course number: me 321 fluid mechanics i fluid statics ...teacher.buet.ac.bd/mmrazzaque/fluid...
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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.
3
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:
5
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.
7
Pressure Variation in a Fluid Body
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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9
Pressure Variation in a Fluid Body
Newton's second law
is written in vector
form for a constant-
mass system as
10
Pressure Variation in a Fluid Body
11
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.
13
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.
14
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
15
Linearly Accelerating Containers
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.
Linearly Accelerating Containers
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.
Linearly Accelerating Containers
18
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.
Linearly Accelerating Containers
19
Linearly Accelerating Containers
20
Linearly Accelerating Containers
21
Rotating Containers
22
23
Rotating Containers
24
Rotating Containers
Note that density or viscosity does not appear in the above equation.
25
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.
26
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
27
Rotating Containers
28
Rotating Containers
29
Rotating Containers
30
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.
31
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.
32
First get the specific weights from
Tables.
Now proceed from A to B,
calculating the pressure change in
each fluid and adding:
33
34
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.
35
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
37
The distance to a
centroid is defined as
The expression for the
force then becomes
Forces on Submerged Planes… … …
38
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.
Forces on Submerged Planes… … …
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.).
Forces on Submerged Planes… … …
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
Forces on Submerged Planes… … …
41
Forces on Submerged Planes… … …
42
Forces on Submerged Planes… … …
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Forces on Submerged Planes… … …
44
Forces on Submerged Planes… … …
45
Forces on Submerged Planes… … …
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Forces on Submerged Planes… … …
47
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
48
Forces on Curved Surfaces
49
Forces on Curved Surfaces
50
Forces on Curved Surfaces
51
Forces on Curved Surfaces
52
Forces on Curved Surfaces
53
Forces on Curved Surfaces
54
Forces on Curved Surfaces
55
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
56
Buoyancy …
Alternatively, we can sum the vertical forces on elemental
slices through the immersed body as shown in the Fig.:
57
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.
58
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
60
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.
62
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
64
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.
65
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.
66
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.
67
Stability of Floating Bodies… …
area DOE minus the subtracted wedge with cross-sectional
area AOB.
To locate the centroid of the composite volume, we take
moments as follows:
68
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
Stability of Floating Bodies… …
69
Stability of Floating Bodies… …
70
Stability of Floating Bodies… …
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Stability of Floating Bodies… …
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