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ES206 Fluid Mechanics
UNIT B: Fluid Statics
ROAD MAP . . .
B-1: Pressure in a Stationary Fluid
B-2: Atmospheric Pressure
B-3: Manometry
B-4: Hydrostatic Force
ES206 Fluid Mechanics
Unit B-1: List of Subjects
Absolute, Gauge, and Vacuum Pressure
Pascal’s Law
Hydrostatic Pressure
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Common Abbreviation for Pressure Units
• 1 “Newton per square meter” = 1 N/m2 = 1 “Pascal” = 1 Pa
• 1 “pound per square inch” = 1 lb/in2 = 1 psi
• 1 “pound per square foot” = 1 lb/ft2 = 1 psf
• psi or psf in “gage” pressure = psig or psfg
• psi or psf in “absolute” pressure = psia or psfa
Absolute or Gage “Zero” Pressure
• The pressure in a perfect vacuum is absolute zero pressure (this is only
“theoretical”): there is no such things of “negative psia”
• “Zero psig” means equal to the local atmospheric pressure (1 atm)
Standard Sea-Level Atmospheric Pressure (1 atm)
• 14.7 psi
• 101 kN/m2 (kPa)
• 760 mmHg (barometric pressure)
• 1 bar
Unit B-1Page 1 of 6
Absolute, Gauge, and Vacuum Pressure (1)
Gage Pressure
• Pressure gage = provides positive gage pressure value (relative to the local
atmospheric pressure)
For example, 5 psig = +5 psi, relative to the local atmospheric pressure
• Vacuum gage = provides negative gage pressure value (relative to the local
atmospheric pressure)
For example, 2 psig (vac.) = 2 psi, relative to the local atmospheric pressure
• If the gage is zero = the pressure is equal to the local atmospheric pressure
(ZERO GAGE pressure)
(IMPORTANT): “negative” gage pressure means vacuum (lower pressure than
the local atmospheric pressure). However, there is no such thing of “negative”
absolute pressure . . .
Unit B-1Page 2 of 6
Absolute, Gauge, and Vacuum Pressure (2)
“ZERO GAGE” Pressure(Local Atmospheric)
Pascal’s Law
• The pressure of a stationary fluid acts equally in all directions: this is called
Pascal’s Law
• Also, the Pascal’s Law states that a pressure in a closed system can be
transmitted within the system: larger force can be created by changing the cross
section of the acting pressure
So, the pressure within the system is everywhere constant?
Unit B-1Page 3 of 6
Pascal’s Law
Unit B-1Page 3 of 6
Pascal’s Law
Pressure = p (constant everywhere?)
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Hydrostatic Pressure Variation
Let us apply the static force equilibrium equation in the direction:
0F : sin 0p A p p A A
sin 0p A A sin 0p
Note that: sinz
0z
p
0p z p z
p
z
For the limit 0z : dp
dz
Unit B-1Page 4 of 6
dd
Hydrostatic Pressure (1)
h: depth(positive downward)
z: height (positive upward)
(0)
(1)
(2)
(pressure = p1)
(pressure = p2)
+
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Hydrostatic Pressure Variation for Incompressible Fluid (Liquid)
Note: specific weight () is a function of height (z) or depth (h)
dp
zdz
pressure “decreases” with height
or dp
hdh
pressure “increases” with depth
For incompressible fluid (liquids): constant
dp
dz dp dz
1 1
2 2
p z
p z
dp dz
1 2 1 2p p z z 2 1 1 2p p p z z z
Alternatively, dp
dh dp dh
2 2
1 1
p h
p h
dp dh
2 1 2 1p p h h 2 1 2 1p p p h h h
The pressure at an arbitrary depth h is:
p h (“gage” pressure) or 0p h p (“absolute” pressure)
Unit B-1Page 5 of 6
Hydrostatic Pressure (2)
zdz
dp 2 1 1 2p p p z z z
(gage)p h
dph
dh
0 (absolute)p h p
2 1 2 1p p p h h h
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At the open standpipe (75 ft above the ground), the pressure is equal to the
atmospheric pressure:
standpipe 0p p
In order to feed water, the fire hydrant (ground level), must maintain the
hydrostatic pressure of:
fire hydrant standpipe 0p p h p h
In terms of gage pressure: 0 0p
2
3
fire hydrant
1 ft62.4 lb/ft 75 ft
12 inp h
= 32.5 psi, gage
Also, in kPa: 2 2
2
lb 4.448 N 12 in 1 ft32.5
in 1 lb 1 ft 0.3048 m
= 224,068 N/m2 (224 kPa), gage
Unit B-1Page 6 of 6
Class Example Problem
Related Subjects . . . “Hydrostatic Pressure”
ES206 Fluid Mechanics
UNIT B: Fluid Statics
ROAD MAP . . .
