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Ch 3 – Fluid Statics - I

Prepared forCEE 3500 – CEE Fluid Mechanics

byGilberto E. Urroz,

August 2005

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Introduction to fluid statics (1)

● Fluid at rest:– No shear stresses – Only normal forces due to pressure

● Normal forces are important:– Overturning of concrete dams– Bursting of pressure vessels– Breaking of lock gates on canals

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Introduction to fluid statics (2)

● For design: compute magnitude and location of normal forces

● Development of instruments that measure pressure

● Development of systems that transfer pressure, e.g., – automobile breaks– hoists

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Introduction to fluid statics (3)

● Average pressure intensity p = force per unit area● Let:

– F = total normal pressure force on a finite area A– dF = normal force on an infinitesimal area dA

● The local pressure on the infinitesimal area is

● If pressure is uniform, p = F/A● BG units: psi (=lb/in2) or psf (=lb/ft2)● SI units: Pa (=N/m2), kPa (=kN/m2)● Metric: bar, millibar; 1 mb = 100 Pa

p=dFdA

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Isotropy of pressure

● Along y: forces cancel each other● Along x: p dy dl cos α – px dy dx = 0 p = px

● Along z: pz dy dx – p dy dl sin α - ½ γ dx dy dz = 0● Neglecting highest term p = py = px (isotropic)

x

z

p dl dy

dx

dz dl p

x dy dz

pz dx dy

γ ½ dx dy dz

α α

dy = normal to paper

dy = dl cos α dx = dl sin α

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Variation of pressure in static fluid (1)

O

dx

dy

dz

x

z

y

p∂ p∂ z⋅dz

2⋅dx⋅dy

p−∂ p∂ z⋅dz

2⋅dx⋅dy

⋅dx⋅dy⋅dz

For the figure at left:

● Differential element shown

● Constant density fluid● Forces acting:

– Body force = γ⋅dx⋅dy⋅dz

– Surface forces = pressure forces

● Fluid at rest Element in

equilibrium● Sum of forces

must be zero

∑ F x=0⇒ ∂ p∂ x=0 ∑ F y=0⇒ ∂ p

∂ y=0 ∑ F y=0⇒ ∂ p

∂ z=−

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Variation of pressure in static fluid (2)

O

dx

dy

dz

x

z

y

p∂ p∂ z⋅dz

2⋅dx⋅dy

p−∂ p∂ z⋅dz

2⋅dx⋅dy

⋅dx⋅dy⋅dz

∑ F x= p−∂ p∂ z⋅dz

2⋅dx⋅dy− p∂ p

∂ z⋅dz

2⋅dx⋅dy−⋅dx⋅dy⋅dz=0

dpdz=−

∑ F x=0⇒ ∂ p∂ x=0

∑ F y=0⇒ ∂ p∂ y=0

∑ F y=0⇒ ∂ p∂ z=−

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Variation of pressure in static fluid (3)

O

dx

dy

dz

x

z

y

p∂ p∂ z⋅dz

2⋅dx⋅dy

p−∂ p∂ z⋅dz

2⋅dx⋅dy

⋅dx⋅dy⋅dzdpdz=−

● For incompressible fluids: γ constant, integrate dp/dz = -γ directly● For compressible fluids: g = f(z) or g = f(p), e.g., atmospheric pressure

(Sample problem 3.1 – pp. 47-49)

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Sample problem 3.1. - Pressure variation in the atmosphere – Solving dp/dz = -γ with p(z1) = p1

(a) Assume air has constant density: p – p1 = -γ (z - z1)

(b) Assume isothermal conditions: pv = const p/γ = p1 /γ1 γ = pγ1 /p1 dp/dz = - pγ1 /p1 dp/p = - (γ1 /p1 ) dz

after integration and simplification:

p/p1 = exp(-(γ1 /p1)(z-z1))

(c) Assume isentropic conditions: pv n = p/ρ ν p/γ n=p1 /γ1n

γ = γ1(p/p1 )1/n dp/p1/n = - (γ1 /p1

1/n ) dz after integration and simplification:

p 1-1/n - p11-1/n = - (1-1/n) (γ1 /p1

1/n )(z - z1)

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Sample problem 3.1. - cont. (2) Solving dp/dz = -γ with p(z1) = p1

(d) Assume air temperature decreasing linearly with height at standard lapse

For temperature variation use:

T = a + bz, with a = 518.67oR, b = - 0.003560 oR/ft

Use gas law ρ = p/RT, together with hydrostatic law dp/dz = -ρ⋅g, to get

dp/p = - g/(R(a+bz)) dz

After integration and simplification:

pp1= ab⋅z

ab⋅z1−g /R⋅b

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See Appendix A, Table A.3, to get: T1 = 59.0oF, p1 = 14.70 psia, γ1 = 0.07648 lb/ft3, z1 = 0 ft. Also, for isentropic process use n = 1.4. And, for standard temperature decrease, a = 518.67oR, b = - 0.003560 oR/ft. The elevation of interest is z = 20 000 ft.

