Civil Engineering Hydraulics
Mechanics of Fluids
Pressure and Fluid Statics
The fastest healing part of the body is the tongue.
Common Units
 
2
In order to be able to discuss and
analyze fluid problems we need to be
able to understand some
fundamental terms commonly used
Pressure
1
Common Units
 
 
The most used term in hydraulics and
fluid mechanics is probably pressure
Pressure is defined as the normal
force exerted by a fluid per unit of
area
 
3
The important part of that definition is
the normal (perpendicular) to the unit
of area
Pressure
Common Units
 
 
4
The Pascal is a very small unit of
pressure so it is most often
encountered with a prefix to allow the
numerical values to be easy to
display
Common prefixes are the Kilopascal
(kPa=103Pa), the Megapascal
(MPa=106Pa), and sometimes the
Gigapascal (GPa=109Pa)
Pressure
2
Common Units
  A
bar is defined as 105 Pa so a
millibar (mbar) is defined as 10-3 bar
so the millibar is 102 Pa
The word bar finds its origin in the Greek word báros, meaning weight.
5
Pressure
Common Units
  Standard
atmospheric pressure or
"the standard atmosphere" (1 atm) is
defined as 101.325 kilopascals (kPa).
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Pressure
3
Common Units
  This
"standard pressure" is a purely
arbitrary representative value for
pressure at sea level, and real
atmospheric pressures vary from
place to place and moment to moment
everywhere in the world.
7
Pressure
Common Units
  Pressure
is usually given in reference
to some datum
  Absolute
pressure is given in
reference to a system with no
pressure whatsoever (a vacuum)
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Pressure
4
Common Units
  Pressure
is usually given in reference
to some datum
  Gage
pressure, the more commonly
used pressure, is the difference
between local atmospheric pressure
and the absolute pressure of the
system being measured
9
Pressure
Common Units
  If
the gage pressure of the system
being measured is less than local
atmospheric pressure, the pressure
may be termed a vacuum pressure
  This does not imply that it has no
pressure, just that the pressure is less
then local atmospheric
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Pressure
5
Common Units
11
Pressure
Common Units
  To
be the most precise, when giving
pressure, you should state if it is an
absolute or a gage pressure
  There is a difference
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Pressure
6
Pressure at a point
  What
can appear as non-intuitive is
that the pressure at any point in a fluid
is the same in all directions
  It would seem to make more sense if
the pressure was greater on the top of
the point than on the bottom and not
at all on the sides but remember we
are talking about a point.
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Pressure
Pressure at a point
  The
statics (yes I said statics) of the
situation can be used to define just
what is happening
14
Pressure
7
Pressure at a point
  We
can start with a fluid at rest and
therefore at static equilibrium
  In that fluid, we can pick a segment
with a triangular cross section and a
unit depth, into the fluid or into the
page, with a thickness of 1 unit
15
Pressure
Pressure at a point
The shape was a
thickness of 1 in the ydirection.
Remember the pressure
is defined as the force
exerted normal to the
surface.
Since we have the FBD
of this wedge, we are
showing forces action on
the wedge.
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Pressure
8
Pressure at a point
We need to include one
more force on this FBD
and that is the weight of
this wedge of fluid.
17
W
Pressure
Pressure at a point
We can start by writing
our expressions for force
balances along each axis.
∑F
∑F
∑F
∑F
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z
=0
z
= −W + P2 ( Δx )(1) − P3 ( l ) (1)( cos θ )
x
=0
x
= P1 ( Δz )(1) − P3 ( l ) (1)( sin θ )
W
Pressure
9
Pressure at a point
The weight of the wedge
can be found by solving
for the volume of the
wedge and then using the
mass density and gravity
to find the weight.
W = ρVg
W = ρg
{(
W
}
1
Δx )( Δz ) 1
2
ρ is the mass density of the
fluid and g is the universal
gravitational constant.
19
Pressure
Pressure at a point
Substituting for W in our
initial expressions we
have
∑F
=0
∑F
= −ρ g
z
z
∑ Fx = 0
∑F
x
{(
W
}
1
Δx )( Δz ) 1 + P2 ( Δx )(1) − P3 ( l ) (1)( cos θ )
2
= P1 ( Δz )(1) − P3 ( l ) (1)( sin θ )
Since the left side of all the
expressions are equal to 0, we
can divide both sides by 1 and
get rid of the 1 s in both
expressions.
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Pressure
10
Pressure at a point
Now we have
{
}
1
( Δx )( Δz ) + P2 ( Δx ) - P3 ( l ) ( cos θ )
2
0 = P1 ( Δz ) − P3 ( l ) ( sin θ )
0 = -ρ g
W
In the second expression, the
term l sin θ is equal to Δz so
we have
{(
}
1
Δx )( Δz ) + P2 ( Δx ) - P3 ( l ) ( cos θ )
2
0 = P1 ( Δz ) − P3 ( Δz )
0 = -ρ g
∴ P1 = P3
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Pressure
Pressure at a point
{(
}
1
Δx )( Δz ) + P2 ( Δx ) - P3 ( l ) ( cos θ )
2
