principles _ hydraulics 1 (hydrodynamics)

8/22/2019 Principles _ Hydraulics 1 (Hydrodynamics)

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Hydraulics 1 1 David Apsley

HYDRAULICS 1 (HYDRODYNAMICS) SPRING 2005

Part 1. Fluid-Flow Principles

1. Introduction

1.1 Definitions1.2 Notation and fluid properties

1.3 Hydrostatics

1.4 Fluid dynamics

1.5 Control volumes

1.6 Visualising fluid flow

1.7 Real and ideal fluids

1.8 Laminar and turbulent flow

2. Continuity (mass conservation)2.1 Flow rate2.2 The steady continuity equation

2.3 Unsteady continuity equation

3. The Equation of Motion3.1 Forms of the equation of motion

3.2 Fluid acceleration

3.3 Bernoullis equation3.4 Application to flow measurement

3.5 Other applications (flow through an orifice; tank-emptying)

4. The Momentum Principle4.1 Steady-flow momentum principle

4.2 Applications (pipe contractions; pipe bends; jets)

5. Energy5.1 Derivation of Bernoullis equation from an energy principle5.2 Fluid head5.3 Departures from ideal flow (discharge coefficients; loss coefficients; momentum & energy coefficients)

Part 2. Applications (Separate Notes)

1. Hydraulic Jump

2. Pipe Flow (Dr Lane-Serff)

Recommended Reading

Hamill, 2001, Understanding Hydraulics, 2nd

Edition, Palgrave, ISBN 0-333-77906-1

Chadwick, Morfett and Borthwick, 2004, Hydraulics in Civil and Environmental

Engineering, 4th

Edition, Spon Press, ISBN 0-415-30609-4

Massey, 1998, Mechanics of Fluids, 7th Edition, (Revised by Ward-Smith, J.), Stanley

Thornes, ISBN 0-748-74043-0

White, 2003, Fluid Mechanics, 5th Edition, McGraw-Hill, ISBN 0-07-240217-2


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1. Introduction and Basic Principles

1.1 Definitions

Afluidis a body of matter that canflow; i.e. continues to deform under a shearing force.

Fluids may be liquids (definite volume; free surface) or gases (expand to fill any container).

Fluids obey the usual laws of Newtonian mechanics, but as a continuum. Unlike rigid bodies,

fluid particles may move relative to each other, interacting via internal forces. These are

usually expressed in terms ofstresses (stress = force / area), the principal ones being:

pressure,p normal stress: pushing or pressing;

shear stress, tangential stress: frictional drag; opposing relative motion.

An ideal fluid has no viscosity. This is never exactly true, but is often a useful approximation.

Fluids may be regarded as incompressible (density or volume not changed by the flow) atspeeds much less than that of sound. This is usually the case in hydraulics.

Fluid flow may be described as laminar(adjacent layers slide smoothly past each other) or

turbulent(irregular, with constant intermingling of adjacent layers).

1.2 Notation and Fluid Properties

The main flow variables are:

p pressure (force/area)

u velocityThese arefield variables because they are functions of position and time t;e.g. u = u(x,t)

Vector quantities like position and velocity are often decomposed into components:

),,( zyxx

),,( wvuu

U(or occasionally V) will be used for the magnitude of velocity.

Important fluid properties are:

density = mass / volume

dynamic viscosity; defined by Newtons viscosity law:

y

u

d

d

=

=

/

is called the kinematic viscosity

surface tension = force / length (or surface energy/area)

lF

=

K bulk modulus = pressure change / volumetric strain

)

(

VVKp

=


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Most gases at normal temperatures and pressures satisfy the ideal gas law, which in fluid

mechanics is usually written as

RTp =

The gas constantR depends on the particular gas (R =R*/m = universal gas constant / molar

mass). It has a value of 287 J kg1

K1

for dry air.

1.3 Hydrostatics

Hydrostatics concerns the balance of forces in a fluid at rest.

The principal forces are:

pressure; p is the normal force per unit area

weight;

g is the specific weight(weight per unit volume)

surface tension;

is the force per unit length

In the interior of a stationary fluid, pressure forces balance weight. As

a result the change in pressure with depth is given by the hydrostatic

equation

zgp = (1)

Pressure increases as height decreases and vice versa. As a result there

is a net buoyancy force on an immersed body equal to the weight of

fluid displaced (Archimedes Principle); i.e. the immersed weightof a

body is less than its actual weight.

If density is constant then

constantgzp =+ (2)

The quantity on the LHS is called thepiezometric pressure.

Hydrostatic principles are used to measure:

Pressure in a flowing liquid, using apiezometer tube.

The pressure (relative to atmospheric pressure) is given by

ghp = (3)

Pressure differences using a U-tubemanometer.

The difference in pressure between points A and B is proportional to

the difference in heights of the manometer fluid in the two arms:

ghp m )(

= (4)

where m is the density of the manometer fluid and the density of the

working fluid.(The latter can be ignored if the working fluid is a gas).

Note that, in incompressible flow, absolute pressure is unimportant: pressure differences

determine the net force. In particular, it is usual to work with the gauge pressure i.e.

difference between actual and atmospheric pressures. The pressure at a free surface is thenp = 0, whilst a sub-atmospheric (suction or vacuum) pressure hasp < 0.

p

h

h

m

p >A pB


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Surface tension is vital to survival if you are a water-hopping

insect but it can usually be ignored in large-scale hydraulics.

