# fluid mechanics - PDF by noorfahizah

VIEWS: 1,137 PAGES: 94

• pg 1
```									                                                                                     Unit 1
CIVE1400: An Introduction to Fluid Mechanics

Dr P A Sleigh
P.A.Sleigh@leeds.ac.uk

Dr CJ Noakes
C.J.Noakes@leeds.ac.uk

January 2008

Module web site:
www.efm.leeds.ac.uk/CIVE/FluidsLevel1

Unit 1: Fluid Mechanics Basics                             3 lectures
Flow
Pressure
Properties of Fluids
Fluids vs. Solids
Viscosity

Unit 2: Statics                                            3 lectures
Hydrostatic pressure
Manometry/Pressure measurement
Hydrostatic forces on submerged surfaces

Unit 3: Dynamics                                           7 lectures
The continuity equation.
The Bernoulli Equation.
Application of Bernoulli equation.
The momentum equation.
Application of momentum equation.

Unit 4: Effect of the boundary on flow                     4 lectures
Laminar and turbulent flow
Boundary layer theory
An Intro to Dimensional analysis
Similarity
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1        Lecture 1       1
Unit 1
Notes For the First Year Lecture Course:
An Introduction to Fluid Mechanics
School of Civil Engineering, University of Leeds.
CIVE1400 FLUID MECHANICS
Dr Andrew Sleigh
January 2008

Contents of the Course

Objectives:
The course will introduce fluid mechanics and establish its relevance in civil engineering.
Develop the fundamental principles underlying the subject.
Demonstrate how these are used for the design of simple hydraulic components.

Civil Engineering Fluid Mechanics
Why are we studying fluid mechanics on a Civil Engineering course? The provision of adequate
water services such as the supply of potable water, drainage, sewerage is essential for the
development of industrial society. It is these services which civil engineers provide.
Fluid mechanics is involved in nearly all areas of Civil Engineering either directly or indirectly.
Some examples of direct involvement are those where we are concerned with manipulating the
fluid:
Sea and river (flood) defences;
Water distribution / sewerage (sanitation) networks;
Hydraulic design of water/sewage treatment works;
Dams;
Irrigation;
Pumps and Turbines;
Water retaining structures.
And some examples where the primary object is construction - yet analysis of the fluid
mechanics is essential:
Flow of air in buildings;
Flow of air around buildings;
Bridge piers in rivers;
Ground-water flow – much larger scale in time and space.
Notice how nearly all of these involve water. The following course, although introducing general
fluid flow ideas and principles, the course will demonstrate many of these principles through
examples where the fluid is water.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                  Lecture 1          2
Unit 1

Module Consists of:
Lectures:
20 Classes presenting the concepts, theory and application.
Worked examples will also be given to demonstrate how the theory is applied. You will be
asked to do some calculations - so bring a calculator.
Assessment:
1 Exam of 2 hours, worth 80% of the module credits.
This consists of 6 questions of which you choose 4.
2 Multiple choice question (MCQ) papers, worth 10% of the module credits (5% each).
These will be for 30mins and set after the lectures. The timetable for these MCQs and
lectures is shown in the table at the end of this section.
1 Marked problem sheet, worth 10% of the module credits.
Laboratories: 2 x 3 hours
These two laboratory sessions examine how well the theoretical analysis of fluid dynamics
describes what we observe in practice.
During the laboratory you will take measurements and draw various graphs according to the
details on the laboratory sheets. These graphs can be compared with those obtained from
theoretical analysis.
You will be expected to draw conclusions as to the validity of the theory based on the
results you have obtained and the experimental procedure.
After you have completed the two laboratories you should have obtained a greater
understanding as to how the theory relates to practice, what parameters are important in
analysis of fluid and where theoretical predictions and experimental measurements may
differ.
The two laboratories sessions are:
1. Impact of jets on various shaped surfaces - a jet of water is fired at a target
and is deflected in various directions. This is an example of the application of
the momentum equation.
2. The rectangular weir - the weir is used as a flow measuring device. Its
accuracy is investigated. This is an example of how the Bernoulli (energy)
equation is applied to analyses fluid flow.
[As you know, these laboratory sessions are compulsory course-work. You must
attend them. Should you fail to attend either one you will be asked to complete
some extra work. This will involve a detailed report and further questions. The
simplest strategy is to do the lab.]
Homework:
Example sheets: These will be given for each section of the course. Doing these will greatly
improve your exam mark. They are course work but do not have credits toward the module.
Lecture notes: Theses should be studied but explain only the basic outline of the necessary
concepts and ideas.
Books: It is very important do some extra reading in this subject. To do the examples you
will definitely need a textbook. Any one of those identified below is adequate and will also
be useful for the fluids (and other) modules in higher years - and in work.
Example classes:
There will be example classes each week. You may bring any problems/questions you have
about the course and example sheets to these classes.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                      Lecture 1        3
Unit 1
Schedule:
Lecture     Month          Date       Week       Day      Time        Unit
Tue
1         January         15           0       s        3.00 pm     Unit 1: Fluid Mechanic Basics                   Pressure, density
2                         16           0       Wed      9.00 am                                                     Viscosity, Flow
Tue
double lecture
Extra                      22           1       s        3.00 pm     Presentation of Case Studies
3                         23           1       Wed      9.00 am                                                     Flow calculations
Tue
4                         29           2       s        3.00 pm     Unit 2: Fluid Statics                           Pressure
5                         30           2       Wed      9.00 am                                                     Plane surfaces
Tue
6         February         5           3       s        3.00 pm                                                     Curved surfaces
7                          6           3       Wed      9.00 am     Design study 01 - Centre vale park
Tue
8                         12           4       s        3.00 pm     Unit 3: Fluid Dynamics                          General
9                         13           4       Wed      9.00 am                                                     Bernoulli
Tue
10                        19           5       s        3.00 pm                                                     Flow measurement
MCQ                                                      4.00 pm     MCQ
11                        20           5       Wed      9.00 am                                                     Weir
surveyin                             Tue
12        g               26           6       s        3.00 pm                                                     Momentum
13                        27           6       Wed      9.00 am     Design study 02 - Gaunless + Millwood
Tue
12        March            4           7       s        3.00 pm                                                     Applications
13                         5           7       Wed      9.00 am     Design study 02 - Gaunless + Millwood
Tue
14                        11           8       s        3.00 pm                                                     Applications
15                        12         8         Wed      9.00 am     problem sheet given out                         Calculation
Vacatio
n
Tue
16        April           15           9       s        3.00 pm     Unit 4: Effects of the Boundary on Flow         Boundary Layer
17                        16           9       Wed      9.00 am                                                     Friction
Tue
18                        22          10       s        3.00 pm                                                     Dim. Analysis
19                        23          10       Wed      9.00 am     problem sheet handed in                         Dim. Analysis
Tue
20                        29          11       s        3.00 pm     Revision
MCQ                                                      4.00 pm     MCQ
21                        30          11       Wed      9.00 am

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                        Lecture 1            4
Unit 1

Books:
Any of the books listed below are more than adequate for this module. (You will probably not
need any more fluid mechanics books on the rest of the Civil Engineering course)
Mechanics of Fluids, Massey B S., Van Nostrand Reinhold.
Fluid Mechanics, Douglas J F, Gasiorek J M, and Swaffield J A, Longman.
Civil Engineering Hydraulics, Featherstone R E and Nalluri C, Blackwell Science.
Hydraulics in Civil and Environmental Engineering, Chadwick A, and Morfett J., E & FN Spon -
Chapman & Hall.

Online Lecture Notes:
http://www.efm.leeds.ac.uk/cive/FluidsLevel1
There is a lot of extra teaching material on this site: Example sheets, Solutions, Exams,
Detailed lecture notes, Online video lectures, MCQ tests, Images etc. This site DOES NOT
REPLACE LECTURES or BOOKS.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1               Lecture 1       5
Unit 1

Take care with the System of Units
As any quantity can be expressed in whatever way you like it is sometimes easy to become
confused as to what exactly or how much is being referred to. This is particularly true in the field
of fluid mechanics. Over the years many different ways have been used to express the various
quantities involved. Even today different countries use different terminology as well as different
units for the same thing - they even use the same name for different things e.g. an American
pint is 4/5 of a British pint!
To avoid any confusion on this course we will always use the SI (metric) system - which you will
already be familiar with. It is essential that all quantities are expressed in the same system or
the wrong solutions will results.
Despite this warning you will still find that this is the most common mistake when you attempt
example questions.

The SI System of units
The SI system consists of six primary units, from which all quantities may be described. For
convenience secondary units are used in general practice which are made from combinations
of these primary units.

Primary Units
The six primary units of the SI system are shown in the table below:

Quantity                   SI Unit    Dimension
Length                   metre, m         L
Mass                  kilogram, kg      M
Time                   second, s        T
Temperature                 Kelvin, K        θ
Current                  ampere, A         I
Luminosity                  candela        Cd

In fluid mechanics we are generally only interested in the top four units from this table.
Notice how the term 'Dimension' of a unit has been introduced in this table. This is not a
property of the individual units, rather it tells what the unit represents. For example a metre is a
length which has a dimension L but also, an inch, a mile or a kilometre are all lengths so have
dimension of L.
(The above notation uses the MLT system of dimensions, there are other ways of writing
dimensions - we will see more about this in the section of the course on dimensional analysis.)

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                   Lecture 1          6
Unit 1
Derived Units
There are many derived units all obtained from combination of the above primary units. Those
most used are shown in the table below:
Quantity                  SI Unit                    Dimension
Velocity                   m/s            ms-1         LT-1
acceleration                 m/s2           ms-2         LT-2
force                      N
kg m/s2         kg ms-2        M LT-2
energy (or work)              Joule J
N m,
kg m2/s2        kg m2s-2       ML2T-2
power                 Watt W
N m/s          Nms-1
kg m2/s3        kg m2s-3       ML2T-3
pressure ( or stress)           Pascal
P,            Nm-2
N/m2,         kg m-1s-2     ML-1T-2
kg/m/s2
density                   kg/m3          kg m-3         ML-3
specific weight              N/m3
kg/m2/s2        kg m-2s-2      ML-2T-2
relative density             a ratio                          1
no units                    no dimension
viscosity               N s/m2          N sm-2
kg/m s         kg m-1s-1     M L-1T-1
surface tension                N/m            Nm-1
kg /s2         kg s-2        MT-2

The above units should be used at all times. Values in other units should NOT be used without
first converting them into the appropriate SI unit. If you do not know what a particular unit means
- find out, else your guess will probably be wrong.
More on this subject will be seen later in the section on dimensional analysis and similarity.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                              Lecture 1       7
Unit 1
Properties of Fluids: Density
There are three ways of expressing density:

1. Mass density:
ρ = mass per unit volume
mass of fluid
ρ=
volume of fluid
(units: kg/m3)

2. Specific Weight:
(also known as specific gravity)
ω = weight per unit volume
ω = ρg
(units: N/m3 or kg/m2/s2)

3. Relative Density:
σ = ratio of mass density to
a standard mass density
ρsubs tan ce
σ=
ρ          o
H2 O( at 4 c)
For solids and liquids this standard mass density is
the maximum mass density for water (which occurs
o
at 4 c) at atmospheric pressure.
(units: none, as it is a ratio)

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                  Lecture 1       8
Unit 1
Pressure

Convenient to work in terms of pressure, p,
which is the force per unit area.

Force
pressure =
Area over which the force is applied
F
p=
A

Units: Newtons per square metre,
N/m2, kg/m s2 (kg m-1s-2).

Also known as a Pascal, Pa, i.e. 1 Pa = 1 N/m2

Also frequently used is the alternative SI unit the bar,
where 1bar = 105 N/m2
Standard atmosphere = 101325 Pa = 101.325 kPa
1 bar = 100 kPa (kilopascals)
1 mbar = 0.001 bar = 0.1 kPa = 100 Pa

Uniform Pressure:
If the pressure is the same at all points on a surface
uniform pressure

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 1       9
Unit 1
Pascal’s Law: pressure acts equally in all
directions.
ps
B
δz

A                                          δs

px
δy                    F                                C

θ
E                                               D
δx

py

No shearing forces :
All forces at right angles to the surfaces

Summing forces in the x-direction:
Force in the x-direction due to px,
Fx x = p x × Area ABFE = p x δx δy
Force in the x-direction due to ps,
Fx s = − ps × Area ABCD × sin θ
δy
= − psδs δz
δs
= − psδy δz
( sin θ =
δy
δs )
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                      Lecture 1      10
Unit 1
Force in x-direction due to py,
Fx y = 0
To be at rest (in equilibrium) sum of forces is zero
Fx x + Fx s + Fx y = 0
p xδxδy + ( − psδyδz ) = 0
p x = ps

Summing forces in the y-direction.
Force due to py,
Fy = p y × Area EFCD = p yδxδz
y

Component of force due to ps,
Fy = − ps × Area ABCD × cosθ
s
δx
= − psδsδz
δs
= − psδxδz
( cos θ        = δx
δs )

Component of force due to px,
Fy x = 0
Force due to gravity,
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1              Lecture 1      11
Unit 1
weight = - specific weight × volume of element
1
= − ρg × δxδyδz
2
To be at rest (in equilibrium)
Fy + Fy + Fy + weight = 0
y       s      x
⎛      1     ⎞
p yδxδy + ( − psδxδz ) + ⎜ − ρg δxδyδz⎟ = 0
⎝      2     ⎠
The element is small i.e. δx, δx, and δz, are small,
so δx × δy × δz, is very small
and considered negligible, hence
p y = ps

We showed above
px = ps
thus

p x = p y = ps

Pressure at any point is the same in all directions.

This is Pascal’s Law and applies to fluids at rest.

Change of Pressure in the Vertical Direction
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 1      12
Unit 1
p2, A
Area A

Fluid density ρ                                 z2

p1, A   z1

Cylindrical element of fluid, area = A, density = ρ

The forces involved are:
Force due to p1 on A (upward) = p1A
Force due to p2 on A (downward) = p2A
Force due to weight of element (downward)
= mg= density × volume × g
= ρ g A(z2 - z1)

Taking upward as positive, we have
p1 A − p2 A − ρgA( z2 − z1 ) = 0
p2 − p1 = − ρg( z2 − z1 )

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                     Lecture 1      13
Unit 1
In a fluid pressure decreases linearly with
increase in height
p2 − p1 = − ρg( z2 − z1 )
This is the hydrostatic pressure change.

