Lecture 2 - Fluid Mechanics
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Transcript of Lecture 2 - Fluid Mechanics
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Fluid Static
Dr. Mohammed Zakria Salih Xoshnaw
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Ch Fluid Statics
• Fluid either at rest or moving in a manner that ther
is no relative motion between adjacent particles.
• No shearing stress in the fluid
• Only pressure (force that develop on the surfaces othe particles)
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Outline
1. Pressure at a Point
2. Basic Equations for the Pressure Field
3. Hydrostatic Condition
4. Standard Atmosphere
5. Manometer and Pressure Measurements
6. Barometer
7. Piezometer 8. Differential manometer
9. Example Problems
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Fluid Mechanics Overview
Gas Liquids Statics Dynamics
Air, He, Ar,
N2, etc.
Water, Oils,
Alcohols,
etc.
0 i F
Viscous/Inviscid
Steady/Unsteady
Compress
Incompres
0 i F
Laminar/
Turbulent
, Flows
Compressibility ViscosityVapor
Pressure
Density
PressureBuoyancy
Stability
Chapter 1: Introduction Chapter 2: Fluid StaticsFluid Dynami
Rest of Cour
Surface
Tension
Fluid Mechanics
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1. Pressure at a point N/m2 (Force/Area)
am F
Y: sin s x P z x p F s y y
ya
z y x
2
Z: z z z z a
z y x s x p y x p F
2
cos
z
a z y x
2
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sinz ; cos s s y
2)(:
2 :
z a p p z
ya p p y
z s z
y s y
What happen at a pt. ?0,, z y x
s z
s y
p p
p p
s z y p p p θ
is arbitrarily chosen
Pressure at a pt. in a fluid at rest, or in motion, is
independent of direction as long as there are no shearing
stresses present. (Pascal’s law)
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2. Basic equation for Pressure Field
Surface & body forces acting on small fluid element
pressure weight
How does the pressure in a fluid which there are no shearing stresses vary
from pt. to pt.?
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Surface forces:
z x y
y
p p z x
y
y
p p F y y
)
2()
2(:
z y x y p F y
Similarly, in z and x directions:
z y x
x
p F x
z y x
z
p F z
z y xk z
p j
y
pi
x
pk F j F i F F z y x s )(
k z
j y
i x
z y x p
)(
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Newton’s second law
W F am F s
z y x
3. Pressure variation in a fluid at rest
ak p
General equation of motion for a fluid in which there
are no shearing stresses.
2Eq.(
0
0
0 0
dz
dp
y
p
x
p
k pa
z y x z y x p F
2 3 1 I ibl
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2.3.1 Incompressible
hγ ) z z ( γ p pdz γdp
const g ργ
z
z
p
p
1221
1
2
2
1
Hydrostatic Distribution21
phγ p *see Fig. 2.2
21 p p
h
pressure head
Ex:
)133()62.4(
518or1.2310
32
21
m KN
ft lb
mmHg ft h p p psi
phγ p
Pressure in a homogeneous, incompressible fluid at rest: ~ reference level,
indep. of size or shape of the container.
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The required equality of pressures at equal elevations
Throughout a system.
1
1
2
2 F A
A F
Transmission of fluid pressure
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2.3.2 Compressible Fluid perfect gas:
RT ρ p
2
1
2
1)z(zconst., ln 21
1
2 p p
Z Z R g
T
dz
R
g
p
p
p
dp RT
gp g
dz
dp
Assume
0
12
12
210
)(
exp
conditionsisothermal ,
RT
z z g
p p
z z over T T
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Troposphere:
00357.0
0065.0
0@
ft
R
ratelaposem
K
z T z T T aa
a
a )
T
z β ( p p 1
2.4 Standard Atmosphere
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vapor atm phγ p
(Mercury barometer)
2.5 Measurement of Pressure
Parameter= measure atmospheric pressure
A
atom B
B A
p p
p p
A ph p
h
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Manometry
1. Piezometer Tube:
2. U-Tube Manometer:
3. Inclined-tube manometer
P h p
11h p A
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gasanotliquid,3.
reasonableish2.
1.
1 a
a
p p
p p
2. U-Tube Manometer:
1122
2211 0
hh phh p
A
A
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)γγ( h p p
p )hh( γhγhγ p p pk ) flowtheof ratevolumethe( Q
p p p Δ , p Δ ,u
B A
B A
B A
B A
122
2112211
Small difference in gas pIf pipes A & B contain a
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θ sinγ
p pl
θ sinl γ p p
B A
B A
2
2
22
Inclined – Tube manometer
2 7 Mechanical and Electronic Pressure Measuring Device
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2.7 Mechanical and Electronic Pressure Measuring Device
.Bourdon pressure gage (elastic structure)
Bourdon Tube
, p curved tube
straight
deformation
dial
.A zero reading on the gage indicates that the measured
pressure
. Pressure transducer-pressure V.S. time
Bourdon tube is connected to a linear variable
differential transformer(LVDT), Fig. 2.14
coil; voltage
.Aneroid barometer-measure atmospheric pressure
(absolute pressure)
This voltage is linear function of the pressure and could
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This voltage is linear function of the pressure, and could
be recorded on an oscillograph, or digitized for storage
or processing on computer.