The capillary rise in a thin-bore tube is given by

cos

gdh = (5)

where d is the tube diameter and

is the contact angle ( 0

for water on glass).

1.4 Fluid Dynamics

Fluid dynamics concerns fluids in motion. Almost all hydraulic problems can be addressed by

the application of one or more of three equations/physical principles:

continuity (mass conservation); momentum;

energy.

Continuity is almost invariably required and is usually applied first to reduce the number of

unknowns.

Continuity (Mass Conservation)

Mass is neither created nor destroyed.

For a steady flow,

mass flow rate in = mass flow rate out

or, if the flow is also incompressible,

volume flow rate in = volume flow rate out

Any mathematical expression of this is called the continuity equation.

Q1

2Q

3Q

Qin = Qout Q1 = Q2 + Q3

h


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Momentum

Based on Newtons Laws of Motion; specifically:

Second Law:

force = rate of change of momentum.

Third Law (Action and reaction):force exerted by a fluid on its containment is equal and opposite to the force

exerted by the containment on the fluid.

force onFLUID

force onPIPE

force onBODY

force onFLUID

wake

Energy

Energy principle:

change of energy = work done + heat input

For incompressible fluids we need only consider the mechanical energy principle:change of kinetic energy = work done

For ideal fluids (no losses) this is expressed asBernoullis equation

constantUgzp =++2

21

(along a streamline)

In particular:

velocity increases pressure decreases

small lossof head

Bernoullis equation is strictly for ideal fluids (no friction or other fluid energy

changes), but it can be amended to account for:

energy put in (by pumps) or taken out (by turbines);

frictional losses (e.g. pipe flow).

Because all forms of fluid energy can be converted to an equivalent gravitational

potential energy it is common in hydraulics to measure energy in height units or fluidhead.


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1.5 Control Volumes

Although it is possible to write the equations of fluid mechanics in terms of differential

equations (describing what is happening at every point), engineers are more often concerned

with the behaviour of fluid in bulk and the usual method of analysis is to apply mass and

momentum principles, etc., to all the fluid in a control volume.

For example, consider suitable control volumes for the two pipe fittings shown.

F

Mass conservation:

0

rate)flowmass(rate)flowmass(

volumecontrolofoutfluxmassNet

=

=AB

Momentum principle:

yin velocitchangerate)flowmass(

momentumofchangeofrate

volumeontrolcinfluidonforceNet

=

=

1.6 Visualising Fluid Flow

Velocity vectors are arrows of

length proportional to magnitude.

Streamlines are everywhere tangential to the instantaneous flow direction. The flow is fastest

where the streamlines are closest together

(why?). Streamlines can never cross (else

the velocity would be 2-valued) and cannot

terminate in the interior of the flow. In

steady flow one streamline always coincides

with a solid boundary or free surface.


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Path lines orparticle paths are the lines followed by individual elements of fluid. (They are

what you would see on a long-exposure photograph).

Streaklines are swept out by all the fluid elements which have passed through a particular

point; e.g. dye injection.

In steady (i.e. not time-varying) flow streamlines, streaklines and particle paths are identical.

Stream tubes are bundles of streamlines (virtual pipes). By

definition, there is no flow across the sides and the same flow

must enter at one end as leaves at the other.

1.7 Real and Ideal Fluids

Ideal fluids have no viscosity there is no internal friction or loss of mechanical energy. No

such fluid exists, but many flows can be approximated as ideal if viscous forces are small anddo not cause major flow phenomena such as boundary-layer separation.

Real fluids have non-zero viscosity. This has two important consequences:

(1) They satisfy the no-slip condition at solid boundaries.

i.e. the (relative) velocity at the boundary is zero.

ideal real

(2) There are frictional forces between adjacent layers of fluid moving at different speeds

and between the fluid and a boundary.

shear stress ofUPPER fluidon LOWER

shear stress ofLOWER fluid

on UPPER

ForNewtonian fluids (which includes most fluids of interest), the stress (force per unit

area) of the upper fluid on the lower fluid is given byNewtons viscosity law:

y

u

d

d

= (6)

An equal and opposite force is exerted by the lower fluid on the upper fluid.


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1.8 Laminar and Turbulent FlowAt low flow speeds viscosity ensures that adjacent layers of fluid slide smoothly over one

another in a steady fashion without intermingling. This flow regime is called laminar.

At higher speeds viscosity is insufficient to smooth out minorperturbations to the flow, which grow rapidly to create

unsteadiness and eddying motions. This flow regime is called

turbulent.

The transition between laminar and turbulent flow can be

observed in the rising plume from a cigarette. There is an initial

laminar flow which wavers and then breaks up into turbulent

eddies as the flow accelerates.