With liquids we normally measure from the
surface.

Measuring h down from the
free surface so that h = -z

z                                           h

y

x

giving p 2 − p1 = ρgh

Surface pressure is atmospheric, patmospheric .
p = ρgh + patmospheric

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1       Lecture 1      14
Unit 1

It is convenient to take atmospheric
pressure as the datum

Pressure quoted in this way is known as
gauge pressure i.e.

Gauge pressure is
pgauge = ρ g h

The lower limit of any pressure is
the pressure in a perfect vacuum.

Pressure measured above
a perfect vacuum (zero)
is known as absolute pressure

Absolute pressure is
pabsolute = ρ g h + patmospheric

Absolute pressure = Gauge pressure + Atmospheric

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1       Lecture 1      15
Unit 1

Pressure density relationship

Boyle’s Law
pV = constant

Ideal gas law
pV = nRT

where
p is the absolute pressure, N/m2, Pa
V is the volume of the vessel, m3
n is the amount of substance of gas, moles
R is the ideal gas constant,
T is the absolute temperature. K

In SI units, R = 8.314472 J mol-1 K-1
(or equivalently m3 Pa K−1 mol−1).

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 1      16
Unit 1

Lecture 2: Fluids vs Solids, Flow

What makes fluid mechanics different
to solid mechanics?

Fluids are clearly different to solids.
But we must be specific.

Need definable basic physical
difference.

Fluids flow under the action of a force,
and the solids don’t - but solids do
deform.

• fluids lack the ability of solids to
resist deformation.
• fluids change shape as long as a
force acts.
Take a rectangular element

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 17
Unit 1

A                      B                        A’               B’    F

F
C                    D                     C                D

Forces acting along edges (faces), such as F,
are know as shearing forces.

A Fluid is a substance which deforms continuously,
or flows, when subjected to shearing forces.

This has the following implications
for fluids at rest:

If a fluid is at rest there are NO shearing forces acting
on it, and
any force must be acting perpendicular to the fluid

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1            Lecture 2 18
Unit 1

Fluids in motion

Consider a fluid flowing near a wall.
- in a pipe for example -

Fluid next to the wall will have zero velocity.

The fluid “sticks” to the wall.

Moving away from the wall velocity increases
to a maximum.

v

Plotting the velocity across the section gives
“velocity profile”

Change in velocity with distance is
dy

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 19
Unit 1

As fluids are usually near surfaces
there is usually a velocity gradient.

Under normal conditions one fluid
particle has a velocity different to its
neighbour.

Particles next to each other with different
velocities exert forces on each other
(due to intermolecular action ) ……

i.e. shear forces exist in a fluid moving
close to a wall.

What if not near a wall?

v

No velocity gradient, no shear forces.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 20
Unit 1

What use is this observation?

It would be useful if we could quantify
this shearing force.

This may give us an understanding of
what parameters govern the forces
different fluid exert on flow.

We will examine the force required to
deform an element.

Consider this 3-d rectangular element,
under the action of the force F.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 21
Unit 1

δx

a                               b
δz

F
A                                B

δy

F
C                               D

under the action of the force F

a     a’                        b   b’

F
A’                       B      B’
A

E

F
C                               D

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1         Lecture 2 22
Unit 1

A 2-d view may be clearer…
A’                      B         B’   F

E x
φ                                     E’

y

F
C                               D
The shearing force acts on the area
A = δz × δx

Shear stress, τ, is the force per unit area:
F
τ =
A

The deformation which shear stress causes is
measured by the angle φ, and is know as
shear strain.

Using these definitions we can amend our
definition of a fluid:

In a fluid φ increases for as long as τ is applied -
the fluid flows
In a solid shear strain, φ, is constant for a fixed
shear stress τ.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1            Lecture 2 23
Unit 1

It has been shown experimentally that the
rate of shear strain is directly
proportional to shear stress

φ
τ∝
time
φ
τ = Constant ×
t

We can express this in terms of the cuboid.

If a particle at point E moves to point E’ in
time t then:

for small deformations
x
shear strain φ =
y
rate of shear strain =

=            =

=
x
(note that          = u is the velocity of the particle at E)
t

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1       Lecture 2 24
Unit 1

So
u
τ = Constant ×
y
u/y is the rate of change of velocity with distance,
du
in differential form this is
dy
The constant of proportionality is known as
the dynamic viscosity, μ.
giving

du
τ =μ
dy
which is know as Newton’s law of viscosity

A fluid which obeys this rule is know as a
Newtonian Fluid

(sometimes also called real fluids)

Newtonian fluids have constant values of μ

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 25
Unit 1

Non-Newtonian Fluids

Some fluids do not have constant μ.
They do not obey Newton’s Law of viscosity.

They do obey a similar relationship and can
be placed into several clear categories
The general relationship is:
n
⎛ δu ⎞
τ = A + B⎜ ⎟
⎝ δy ⎠
where A, B and n are constants.

For Newtonian fluids A = 0, B = μ and n = 1

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 26
Unit 1

This graph shows how μ changes for different fluids.

Bingham plastic                 Pseudo plastic
plastic
Newtonian
Shear stress, τ

Dilatant

Ideal, (τ=0)

Rate of shear, δu/δy
• Plastic: Shear stress must reach a certain minimum before
flow commences.
• Bingham plastic: As with the plastic above a minimum shear
stress must be achieved. With this classification n = 1. An
example is sewage sludge.
• Pseudo-plastic: No minimum shear stress necessary and the
viscosity decreases with rate of shear, e.g. colloidal
substances like clay, milk and cement.
• Dilatant substances; Viscosity increases with rate of shear
e.g. quicksand.
• Thixotropic substances: Viscosity decreases with length of
time shear force is applied e.g. thixotropic jelly paints.
• Rheopectic substances: Viscosity increases with length of
time shear force is applied

• Viscoelastic materials: Similar to Newtonian but if there is a
sudden large change in shear they behave like plastic
Viscosity
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                Lecture 2 27
Unit 1

There are two ways of expressing viscosity

Coefficient of Dynamic Viscosity
τ
μ=
du
dy

Units: N s/m2 or Pa s or kg/m s
The unit Poise is also used where 10 P = 1 Pa·s

Water µ = 8.94 × 10−4 Pa s
Mercury µ = 1.526 × 10−3 Pa s
Olive oil µ = .081 Pa s
Pitch µ = 2.3 × 108 Pa s
Honey µ = 2000 – 10000 Pa s
Ketchup µ = 50000 – 100000 Pa s (non-newtonian)

Kinematic Viscosity

ν = the ratio of dynamic viscosity to mass density
μ
ν=
ρ
Units m2/s
Water ν = 1.7 × 10−6 m2/s.
Air ν = 1.5 × 10−5 m2/s.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 28
Unit 1

Flow rate

Mass flow rate

dm                mass
m=
&     =
dt time taken to accumulate this mass

A simple example:
An empty bucket weighs 2.0kg. After 7 seconds of
collecting water the bucket weighs 8.0kg, then:

mass of fluid in bucket
mass flow rate = m =
&
time taken to collect the fluid
8.0 − 2.0
=
7
= 0.857kg / s

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 29
Unit 1

Volume flow rate - Discharge.

More commonly we use volume flow rate
Also know as discharge.

The symbol normally used for discharge is Q.

volume of fluid
discharge, Q =
time

A simple example:
If the bucket above fills with 2.0 litres in 25 seconds,
what is the discharge?

2.0 × 10 − 3 m3
Q=
25 sec
= 0.0008 m3 / s
= 0.8 l / s

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 30
Unit 1

Discharge and mean velocity

If we know the discharge and the diameter of a
pipe, we can deduce the mean velocity

um t

x                                                  area A
Pipe                                  Cylinder of fluid

Cross sectional area of pipe is A
Mean velocity is um.

In time t, a cylinder of fluid will pass point X with
a volume A× um × t.

The discharge will thus be

volume A × um × t
Q=        =
time      t
Q = Aum

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                Lecture 2 31
Unit 1

A simple example:
If A = 1.2×10-3m2
And discharge, Q is 24 l/s,
mean velocity is
Q
um =
A
2.4 × 10 − 3
=
12 × 10 − 3
.
= 2.0 m / s

Note how we have called this the mean velocity.

This is because the velocity in the pipe is not
constant across the cross section.

x

u
um   umax

This idea, that mean velocity multiplied by the
area gives the discharge, applies to all situations
- not just pipe flow.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 32
Unit 1

Continuity
This principle of conservation of mass says matter
cannot be created or destroyed

This is applied in fluids to fixed volumes, known as
control volumes (or surfaces)

Mass flow in
Control
volume

Mass flow out

For any control volume the principle of conservation of
mass says
Mass entering =                    Mass leaving + Increase
per unit time                      per unit time  of mass in
control vol
per unit time

For steady flow there is no increase in the mass within
the control volume, so
Mass entering =                             Mass leaving

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 33
Unit 1

In a real pipe (or any other vessel) we use the
mean velocity and write

ρ1 A1um1 = ρ2 A2 um2 = Constant = m
&

For incompressible, fluid ρ1 = ρ2 = ρ
(dropping the m subscript)

A1u1 = A2 u2 = Q

This is the continuity equation most often used.

This equation is a very powerful tool.

It will be used repeatedly throughout the rest of
this course.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 2 34
Unit 1

Lecture 3: Examples from
Unit 1: Fluid Mechanics Basics

Units
1.
A water company wants to check that it will have sufficient water if there is a prolonged drought
in the area. The region it covers is 500 square miles and various different offices have sent in
the following consumption figures. There is sufficient information to calculate the amount of
water available, but unfortunately it is in several different units.
Of the total area 100 000 acres are rural land and the rest urban. The density of the urban
population is 50 per square kilometre. The average toilet cistern is sized 200mm by 15in by
0.3m and on average each person uses this 3 time per day. The density of the rural population
is 5 per square mile. Baths are taken twice a week by each person with the average volume of
water in the bath being 6 gallons. Local industry uses 1000 m3 per week. Other uses are
estimated as 5 gallons per person per day. A US air base in the region has given water use
figures of 50 US gallons per person per day.
The average rain fall in 1in per month (28 days). In the urban area all of this goes to the river
while in the rural area 10% goes to the river 85% is lost (to the aquifer) and the rest goes to the
one reservoir which supplies the region. This reservoir has an average surface area of 500
acres and is at a depth of 10 fathoms. 10% of this volume can be used in a month.
a) What is the total consumption of water per day?
b) If the reservoir was empty and no water could be taken from the river, would there be
enough water if available if rain fall was only 10% of average?

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1             Lecture 1      35
Unit 1

Fluid Properties
1. The following is a table of measurement for a fluid at constant temperature.
Determine the dynamic viscosity of the fluid.
du/dy (s-1)    0.0 0.2 0.4 0.6 0.8
τ (N m )
-2
0.0 1.0 1.9 3.1 4.0

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 1      36
Unit 1

2. The density of an oil is 850 kg/m3. Find its relative density and
Kinematic viscosity if the dynamic viscosity is 5 × 10-3 kg/ms.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 1      37
Unit 1

3. The velocity distribution of a viscous liquid (dynamic viscosity μ = 0.9 Ns/m )
2

flowing over a fixed plate is given by u = 0.68y - y2 (u is velocity in m/s and y is
the distance from the plate in m).
What are the shear stresses at the plate surface and at y=0.34m?

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1     Lecture 1       38
Unit 1

4. 5.6m3 of oil weighs 46 800 N. Find its mass density, ρ and relative density, γ.

5.       From table of fluid properties the viscosity of water is given as 0.01008
poises. What is this value in Ns/m2 and Pa s units?

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1    Lecture 1      39
Unit 1

6.       In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125m/s.
The fluid has absolute viscosity 0.048 Pa s and relative density 0.913. What is the
velocity gradient and shear stress at the boundary assuming a linear velocity distribution.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1              Lecture 1      40
Unit 1

Continuity

Section 1                            Section 2

A liquid is flowing from left to right.

By continuity
A1u1ρ1 = A2 u2 ρ2

As we are considering a liquid (incompressible),
ρ1 = ρ2 = ρ
Q1 = Q2
A1u1 = A2u2

If the area A1=10×10-3 m2 and A2=3×10-3 m2
And the upstream mean velocity u1=2.1 m/s.

What is the downstream mean velocity?

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1               Lecture 1      41
Unit 1

Now try this on a diffuser, a pipe which expands or diverges
as in the figure below,

Section 1                 Section 2

If d1=30mm and d2=40mm and the velocity u2=3.0m/s.

What is the velocity entering the diffuser?

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1               Lecture 1      42
Unit 1

Velocities in pipes coming from a junction.

2

1

3

mass flow into the junction = mass flow out

ρ1Q1 = ρ2Q2 + ρ3Q3

When incompressible

Q1 = Q2 + Q3

Α1u1 = Α2u2 + Α3u3

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1           Lecture 1      43
Unit 1

If pipe 1 diameter = 50mm, mean velocity 2m/s, pipe 2 diameter
40mm takes 30% of total discharge and pipe 3 diameter 60mm.
What are the values of discharge and mean velocity in each pipe?

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lecture 1      44
CIVE1400: Fluid Mechanics                                              Section 2: Statics   CIVE1400: Fluid Mechanics                                           Section 2: Statics

We have the vertical pressure relationship                                                          It is convenient to take atmospheric
dp                                                                      pressure as the datum
g,
dz
integrating gives                                                                               Pressure quoted in this way is known as
p = - gz + constant                                                                       gauge pressure i.e.