B-1: Pressure in a Stationary Fluid
B-2: Atmospheric Pressure
B-3: Manometry
B-4: Hydrostatic Force
ES206 Fluid Mechanics
Unit B-2: List of Subjects
Isothermal Pressure Variation
U.S. Standard Atmosphere
Mercury Barometer
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Hydrostatic Pressure Variation for Compressible Fluid (Gas)
Now, let us consider hydrostatic pressure variation in gas: z
dp
zdz
=> dp
z g zdz
(Note that: g )
(1) let’s assume that constantg (near sea-level, low altitude):
dp pg
dz RT =>
dp g dz
p R T
Integrating the equation: 2 2
1 1
p z
p z
dp g dz
p R T =>
2
1
2
1
ln
z
z
p g dz
p R T
The solution of this equation depends on the variation of temperature:
(2) if we assume constant temperature (isothermal condition: 0T T )
2 1
0
2 1
g z z
RTp p e
Unit B-2Page 1 of 7
Isothermal Pressure Variation
0
1212
)(exp
RT
zzgpp
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If the temperature is constant (“isothermal”):
2 1
0
2 1
g z z
RTp p e
If the air is assumed to be “incompressible”: 1 2p p h
(a) Assuming the air to be at a common temperature of 59 F (isothermal)
2
2 12
o
1 0
32.2 ft/s 1,250 ftexp exp
1,716 ft lb/slug R 59 F 460 R
g z zp
p RT
= 0.956
(b) Assuming the air to be incompressible
1 2p p h (hydrostatic equation)
or, 2 1 2 1p p z z
Therefore,
3
2 12
2
1 1 2
0.0765 lb/ft 1,250 ft1 1
12 in14.7 lb/in
1 ft
z zp
p p
= 0.955
Unit B-2Page 2 of 7
EXAMPLE 2.2
Incompressible and Isothermal Pressure–Depth Variations
Textbook (Munson, Young, and Okiishi), page 46
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Troposphere (Sea-Level => 11 km)
Temperature decreases linearly with altitude (z):
aT T z
From hydrostatic pressure variation: dp dz gdz
dp gdz gdz
p RT RT
=>
0a
p z
p
dp g dz
p R T
Note that: aT T z means dT dz => dT
dz
, thus:
a a
p T
p T
dp g dT
p R T => ln ln
a a
p g T
p R T =>
g
R
a a
p T
p T
Substituting the temperature variation ( aT T z ) yields:
1
g
R
a
a
zp p
T
, where:
aT & ap = Temperature & pressure at Sea-Level
= Lapse Rate (0.0065 K/m or 0.00357 R/ft)
Unit B-2Page 3 of 7
U.S. Standard Atmosphere (1)
Stratosphere (11 km => 20.1 km)
• Temperature remains constant (isothermal condition: 0T T ) 2 1
0
2 1
g z z
RTp p e
Cruise Altitude of a Jet Airliner
• At the edge of the troposphere (11 km or 36,000 ft)
• Temperature / Pressure = 56.6 C / 22.6 kPa
An example of the pressure calculation, based on the U.S. standard atmosphere
model involves:
(1) Starting from standard sea-level conditions: pa = 101.3 kPa / Ta = 15 C
(2) Apply Troposphere equation (0 => 11 km)
1 1aT T z 11 1
g
R
a
a
zp p
T
(determine T1 and p1 at 11 km altitude)
(3) Apply Stratosphere equation (11 km => 20.1 km):
2 1
0
2 1
g z z
RTp p e
Unit B-2Page 4 of 7
U.S. Standard Atmosphere (2)
“Gradient”
Region
“Isothermal”
Region
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The stratosphere is the isothermal region ( o56.5 CT : constant):
2 12
1 0
expg z zp
p RT
At the edge of the troposphere:
1 11 kmz , 1 22.6 kPap
Also, for standard air: 287 J/kg KR
From the textbook table C.2 in Appendix C (page 765):
12.11 kPap and 30.1948 kg/m
Therefore:
2 1
2 1
0
expg z z
p pRT
2
3
o
9.77 m/s 15,000 m 11,000 m22.6 10 Pa exp
287 J/kg K 56.5 C 273 K
= 12.05103 Pa (12.05 kPa)
Also, applying the equation of state: p
RT
3
22 o
2
12.05 10 Pa
287 J/kg K 56.5 C 273 K
p
RT
= 0.1939 kg/m3
Unit B-2Page 5 of 7
Class Example Problem
Related Subjects . . . “U.S. Standard Atmosphere”
Barometric Pressure
• It is conventional to measure and specify atmospheric pressure in terms of the
height of liquid, called Barometric Pressure (head)