The pioneering experiments determining the occurrence of laminar or turbulent flow in pipes

were first performed by Osborne Reynolds (Professor of Engineering at Owens College soon to become the University of Manchester). He demonstrated that which regime occurred

depended on a dimensionless parameter later to become known as theReynolds number:

or

ReULUL

= (7)

where Uand L are typical velocity and length scales of the flow, is the density, is the

dynamic viscosity and

(=

/

) the kinematic viscosity. The Reynolds number is probably

the single most important parameter in fluid mechanics!

am nar

turbulent

The Reynolds number can be regarded as a measure of the ratio of the total force

(= mass acceleration) to the viscous force. For a block of fluid, volume L3

, using order-of-magnitude estimates:

223

/)( LU

UL

ULonacceleratimass =

whilst

ULLL

UAforceviscous

2=

Hence,

Re

22

== UL

UL

LU

forceviscous

onacceleratimass

When the Reynolds number is large the effects of viscosity are small and conversely. Ingeneral,

laminar

turbulent


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low Re laminar flow

high Re

turbulent flow

Note that low and high is all relative and depends on the particular flow and the choice of

velocity and length scales (which should be stated). For pipes it is conventional to take Uas

average velocity andL as the diameter; then:Re < 2000 (laminar flow)

Re > 4000 (turbulent flow)

(The commonly-accepted critical Reynolds number for transition in a round pipe is 2300).

These particular numerical values are for pipe flow only, with a particular choice of velocity

and length scales. For example, choosing radius rather than diameter as a length scale would

immediately halve the Reynolds number, but wouldnt affect the flow regime.

The numerical size of viscosity for the common fluids, air and water ( air 1.510

5m

2s

1;

water

1.0

10

6

m

2

s

1

), means that most civil-engineering flows are fully turbulent.

Even when the flow is turbulent it usually consists of relatively small fluctuations about a

mean flow, which is what we are actually interested in. In turbulent flow, however, most of

the net transfer of momentum and energy between layers of fluid occurs by the eddying

motions mingling fluid elements and so, as far as the mean flow is concerned, the mean stress is no longer given by Newtons viscosity law ( =

du/dy) but is substantially increased by

the net effect of turbulent eddies. This mixing of fluid streams leads, for example, to a

turbulent velocity profile in a pipe being much more uniform than a corresponding laminar

velocity profile.


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2. Continuity (Mass Conservation)

In fluid mechanics any mathematical expression of the conservation of mass is called the

continuity equation.

2.1 Flow Rate

Consider fluid passing through an area at uniform velocity u.

In time

tthe fluid moves a distance u

t, and hence the volume of

fluid that has passed through that area is

tuA

The volumetric flow rateQ (aka volume flux, quantity of flow or discharge) is given by

t

tuA

time

volumeQ

==

i.e.

uAQ = (8)

It may be measured in m3

s1

or L s1

(1 L = 103

m3

s1

).

The mass flow rate (or mass flux) through a section, m , is given by

time

volumedensity

time

massm

==

i.e.

uAQm

==

(9)It may be measured in kg s

1.

Example. A pipe of internal diameter 200 mm carries water (density 1000 kg m3

) at average

velocity 2.5 m s1

. Find the volumetric flow rate and mass flow rate.

Note. In general the flow may be at any angle to the areaA. In general the Uwhich appears inthe formula for Q or m should be the component normal to the area.

u

A

A

u t


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2.2 The Steady Continuity Equation

The continuity equation is invariably applied when either:

there is a change of cross section;

there is a junction.

Change of Cross Section

Consider a stream tube with a change of cross section fromA1 toA2.

A1

A2u1 u2

1 2

If the flow is steady (i.e. doesnt change with time) then no mass can accumulate between the

two cross sections and hence

222111

21

)()(

AuAu

fluxmassfluxmass

=

=(10)

In hydraulics the fluid can usually be treated as incompressible (density constant along a

streamline) and hence the volume flow rate is constant:

2211 AuAuQ == (11)

Junctions

At a junction of more than two pipes, we require that

total flow into junction = total flow out of junction

e.g.

Q1

2Q

3Q

Q1 = Q2 + Q3


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Example. A circular pipe carrying water undergoes a smooth contraction from internal

diameter 200 mm to one of diameter 80 mm. If the velocity in the larger cross section is

0.6 m s1

find:

(a) the velocity in the smaller-diameter pipe;

(b) the volume and mass flow rates.

Example.A manifold splits the air supply (unequally) between two components of an engine. The inlet

duct has square section with side 150 mm. The outlet ducts have circular cross-section with

diameters 80 mm and 120 mm. The air velocity in the intake duct is 2 m s1

and that in the

smaller outlet duct is 4 m s

1

. Find the volumetric flow rate and velocity in the larger outletduct.

150 mm

120 mm

80 mm

2 m/s

4 m/s

Example. Liquid of controlled density can be supplied by injecting saturated brine (s.g. 1.20)into a freshwater stream. If brine is injected at 2 L s

1into a pipe of internal diameter 100 mm

carrying fresh water at 3 L s1

find, in the well-mixed region downstream:

(a) the total quantity of flow;

(b) the mass flow rate;

(c) the average velocity;

(d) the density of fluid.


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2.3 Unsteady Continuity Equation

Mass conservation applied to the fluid in a control volume which is either:

(a) moving; or

(b) changing density;

gives

outin fluxmassfluxmassmasst

)()()(d

d= (12)

In many cases the fluid is incompressible and continuity can equally well be applied to

volume rather than mass:

outinQQ

t

V=

d

d(13)

i.e.

rate of change of volume =flow rate in flow rate out

Example. A cylinder of diameter 0.2 m contains water and has intake and exit pipes both with

cross sections of 4103

m2

and fitted with non-return valves. A piston oscillates sinusoidally

up and down in the cylinder with amplitude 80 mm and frequency 2 Hz. Find the maximum

velocity of water in the exit pipe and the volume discharged over a cycle.


