Post on 10-Dec-2015
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Fluid statics- Lecture outline
1 Introduction to fluid statics
2 Pressure
3 Pressure measurement
4 Hydrostatic forcesa) on submerged plane surfaces
b) on submerged curved surfaces
5 Buoyancy and stability
6 Fluids in rigid-body motion
Engineering Fluid Mechanics- Fluid statics2
Introduction to fluid statics
Engineering Fluid Mechanics- Fluid statics3
• Fluid statics deals with stationary fluids (at rest, not moving).
• Because fluid is not moving, there is no relative motion between fluid layers; thus, no shear stress.
• In statics, we are only concerned with the normal stress, which is pressure.
The normal stress and shear stress at the surface of a fluid element. For fluids at rest, the shear stress is zero and pressure is the only normal stress.Hydrostatics are used to analyze
forces on dams.
Pressure basics
Engineering Fluid Mechanics- Fluid statics4
• SI units• 1 Pa = 1 Nm-2
• other common units• 1 bar = 100 kPa
• 1 atm = 101.325 kPa
• 760 mm Hg = 1 atm
• English units• psi (1 atm = 14.696 psi)
Absolute, gauge, and vacuum pressure are related.
𝑃𝑔𝑎𝑢𝑔𝑒 = 𝑃𝑎𝑏𝑠 − 𝑃𝑎𝑡𝑚
𝑃𝑣𝑎𝑐 = 𝑃𝑎𝑡𝑚 − 𝑃𝑎𝑏𝑠
Pressure (P) is a normal force exerted by a fluid per unit area
Pressure is scalar; at any
point in a fluid, pressure is
the same from all directions.
Pressure basics
Engineering Fluid Mechanics- Fluid statics5
∆𝑃 = 𝑃1 − 𝑃2 = 𝜌𝑔𝑧1 − 𝜌𝑔𝑧2 = 𝜌𝑔 𝑧1 − 𝑧2 = 𝛾𝑠∆𝑧
Pressure in a static fluid increases linearly with depth• As depth increases, more fluid rests on the deeper layers extra weight
increases the pressure at depth*
𝑃𝑏𝑒𝑙𝑜𝑤 = 𝑃𝑎𝑏𝑜𝑣𝑒 + 𝜌g ∆𝑧 = 𝑃𝑎𝑏𝑜𝑣𝑒 + 𝛾𝑠 ∆𝑧
To note:• Difference in pressure between two points is
proportional to h and the density of the fluid; • Greater h greater ΔP• Greater ρ greater ΔP(think about pressure changes in air vs. water)
*assuming constant density of fluid
𝑃𝑔𝑎𝑢𝑔𝑒 = 𝜌𝑔ℎ
If “above” point is at a free surface (where the pressure is atmospheric), relationship simplifies to:
Pabove= atm
Pbelow= atm +ρgh
h
𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝜌𝑔ℎ
Pressure basics
Engineering Fluid Mechanics- Fluid statics6
Under hydrostatic conditions, the pressure at a given depth is the same everywhere within the same fluid.
Why is the PH≠PI?
Pressure basics
Engineering Fluid Mechanics- Fluid statics7
Pascal’s law: pressure applied to a confined fluid increases pressure equivalently throughout.
Application: • A small force applied to small area can
exert a large force over a larger area when the two areas are hydraulically connected
• Example: hydraulic lift
𝑃1 = 𝑃2 →𝐹1
𝐴1=
𝐹2
𝐴2→
𝐹2
𝐹1=
𝐴2
𝐴1
Pressure measurement
Engineering Fluid Mechanics- Fluid statics8
Barometer• Used to measure atmospheric conditions• 760 mm Hg = 1 atm = 101.325 kPa• Dimensions of tube (height, diameter) have no effect on reading (as
long as tube is large enough to avoid capillary effects
Pressure measurement
Engineering Fluid Mechanics- Fluid statics9
Manometer• Used to measure small pressure differences.• Orange section is manometer fluid- could be
water, oil, air, mercury, ect.• Manometer fluid :
• must be a different fluid than in the tank.• cannot mix with tank fluid- the two fluids
must be immiscible.• must be denser than working fluid.
How it works:• Pressure at 1 is pressure of the tank.• Because elevation of 1 and 2 are equal, the
pressure at 1 = pressure at 2.• Pressure at top of h is atmospheric.
• ρ is density of manometer fluid.• Diameter of the tube should be large enough to
avoid capillary rise.
𝑃𝑡𝑎𝑛𝑘 = 𝑃1 = 𝑃2 = 𝑃𝑎𝑡𝑚 + 𝜌𝑔ℎ
Pressure measurement
Engineering Fluid Mechanics- Fluid statics10
Computing pressure difference across multiple immiscible static fluids:• Start at a point of known pressure (like
a free surface)• Add or subtract ρgh terms as you
move towards the point of interest:
𝑃1 = 𝑃𝑎𝑡𝑚 + 𝜌1𝑔ℎ1 + 𝜌2𝑔ℎ2 + 𝜌3𝑔ℎ3
Hydrostatic forces on a plane
Engineering Fluid Mechanics- Fluid statics11
• Submerged objects are subject to fluid pressure, which varies with depth.
• We often wish to know the magnitude of hydrostatic force and where it acts (center of pressure).
𝐹𝑅 = 𝑃𝐶𝐴
Because atmospheric pressure
acts on both sides of the plane, we
may remove it and work in gauge
pressure.
𝑃𝐶 = 𝑃𝑎𝑡𝑚 + 𝜌𝑔ℎ𝑐
• Magnitude of force is given by pressure ∙ area of the object
• But pressure varies with depth- which pressure do we use?
• Compute pressure at the object’s centroid (Pc).
Hydrostatic forces on a plane
Engineering Fluid Mechanics- Fluid statics12
• Line of action of resultant force acts at the center of pressure (not necessarily at the centroid).
• Location of the center of pressure is derived by setting the moment of the resultant force equal to the moment of the distributed pressure force, using a moment of inertia around the centroid (applying the parallel axis theorem to move it to the center of pressure).
𝑦𝑃 = 𝑦𝐶 +𝐼𝑥𝑥,𝑐
𝑦𝐶𝐴
ℎ𝑝 = 𝑦𝑃 sin 𝜃
𝑦𝑃 = distance to center of pressure from 0𝑦𝐶 = distance to centroid from 0𝐼𝑥𝑥,𝑐=second moment of area passing through the centroid
𝒉𝑷 =depth to center of pressure
Hydrostatic forces on a plane
Engineering Fluid Mechanics- Fluid statics13
Consider a rectangular plane (height b, width a), tilted at θ, top edge is distance sfrom a free surface.
𝑦𝑃 = 𝑠 +𝑏
2+
𝑏2
12𝑠 +
𝑏2
+ 𝑃𝑂
𝜌𝑔 sin 𝜃
𝒉𝒑 = 𝒚𝑷 𝐬𝐢𝐧 𝜽
𝐹𝑅 = 𝑃𝐶𝐴 = 𝑃𝑂 + 𝜌𝑔(𝑠 +𝑏
2) sin 𝜃 ab
Hydrostatic forces on a plane
Engineering Fluid Mechanics- Fluid statics14
Consider a vertical rectangular plane (height b, width a), top edge is distance sfrom a free surface.
If s=0 and we disregard P0:
ℎ𝑝 = 𝑦𝑝 =2
3𝑏
𝐹𝑅 = 𝑃𝐶𝐴 = 𝑃𝑂 + 𝜌𝑔(𝑠 +𝑏
2) ab
𝐹𝑅 =𝜌𝑔𝑎𝑏2
2