Transcript MECH 221 FLUID MECHANIC

MECH 221 FLUID MECHANICS
(Fall 06/07)
Chapter 2: FLUID STATICS
Instructor: Professor C. T. HSU
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MECH 221 – Chapter 2
2.1. Hydrostatic Pressure

Fluid mechanics is the study of fluids in
macroscopic motion. For a special static case:
No Motion at All

Recall that by definition, a fluid moves and
deforms when subjected to shear stress and,
conversely, a fluid that is static (at rest) is not
subjected to any shear stress. Otherwise it
will move.  No shear stress, i.e., Normal
stress only
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MECH 221 – Chapter 2
2.1. Hydrostatic Pressure

The force/stress on any given surface immersed
in a fluid at rest, is always perpendicular (normal)
to the surface. This normal stress is called
“pressure”



Fluid statics is to determine the pressure field
At any given point in a fluid at rest, the
normal stress is the same in all directions
(hydrostatic pressure)

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MECH 221 – Chapter 2
2.1. Hydrostatic Pressure

Proof:

Take a small, arbitrary, wedged shaped
element of fluid

Fluid is in equilibrium, so ∑F = 0

Let the fluid element be sufficiently small so
that we can assume that the pressure is
constant on any surface (uniformly
distributed).
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MECH 221 – Chapter 2
2.1. Hydrostatic Pressure

F1=p1 A1

F2=p2 A2

F3=p3 A3

 m =  V


Fluid Density : 
Fluid Volume :
 V=  x  y  z/2
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MECH 221 – Chapter 2
2.1. Hydrostatic Pressure
Look at the side view
∑Fx = 0 :
F1 cos  - F2 = 0

p1 A1cos  - p2 A2 = 0
Since  A1cos  =  A2 =  y  z
 p1 = p2
∑Fz = 0 :
F1 sin +  m.g = F3
p1  A1 sin  +   V g = p3  A3
p1( x/sin )  y sin  +  g  x  y  z/2 = p3  x  y
p1 +  g  z/2 = p3
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MECH 221 – Chapter 2
2.1. Hydrostatic Pressure

Shrink the element down to an infinitesimal point, so
that  z0, then p1 = p3 p1 = p2 = p3

Notes:
• Normal stress at any point in a fluid in equilibrium is
the same in all directions.
• This stress is called hydrostatic pressure.
• Pressure has units of force per unit area.
P = F/A
[N/m2]
• The objective of hydrostatics is to find the pressure
field (distribution) in a given body of fluid at rest.
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MECH 221 – Chapter 2
2.2. Vertical Pressure Variation

Take a fluid element of small
control volume in a tank at rest
Force balance:
(P+  P) A +  g A  y = P A
 P/  y = -  g


Negative sign indicates that P
decreases as y increases
A
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MECH 221 – Chapter 2
2.2. Vertical Pressure Variation

For a constant density fluid, we can integrate
for any 2 vertical points in the fluid (1) & (2).
P2 - P1 = -  g (y2 - y1)
If  = (y), then: ∫dP = -g ∫ (y)dy
If  = (p,y)
such as for ideal gas P =  RT where T=T (y)
∫dP/P = -(g/R) ∫dy/T(y)
The integration at the right hand side
depends on the distribution of T(y).
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MECH 221 – Chapter 2
2.3. Horizontal Pressure Variation

Take a fluid element of small control volume
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Force balance:
P 1 A = P2 A
P 1 = P2
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Static pressure is constant in any horizontal plane.

Having the vertical & horizontal variations, it is
possible to determine the pressure at any point in
a fluid at rest.
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MECH 221 – Chapter 2
2.3. Horizontal Pressure Variation

Absolute Pressure v.s. Gage Pressure

Absolute pressure:


Gage pressure:


Measured from absolute zero
Measured from atmospheric pressure
If negative, it is called vacuum pressure

Pabs = Patm + Pgage
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces

For constant density fluid:
* The pressure varies with depth, P= gh.
* The pressure acts perpendicularly to an
immersed surface
2.4.1. Plane Surface
•
•
Let the surface be infinitely thin, i.e. NO
volume
Plate has arbitrary plan form, and is set at an
arbitrary angle, , with the horizontal.
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces

Looking at the top plate surface only, the pressure
acting on the plate at any given h is:
P = Patm + gh

So, the pressure distribution on the surface is,
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces



To find the total force on the top surface, integrate
P over the area of the plate,
F = ∫P dA = PatmA + g ∫h dA
Note that h = y sin, therefore:
F = PatmA + g sin ∫y dA
Recall that the location of c.g.(center of gravity) in y is:
yc.g. = (1/A) ∫y dA
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces






So, F = PatmA + g sin yc.g.A
Or, F = PatmA + ghc.g.A = (Patm + ghc.g)A
If Pc.g=Patm + ghc.g , then the pressure acting at
c.g. is: F = P c.g. A
In a fluid of uniform density, the force
on a submerge plane surface is equal to the
pressure at the c.g. of the plane multiplied by the
area of the plane.
F is independent of .
The shape of the plate is not important
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces

Where does the total/resultant force act?
Similar to c.g., the point on the surface where the
resultant force is applied is called the Center of
Pressure, c.p.
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces