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3. The Equation of Motion

3.1 Forms of the Equation of Motion

The equation of motion is any mathematical expression of Newtons Second Law.

As you have seen in the Mechanics module, this can be written down in various ways; e.g.

F= ma

or, equivalently, as a momentum principle:

force = rate of change of momentum

or as a mechanical energy principle:

work done = change of kinetic energy

In fluid mechanics, also, the equation of motion can be expressed in different ways. In

Section 3 we examine the first and third of the forms above, leading to an important result

known as Bernoullis equation. In Section 4 we apply the momentum principle using control

volumes in order to deduce forces on structures.

3.2 Fluid Acceleration

In order to apply Newtons Second Law (F= ma) to fluid motion one must know the fluid

acceleration.

Suppose the position of a particle at time tis given byx(t). Its velocity u is given by

t

xu

d

d=

Its acceleration is defined ast

ua

d

d= , but you will know from Mechanics that this can also be

written in terms ofdisplacementx as

rule)chainby the(d

d

d

d

d

d

t

x

x

u

t

ua ==

i.e.

x

uua

d

d=

Consider now the fluid flow through a converging section as shown.

u1 u2 u3

a

If the flow is steady then the velocity at any particular position does not change with time; i.e.

0/ = tu . However the fluid is accelerating because as you move with any particular fluidelement along the centreline you are moving with faster velocity. The fluid acceleration i.e.


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the acceleration of a particular element of fluid is given in this instance by xuu / .

In general, the acceleration of a fluid element in a flow u(x,t) is the sum of two parts:

t

u

(temporal derivative) because of changes with time at a point;

x

uu

(advective derivative) because of changes as you move with the flow.

The latter usually dominates.

When the velocity is also a function of the other space components y and z the fluid

acceleration can be written as the total (or material, or substantive) derivative

zw

yv

xu

tt

+

+

+

uuuuua

D

D(14)

Fortunately, many flows that you will deal with are steady and quasi-1-dimensional, and this

can often be simplified to

x

uua

= (15)

Example. An air supply system consists of ducts with rectangular cross section. The upstream

part of the ducting has a width of 0.4 m and height 0.5 m, whilst the downstream part has the

same width (0.4 m) but a height of 0.2 m. These two sections are connected by a simple

transition duct where the height varies linearly over a distance of 2 m.

0.5 m 0.2 m

0.4 m

x

(a) If the volumetric flow rate along the duct is 0.6 m3

s1

, find an expression for the

speed of the flow as a function of the distance from the start of the contraction,x.

(b) Find an expression for the acceleration of the air as a function of distance along the

contraction.

Assume that the flow is quasi-1d.

(Answer: withx in m, u in m s1

, a in m s2

, (a)x

u3.01

3

= ; (b)

3)3.01(

7.2

xa

= )

What force provides this acceleration?


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3.3 Bernoullis Equation

A (better) derivation of Bernoullis equation from

general energy considerations will be given later.

For now we consider the forces on a short fluid

element, length

s, uniform cross-section A andaligned locally with the flow.

We assume that the flow is inviscid (no friction),

steady (no time derivative) and incompressible (

constant along a streamline). Resolving forces in the

direction of flow:

mamgApppA =+ sin)(

Writing:

s

UUa

d

d= for the acceleration (Uis the magnitude of velocity, s is the distance)

)

( sAm = for the mass

s

z

sin =

gives

s

UsUAzgApA

d

d

=

and, on dividing by the volume (A

s) and noting that )(

d

d

d

d 22

1 U

ss

UU = :

)(d

d

22

1 Uss

zg

s

p=

Since the flow is incompressible,

is constant along the streamline and can be taken inside

the differential. In the limit as

s 0:

0)(d

d 221 =++ Ugzp

s

i.e.

constantUgzp =++2

21 (16)

Remember the qualifiers:inviscid(no friction)

steady

incompressible

along a streamline

Some useful things to remember when applying Bernoullis theorem are:

use continuity first to reduce the number of unknowns;

at a free surface the pressure is atmospheric:p = 0;

for a free discharge to atmosphere the pressure is atmospheric:p = 0; for a large tank/reservoir velocity at the surface is unaffected by the discharge: U= 0.

(p+ p)A

pA

mg

s


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Example (Exam, May 2003 *** modified***)

Air (density 1.2 kg m3

) is flowing through a smooth contraction from a circular pipe of

diameter 120 mm to a tube of diameter 60 mm as shown in the figure. There are tappings at

four locations along the contraction, where the diameter is 120 mm, 100 mm, 80 mm and

60 mm (labelled A, B, C and D respectively).

(a) If the air flow along the duct is 0.2 m3

s1

, find the speed of the flow at each location

A, B, C and D.

(b) Assuming no energy losses, calculate the pressure difference between locations A and

B, and the pressure difference between locations C and D.

