2 Fluid Statics - Bangladesh University of Engineering and ...teacher.buet.ac.bd/mmrazzaque/ME 221/2...

COURSE NUMBER: ME 221COURSE NUMBER: ME 221

Elements of Fluid Mechanics and Machinery

3 credit hour

http://teacher.buet.ac.bd/mmrazzaque/lecturenotes.html

Course teacher

Dr M Mahbubur RazzaqueDr. M. Mahbubur Razzaque

Professor

Department of Mechanical EngineeringDepartment of Mechanical Engineering

BUET1

Fluid statics

Fluid statics is the study of fluids in which there is no relativemotion between fluid particles.

If there is no relative motion; no shearing stresses exist. The onlystress that exists is a normal stress, the pressure. So, it is the pressurethat is of primary interest in fluid statics.

When the fluid velocity is zero, denoted as the hydrostatic condition,the pressure variation is due only to the weight. For a known fluid ina given gravity field, the pressure may easily be calculated byintegration.

If the fluid is moving in rigid-body motion, such as a tank of liquidwhich has been spinning for a long time, the pressure can be easilyp g g , p ycalculated, because the fluid is free of shear stress.

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Fluid statics

Three situations are usually investigated in fluid statics. Thesei l d ( ) fl id t t h li id i i (b) fl idinclude (a) fluids at rest, such as liquid in a reservoir, (b) fluidscontained in devices that undergo linear acceleration, and (c)fluids contained in rotating cylinders.

In addition to the examples shown above, we considerinstruments called manometers and investigate the forces ofinstruments called manometers and investigate the forces ofbuoyancy and the stability of floating bodies.

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PRESSURE AT A POINT

Pressure at a point is defined as being the infinitesimal normalcompressive force divided by the infinitesimal area over whichp yit acts. Now does the pressure, at a given point, vary as thenormal to the area changes direction?

To show that this is not the case, even for fluids in motion,consider the wedge-shaped element of unit depth (in the z-consider the wedge shaped element of unit depth (in the zdirection) shown in Fig. 2.2.

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Assume that a pressure p acts on the hypotenuse and that adifferent pressure acts on each of the other areas as showndifferent pressure acts on each of the other areas, as shown.Since the forces on the two end faces are in the z-direction,we have not included them on the element.

Now, let us apply Newton's second law to the element, forboth the x- and y-directions:

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Where we have used The pressures shown are dueto the surrounding fluid and are the average pressure on the areas.to the surrounding fluid and are the average pressure on the areas.Substituting

Eqs 2 2 1 take the formEqs. 2.2.1 take the form

As the element shrinks to a point ∆ x → 0 and ∆ y → 0 HenceAs the element shrinks to a point, ∆ x → 0 and ∆ y → 0. Hencethe right-hand sides in the equations above go to zero, even forfluids in motion, providing with the result that, at a point,

Since θ is arbitrary, this relationship holds for all angles at a point.Thus we conclude that the pressure in a fluid is constant at a point.That is, pressure is a scalar function. It acts equally in all directionsat a given point for both a static fluid and a fluid that is in motion.

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Pressure Variation in a Fluid Body

To determine the pressure variation in fluids at rest or fluidsundergoing an acceleration while the relative position of fluidelements to one another remains the same, consider the,infinitesimal element displayed in Fig.2.4, where the z-axis isin the vertical direction.

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If we assume that p is the pressure at the center of this element,the pressures at any other point can be expressed by using ap y p p y gfirst-order Taylor series expansion with p(x, y, z):

lf we move from the center to a face a distance (dx/2) away, weh h isee that the pressure is

The pressures at all faces are expressed in this manner, asshown in Fig 2 4shown in Fig. 2.4.

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Newton's second lawis written in vectorform for a constant-mass system as

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Fluids at Rest

A fluid at rest does not undergo any acceleration. Therefore, Eq2.3.6 reduces to

or

This equation implies that there is no pressure variation in the x andy directions, that is, in the horizontal plane. The pressure varies inh d lthe z direction only.

