fluid dynamics1
TRANSCRIPT
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FLUID MECHANICS FOR CIVIL
ENGINEERING
CHAPTER-5
FLUID-DYNAMICS
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FLUID DYNAMICS
Fluid in
motionThat Involves Forces ofACTIONS &
REACTIONS
Forces which cause acceleration
and force which resist acceleration
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Governed by:
1. Eulers Equation (momentum principle)
2. Bernoullis Equation (Energy Principle)
ds
dp
ds
dyg
ds
dp
ds
duu
1
tconsgyPv
tan2
2
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gdAdsW
dy
dy
dx
W
dA = cross section area of fluid
elementdS = length of fluid element
Centroid of the downstream
face lies at a level dy higher
than the centroid of the
upstream face
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NORMAL FORCE DUE TO PRESSURE
P= Pressure intensity at the upstream and (P + dP) is the pressure intensity downstream
Net pressure force acting on the element in the direction of motion is given by
dPdAdAdPPPdA )(
TANGENTIAL FORCE DUE TO VISCOUS FORCE
If the fluid element has a mean perimeter dp then shear force on the element is
dpdsdFs Is the frictional surface force per unit area acting onthe walls of the stream tube
BODY FORCE SUCH AS GRAVITY ACTING IN THE DIRECTION OF GRAVITATIONAL
FIELD gdAdygdAds sinTHE RESULTANT FORCE IN THE DIRECTION OF MOTION MUST EQUAL THE
PRODUCT OF MASS & ACCELERATION IN THAT DIRECTION
sdAdsadpdsgdAdydPdA
[1]
[2]
[3]
[4]
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Velocity of an elementary fluid particle along a streamline is a function of position
and time, u = f (s,t)
t
u
s
uua
t
u
dt
ds
s
u
dt
du
dttuds
sudu
s
In a steady flow
0
t
uas
ds
duuas
As partial derivative becomes
total derivative
[5]
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Substituting eqn [5] in [4] gives
dAududpdsgdAdydPdA
Dividing throughout by a fluid mass and rearrangingdAds
dA
dp
ds
dyg
ds
dP
ds
duu
1
ds
duu
Is a measure of convective acceleration experienced by the fluid as it
moves from a region of one velocity to another region of different velocity. It
represents a change in K.E
i.
ii. Represents the force per unit mass caused by pressure distributionds
dP
1
iii. Represents the force per unit mass resulting from gravitational pullds
dyg
iv. The term is the force per unit mass caused by frictiondA
dp
[6]
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For ideal fluid and the equation reduces to :0 01
ds
dyg
ds
dP
ds
duu
Hence 0 gdydpudu
Integrating Eulers equation, we gettconsgy
Putan
2
2
Dividing throughout by g We get tconsyP
g
utan
2
2
[7]
g
u
2
2
Velocity head
PPressure head of static head
y Elevation head, position head, potential head or geodetic head
HyP
g
u
2
2
Where H is total or hydrodynamic head
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Steady flow The first limitation on the Bernoulli equation is that it isapplicable to steady flow.
Frictionless flow Every flow involves some friction, no matter how small,andfrictional effects may or may not be negligible.
No shaft work The Bernoulli equation was derived from a force balance ona particle moving along a streamline.
Incompressible flow One of the assumptions used in the derivation of theBernoulli equation is that = constant and thus the flow is incompressible.
No heat transfer The density of a gas is inversely proportional totemperature, and thus the Bernoulli equation should not be used for flowsections that involve significant temperature change such as heating orcooling sections.
Strictly speaking, the Bernoulli equation is applicable along a streamline,and the value of the constant C, in general, is different for differentstreamlines. But when a region of the flow is irrotational, and thus there isno vorticityin the flow field, the value of the constant Cremains the samefor all streamlines, and, therefore, the Bernoulli equation becomesapplicable across streamlines as well.
Limitations on the Use of the Bernoulli
Equation
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EXAMPLE 22Spraying Water into the Air
Water is flowing from a hose attached to a water main at 400 kPa gage (Fig.
below). A child places his thumb to cover most of the hose outlet, causing a
thin jet of high-speed water to emerge. If the hose is held upward, what is the
maximum height that the jet could achieve?
This problem involves the conversion of flow, kinetic, and potential energies
to each other without involving any pumps, turbines, and wasteful
components with large frictional losses, and thus it is suitable for the use of
the Bernoulli equation. The water height will be maximum under the stated
assumptions. The velocity inside the hose is relatively low (V1 = 0) and we
take the hose outlet as the reference level (z1= 0). At the top of the water
trajectory V2 = 0, and atmospheric pressure pertains. Then the Bernoulli
equation simplifies to
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EXAMPLE 2-3Water Discharge from a Large Tank
A large tank open to the atmosphere is filled with water to a height of 5 m from
the outlet tap (Fig. below). A tap near the bottom of the tank is now opened, and
water flows out from the smooth and rounded outlet. Determine the water velocity
at the outlet.This problem involves the conversion of flow, kinetic, and potential energies to
each other without involving any pumps, turbines, and wasteful components
with large frictional losses, and thus it is suitable for the use of the Bernoulli
equation. We take point 1 to be at the free surface of water so that P1= Patm
(open to the atmosphere), V1 = 0 (the tank is large relative to the outlet), and z1=
5 m and z2 = 0 (we take the reference level at the center of the outlet).