(c) A water manometer is connected across tappings A and B, and another manometer is

connected across C and D (see the Figure). What are the water-level differences (

z1

and

z2) in the two manometers.

z1 z2

120 mm 100 mm 80 mm 60 mm

A B C D


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Example. Water is being siphoned as shown. The internal diameter of the pipe is 20 mm and

that of the nozzle is 15 mm. Assuming no losses find:

(a) the volumetric flow rate;

(b) the pressure at the top.

2.5 m

1.5 m

Example (Exam, June 2004)

A pipe carrying water with a flow rate of 0.04 m3

s1

contracts from a diameter of 100 mm to

60 mm over a distance of 0.5 m, with the diameter changing linearly with distance.

(a) What is the flow speed as a function of distance along the pipe?

(b) What is the acceleration of the flow as a function of distance along the pipe?

(c) If the pressure where the diameter is 60 mm is denoted byp0, what is the pressure as a

function of distance along the pipe, assuming no energy loss?

(Answer: Withx in m, u in m s1

, Q in m3

s1

,p in Pa,

(a)2)8.01(

16

xu

= ; (b)

52 )8.01(6.409

xQ

= ; (c) )

)8.01(

1

6.0

1()

16(

44

2

21

0x

pp

+= )

Example. A river may be approximated by a rectangular channel of width 10 m. The depth of

water is 1.4 m. The abutments of a bridge reduce the channel width to 7 m and the depth ofwater to 1.2 m. Find the upstream velocity and quantity of flow.

Explain, physically, why the depth of water decreases between the abutments.


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3.4 Application to Flow Measurement

Some of the most important applications of Bernoullis equation are to flow measurement.

The main reason is that velocity changes can be related to pressure changes and pressure is

usually cheap and easy to measure with a manometer or piezometer tube.

Ideal-flow theory is used to compute a theoretical discharge Q. This can then be corrected for

friction, turbulence, etc. by applying a discharge coefficientcD (see Section 5.3).

3.4.1 Pitot and Pitot-Static Tubes

A Pitot tube is used for measuring velocity in the

flow. It works by converting the kinetic energy into

pressure energy.

Flow is brought to rest at the stagnation pointat thefront. By Bernoullis theorem this raises the pressure

to the stagnation or Pitot pressure2

21

0

UPP += (17)

where P and U are the pressure and velocity in the

undisturbed flow. If one also measures the static

pressure P then the difference PP 0 is equal to2

21 U , from which the velocity can be deduced.

In a pitot-static tube both pitot and static pressures are measured with the same instrument.The pressure difference can be measured with a manometer: ghPP m )(0 = .

up0

p

to manometer

h

p0

p

ghUPP m )( 2

21

0 ==

Example. The velocity of water in a conduit is to be measured using a piezometer and Pitot

tube. If the water level in the piezometer rises to 0.3 m and that in the Pitot tube to 0.5 m find

the excess pressure and the flow velocity at the measuring point.

How do you expect the height in the pitot tube to change as the measuring point istraversed across the conduit?

u

p ezometer tot tu e

g

p

0

g

p


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Example. The speed of air is measured with a pitot-static tube connected to a U-tube

manometer containing alcohol (s.g. 0.79).

(a) Find the air velocity if the difference in fluid levels in the two arms of the manometer

is 150 mm.

(b) What difference in levels would be expected for a flow speed of 5 m s1

?

How could the accuracy of the reading be enhanced for low flow speeds?

3.4.2 Venturi Meter

A venturi is a smooth contraction in a pipe. The idea is that the contraction causes an increase

in velocity and hence a decrease in pressure, which can be measured.

Continuity and Bernoulli give two equations for two unknowns (U1 and U2) and hence either

can be found and multiplied by the corresponding area to give the volumetric flow rate Q.

u1

p2

u2

p1

A1A2

Apply Bernoulli between sections 1 and 2:2

221

2

2

121

1

UpUp +=+

/)(2 212

1

2

2 ppUU = (18)

Apply continuity between sections 1 and 2:

2211 AUAU =

1

2

1

2 UA

A

U=

(19)

Substitution in (18) yields an expression for U1, which, when multiplied by A1, gives the

volumetric flow rate (exercise: complete the analysis):

2/1

2

21

1

1)/(

/

2pconstant

AA

pAQ =

=

Notes

1. This is the ideal flow rate. To account for non-ideal behaviour it can be multiplied by

a discharge coefficient cD, so that

idealDactual QcQ=

However, for a well-defined venturi meter, cD is 0.98 or more, so is often neglected.


21/33


2. The downstream expansion plays no role in the analysis, but should be at a small

angle so as not to provoke flow separation and cause energy loss.

3. The same principle is used in an orifice meter. This is a sharp contraction in a pipe.

The downstream pressure at the walls is assumed to be the same as at the orifice, dueto the recirculating flow region downstream.

p2

A12A

p1

An orifice meter is much cheaper to manufacture and install but causes substantialenergy loss. If the Bernoulli analysis is used to predict an ideal flow rate then a

discharge coefficient with a value of the order 0.6 is needed, mainly due to the

continued convergence of streamlines after passing through the orifice.

