fluid dynamics1

7/29/2019 Fluid Dynamics1

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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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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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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


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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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.

fluid dynamics1

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