Chapter 3: Pressure and Fluid Statics
Eric G. Paterson
Department of Mechanical and Nuclear Engineering
The Pennsylvania State University
Spring 2005
Note to Instructors
These slides were developed1 during the spring semester 2005, as a teaching aid
for the undergraduate Fluid Mechanics course (ME33: Fluid Flow) in the Department of
Mechanical and Nuclear Engineering at Penn State University. This course had two
sections, one taught by myself and one taught by Prof. John Cimbala. While we gave
common homework and exams, we independently developed lecture notes. This was
also the first semester that Fluid Mechanics: Fundamentals and Applications was
used at PSU. My section had 93 students and was held in a classroom with a computer,
projector, and blackboard. While slides have been developed for each chapter of Fluid
Mechanics: Fundamentals and Applications, I used a combination of blackboard and
electronic presentation. In the student evaluations of my course, there were both positive
and negative comments on the use of electronic presentation. Therefore, these slides
should only be integrated into your lectures with careful consideration of your teaching
style and course objectives.
Eric Paterson
Penn State, University Park
August 2005
1 These
slides were originally prepared using the LaTeX typesetting system (http://www.tug.org/)
and the beamer class (http://latex-beamer.sourceforge.net/), but were translated to PowerPoint for
wider dissemination by McGraw-Hill.
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Chapter 3: Pressure and Fluid Statics
Pressure
Pressure is defined as a normal force
exerted by a fluid per unit area.
Units of pressure are N/m2, which is called
a pascal (Pa).
Since the unit Pa is too small for pressures
encountered in practice, kilopascal (1 kPa
= 103 Pa) and megapascal (1 MPa = 106
Pa) are commonly used.
Other units include bar, atm, kgf/cm2,
lbf/in2=psi.
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Chapter 3: Pressure and Fluid Statics
Absolute, gage, and vacuum pressures
Actual pressure at a give point is called
the absolute pressure.
Most pressure-measuring devices are
calibrated to read zero in the atmosphere,
and therefore indicate gage pressure,
Pgage=Pabs - Patm.
Pressure below atmospheric pressure are
called vacuum pressure, Pvac=Patm - Pabs.
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Chapter 3: Pressure and Fluid Statics
Absolute, gage, and vacuum pressures
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Chapter 3: Pressure and Fluid Statics
Pressure at a Point
Pressure at any point in a fluid is the same
in all directions.
Pressure has a magnitude, but not a
specific direction, and thus it is a scalar
quantity.
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Chapter 3: Pressure and Fluid Statics
Variation of Pressure with Depth
In the presence of a gravitational
field, pressure increases with
depth because more fluid rests
on deeper layers.
To obtain a relation for the
variation of pressure with depth,
consider rectangular element
Force balance in z-direction gives
F
z
 maz  0
P2 Dx  P1Dx   g DxDz  0
Dividing by Dx and rearranging
gives
DP  P2  P1   gDz   s Dz
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Chapter 3: Pressure and Fluid Statics
Variation of Pressure with Depth
Pressure in a fluid at rest is independent of the
shape of the container.
Pressure is the same at all points on a horizontal
plane in a given fluid.
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Chapter 3: Pressure and Fluid Statics
Scuba Diving and Hydrostatic Pressure
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Chapter 3: Pressure and Fluid Statics
Scuba Diving and Hydrostatic Pressure
Pressure on diver at
100 ft?
1
 1m 
kg 
m

Pgage ,2   gz   998 3  9.81 2  100 ft  

m 
s 

 3.28 ft 
100 ft
Pabs ,2
1atm


 298.5kPa 
  2.95atm
101.325
kPa


 Pgage ,2  Patm  2.95atm  1atm  3.95atm
Danger of emergency
ascent?
2
PV
1 1  PV
2 2
If you hold your breath on ascent, your lung
volume would increase by a factor of 4, which
would result in embolism and/or death.
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Boyle’s law
V1 P2 3.95atm
 
4
V2 P1
1atm
Chapter 3: Pressure and Fluid Statics
Pascal’s Law
Pressure applied to a
confined fluid increases
the pressure throughout
by the same amount.
In picture, pistons are at
same height:
F1 F2
F2 A2
P1  P2  