The moment of the resultant force about the xaxis should equal the moment of the original
distributed pressure about the x-axis
yc.p.F = ∫y dF =  g sin  ∫y2 dA +Patm ∫y dA
Recall that the moment of inertia about the x-axis,
Iox, is by definition:
Iox = ∫y2 dA = y2c.g.A + Ic.g.x
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces
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

Ic.g.x - moment of inertia about the x-axis at c.g.
yc.p.F = g sin Iox +Patm yc.g.A
= g sin (y2c.g.A + Ic.g.x) +Patm yc.g.A
=(g sin yc.g.A + PatmA) yc.g.+ g sin Ic.g.x
yc.p. = yc.g. + (g sin Ic.g.x) / (Pc.g.A)
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces

Similarly,
xc.p. = xc.g. + (g sin Ic.g.y) / (Pc.g.A)
Ic.g.y - moment of inertia about the y-axis at c.g.
* Tables of Ic.g. for common shapes are
available
* For simple pressure distribution profiles, the
c.p. is usually at "c.g." of the profile
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces
2.4.2. Curved Surface

Suppose a warped plate is submerged in
water, what is the resulting force on it?

The problem can be simplified by examining the
horizontal and vertical components separately.
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces
2.4.2.1. Horizontal Force


Zoom on an arbitrary point 'a'.
Locally, it is like a flat plate


The force due to the pressure at 'a' is:

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Pa is the pressure acting at 'a', and it is normal to the
surface.
Fa = Pa Aa, which acts along the same direction as Pa
Its horizontal component is:

FaH = Fa sin = Pa.Aa sin
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces

But, Aasin is the vertical projection of 'a', so that
the horizontal force at 'a' due to pressure is equal
to the force that would be exerted on a plane,
vertical projection of 'a'. This can be generalized
for the entire plane

The horizontal force on a curved surface equals
the force on the plane area formed by the
projection of the curved surface onto a vertical
plane

The line of action on a curved surface is the same
as the line of action on a projected plane
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces

This is true because for every point on the
vertical projection there is a corresponding
point on the warped plate that has the same
pressure.
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces
2.4.2.2. Vertical Force



Similar to the previous approach,
FaV = Fa cos = Pa Aacos 
Aacos is the horizontal projection of 'a', but this is
only at a point!
Notice that if one looks at the entire plate, the
pressures on the horizontal projection are not equal
to the pressures on the plate
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces

Note:

Pa= gha  FaV = ghaAa cos

In general, Pa ≠ Pa'

Consequently, one needs to integrate along
the curved plate

This is not difficult if the shape of the plate is
given in a functional form
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces

The ultimate result is:

The vertical component of the force on a curved surface is
equal to the total weight of the volume of fluid above it

The line of action is through the c.g. of the volume

If the lower side of a surface is exposed while the
upper side is not, the resulting vertical force is
equal to the weight of the fluid that would be
above the surface
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MECH 221 – Chapter 2
2.4. Forces on Immersed Surfaces

So far, only surfaces (not volumes) have been
discussed

In fact, only one side of the surface has been
considered

Note that for a surface to be in equilibrium, there
has to be an equal and opposite force on the other
side
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MECH 221 – Chapter 2
2.5. Bodies with Volume (Buoyancy)

The volume can be constructed from two curved surfaces put
together, and thus utilize the previous results.

Since the vertical projections of both plates are the same,
FHab = FHcd,
Where FVab =g (vol. 1-a-b-2-1), FVcd =g (vol. 1'-d-c-2'-1')

*Note that this is true regardless of whether
there isn't any fluid above c-d.
there is or
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MECH 221 – Chapter 2
2.5. Bodies with Volume (Buoyancy)

Join the two plates together
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Total force: FB=FVcd-FVab= g(vol. a-b-c-d)

This force FB is called Buoyancy Force
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MECH 221 – Chapter 2
2.6. Archimedes' Principle

The net vertical force on an immersed body of arbitrary
shape due to the pressure forces acting on the surfaces
of the body is equal to the weight of the displaced fluid
* The line of action is through the center of
the mass of the displaced fluid volume
* Direction of buoyant force is upward

If a body immersed in a fluid is in equilibrium, then:
W = FB

W is the weight of the body.
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MECH 221 – Chapter 2
2.6. Archimedes' Principle

For a body in a fluid of varying density, e.g. ocean,
the body will sink or rise until it is at a height
where its density is equal to the density of the
fluid

For a body in a constant density fluid, the body
will float at a level such that the weight of the
volume of fluid it displaces is equal to its own
weight
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MECH 221 – Chapter 2
2.7. Pressure Variation with Rigid-Body Motion

The variation of pressure with distance is balanced by the
total accelerations that may be due to gravitational
acceleration g, constant linear acceleration al and constant
rotational acceleration ar. Generally,
a = -(g + al + ar)

For g in the vertical y direction, g = gj

For linear acceleration in the x and y directions,
al = axi + ayj

For fluid rotates rigidly at a constant angular velocity ω, the
acceleration ar is in the radial r direction, i.e.,
ar = -rω2er where er is the unit vector in r direction
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