Transcript Slide 1

Application of Environment Spatial Information System
CHAPTER 2
FLUID STATICS
Minkasheva Alena
Thermal Fluid Engineering Lab.
Department of Mechanical Engineering
Kangwon National University
• 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.
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 δA 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)
Figure 2.1 Free-body diagram of wedge-shaped
particle
• 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, γ is the unit gravity
force of the fluid, ρ 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 θ, 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.
• When divided by δy and δx, respectively, the equations can be combined:
(2.1.1)
• Since θ 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,
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 - γ δx δy δz 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 δy/2 is the distance from center to a face normal to y.
Figure 2.2
Rectangular
parallelepiped
element of fluid
at rest
• 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 δF is given by
• If the element is reduced to zero size, alter dividing through by δx δy δz
= δV, 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)
• 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 - jγ is the resultant fluid
force when gravity is the only body force acting.
• 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, γ 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.
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
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 γ in Eq. (2.2.7) is replaced by ρg and ρ is eliminated
between Eqs. (2.2.7) and (2.2.9),
(2.2.10)
• If P = P0 when ρ = ρ0, integration between limits
(2.2.11)
(2.2.12)
which is the equation for variation of pressure with elevation in an
isothermal gas.
•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)
Example 2.2
Assuming isothermal conditions to prevail in the atmosphere,
compute the pressure and density at 2000 m elevation if P = 105 Pa, ρ
= 1.24 kg/m3 at sea level.
From Eq. (2.2.12):
Then, from Eq. (2.2.9):
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.
• 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.
Figure 2.3 Bourdon gage
• 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 = γh [Eq.
(2.2.8)] shows the relation between head h, in length of a fluid column
of unit gravity force γ, and the pressure p (p is in pascals, γ 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 γw may be taken as 9806 N/m3.
Figure 2.4 Units and scales for pressure measurement
• 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
mm Hg
Figure 2.5
Mercury
barometer
• 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-pressure 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.
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 β = -0.00651oC per meter and
t0 = 20oC.
(b) At an inverted lapse rate β = 0.00365oC per meter.
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 β = -0.00651oC per metre, R = 287 m·N/(kg·K), a = 2.002, and y = 3201
m. For the atmospheric temperature inversion β = -0.00365oC per metre, a =
-0.2721, and y = 809.2 m.
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.
Figure 2.6 Simple manometers
• 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.
•
A general procedure should be followed in working all manometer problems:
1. 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.
2. 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.
3.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,
• 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.
Figure 2.7 Differential manometers
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)
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.
Figure 2.8 Micromanometer using two
gage liquids
• 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 γ1, γ2 and γ3 are the unit gravity force. Simplifying for Δy:
(2.4.1)
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 γ1(a/A) may be neglected. Substituting into Eq. (2.4.1) gives
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.
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.
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.
Figure 2.10 Notation for determining the line
of action of a force
Inclined Surfaces
•
•
In Fig. 2.11 a plane surface is indicated by its trace A'B'; it is inclined θo 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 δA:
•
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,
(2.5.1)
(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.
Figure 2.11 Notation for force of liquid on one
side of a plane inclined area
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)
Eq`ns may be evaluated conveniently through graphical integration; for
simple areas they may be transformed into general formulas:
(2.5.7)
(2.5.8)
•When either of the centroidal axes is an axis of symmetry for the surface, I xy
vanishes and the pressure center lies on x = x-. Since
I xy 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),
1
yp 
yA sin 
1
Ix
2
A yy sin dA  yA A y dA  yA
(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.
If Ix is eliminated from Eq. (2.5.9)
(2.5.10)
or
(2.5.11)
IG is always positive; hence, yp – yˉ is always positive and the pressure
center is always below the centroid of the surface.
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
(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 γ sinθ leads to
By Eq. (2.5.2)
(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,
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 = γh.
• h is the vertical distance to the free surface (see Fig. 2.13). In the figure, γh
may be laid off to any convenient scale such that its trace is OM.
• The force acting on an elemental area δA 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.
Figure 2.13 Pressure prism
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 = γh  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.
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 2.5γ and γ 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 = 5000γ.
– The resultant upward force on the base equals the gravity force of the dam less
the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N. The position of Ry
is such that the free body is in equilibrium. For moments around O,
Figure 2.15
Concrete gravity
dam
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 σmax, σmin 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 σmax
and σmin,
Simplifying gives:
When the resultant falls within the middle third of the base of the dam, σmin 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.