Also, P2 = Patm (water discharges into the atmosphere).
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Then the Bernoulli equation simplifies to
Solving forV2 and substituting
The relation is called the Toricelli equation.
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
Graphical interpretations of the energy along a pipeline may be obtained
through the EGL and HGL:
EGL
p V
g z
2
2
HGLp
z
EGL and HGL may be obtained via a pitot tube and a piezometer tube,
respectively
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
EGLp V
gz
2
2HGL
pz
h hL f - head loss, say,
due to friction
piezometer
tube
V
g
2
2
2
z2
z1
pitot tube
( )z 0
EGL
HGL hL
p2 /
Datum
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
Large V2/2gbecause
smaller pipe here
Steeper EGL and HGL
because greaterhLper length of pipe
Head loss at
submerged discharge
EGL
HGL
p/
z
z 0
HGLp
z
EGLp V
gz
2
2h hL f
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Positive
Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
If then and cavitation may be possibleP
0HGL z
HGL
EGLp/
Positive
V
g
2
2
p
Negativep
z
z 0
EGLp V
gz
2
2
HGLp
z
h hL f
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
Helpful hints when drawing HGL and EGL:
1. EGL = HGL + V2/2g, EGL = HGL forV=0
2. Ifp=0, then HGL=z
3. A change in pipe diameter leads to a change in V(V2/2g) due to continuity
and thus a change in distance between HGL and EGL
4. A change in head loss (hL) leads to a change in slope of EGL and HGL
5. If then and cavitation may be possibleHGL zP
0
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Principle of linear impulse and momentum
The impulse may be determined
by direct integration.Graphically, it can be
represented by the area under
the force versus time curve. IfF
is constant, then I= F(t2 t1) .
Linear impulse: The integral Fdt is the linear impulse,
denoted I. It is a vector quantity measuring the effect of aforce during its time interval of action. Iacts in the same
direction as Fand has units of Ns
Linear momentum: The vector mvis called the linear
momentum, denoted as L . This vector has the samedirection as v. The linear momentum vector has units of
(kgm)/s.
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This principle is useful for solving problems that
involve force, velocity, and time. It can also be used
to analyze the mechanics of impact.
This is known as Impulse momentum theorem and is
stated as:
The quantity Fdt (product of force and the timeincrement during which it acts) represents the impulse
of applied force, while the quantity mdV represents the
change in momentum & [Fdt = mdV]
The impulse due to Force acting on a fluid mass in a small interval of
time is equal to change in momentum of the fluid mass
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a
b
a
b
c
d
c
d
1
2
V1
V2
a-a = ds1 = V1dt
d-d = ds2 = V2dt
1
1
2
Under the effect of external forces on stream, the
mass of fluid in region abcd shifts to abcd after a
time interval dt.A1V2, hence mv1>mv2, since abcd area is
common to both the regions, abcd and abcd will not
experience any momentum change so we have to
consider change of momentum between ab-ab and
cd-cd
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Fluid mass within abab = fluid mass within adad [by conservation of mass]
222111 dsAdsA
Momentum of fluid mass contained in region abab = 11111111 )()( VdtVAVdsA
Momentum of fluid mass contained in region cdcd =22222222
)()( VdtVAVdsA
Change in Momentum = 11112222 )()( VdtVAVdtVA
For steady incompressible flow 21 and from continuity equation A1V1=A2V2=Q
Change in momentum is dtVVQ )( 12 = dtVVg
Q)( 12
By impulse momentum principle :dtVV
g
QFdt )( 12
)( 12 VVg
Q
F
g
Q
is called Momentum flux
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Force F and velocities V1 & V2 are vector quantities and can be resolved into
components in the direction of the coordinate x and y. If represents the
inclination, with horizontal, of the centerline of the pipe at ab and cd, then components of
V1 and V2 along x-axis are:
21 and
2211
coscos andVV
And along y-axis are2211 sin SinandVV
And Force F along X & Y axis are:
)coscos( 1122
VVg
QFx
)( 1122
SinVSinVg
QFy
This eqn. represents the force component
exerted by pipe bend on the fluid mass
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The dynamic force must be supplemented by static pressure forces acting over the
inlet and outlet sections
Force exerted by fluid on the pipe bend would be
)coscos( 2211
VVg
QFx
)( 2211
SinVSinVg
QFy
)(tantan 1
x
y
x
y
F
F
F
F 2
1
22 ])()[( yx FFF
Magnitude of resultant force and direction are:
and
2221112211 coscos)coscos(
APAPVVg
QFx
2221112211 sinsin)sinsin(
APAPVVg
QFy
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Application: It provides a bulk solution for :
1. Force on pipe bends and transition
2. Force exerted by fluid jet striking against fixed or moving vanes3. Force on propeller blades
4. Jet propulsion
5. Head loss due to sudden enlargement or contraction in a pipe system and due to
Hydraulic jump in open channel
FlowP1 Flow
30cm
7.5cm
Water flow through the pipe, Q = 0.15m3/sec
Force exerted by fluid on the nozzle =?.