Gauge pressure is
measuring z from the free surface so that z = -h                                                                             pgauge =        gh

z                                    h
The lower limit of any pressure is
y
the pressure in a perfect vacuum.
x

Pressure measured above
p           gh constant                                                            a perfect vacuum (zero)
is known as absolute pressure
surface pressure is atmospheric, patmospheric .
Absolute pressure is
patmospheric             constant                                                                pabsolute =        g h + patmospheric
so
Absolute pressure = Gauge pressure + Atmospheric
p         gh        patmospheric

CIVE1400: Fluid Mechanics                                        Section 2: Statics 35      CIVE1400: Fluid Mechanics                                       Section 2: Statics 36

CIVE1400: Fluid Mechanics                                              Section 2: Statics   CIVE1400: Fluid Mechanics                                           Section 2: Statics

A gauge pressure can be given                                                      Pressure Measurement By Manometer
using height of any fluid.
p      gh                                            Manometers use the relationship between pressure
This vertical height is the head.

The Piezometer Tube Manometer
If pressure is quoted in head,
the density of the fluid must also be given.                                                  The simplest manometer is an open tube.
Example:                                                                                     This is attached to the top of a container with liquid
What is a pressure of 500 kNm-2 in                                                               at pressure. containing liquid at a pressure.
head of water of density, = 1000 kgm-3
Use p = gh,
p        500 103                                                                                    h1                         h2
h                                         50.95m of water
g       1000 9.81
A
In head of Mercury density                         = 13.6 103 kgm-3.
3
500 10                                                                                                    B
h                                        3.75m of Mercury
3
13.6 10                 9.81
In head of a fluid with relative density = 8.7.
remember            =          water)
The tube is open to the atmosphere,
The pressure measured is relative to
500 103
h                      586m of fluid
.                                          = 8.7                  atmospheric so it measures gauge pressure.
8.7 1000 9.81

CIVE1400: Fluid Mechanics                                        Section 2: Statics 37      CIVE1400: Fluid Mechanics                                       Section 2: Statics 38
CIVE1400: Fluid Mechanics                                                    Section 2: Statics   CIVE1400: Fluid Mechanics                                            Section 2: Statics

An Example of a Piezometer.
Pressure at A = pressure due to column of liquid h1                                         What is the maximum gauge pressure of water that
can be measured by a Piezometer of height 1.5m?
And if the liquid had a relative density of 8.5 what
pA =         g h1                                                would the maximum measurable gauge pressure?

Pressure at B = pressure due to column of liquid h2

pB =         g h2

Problems with the Piezometer:

1. Can only be used for liquids

2. Pressure must above atmospheric

3. Liquid height must be convenient
i.e. not be too small or too large.

CIVE1400: Fluid Mechanics                                              Section 2: Statics 39      CIVE1400: Fluid Mechanics                                      Section 2: Statics 40

CIVE1400: Fluid Mechanics                                                    Section 2: Statics   CIVE1400: Fluid Mechanics                                            Section 2: Statics

Equality Of Pressure At
The Same Level In A Static Fluid
P                                                 Q

Fluid density ρ
Area A

z                                              z
pl, A                                                pr, A

Face L                          Face R                                                L                                                R
weight, mg

Horizontal cylindrical element
We have shown
cross sectional area = A
mass density =                                                                                            pl = pr
left end pressure = pl                                                For a vertical pressure change we have
right end pressure = pr                                                                                   pl    pp        gz
and
For equilibrium the sum of the                                                                                     pr    pq        gz
forces in the x direction is zero.
so
pl A = pr A                                                                              pp        gz    pq        gz
pp    pq
pl = pr

Pressure in the horizontal direction is constant.                                                   Pressure at the two equal levels are the same.

This true for any continuous fluid.

CIVE1400: Fluid Mechanics                                              Section 2: Statics 31      CIVE1400: Fluid Mechanics                                      Section 2: Statics 32
CIVE1400: Fluid Mechanics                                                 Section 2: Statics   CIVE1400: Fluid Mechanics                            Section 2: Statics

The “U”-Tube Manometer                                                                                       We know:

“U”-Tube enables the pressure of both liquids                                                          Pressure in a continuous static fluid
and gases to be measured                                                                      is the same at any horizontal level.
“U” is connected as shown and filled with
manometric fluid.                                                                                 pressure at B = pressure at C
pB = pC
Important points:
1. The manometric fluid density should be                                                 For the left hand arm
greater than of the fluid measured.                                                           pressure at B = pressure at A + pressure of height of
man >                                                                                                              liquid being measured

2. The two fluids should not be able to mix                                                                       pB = pA + gh1
they must be immiscible.
For the right hand arm
Fluid density ρ                                                                             pressure at C = pressure at D + pressure of height of
D
manometric liquid
pC = patmospheric +      man gh2
h2
A
h1                               We are measuring gauge pressure we can subtract
B                             C           patmospheric giving
pB = pC
Manometric fluid density ρ
man
pA =    man gh2   - gh1

CIVE1400: Fluid Mechanics                                       Section 2: Statics 41          CIVE1400: Fluid Mechanics                        Section 2: Statics 42

CIVE1400: Fluid Mechanics                                                 Section 2: Statics   CIVE1400: Fluid Mechanics                            Section 2: Statics

An example of the U-Tube manometer.
What if the fluid is a gas?                                                    Using a u-tube manometer to measure gauge
pressure of fluid density = 700 kg/m3, and the
manometric fluid is mercury, with a relative density
Nothing changes.                                                      of 13.6.
What is the gauge pressure if:
a) h1 = 0.4m and h2 = 0.9m?
The manometer work exactly the same.                                                         b) h1 stayed the same but h2 = -0.1m?

BUT:

As the manometric fluid is liquid
(usually mercury , oil or water)

And Liquid density is much
greater than gas,

man    >>

gh1 can be neglected,

and the gauge pressure given by

pA =     man gh2

CIVE1400: Fluid Mechanics                                       Section 2: Statics 43          CIVE1400: Fluid Mechanics                        Section 2: Statics 44
CIVE1400: Fluid Mechanics                                                              Section 2: Statics   CIVE1400: Fluid Mechanics                                           Section 2: Statics

Pressure difference measurement                                                                                  pressure at C = pressure at D
Using a “U”-Tube Manometer.
pC = pD
The “U”-tube manometer can be connected
at both ends to measure pressure difference between                                                                                     pC = pA +             g ha
these two points

pD = pB +                g (hb + h) +           man g     h
B

pA +         g ha = pB +                g (hb + h) +             man g      h
Fluid density ρ
Giving the pressure difference
hb
E

pA - pB =              g (hb - ha) + (           man    - )g h
h

A

ha                                                                          Again if the fluid is a gas man >> , then the terms
C                  D                            involving can be neglected,

pA - pB =         man   gh
Manometric fluid density ρman

CIVE1400: Fluid Mechanics                                                       Section 2: Statics 45       CIVE1400: Fluid Mechanics                                      Section 2: Statics 46

CIVE1400: Fluid Mechanics                                                              Section 2: Statics   CIVE1400: Fluid Mechanics                                           Section 2: Statics

An example using the u-tube for pressure                                                                 Advances to the “U” tube manometer
difference measuring
In the figure below two pipes containing the same
fluid of density = 990 kg/m3 are connected using a                                                                     Problem: Two reading are required.
u-tube manometer.                                                                                                      Solution: Increase cross-sectional area
What is the pressure between the two pipes if the                                                                                of one side.
manometer contains fluid of relative density 13.6?
Fluid density ρ                                                                                              Result:   One level moves
much more than the other.
Fluid density ρ
A

p1                                           p2
B

ha = 1.5m
E

hb = 0.75m
diameter D
h = 0.5m
diameter d
z2
Datum line
C              D                                                                    z1

Manometric fluid density ρman = 13.6 ρ

If the manometer is measuring the pressure
difference of a gas of (p1 - p2) as shown,
we know
p1 - p2 = man g h

CIVE1400: Fluid Mechanics                                                       Section 2: Statics 47       CIVE1400: Fluid Mechanics                                      Section 2: Statics 48
CIVE1400: Fluid Mechanics                                                              Section 2: Statics   CIVE1400: Fluid Mechanics                            Section 2: Statics

volume of liquid moved from
the left side to the right
= z2 ( d2 / 4)                                                                               Problem: Small pressure difference,

The fall in level of the left side is
Volume moved
z1                                                                                   Solution 1: Reduce density of manometric
Area of left side                                                                        fluid.
z2 d 2 / 4
D2 / 4                                                                Result:           Greater height change -
2
z2
D
Putting this in the equation,
2
d                                      Solution 2: Tilt one arm of the manometer.
p1           p2            g z2              z2
D
2
d
gz 2 1                                                            Result:           Same height change - but larger
D
movement along the
If D >> d then (d/D)2 is very small so                                                                                             manometer arm - easier to read.
p1          p2                gz2

Inclined manometer
CIVE1400: Fluid Mechanics                                                          Section 2: Statics 49    CIVE1400: Fluid Mechanics                        Section 2: Statics 50

CIVE1400: Fluid Mechanics                                                              Section 2: Statics   CIVE1400: Fluid Mechanics                            Section 2: Statics
Example of an inclined manometer.
p1
p2                    An inclined manometer is required to measure an air
diameter d
pressure of 3mm of water to an accuracy of +/- 3%.
The inclined arm is 8mm in diameter and the larger
diameter D
er
x                                     arm has a diameter of 24mm. The manometric fluid
eR
e
has density man = 740 kg/m3 and the scale may be
al                           z2
Datum line
z1                                                                                 What is the angle required to ensure the desired
accuracy may be achieved?
θ

The pressure difference is still given by the
height change of the manometric fluid.

p1          p2              gz2
but,
z2          x sin
p1          p2           gx sin

The sensitivity to pressure change can be increased
further by a greater inclination.

CIVE1400: Fluid Mechanics                                                          Section 2: Statics 51    CIVE1400: Fluid Mechanics                        Section 2: Statics 52
CIVE1400: Fluid Mechanics                              Section 2: Statics   CIVE1400: Fluid Mechanics                  Section 2: Statics

Choice Of Manometer                                      Forces on Submerged Surfaces in Static Fluids

Take care when fixing the manometer to vessel                              We have seen these features of static fluids
Burrs cause local pressure variations.
Hydrostatic vertical pressure distribution
Disadvantages:                                     Pressures at any equal depths in a continuous
Slow response - only really useful for very slowly                          fluid are equal
varying pressures - no use at all for fluctuating
pressures;                                                                  Pressure at a point acts equally in all
directions (Pascal’s law).
For the “U” tube manometer two measurements
must be taken simultaneously to get the h value.                            Forces from a fluid on a boundary acts at right
It is often difficult to measure small variations in                        angles to that boundary.
pressure.
It cannot be used for very large pressures unless
several manometers are connected in series;
Fluid pressure on a surface
For very accurate work the temperature and
relationship between temperature and must be                                   Pressure is force per unit area.
known;                                                                   Pressure p acting on a small area A exerted
force will be
They are very simple.
F=p   A
No calibration is required - the pressure can be
calculated from first principles.
Since the fluid is at rest the force will act at
right-angles to the surface.

CIVE1400: Fluid Mechanics                          Section 2: Statics 53    CIVE1400: Fluid Mechanics             Section 2: Statics 54

CIVE1400: Fluid Mechanics                              Section 2: Statics   CIVE1400: Fluid Mechanics                  Section 2: Statics

General submerged plane                                                    Horizontal submerged plane
F =p δA1
1 1
F =p δA
2 2 2
The pressure, p, will be equal at all points of
F =p δA
n n n                                                                          the surface.
The resultant force will be given by
R pressure      area of plane
The total or resultant force, R, on the                                          R = pA
plane is the sum of the forces on the
small elements i.e.                                                           Curved submerged surface
R      p1 A1       p 2 A2        p n An       p A
and                                                          Each elemental force is a different
This resultant force will act through the                                    magnitude and in a different direction (but
centre of pressure.                                                      still normal to the surface.).

It is, in general, not easy to calculate the
For a plane surface all forces acting
resultant force for a curved surface by
can be represented by one single                                                combining all elemental forces.
resultant force,
acting at right-angles to the plane                                     The sum of all the forces on each element
through the centre of pressure.                                         will always be less than the sum of the
individual forces, p A .

CIVE1400: Fluid Mechanics                          Section 2: Statics 55    CIVE1400: Fluid Mechanics             Section 2: Statics 56
CIVE1400: Fluid Mechanics                                                     Section 2: Statics   CIVE1400: Fluid Mechanics                                   Section 2: Statics

Resultant Force and Centre of Pressure on a                                                                                 z A is known as
general plane surface in a liquid.
O
the 1st Moment of Area of the
O
θ
Fluid                                     Q                                  elemental
density ρ                         z
area δA
Resultant
z
s                                                          plane PQ about the free surface.
Force R D
G
area δA                               G
x

C                      Sc                area A      d
And it is known that
P
x                                                     z A      Az

Take pressure as zero at the surface.
A is the area of the plane
Measuring down from the surface, the pressure on                                                       z is the distance to the centre of gravity
an element A, depth z,
(centroid)
p = gz
In terms of distance from point O
z A      Ax sin
So force on element
F = gz A                                                                                   = 1st moment of area                sin
Resultant force on plane
(as   z     x sin )

R             g         z A
The resultant force on a plane
(assuming             and g as constant).                                                                                       R      gAz
gAx sin
CIVE1400: Fluid Mechanics                                             Section 2: Statics 57        CIVE1400: Fluid Mechanics                              Section 2: Statics 58

CIVE1400: Fluid Mechanics                                                     Section 2: Statics   CIVE1400: Fluid Mechanics                                   Section 2: Statics

This resultant force acts at right angles                                                                 Sum of moments                  g sin      s2 A
through the centre of pressure, C, at a depth D.

Moment of R about O =              R      Sc =    gAx sin S c
How do we find this position?

