) (b) the location of pressure center
(c) the force P needed to open the gate.
Figure 2.12 Triangular gate
46
  • (a) By integration with reference to Fig. 2.12,
  • When y 4, x 0, and when y 6.5, x 3, with
    x varying linearly with y thus
  • in which the coordinates have been substituted
    to find x in terms of y. Solving for a and b
    gives
  • Similarly, y 6.5, x 3 y 9, x 0 and x
    6/5(9 - y). Hence,
  • Integrating and substituting for ? sin? leads to
  • By Eq. (2.5.2)

47
  • (b) With the axes as shown,
  • In Eq. (2.5.8)
  • I-xy is zero owing to symmetry about the
    centroidal axis parallel to the x axis hence
  • In Eq. (2.5.11),
  • i.e., the pressure center is 0.16 m below the
    centroid, measured in the plane of the area.
  • (c) When moments about CD are taken and the
    action of the oil is replaced by the resultant,

48
The Pressure Prism
  • Another approach to the problem of determine the
    resultant force and line of action of the force
    on a plane surface is given by the concept of a
    pressure prism. It is a prismatic volume with its
    base the given surface area and with altitude at
    any point of the base given by p ?h.
  • h is the vertical distance to the free surface
    (see Fig. 2.13). In the figure, ?h may be laid
    off to any convenient scale such that its trace
    is OM.
  • The force acting on an elemental area dA is
  • (2.5.12)
  • which is an element of volume of the pressure
    prism. After integrating F V.
  • From Eqs. (2.5.5) and (2.5.6),
  • (2.5.13)
  • which show that xp, yp are distances to the
    centroid of the pressure prism. Hence, the line
    of action of the resultant passes through the
    centroid of the pressure prism.

49
Figure 2.13 Pressure prism
50
Effects of Atmospheric Pressure on Forces on
Plane Areas
  • In the discussion of pressure forces the pressure
    datum was not mentioned. The pressure were
    computed by p ?h ? the datum taken was gage
    pressure zero, or the local atmospheric pressure.
  • When the opposite side of the surface is open to
    the atmosphere, a force is exerted on it by the
    atmosphere equal to the product of the
    atmospheric pressure P0 and the area, or P0A,
    based on absolute zero as datum. On the liquid
    side the force is
  • The effect P0A of the atmosphere acts equally on
    both sides and in no way contributes to the
    resultant force or its location.
  • So long as the same pressure datum is selected
    for all sides of a free body, the resultant and
    moment can be determined by constructing a free
    surface at pressure zero on this datum and using
    the above methods.

51
  • Example 2.7
  • An application of pressure forces on plane areas
    is given in the design of a gravity dam. The
    maximum and minimum compressive stresses in the
    base of the dam are computed from the forces
    which act on the dam.
  • Figure 2.15 shows a cross section through a
    concrete dam where the unit gravity force of
    concrete has been taken as 2.5? and ? is the unit
    gravity force of water. A 1 m section of dam is
    considered as a free body the forces are due to
    the concrete, the water, the foundation pressure,
    and the hydrostatic uplift.
  • Determining amount of hydrostatic uplift is
    beyond the scope of this treatment, but it will
    be assumed to be one-half the hydrostatic head at
    the upstream edge, decreasing linearly to zero at
    the downstream edge of the dam. Enough friction
    or shear stress must be developed at the base of
    the dam to balance the thrust due to the water
    that is Rx 5000?.
  • The resultant upward force on the base equals the
    gravity force of the dam less the hydrostatic
    uplift Ry 6750? 2625? - 1750? 7625? N. The
    position of Ry is such that the free body is in
    equilibrium. For moments around O,

52
Figure 2.15 Concrete gravity dam
53
  • It is customary to assume that the foundation
    pressure varies linearly over the base of the
    dam, i.e., that the pressure prism is a trapezoid
    with a volume equal to Ry thus
  • in which smax, smin are the maximum and minimum
    compressive stresses in pascals. The centroid of
    the pressure prism is at the point where x 44.8
    m. By taking moments about O to express the
    position of the centroid in terms of smax and
    smin,
  • Simplifying gives
  • When the resultant falls within the middle third
    of the base of the dam, smin will always be a
    compressive stress. Owing to the poor tensile
    properties of concrete, good design requires the
    resultant to fall within the middle third of the
    base.

