COURSE NUMBER: ME 321 Fluid Mechanics I Fluid statics ...teacher.buet.ac.bd/mmrazzaque/Fluid...

COURSE NUMBER: ME 321

Fluid Mechanics I

Fluid statics

Course teacher

Dr. M. Mahbubur Razzaque

Professor

Department of Mechanical Engineering

BUET

1

Fluid statics

Fluid statics is the study of fluids in which there is no relative

motion between fluid particles.

If there is no relative motion; no shearing stresses exist. The only

stress that exists is a normal stress, the pressure. So, it is the pressure

that 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 in

a given gravity field, the pressure may easily be calculated by

integration.

If the fluid is moving in rigid-body motion, such as a tank of liquid

which has been spinning for a long time, the pressure can be easily

calculated, because the fluid is free of shear stress.

2

Fluid statics

Three situations are usually investigated in fluid statics. Theseinclude (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 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 normal

compressive force divided by the infinitesimal area over which

it acts. Now does the pressure, at a given point, vary as the

normal 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-

direction) shown in Fig. 2.2.

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Assume that a pressure p acts on the hypotenuse and that a

different 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, for

both the x- and y-directions:

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Where we have used The pressures shown are due

to the surrounding fluid and are the average pressure on the areas.

Substituting

Eqs. 2.2.1 take the form

As the element shrinks to a point, D x → 0 and D y → 0. Hence

the right-hand sides in the equations above go to zero, even for

fluids in motion, providing with the result that, at a point,

Since q 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 directions

at a given point for both a static fluid and a fluid that is in motion.6

Pressure Variation in a Fluid Body

To determine the pressure variation in fluids at rest or fluids

undergoing an acceleration while the relative position of fluid

elements to one another remains the same, consider the

infinitesimal element displayed in Fig.2.4, where the z-axis is

in 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 a

first-order Taylor series expansion with p(x, y, z):

lf we move from the center to a face a distance (dx/2) away, we

see that the pressure is

The pressures at all faces are expressed in this manner, as

shown in Fig. 2.4.

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Newton's second law

is written in vector

form for a constant-

mass system as

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Fluids at Rest

A fluid at rest does not undergo any acceleration. Therefore, Eq

2.3.6 reduces to

or

This equation implies that there is no pressure variation in the x and

y directions, that is, in the horizontal plane. The pressure varies in

the z direction only.

Also note that dp is negative if dz is positive; that is, the pressure

decreases 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 in water and therefore have identical pressures.

Points A, B, and C are also at equal depths in water and have

identical pressures higher than a, b, c, and d.

Potnt D has a different pressure from A, B, and C because it is not

connected to them by a water Path.

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Pressures in Liquids at Rest

If the density can be assumed constant (which is the case with

liquids), Eq.2.4.2 may be integrated to yield

so that pressure increases with depth. Note that z is positive in the

upward direction. The quantity (p/g + z) is often referred to as the

piezometric head.

If the point of interest were a distance h below a free surface Eq.

2.4.3 would result in

This equation is quite useful

in converting pressure to an

equivalent height of liquid.

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Gage Pressure and Vacuum Pressure

Engineers usually specify pressures as (1) the absolute or (2) the

gage pressure ( pressure relative to the local ambient atmosphere).

The second case occurs because many pressure instruments are of

dffirential type and record, not an absolute magnitude, but the

difference between the fluid pressure and the atmosphere.

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) Regardless of the

shape of the mug, the

free surface tilts at an

angle q given by

as az = 0.


17

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 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

of coffee is 1010 kg/m3.


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If 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 q) = 2.14 cm.

This 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

pA = rg(zsur - zA) = (1010 kg/m3)(9.81 m/s2)(0.07 m) = 694 Pa.

During acceleration, the pressure at A becomes

pA = rg(zsur - zA) = (1010 kg/m3)(9.81 m/s2)(0.0914 m) = 906 Pa,

which is 31% higher than the pressure at rest.


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Rotating Containers

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Rotating Containers

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Rotating Containers

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

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Rotating Containers

If r2 = R = the

container radius, then

the total rise of water

level at the wall,

Since the volume of a paraboloid is one-half the base area time

height, the still –water level is exactly halfway between the high

and low points of the free surface.

That is the center of the fluid drops an amount , and

the edges rise an equal amount.

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A 10 cm deep coffee mug contains 7 cm coffee of density 1010 kg/m3. It is given a rigid body rotation about its central axis. Find (a) the angular velocity which will cause the coffee to just reach the lip of the cup and (b) the gage pressure at point A for this condition.

Rotating Containers

=> 0.03 m = 2(0.03m)2/4(9.81m/s2)=> = 36.2 rad/s = 345 rpm

Pressure at A may be determined by eq. 2.6.4 or even by PA = ghA = (1010 *9.81 N/m3)(0.1m)= 990 Pa

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Rotating Containers

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Rotating Containers

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Rotating Containers

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Manometers

Manometers are instruments that use columns of liquids to measure

pressures.

The figure shows a simple open manometer for measuring pA in a

closed chamber relative to atmospheric pressure, pa.

The manometric fluid (r2) is chosen different from the chamber

fluid (r1) to isolate the chamber fluid from the environment and to

suitably scale the length of the open tube.

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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 then

use the basic hydrostatic formula “up” to level z2.

The physical reason that we can ‘jump across" at section I is that a

continuous length of the same fluid connects these two equal

elevations.

Any two points at the same elevation in a continuous mass

of the same static fluid will be at the same pressure.

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First get the specific weights from

Tables.

