Fluid statics

Engineering Fluid Mechanics, Lecture 4

September 9, 2015

Dr. Kelly Kibler

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

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∆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

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Under hydrostatic conditions, the pressure at a given depth is the same everywhere within the same fluid.

Why is the PH≠PI?

Pressure basics

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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

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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

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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

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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

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• 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

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• 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

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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

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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

Hydrostatic forces on a plane

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Consider a horizontal rectangular plane surface (height b, width a), distance h from a free surface.

FR acts at centroid of the plate

𝐹𝑅 = 𝑃𝑂 + 𝜌𝑔ℎ ab