Also note that dp is negative if dz is positive; that is, the pressuredecreases as we move up and increases as we move down,

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Fluids at Rest

Hydrostatic-pressure distribution. Points a, b, c, and d are at equal depths inwater and therefore have identical pressures.

Points A, B, and C are also at equal depths in water and have identicalpressures higher than a, b, c, and d.

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Point D has a different pressure from A, B, and C because it is notconnected to them by a water Path.

Pressures in Liquids at Rest

If the density can be assumed constant (which is the case withliquids), Eq.2.4.2 may be integrated to yield

so that pressure increases with depth. Note that z is positive in thed di i Th i ( / + ) i f f d hupward direction. The quantity (p/γ + z) is often referred to as the

piezometric head.

If h f d h b l f f EIf the point of interest were a distance h below a free surface Eq.2.4.3 would result in

This equation is quite usefulin converting pressure to anequivalent height of liquid.

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q g q

Gage Pressure and Vacuum Pressure

Engineers usually specify pressures as (1) the absolute or (2) thegage pressure ( pressure relative to the local ambient atmosphere).

The second case occurs because many pressure instruments are ofdffirential type and record, not an absolute magnitude, but thediff b t th fl id d th t hdifference between the fluid pressure and the atmosphere.

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Linearly Accelerating Containers

Consider a fluid at rest relative to a reference frame that is linearly accelerating with a horizontal component ax and a vertical component az. Then Eq. 2.3.6 simplifies to

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16Note that density or viscosity does not appear in the above equation.


A coffee mug on a horizontal tray is accelerated at 7 m/s2.The mug is 10 cm deep and 6 cm in diameter and contains coffee 7 cm deep at rest (a) Assuming rigid body acceleration of thecm deep at rest. (a) Assuming rigid body acceleration of the coffee, determine whether it will spill out of the mug,(b) Calculate the gage pressure in the corner at point A if the density

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(a) Regardless of the

of coffee is 1010 kg/m3.

shape of the mug, the free surface tilts at an angle θ given by g g y

as az = 0.17

Linearly Accelerating Containers If the mug is symmetric about itsIf the mug is symmetric about its central axis, the volume of coffee is conserved if the tilted face intersects the original rest surface exactly at the centerline, as shown.

Thus the deflection at the left side of the mug is z = (3 cm)(tan θ) = 2.14 cm. This is less than the 3 cm clearanceThis is less than the 3-cm clearance available, so the coffee will not spill.

(b) When at rest the gage pressure at point A is given by(b) When at rest, the gage pressure at point A is given by pA = ρg(zsur - zA) = (1010 kg/m3)(9.81 m/s2)(0.07 m) = 694 Pa.During acceleration, the pressure at A becomes

3 2

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pA = ρg(zsur - zA) = (1010 kg/m3)(9.81 m/s2)(0.0914 m) = 906 Pa, which is 31% higher than the pressure at rest.

Manometers

Manometers are instruments that use columns of liquids to measure pressures.

The figure shows a simple open manometer for measuring pA in ag p p g pAclosed chamber relative to atmospheric pressure, pa.

The manometric fluid (ρ2) is chosen different from the chamber

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The manometric fluid (ρ2) is chosen different from the chamberfluid (ρ1) to isolate the chamber fluid from the environment and tosuitably scale the length of the open tube.

Manometers

Here, one can begin at A, apply the basic hydrostatic formula“down” to z1, jump across fluid 2 to the same pressure p1, and thenuse the basic hydrostatic formula “up” to level z2.

The physical reason that we can ‘jump across" at section I is that acontinuous length of the same fluid connects these two equalelevations.

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Any two points at the same elevation in a continuous massof the same static fluid will be at the same pressure.

First get the specific weights fromTables.