A1 A2
F1 A1
Ratio A2/A1 is called ideal
mechanical advantage
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Chapter 3: Pressure and Fluid Statics
The Manometer
An elevation change of
Dz in a fluid at rest
corresponds to DP/g.
A device based on this is
called a manometer.
A manometer consists of
a U-tube containing one
or more fluids such as
mercury, water, alcohol,
or oil.
Heavy fluids such as
mercury are used if large
pressure differences are
anticipated.
P1  P2
P2  Patm   gh
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Chapter 3: Pressure and Fluid Statics
Mutlifluid Manometer
For multi-fluid systems
Pressure change across a fluid
column of height h is DP = gh.
Pressure increases downward, and
decreases upward.
Two points at the same elevation in a
continuous fluid are at the same
pressure.
Pressure can be determined by
adding and subtracting gh terms.
P2  1gh1  2 gh2  3 gh3  P1
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Chapter 3: Pressure and Fluid Statics
Measuring Pressure Drops
Manometers are well-suited to measure
pressure drops across
valves, pipes, heat
exchangers, etc.
Relation for pressure
drop P1-P2 is obtained by
starting at point 1 and
adding or subtracting gh
terms until we reach point
2.
If fluid in pipe is a gas,
2>>1 and P1-P2= gh
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Chapter 3: Pressure and Fluid Statics
The Barometer
Atmospheric pressure is
measured by a device called a
barometer; thus, atmospheric
pressure is often referred to as
the barometric pressure.
PC can be taken to be zero
since there is only Hg vapor
above point C, and it is very
low relative to Patm.
Change in atmospheric
pressure due to elevation has
many effects: Cooking, nose
bleeds, engine performance,
aircraft performance.
PC   gh  Patm
Patm   gh
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Chapter 3: Pressure and Fluid Statics
Fluid Statics
Fluid Statics deals with problems associated
with fluids at rest.
In fluid statics, there is no relative motion
between adjacent fluid layers.
Therefore, there is no shear stress in the fluid
trying to deform it.
The only stress in fluid statics is normal stress
Normal stress is due to pressure
Variation of pressure is due only to the weight of the
fluid → fluid statics is only relevant in presence of
gravity fields.
Applications: Floating or submerged bodies,
water dams and gates, liquid storage tanks, etc.
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Chapter 3: Pressure and Fluid Statics
Hoover Dam
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Chapter 3: Pressure and Fluid Statics
Hoover Dam
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Chapter 3: Pressure and Fluid Statics
Hoover Dam
Example of elevation
head z converted to
velocity head V2/2g.
We'll discuss this in
more detail in Chapter
5 (Bernoulli equation).
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Chapter 3: Pressure and Fluid Statics
Hydrostatic Forces on Plane Surfaces
On a plane surface, the
hydrostatic forces form a
system of parallel forces
For many applications,
magnitude and location of
application, which is
called center of
pressure, must be
determined.
Atmospheric pressure
Patm can be neglected
when it acts on both sides
of the surface.
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Chapter 3: Pressure and Fluid Statics
Resultant Force
The magnitude of FR acting on a plane surface of a
completely submerged plate in a homogenous fluid
is equal to the product of the pressure PC at the
centroid of the surface and the area A of the
surface
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Chapter 3: Pressure and Fluid Statics
Center of Pressure
Line of action of resultant force
FR=PCA does not pass through
the centroid of the surface. In
general, it lies underneath
where the pressure is higher.
Vertical location of Center of
Pressure is determined by
equation the moment of the
resultant force to the moment
of the distributedIpressure
force.y  y  xx ,C
p
C
yc A
$Ixx,C is tabulated for simple
geometries.
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Chapter 3: Pressure and Fluid Statics
Hydrostatic Forces on Curved Surfaces
FR on a curved surface is more involved since it
requires integration of the pressure forces that
change direction along the surface.
Easiest approach: determine horizontal and
vertical components FH and FV separately.
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Chapter 3: Pressure and Fluid Statics
Hydrostatic Forces on Curved Surfaces
Horizontal force component on curved surface:
FH=Fx. Line of action on vertical plane gives y
coordinate of center of pressure on curved
surface.
Vertical force component on curved surface:
FV=Fy+W, where W is the weight of the liquid in
the enclosed block W=gV. x coordinate of the
center of pressure is a combination of line of
action on horizontal plane (centroid of area) and
line of action through volume (centroid of
volume).
Magnitude of force FR=(FH2+FV2)1/2
Angle of force is a = tan-1(FV/FH)
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Chapter 3: Pressure and Fluid Statics
Buoyancy and Stability
Buoyancy is due to the fluid displaced by a
body. FB=fgV.
Archimedes principal : The buoyant
force acting on a body immersed in a fluid
is equal to the weight of the fluid displaced
by the body, and it acts upward through
the centroid of the displaced volume.
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Chapter 3: Pressure and Fluid Statics
Buoyancy and Stability
Buoyancy force FB is
equal only to the displaced
volume fgVdisplaced.
Three scenarios possible
1. body<fluid: Floating body
2. body=fluid: Neutrally buoyant
3. body>fluid: Sinking body
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Chapter 3: Pressure and Fluid Statics
Example: Galilean Thermometer
Galileo's thermometer is made of a sealed
glass cylinder containing a clear liquid.
Suspended in the liquid are a number of
weights, which are sealed glass containers
with colored liquid for an attractive effect.
As the liquid changes temperature it changes
density and the suspended weights rise and
fall to stay at the position where their density is
equal to that of the surrounding liquid.
If the weights differ by a very small amount and
ordered such that the least dense is at the top
and most dense at the bottom they can form a
temperature scale.
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Chapter 3: Pressure and Fluid Statics
Example: Floating Drydock
Auxiliary Floating Dry Dock Resolute
(AFDM-10) partially submerged
Submarine undergoing repair work on
board the AFDM-10
Using buoyancy, a submarine with a displacement of 6,000 tons can be lifted!
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Chapter 3: Pressure and Fluid Statics
Example: Submarine Buoyancy and Ballast
Submarines use both static and dynamic depth
control. Static control uses ballast tanks
between the pressure hull and the outer hull.
Dynamic control uses the bow and stern planes
to generate trim forces.
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Chapter 3: Pressure and Fluid Statics
Example: Submarine Buoyancy and Ballast
SSN 711 nose down after accident
which damaged fore ballast tanks
Normal surface trim
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Chapter 3: Pressure and Fluid Statics
Example: Submarine Buoyancy and Ballast
Damage to SSN 711
(USS San Francisco)
after running aground on
8 January 2005.
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Chapter 3: Pressure and Fluid Statics
Example: Submarine Buoyancy and Ballast
Ballast Control Panel: Important station for controlling depth of submarine
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Chapter 3: Pressure and Fluid Statics
Stability of Immersed Bodies
Rotational stability of immersed bodies depends upon
relative location of center of gravity G and center of
buoyancy B.
G below B: stable
G above B: unstable
G coincides with B: neutrally stable.
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Chapter 3: Pressure and Fluid Statics
Stability of Floating Bodies
If body is bottom heavy
(G lower than B), it is
always stable.
Floating bodies can be
stable when G is higher
than B due to shift in
location of center
buoyancy and creation of
restoring moment.
Measure of stability is the
metacentric height GM. If
GM>1, ship is stable.
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Chapter 3: Pressure and Fluid Statics
Rigid-Body Motion
There are special cases where a body of fluid can undergo rigidbody motion: linear acceleration, and rotation of a cylindrical
container.
In these cases, no shear is developed.
Newton's 2nd law of motion can be used to derive an equation of
motion for a fluid that acts as a rigid body
P   gk    a
In Cartesian coordinates: P    ax , P    a y , P     g  ax 
x
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y
z
Chapter 3: Pressure and Fluid Statics
Linear Acceleration
Container is moving on a straight path
ax  0, a y  az  0
P
P
P
  ax ,
 0,
  g
x
y
z
Total differential of P
dP    ax dx   gdz
Pressure difference between 2 points
P2  P1    ax  x2  x1    g  z2  z1 
Find the rise by selecting 2 points on
free surface P2 = P1
a
Dzs  zs 2  zs1   x  x2  x1 
g
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Chapter 3: Pressure and Fluid Statics
Rotation in a Cylindrical Container
Container is rotating about the z-axis
ar  r 2 , a  az  0
P
P
P
  r 2 ,
 0,
  g
r