Now proceed from A to B,

calculating the pressure change in

each fluid and adding:

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First identify the relevant points

1, 2, 3, 4, and 5 as shown in the

figure. For simplicity, neglect

the weight of the air and assume

the pressure at point 3 is equal

to the pressure at point 4.

Solution

Forces on Submerged Planes

In the design of submerged devices and objects, such as dams,

flow barriers, ships, and holding tanks, it is necessary to

calculate the magnitudes and locations of forces that act on

their surfaces, both plane and curved.

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Here, we consider

only plane surfaces,

such as the plane

surface of general

shape shown in the

Fig. Note that a side

view is given as

well as a view

showing the shape

of the plane.

Forces on Submerged Planes…

The total force of the liquid on the plane surface is found by

integrating the pressure over the area, that is,

36

The x and y coordinates are

in the plane of the plane

surface, as shown. Assuming

p = 0 at h = 0, we know that

where h is measured vertically

down from the free surface to the

elemental area dA and y is

measured from point 0 on the free

surface.

Forces on Submerged Planes… …

The force may then be expressed as

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The distance to a

centroid is defined as

The expression for the

force then becomes

Forces on Submerged Planes… … …

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Where is the vertical

distance from the free

surface to the centroid

of 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.

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


39

The force does not, in

general, act at the

centroid. The location

of the resultant force is

found by taking the

sum of the moments of

all the infinitesimal

pressure forces acting

on the area equal to the

moment of the resultant

force.

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

(c.p.).


40

The value of yp can be

obtained 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 rectangular

gate shown.

Sol hints:

F = = 9810*(5sin40o/2)*(5x3) N

= 2.5 + m

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

5.2*)3x5(

125*3 3

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 force

equal to the weight of the fluid it displaces.

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

floats.

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 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.:

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This result is

identical to the

previous one and

equivalent to law I

above.

Here, it is assumed

that the fluid has

uniform specific

weight.

Buoyancy … …

The line of action of the buoyant force passes through the

centroid of the displaced liquid volume only if it has uniform

density.

This 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.

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Buoyancy … …

Gases also exert buoyancy on any body immersed in them.

For example, human beings have an average specific weight of

about 60 lbf/ft3. If the weight of a person is 180 lbf, the

person's total volume will be 3.0 ft3.

However, in so doing we are neglecting the buoyant force of

the air surrounding the person. At standard conditions, the

specific weight of air is 0.0763 lbf/ft3; hence the buoyant force

is approximately 0.23 lbf. If measured in vacuo, the person

would weigh about 0.23 lbf more.

For balloons, the buoyant force of air, instead of being

negligible, is the controlling factor in the design.59

Buoyancy of Floating Bodies

Floating bodies are a special case; only a portion of the body is

submerged, with the remainder poking up out of the free

surface. This is illustrated in the Fig. From a static force

balance, it may be derived that

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Buoyancy of Floating Bodies…

Not only does the buoyant force equal the body weight but also

they are collinear since there can be no net moments for static

equilibrium. The above equation is the mathematical

equivalent of Archimedes' law 2.

Occasionally, a body will have exactly the right weight and

volume for its ratio to equal the specific weight of the fluid. If

so, the body will be neutrally buoyant and remain at rest at

any point where it is immersed in the fluid. Small neutrally

buoyant particles are sometime used for flow visualization.

A submarine can achieve positive, neutral, or negative

buoyancy by pumping water in or out of its ballast61

Stability of Floating Bodies

If a floating object is raised a small distance, the buoyant

force decreases and the object's weight returns the object to

its original position.

Conversely, if a floating object is lowered slightly, the

buoyant force increases and the larger buoyant force

returns the object to its original position.

Thus a floating object has vertical stability since a small

departure from equilibrium results in a restoring force.

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Rotational Stability of Submerged Bodies

63

Let us now consider the rotational

stability of a submerged body.

If the center of gravity G of the body

is above the centroid C (also referred

to as the center of buoyancy) of the

displaced volume and a small angular

rotation results in a moment that will

continue to increase the rotation; the

body is unstable and overturning

would result.

Engineers must design to avoid

floating instability.

Rotational Stability of Submerged Bodies

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If the center of gravity G is below the

centroid C, a small angular rotation

provides a restoring moment and the body

is stable.

If the center of gravity and the

centroid coincide, the body is said

to be neutrally stable, a situation

that is encountered whenever the

density is constant throughout the

floating body.

Stability of Floating Bodies

If the center of gravity is below the centroid, the body is

always stable, as with submerged bodies.

The body may be stable, though, even if the center of gravity is

above the centroid, as sketched.

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Stability of Floating Bodies…

This is determined by the metacentric height GM.

Metacentre M is the point of intersection of the buoyant force

before rotation with the buoyant force after rotation.

If GM is positive, as shown, the body is stable; if GM is

negative (M lies below G), the body is unstable.

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When the body rotates the centroid

of the volume of displaced liquid

moves to the new location C'. If the

centroid C' moves sufficiently far, a

restoring moment develops and the

body is stable.

Stability of Floating Bodies… …

To determine a quantitative relationship for the distance GM

consider the sketch, which shows the uniform cross section of

the floating body in rotated condition.

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area DOE minus the subtracted wedge with cross-sectional

area AOB.

To locate the centroid of the composite volume, we take

moments as follows:

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An expression for , the x-

coordinate of the centroid of

the displaced volume can be

found by considering the

volume to be the original

volume plus the added

wedge with cross-sectional


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COURSE NUMBER: ME 321 Fluid Mechanics I Fluid statics ...teacher.buet.ac.bd/mmrazzaque/Fluid...

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Transcript of COURSE NUMBER: ME 321 Fluid Mechanics I Fluid statics ...teacher.buet.ac.bd/mmrazzaque/Fluid...