Ch 3 – Fluid Statics - I
Introduction to fluid statics (1)
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Fluid at rest:
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Prepared for
CEE 3500 – CEE Fluid Mechanics
by
Gilberto E. Urroz,
August 2005
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Normal forces are important:
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Overturning of concrete dams
Bursting of pressure vessels
Breaking of lock gates on canals
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Introduction to fluid statics (2)
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Introduction to fluid statics (3)
For design: compute magnitude and location of
normal forces
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Development of instruments that measure pressure
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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
Development of systems that transfer pressure,
e.g.,
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–
p=
automobile breaks
hoists
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dF
dA
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
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Variation of pressure in static fluid (1)
Isotropy of pressure
dy = normal
to paper
dx = dl sin α
dy = dl cos α
z
x
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No shear stresses
Only normal forces due to pressure
z
p dl dy
α
α
dl
dz
px dy dz
∂ p dz
⋅ ⋅dx⋅dy
∂z 2
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dx
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⋅dx⋅dy⋅dz
dx
pz dx dy
For the figure at left:
 p
dz
γ ½ dx dy dz
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dy
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)
∂ p dz
 p− ⋅ ⋅dx⋅dy
∂z 2
x
∂p
∑ F x =0 ⇒ ∂ x =0
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y
O
∂p
∑ F y =0 ⇒ ∂ y =0
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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
∂p
∑ F y =0 ⇒ ∂ z =−
Variation of pressure in static fluid (2)
Variation of pressure in static fluid (3)
z
 p
∂p
∑ F x =0 ⇒ ∂ x =0
 p
dx
∂p
∑ F y =0 ⇒ ∂ y =0
⋅dx⋅dy⋅dz
⋅dx⋅dy⋅dz
dz
dy
dy
y
O
∂ p dz
 p− ⋅ ⋅dx⋅dy
∂z 2
x
∂p
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dp
=−
dz
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y
∂ p dz
 p− ⋅ ⋅dx⋅dy
∂z 2
x
∂ p dz
∂ p dz
F x = p−
⋅ ⋅dx⋅dy− p
⋅ ⋅dx⋅dy−⋅dx⋅dy⋅dz =0
∂z 2
∂z 2
∑ F y =0 ⇒ ∂ z =−
∂ p dz
⋅ ⋅dx⋅dy
∂z 2
dx
dp
=−
dz
dz
O
∑
z
∂ p dz
⋅ ⋅dx⋅dy
∂z 2
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
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
(a) Assume air has constant density: p – p1 = -γ (z - z1)
For temperature variation use:
(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:
T = a + bz,
with a = 518.67oR, b = - 0.003560 oR/ft
p/p1 = exp(-(γ1 /p1)(z-z1))
Use gas law ρ = p/RT, together with hydrostatic law dp/dz = -ρ⋅g, to get
(c) Assume isentropic conditions: pv n = p/ρ ν
p/γ n=p1 /γ1n
1/n
1/n
1/n
γ = γ1(p/p1 )
dp/p = - (γ1 /p1 ) dz
after integration and simplification:
dp/p = - g/(R(a+bz)) dz
After integration and simplification:
−g / R⋅b
p
ab⋅z
=

p 1 ab⋅z 1
p 1-1/n - p11-1/n = - (1-1/n) (γ1 /p11/n )(z - z1)
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Sample problem 3.1. - cont. (3)
Sample problem 3.1. - cont. (4)
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.
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
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
(d) From page 22, for air R = 1715 ft⋅lb/(slug⋅ oR)
32.2
−g
=−
=5.27
R⋅b
1715⋅−0.003560
(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 /p10.714 )(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
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p= p 1⋅
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−g / R⋅b
ab⋅z

ab⋅z 1
=14.70⋅
5.27
518.67−0.003560⋅20000

518.670
=6.75 psia
Sample problem 3.1. - cont. (5) - plots
Pressure variation for incompressible fluids (1)
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From Sample Problem 3.1:
p – p1 = -γ (z – z1)
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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)
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Pressure variation for incompressible fluids (3)
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=γh
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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 as fluid height (1)
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Pressure as fluid height (2)
For constant density fluids, and taking the freesurface pressure as zero, p = γ h.
Thus
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p1
p
z= z 1=constant