z
Total differential of P
dP   r 2 dr   gdz
On an isobar, dP = 0
dzisobar r 2
2 2

 zisobar 
r  C1
dr
g
2g
Equation of the free
surface
2
zs  h0 
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
R

4g
2
 2r 2 
Chapter 3: Pressure and Fluid Statics
Examples of Archimedes Principle
The Golden Crown of Hiero II, King of
Syracuse
Archimedes, 287-212 B.C.
Hiero, 306-215 B.C.
Hiero learned of a rumor where
the goldsmith replaced some of
the gold in his crown with silver.
Hiero asked Archimedes to
determine whether the crown was
pure gold.
Archimedes had to develop a
nondestructive testing method
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Chapter 3: Pressure and Fluid Statics
The Golden Crown of Hiero II, King of
Syracuse
The weight of the crown and
nugget are the same in air: Wc =
cVc = Wn = nVn.
If the crown is pure gold, c=n
which means that the volumes
must be the same, Vc=Vn.
In water, the buoyancy force is
B=H2OV.
If the scale becomes unbalanced,
this implies that the Vc ≠ Vn,
which in turn means that the c ≠
n
Goldsmith was shown to be a
fraud!
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Chapter 3: Pressure and Fluid Statics
Hydrostatic Bodyfat Testing
What is the best way to
measure body fat?
Hydrostatic Bodyfat Testing
using Archimedes Principle!
Process
Measure body weight
W=bodyV
Get in tank, expel all air, and
measure apparent weight Wa
Buoyancy force B = W-Wa =
H2OV. This permits
computation of body volume.
Body density can be
computed body=W/V.
Body fat can be computed
from formulas.
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Chapter 3: Pressure and Fluid Statics
Hydrostatic Bodyfat Testing in Air?
Same methodology as
Hydrostatic testing in water.
What are the ramifications of
using air?
Density of air is 1/1000th of
water.
Temperature dependence of
air.
Measurement of small volumes.
Used by NCAA Wrestling (there
is a BodPod on PSU campus).
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Chapter 3: Pressure and Fluid Statics