Take moments of the forces.                                                      Equating
gAx sin S c           g sin       s2 A
As the plane is in equilibrium:
The moment of R will be equal to the sum of the
The position of the centre of pressure along the
moments of the forces on all the elements A                                                       plane measure from the point O is:
s2 A
Sc
It is convenient to take moment about O                                                                                              Ax

The force on each elemental area:                                                                                           How do we work out
Force on A                        gz A                                                                the summation term?
g s sin    A
This term is known as the
the moment of this force is:
2nd Moment of Area , Io,
Moment of Force on A about O                                   g s sin             A       s                               of the plane
2
g sin              As                                (about the axis through O)
, g and are the same for each element, giving the
total moment as

CIVE1400: Fluid Mechanics                                             Section 2: Statics 59        CIVE1400: Fluid Mechanics                              Section 2: Statics 60
CIVE1400: Fluid Mechanics                                          Section 2: Statics   CIVE1400: Fluid Mechanics                                          Section 2: Statics

2nd moment of area about O                         Io       s2 A
How do you calculate the 2nd moment of
area?
It can be easily calculated
for many common shapes.
2nd moment of area is a geometric property.

It can be found from tables -
The position of the centre of pressure
along the plane measure from the point O is:
an axis through its centroid = IGG.

2 nd Moment of area about a line through O
Sc                                                                                        Usually we want the 2nd moment of area
1st Moment of area about a line through O

and
Through O in the above examples.

Depth to the centre of pressure is                                                                             We can use the
parallel axis theorem
D S c sin                                                                     to give us what we want.

CIVE1400: Fluid Mechanics                                      Section 2: Statics 61    CIVE1400: Fluid Mechanics                                      Section 2: Statics 62

CIVE1400: Fluid Mechanics                                          Section 2: Statics   CIVE1400: Fluid Mechanics                                          Section 2: Statics
nd
The 2 moment of area about a line
The parallel axis theorem can be written                                                through the centroid of some common
Io     I GG       Ax 2
shapes.

Shape                                  Area A    2nd moment of area, I GG ,
We then get the following                                                                                                  an axis through the centroid
equation for the                                                     Rectangle
b
position of the centre of pressure                                                                                    bd                 bd 3
h                                                       12
G                                 G
I GG
Sc                 x
Ax
I GG                                       Triangle
D         sin             x                                                                         bd                 bd 3
h

Ax                                        G
h/3
G

b
2                  36
Circle
(In the examination the parallel axis theorem
G
R
G      R2                  R4
and the I GG will be given)
4
Semicircle
R2
01102 R 4
R
G
(4R)/(3π)
2               .

CIVE1400: Fluid Mechanics                                      Section 2: Statics 63    CIVE1400: Fluid Mechanics                                      Section 2: Statics 64
CIVE1400: Fluid Mechanics                                  Section 2: Statics   CIVE1400: Fluid Mechanics                                      Section 2: Statics

An example:
Find the moment required to keep this triangular                                               Submerged vertical surface -
gate closed on a tank which holds water.                                                           Pressure diagrams

1.2m
For vertical walls of constant width
D
2.0m
it is possible to find the resultant force and
centre of pressure graphically using a
G                            1.5m                                              pressure diagram.
C

We know the relationship between
pressure and depth:
p = gz

So we can draw the diagram below:

z                    ρgz

H            2H
3
R
p

ρgH

This is know as a pressure diagram.

CIVE1400: Fluid Mechanics                            Section 2: Statics 65      CIVE1400: Fluid Mechanics                                  Section 2: Statics 66

CIVE1400: Fluid Mechanics                                  Section 2: Statics   CIVE1400: Fluid Mechanics                                      Section 2: Statics

Pressure increases from zero at the
surface linearly by p = gz, to a
maximum at the base of p = gH.                                            For a triangle the centroid is at 2/3 its height
i.e. the resultant force acts
2
The area of this triangle represents the                                         horizontally through the point z        H.
3
resultant force per unit width on the
vertical wall,                                                               For a vertical plane the
depth to the centre of pressure is given by
Units of this are Newtons per metre.
1                                                                                     2
Area       AB BC                                                                    D          H
2                                                                                     3
1
H gH
2
1
gH 2
2
Resultant force per unit width
1
R        gH 2   ( N / m)
2

The force acts through the centroid of
the pressure diagram.
CIVE1400: Fluid Mechanics                            Section 2: Statics 67      CIVE1400: Fluid Mechanics                                  Section 2: Statics 68
CIVE1400: Fluid Mechanics                                     Section 2: Statics   CIVE1400: Fluid Mechanics                                           Section 2: Statics

Check this against                                          The same technique can be used with combinations
the moment method:                                           of liquids are held in tanks (e.g. oil floating on water).
For example:

The resultant force is given by:                                                                   oil ρo             0.8m
D
R           gAz       gAx sin
1.2m
H                                                 water ρ                        R
g H 1       sin
2
1
gH 2                                                                                             ρg0.8       ρg1.2

2                                                  Find the position and magnitude of the resultant
and the depth to the centre of pressure by:                                        force on this vertical wall of a tank which has oil
Io                                   floating on water as shown.
D     sin
Ax
and by the parallel axis theorem (with width of 1)
Io         I GG     Ax 2
2
1 H3     H                   H3
1 H
12      2                   3
Depth to the centre of pressure

H3 / 3        2
D         2
H
H /2          3

CIVE1400: Fluid Mechanics                                 Section 2: Statics 69    CIVE1400: Fluid Mechanics                                    Section 2: Statics 70

CIVE1400: Fluid Mechanics                                     Section 2: Statics   CIVE1400: Fluid Mechanics                                           Section 2: Statics

Submerged Curved Surface
In the diagram below liquid is resting on
If the surface is curved the resultant force                                                top of a curved base.
must be found by combining the elemental
forces using some vectorial method.                                                                         E                        D

Calculate the                                                                       C
B

horizontal and vertical                                                                                  G

FAC           O                     RH
components.
A

Combine these to obtain the resultant                                                                                  Rv                   R

force and direction.
The fluid is at rest – in equilibrium.

So any element of fluid
(Although this can be done for all three                                           such as ABC is also in equilibrium.
dimensions we will only look at one vertical
plane)

CIVE1400: Fluid Mechanics                                 Section 2: Statics 71    CIVE1400: Fluid Mechanics                                    Section 2: Statics 72
CIVE1400: Fluid Mechanics                                     Section 2: Statics   CIVE1400: Fluid Mechanics                         Section 2: Statics

Consider the Horizontal forces
The resultant horizontal force of a fluid
above a curved surface is:
The sum of the horizontal forces is zero.
C
RH = Resultant force on the projection of the
B
curved surface onto a vertical plane.
FAC                               RH
We know
A                                                   1. The force on a vertical plane must act
horizontally (as it acts normal to the plane).
No horizontal force on CB as there are                                        2. That RH must act through the same point.
no shear forces in a static fluid
So:
Horizontal forces act only on the faces                                      RH acts horizontally through the centre of
AC and AB as shown.                                                        pressure of the projection of
the curved surface onto an vertical plane.
FAC, must be equal and opposite to RH.
We have seen earlier how to calculate
AC is the projection of the curved surface                                              resultant forces and point of action.
AB onto a vertical plane.
Hence we can calculate the resultant
horizontal force on a curved surface.

CIVE1400: Fluid Mechanics                                 Section 2: Statics 73    CIVE1400: Fluid Mechanics                     Section 2: Statics 74

CIVE1400: Fluid Mechanics                                     Section 2: Statics   CIVE1400: Fluid Mechanics                         Section 2: Statics

Consider the Vertical forces                                                                      Resultant force
The sum of the vertical forces is zero.
E                D
The overall resultant force is found by
combining the vertical and horizontal
C
B
components vectorialy,
G

Resultant force
2     2
A
R    RH    RV
Rv

There are no shear force on the vertical edges,                                   And acts through O at an angle of .
so the vertical component can only be due to
the weight of the fluid.                                              The angle the resultant force makes to the
horizontal is
So we can say                                                                              R
The resultant vertical force of a fluid above a                                                    tan 1 V
RH
curved surface is:

RV = Weight of fluid directly above the curved
surface.                                                        The position of O is the point of interaction of
the horizontal line of action of R H and the
It will act vertically down through the centre of                                           vertical line of action of RV .
gravity of the mass of fluid.

CIVE1400: Fluid Mechanics                                 Section 2: Statics 75    CIVE1400: Fluid Mechanics                     Section 2: Statics 76
CIVE1400: Fluid Mechanics                                       Section 2: Statics   CIVE1400: Fluid Mechanics                                Section 2: Statics
A typical example application of this is the
determination of the forces on dam walls or curved
What are the forces if the fluid is below the
sluice gates.
curved surface?
Find the magnitude and direction of the                                                     This situation may occur or a curved sluice gate.
resultant force of water on a quadrant gate as
shown below.                                                                                                      C
Gate width 3.0m                                                                       B
G
1.0m

FAC             O           RH

Water ρ = 1000 kg/m3

A

Rv          R

The force calculation is very similar to
when the fluid is above.

CIVE1400: Fluid Mechanics                                   Section 2: Statics 77    CIVE1400: Fluid Mechanics                            Section 2: Statics 78

CIVE1400: Fluid Mechanics                                       Section 2: Statics   CIVE1400: Fluid Mechanics                                Section 2: Statics

Horizontal force                                                                     Vertical force
C
B
G
B

FAC              O                   RH

A

A                    A’                                                                    Rv

What vertical force would
The two horizontal on the element are:
The horizontal reaction force RH                                                      keep this in equilibrium?
The force on the vertical plane A’B.
If the region above the curve were all
The resultant horizontal force, RH acts as shown in                                         water there would be equilibrium.
the diagram. Thus we can say:

Hence: the force exerted by this amount of fluid
must equal he resultant force.
The resultant horizontal force of a fluid below a
curved surface is:
RH = Resultant force on the projection of the
curved surface onto a vertical plane.                                             The resultant vertical force of a fluid below a
curved surface is:
Rv =Weight of the imaginary volume of fluid
vertically above the curved surface.

CIVE1400: Fluid Mechanics                                   Section 2: Statics 79    CIVE1400: Fluid Mechanics                            Section 2: Statics 80
CIVE1400: Fluid Mechanics                      Section 2: Statics   CIVE1400: Fluid Mechanics                       Section 2: Statics

The resultant force and direction of application                   An example of a curved sluice gate which
are calculated in the same way as for fluids                      experiences force from fluid below.
above the surface:                                  A 1.5m long cylinder lies as shown in the figure,
holding back oil of relative density 0.8. If the cylinder
has a mass of 2250 kg find
a) the reaction at A    b) the reaction at B
Resultant force                                                                                 E
C
2    2
R    RH   RV
A
D

B

And acts through O at an angle of .
The angle the resultant force makes to the horizontal
is
R
tan 1 V
RH

CIVE1400: Fluid Mechanics                  Section 2: Statics 81    CIVE1400: Fluid Mechanics                   Section 2: Statics 82
Unit 3: Fluid Dynamics                                                                    Unit 3: Fluid Dynamics

CIVE1400: An Introduction to Fluid Mechanics                                                                                                    Fluid Dynamics

Unit 3: Fluid Dynamics                                                                                                      Objectives

Dr P A Sleigh:             P.A.Sleigh@leeds.ac.uk
Dr CJ Noakes: C.J.Noakes@leeds.ac.uk
1.Identify differences between:
January 2008
uniform/non-uniform
Module web site: www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                           compressible/incompressible flow
Unit 1: Fluid Mechanics Basics                                             3 lectures
Flow                                                                                         2.Demonstrate streamlines and stream tubes
Pressure
Properties of Fluids
Fluids vs. Solids
Viscosity                                                                                    3.Introduce the Continuity principle
Unit 2: Statics                                                            3 lectures
Hydrostatic pressure                                                                         4.Derive the Bernoulli (energy) equation
Manometry / Pressure measurement
Hydrostatic forces on submerged surfaces

Unit 3: Dynamics                                                           7 lectures              5.Use the continuity equations to predict pressure
The continuity equation.                                                                      and velocity in flowing fluids
The Bernoulli Equation.
Application of Bernoulli equation.
The momentum equation.
Application of momentum equation.                                                            6.Introduce the momentum equation for a fluid
Unit 4: Effect of the boundary on flow                                     4 lectures
Laminar and turbulent flow                                                                   7.Demonstrate use of the momentum equation to
Boundary layer theory
An Intro to Dimensional analysis                                                              predict forces induced by flowing fluids
Similarity

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                      Lecture 8         98    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8         99

Unit 3: Fluid Dynamics                                                                    Unit 3: Fluid Dynamics

Fluid dynamics:                                                                                        Flow Classification
Fluid flow may be
The analysis of fluid in motion                                                                  classified under the following headings

Fluid motion can be predicted in the                                                                          uniform:
same way as the motion of solids                                                       Flow conditions (velocity, pressure, cross-section or
depth) are the same at every point in the fluid.
By use of the fundamental laws of physics and the                                                                             non-uniform:
physical properties of the fluid                                                                  Flow conditions are not the same at every point.

Some fluid flow is very complex:                                                                              steady
e.g.                                                                        Flow conditions may differ from point to point but
Spray behind a car                                                                           DO NOT change with time.
waves on beaches;
Flow conditions change with time at any point.
any other atmospheric phenomenon

Fluid flowing under normal circumstances
All can be analysed
- a river for example -
with varying degrees of success                                                                      conditions vary from point to point
(in some cases hardly at all!).                                                                          we have non-uniform flow.

There are many common situations                                                                    If the conditions at one point vary as time passes
which analysis gives very accurate predictions                                                                      then we have unsteady flow.

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                      Lecture 8        100    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        101
Unit 3: Fluid Dynamics                                                                    Unit 3: Fluid Dynamics

Combining these four gives.                                                               Compressible or Incompressible Flow?

Steady uniform flow.                                                                              All fluids are compressible - even water.
Conditions do not change with position
in the stream or with time.                                                                          Density will change as pressure changes.
E.g. flow of water in a pipe of constant diameter at
- provided that changes in pressure are small - we
Steady non-uniform flow.                                                                            usually say the fluid is incompressible
Conditions change from point to point in the stream but                                                                    - it has constant density.
do not change with time.
E.g. Flow in a tapering pipe with constant velocity at the
inlet.                                                                                                         Three-dimensional flow
In general fluid flow is three-dimensional.
At a given instant in time the conditions at every point are                                                     Pressures and velocities change in all directions.
the same, but will change with time.
