good hydraulics book

7/29/2019 Good Hydraulics Book

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Basic hydraulicslecture notesNigel Wright

UNESCO-IHE Institute for Water Education

24/10/2007


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Contents1 Acceleration in a fluid ......................................................................................... 22 Pressure below a free surface ............................................................................. 3

3 Conservation of mass .......................................................................................... 3

4 Examples of Mass Conservation ........................................................................ 4

5 Conservation of energy (Bernoullis equation)................................................. 7

6 Examples of the use of Bernoullis Equation .................................................... 8

7 Momentum (Newtons Second Law)................................................................ 11

8 Examples of the Use of the Momentum Principle.......................................... 12

9 Further examples............................................................................................... 19

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Acceleration in a fluidThere are a number of basic concepts that we need to review before going into the main part of the

module. Many of you will have met these before, but you need to review them and make sure you

understand them. If you have not met these before you need to do some additional reading to cover

this. Speak to your lecturers for more assistance.

Acceleration of a fluid in a fluid there are two different contributions to a change in velocity as the

observer moves from one point to another. This reflects that fact that the velocity varies in both time

and position. We can calculate the total acceleration as follows.

x

y

z

A

B

Veloci

ty at A ( , , , )V x y z t =

( , , , ) ( , , ,V x x y y z z t t V x y z t = + + + + )Velocity at B

So the difference between A and B which is:

( , , , ) ( , , ,B A

V V V x x y y z z t t V x y z t = + + + + )

Becomes

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V V V V V

x y z tx y z t

= + + + +

V

V V V V

x y z tx y z t

= + + +

Dividing by tgives the acceleration:

V V V V x y z

xt

y z t

t t t t

= + + +

Using the fact thatDx/Dtis the x-component of velocity u gives the following expression for the total

acceleration, which we callDV/Dt:

tVzVwyVvxVuDtDV +++=

2

h

Pressure below a free surfaceThe pressure in a fluid below a free surface is given by:

p g=

Whereis the fluid density, g is the acceleration due to gravity andh is the distance below the free

surface. It is important to note that the pressure is not dependent on the width of the fluid body only

the depth.

Conservation of massConservation of mass is based on the physical law that mass cannot be destroyed. Based onthis we take a control volume (an area of bounded space) and state that:

Mass of fluid in per second - Mass of fluid out per second = Mass stored per second

Note:

Mass of fluid flowing in or out in 1 second is the mass flow rate or mass flux, usually denoted

bydt

dm- kg/s.

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A density, Control volume

Velocity, V

In one second a length Vof fluid enters control volume

dm= AV

dt

Mass can be stored within the control volume in one of two ways

by a change in density by a change in the size of the control volume

/sm3

AVmQ ==

The volume flow rate or discharge . If flow is steady and density is

constant then Qout = Qin

Examples of Mass Conservationa)

Nozzle

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0.5m/s10cm 3cm

V

The fluidis water with density 1000 kg/m3. FindQ, Vand m

cumecs)called(sometimes/s0.00393m0.1

0.5 32

=

=4

Qin

/s0.00393m=4

0.03== 3

2

VQQ inout

m/s56.5=V

kg/s93.300393.01000 ====outin QQm

b) branching pipe - water, density 1000 kg/m3

1

2

3

D = 30 cm D = 10 cm D = 20 cm1 2 3

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Q1 = 0.01 m3/s V = 0.4 m/s2

FindQ , Q , V andV2 3 1 3

/sm0.00314=0.44

0.1

=V4=

32

2

2

2

2

D

Q

Conservation of mass implies:

1 2Q Q Q

3= +

3

30.01 0.00314 0.00686 /Q m= = s

1

2

11

4

= V

DQ

m/s0.142=/40.3

0.01=V 21

3

2

3

3 V4

D=Q

( )m/s0.218=

/40.2

0.00686=V

23

c) Reservoir

I H

Q

DamPlan area A

What is the relationship between the inflowI(m3/s), outflow Q (m3/s) and the depth over the

dam crestH?

Conservation of mass

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dt

)d (VolumeI - Q =

HAVolume =)(

dtdHAQI =Thus

IQ

t

In practiceHis a simple function ofQ and the variation ofQ with tcan be found if thevariation ofIwith tis known (e.g. for a flood wave).

Conservation of energy (Bernoullis equation)The total energy of a given mass of fluid does not change over time as long as no work is put

in nor taken out of the fluid. If there is friction in a fluid then energy will be taken out.

Therefore for energy to be conserved we have to consider the fluid to have no viscosity i.e. it

is inviscid.