p
h=

Pressure related to the height, h, of a fluid
column.
Referred to as the pressure head
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h(ft of H20) = 144 psi/62.4 = 2.308 psi
h(m of H20) = kPa/9.81 = 0.1020 kPa
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Equation p – p1 = -γ (z – z1) can be rearranged as:
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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.
Pressure as fluid height (3)
Absolute and gage pressures (1)
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Pressure measured:
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–
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If p < patm, we call it a vacuum, its gage value =
how much below atmospheric
Absolute pressure values are all positive
Gage pressures:
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pA

z A =
pB

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z B =constant
Positive: if above atmospheric
Negative: if below atmospheric
Relationship:
pabs = patm + pgage
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Absolute and gage pressures (2)
Absolute and gage pressure (3)
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Gage
pressure
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Pressure
Atmospheric pressure
Absolute
pressure
Atmospheric
pressure
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Atmospheric pressure is also called barometric
pressure
Atmospheric pressures varies:
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Absolute
pressure
Vacuum = negative
gage pressure
Absolute zero
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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 (1)
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Barometer
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Bourdon gage
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Pressure transducer
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Piezometer column
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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
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Relative to absolute zero (perfect vacuum): absolute
Relative to atmospheric pressure: gage
With negligible pvapor
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Simple manometer
patm = γ⋅y
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Mercury barometer diagram
Mercury
barometer
photograph
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Aneroid barometer photograph
Barometer (2)
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Aneroid
barometer: uses
elastic diaphragm
to measure
atmospheric
pressure
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Barometer (2)
Bourdon gage (1)
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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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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
Bourdon gage (2)
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Bourdon gage (3)
Pressure indicated at
center of gage
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If tube filled with same
fluid as in A and pressure
graduated in psi
InHg vacuum at A =gage reading(inHg
vacuum) – γ⋅h/144 (29.92/14.70)
pA(psi)=gage reading(psi) + γ⋅h/144
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Bourdon gage (3)
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Bourdon gage (5)
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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Vacuum gage (negative
pressures) graduated in
millimiters or inches of
mercury
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
Piezometer column (1)
To measure
moderate pressures
of liquids
Sufficiently long
tube where fluid
rises w/o
overflowing
Height in tube is
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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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How to write a manometer's equation
Piezometer columns in Venturi meter
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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
Vacuum pressure (1)
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:
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0 + sM⋅γW ⋅Rm + sF⋅γW ⋅h = pA
Divide by γ = sF⋅γW, then
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pA/γ = h + (sM/sF)⋅Rm
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Manometer equation:
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Divide by γ = sF⋅γW, then:
0 - sM⋅γW ⋅Rm + sF⋅γW ⋅h = pA
pA/γ = h-(sM/sF)⋅Rm
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Vacuum pressure (2)
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Differential Manometer (1):
Manometer equation:
U-tube manometer for
Venturi meter in pipeline
0 - sM⋅γ W⋅Rm - sF⋅γW⋅h = pA
pA/γ = -h-(sM/sF)⋅ Rm
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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 (2)
Differential manometer (3)
Manometer equation:
pA – sF⋅ γW ⋅hA- sM⋅ γW ⋅Rm
+ sF⋅ γW ⋅hB = pB
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Divide by γ = sF⋅ γW
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pA /γ - pB /γ = hA-hB+(sM/sF)Rm
Also, hA-hB= (zA-zB)-Rm
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pA /γ -pB /γ = zA-zB+(sM/sF-1)⋅Rm
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Δ(p /γ +z)A-B = (sM/sF-1)⋅Rm
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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
Differential manometer (4)
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Differential manometer (4)
Manometer equation:
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pA – sF⋅ γW ⋅(zA-zB+x+Rm)
- sM⋅ γW ⋅Rm+ sF⋅ γW ⋅ x = pB
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Simplify and divide by γ = sF⋅ γW
pA /γ -pB /γ = zA-zB+(sM/sF-1)⋅Rm
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or,
Δ(p /γ +z)A-B = (sM/sF-1)⋅Rm
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Pressure transducers
integrated into a digital
differential manometer
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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