1.
KINETIC ENERGY CORRECTION FACTOR
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dAVelocity Profile
Local velocity, u
Real Fluid: Velocity is different for different fluid cross section & thus 1 dimensional
uniform flow is no longer applicable.
Velocity across section depends upon:
1. Nature of flow (Laminar or Turbulent)
2. Smoothness/Roughness of the pipe surface
KINETIC ENERGY CORRECTION FACTOR
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Kinetic Energy of the fluid contained in the infinitesimal area dA which carries
mass dm is
dAu
udA
udAu
udm
3
322
2
222
If the velocity distribution across the
cross is presumed, velocity can be
evaluated.
Total kinetic energy is generally prescribed in terms of average velocity V and Called Kinetic Energy Correction factor
The average velocity can be obtained by noting the discharge per unit time and dividing
it by cross section area OR using expression
,1
222)(
2)(1
3
3
3322
dAuAV
AVVAudAuVmassudAA
V
Thus kinetic energy in Bernoullis equation becomes equal to
]2
[2
g
V Value of = 2 for laminar flow & 1.02 to 1.15 for
turbulent flow
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MOMENTUM CORRECTION FACTOR
CASE: NON-UNIFORM FLOW dA Velocity
Profile
Local velocity, u
Momentum flux per unit time through
the elementary area dA would be :
2
2
*)(
dAu
dAuudAu
Momentum calculated on the basis of average velocity V and the
momentum correction factor would be:AVVVA
2
)(
Momentum flux =
dAuAV2
2
1
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In the flow of real fluid, the impulse momentum equation would get modified and
take the form:
)coscos( 222111
VVg
Q
Fx
)sinsin( 222111 VVFy
Practical use of are:
1. = 1.33 for laminar flow with parabolic velocity distribution
2. = 1.01 to 1.07 for general turbulent flow
Since majority of flow situations are turbulent in character, the usual practice is to
assign unit values for and
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h2
D
h1
0.2m
0.1m 12
3
42
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3. Water enters a 60mm pipe under a pressure of 5MN/m2 and leaves by a 30mm
pipe with pressure 4.5MN/m2. A vertical distance of 4m separate the center of two
pipes at section where pressure have been measured; the exit pipe lies at higherlevel. If discharge through pipe lines is 42.4 x 10-4 m3/s, work out the power
expended in overcoming the frictional loss. Specific weight of water is 9.8KN/m3.
4. How do you account for friction loss when applying Bernoullis equation to real fluid flows?
A constant diameter pipeline laid in inclined position is filled with water and has pressure tappingboth at upstream (1) and downstream (2) sections. When the valve on the down stream end is
closed, the differential pressure gauge indicates a pressure difference, (p2-p1)=68kPa.
Subsequently the valve is opened, the water flows at the rate of 0.27m3/s and the differential
pressure gauge reads (p1-p2)=230 kPa. Calculate the head loss between the two section for the
condition mentioned above.
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6. During a process of steam raising, it is required to supply 50 liters of water per minute into
a boiler in which pressure is 9.81 bar. Calculate the power expended
7. The rate of discharge through a vertical conical draft tube of a Kaplan is 17.5m3/s. The
diameter of the draft tube on the side connected to the outlet of the turbine runner is 2.5m
and the average velocity at the exit is 1.5m/s. If the pressure at inlet to the tube is not to beless than -0.7 bar, how far the tube should extend above the tail race. Neglect friction effects
and presume that exit of the draft tube lies 1.2m below the tail water level.
5. Bernoullis equation for steady flow may be expressed as tConsyP
g
Vtan
2
2
Each of the term on the LHS is an energy term of different type. What are they and explain
how they represent energy form?
8. Explain the difference between the Energy line and the hydraulic gradient line in a pipe flow.
A pipe, 25cm dia X 250m long, carries water from station A to station B which is located at a
level 10m higher. If the shear stress between the liquid and the pipe wall is 25N/m2, calculatethe pressure change in the pipe and the head lost.
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9. State and explain a few engineering applications of the momentum equations
A fireman holds a water hose ending into a nozzle that issues a 20mm diameter jet
Of water. If the pressure of water in the 60mm diameter hose is 700 kPa, find the force
experienced by the fireman.
10. A 30 cm diameter pipe carries water under a head of 20m with a velocity of 3.5m/s. If the
axis of the pipe turns through 45o, find the magnitude and direction of the resultant force on
the bend.
11. The discharge of water through a 140 bend, shown in Figure 1, is 30 litres/s. The bend islying in the horizontal plane and the diameters at the entrance and exit are 200mm and 100mm
respectively. The pressure measured at the entrance is 100 kN/m2, what is the magnitude and
direction of the force exerted by the water on the bend?
Comment on how frictional losses might be included in the above analysis.