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1
Chapter 3: Fluid Statics
By
Dr Ali Jawarneh
2
Outline
We will discuss:
• The definition of pressure.
• Pascal’s principle.
• The concepts of absolute, gage and vacuum pressures.
• Pressure variation with elevation for uniform density fluids.
3
Outline
• We will become familiar with different pressure measurement techniques and devices:
– Simple and differential manometers
– Bourdon-tube gages
– Pressure Transducers
4
Outline
• become familiar with hydrostatic forces on plane surfaces.
• Calculate the magnitude of the resultant hydrostatic force.
• Locate the line of action of the resultant hydrostatic force.
• Analyse hydrostatic forces on curved surfaces.
• Discuss the principle of buoyancy.
5
3.1: Pressure, p
• For fluids at rest, only normal forces exist, which are basically known as pressure forces.
• At every point in a static fluid, a pressure defined as follows exists:
• Pressure is a scalar quantity. It is unit: N/m2
(pa), Ibf/in2 (psi), or Ibf/ft2 (psf)
• Pressure at a point in a static fluid acts with the same magnitude in all directions.
A
Fp
6
Pressure is defined as a normal force exerted by a fluid per unit area. We speak of pressure only when
we deal with a gas or a liquid. The counterpart of
pressure in solids is normal stress. Since pressure is
defined as force per unit area, it has the unit of
newtons per square meter (N/m2), which is called a
pascal (Pa).
7
Pressure
Considering an element of fluid in
equilibrium:
8
Pressure
• The summation of forces in the x-direction:
• The summation of forces in the z-direction:
0)sin(sin)( lyplyp xn
0sincos2
1)cos(cos)( ylllyplyp zn
xn pp
zn pp
9
Pressure Transmission• Pascal’s principle states that a
pressure change produced at one point in the system will be transmitted throughout the entire system.
• A consequence of the pressure in a fluid remaining constant in the horizontal direction is that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is called Pascal’s law, after Blaise Pascal(1623–1662).
• This principle is utilised in the design and development of hydraulic controls used in a wide range of applications.
Pascal’s law states
that increase in
pressure on
the surface of a
confined fluid is
transmitted
undiminished
throughout the
confining vessel or
system
the pressure at
a point in a fluid at
rest, or in motion, is
independent of
direction as long as
there are no
shearing stresses
present. This important
result is known as
Pascal’s law named in
honor of
Blaise Pascal
10
Pressure Transmission
11
12
Absolute, Gage, and Vacuum Pressure
• Pabs=pgage+ patm
• Pressure values measured with reference
to the atmospheric pressure are referred
to as gage pressures.
• Most pressure measurement devices
measure gage pressure, such as the
Bourdon-tube gage.
13
Absolute and Gage Pressure
• Gage pressures can take negative values.
• Negative gage pressures are also referred
to as vacuum pressures.
• If the atmospheric pressure is 101.3 kPa
which is measured at sea level at T=23 oC:
• 50 kPa gage = 151.3 kPa absolute.
• - 50 kPa gage = 51.3 kPa absolute.
14
3.2: Pressure Variation with
Elevation
• The general equation
for pressure variation
in the fluid element
shown:
dl
dz
dl
dp
dz
dpThe pressure changes
inversely with elevation. If
one travels upward in the
fluid, the pressure
decreases
15
Pressure Variation with Elevation
• Pressure variation occurs only along a
vertical path through the fluid.
• No pressure variation occurs along a
horizontal path through the fluid.
• Pressure changes inversely with elevation;
i.e.: going down increases the pressure
and going up decreases the pressure.
16
Pressure Variation with Elevation
• For a fluid with uniform density, integrating the
basic equation results the following:
• This sum is called the piezometric pressure.
• Another form is:
• This sum is called the piezometric head.
constant zp
constant zp
17
Pressure Variation with Elevation
• Therefore, the pressure and elevation at
one point can be related to the pressure
and elevation at another point according to
the following:
2211 zpzp
18
- Pressure Variation for
Compressible Fluids
• For Ideal gas: p=RT or =p/RT
• Multiply by g: g=pg/RT, then
=pg/RT
• =fn (p, T)
• dp/dz = - = fn (p, T)
19
Figure 3.5
Temperature
variation with
altitude for the
U.S. standard
atmosphere in
July (2).