• Applying hydrostatic equation: vapor Mercury atmp h p
atm vapor
Mercury
p ph
=> atm
Mercury
ph
( atm vaporp p )
• 1 atm is approximately 760 mm Hg or 29.9 inches Hg
(standard sea-level barometric pressure)
Unit B-2Page 6 of 7
Mercury Barometer
(pvapor = 0.000023 lb/in.2 (abs) at 68 F)
atm vapor atm
Mercury Mercury
p p ph
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A barometric pressure of 29.4 in means: atm
Hg
29.4 in Hg (barometric pressure)p
This corresponds to:
2
3
atm Hg 2
1 ft lb 1 ft29.4 in 29.4 in 847 lb/ft 2,075.15
12 in ft 12 inp
= 14.41 psi
In pascals, 99.35 kPa
Also, in terms of column of water:
2
2
2
atm
3
H O
12 in14.41 lb/in
1 ft
62.4 lb/ft
p
= 33.26 ft in H2O
Unit B-2Page 7 of 7
Class Example Problem
Related Subjects . . . “Mercury Barometer”
ES206 Fluid Mechanics
UNIT B: Fluid Statics
ROAD MAP . . .
B-1: Pressure in a Stationary Fluid
B-2: Atmospheric Pressure
B-3: Manometry
B-4: Hydrostatic Force
ES206 Fluid Mechanics
Unit B-3: List of Subjects
Concept of Manometers
Differential U-Tube Manometer
Differential Manometer
Inclined U-Tube Differential Manometer
Piezometer Tube
• The simplest type of manometer: the pressure at point A can be given by a series
of hydrostatic pressure variation applications as follows:
Starting from the location (A): pressure is Ap
Location (1): pressure, relative to A is: 1 Ap p
Open-end of the tube: pressure is “zero gauge,” and relative to 1 is: 1 1 10 p h
1 1Ap h
U-Tube Manometer
• The similar application of hydrostatic pressure variation, starting from the open
end (open to the local atmosphere: zero gage) yields:
2 2 1 10 Ah h p 2 2 1 1Ap h h
• If the fluid ( 1 ) is gas, since 1 2gas liquid :
2 2Ap h
Unit B-3Page 1 of 6
Concept of Manometers
11hpA
PiezometerTube
1122 hhpA
U-Tube Manometer
Differential Manometer
• This manometer cannot measure actual pressure: only measures pressure
difference ( p )
• Starting from one end of the manometer:
1 1 2 2 3 3A Bp h h h p 2 2 3 3 1 1A Bp p h h h
Manometer Equations
• It is clear now, that each manometer, depending on the shape and type of the
choices of fluids, comes with a unique equation to describe the pressure being
measured: this is called the “manometer equation.”
• Manometer equation is purely dependent upon the hydrostatic pressure
variation.
Unit B-3Page 2 of 6
Differential U-Tube Manometer
113322 hhhpp BA
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Recall the hydrostatic pressure variation: p h
Applying the hydrostatic pressure equation from one end to the other end:
Oil Mercury Water4 in 3 in 12 in 12 in 3 inA Bp p
Hence,
3 31 ft 1 ft57 lb/ft 4 in 3 in 847 lb/ft 12 in
12 in 12 inAp
3 1 ft62.4 lb/ft 12 in 3 in
12 inBp
The pressure difference is:
A Bp p = 802.25 lb/ft2
Unit B-3Page 3 of 6
Class Example Problem
Related Subjects . . . “U-Tube Manometer”
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Note: pressure differential (1 2p p ) is given: 0.5 lb/in2.
Applying the hydrostatic pressure equation from 1p to
2p :
2 21 H O 1 gf H O 1 2p h h h h p
or, 21 H O gf 2p h h p
Therefore,
2
2
2
1 2
3 3
gf H O
12 in0.5 lb/in
1 ft
112 lb/ft 62.4 lb/ft
p ph
= 1.452 ft
Unit B-3Page 4 of 6
Class Example Problem
Related Subjects . . . “U-Tube Manometer”
Inclined U-Tube Differential Manometer
• One leg of the manometer is inclined at an angle , and differential reading 2 is
measured along the inclined tube:
2 2 3 3 1 1sinA Bp p h h
• If A and B are gases, 2 2 sinA Bp p
• Inclined U-tube manometers can have a “higher measurement resolution” than
regular (vertical tube) U-tube manometers.