54
2.6 FORCE COMPONENTS ON CURVED SURFACES
  • When the elemental forces p dA vary in direction,
    as in the case of a curved surface, they must be
    added as vector quantities
  • their components in three mutually perpendicular
    directions are added as scalars, and then the
    three components are added vectorially.
  • With two horizontal components at right angles
    and with the vertical component - all easily
    computed for a curved surface - the resultant can
    be determined.
  • The lines of action of the components also are
    readily determined.

55
Horizontal Component of Force on a Curved Surface
  • The horizontal component  pressure force on a
    curved surface is equal to the pressure force
    exerted on a projection of the curved surface.
    The vertical plane of projection is normal to the
    direction of the component.
  • The surface of Fig. 2.16 represents any
    three-dimensional surface, and dA an element of
    its area, its normal making the angle ? with the
    negative x direction. Then
  • (2.6.1)
  • Projecting each element on a plane perpendicular
    to x is equivalent to projecting the curved
    surface as a whole onto the vertical plane.
  • Hence, force acting on this projection of the
    curved surface is the horizontal component of
    force exerted on the curved surface in the
    direction normal to the plane of projection.
  • To find the horizontal component at right angles
    to the x direction, the curved surface is
    projected onto a vertical plane parallel to x and
    the force on the projection is determined.

56
Figure 2.16 Horizontal component of force on a
curved surface
Figure 2.17 Projections of area elements on
opposite sides of a body
57
  • When the horizontal component of pressure force
    on a closed body is to be found, the projection
    of the curved surface on a vertical plane is
    always zero, since on opposite sides of the body
    the area-element projections have opposite signs
    (see Fig. 2.17).
  • Let a small cylinder of cross section dA with
    axis parallel to x intersect the closed body at B
    and C. If the element of area of the body cut by
    the prism at B is dAB and at C is dAC, then
  • and similarly for all other area elements.
  • To find the line of action of a horizontal
    component of force on a curved surface, the
    resultant of the parallel force system composed
    of the force components from each area element is
    required.
  • This is exactly the resultant of the force on the
    projected area, since the two force systems have
    an identical distribution of elemental horizontal
    force components.

58
  • Example 2.8
  • The equation of an ellipsoid of revolution
    submerged in water is x2/4 y2/4 z2/9 1. The
    center of the body is located 2 m below the free
    surface. Find the horizontal force components
    acting on the curved surface that is located in
    the first octant.
  • Consider the xz plane to be horizontal and y to
    be positive upward.
  • Projection of the surface on the yz plane has an
    area of
  • Its centroid is located m below the free
    surface.
  • Hence,

59
Vertical Component of Force on a Curved Surface
  • The vertical component of pressure force on a
    curved surface is equal to the weight surface and
    extending up to the free surface. Can be
    determined by summing up the vertical components
    of pressure force on elemental areas dA of the
    surface.
  • In Fig. 2.18 an area element is shown with the
    force p dA acting normal to it. Let ? be the
    angle the normal to the area element makes with
    the vertical. Then the vertical component of
    force acting on the area element is p cos ? dA,
    and the vertical component of force on the curved
    surface is given by
  • (2.6.2)
  • When p replaced by its equivalent ?h, and it is
    noted that cos ? dA is the projection of dA on a
    horizontal plane ? Eq. (2.6.2) becomes
  • (2.6.3-2.6.4)
  • in which dV is the volume of the prism of
    height h and base cos ? dA, or the volume of
    liquid vertically above the area element.

60
Figure 2.18 Vertical component of force on a
curved surface
Figure 2.19 Liquid with equivalent free surface
61
  • When the liquid is below the curved surface (Fig.
    2.19) and the pressure magnitude is known at some
    point (e.g., O), an imaginary or equivalent free
    surface s-s can be constructed p/? above O, so
    that the product of unit gravity force and
    vertical distance to any point in the tank is the
    pressure at the point.
  • The weight of the imaginary volume of liquid
    vertically above the curved surface is then the
    vertical component of pressure force on the
    curved surface.
  • In constructing an imaginary free surface, the
    imaginary liquid must be of the same unit gravity
    force as the liquid in contact with the curved
    surface otherwise, the pressure distribution
    over the surface will not be correctly
    represented.
  • With an imaginary liquid above a surface, the
    pressure at a point on the curved surface is
    equal on both sides, but the elemental force
    components in the vertical direction are opposite
    in sign. Hence, the direction of the vertical
    force component is reversed when an imaginary
    fluid is above the surface.
  • In some cases a confined liquid may be above the
    curved surface, and an imaginary liquid must be
    added (or subtracted) to determine the free
    surface.

