Download - Ch 3 – Fluid Statics - I
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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)
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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)
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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 /p11/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=− 32.2
1715⋅−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
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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
∆(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,
∆(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 ∆(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