E.g. A pipe of constant diameter connected to a pump
pumping at a constant rate which is then switched off.                                                        In many cases the greatest changes only occur in
two directions or even only in one.
Every condition of the flow may change from point to                                                        Changes in the other direction can be effectively
point and with time at every point.                                                                 ignored making analysis much more simple.
E.g. Waves in a channel.

This course is restricted to Steady uniform flow
- the most simple of the four.

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                       Lecture 8        102    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        103

Unit 3: Fluid Dynamics                                                                    Unit 3: Fluid Dynamics

One dimensional flow:                                                                                         Two-dimensional flow

Conditions vary only in the direction of flow                                                             Conditions vary in the direction of flow and in
not across the cross-section.                                                                          one direction at right angles to this.

The flow may be unsteady with the parameters                                                            Flow patterns in two-dimensional flow can be shown
varying in time but not across the cross-section.                                                                   by curved lines on a plane.
E.g. Flow in a pipe.

Below shows flow pattern over a weir.
But:
Since flow must be zero at the pipe wall
- yet non-zero in the centre -
there is a difference of parameters across the
cross-section.

Pipe              Ideal flow   Real flow

In this course we will be considering:
Should this be treated as two-dimensional flow?
Possibly - but it is only necessary if very high                                                                                           steady
accuracy is required.                                                                                                      incompressible
one and two-dimensional flow

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                       Lecture 8        104    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        105
Unit 3: Fluid Dynamics                                                                    Unit 3: Fluid Dynamics

It is useful to visualise the flow pattern.                                        Close to a solid boundary, streamlines are parallel
Lines joining points of equal velocity - velocity                                                       to that boundary
contours - can be drawn.
The direction of the streamline is the direction of
the fluid velocity
These lines are know as streamlines
Fluid can not cross a streamline
Here are 2-D streamlines around a cross-section of
an aircraft wing shaped body:
Streamlines can not cross each other

Any particles starting on one streamline will stay
on that same streamline

In unsteady flow streamlines can change position
with time
Fluid flowing past a solid boundary does not flow
into or out of the solid surface.                                                   In steady flow, the position of streamlines does
not change.

Very close to a boundary wall the flow direction
must be along the boundary.

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        106    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        107

Unit 3: Fluid Dynamics                                                                    Unit 3: Fluid Dynamics

A circle of points in a flowing fluid each                                           The “walls” of a streamtube are streamlines.
has a streamline passing through it.
Fluid cannot flow across a streamline, so fluid
These streamlines make a tube-like shape known                                                         cannot cross a streamtube “wall”.
as a streamtube
A streamtube is not like a pipe.
Its “walls” move with the fluid.

In unsteady flow streamtubes can change position
with time

In steady flow, the position of streamtubes does
not change.

In a two-dimensional flow the streamtube is flat (in
the plane of the paper):

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        108    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        109
Unit 3: Fluid Dynamics                                                                                     Unit 3: Fluid Dynamics

Flow rate                                                                           Discharge and mean velocity

Mass flow rate                                                                         Cross sectional area of a pipe is A
Mean velocity is um.

dm                           mass                                                                                                      Q = Au m
m
dt            time taken to accumulate this mass

We usually drop the “m” and imply mean velocity.

Volume flow rate - Discharge.                                                                                                      Continuity                               Mass flow in
Control
volume

Mass flow out
Mass entering = Mass leaving                                        +    Increase
More commonly we use volume flow rate
per unit time   per unit time                                            of mass in
Also know as discharge.                                                                                                                                           control vol
per unit time
The symbol normally used for discharge is Q.
For steady flow there is no increase in the mass within
the control volume, so
volume of fluid
time                                                                    Mass entering = Mass leaving
per unit time   per unit time

Q1 = Q2 = A1u1 = A2u2

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                       Lecture 8        110    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                        Lecture 8                  111

Unit 3: Fluid Dynamics                                                                                     Unit 3: Fluid Dynamics

Applying to a streamtube:                                                      In a real pipe (or any other vessel) we use the mean
velocity and write
Mass enters and leaves only through the two ends
(it cannot cross the streamtube wall).                                                                                     1 A1um1                    2 A2 um2               Constant           m
ρ2
u2
A2

For incompressible, fluid                             1   =   2   =
(dropping the m subscript)
ρ1

u1

A1

A1u1             A2 u2        Q
Mass entering =                           Mass leaving
per unit time                             per unit time
This is the continuity equation most often used.

1     A1u1             2   A2u2

This equation is a very powerful tool.
It will be used repeatedly throughout the rest of this
dm                                                course.
1   A1u1            2   A2 u2          Constant     m
dt

This is the continuity equation.

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                       Lecture 8        112    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                        Lecture 8                  113
Unit 3: Fluid Dynamics                                                                                      Unit 3: Fluid Dynamics

Some example applications of Continuity
Water flows in a circular pipe which increases in diameter
1. What is the outflow?                                                                                 from 400mm at point A to 500mm at point B. Then pipe
then splits into two branches of diameters 0.3m and 0.2m
discharging at C and D respectively.

1.5 m3/s             If the velocity at A is 1.0m/s and at D is 0.8m/s, what are
the discharges at C and D and the velocities at B and C?

Solution:
Draw diagram:
Qin = Qout                                                                                                                                                            C
A                                dB=0.5m                                  dC=0.3m
1.5 + 1.5 = 3                                                                                                                        B
Qout = 3.0 m3/s
dA=0.4m
2. What is the inflow?
vA=1.0m/s
D
u = 1.5 m/s                                                                                                                                                                              dD=0.2m
A = 0.5 m2                                                                                                                                                                               vD=0.8m/s
u 3. 0.2 m/s
=                    Make a table and fill in the missing values
u = 1.0 m/s                                                                           A 4. 1.3 m2
=
A = 0.7 m2                                                                                 5.
Point              Velocity m/s                Diameter m         Area m²         Q m³/s

A                     1.00                         0.4          0.126           0.126
Q = 2.8 m3/s                        Q
B                     0.64                         0.5          0.196           0.126
C                     1.42                         0.3          0.071           0.101
Q = Area               Mean Velocity = Au
D                     0.80                         0.2          0.031           0.025
Q + 1.5 0.5 + 1 0.7 = 0.2 1.3 + 2.8
Q = 3.72 m3/s
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                      Lecture 8        114    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                        Lecture 8      115

Unit 3: Fluid Dynamics                                                                                      Unit 3: Fluid Dynamics

Restrictions in application
Lecture 9: The Bernoulli Equation                                                                                          of Bernoulli’s equation:
Unit 3: Fluid Dynamics
The Bernoulli equation is a statement of the
principle of conservation of energy along a                                                            Density is constant (incompressible)
streamline
Friction losses are negligible
It can be written:
2                                                                    It relates the states at two points along a single
p1          u1
z1                    H = Constant                                 streamline, (not conditions on two different
g          2g
streamlines)

These terms represent:
All these conditions are impossible to satisfy at any
instant in time!
Pressure     Kinetic    Potential                                                      Total
energy per energy per energy per                                                     energy per                Fortunately, for many real situations where the
unit weight unit weight unit weight                                                  unit weight             conditions are approximately satisfied, the equation
gives very good results.
These term all have units of length,
they are often referred to as the following:
p                                 u2
g                                 2g
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                      Lecture 8        116    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                        Lecture 8      117
Unit 3: Fluid Dynamics                                                                                                  Unit 3: Fluid Dynamics

The derivation of Bernoulli’s Equation:                                                                                                   m
Cross sectional area a                                     distance AA’ =
B
B’
a
A                                                                                 work done = force                        distance AA’
z                      A’
m            pm
mg                                                                                                  =    pa
a
An element of fluid, as that in the figure above, has potential
energy due to its height z above a datum and kinetic energy
p
work done per unit weight =
due to its velocity u. If the element has weight mg then                                                                                                                                g
potential energy = mgz                                                                                            This term is know as the pressure energy of the flowing stream.
potential energy per unit weight =                                 z                                          Summing all of these energy terms gives
1 2
kinetic energy =                   mu                                                                                                          Pressure          Kinetic           Potential       Total

2                                                                                                           energy per energy per energy per
unit weight unit weight unit weight
energy per
unit weight

u2
kinetic energy per unit weight =                                                                              or
2g
At any cross-section the pressure generates a force, the fluid
p          u2
z                    H
will flow, moving the cross-section, so work will be done. If the                                                                                                    g          2g
pressure at cross section AB is p and the area of the cross-
section is a then
force on AB = pa                                                                                                 By the principle of conservation of energy, the total energy in
the system does not change, thus the total head does not
when the mass mg of fluid has passed AB, cross-section AB                                                              change. So the Bernoulli equation can be written
will have moved to A’B’
mg            m                                                                                   p          u2
volume passing AB =
z                     H Constant
g                                                                                                g          2g
therefore

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                                    Lecture 8     118    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                                  Lecture 8        119

Unit 3: Fluid Dynamics                                                                                                  Unit 3: Fluid Dynamics

The Bernoulli equation is applied along                                                                             Practical use of the Bernoulli Equation
_______________
like that joining points 1 and 2 below.                                                                  The Bernoulli equation is often combined with the
2
continuity equation to find velocities and pressures
at points in the flow connected by a streamline.

Example:
1                                                                                     Finding pressures and velocities within a
total head at 1 = total head at 2                                                             contracting and expanding pipe.
or
2                           2
p1        u1                       p2 u2                                                         u1                                                                                       u2
z1                          z2                                                      p1                                                                                       p2
g        2g                        g 2g

section 1
This equation assumes no energy losses (e.g. from friction) or                                                                                                                                                 section 2
3
energy gains (e.g. from a pump) along the streamline. It can be                                                        A fluid, density = 960 kg/m is flowing steadily through
expanded to include these simply, by adding the appropriate                                                           the above tube.
energy terms:
The section diameters are d1=100mm and d2=80mm.
Total                Total              Loss     Work done       Energy
The gauge pressure at 1 is p1=200kN/m2
energy per         energy per unit per unit           per unit       supplied
unit weight at 1        weight at 2           weight    weight     per unit weight                  The velocity at 1 is u1=5m/s.
The tube is horizontal (z1=z2)
2                           2
p1          u1                       p2 u2
z1                          z2 h w q
g          2g                        g 2g                                                                            What is the gauge pressure at section 2?

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                                    Lecture 8     120    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                                  Lecture 8        121
Unit 3: Fluid Dynamics                                                                                    Unit 3: Fluid Dynamics

Apply the Bernoulli equation along a streamline joining                                                         We have used both the Bernoulli equation and the
section 1 with section 2.                                                                         Continuity principle together to solve the problem.
2              2
p1 u1            p2 u2
z1               z2
g        2g                         g 2g                                        Use of this combination is very common. We will be
seeing this again frequently throughout the rest of
2     2                                                         the course.
p2           p1                (u1    u2 )
2

Use the continuity equation to find u2                                                                                         Applications of the Bernoulli Equation
A1u1               A2u2
2                                              The Bernoulli equation is applicable to many
A1u1            d1
u2                                     u1                                              situations not just the pipe flow.
A2             d2
7.8125 m / s                                                       Here we will see its application to flow
So pressure at section 2                                                                                           measurement from tanks, within pipes as well as in
p2           200000 17296.87                                                                   open channels.

182703 N / m2
182.7 kN / m2

Note how
the velocity has increased
the pressure has decreased

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                        Lecture 8             122   CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                Lecture 8        123

Unit 3: Fluid Dynamics                                                                                    Unit 3: Fluid Dynamics

Applications of Bernoulli: Flow from Tanks                                                                 Apply Bernoulli along the streamline joining point 1 on the
Flow Through A Small Orifice                                                                surface to point 2 at the centre of the orifice.

At the surface velocity is negligible (u1 = 0) and the pressure
Flow from a tank through a hole in the side.                                                                              atmospheric (p1 = 0).

At the orifice the jet is open to the air so
1
Aactual                              again the pressure is atmospheric (p2 = 0).

h
If we take the datum line through the orifice
then z1 = h and z2 =0, leaving
2                                      Vena contractor                                                                     2
u2
h
2g
The edges of the hole are sharp to minimise frictional losses by                                                                                                    u2             2 gh
minimising the contact between the hole and the liquid.

The streamlines at the orifice                                                        This theoretical value of velocity is an overestimate as
friction losses have not been taken into account.
contract reducing the area of flow.

This contraction is called the vena contracta                                                   A coefficient of velocity is used to correct the theoretical
velocity,
The amount of contraction must                                                                                        uactual                Cv utheoretical
be known to calculate the flow
Each orifice has its own coefficient of velocity, they
usually lie in the range( 0.97 - 0.99)

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                        Lecture 8             124   CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                Lecture 8        125
Unit 3: Fluid Dynamics                                                                                             Unit 3: Fluid Dynamics

The discharge through the orifice                                                                                   Time for the tank to empty
is                                                                              We have an expression for the discharge from the tank
jet area jet velocity                                                                                                       Q         Cd Ao 2 gh

The area of the jet is the area of the vena contracta not
We can use this to calculate how long
the area of the orifice.                                                                                              it will take for level in the to fall

We use a coefficient of contraction                                                               As the tank empties the level of water falls.
to get the area of the jet                                                                           The discharge will also drop.

Aactual                  Cc Aorifice
h1
Giving discharge through the orifice:                                                                                                                                    h2

Q           Au
Qactual                   Aactual uactual
Cc Cv Aorifice utheoretical                                                     The tank has a cross sectional area of A.