62
  • The line of action of the vertical component is
    determined by equating moments of the elemental
    vertical components about a convenient axis with
    the moment of the resultant force.
  • With the axis at O (Fig. 2.18),
  • in which is the distance from O to the
    line of action.
  • Since Fv ?V
  • the distance to the centroid of the volume.
  • Therefore, the line of action of the vertical
    force passes through the centroid of the volume,
    real or imaginary, that extends above the curved
    surface up to the real or imaginary free surface.

63
  • Example 2.9
  • A cylindrical barrier (Fig. 2.20) holds water as
    shown. The contact between cylinder and wall is
    smooth. Considering a 1-m length of cylinder,
    determine (a) its gravity force and (b) the force
    exerted against the wall.
  • (a) For equilibrium the weight of the cylinder
    must equal the vertical component of force
    exerted on it by the water. (The imaginary free
    surface for CD is at elevation A.) The vertical
    force on BCD is
  • The vertical force on AB is
  • Hence, the gravity force per metre of length is
  • (b) The force exerted against the wall is the
    horizontal force on ABC minus the horizontal
    force on CD. The horizontal components of force
    on BC and CD cancel the projection of BCD on a
    vertical plane is zero ?,
  • since the projected area is 2 m2 and the
    pressure at the centroid of the projected area is
    9806 Pa.

64
Figure 2.20 Semifloating body
65
Tensile Stress in a Pipe and Spherical Shell
  • A circular pipe under the action of an internal
    pressure is in tension around its periphery.
    Assuming that no longitudinal stress occurs, the
    walls are in tension, as shown in Fig. 2.21.
  • Consider a section of pipe of unit length (the
    ring between two planes normal to the axis and
    unit length apart). If one-half of this ring is
    taken as a free body, the tensions per unit
    length at top and bottom are respectively T1 and
    T2 (as shown in the figure).
  • The horizontal component of force acts through
    the pressure center of the projected area and is
    2pr, in which p is the pressure at the centerline
    and r is the internal pipe radius.
  • For high pressures the pressure center may be
    taken at the pipe center then T1 T2, and
  • (2.6.5)
  • in which T is the tensile force per unit length.
    For wall thickness e, the tensile stress in the
    pipe wall is
  • (2.6.6)

66
Figure 2.21 Tensile stress in pipe
67
  • Example 2.10
  • A 100 mm-ID steel pipe has a 6 mm wall
    thickness. For an allowable tensile stress of 70
    MPa, what is the maximum pressure?
  • From Eq. (2.6.6)

68
2.7 BUOYANT FORCE
  • Buoyant force - the resultant force exerted on a
    body by a static fluid in which it is submerged
    or floating.
  • Always acts vertically upward. There can be no
    horizontal component of the resultant because the
    projection of the submerged body or submerged
    portion of the floating body on a vertical plane
    is always zero.
  • The buoyant force on a submerged body is the
    difference between the vertical component of
    pressure force on its underside and the vertical
    component of pressure force on its upper side.

69
  • In Fig. 2.22 the upward force on the bottom is
    equal to the gravity force of liquid, real or
    imaginary, which is vertically above the surface
    ABC, indicated by the gravity force of liquid
    within ABCEFA. The downward force on the upper
    surface equals the gravity force of liquid
    ADCEFA.
  • The difference between the two forces is a force,
    vertically upward, due to the gravity force of
    fluid ABCD that is displaced by the solid. In
    equation form
  • (2.7.1)
  • FB is buoyant force, V is the volume of fluid
    displaced, and ? is the unit gravity force of
    fluid.
  • The same formula holds for floating bodies when V
    is taken as the volume of liquid displaced.