If we take these assumptions and analyse fluid flow using the laws of thermodynamics we get

the Bernoulli equation for the flow along a streamline of a steady, incompressible, inviscid

fluid.

constant2

2

=++ gzVp

The three terms in this equation represent the three different forms of energy in a fluid.

gz - the potential energy per kg

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V2/2 - the kinetic energy per kg

/p - the pressure energy or more correctly the work done by the moving fluid per

kg

It is the last term that distinguishes the mechanics of fluids from the mechanics of solid

particles.

Bernoullis equation can also be expressed as

constant2

2

=++g

p

g

Vz

All these terms have dimensions of length or head.

z = elevation head

g

v

2

2

- velocity head

gp

- pressure head

The Bernoulli sum is the total energy/unit weight. It is sometimes referred to as the specific

energy equation.

Alternatively the Bernoulli equation can be derived more rigorously from Newtons Second

Law. Try this yourself.

Examples of the use of Bernoullis Equationa) Nozzle

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0.5m/s 5.56m/s

1

2

If 2 is at atmospheric pressure, find the pressure at 1

2

2

221

2

11

22z

g

V

g

pz

g

V

g

p++=++

0, 221 == pzz

g

V

g

V

g

p

22

2

2

2

11 =+

kPa33.152

1000)5.056.5( 22

1 =

=p

This is the gauge pressure - pressure above atmospheric.

b) Branching pipe

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0.142m/s

0.4m/s

0.218m/s

12

3

If the pipe junction is horizontal, = 0.25kPa gauge and = 1000kg/m31p m , find and2p 3p

p

g

V

g

p

g

V

g

p

g

V

g

1 1

2

2 2

2

3 3

2

2 2 + = + +=

2

81.92

218.0

81.9100081.92

4.0

81.9100081.92

142.0

81.91000

250 232

2

2

+=+=+

pp

So,

= 0.180 kPap2

p = 0.236 kPa3

c) Vertical jet

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D

0.1m

30m

0.25m3

/s

1

2

V

Find V and D

sm /8.3141.0

25.02

=

At 1, velocity =

Applying conservation of energy from 1 to 2, assuming atmospheric pressure at both sections,

section 1 at datum level and section 2, 30 m above the datum

30+

9.812

V=

9.812

31.8 22

V = 20.56 m/s.

By continuity

56.204

25.02

=

D

m124.0=D

Momentum (Newtons Second Law)For solid body dynamics Newtons 2ndLaw is written:

Force on body = Rate of change of momentum

For a control volume of fluid, Newtons 2ndlaw may be written:

Force on control volume = Momentum per second out of control volume

- Momentum per second into control volume

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Remember: momentum is a vector quantity and therefore this law can be applied in

perpendicular directions.

In one second length Vof fluid enters control volume. Therefore the momentum entering the

control volume in this time is

2AVVAV =

More precisely

AV V

or

Q V

A density, Control volume

Velocity, V

A similar analysis applies for the momentum leaving the control volume.

8 Examples of the Use of the MomentumPrinciple

a) Nozzle

What is the force exerted by this flow of water on the nozzle?

Consider forces on fluid and momentum inflows and outflows

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0.5m/s. p=15.33kPa10cm 3cm

V=5.56m/s

p=0

M1 M2

F1 F2

FN

FN = force exerted on fluid by nozzle

N1.96=0.54

0.11000=nozzle,theintomomentumofRate 2

2

1

M

N21.85=56.54

0.031000=nozzle,ofoutmomentumofRate

22

2

M

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N120.4=4

0.115330=1,atpressuretodueForce

2

1

F

0=2,atpressuretodueForce 2F (atmospheric pressure at 2)

Newtons Second Law

NFFFMM = 2112

NF= 04.12096.185.21

N5.10085.214.12096.1 =+=NF

Force exerted on the nozzle is equal and opposite to this

b) Branching pipe

1

2

3

V = 0.40 m/s,p2 2 = 0.180 kPa,D = 10 cm2

= 0.142 m/s,p = 0.25 kPa,DV = 30 cm1 1 1= 0.218 m/s,p = 0.236 kPa,DV = 20 cm3 3 3

Find theXandYcomponents of the forces on the pipe junction

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M1

F1

M2

F3M3

F2

X

Y

Consider the equilibrium of the control volume of water

N43.1142.04

3.01000 2

2

1 =

=

M

N26.140.04

1.01000 2

2

2 =

=

M

2

3

0.21000 0.218 1.49N

4M

=

-0.218 =

N.B. the velocity (=-0.218) is negative as it in the opposite direction to they axis. The

discharge (20.2

0.2184

) is a scalar and has no direction therefore it is positive.