N d f A t BNow proceed from A to B,calculating the pressure change ineach fluid and adding:

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Solution

First identify the relevant pointsFirst identify the relevant points1, 2, 3, 4, and 5 as shown in thefigure. For simplicity, neglectth i ht f th i dthe weight of the air and assumethe pressure at point 3 is equalto the pressure at point 4.

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Forces on Submerged Planes

In the design of submerged devices and objects, such as dams,flow barriers, ships, and holding tanks, it is necessary tocalculate the magnitudes and locations of forces that act oncalculate the magnitudes and locations of forces that act ontheir surfaces, both plane and curved.

Here, we consideronly plane surfaces,such as the planesuc as t e p a esurface of generalshape shown in theFig Note that a sideFig. Note that a sideview is given aswell as a view

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showing the shapeof the plane.

Forces on Submerged Planes…

The total force of the liquid on the plane surface is found byintegrating the pressure over the area, that is,

The x and y coordinates are i h l f h lin the plane of the plane surface, as shown. Assuming p = 0 at h = 0, we know thatp ,

h h i d ti llwhere h is measured vertically down from the free surface to the elemental area dA and y is

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measured from point 0 on the free surface.

Forces on Submerged Planes… …

The force may then be expressed as

Th di t tThe distance to a centroid is defined as

The expression for theThe expression for the force then becomes

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Forces on Submerged Planes… … …

Where is the verticalWhere is the vertical distance from the free surface to the centroidof the area and p is theof the area and pc is the pressure at the centroid.

Thus the magnitude of the force on a submerged plane surface is the pressure at the centroid multiplied by the area.

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p p y

The force does not depend on the angle of inclination, α.


The force does not, in general, act at the centroid The locationcentroid. The location of the resultant force is found by taking the sum of the moments of all the infinitesimal pressure forces acting p gon the area equal to the moment of the resultant forceforce.

Let the force F act at the point (xp,yp), the center of pressure(c p )

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(c.p.).


The value of yp can be bt i d b tiobtained by equating

moments about the x-axis:

where the second moment of the area about the x-axis is

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Problem: Find the force P needed to hold the 3 m wide rectangulargate showngate shown.

Sol hints:F 9810*(5sin40o/2)*(5x3) NF = = 9810*(5sin40o/2)*(5x3) N

= 2.5 + m5.2*)3x5(

125*3 3

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: 7*Psin40o = (5 - yp)*F)(

Forces on Curved Surfaces

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Buoyancy

Two laws of buoyancy discovered by Archimedes in the third century B.C.:1. A body immersed in a fluid experiences a vertical buoyant force1. A body immersed in a fluid experiences a vertical buoyant force

equal to the weight of the fluid it displaces.2. A floating body displaces its own weight in the fluid in which it

floatsfloats.These two laws are easily derived by referring to the Fig. The body lies between an upper curved surface 1 and a lower curved surface 2. The body experiences a net upward forceThe body experiences a net upward force

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Buoyancy …

Alternatively, we can sum the vertical forces on elemental slices through the immersed body as shown in the Fig.:g y g

This result is identical to the

previous one andprevious one and equivalent to law I

above.H it i dHere, it is assumed that the fluid has uniform specific

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weight.

Buoyancy … …

The line of action of the buoyant force passes through the centroid of the displaced liquid volume only if it has uniformcentroid of the displaced liquid volume only if it has uniform density.

This point through which F acts is called the center ofThis point through which FB acts is called the center of buoyancy.

Of course, the center of buoyancy may or may not correspond to the actual center of mass of the body's own material, which may have variable density. y y

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Buoyancy … …

Gases also exert buoyancy on any body immersed in them.

For example, human beings have an average specific weight ofabout 60 lbf/ft3. If the weight of a person is 180 lbf, theperson's total volume will be 3 0 ft3person's total volume will be 3.0 ft3.

However, in so doing we are neglecting the buoyant force ofthe air surrounding the person. At standard conditions, thespecific weight of air is 0.0763 lbf/ft3; hence the buoyant forceis approximately 0.23 lbf. If measured in vacuo, the personis approximately 0.23 lbf. If measured in vacuo, the personwould weigh about 0.23 lbf more.