70
Figure 2.22 Buoyant force on floating and
submerged bodies
71
  • In Fig. 2.23 the vertical force exerted on an
    element of the body in the form of a vertical
    prism of cross section dA is
  • dV is the volume of the prism. Integrating over
    the complete body gives
  • when ? is considered constant throughout the
    volume.
  • To find the line of action of the buoyant force,
    moments are taken about a convenient axis O and
    are equated to the moment of the resultant, thus,
  • x is the distance from the axis to the line of
    action.
  • This equation yields the distance to the centroid
    of the volume ? the buoyant force acts through
    the centroid of the displaced volume of fluid.
    This holds for both submerged and floating
    bodies.
  • The centroid of the displaced volume of fluid is
    called the center of buoyancy.

72
Figure 2.23 Vertical force components on element
of body
73
  • Determining gravity force on an odd-shaped object
    suspended in two different fluids yields
    sufficient data to determine its gravity force,
    volume, unit gravity force, and relative density.
  • Figure 2.24 shows two free-body diagrams for the
    same object suspended and gravity force
    determined in two fluids, F1 and F2 ?1 and ?2
    are the unit gravity forces of the fluids. W and
    V, the gravity force and volume of the object,
    are to be found.
  • The equations of equilibrium are written and
    solved

74
Figure 2.24 Free body diagrams for body
suspended in a fluid
75
  • A hydrometer uses the principle of buoyant force
    to determine relative densities of liquids.
  • Figure 2.25 shows a hydrometer in two liquids
    with a stem of prismatic cross section a.
  • Considering the liquid on the left to be
    distilled water (unit relative density S 1.00),
    the hydrometer floats in equilibrium when
  • (2.7.2) 
  • V0 is the volume submerged, ? is the unit
    gravity force of water, and W is the gravity
    force of hydrometer.
  • The position of the liquid surface is marked 1.00
    on the stem to indicate unit relative density S.
    When the hydrometer is floated in another liquid,
    the equation of equilibrium becomes
  • (2.7.3)
  • where ?V a?h. Solving for ?h with Eqs. (2.7.2)
    and (2.7.3)
  • (2.7.4)

76
Figure 2.25 Hydrometer in water and in liquid of
relative density
77
  • Example 2.11
  • A piece of ore having a gravity force of 1.5 N
    in air is found to have a gravity force 1.1 N
    when submerged in water. What is its volume, in
    cubic centimetres, and what is its relative
    density?
  • The buoyant force due to air may be neglected.
    From Fig. 2.24

78
2.8 STABILITY OF FLOATING AND SUBMERGED BODIES
  • A body floating in a static liquid has vertical
    stability. A small upward displacement decreases
    the volume of liquid displaced, resulting in an
    unbalanced downward force which tends to return
    the body to its original position.
  • Similarly, a small downward displacement results
    in a greater buoyant force, which causes an
    unbalanced upward force.
  • A body has linear stability when a small linear
    displacement in any direction sets up restoring
    forces tending to return it to its original
    position.
  • A body has rotational stability when a restoring
    couple is set up by any small angular
    displacement.

79
  • Methods for determining rotational stability are
    developed in the following discussion.
  • A body may float in
  • stable equilibrium
  • unstable equilibrium (any small angular
    displacement sets up a couple that tends to
    increase the angular displacement)
  • neutral equilibrium (any small angular
    displacement sets up no couple whatever).
  • Figure 2.26 illustrates the three cases of
    equilibrium
  • a light piece of wood with a metal mass at its
    bottom is stable
  • when the metal mass is at the top, the body is in
    equilibrium but any slight angular displacement
    causes it to assume the position in a
  • a homogeneous sphere or right-circular cylinder
    is in equilibrium for any angular rotation i.e.,
    no couple results from an angular displacement.

80
Figure 2.26 Examples of stable, unstable, and
neutral equilibrium
81
  • A completely submerged object is rotationally
    stable only when its center of gravity is below
    the center of buoyancy (Fig. 2.27a)
  • When the object is rotated counterclockwise, the
    buoyant force and gravity force produce a couple
    in the clockwise direction (see Fig. 2.27b)

Figure 2.27 Rotationally stable submerged body
82
  • Normally, when a body is too heavy to float, it
    submerges and goes down until it rests on the
    bottom.
  • Although the unit gravity force of a liquid
    increases slightly with depth, the higher
    pressure tends to cause the liquid to compress
    the body or to penetrate into pores of solid
    substances and thus decrease the buoyancy of the
    body.
  • A ship, for example, is sure to go to the bottom
    once it is completely submerged, owing to
    compression of air trapped in its various parts.

