MECH 221 FLUID MECHANICS Fall 0607 Chapter 2: FLUID STATICS

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MECH 221 FLUID MECHANICS(Fall 06/07)Chapter 2
FLUID STATICS

• Instructor Professor C. T. HSU

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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

• F1p1? A1
• F2p2? A2
• F3p3? 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 ? 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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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 TT (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.gPatm ?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. 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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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
• 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. 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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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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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 FBFVcd-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 ?, the acceleration ar is in the radial
  r direction, i.e.,
• ar -r?2er where er is the unit vector in r
  direction