Example: Find the pressure at the tank bottom.
Solution: =9790 N.m3 (Table A.5)
Another practical and fast way:
or
20
22
11 z
γ
pz
γ
p
)gauge(kpa.p)(γ
p5819200 2
2
kpa.pp))((
pγhp
5819297900 22
21
1212 pγhppγhp
Example: Find the pressure at the tank bottom
Solution:
Second way:
21
kpa.p).()(.
p
zγ
pz
γ
p
oiloil
924350981080
00 22
22
11
kpa.p)(p
zγ
pz
γ
p
waterwater
5042329790
09790
39243
3
33
22
kpa.pp)().)((.
phγhγp wateroil
5042329790509810800 33
3211
Example: Find the pipe pressure if h1=1.2 m,
h2=1 m, and h3=3 m
Solution:
22
21231 phγhγhγp waterHgoil
002198101981061339810901 ....p
kpa.pp pipe 157951
Example: Given F1= 200 N, Find the F2. Neglect
the weights of the pistons
Solution:
23
21 pγhp
kpa.
.A
Fp 23159
0404
200
21
11
kpa.).(p 558142298108501592302
kN..APF 1191104
1425582
222
24
Manometry• Pressure is one of the most important
variables in fluid mechanics.
• Numerous instruments and devices have
been developed for its measurement.
• One of the simplest and most common
measurement techniques is based on
hydrostatics or manometry.
• The principle is basically to utilise the
change in pressure with elevation to
determine the value of the pressure.
3.3: Pressure Measurements
25
a: Piezometer
• A simple manometer,
or a piezometer can
be attached to a pipe
and the height of the
liquid’s column is an
indicator of the
pressure in the pipe:
hp
26
b: U-tube Manometers
• A simple manometer
becomes impractical
in the case of gases
or in the case of
high pressures.
• For such cases, a
U-tube filled with
another liquid can
be used. 4 m fp h L
27
c: Differential Manometers
• A differential
manometer can also
be used to measure
the pressure
difference between
two points in a pipe,
according to:
hp fm )(
Example: Find the pipe pressure.
Solution:
28
015131 HgairoilA γ.γ.γp
019810513031981090 ...pA
kpa.pA 912143
0.0
• Example: Find the pipe pressure
• Solution:
29
0102050101033 L.pB
m.L.
L.sin 20
50
10
5
3
kPapB 1
30
- Bourdon-Tube Gage
• The Bourdon-tube
gage is a very
common device that
utilises the deflection
in a spring tube to
measure the
pressure.
• Bourdon gages need
to be periodically
calibrated.
31
- Pressure Transducers
• Similarly, pressure transducers utilise the
deflection of a diaphragm to produce an electrical
signal that can be related to the pressure.
• Pressure transducers are widely used with
applications requiring extensive data acquisition
and processing.
32
3.4:Hydrostatic Forces on Plane
Surfaces
• For a horizontal surface in a fluid, the
pressure is uniformly distributed and
the resulting force is simply equal to
the product of the pressure and the
area.
• For a vertical or an inclined surface the
pressure is not uniformly distributed
and thus some derivation is required to
determine the resulting force.
33
34
35
Hydrostatic Forces on Plane Surfaces
36
Hydrostatic Forces on Plane
Surfaces
• The pressure on the differential area is
equal to:
• Consequently, the differential force on the
differential area is equal to:
• And the total force on the area is equal to:
sinyp
dAydF sin
dAyFA
sin
37
Hydrostatic Forces on Plane
Surfaces
• Rewriting,
• Integrating:
• In short, the magnitude of the resultant
hydrostatic force on a plane surface is the
product of the pressure at the centroid of
the surface and the area of that surface.
dAyFA
sin
ApAyF sin
38
Location of Resultant Hydro. Force
• The location of the line of action of the
resultant hydrostatic force lies at a point
called the center of pressure.