83
Determination of Rotational Stability of Floating
Objects
  • Any floating object with center of gravity below
    its center of buoyancy (centroid of displaced
    volume) floats in stable equilibrium (see Fig.
    2.26a). Certain floating objects, however, are in
    stable equilibrium when their center of gravity
    is above the center of buoyancy.
  • Figure 2.28a is a cross section of a body with
    all other parallel cross sections identical. The
    center of buoyancy is always at the centroid of
    the displaced volume, which is at the centroid of
    the cross-sectional area below liquid surface in
    this case.

84
Figure 2.28 Stability of a prismatic body
85
  • Hence, when the body is tipped (Fig. 2.28b), the
    center of buoyancy is at the centroid B' of the
    trapezoid ABCD, the buoyant force acts upward
    through B', and the gravity force acts downward
    through G, the center of gravity of the body.
  • When the vertical through B' intersects the
    original centerline above G, as at M, a restoring
    couple is produced and the body is in stable
    equilibrium.
  • The intersection of the buoyant force and the
    centerline is called the metacenter (M).
  • When M is above G, the body is stable when below
    G, it is unstable and when at G, it is in
    neutral equilibrium.
  • The distance MG is called the metacentric height
    and is a direct measure of the stability of the
    body. The restoring couple is
  • in which ? is the angular displacement and W the
    gravity force of the body.

86
  • Example 2.12
  • In Fig. 2.28 a scow 6 m wide and 20 m long has a
    gross mass of 200 Mg. Its center of gravity is 30
    cm above the water surface. Find the metacentric
    height and restoring couple when ?y 30 cm.
  • The depth of submergence h in the water is
  • The centroid in the tipped position is located
    with moments about AB and BC,
  • By similar triangles AEO and B'PM,

87
  • G is 1.97 m from the bottom hence
  • The scow is stable, since is positive
    the righting moment is

88
Nonprismatic Cross Sections
  • For a floating object of variable cross section
    (e.g., a ship) (Fig. 2.29a), a convenient formula
    can be developed for determination of metacentric
    height for very small angles of rotation ?.
  • The horizontal shift in center of buoyancy r
    (Fig. 2.29b) is determined by the change in
    buoyant forces due to the wedge being submerged,
    which causes an upward force on the left, and by
    the other wedge decreasing the buoyant force by
    an equal amount ?FB on the right.
  • The force system, consisting of the original
    buoyant force at B and the couple ?FB s due to
    the wedges, must have as resultant the equal
    buoyant force at B'. With moments about B to
    determine the shirt r,
  • (2.8.1)

89
Figure 2.29 Stability relations in a body of
variable cross section
90
  • The amount of the couple can be determined with
    moments about O, the centerline of the body at
    the liquid surface.
  • For an element of area dA on the horizontal
    section through the body at the liquid surface,
    an element of volume of the wedge is x? dA. The
    buoyant force due to this element is ?x? dA, and
    its moment about O is ?x2? dA, in which ? is the
    small angle of tip in radians.
  • By integrating over the complete original
    horizontal area at the liquid surface, the couple
    is determined to be
  • (2.8.2)
  • I is the moment of inertia of the area about the
    axis y - y (Fig. 2.29a) Substitution into Eq.
    (2.8.1) produces
  • V is the total volume of liquid displaced.
  • Since ? is very small,
  • and
  • (2.8.3)

91
  • Example 2.13
  • A barge displacing 1 Gg has the horizontal cross
    section at the waterline shown in Fig. 2.30. Its
    center of buoyancy is 2.0 m below the water
    surface, and its center of gravity is 0.5 m below
    the water surface. Determine its metacentric
    height for rolling (about y - y axis) and for
    pitching (about x - x axis).
  • GB 2 0.5 1.5 m
  • For rolling
  • For pitching