Disadvantage-elastic sensing element
meas. pressure are static or only changing
slowly(quasistatic).
relatively mass of Bourdon tube
<diaphragm>
*strain-gage pressure transducer *
Fig. 2.15 (arterial blood pressure)
piezo-electric crystal. (Refs. 3, 4, 5 )
1Hz
Application Examples
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Application Examples
Feeder Gates for Canal
Gate Valves fo
Spillway Contr
Applications (cont.)
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Applications (cont.)
Spillway Drum Gates
hollow inside, use
buoyancy to control
position of the gate.
2 8 Hydrostatic Force on a Plane Surface
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2.8 Hydrostatic Force on a Plane Surface
pA F R
Storage tanks, ships
Fig. 2.16 Pressure and resultants hydrostatic force
developed on the bottom of an open tank.
. For fluid at rest we know that the force must be
perpendicular to the surface, since there are no shearing
stress present.
H d t ti F I li d Pl S f
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Hydrostatic Force on an Inclined Plane Surface
h y
dF
sin
Integrate over the entire surf
sin
Define centroid of the area y
1, so that
sin
In order to find equilavent sy
need to make sure that
R
C
R C C
dF PdA hdA ghdA
gy dA
F dF g ydA
y ydA A
F gAy gAh
the m
of the resultant force must eq
the moment of the distribute
x
y
x
Assume atmoshperic condition on the other
side of the surface
Free surface
Hydrostaic forces
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y
FR
O
O
R
2
2
xx
C
Taking Mmoment about the x-axis: y'F
' sin sin s
Recognize that y (area mome
ITherefore, y'=
Ay
Also, from parallel axis theorem, we can
C
A
xx A
y g y A gy dA g
dA I
2 ˆ ˆxxˆ ˆxx xx C
ˆ ˆxy xy
C
C
moment of inertia about the centroid of
found in table)
II = I , therefore, y'=y
I ISimilarly, x'= x
Ay
C
C
C
Ay Ay
Ay
z
y
y
x
FR
z
y’
yc
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Example
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hinge
The square flood gate (2m by 2m) is hinged along its bott
shown. Determine the moment at the hinge in order to
the gate steady.
R
3
ˆ ˆ
C
First, find the resultant force:
F (1000)(9.8)(1)(2 2) 39200(Then, determine the point of action:
1 (2)(2) 1 412y'=y (1) 1 ((2 2)(1) 3 3
As expected, it falls at a depth 2/3 of the tot
C
xx
C
gh A N
I m
Ay
O
al
The holding moment (M) on the hinge O will b
4M (2 ) 0,
3
18479( . )
R M F
M N m
y
x
O
2my’
Example (cont.)
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2my’
If the square gate is replaced by a circular
shaped gate as shown, recalculate the ho
moment.
x
y
2
R
4
ˆ ˆ
C 2
O
Again, find the resultant force first:
F (1000)(9.8)(1) (1) 30772
Next, the line of action:
11 54y'=y 1 1 ( )
(1) 4 4
The holding moment:
5 3M (2 ) 0
4 4
23079( . )
C
x x
C
R R
gh A
R I m
Ay R
M F M F
M N m
Example (cont.)45°
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2m
y
y’
45
If the square gate is placed at an angle of 45° as
recalculate the holding moment again. Note: th
has been redefined to follow the gate for conve
R
ˆˆ
C
First, calculate the resultant force:
F (1000)(9.8)(1)(2 2 2 2) 78400(
Note: the h stays the same and is independent
of the incline angle, however, the gate area increases
1 (22 2 12y'=y
2
C
xx
C
gh A N
I Ay
3
R
2)(2 2)
2 2(2 2 2 2)2
2 4 2' 2
3 3
4 2The holding moment: M=F (2 2 ) 73916(3
y
An interesting observation
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When the gas tank is low, the low fuel light will lit to warn the driver. Have you n
that the light will not always stay on for a period of time. It turns off when either
accelerate (decelerate) or climb (descend) on a sloped road. Can you explain this
phenomenon by using the principle of fluid statics.
Fuel level transducer
Accelerating (climbing) Decelerating (descending)
Hydrostatic balance can be applied to a small fluid element as shown
d
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( ) , , integrate from fluid elemen
the free surface ( )
dp pA p dp A mg Agdy g
dy
p h
p gh
p
p+dp h
Free surface, p=p
Example: If a container of fluid is accelerating
with an acceleration of ax to the right as shown belothe free surface of the fluid will incline with an ang
shown.
ax
p p+dp
1
( ) ,
tan( ) , tan
x x
x
x
dp pA p dp A ma Adxa dx
dpa dy g
dpdx g a
a a
a
dx
dy
x
y
2.9 Pressure Prism
the pressure varies linearly with depth
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the pressure varies linearly with depth.