Cd Aorifice utheoretical                                                                   In a time t the level falls by h
Cd Aorifice 2 gh                                                                              The flow out of the tank is
Q Au
h
Cd is the coefficient of discharge,                                                                                                         Q          A
t
Cd = Cc Cv                                                                 (-ve sign as h is falling)

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                       Lecture 8        126    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                             Lecture 8        127

Unit 3: Fluid Dynamics                                                                                             Unit 3: Fluid Dynamics

This Q is the same as the flow out of the orifice so                                                                          Submerged Orifice
What if the tank is feeding into another?
h                                                                        Area A1
Cd Ao 2 gh                            A
t                                                                                                                 Area A2

h1
h2
h
A
t
Cd Ao 2 g h
Orifice area Ao

Apply Bernoulli from point 1 on the surface of the deeper
Integrating between the initial level, h1, and final level, h2,                                                      tank to point 2 at the centre of the orifice,
gives the time it takes to fall this height                                                                             2                 2
A                     h                                                      p1 u1             p2 u2
h2                                                                       z1                z2
t                                     h1                                                                              g        2g                          g     2g
Cd Ao 2 g                          h
2
gh2        u2
0 0 h1                                        0
1                                                                                                                                             g         2g
1/ 2           1/ 2
h               2h            2 h
h                                                                                                                          u2               2 g (h1      h2 )
And the discharge is given by
A                         h                                                                       Q          Cd Ao u
t                  2 h h2
Cd Ao 2 g      1                                                                                                          Cd Ao 2 g (h1 h2 )
2A
h2    h1                                                                 So the discharge of the jet through the submerged orifice
Cd Ao 2 g
depends on the difference in head across the orifice.

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                       Lecture 8        128    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                             Lecture 8        129
Unit 3: Fluid Dynamics                                                                                        Unit 3: Fluid Dynamics

Using the Bernoulli equation we can calculate the
Lecture 10: Flow Measurement Devices                                                                                pressure at this point.
Unit 3: Fluid Dynamics
Along the central streamline at 1: velocity u1 , pressure p1
At the stagnation point (2): u2 = 0. (Also z1 = z2)
2
Pitot Tube                                                                               p1 u1                    p2
The Pitot tube is a simple velocity measuring device.                                                                                            2
1 2
Uniform velocity flow hitting a solid blunt body, has
p2            p1     u1
2
streamlines similar to this:

How can we use this?
1                                      2

The blunt body does not have to be a solid.
It could be a static column of fluid.
Some move to the left and some to the right.
The centre one hits the blunt body and stops.                                           Two piezometers, one as normal and one as a Pitot tube
within the pipe can be used as shown below to measure
velocity of flow.
At this point (2) velocity is zero

The fluid does not move at this one point.
This point is known as the stagnation point.

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1               Lecture 8        130    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                        Lecture 8        131

Unit 3: Fluid Dynamics                                                                                        Unit 3: Fluid Dynamics

Pitot Static Tube
The necessity of two piezometers makes this
h1                h2                                                   arrangement awkward.

The Pitot static tube combines the tubes and they
1
2                                       can then be easily connected to a manometer.

1
We have the equation for p2 ,                                                                          2

1 2
p2             p1       u1                                                                             1
X
2                                                                                                          h

1 2                                                                                                A           B
gh2               gh1        u1
2
u           2 g (h2 h1 )                             [Note: the diagram of the Pitot tube is not to scale. In reality its diameter
is very small and can be ignored i.e. points 1 and 2 are considered to
be at the same level]
We now have an expression for velocity from two
pressure measurements and the application of the
Bernoulli equation.
The holes on the side connect to one side of a
manometer, while the central hole connects to the other
side of the manometer

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1               Lecture 8        132    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                        Lecture 8        133
Unit 3: Fluid Dynamics                                                                                    Unit 3: Fluid Dynamics

Using the theory of the manometer,                                                                               Pitot-Static Tube Example
pA             p1              g X       h    man gh
pB             p2              gX                                         A pitot-static tube is used to measure the air flow at
the centre of a 400mm diameter building ventilation
pA             pB                                                         duct.
p2             gX             p1              g X       h                                If the height measured on the attached manometer is
man gh
10 mm and the density of the manometer fluid is 1000
1 2                                                kg/m3, determine the volume flow rate in the duct.
We know that                    p2           p1            u1 , giving                                      Assume that the density of air is 1.2 kg/m3.
2
2
u1
p1 hg man                                       p1
2
2 gh( m          )
u1

The Pitot/Pitot-static is:

Simple to use (and analyse)

Gives velocities (not discharge)

May block easily as the holes are small.

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                      Lecture 8        134    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                    Lecture 8        135

Unit 3: Fluid Dynamics                                                                                    Unit 3: Fluid Dynamics

Venturi Meter                                                 Apply Bernoulli along the streamline from point 1 to point 2
2                            2
p1          u1                     p2    u2
The Venturi meter is a device for measuring
z1                           z2
g          2g                      g    2g
discharge in a pipe.
By continuity
Q          u1 A1       u2 A2
It is a rapidly converging section which increases the
velocity of flow and hence reduces the pressure.                                                                                                            u1 A1
u2
A2
It then returns to the original dimensions of the pipe by a
gently diverging ‘diffuser’ section.                                                          Substituting and rearranging gives

2              2
p1       p2                                 u1         A1
z1         z2                              1
2  2    2
u1 A1 A2
2
2g     2
A2
1

p1    p2
2g               z1          z2
z2                                                                                                          g
u1          A2                   2     2
z1
h                                                                                                                   A1    A2
datum

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                      Lecture 8        136    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                    Lecture 8        137
Unit 3: Fluid Dynamics                                                                    Unit 3: Fluid Dynamics

The theoretical (ideal) discharge is u A.                                                                                     Venturimeter design:

Actual discharge takes into account the losses due to friction,                                               The diffuser assures a gradual and steady deceleration after
we include a coefficient of discharge (Cd 0.9)                                                        the throat. So that pressure rises to something near that
Qideal               u1 A1                                                                          before the meter.

Qactual                 Cd Qideal                    Cd u1 A1
The angle of the diffuser is usually between 6 and 8 degrees.
p1     p2
2g                     z1 z2                       Wider and the flow might separate from the walls increasing
g                                     energy loss.
Qactual                 Cd A1 A2
2      2
A1     A2
If the angle is less the meter becomes very long and pressure
losses again become significant.
In terms of the manometer readings
p1             gz1            p2         man gh          g ( z2       h)             The efficiency of the diffuser of increasing pressure back to
the original is rarely greater than 80%.
p1        p2                                            man
z1 z2                  h                 1                                     Care must be taken when connecting the manometer so that
g
no burrs are present.
Giving

2 gh       man     1
Qactual                 Cd A1 A2
2      2
A1     A2
This expression does not include any
elevation terms. (z1 or z2)

When used with a manometer
The Venturimeter can be used without knowing its angle.
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                       Lecture 8        138    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        139

Unit 3: Fluid Dynamics                                                                    Unit 3: Fluid Dynamics

Venturimeter Example
Lecture 11: Notches and Weirs
A venturimeter is used to measure the flow of water                                                                                 Unit 3: Fluid Dynamics
in a 150 mm diameter pipe. The throat diameter of the
venturimeter is 60 mm and the discharge coefficient
is 0.9. If the pressure difference measured by a
A notch is an opening in the side of a tank or reservoir.
manometer is 10 cm mercury, what is the average
velocity in the pipe?
Assume water has a density of 1000 kg/m3 and                                                                                             It is a device for measuring discharge
mercury has a relative density of 13.6.
A weir is a notch on a larger scale - usually found in rivers.

It is used as both a discharge measuring device and a device
to raise water levels.

There are many different designs of weir.
We will look at sharp crested weirs.

Weir Assumptions
velocity of the fluid approaching the weir is small so we
can ignore kinetic energy.
The velocity in the flow depends only on the depth below the
free surface. u    2 gh

These assumptions are fine for tanks with notches or reservoirs
with weirs, in rivers with high velocity approaching the weir is
substantial the kinetic energy must be taken into account

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                       Lecture 8        140    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        141
Unit 3: Fluid Dynamics                                                                                         Unit 3: Fluid Dynamics

A General Weir Equation                                                                                       Rectangular Weir

Consider a horizontal strip of                                                          The width does not change with depth so
width b, depth h below the free surface
b          constant                 B
b                       h
H
δh                                                                               B

H

velocity through the strip, u                                 2 gh
discharge through the strip, Q                                  Au b h 2 gh                    Substituting this into the general weir equation gives
H
Integrating from the free surface, h=0, to the weir crest,                                                      Qtheoretical B 2 g h1/ 2 dh
h=H, gives the total theoretical discharge                                                                                      0
H                                                                                                                                 2
Qtheoretical    2 g bh1/ 2 dh                                                                                                                          B 2 gH 3/ 2
3
0
To get the actual discharge we introduce a coefficient of
discharge, Cd, to account for
This is different for every differently                                                   losses at the edges of the weir
shaped weir or notch.                                                           and contractions in the area of flow,
2
We need an expression relating the width of flow across
Qactual                Cd              B 2 gH 3 / 2
3
the weir to the depth below the free surface.
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1              Lecture 8        142    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                         Lecture 8        143

Unit 3: Fluid Dynamics                                                                                         Unit 3: Fluid Dynamics

Rectangular Weir Example                                                                                            ‘V’ Notch Weir
The relationship between width and depth is dependent
Water enters the Millwood flood storage area via a                                                                    on the angle of the “V”.
rectangular weir when the river height exceeds the
weir crest. For design purposes a flow rate of 162
litres/s over the weir can be assumed                                                                                                                                b                h

H
θ

1. Assuming a height over the crest of 20cm and
Cd=0.2, what is the necessary width, B, of the weir?
The width, b, a depth h from the free surface is

b        2 H                 h tan
2
So the discharge is
H
Qtheoretical                      2 2 g tan                            H h h1/ 2 dh
2 0

2. What will be the velocity over the weir at this                                                                                                                           2 3/ 2       2 5/ 2 H
2 2 g tan                          Hh           h
design?                                                                                                                                                                2   3            5      0
8
2 g tan   H 5/ 2
15         2
The actual discharge is obtained by introducing a
coefficient of discharge
8
Qactual                Cd         2 g tan   H 5/ 2
15         2
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1              Lecture 8        144    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                         Lecture 8        145
Unit 3: Fluid Dynamics                                                                          Unit 3: Fluid Dynamics

‘V’ Notch Weir Example
Water is flowing over a 90o ‘V’ Notch weir into a tank                                                    Lecture 12: The Momentum Equation
with a cross-sectional area of 0.6m2. After 30s the                                                              Unit 3: Fluid Dynamics
depth of the water in the tank is 1.5m.
If the discharge coefficient for the weir is 0.8, what is                                                                       We have all seen moving
the height of the water above the weir?
fluids exerting forces.

The lift force on an aircraft is exerted by the air
moving over the wing.

A jet of water from a hose exerts a force on
whatever it hits.

The analysis of motion is as in solid mechanics: by
use of Newton’s laws of motion.

The Momentum equation
is a statement of Newton’s Second Law

It relates the sum of the forces
to the acceleration or
rate of change of momentum.

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1         Lecture 8        146    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1          Lecture 8        147

Unit 3: Fluid Dynamics                                                                          Unit 3: Fluid Dynamics

From solid mechanics you will recognise                                                            In time t a volume of the fluid moves
F = ma                                                                                 from the inlet a distance u1 t, so

What mass of moving fluid we should use?                                            volume entering the stream tube = area                                        distance
= A 1u1 t
We use a different form of the equation.
The mass entering,
Consider a streamtube:                                          mass entering stream tube = volume density
= 1 A1 u1 t

A2
And momentum
u2                            momentum entering stream tube = mass                                           velocity
A1
u1                                   ρ2                                                                                             = 1 A1 u1 t u1
ρ1

u1 δt                                           Similarly, at the exit, we get the expression:
momentum leaving stream tube =                                      2 A 2 u2      t u2

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1         Lecture 8        148    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1          Lecture 8        149
Unit 3: Fluid Dynamics                                                                             Unit 3: Fluid Dynamics
nd
By Newton’s 2                      Law.                                                           An alternative derivation
From conservation of mass
Force = rate of change of momentum                                                                      mass into face 1 = mass out of face 2

( 2 A2u2 t u2                                                    we can write
F=                                        1 A1u1   t u1 )
t                                                                                                              dm
rate of change of mass                               m
dt
1 A1u1  2 A2 u2
We know from continuity that
The rate at which momentum enters face 1 is
Q A1u1 A2 u2                                                                                                 1 A1u1u1         mu1

And if we have a fluid of constant density,                                                 The rate at which momentum leaves face 2 is
i.e. 1     2     , then                                                                                                          2 A2 u2 u2        mu2

F Q (u2                  u1 )                                 Thus the rate at which momentum changes across
the stream tube is
2 A2 u2 u2                    1 A1u1u1    mu2 mu1
So

Force = rate of change of momentum
F m ( u2 u1 )

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                    Lecture 8       150    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1             Lecture 8        151

Unit 3: Fluid Dynamics                                                                             Unit 3: Fluid Dynamics

The previous analysis assumed the inlet and outlet
So we have these two expressions,                                                                      velocities in the same direction
either one is known as the momentum equation                                                                   i.e. a one dimensional system.

What happens when this is not the case?
u2
F m ( u2 u1 )
θ2

F Q ( u2 u1)

θ1
The Momentum equation.
u1

This force acts on the fluid
in the direction of the flow of the fluid.                                                         We consider the forces by resolving in the
directions of the co-ordinate axes.

The force in the x-direction
Fx           m u2 cos 2 u1 cos 1
m u2 x u1 x
or
Fx              Q u2 cos 2 u1 cos 1
Q u2 x u1 x
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                    Lecture 8       152    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1             Lecture 8        153
Unit 3: Fluid Dynamics                                                                                 Unit 3: Fluid Dynamics

And the force in the y-direction
Fy          m u2 sin 2                    u1 sin 1                                                               In summary we can say:

m u2 y                   u1 y                                                             Total force                                 rate of change of
on the fluid                     =          momentum through
or
the control volume
Fy              Q u2 sin 2                 u1 sin 1

Q u2 y              u1 y                                                                                F         m uout            uin
or

The resultant force can be found by combining                                                                                                  F             Q uout          uin
these components
Fy
FResultant

Remember that we are working with vectors so F is
φ
in the direction of the velocity.