• Also, the inclination angle ( ) can be “adjusted,” so that the measured length
(gage reading) can conveniently be interpreted as a specific standard liquid’s
liquid column height (equivalent).
(for example . . . 10 inches of water)
• Inclined U-tube manometers are commonly used (combined with “Pitot-static
probe”) for airspeed measurement.
Unit B-3Page 5 of 6
Inclined U-Tube Differential Manometer (1)
sin22 BA pp
Airspeed Measurements
• Inclined U-Tube Differential Manometer is typically used to measure small
pressure differences in gas.
• Pitot-static probe is usually combined with this type of manometer for airspeed
measurement (for example, the figure shows the setup in a wind tunnel test
section airspeed measurement).
• pt: the “total” pressure – this is the pressure measured at the location of zero
airspeed (called, the “stagnation point.”).
• ps: the “static” pressure – this is the pressure of a local atmosphere given at
the condition of the wind tunnel test section.
• t sp p p : the pressure difference between above two is called the
“dynamic” pressure (q):
21
2q V
• Obviously, this pressure difference can be determined by using the
differential U-tube (usually inclined) manometer. sinp
Unit B-3Page 6 of 6
Inclined U-Tube Differential Manometer (2)
ES206 Fluid Mechanics
UNIT B: Fluid Statics
ROAD MAP . . .
B-1: Pressure in a Stationary Fluid
B-2: Atmospheric Pressure
B-3: Manometry
B-4: Hydrostatic Force
ES206 Fluid Mechanics
Unit B-4: List of Subjects
Hydrostatic Force
Hydrostatic Force on an Inclined Plane Surface of Arbitrary Shape
Geometric Properties of Some Shapes
Unit B-4Page 1 of 8
Hydrostatic Force
UNIT BUNIT B--44SLIDE SLIDE 11
Hydrostatic ForceHydrostatic Force
Textbook (Munson, Young, and Okiishi), page 58
➢ The resultant force of a static fluid on a plane surface is due to the hydrostatic pressure distribution on the surface
➢ The resultant force acts through the centroid of the surface
pAFR hp (Hydrostatic Pressure) (Resultant Force)
?UNIT BUNIT B--44SLIDE SLIDE 11
Hydrostatic ForceHydrostatic Force
Textbook (Munson, Young, and Okiishi), page 58
➢ The resultant force of a static fluid on a plane surface is due to the hydrostatic pressure distribution on the surface
➢ The resultant force acts through the centroid of the surface
pAFR hp (Hydrostatic Pressure) (Resultant Force)
?h
3/h
1
2RF h A
RF h A
The resultant force does not act through the centroid of the surface
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Differential force (dF) developed on a differential area (dA) is: dF hdA
Also, the relationship between h and y is: sinh y or siny h
Unit B-4Page 2 of 8
Hydrostatic Force on an Inclined Plane Surface of Arbitrary Shape (1)
Reference Line
UNIT BUNIT B--44SLIDE SLIDE 44
Hydrostatic Force on an Inclined Hydrostatic Force on an Inclined
Plane Surface of Arbitrary ShapePlane Surface of Arbitrary Shape (2)(2)
➢ Resultant force due to hydrostatic pressure is:
➢ Define the first moment of the area:
➢ Define the hydrostatic pressure at centroid:
AAA
R ydAdAyhdAF sinsin
AyydA c
A
AhAyF ccR sin
cc hp
ApAhF ccR (Hydrostatic Resultant Force)
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Taking the moment at point O: 2sinR R
A A
F y ydF y dA
(moment around the x-axis)
=>
2 2 2sin sin sin
sinA A A
R
R c c
y dA y dA y dA
yF h A y A
Dividing by sin : =>
2
AR
c
y dA
yy A
Unit B-4Page 3 of 8
Hydrostatic Force on an Inclined Plane Surface of Arbitrary Shape (2)
Hydrostatic
Resultant Force
ApF cR
UNIT BUNIT B--44SLIDE SLIDE 66
Hydrostatic Force on an Inclined Hydrostatic Force on an Inclined
Plane Surface of Arbitrary ShapePlane Surface of Arbitrary Shape (4)(4)