Ah
bhh
e prismof pressur volume F
Ah
A P F
R
Ave R
2))((
2
1
)
2
(
No matter what the shape of the pressure prism is, the resultaforce is still equal in magnitude to the volume of the pressure
Prism, and it passes through the centroid of the volume.
First, draw the pressure prism out. dz
dp
0 p z p
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Hydrostatic Force on a Curved Surface• General theory of plane surfaces does not apply to curved surfaces
• Many surfaces in dams, pumps, pipes or tanks are curved
• No simple formulas by integration similar to those for plane surfaces
• A new method must be used
Isolated Volume
Bounded by AB an AC and
BC
Then we mark a F.B.D. for the volume:
F1 and F2 is the hydrostatic force on each
planar face
FH and FV is the component of the result
force on the curved surface.
W is the weight of the fluid volume.
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Hydrostatic Force on a Curved Surface
Now, balancing the forces for the Equilibrium condition:
Horizontal Force:
Vertical Force:
Resultant Force:
The location of the Resultant Force is through O by sum of Moments:
H H
V V c
x F x F
x F Wx x F
22
11Y-axis:
X-axis:
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Buoyancy: Archimedes’Principle
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c p e
Archimedes (287-212 BC) Story
•Buoyant force is a force that results from a floating or submerged body in a fluid.
•The force results from different pressures on the top and bottom of the object
•The pressure forces acting from below are greater than those on top
Now, treat an arbitrary submerged object as a planar surface:
Arbitrary Shape
V
Forces on the Fl
Archimedes’ Principle states that the buoyant
force has a magnitude equal to the weight of the
fluid displaced by the body and is directed
vertically upward.
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Buoyancy and Flotation: Archimedes’ Principle
Balancing the Forces of the F.B.D. in the vertical Direction:
V AhhW 12
W is the weight of the shaded area
F1 and F2 are the forces on the plane surfaces
FB is the bouyant force the body exerts on the fluid
Then, substituting:
Simplifying,
The force of the fluid on the body is opposite, or vertica
upward and is known as the Buoyant Force.
The force is equal to the weight of the fluid it displaces.
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Buoyancy and Flotation: Archimedes’ Principle
Sum the Moments about the z-axis:
Find where the Buoyant Force Acts by Summing Moments:
We find that the buoyant forces acts through
the centroid of the displaced volume.
The location is known as the center of buoyancy.
VT is the total volume of the parallelpiped
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Buoyancy and Flotation: Archimedes’ Principle
We can apply the same principles to floating objects:
If the fluid acting on the upper surfaces has very small specific weight (air), the centroid
is simply that of the displaced volume, and the buoyant force is as before.
If the specific weight varies in the fluid the buoyant force does not pass through the
centroid of the displaced volume, but through the center of gravity of the displaced
volume.
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Stability: Submerged Object
Stable Equilibrium: if when displaced returns to equilibrium position.
Unstable Equilibrium: if when displaced it returns to a new equilibrium position
Stable Equilibrium: Unstable Equilibrium:
C > CG, “Higher” C < CG, “Lower”
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Buoyancy and Stability: Floating Object
Slightly more complicated as the location of the center buoyancy can change:
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Pressure Variation, Rigid Body Motion: Linear Motion
Governing Equation with no Shear (Rigid Body Motion):
The equation in all three directions are the following:
Consider, the case of an open container of liquid with a constant acceleration:
Estimating the pressure between two closely spaced points apart some dy, dz:
Substituting the partials
Along a line of constant pressure, dp = 0: Inclined free
surface for ay≠ 0
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Pressure Variation, Rigid Body Motion: Linear Motion
Now consider the case where ay = 0, and az ≠ 0:
0
x
pRecall, already:
z
a g z
p
y
p
0Then,
So, Non-Hydrostatic
Pressure will vary linearly with depth, but variation is the combination of gravity and externally
developed acceleration.
A tank of water moving upward in an elevator will have slightly greater pressure at the bottom.
If a liquid is in free-fall az = -g, and all pressure gradients are zero—surface tension is all that keep
the blob together.
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Pressure Variation, Rigid Body Motion: Rotatio
Governing Equation with no Shear (Rigid Body Motion):
Write terms in cylindrical coordinates for convenience:
Pressure Gradient:
Accceleration Vector:
Motion in a Rotating Tank:
P V i ti Ri id B d M ti R t ti
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Pressure Variation, Rigid Body Motion: Rotatio
The equation in all three directions are the following:
Estimating the pressure between two closely spaced points apart some dr, dz:
Substituting the partials
Along a line of constant pressure, dp = 0:
Equation of constant pressure surfaces:
The surfaces of constant pressure are parabolic
P V i ti Ri id B d M ti R t ti
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Pressure Variation, Rigid Body Motion: Rotatio
Now, integrate to obtain the Pressure Variation:
Pressure varies hydrostaticly in the vertical, and increases radialy
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