92
Figure 2.30 Horizontal cross section of a ship
at the waterline
93
2.9 RELATIVE EQUILIBRIUM
  • In fluid statics the variation of pressure is
    simple to compute, thanks to the absence of shear
    stresses. For fluid motion such that no layer
    moves relative to an adjacent layer, the shear
    stress is also zero throughout the fluid.
  • A fluid with a translation at uniform velocity
    still follows the laws of static variation of
    pressure.
  • When a fluid is being accelerated so that no
    layer moves relative to an adjacent one (when the
    fluid moves as if it were a solid), no shear
    stresses occur and variation in pressure can be
    determined by writing the equation of motion for
    an appropriate free body.
  • Two cases are of interest, a uniform linear
    acceleration and a uniform rotation about a
    vertical axis. When moving thus, the fluid is
    said to be in relative equilibrium.

94
Uniform Linear Acceleration
  • A liquid in an open vessel is given a uniform
    linear acceleration a as in Fig. 2.31. After some
    time the liquid adjusts to the acceleration so
    that it moves as a solid i.e., the distance
    between any two fluid particles remains fixed,
    and no shear stresses occur.
  • By selecting a cartesian coordinate system with y
    vertical and x such that the acceleration vector
    a is in the xy plane (Fig. 2.31a), the z axis is
    normal to a and there is no acceleration
    component in that direction. Eqn (2.2.5) applies
    to this situation,
  • (2.2.5)
  • Fig. 2.31b shows the pressure gradient ?p is then
    the vector sum of -?a and -j?.
  • Since ?p is in the direction of maximum change in
    p (the gradient), at right angles to ?p there is
    no change in p. Surfaces of constant pressure,
    including the free surface, must therefore be
    normal to ?p.

95
Figure 2.31 Acceleration with free surface
96
  • To obtain a convenient algebraic expression for
    variation of p with x, y, and z, that is, p
    p(x, y, z), Eq. (2.2.5) is written in component
    form
  • Since p is a function of position (x, y, z), its
    total differential is
  • Substituting for the partial differentials gives
  • (2.9.1)
  • which can be integrated for an incompressible
    fluid,

97
  • To evaluate the constant of integration c, let x
    0, y 0, p p0 then c  p0 and
  • (2.9.2)
  • When the accelerated incompressible fluid has a
    free surface, its equation is given by setting p
    0 in the above Eq. (2.9.2). Solving it for y
    gives
  • (2.9.3)
  • The lines of constant pressure, p const, have
    the slope
  • and are parallel to the free surface. The y
    intercept of the free surface is

98
  • Example 2.14
  • The tank in Fig. 2.32 is filled with oil,
    relative density 0.8, and accelerated as shown.
    There is a small opening in the rank at A.
    Determine the pressure at B and C and the
    acceleration ax required to make the pressure at
    B zero.
  • By selecting point A as origin and by applying
    Eq. (2.9.2) for ay 0
  • At B, x 1.8 m, y - 1.2 m, and p 2.35 kPa.
  • At C, x -0.15 m, y -1.35 m, and p 11.18
    kPa.
  • For zero pressure at B, from Eq. (2.9.2) with
    origin at A,

99
Figure 2.32 Tank completely filled with liquid
100
  • Example 2.15
  • A closed box with horizontal base 6 by 6 units
    and a height of 2 units is half-filled with
    liquid (Fig. 2.33). It is given a constant linear
    acceleration ax g/2, ay -g/4. Develop an
    equation for variation of pressure along its
    base.
  • The free surface has the slope
  • hence, the free surface is located as shown in
    the figure. When the origin is taken at 0, Eq.
    (2.9.2) becomes
  • Then, for y 0, along the bottom,

101
Figure 2.33 Uniform linear acceleration of
container
102
Uniform Rotation about a Vertical Axis
  • Rotation of a fluid, moving as a solid, about an
    axis is called forced-vortex motion.
  • Every particle of fluid has the same angular
    velocity.
  • This motion is to be distinguished from
    free-vortex motion, in which each particle moves
    in a circular path with a speed varying inversely
    as the distance from the center.
  • A liquid in a container, when rotated about a
    vertical axis at constant angular velocity, moves
    like a solid alter some time interval.
  • No shear stresses exist in the liquid, and the
    only acceleration that occurs is directed
    radially inward toward the axis of rotation.
  • By selecting a coordinate system (Fig. 2.34a)
    with the unit vector i in the r direction and j
    in the vertical upward direction with y the axis
    of rotation, the following equation may be
    applied to determine pressure variation
    throughout the fluid
  • (2.2.5)