Fx

2           2
Fresultant                     Fx          Fy

And the angle of this force

Fy
tan 1
Fx
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                       Lecture 8        154    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                 Lecture 8        155

Unit 3: Fluid Dynamics                                                                                 Unit 3: Fluid Dynamics

This force is made up of three components:
FR = Force exerted on the fluid by any solid body                                                      Application of the Momentum Equation
touching the control volume
Forces on a Bend
FB = Force exerted on the fluid body (e.g. gravity)

FP = Force exerted on the fluid by fluid pressure
Consider a converging or diverging pipe bend lying
outside the control volume
in the vertical or horizontal plane
turning through an angle of .
So we say that the total force, FT,
is given by the sum of these forces:                                                                Here is a diagram of a diverging pipe bend.
y                                          p2 u
2 A2
FT = FR + FB + FP
x

1m
The force exerted                                                                       p1

u1                                               45°

A1
by the fluid
on the solid body

touching the control volume is opposite to FR.

So the reaction force, R, is given by
R = -FR

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                       Lecture 8        156    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                 Lecture 8        157
Unit 3: Fluid Dynamics                                                                                         Unit 3: Fluid Dynamics

Why do we want to know the forces here?                                                                        An Example of Forces on a Bend

The outlet pipe from a pump is a bend of 45 rising in the vertical plane (i.e. and
As the fluid changes direction                                                  internal angle of 135 ). The bend is 150mm diameter at its inlet and 300mm diameter
a force will act on the bend.                                                  at its outlet. The pipe axis at the inlet is horizontal and at the outlet it is 1m higher. By
neglecting friction, calculate the force and its direction if the inlet pressure is 100kN/m2
and the flow of water through the pipe is 0.3m3/s. The volume of the pipe is 0.075m3.
[13.95kN at 67 39’ to the horizontal]
This force can be very large in the case of water
supply pipes. The bend must be held in place
to prevent breakage at the joints.                                                            1&2 Draw the control volume and the axis
system
y                                         p2 u
We need to know how much force a support                                                                                                                                2 A2

(thrust block) must withstand.                                                                                               x

p1                                                                  1m
Step in Analysis:
u1                                              45°

A1
1.Draw a control volume
2.Decide on co-ordinate axis system
3.Calculate the total force
4.Calculate the pressure force                                                                    p1 = 100 kN/m2,
5.Calculate the body force                                                                        Q = 0.3 m3/s
6.Calculate the resultant force                                                                      = 45

d1 = 0.15 m                                  d2 = 0.3 m

A1 = 0.177 m2                                A2 = 0.0707 m2

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                     Lecture 8        158    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                         Lecture 8        159

Unit 3: Fluid Dynamics                                                                                         Unit 3: Fluid Dynamics

3 Calculate the total force                                                                                4 Calculate the pressure force.
in the x direction                                                                                                       FP           pressure force at 1 - pressure force at 2
1      0,                      2

FT x              Q u2 x         u1 x
FP x          p1 A1 cos 0             p 2 A2 cos         p1 A1    p 2 A2 cos
Q u2 cos               u1
FP y          p1 A1 sin 0            p 2 A2 sin            p 2 A2 sin
by continuity A1u1                                  A2 u2          Q , so
We know pressure at the inlet
but not at the outlet
0.3
u1                                      16.98 m / s
0152 / 4
.                                                                                we can use the Bernoulli equation
0.3                                                                                 to calculate this unknown pressure.
u2                                  4.24 m / s
0.0707
2                              2
p1        u1                      p2     u2
z1                        z2 h f
FT x            1000 0.3 4.24 cos 45 16.98                                                                         g        2g                       g     2g
4193.68 N
and in the y-direction                                                                                                     where hf is the friction loss
In the question it says this can be ignored, hf=0
FT y              Q u2 y            u1 y
Q u2 sin                 0                                                      The height of the pipe at the outlet
1000 0.3 4.24 sin 45                                                                     is 1m above the inlet.
Taking the inlet level as the datum:
899.44 N
z1 = 0           z2 = 1m
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                     Lecture 8        160    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                         Lecture 8        161
Unit 3: Fluid Dynamics                                                                                              Unit 3: Fluid Dynamics

6 Calculate the resultant force
So the Bernoulli equation becomes:
100000   16.982        p2       4.24 2                                                                                                       FT x            FR x         FP x             FB x
0                      .
10
1000 9.81 2 9.81    1000 9.81 2 9.81                                                                                                          FT y            FR y             FP y         FB y
p2 2253614 N / m2
.
FR x           FT x           FP x             FB x
FP x          100000 0.0177 2253614 cos 45 0.0707
.                                                                                                       4193.6 9496.37
1770 11266.34                                  9496.37 kN                                                                  5302.7 N

FP y              2253614 sin 45 0.0707
.                                                                                                 FR y            FT y           FP y             FB y
11266.37                                                                                                               899.44 11266.37 735.75
1290156 N
.
5 Calculate the body force
And the resultant force on the fluid is given by
The only body force is the force due to gravity. That                                                                                                 FRy
FResultant
is the weight acting in the -ve y direction.
FB y                  g volume
φ
1000 9.81 0.075
FRx
1290156 N
.                                                                                                      2            2
FR               F     Rx           F
Ry
There are no body forces in the x direction,
FB x       0                                                                                     5302.7 2 12901562
.
13.95 kN
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                  Lecture 8        162    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                              Lecture 8        163

Unit 3: Fluid Dynamics                                                                                              Unit 3: Fluid Dynamics

And the direction of application is                                                                     Lecture 14: Momentum Equation Examples
FR y                                              Unit 3: Fluid Dynamics
tan 1
FR x
1290156
.                                                                          Impact of a Jet on a Plane
tan 1
5302.7
67.66              67 39'                                 A jet hitting a flat plate (a plane) at an angle of 90

The force on the bend is the same magnitude but in                                                           We want to find the reaction force of the plate.
the opposite direction                                                                                     i.e. the force the plate will have to apply to stay in
the same position.
R             FR            13.95 kN
1 & 2 Control volume and Co-ordinate axis are
shown in the figure below.
y                           u2

x

Lecture 13: Design Study 2                                                                                              u1

See Separate Handout

u2

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                  Lecture 8        164    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                              Lecture 8        165
Unit 3: Fluid Dynamics                                                                                              Unit 3: Fluid Dynamics

3 Calculate the total force                                                                           6 Calculate the resultant force
In the x-direction
FT x             FR x               FP x       FB x
FT x                Q u2 x u1 x                                                                   FR x             FT x           0 0
Qu1 x                                                                                                Qu1 x
Exerted on the fluid.
The system is symmetrical
the forces in the y-direction cancel.                                            The force on the plane is the same magnitude but in
the opposite direction
FT y          0                                                                                     R             FR x
If the plane were at an angle
4 Calculate the pressure force.                                                                                        the analysis is the same.
The pressures at both the inlet and the outlets                                                     But it is usually most convenient to choose the axis
system normal to the plate.
to the control volume are atmospheric.                                                                                                                   y
The pressure force is zero                                                                                                                            x                    u2

FP x               FP y   0

u1
5 Calculate the body force
θ
As the control volume is small
we can ignore the body force due to gravity.
u3
FB x               FB y   0

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                   Lecture 8     166    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                              Lecture 8        167

Unit 3: Fluid Dynamics                                                                                              Unit 3: Fluid Dynamics

Force on a curved vane                                              3 Calculate the total force
in the x direction
This case is similar to that of a pipe, but the
analysis is simpler.                                                                                               FT x                Q u2             u1 cos

Pressures at ends are equal at atmospheric
Q
by continuity u1                            u2             , so
A
Both the cross-section and velocities
(in the direction of flow) remain constant.
Q2
FT x                        1 cos
A
u2
y
and in the y-direction
x

FT y                Q u2 sin                  0
u1
Q2
θ
A
4 Calculate the pressure force.
The pressure at both the inlet and the outlets to the
control volume is atmospheric.
1 & 2 Control volume and Co-ordinate axis are
shown in the figure above.                                                                                                                          FP x               FP y        0

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                   Lecture 8     168    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                              Lecture 8        169
Unit 3: Fluid Dynamics                                                                                     Unit 3: Fluid Dynamics

5 Calculate the body force
2          2
FR                FR x       FR y
No body forces in the x-direction, FB x = 0.

And the direction of application is
In the y-direction the body force acting is the weight
of the fluid.
If V is the volume of the fluid on the vane then,                                                                                                                FR y
tan 1
FB x             gV                                                                                               FR x
exerted on the fluid.
(This is often small as the jet volume is small and
sometimes ignored in analysis.)                                                             The force on the vane is the same magnitude but in
the opposite direction
6 Calculate the resultant force
R           FR
FT x            FR x            FP x     FB x
Q2
FR x             FT x                      1 cos
A

FT y            FR y           FP y   FB y

Q2
FR y             FT y
A
And the resultant force on the fluid is given by
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                   Lecture 8        170    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                     Lecture 8        171

Unit 3: Fluid Dynamics                                                                                     Unit 3: Fluid Dynamics

SUMMARY                                                                  We work with components of the force:
u2

θ2
The Momentum equation
is a statement of Newton’s Second Law

For a fluid of constant density,
θ1

Total force                                 rate of change of                                           u1

on the fluid                     =          momentum through
Fx               Q u2 x               u1x                Q u2 cos 2 u1 cos 1
the control volume

F          m uout                uin           Q uout     uin                               Fy              Q u2 y               u1 y               Q u2 sin 2 u1 sin 1

This force acts on the fluid                                                            The resultant force can be found by combining
in the direction of the velocity of fluid.                                                                  these components
Fy
FResultant

This is the total force FT where:
FT = FR + FB + FP                                                                                     φ
FR = External force on the fluid from any solid body                                                                                                                 2         2
Fx           Fresultant             Fx        Fy
touching the control volume
FB = Body force on the fluid body (e.g. gravity)                                                                   And the angle this force acts:
FP = Pressure force on the fluid by fluid pressure                                                                                                  Fy
outside the control volume                                                                                                               tan 1
Fx
CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                   Lecture 8        172    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1                     Lecture 8        173
Unit 3: Fluid Dynamics                                                                    Unit 3: Fluid Dynamics

2. A 600mm diameter pipeline carries water under a head of
30m with a velocity of 3m/s. This water main is fitted with a
Lecture 15: Calculations                                                 horizontal bend which turns the axis of the pipeline through
Unit 3: Fluid Dynamics                                                  75 (i.e. the internal angle at the bend is 105 ). Calculate
the resultant force on the bend and its angle to the
horizontal.

1. The figure below shows a smooth curved vane attached to
a rigid foundation. The jet of water, rectangular in section,
75mm wide and 25mm thick, strike the vane with a velocity
of 25m/s. Calculate the vertical and horizontal components
of the force exerted on the vane and indicate in which
direction these components act.

45
25

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1         Lecture 8        174    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        175

Unit 3: Fluid Dynamics                                                                    Unit 3: Fluid Dynamics

3. A 75mm diameter jet of water having a velocity of 25m/s
strikes a flat plate, the normal of which is inclined at 30 to                              4. In an experiment a jet of water of diameter 20mm is fired
the jet. Find the force normal to the surface of the plate.                                    vertically upwards at a sprung target that deflects the water
at an angle of 120° to the horizontal in all directions. If a
500g mass placed on the target balances the force of the
jet, was is the discharge of the jet in litres/s?

5. Water is being fired at 10 m/s from a hose of 50mm
diameter into the atmosphere. The water leaves the hose
through a nozzle with a diameter of 30mm at its exit. Find
the pressure just upstream of the nozzle and the force on
the nozzle.

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1         Lecture 8        176    CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1    Lecture 8        177
Unit 4                                                                                               Unit 4
CIVE1400: An Introduction to Fluid Mechanics                                                                                                 Real fluids

Dr P A Sleigh                                                                       Flowing real fluids exhibit
P.A.Sleigh@leeds.ac.uk                                                                    viscous effects, they:

Dr CJ Noakes                                                                     “stick” to solid surfaces
C.J.Noakes@leeds.ac.uk                                                             have stresses within their body.

January 2008
From earlier we saw this relationship between
Module web site:                                                          shear stress and velocity gradient:
www.efm.leeds.ac.uk/CIVE/FluidsLevel1
du
Unit 1: Fluid Mechanics Basics
Flow
3 lectures                                                                                     dy
Pressure
Properties of Fluids                                                                        The shear stress, , in a fluid
Fluids vs. Solids                                                                      is proportional to the velocity gradient
Viscosity
- the rate of change of velocity across the flow.
Unit 2: Statics                                            3 lectures
Hydrostatic pressure
Manometry/Pressure measurement
Hydrostatic forces on submerged surfaces                                                       For a “Newtonian” fluid we can write:
Unit 3: Dynamics                                           7 lectures                                                                                     du
The continuity equation.
The Bernoulli Equation.                                                                                                                             dy
Application of Bernoulli equation.
The momentum equation.                                                                            is coefficient of viscosity
where
Application of momentum equation.
(or simply viscosity).
Unit 4: Effect of the boundary on flow                     4 lectures
Laminar and turbulent flow
Here we look at the influence of forces due to
Boundary layer theory                                                                    momentum changes and viscosity
An Intro to Dimensional analysis
Similarity                                                                                       in a moving fluid.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1    Lectures 16-19     178    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1          Lectures 16-19     179

Unit 4                                                                                               Unit 4

Laminar and turbulent flow                                                                          All three would happen -
but for different flow rates.

Injecting a dye into the middle of flow in a pipe,                                                                                   Top: Slow flow
what would we expect to happen?                                                                                             Middle: Medium flow
This                                                                                                                                 Bottom: Fast flow

Top:                                 Laminar flow
Middle:                              Transitional flow
Bottom:                              Turbulent flow

Laminar flow:
this                                                                                              Motion of the fluid particles is very orderly
all particles moving in straight lines
parallel to the pipe walls.

Turbulent flow:
Motion is, locally, completely random but the
or this                                                                                           overall direction of flow is one way.

But what is fast or slow?
At what speed does the flow pattern change?
And why might we want to know this?

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1    Lectures 16-19     180    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1          Lectures 16-19     181
Unit 4                                                                                                   Unit 4
The was first investigated in the 1880s                                                                       After many experiments he found this
by Osbourne Reynolds                                                                                             expression
in a classic experiment in fluid mechanics.
ud
A tank arranged as below:

= density,                                  u = mean velocity,
d = diameter                                    = viscosity

This could be used to predict the change in
flow type for any fluid.