3.4.3 Sharp-Crested Weir

Sharp-crested weirs are slotted plates used to provide accurate measurements of quantity of

flow in small open channels for example, in hydraulics laboratories. (Broad-crested weirs

are used for larger channels and work by a different principle see open-channel flow in

Hydraulics 3).

nappecrest, or sill

air

Q

H

The quantity of flow Q is related to the height over the weir, H. For a given rectangular weir,

Bernoullis theorem predicts2/3

HQ

Key to operation is that air at atmospheric pressure reaches the underside of the weir (so that

the pressure across the nappe is approximately atmospheric,p = 0.

the weir has a sharp crest (so that the flow breaks away on the downstream side);

the backwater level does not approach the crest of the weir;

the notch does not span the full width of the channel (contracted weir) or, if it is fullwidth, then a vent pipe is used to ensure atmospheric pressure underneath the nappe.


22/33


Idealised Analysis

Velocity is not uniform over the weir but is a

function of height. Let h be the distance from thefree surface. Apply Bernoulli between points 1

upstream and 2 over the crest of the weir:2

221

22

2

121

11

UgzpUgzp ++=++

221

0

2

121

101

)(0)( UhHgUhHggh ++=++

212 2 UghU +=

As a first approximation it is usually assumed that the approach flow U1 is small; (the notch

of the weir is considerably smaller than the cross-section of the channel). Then2/1)2( ghU= , 0 < h


23/33


3.5 Other Applications

3.5.1 Discharge Through an Orifice

Consider water discharging from a small hole at depth h

below the surface of a large tank. Apply Bernoullis theorembetween free surface (1) and exit point (2):

2

2

21

1

2

21 )

()

( UgzpUgzp ++=++

At the free surface, p1 = 0 (free surface) and U1 = 0 (large

reservoir).

At the exit point,p2 = 0 (free discharge).

For convenience, take all heights relative to the orifice. Then2

21 0000 exitUgh ++=++

HenceghUexit 2= (21)

Notes

1. This result is called Torricellis formula and it is basically a statement of energy

conservation (potential energy

kinetic energy).

2. If the orifice is not small then the depth below the free surface and hence the exit

velocity will vary. The quantity of flow must then be obtained by integration.

3. The exit velocity can be multiplied by the exit area to compute discharge. However, in

practice, this ideal volumetric flow rate must be multiplied by a discharge coefficient

(Section 5.3) to account for losses and flow convergence after exit.

Example. Water flows from a 3-cm-diameter hole 2 m below the surface of a large tank.

(a) Assuming no losses calculate the exit velocity and volumetric flow rate.

(b) Using the results of (a) determine the subsequent path of the jet.

(c) If a discharge coefficient cD = 0.6 is applied, calculate the discharge.

h

1

2uexit


24/33


3.5.2 Tank EmptyingBy equating the rate of change of fluid volume in a tank to the volumetric flow rate,

outQt

V=

d

d(22)

the time to empty the tank can be found.

Using Torricellis formula for the exit velocity, and allowing for a discharge coefficient cD

the RHS of (22) can be expanded to give

ghAct

VoutD 2

d

d= (23)

The volume of fluid V is a function of the depth of water h, so this becomes a differential

equation in h.

Note that one assumption of Bernoullis theorem that of steady-state conditions is strictly

not met here!

Example. A tank has a rectangular base with sides 2 m by 3 m and initially contains water to

a depth of 0.5 m. When the plug is pulled out water exits from the base through a hole of area

103

m2. Assuming no losses, estimate the time to empty the tank.


25/33


4. The Momentum Principle

Consideration of momentum is necessary whenever one is considering forces. Two

mechanical principles are involved.

MOMENTUM PRINCIPLE (Newtons 2nd

Law):Force =Rate of change of momentum

ACTION/REACTION (Newtons 3rd

Law):

The force exerted by a fluid on its containment is equal and opposite to that exerted

by its containment on the fluid.

These principles are applied to the whole body of fluid in a control volume. Only external

forces need be considered (since internal forces cancel in action/reaction pairs).

4.1 Steady-Flow Momentum Principle

Consider a control volume consisting of a (thin) stream

tube carrying mass flow rate m .

Since every element of fluid has its velocity changed from

u1 to u2 the net rate of change of momentum in this control

volume is )( 12 uu m . Hence,

)( 12 uuF = m (24)

Note:

1. F is the sum of all external forces on the control volume, including reaction from solid boundaries

fluid forces from adjacent fluid (pressure, viscous forces etc.)

2. Force, velocity and momentum are all vector quantities and each direction must be

considered.

(*) does not generalise easily to non-uniform flow (where velocity varies over a cross-

section). Since mass flux is constant along the stream tube (continuity) a better way of

writing it is

12 )()( uuF mm =

or

force = (momentum flux)out (momentum flux)in

where

u)( uA

velocityfluxmassfluxmomentum

=

=

Can you see how this would generalise to non-uniform inflow or outflow velocity profilesand to time-dependent flow? (These will be addressed in Hydraulics 2).

F

u1

u2


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4.2 Applications

(i) Pipe contraction

(ii) Forces on pipe bends

(iii) Jets and nozzles

(iv) Hydraulic jump (considered as a special application later).

The basic technique is to apply the steady-state momentum principle:

)(

12 uuF = Q

to the fluid in a control volume.