N67.174

3.0250

2

1 =

=

F

N41.14

1.0

180

2

2 =

=

F

N41.74

2.0236

2

3 =

=

F

Newtons 2ndLaw horizontally

XFM += 110

Newtons 2ndLaw vertically

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2 3 3 20M M F F+ = + Y

Y1.41-7.411.49-1.26 +=

The force on the junction is opposite to the force on the fluid so it is 19.1N horizontally and

6.23N vertically.

Note: in the diagrams the double arrows represent the direction of flow, not the direction in

which momentum is considered.

d) Vertical jetFlat plate target

F

20.56m/s

0.124m

Find the force on the fluid, F, assuming no friction losses

2

2 2

=0-

0.124= 1000 20.564

= 5.10 kN

F AV

F

Cap target

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F

If the inlet parameters are as before F can be found by assuming complete reversal of fluid

with no loss of velocity and same flow area.

kN2.10

2

)

2

2(

2

=

=

=

AVF

AVAVF

e) Hydraulic jump

A hydraulic jump occurs at a transition between fast and slow flow. It is a useful phenomena

in the design of river or dam works.

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V1

V2

y1 y2

The channel has width, b.

The force is equal to the change in momentum

1221 MMFF =

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2

1gy is the average pressure on the vertical plane and is given by so:2Av 2Vyb

2

11

2

2222

11

22bVybVyby

gyby

gy

=

From continuity

2211 yVyV =

and rearranging givessubstituting for V2

2

1

1

2

1

1

21

2

1

2

1Fr

gy

V

y

yy

y

y==

+

Further examples

9.1 Orifice Plate in Pipe

Pressure tappings to manometer

The analysis is the same as for a venturimeter

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Q CA

Z g

A

A

o

m

m

o

=

2 1

1

2

1 2

/

Ao = Orifice Area

A = Pipe Area

C = Plate Coefficient ( 0.61)

9.2 Large Orifice

y

b

h1y h2

We can assume that the free surface velocity is small.

Apply energy equation from free surface (datum) for streamline to shaded strip

0

2

2

= v

g

y

v g= 2 y Therefore the discharge for the strip is given by

Q b y gy= 2

To get the total discharge, Q, we have to integrate as y varies significantly across the orifice,

so

dygybdQQh

h ==2

1

2

= 2

3

2 1 2 23 2

1

3 2b g h h( ) ( )/ / /

We would need to apply a discharge coefficient to account for streamline curvature etc..

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9.3 Flow around a horizontal pipe bend

Q

F

Q

1

2

Q = 0.1m

3/s D = 300mm1D2= 200mm = 60

0

p1 = 140 kN/m

2

Find F.

Applying continuity gives

4

2.0

4

3.01.0

2

2

2

1

VV ==

V1 = 1.415 m/s

V2 = 3.184 m/s

The energy equation givesp

g

V

g

p

g

V

g

1 1

2

2 2

2

2 2 + = +

p p V V2 1 12

2

2 2= + ( )/

= + 140 10 1415 3184 10 23 2 2 3( . . ) /

= 13593. kN / m2

From Newtons 2nd

Law perpendicular to the inlet:

cosfluidonForce 2211 ApFAp =

2 1


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Rate of change of momentum

= V A V A22

2 1

2

1cos

)cos( 12 VVQ =

Therefore

)cos(cos 122211 VVQApFAp =

o60cos

4

22.031093.1354

23.0310140 F

)415.160cos184.3(101.0 03 = F = 7 74. kN

9.4 Orifice Flow from a tank

v

hV

D

d

Cd

c

2

4The flow area at the orifice =

Cc = coefficient of contraction ( 0 6. )

from the energy equation, ideal flow velocity at orifice (v ) is given byi

g

vh

g

V i

22

22

=

+

v V gi = +2

2 h But v C vv i=

2 2


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C = Coefficient of velocity ( 0 99. )vTherefore

v C V ghv= +2 2

is given byThe ideal discharge Qi

Qd

vi i= 2

4.

The real discharge Q is given byi

Q Cd

C vc v= 2

4. . i

v

But Q C where CQd i= dis the discharge coefficient.

Therefore C C Cd c=How long will it take for tank to empty?

From continuity

V D v d Cc

2 2

4 4= (1)

At free surface

Vdh

dt= (2)

Also from above

)2( 222 ghVCv v += (3)combining (1) and (3)

( )V DC d

C V ghc

v

24

2 4

2 2 2

= +

VD

C C dgh

v c

24

2 2 41 2

=

Assume D d>>

Therefore

V

d

D C C ghv c=

2

2 2 using (2)

=dh

dt

d

DC C gh

v c

2

22

Integrate between t = 0 h = H

t = T h = 0

Hv c

Tdh

h

d

DC C g dt

02

2 02 =

2 3


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TD

d C C

H

gv c

=2

2

2

2

good hydraulics book

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