Expressing p1 in standard BG units: p1 = 14.70×144 = 2116.8 lb/ft2

(a) p = p1 – γ (z-z1) = 2116.8 – 0.07648× (20 000 - 0) = 587.20 lb/ft2 = 587.20/144 psia = 4.08 psia

(b) p = p1 exp(-(γ1/p1)(z-z1)) = 14.70 exp(-(0.07648/2116.8)(20 000 – 0)) = 7.14 psia

(c) n = 1.4, 1/n = 0.714, 1-1/n = 0.286,p 0.286 = p1 0.286 -0.286 (γ1 /p1

0.714 )(z – z1) = 2116.8 0.286 – 0.286(0.07648/2116.8 0.714)(20 000 – 0) = 7.0892

p = (7.0892)1 / 0.286 = 942.17 psfa = 942.17/144 psia = 6.54 psia

Sample problem 3.1. - cont. (3)

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See Appendix A, Table A.3, to get: T1 = 59.0oF, p1 = 14.70 psia, γ1 = 0.07648 lb/ft3, z1 = 0 ft. Also, for isentropic process use n = 1.4. And, for standard temperature decrease, a = 518.67oR, b = - 0.003560 oR/ft. The elevation of interest is z = 20 000 ft.

Expressing p1 in standard BG units: p1 = 14.70×144 = 2116.8 lb/ft2

(d) From page 22, for air R = 1715 ft⋅lb/(slug⋅ oR)

Sample problem 3.1. - cont. (4)

p= p1⋅ab⋅zab⋅z1

−g /R⋅b

=14.70⋅ 518.67−0.003560⋅20000518.670

5.27

=6.75 psia

−gR⋅b

=− 32.21715⋅−0.003560

=5.27

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Sample problem 3.1. - cont. (5) - plots

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Pressure variation for incompressible fluids (1)

● From Sample Problem 3.1:

p – p1 = -γ (z – z1)

● Applies to liquids – no need to consider compressibility unless dealing with large changes in z (e.g., deep in the ocean).

● Applies to gases for small changes in z only

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Pressure variation for incompressible fluids (2)

● If measuring depth h = z1 - z from the free surface (z = z1), with p1 = 0 , arbitrarily set:

p – p1 = -γ (z – z1) p – 0 = -γ (-h)

p = γ h

● Pascal's law: all points in a connected body of a constant-density fluid at rest are under the same pressure if they are a the same depth below the liquid surface.

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Pressure variation for incompressible fluids (3)

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Pressure as fluid height (1)● For constant density fluids, and taking the free-

surface pressure as zero, p = γ h.● Thus

● Pressure related to the height, h, of a fluid column.

● Referred to as the pressure head

h(ft of H20) = 144 psi/62.4 = 2.308 psih(m of H20) = kPa/9.81 = 0.1020 kPa

h= p

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Pressure as fluid height (2)● Equation p – p1 = -γ (z – z1) can be rearranged as:

● Terms: z = elevation, p/γ = pressure head

● Thus, in a liquid at rest, an increase in the elevation (z) means a decreases in pressure head (p/γ), and vice versa.

pz=

p1

z1=constant

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Pressure as fluid height (3)

p A

z A=

pB

z B=constant

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Absolute and gage pressures (1)

● Pressure measured:– Relative to absolute zero (perfect vacuum): absolute– Relative to atmospheric pressure: gage

● If p < patm, we call it a vacuum, its gage value = how much below atmospheric

● Absolute pressure values are all positive● Gage pressures:

– Positive: if above atmospheric– Negative: if below atmospheric

● Relationship: pabs = patm + pgage

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Absolute and gage pressures (2)

Absolutepressure

Atmosphericpressure

Atmospheric pressure

Vacuum = negative gage pressure

Absolutepressure

Gagepressure

Absolute zero

Pres

sure

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Absolute and gage pressure (3)

● Atmospheric pressure is also called barometric pressure

● Atmospheric pressures varies:– with elevation– with changes in meteorological conditions

● Use absolute pressure for most problems involving gases and vapor (thermodynamics)

● Use gage pressure for most problems related to liquids

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Measurement of pressure

● Barometer

● Bourdon gage

● Pressure transducer

● Piezometer column

● Simple manometer

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Barometer (1)● Measures the absolute

atmospheric pressure● Tube barometer shown● Tube must be long enough● Vapor pressure at top of tube● Liquid reached maximum height

in tube

pO = γ⋅y+pvapor = patm

● With negligible pvapor

●

patm = γ⋅y

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Mercury barometer diagram

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Mercury barometer

photograph

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Barometer (2)

● Aneroid barometer: uses elastic diaphragm to measure atmospheric pressure

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Aneroid barometer photograph

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Barometer (2)

● Values of standard sea-level atmospheric pressure:

14.696 psia = 2116.2 psfa = 101.325 kPa abs = 1013.25 mb abs = 29.92 in Hg = 760 mm Hg =

33.19 ft H20 = 10.34 m H20

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Bourdon gage (1)

● Curved tube of elliptical cross-section changes curvature with changes in pressure

● Moving end of tube rotates a hand on a dial through a linkage system

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Bourdon gage (2)

● Pressure indicated at center of gage

● If tube filled with same fluid as in A and pressure graduated in psi

pA(psi)=gage reading(psi) + γ⋅h/144

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Bourdon gage (3)

● Vacuum gage (negative pressures) graduated in millimiters or inches of mercury

InHg vacuum at A =gage reading(inHg vacuum) – γ⋅h/144 (29.92/14.70)

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Bourdon gage (3)

● Note: h < 0 if Bourdon gage is below measuring point

● In pipes, pressure is typically measuredat centerline

● For measurements in gas pipes, elevation correction is negligible

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Bourdon gage (5)

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Pressure transducer (1)● Transducer: a device that transfer energy from system to

another (e.g., Bourdon gage transfers pressure to displacement)

● Electrical pressure transducer converts displacement of a diaphragm to an electrical signal.

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Tip of submergence pressure transducer

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Piezometer column (1)

● To measure moderate pressures of liquids

● Sufficiently long tube where fluid rises w/o overflowing

● Height in tube is

h = p/γ

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Piezometer tubes in orifice meter (1)

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Piezometer columns in orifice meter (2)

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Piezometer columns in Venturi meter

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How to write a manometer's equation

● Start at point of know pressure (pgage = 0 at open end), write down that pressure.

● Follow the path of the manometer in a given direction, moving from one meniscus to the next in the proper order.

● Add γ⋅h if moving downwards to next meniscus or point of interest. Use proper value of γ.

● Subtract γ⋅h if moving upwards to next meniscus or point of interest. Use proper value of γ.

● Make equation equal to pressure of end point

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Simple manometer● Mercury U tube shown● Determine gage pressure at

A● Gage or manometer equation● s = specific gravity

– sM = for manometer fluid– sF = for the “fluid”

● γW = specific weight of water● Manometer equation:

0 + sM⋅γW ⋅Rm + sF⋅γW ⋅h = pA

● Divide by γ = sF⋅γW, then

pA/γ = h + (sM/sF)⋅Rm 44

Vacuum pressure (1)

● Manometer equation: 0 - sM⋅γW ⋅Rm + sF⋅γW ⋅h = pA

● Divide by γ = sF⋅γW, then: pA/γ = h-(sM/sF)⋅Rm

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Vacuum pressure (2)

● Manometer equation:

0 - sM⋅γ W⋅Rm - sF⋅γW⋅h = pA

pA/γ = -h-(sM/sF)⋅ Rm

● If absolute pressure is sought, replace 0 with patm in the previous equations

● If the “fluid” is a gas, sF ≈ 0, and thus, pressure contributions due to the gas are negligible.

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Differential Manometer (1):

U-tube manometer for Venturi meter in pipeline

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Differential manometer (2)

● Manometer equation:pA – sF⋅ γW ⋅hA- sM⋅ γW ⋅Rm

+ sF⋅ γW ⋅hB = pB

Divide by γ = sF⋅ γW

pA /γ - pB /γ = hA-hB+(sM/sF)Rm

Also, hA-hB= (zA-zB)-Rm

pA /γ -pB /γ = zA-zB+(sM/sF-1)⋅Rm

Δ(p /γ +z)A-B = (sM/sF-1)⋅Rm

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Differential manometer (3)

● Fluids in A and B are the same● Common mistake: omitting the (sM/sF-1)⋅ factor in

equation● When the manometer fluid is mercury (sM = 13.56),

the differential manometer is suitable for measuring large pressure differences

● For smaller pressure differences, use oil (e.g., sM = 1.6, sM = 0.8)

● Manometer fluid should not mix with the fluid whose pressure difference is being measured

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Differential manometer (4)

● Manometer equation:

pA – sF⋅ γW ⋅(zA-zB+x+Rm) - sM⋅ γW ⋅Rm+ sF⋅ γW ⋅ x = pB

Simplify and divide by γ = sF⋅ γW

pA /γ -pB /γ = zA-zB+(sM/sF-1)⋅Rm

or,

Δ(p /γ +z)A-B = (sM/sF-1)⋅Rm

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Differential manometer (4)

● In this case, sM/sF < 1.● As sM -> sF, (1-sM/sF) -> 0,

larger values of Rm, i.e., increased sensitivity of manometer

● To measure Δ(p/γ+z) in liquids we often use air for the manometer fluid

● If needed, air can be pumped through valve V until the pressure is enough to bring liquid columns to suitable levels

● An alternative for increasing manometer sensitivity: incline the gage tube

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Pressure transducers integrated into a digital differential manometer