• Writing the moment equation:
• Or:
dApydFyFycp
dAyFycp sin2
39
Location of Resultant Hydro. Force
• Re-writing:
• Substituting with the area moment of
Inertia ( I ):
• Applying the parallel-axis theorem:
dAyFycp
2sin
ocp IFy sin
)(sin 2 AyIFycp
40
Location of Resultant Hydro. Force• Substituting the value of F:
• Reducing:
• where is the 2nd moment of area (area moment of inertia): see Fig. A.1
• This equation implies that the center of pressure is always below the centroid, but comes closer to the centroid as the depth increases.
Ay
Iyycp
I
ApAyF sin
41
Fig. A.1
Centroids
and
moments
of inertia
of plane
areas
42
• Example: The gate shown is rectangular and has
dimensions 6 m x 4 m. What is the reaction at point
A? Neglect the weight of the gate.
• Solution:
43
m.LL
cos 46433
30
44
sinyP
APF
m..Ly 4646464333
pa.sin.P 26549166046469810
N][.F 1317990462654916
Ay
Iyycp 72
12
64
12
33
hb
I
m.yor
.][.Ay
Iyy
cp
cp
9286
4640644646
72
kNR
R).(
R).(F
.M stop
557
6464031317990
646403
00
45
3.5: Hydrostatic Forces on Curved
Surfaces
• The hydrostatic forces
on curved forces can
be found by
equilibrium concepts
on systems
comprised of the fluid
in contact with the
curved surface.
46
Forces on Curved Surfaces• For the free body
diagram:
ApFF AChorizontal
CBvertical FWF The resultant hydrostatic force acting on the
curved solid surface is then equal and opposite
to the force acting on the curved liquid surface
(Newton’s third law).
Horizontal force component on curved
surface:
Vertical force component on curved
surface:
FR=F
47
• Example: Find the vertical and horizontal
forces on the given gate
(L=1 m).
• Solution:
48
tV cV
49
N)()L(hAPF 1962011298101111
N.))L(
(
)VV(VVWWW ctctct
15210914
2
41119810
2
kN..WFFvertical 5117152109196201
2222 AsinyAPF m.LL
y 5112
1
2
190sinsin 2
2 111 mA
kN..F 71514115198102
2Ay
Iyycp 121
12
11
12
33
/bh
I
m.ycp 5551
kN.FFhorizontal 715142
50
3.6: Buoyancy
• Considering the
submerged body in
the figure:
• The force acting on
the lower surface of
the body is equal to
the weight of the
liquid needed to fill
the volume above this
surface.
51
Buoyancy
• The force acting on
the upper surface of
the body is equal to
the weight of the
liquid above this
surface.
• Summing the forces:
B up down fluid bodyF F F V
A body completely submerged in a liquid
52
A body partially submerged in a liquid
B fluid DF V
53
Buoyancy
• Consequently, the buoyant force on a
submerged body is equal to the weight of
the liquid that would be needed to occupy
the volume of the body.
• Similarly, for a floating body, the buoyant
force is equal to the weight of the liquid
that would be needed to occupy the
submerged volume of the body.
54
Buoyancy
• This principle of buoyancy is basically
what is commonly known as the
Archimedes’ principle: “For an object
partially or completely submerged in a
fluid, there is a net upward force equal
to the weight of the displaced fluid.”
• The buoyancy force is an upward force
that passes through the centroid of the
displaced volume.
55
Hydrometry• Device [glass bulb] to measure the or S of a
liquid based on the principle of buoyancy
• Example: Find the tension in the given figure.
• Solution:
56
kN.T
kN.dπ
γVγF
WFT
B
B
359
85176
3
• Example: find the specific gravity of the given
unknown fluid, the hydrometry weight is 0.015 N.
• Solution:
57
3
2232
9202
0150100
361010101
0150
m/N
.).
}{.}{(VF
N.WF
oil
oiloilB
B
93809810
9202.S
water
oil
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