For balloons the buoyant force of air instead of beingFor balloons, the buoyant force of air, instead of beingnegligible, is the controlling factor in the design.

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Buoyancy of Floating Bodies

Floating bodies are a special case; only a portion of the body isb d ith th i d ki t f th fsubmerged, with the remainder poking up out of the free

surface. This is illustrated in the Fig. From a static forcebalance, it may be derived that

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Buoyancy of Floating Bodies…

Not only does the buoyant force equal the body weight but alsoNot only does the buoyant force equal the body weight but alsothey are collinear since there can be no net moments for staticequilibrium. The above equation is the mathematicalequivalent of Archimedes' law 2equivalent of Archimedes' law 2.

Occasionally, a body will have exactly the right weight andvolume for its ratio to equal the specific weight of the fluid. Ifso, the body will be neutrally buoyant and remain at rest atany point where it is immersed in the fluid. Small neutrallyany point where it is immersed in the fluid. Small neutrallybuoyant particles are sometime used for flow visualization.

A submarine can achieve positive neutral or negativeA submarine can achieve positive, neutral, or negativebuoyancy by pumping water in or out of its ballast

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Stability of Floating Bodies

If a floating object is raised a small distance the buoyantIf a floating object is raised a small distance, the buoyantforce decreases and the object's weight returns the object toits original position.

Conversely, if a floating object is lowered slightly, thebuoyant force increases and the larger buoyant forcey g yreturns the object to its original position.

Thus a floating object has vertical stability since a smallThus a floating object has vertical stability since a smalldeparture from equilibrium results in a restoring force.

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Rotational Stability of Submerged Bodies

Let us now consider the rotationalstability of a submerged body.y g y

If the center of gravity G of the bodyis above the centroid C (also referredis above the centroid C (also referredto as the center of buoyancy) of thedisplaced volume and a small angular

i l i h illrotation results in a moment that willcontinue to increase the rotation; thebody is unstable and overturningy gwould result.

Engineers must design to avoid

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Engineers must design to avoidfloating instability.

Rotational Stability of Submerged Bodies

If the center of gravity G is below thecentroid C a small angular rotationcentroid C, a small angular rotationprovides a restoring moment and the bodyis stable.

If the center of gravity and theIf the center of gravity and thecentroid coincide, the body is saidto be neutrally stable, a situationthat is encountered whenever thethat is encountered whenever thedensity is constant throughout thefloating body.

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Stability of Floating Bodies

If the center of gravity is below the centroid, the body isalways stable, as with submerged bodies.w ys s b e, s w sub e ged bod es.

The body may be stable, though, even if the center of gravity isabove the centroid as sketchedabove the centroid, as sketched.

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Stability of Floating Bodies…

When the body rotates the centroidof the volume of displaced liquidof the volume of displaced liquidmoves to the new location C'. If thecentroid C' moves sufficiently far, arestoring moment de elops and therestoring moment develops and thebody is stable.

This is determined by the metacentric height GM.Metacentre M is the point of intersection of the buoyant forcebefore rotation with the buoyant force after rotation.be o e otat o w t t e buoya t o ce a te otat o .

If GM is positive, as shown, the body is stable; if GM isnegative (M lies below G) the body is unstablenegative (M lies below G), the body is unstable.

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Stability of Floating Bodies… …

To determine a quantitative relationship for the distance GMconsider the sketch, which shows the uniform cross section ofthe floating body in rotated conditionthe floating body in rotated condition.

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An expression for , the x-coordinate of the centroid ofthe displaced volume can befound by considering thevolume to be the originalvolume to be the originalvolume plus the addedwedge with cross-sectional

area DOE minus the subtracted wedge with cross-sectionalarea AOB.

To locate the centroid of the composite volume, we takemoments as follows:

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2 Fluid Statics - Bangladesh University of Engineering and ...teacher.buet.ac.bd/mmrazzaque/ME 221/2...

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