lecture 4 statics

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


• 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


• 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


∆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


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

Why is the PH≠PI?

Pressure basics


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


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



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



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


• 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).



• 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



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



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



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