Technique

1. Choose a suitable control volume; (usually coincides with boundaries, so that there is

only one inflow and outflow; cuts fluid flow where it is simple uniform, if possible).

2. Mark in all the forces on the fluid(equal and opposite to that on the boundary; usually

use gauge pressures so that atmospheric pressure is zero).

3. Mark inflow and outflow velocities (and their directions) and use continuity to relate

them if possible.

4. Apply the momentum principle in each relevant direction.

4.2.1 Pipe Contraction

Example. For the pipe-reducing section shown, the greater and smaller diameters are 100 mmand 60 mm, respectively and the water discharges to the atmosphere at section 2. When the

volumetric flow rate is 25 L s1

the differential height in the mercury manometer is

h = 700 mm. Calculate the total force resisted by the flange bolts. (Mercury has specific

gravity 13.6).

h

mercury

water

12


27/33


4.2.2 Forces on Pipe Bends

Example (Exam, June 2004)

Water is flowing at a rate of 0.001 m3

s1

through a reducing bend, with the diameter

contracting from 20 mm to 15 mm and the flow direction changing by 45 as shown in the

figure. The bend is in the horizontal plane and the flow exits to the atmosphere at the narrowend. You may ignore energy losses.

(a) What is the pressure (relative to atmospheric pressure) in the 20 mm pipe?

(b) What is the force on the pipe bend (expressed in thex-y coordinate system shown)?

45ox

y

Example (Exam, May 2003 *** slightly modified ***)

Water is flowing through a horizontal pipe of diameter 15 mm with a flow rate of

0.003 m3

s-1

. At the end of the pipe there is a T-junction (see Figure) with both armshorizontal and each of diameter 10 mm. The water exits the T-junction directly into the

atmosphere.

(a) Calculate the velocity of the water in the 15 mm pipe (u1) and in the two branches of

the T-junction (u2).

(b) If the head-loss coefficient for the T-junction (based on the inflow speed u1) is

K= 1.2, calculate the pressure in the pipe just upstream of the T-junction (ignoring

any other energy losses).

(c) What is the force exerted by the water on the T-junction?

u1

u2

u2


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4.2.3 Jets

The most important aspect of these calculations is that the jet is free and the pressure

throughout is atmospheric.

Example. A jet of diameter 80 mm and carrying a flow of 50 L s1 impacts normally upon aplate. Calculate the force on the plate.

Example. (from White, 2003) A liquid jet of velocity Vand diameterD strikes a fixed hollowcone and deflects back as a conical sheet at the same speed. Find the cone angle

for which

the restraining force 223 AVF= , whereA is the cross-sectional area of the jet.

V

V


29/33


5. Energy

Principle of Conservation of Energy (1st

Law of Thermodynamics):

Change of energy = work done + heat input

This is an important equation for mechanical engineers. For incompressible fluids, however,we do not need the thermal elements of this equation and can work with the much simpler

Mechanical Energy Principle:

Change of kinetic + potential energy = work done by non-conservative forces

You will remember, from Mechanics, that:

work done = force displacement

rate of doing work (i.e.power) = force velocity

Remember that displacement and velocity both refer to components in the direction of

the force.

5.1 Derivation of Bernoullis Equation From an Energy Principle

Consider steady, frictionless flow through a

length of a thin stream tube.

Each mass m of fluid has kinetic energy2

21 mU and potential energy mgz. The rate at which energy passes section 1 or 2 is,

therefore, given by2,1

2

21 )( gzUm +

and hence the rate at which (kinetic + potential) energy is created in this section is given by

1

2

21

2

2

21 )()( gzUmgzUm ++

where

Qm

=

is the mass flow rate. m is the same at both sections since, by definition, there is no flow

through the long sides of the tube.

In the absence of viscosity, work is done only by pressure forces acting on the two ends.

222111 UApUAp

velocityforceworkingofRate

=

=

(No work is done on the long sides of the streamtube because the pressure force is

perpendicular to velocity there).

Hence, applying

rate of working = rate of change of(kinetic + potential) energy

1

2

21

2

2

21

222111 )()( gzUmgzUmUApUAp ++=

Dividing by )( 222111 AUAUm == ,

1221222

1

2

2

1

1 )()(

gzUgzUpp

++=

u1 u21 2


30/33


i.e.

2

2

21

1

2

21 )

()

( gzUp

gzUp

++=++

For incompressible fluids,

is also constant along a streamline and hence

)( 221 streamlineaalongconstantgzUp =++ (25)

Notes

(1) Reminder of the qualifiers:

inviscid(we shall examine how to include losses shortly)

steady (for the unsteady Bernoulli equation see Hydraulics 3)

incompressible (but see note 2 below)

along a streamline

(2) An advantage of the energy-related derivation is that it is easy to incorporate (i) thermal

effects and (ii) frictional losses (or energy input/extraction by pumps/turbines). For thermalflows the total energy is supplemented by the internal energy (per unit mass) e and equation

generalised to read

inputheatdoneworkgzUh +=++2

21 (26)

where h = e +p/

is the enthalpy and the RHS represents energy transferred to the fluid (per

unit mass) by mechanical means (positive for pumps, negative for frictional forces and

turbines) or by heat transfer. The incompressible assumption can be dropped.

5.2 Fluid Head

In fluid mechanics, energy is often measured in length units, referred to as fluid head.

