1 MECH 221 FLUID MECHANICS (Fall 06/07) Chapter 2: FLUID STATICS Instructor: Professor C. T. HSU.

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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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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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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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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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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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2.1. Hydrostatic Pressure Shrink the element down to an infinitesimal point, so

that z0, 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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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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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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2.3. Horizontal Pressure Variation Take a fluid element of small control volume

Force balance: P1A = P2A

P1 = P2

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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2.3. Horizontal Pressure Variation

Absolute Pressure v.s. Gage Pressure

Absolute pressure: Measured from absolute zero

Gage pressure: Measured from atmospheric pressure

If negative, it is called vacuum pressure Pabs = Patm + Pgage

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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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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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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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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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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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2.4. Forces on Immersed Surfaces The moment of the resultant force about the x-axis

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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2.4. Forces on Immersed Surfaces 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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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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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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2.4.2.1. Horizontal Force

Zoom on an arbitrary point 'a'. Locally, it is like a flat plate

Pa is the pressure acting at 'a', and it is normal to the surface.

The force due to the pressure at 'a' is: 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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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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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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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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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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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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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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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 is or there isn't any fluid above c-d.

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2.5. Bodies with Volume (Buoyancy) Join the two plates together

Total force: FB=FVcd-FVab= g(vol. a-b-c-d)

This force FB is called Buoyancy Force

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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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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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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

1 MECH 221 FLUID MECHANICS (Fall 06/07) Chapter 2: FLUID STATICS Instructor: Professor C. T. HSU.

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Transcript of 1 MECH 221 FLUID MECHANICS (Fall 06/07) Chapter 2: FLUID STATICS Instructor: Professor C. T. HSU.