0 = P1 ( Δz ) − P3 ( Δz )
0 = -ρ g
W
∴ P1 = P3
In the first expression, the term
l cos θ is equal to Δx so we
have
"1
%
0 = - ! g # ( !x ) ( !z )& + P2 ( !x ) - P3 ( !x )
$2
'
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Pressure
11
Pressure at a point
0 = -ρ g
{(
}
1
Δx )( Δz ) + P2 ( Δx ) - P3 ( Δx )
2
W
P1 = P3
Dividing all the terms by Δx we
have
"1
%
0 = - ! g # ( !z )& + P2 - P3
$2
'
23
Pressure
Pressure at a point
Remember that we are talking
about pressure at a point in the
fluid.
W
We can reduce our wedge to a
point by allowing the Δx and Δz
dimensions to approach 0.
As Δz approaches 0, the
weight term also approaches
0.
"1
%
0 = - ! g # ( !z )& + P2 - P3
$2
'
0 = P2 - P3
This reduces our expression to
P1 = P3
( P2 = P3 = P1
24
Pressure
12
Pressure at a point
Notice that the value for θ was
not critical for our derivation
and the density of the fluid did
not enter into our calculation at
the end.
W
At any point in a
fluid, the magnitude
of the pressure is the
same in all
0 = P2 - P3
directions.
P1 = P3
∴ P2 = P3 = P1
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Pressure
Pressure at a point
Note that this statement is
made about any point, it not
made about any two points
having the same pressure.
That is a different problem and
not covered by the
assumptions that we just
made.
W
0 = P2 - P3
P1 = P3
∴ P2 = P3 = P1
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Pressure
13
Pressure Variation with
Depth
  If
we consider a fluid with a constant
density over a depth
  We
can consider most gasses as
having a constant density with depth
over reasonable depths and most
liquids also
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Pressure
Pressure Variation with
Depth
  We
can start by remembering that
pressure is the force exerted per unit
of area.
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Pressure
14
Pressure Variation with
Depth
We can start with a cylinder of
diameter d and find the pressure
at some depth h in the cylinder.
d
The fluid has a mass density of
ρ and the pressure at the top of
the cylinder is patm (atmospheric
pressure).
h
The pressure at the depth h is
going to be the sum of the patm
and the pressure exerted by the
weight of the fluid about the
depth h.
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Pressure
Pressure Variation with
Depth
The weight of the fluid can be
determined by taking the volume
of the fluid and multiplying that
by the product of the mass
density ρ and the gravitational
constant g.
d
h
V=
π d2
4
h
F = Vρ g =
30
π d2
4
hρ g
Pressure
15
Pressure Variation with
Depth
The area that the force is acting
normal to is the cross sectional
area of the cylinder.
d
V=
h
π d2
4
h
F = Vρ g =
A=
31
π d2
4
hρ g
π d2
4
Pressure
Pressure Variation with
Depth
So the added pressure of the
fluid is the force exerted by the
fluid divided by the area it is
acting over.
d
F = V!g =
h
A=
The development in the text uses z for
the depth of the fluid rather than h.
32
" d2
h! g
4
" d2
4
! d2
h! g
F
4
p= =
= h! g
" d2
A
4
Pressure
16
Pressure Variation with
Depth
We can check for dimensional
consistency.
d
p = h! g
mass ! length
mass length
time 2
= length !
!
2
length
length 3 time 2
h
33
Pressure
Pressure Variation with
Depth
The pressure is not dependent
on the area.
d
If we assume that density
remains constant with depth and
that the gravitational constant
also remains constant with
depth the pressure becomes a
linear function of depth.
h
As you go deeper in a fluid, the
pressure increases linearly.
p = h! g
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Pressure
17
Pressure Variation with
Depth
p = h! g
In your text, the symbol z is
used for depth because h is
often also used as a variable in
thermodynamics.
We may use both depending on
the problem. The important thing
is to take the variable as
measuring the depth below the
surface in the fluid.
35
Pressure
Pressure Variation with
Depth
p = h! g
  In
the previous slide, we made two
assumptions
  The
first was that we were starting from 0
gage pressure at the top of the fluid which
was the reference for our depth
measurement
  The second was that the density of the fluid
did not change with depth
36
Pressure
18
Pressure Variation with
Depth
  A
more formal expression for change of
pressure with changing depth would be
z2
h2
z2
h2
z1
h1
z1
h1
p2 ! p1 = " ! g dz = " ! g dh = " ! dz = " ! dh
37
Pressure
Pressure Variation with
Depth
  This
would allow for a fluid which might
change density as a function of depth
z2
h2
z2
h2
z1
h1
z1
h1
p2 ! p1 = " ! g dz = " ! g dh = " ! dz = " ! dh
38
Pressure
19
Problem
39
Pressure
pressure increases downward in a
given fluid and decreases upward
Pressure
pressure increases downward in a
given fluid and decreases upward
Problem
40
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Reading
 
41
Sections 2-2 and 2-3
Pressure
Homework
42
Pressure
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Homework
43
Pressure
Homework
44
Pressure
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