2.6 FORCE COMPONENTS ON CURVED SURFACES
• When the elemental forces p δA 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.
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 δA
an element of its area, its normal making the angle θ 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.
Figure 2.16 Horizontal
component of force on a
curved surface
Figure 2.17 Projections of
area elements on
opposite sides of a body
• 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 δA 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 δAB and at C is δAC, 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.
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,
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 δA of
the surface.
In Fig. 2.18 an area element is shown with the force p δA acting normal to it.
Let θ 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 θ δA, and
the vertical component of force on the curved surface is given by
(2.6.2)
When p replaced by its equivalent γh, and it is noted that cos θ δA is the
projection of δA on a horizontal plane  Eq. (2.6.2) becomes:
(2.6.3-2.6.4)
in which δV is the volume of the prism of height h and base cos θ δA, or the
volume of liquid vertically above the area element.
Figure 2.18 Vertical
component of force on a
curved surface
Figure 2.19 Liquid with
equivalent free surface
• 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/γ 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.
• 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 = γV
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.
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.
Figure 2.20 Semifloating body
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)
Figure 2.21 Tensile stress in pipe
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)
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.
• 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 γ is the
unit gravity force of fluid.
• The same formula holds for floating bodies when V is taken as the
volume of liquid displaced.
Figure 2.22 Buoyant force on floating and
submerged bodies
• In Fig. 2.23 the vertical force exerted on an element of the body in the form
of a vertical prism of cross section δA is
• δV is the volume of the prism. Integrating over the complete body gives
when γ 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,
xˉ 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.
Figure 2.23 Vertical force components
on element of body
• 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; γ1
and γ2 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:
Figure 2.24 Free body diagrams for body suspended in a fluid
• 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, γ 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 ΔV = aΔh. Solving for Δh with Eqs. (2.7.2) and (2.7.3)
(2.7.4)
Figure 2.25 Hydrometer in water and in liquid of relative
density
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
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.
•
•
•
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. a light piece of wood with a metal mass at its bottom is stable;
b. 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;
c. a homogeneous sphere or right-circular cylinder is in equilibrium for
any angular rotation; i.e., no couple results from an angular
displacement.
Figure 2.26 Examples of stable, unstable, and neutral
equilibrium
•
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
•
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.
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 crosssectional area below liquid surface in this case.
Figure 2.28 Stability of a prismatic body
• 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 θ is the angular displacement and W the gravity force of the body.
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 Δy = 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,
G is 1.97 m from the bottom; hence
The scow is stable, since
is positive; the righting moment is
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 θ.
• 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 ΔFB on the right.
• The force system, consisting of the original buoyant force at B and the
couple ΔFB × 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)
Figure 2.29 Stability relations in a body of
variable cross section
• 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 δA on the horizontal section through the body at the
liquid surface, an element of volume of the wedge is xθ δA. The buoyant force
due to this element is γxθ δA, and its moment about O is γx2θ δA, in which θ 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 θ is very small,
and
(2.8.3)
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:
Figure 2.30
Horizontal cross
section of a ship
at the waterline
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.
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. Eq`n (2.2.5) applies
to this situation,
(2.2.5)
• Fig. 2.31b shows the pressure gradient ∇p is then the vector sum of -ρa and jγ.
• 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.
Figure 2.31 Acceleration with free surface
• 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,
• 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
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,
Figure 2.32 Tank completely filled with liquid
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,
Figure 2.33 Uniform linear acceleration of container
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)
Figure 2.34 Rotation of a fluid about a vertical axis
• For constant angular velocity ω, any particle of fluid P has an acceleration
ω2r directed radially inward (a = -iω2r).
• Vector addition of -jγ and -ρa (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 (γ ≈ const) integration yields
in which c is the constant of integration.
• 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 γ,
(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.
• 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 ω. 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 ω2r02/2g (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 ω, r0, and depth before
rotation are given.
Figure 2.35
Rotation of circular
cylinder about its
axis
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 ω = 200 × 2π/60 = 20.95 rad/s, γ = 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
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 zeropressure 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
Figure 2.36
Rotation of inclined
tube of liquid about
a vertical axis
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.