➢ Resultant force do not act through the centroid:
➢ Since resultant force is given as:
➢ Define the second moment of the area (moment of inertia):
sincR AyF
Ay
dAy
Ay
dAy
yc
A
c
AR
22
sin
sin
AA
RR dAyydFyF 2 sin
A
x dAyI 2
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Unit B-4Page 4 of 8
Hydrostatic Force on an Inclined Plane Surface of Arbitrary Shape (3)
Location of Resultant Force
(Center of Pressure)
Watch Out!Resultant force do not act through the centroid, but
Center of Pressure
Ay
Iyy
c
xccR
UNIT BUNIT B--44SLIDE SLIDE 77
Hydrostatic Force on an Inclined Hydrostatic Force on an Inclined
Plane Surface of Arbitrary ShapePlane Surface of Arbitrary Shape (5)(5)
➢ The location of resultant force is:
➢ Using parallel axis theorem: 2
cxcx AyII
Ay
AyIy
c
cxcR
2
Ay
I
Ay
dAy
yc
x
c
AR
2
Ay
Iyy
c
xccR
Location of Resultant Force (Center of Pressure)
Taking, once again, the moment at point O (moment around the y-axis):
sinR R
A A
F x xdF xydA
Unit B-4Page 5 of 8
Hydrostatic Force on an Inclined Plane Surface of Arbitrary Shape (4)
Location of
Resultant Force
(Center of
Pressure)
Watch Out!If cross section is
symmetric along vertical
(y) axis (Ixyc = 0), centroid
and center of pressure
will coincide
Ay
Ixx
c
xyc
cR
UNIT BUNIT B--44SLIDE SLIDE 99
Hydrostatic Force on an Inclined Hydrostatic Force on an Inclined
Plane Surface of Arbitrary ShapePlane Surface of Arbitrary Shape (7)(7)
➢ Similarly, the x-coordinate location of resultant force can be determined as:
➢ Using parallel axis theorem:
A
RR xydAxF sinAy
I
Ay
xydA
xc
xy
c
AR
ccxycxy yAxII
Ay
yAxIx
c
ccxyc
R
Ay
Ixx
c
xyc
cR
Location of Resultant Force (Center of Pressure)
Centroid (C) is a geometric center of an object
Centroid = center of gravity (CG), if the object is perfectly homogeneous
Center of Pressure (CP) is where a resultant force (due to the distribution of
hydrostatic pressure) acts on an object
CP location is usually not equal to the centroid (only exception is the case that the
plane surface is horizontally submerged under the liquid: means that = 0)
Unit B-4Page 6 of 8
Geometric Properties of Some Shapes
UNIT BUNIT B--44SLIDE SLIDE 11
Hydrostatic ForceHydrostatic Force
Textbook (Munson, Young, and Okiishi), page 58
➢ The resultant force of a static fluid on a plane surface is due to the hydrostatic pressure distribution on the surface
➢ The resultant force acts through the centroid of the surface
pAFR hp (Hydrostatic Pressure) (Resultant Force)
?
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(a) The magnitude and location of the resultant force exerted on the gate by the
water
23 39.80 10 N/m 10 m 4 m
4R c cF p A h A
= 1,230103 N (1.23 MN)
xcR c
c
Iy y
y A where,
4
4xc
RI
, and
10 m
sin 60cy
Hence,
4
2
2 m 4 10 m
sin 6010 m sin 60 4 4 mRy
= 11.63 m
(b) The moment that would have to be applied to the shaft to open the gate
0cM : 3 10 m1,230 10 N 11.63 m
sin 60R R cM F y y
= 102,083 Nm
Unit B-4Page 7 of 8
EXAMPLE 2.6Textbook (Munson, Young, and Okiishi), page 61
Hydrostatic Pressure Force on a Plane Surface
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Hydrostatic resultant force is:
3 3 4 m9.80 10 N/m 2 m 1 m
sin 51.3R cF h A
= 100103 N
Static equilibrium 0xF :
osin 51.3R fF F N (eqn. 1)
Also, 0yF :
cos 51.3RN W F (eqn. 2)
where, concrete concreteW V and
3
concrete
4 m 5 m2 m 5 m 1 m 20 m
2V
Hence, from eqn. 2:
3 3 3 323.6 10 N/m 20 m 100 10 N cos 51.3N = 534,524 N
From eqn. 1: 3100 10 N sin 51.3sin 51.3
534,524 N
RF
N
= 0.146
Unit B-4Page 8 of 8
Class Example Problem
Related Subjects . . . “Hydrostatic Force”
Unit B-4Page 8 of 8
Class Example Problem
Related Subjects . . . “Hydrostatic Force”
W FR
N
Ff
1 o5tan 51.3
4