If you lift mass m through height Hyou give it (potential) energy mgH. Thus, the height to

which you lift something is a measure of its energy:

weightunitperenergymg

energypotentialH ==

The same applies in fluid mechanics. An obvious example is a pumped-

storage power station (e.g. Dinorwig) which pumps water to the top of a

hill during the night, storing energy for release to the turbines during the

day. Another example is the constant-head tank which pressurises the hot-

water supply in your house: it is basically the available head which

determines how fast hot water comes out of the taps.

Pressure also represents stored energy, and it is common to measure it in length units; e.g

mm Hg (millimeters of mercury) or metres of water. Atmospheric pressure is about

760 mm Hg or 10 m of water.

Consider Bernoullis equation:

)the(, 02

21 pressuretotalpconstantUgzp =++

Each term represents a form of energy per unit volume. If we divide by the weight per unitvolume (the specific weight

g) we obtain an equivalent form

h


31/33


constantHg

Uz

g

p,

2

2

=++ (27)

His called the total head.

p/

g is thepressure headandp/

g +z is thepiezometric head.U

2/2g is the dynamic head.

Power

For mass m, energy stored = mgH.

gHtime

mass

time

energyPower ==

Hence, in fluid flow if a mass flow rate

Q experiences a change of head Hthen the power

involved is given by

QgHpower

=

This formula is applicable to both pumps and turbines. It may be multiplied by an efficiency

to account for energy losses.

Example. The Hoover dam in Colorado holds back water to a depth of 180 m. If its Francis-

type hydroelectric turbines can pass 1400 cubic metres of water per second with a generating

efficiency of 80%, calculate the power available. (Dont quote me on the flow rate or

efficiency please!)

Key Points

1. In fluid mechanics pressure and energy are commonly measured in length units,

referred to as fluid head.

2. Pressure and head are connected (as in hydrostatics) by

ghp

=

3. Bernoullis equation describes an interchange between different forms of energy:

pressure energy gravitational potential energy kinetic energy

4. In

Hg

Uz

g

p=++

2

2

H is the total head, and is constant unless there are frictional losses (or energy is

input/extracted by pumps/turbines).Hcan be plotted graphically as the energy line.

5. gU 2/2 is the dynamic head. Many energy losses are quantified as multiples of the

dynamic head (see Section 5.3).

6. Power = efficiency

gQH.


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5.3 Departures From Ideal Flow

Many theoretical results are derived for ideal fluids, in particular assuming no frictional

losses and uniform velocity profiles.

In practice, compensation is necessary for non-ideal flow. Important examples include: discharge coefficients to correct the quantity of flow deduced by Bernoullis equation;

loss coefficients to quantify energy losses in pipelines;

momentum and energy coefficients to account for non-uniform velocity profiles.

Such factors are usually determined experimentally and documented in official standards.

5.3.1 Discharge Coefficients

A discharge coefficientcD is the ratio of the actual quantity of flow, Q, to that deduced from

Bernoullis equation on the basis of ideal flow. i.e.

idealDQcQ = (28)

cD can be nearly 1.0 for a well-designed venturi meter, but as low as 0.6 for a weir or sharp-

edged orifice.

5.3.2 Loss Coefficients

These are used to quantify energy losses, particularly in pipelines (Dr Lane-Serff will cover

this in detail in the section on Pipe Flow).

A loss coefficientKis the ratio of energy loss to the kinetic energy, or head loss h to dynamic

head ( )2/2 gV :

g

VKh

2

2

= (29)

For pipe flow there is a gradual energy loss due to pipe friction over a length L of pipe. This

is quantified in terms of afriction factor

:

g

V

D

Lh

2

2

= , i.e. D

LK = (30)

Example. A parks water feature is supplied with water from a large tank via a uniform pipeof length 35 m and diameter 150 mm. There is a free discharge at a point 10 m below the

water level in the tank. Find how long it takes to deliver 2000 L of water:

(a) assuming no losses;

(b) allowing for an entry loss coefficient K= 1.0 and pipe friction factor = 0.05.


33/33

5.3.3 Momentum and Energy CoefficientsThese may be applied to account for non-uniform velocity profiles in the application of theMomentum Principle or Bernoulli equation.

In the control-volume application of the Momentum Principle momentum fluxes are

determined as

velocityfluxmassfluxmomentum =

If the velocity is uniform:

AuuuAfluxmomentum2)

( ==

If the velocity is not uniform then one has to sum over individual small contributions:

= Aufluxmomentum d 2

(For continuous velocity profiles this sum becomes an integral see Hydraulics 2).

Since it is inconvenient to write down a sum or integral each time it is common to introduce a

momentum coefficient

as the ratio of the actual momentum flux to that calculated by

assuming a uniform velocity equal to the average velocity uav; i.e.

)( 2

Aufluxmomentum av= (31)

For laminar pipe flow (see later)

is 4/3. For turbulent pipe flow, however, the velocity

profile is much more uniform and

is only slightly more than 1.

Since the kinetic energy flow past a point is given by )()( 221 uuA a similar energy

coefficient

may be defined, which is the ratio of the average value of3

u to that assuming a

uniform velocity,3

avu .

principles _ hydraulics 1 (hydrodynamics)

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