103
Figure 2.34 Rotation of a fluid about a vertical
axis
104
  • For constant angular velocity ?, any particle of
    fluid P has an acceleration ?2r directed radially
    inward (a -i?2r).
  • Vector addition of -j? and -?a (Fig. 2.34b)
    yields ?p, the pressure gradient. The pressure
    does not vary normal to this line at a point ? if
    P is taken at the surface, the free surface is
    normal to ?p.
  • Expanding Eq. (2.2.5)
  • k is the unit vector along the z axis (or
    tangential direction). Then
  • Since p is a function of y and r , the total
    differential dp is
  • For a liquid (? const) integration yields
  • in which c is the constant of integration.

105
  • If the value of pressure at the origin (r 0, y
    0) is p0, then c p0 and
  • (2.9.5)
  • When the particular horizontal plane (y 0) for
    which p0 0 is selected and the above equation
    is divided by ?,
  • (2.9.6)
  • which shows that the head, or vertical depth,
    varies as the square of the radius.
  • The surfaces of equal pressure are paraboloids
    of revolution.

106
  • When a free surface occurs in a container that is
    being rotated, the fluid volume underneath the
    paraboloid of revolution is the original fluid
    volume.
  • The shape of the paraboloid depends only upon the
    angular velocity ?. For a circular cylinder
    rotating about its axis (Fig. 2.35) the rise of
    liquid from its vertex to the wall of the
    cylinder is ?2r02/2g (from Eq. (2.9.6)).
  • Since a paraboloid of revolution has a volume
    equal to one-half its circumscribing cylinder,
    the volume of the liquid above the horizontal
    plane through the vertex is
  • When the liquid is at rest, this liquid is also
    above the plane through the vertex to a uniform
    depth of
  • Hence, the liquid rises along the walls the same
    amount as the center drops, thereby permitting
    the vertex to be located when ?, r0, and depth
    before rotation are given.

107
Figure 2.35 Rotation of circular cylinder about
its axis
108
  • Example 2.16
  • A liquid, relative density 1.2, is rotated at
    200 rpm about a vertical axis. At one point A in
    the fluid 1 m from the axis, the pressure is 70
    kPa. What is the pressure at a point B which is 2
    m higher than A and 1.5 m from the axis?
  • When Eq. (2.9.5) is written for the two points,
  • Then ? 200 2p/60 20.95 rad/s, ? 1.2
    9806 11.767 N/m3, rA 1 m, and rB 1.5 m.
  • When the second equation is subtracted from the
    first and the values are substituted,
  • Hence

109
  • Example 2.17
  • A straight tube 2 m long, closed at the bottom
    and filled with water, is inclined 30o with the
    vertical and rotated about a vertical axis
    through  its midpoint 6.73 rad/s. Draw the
    paraboloid of zero pressure, and determine the
    pressure at the bottom and midpoint of the tube.
    In Fig. 2.36, the zero-pressure paraboloid passes
    through point A. If the origin is taken at the
    vertex, that is, p0 0, Eq. (2.9.6) becomes
  • which locates the vertex at O, 0.577 m below A.
    The pressure at the bottom of the tube is or
  • At the midpoint, 0.289 m and

110
Figure 2.36 Rotation of inclined tube of liquid
about a vertical axis
111
Fluid Pressure Forces in Relative Equilibrium
  • The magnitude of the force acting on a plane area
    in contact with a liquid accelerating as a rigid
    body can be obtained by integrating over the
    surface
  • The nature of the acceleration and orientation of
    the surface governs the particular variation of p
    over the surface.
  • When the pressure varies linearly over the plane
    surface (linear acceleration), the magnitude of
    force is given by the product of pressure at the
    centroid and area, since the volume of the
    pressure prism is given by pG A.
  • For nonlinear distributions the magnitude and
    line of action can be found by integration.
Write a Comment
User Comments (0)
About PowerShow.com