This value is known as the
Reynolds number, Re:

ud
Re

Laminar flow:                                             Re < 2000
Transitional flow:                                 2000 < Re < 4000
Turbulent flow:                                           Re > 4000

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1               Lectures 16-19     182    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1              Lectures 16-19     183

Unit 4                                                                                                   Unit 4
What are the units of Reynolds number?                                                             At what speed does the flow pattern change?

We can fill in the equation with SI units:                                                         We use the Reynolds number in an example:

kg / m3 , u                m / s,   d   m                                                       A pipe and the fluid flowing
Ns / m2            kg / m s                                                                   have the following properties:

ud         kg m m m s                                                    water density                                                 = 1000 kg/m3
Re                                  1
m3 s 1 kg                                                     pipe diameter                                               d = 0.5m
(dynamic) viscosity,                                          = 0.55x103 Ns/m2
It has no units!

A quantity with no units is known as a                                                             What is the MAXIMUM velocity when flow is
non-dimensional (or dimensionless) quantity.                                                          laminar i.e. Re = 2000

(We will see more of these in the section on
dimensional analysis.)                                                                                                            ud
Re                            2000

The Reynolds number, Re,                                                                                            2000     2000 0.55 10 3
u
is a non-dimensional number.                                                                                             d         1000 0.5
u        0.0022 m / s

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1               Lectures 16-19     184    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1              Lectures 16-19     185
Unit 4                                                                                            Unit 4
What is the MINIMUM velocity when flow is                                                                 What does this abstract number mean?
turbulent i.e. Re = 4000
We can give the Re number a physical meaning.
ud
Re                       4000
This may help to understand some of the
u         0.0044 m / s                                          reasons for the changes from laminar to
turbulent flow.

In a house central heating system,
typical pipe diameter = 0.015m,                                                                                                          ud
Re

limiting velocities would be,                                                                                                   inertial forces
0.0733 and 0.147m/s.                                                                                                       viscous forces

Both of these are very slow.                                                      When inertial forces dominate
(when the fluid is flowing faster and Re is larger)
In practice laminar flow rarely occurs                                                              the flow is turbulent.
in a piped water system.
When the viscous forces are dominant
Laminar flow does occur in                                                                             (slow flow, low Re)
fluids of greater viscosity                                                                   they keep the fluid particles in line,
e.g. in bearing with oil as the lubricant.                                                                    the flow is laminar.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1          Lectures 16-19     186    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1       Lectures 16-19     187

Unit 4                                                                                            Unit 4
Laminar flow                                             Pressure loss due to friction in a pipeline
Re < 2000
‘low’ velocity                                                                  Up to now we have considered ideal fluids:
Dye does not mix with water                                                            no energy losses due to friction
Fluid particles move in straight lines
Simple mathematical analysis possible                                                       Because fluids are viscous,
Rare in practice in water systems.                                                 energy is lost by flowing fluids due to friction.

Transitional flow                                                             This must be taken into account.
2000 > Re < 4000
‘medium’ velocity                                                                         The effect of the friction shows itself as a
Dye stream wavers - mixes slightly.                                                                pressure (or head) loss.

Turbulent flow
Re > 4000                                                                                          In a real flowing fluid shear stress
‘high’ velocity                                                                                               slows the flow.
Dye mixes rapidly and completely
Particle paths completely irregular                                                                            To give a velocity profile:
Average motion is in flow direction
Cannot be seen by the naked eye
Changes/fluctuations are very difficult to
detect. Must use laser.
Mathematical analysis very difficult - so
experimental measures are used
Most common type of flow.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1          Lectures 16-19     188    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1       Lectures 16-19     189
Unit 4                                                                                                            Unit 4
Attaching a manometer gives                                                                            Consider a cylindrical element of
pressure (head) loss due to the energy lost by                                                               incompressible fluid flowing in the pipe,
the fluid overcoming the shear stress.                                                                                                         τw

L
το
το
τw                area A

w is the mean shear stress on the boundary
Upstream pressure is p,
Δp
Downstream pressure falls by p to (p- p)

The driving force due to pressure

The pressure at 1 (upstream)                                                      driving force = Pressure force at 1 - pressure force at 2
is higher than the pressure at 2.                                                                                                                                     d2
pA             p            p A             pA           p
4
How can we quantify this pressure loss
in terms of the forces acting on the fluid?
The retarding force is due to the shear stress

shear stress area over which it acts
= w area of pipe wall
= w dL

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1          Lectures 16-19     190    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                       Lectures 16-19     191

Unit 4                                                                                                            Unit 4
As the flow is in equilibrium,                                                                       What is the variation of shear stress in the flow?

driving force = retarding force                                                                                                                                          τw
R

2                                                                                                                       r
d
p                   w dL
4
p        w4L                                                                                                                            τw
d                                                                                  At the wall
R p
Giving pressure loss in a pipe in terms of:                                                                                                       w
2 L

pipe diameter
shear stress at the wall                                                                                                             r p
2 L
r
w
R
A linear variation in shear stress.

This is valid for:
laminar flow
turbulent flow

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1          Lectures 16-19     192    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                       Lectures 16-19     193
Unit 4                                                                                                     Unit 4
Shear stress and hence pressure loss varies                                                                         Pressure loss during laminar flow in a pipe
with velocity of flow and hence with Re.
In general the shear stress w. is almost
Many experiments have been done                                                                                                impossible to measure.
with various fluids measuring
the pressure loss at various Reynolds numbers.                                                                                 For laminar flow we can calculate
a theoretical value for
A graph of pressure loss and Re look like:                                                                        a given velocity, fluid and pipe dimension.

In laminar flow the paths of individual particles
of fluid do not cross.

Flow is like a series of concentric cylinders
sliding over each other.

And the stress on the fluid in laminar flow is
entirely due to viscose forces.
As before, consider a cylinder of fluid, length L,

This graph shows that the relationship between
pressure loss and Re can be expressed as

laminar                         p         u
turbulent                       p         u1.7 ( or   2 .0 )

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                       Lectures 16-19     194    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                Lectures 16-19     195

Unit 4                                                                                                     Unit 4

δr                                         In an integral form this gives an
expression for velocity,
r               r
p 1
R                                                        u                            r dr
L 2
The fluid is in equilibrium,                                                                                               The value of velocity at a
shearing forces equal the pressure forces.                                                                                      point distance r from the centre
2 r L                   pA               p r2                                                                                           p r2
ur                           C
p r                                                                                                             L 4
L 2                                                      At r = 0, (the centre of the pipe), u = umax, at
r = R (the pipe wall) u = 0;

du                                                                                    p R2
Newtons law of viscosity says                                                     ,
C
dy                                                                                    L 4
At a point r from the pipe centre when the flow is
laminar:
We are measuring from the pipe centre, so
p 1
du                                                                                   ur                R2                 r2
L 4
dr
This is a parabolic profile
Giving:
(of the form y = ax2 + b )
p r                  du                                          so the velocity profile in the pipe looks similar to
L 2                  dr
du                  p r
dr                  L 2

v
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                       Lectures 16-19     196    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                Lectures 16-19     197
Unit 4                                                                                               Unit 4
To get pressure loss (head loss)
What is the discharge in the pipe?                                                                   in terms of the velocity of the flow, write
pressure in terms of head loss hf, i.e. p = ghf
The flow in an annulus of thickness r
Q          ur Aannulus                                                       Mean velocity:
u        Q/ A
Aannulus                     (r r ) 2                    r2   2 r r
p 1                                                                                                                          gh f d 2
Q                 R2                     2
r 2 r r                                                                              u
L 4                                                                                                                            32 L
R
p
Q                  R 2 r r 3 dr                                                     Head loss in a pipe with laminar flow by the
L 2 0
Hagen-Poiseuille equation:
p R4                     p d4
L 8                     L128                                                                                                   32 Lu
hf
gd 2
So the discharge can be written
Pressure loss is directly proportional to the
4                                                    velocity when flow is laminar.
p d
Q
L 128
It has been validated many time by experiment.
It justifies two assumptions:
This is the Hagen-Poiseuille Equation
1.fluid does not slip past a solid boundary
for laminar flow in a pipe
2.Newtons hypothesis.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                       Lectures 16-19     198    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1          Lectures 16-19     199

Unit 4                                                                                               Unit 4
Boundary Layers                                                                                 Considering a flat plate in a fluid.

by Douglas J F, Gasiorek J M, and Swaffield J A.
Longman publishers. Pages 327-332.                                                                               Upstream the velocity profile is uniform,
This is known as free stream flow.

Fluid flowing over a stationary surface,
e.g. the bed of a river, or the wall of a pipe,
is brought to rest by the shear stress to                                                                            Downstream a velocity profile exists.
This gives a, now familiar, velocity profile:                                                                          This is known as fully developed flow.

umax
Free stream flow

zero velocity             τo
Fully developed flow
Wall

Zero at the wall
A maximum at the centre of the flow.

The profile doesn’t just exit.
Starting when it first flows past the surface
e.g. when it enters a pipe.                                                                             How do we get to the fully developed state?
Are there any changes in flow as we get there?
Are the changes significant / important?
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                       Lectures 16-19     200    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1          Lectures 16-19     201
Unit 4                                                                                        Unit 4
Understand this Boundary layer growth diagram.                                                                      Boundary layer thickness:

= distance from wall to where u = 0.99 umainstream

increases as fluid moves along the plate.
It reaches a maximum in fully developed flow.

The increase corresponds to a
drag force increase on the fluid.

As fluid is passes over a greater length:

* more fluid is slowed
* by friction between the fluid layers
*        the thickness of the slow layer increases.

Fluid near the top of the boundary layer drags the
fluid nearer to the solid surface along.

The mechanism for this dragging
may be one of two types:

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     202    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     203

Unit 4                                                                                        Unit 4
First: viscous forces                                                                              Second: momentum transfer
(the forces which hold the fluid together)
If the viscous forces were the only action
When the boundary layer is thin:                                                           the fluid would come to a rest.
Viscous shear stresses have held the fluid
by Newton’s law of viscosity                                                      particles in a constant motion within layers.
shear stress, = (du/dy), is large.                                                       Eventually they become too small to
hold the flow in layers;
The force may be large enough to
drag the fluid close to the surface.                                                                       the fluid starts to rotate.

As the boundary layer thickens
shear stress decreases.

Eventually it is too small
to drag the slow fluid along.

Up to this point the flow has been laminar.
The fluid motion rapidly becomes turbulent.
Newton’s law of viscosity has applied.                                            Momentum transfer occurs between fast moving
main flow and slow moving near wall flow.
This part of the boundary layer is the                                            Thus the fluid by the wall is kept in motion.
laminar boundary layer                                                   The net effect is an increase in momentum in the
boundary layer.
This is the turbulent boundary layer.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     204    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     205
Unit 4                                                                                                         Unit 4
Close to boundary velocity gradients are very large.                                         Use Reynolds number to determine which state.
Viscous shear forces are large.                                                                              ud
Possibly large enough to cause laminar flow.                                                                  Re
This region is known as the laminar sub-layer.                                                              Laminar flow:                                 Re < 2000
Transitional flow: 2000 <                           Re < 4000
This layer occurs within the turbulent zone                                                         Turbulent flow:                               Re > 4000
it is next to the wall.
It is very thin – a few hundredths of a mm.

Surface roughness effect

Despite its thinness, the laminar sub-layer has vital
role in the friction characteristics of the surface.

In turbulent flow:
Roughness higher than laminar sub-layer:
increases turbulence and energy losses.

Laminar flow: profile parabolic (proved in earlier lectures)
In laminar flow:                                                 The first part of the boundary layer growth diagram.
Roughness has very little effect

Boundary layers in pipes                                                                   Turbulent (or transitional),
Initially of the laminar form.                                                   Laminar and the turbulent (transitional) zones of the
boundary layer growth diagram.
It changes depending on the ratio of inertial and
viscous forces;                                                               Length of pipe for fully developed flow is
the entry length.
i.e. whether we have laminar (viscous forces high) or
turbulent flow (inertial forces high).                                                                          Laminar flow                            120   diameter
Turbulent flow                          60    diameter

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1    Lectures 16-19     206    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                    Lectures 16-19     207

Unit 4                                                                                                         Unit 4
Boundary layer separation                                                                Boundary layer separation:
* increases the turbulence
Divergent flows:
*        increases the energy losses in the flow.
Pressure increases in the direction of flow.
Separating / divergent flows are inherently
unstable
The fluid in the boundary layer has so little
momentum that it is brought to rest,
Convergent flows:
and possibly reversed in direction.
Reversal lifts the boundary layer.                                                                           Negative pressure gradients

Pressure decreases in the direction of flow.
u1                                           u2
p1
Fluid accelerates and the boundary layer is thinner.
p2

p1 < p2              u1 > u2                                                      u1
u2
p2
p1

p1 > p2              u1 < u2

Flow remains stable

Turbulence reduces.
This phenomenon is known as
boundary layer separation.                                                        Boundary layer separation does not occur.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1    Lectures 16-19     208    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                    Lectures 16-19     209
Unit 4                                                                                        Unit 4
Examples of boundary layer separation                                        Tee-Junctions

A divergent duct or diffuser
velocity drop
(according to continuity)
pressure increase
(according to the Bernoulli equation).

Assuming equal sized pipes),
Velocities at 2 and 3 are smaller than at 1.
Pressure at 2 and 3 are higher than at 1.
Causing the two separations shown

Y-Junctions
Tee junctions are special cases of the Y-junction.
Increasing the angle increases the probability of
boundary layer separation.

Venturi meter
A balance between:
* length of meter
* danger of boundary layer separation.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     210    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     211

Unit 4                                                                                        Unit 4
Bends                                                                                         Flow past a cylinder
Slow flow, Re < 0.5 no separation:

Moderate flow, Re < 70, separation
vortices form.

Two separation zones occur in bends as shown
above.
Pb > Pa causing separation.
Fast flow Re > 70
Pd > Pc causing separation
vortices detach alternately.
Form a trail of down stream.
Karman vortex trail or street.
Localised effect
(Easily seen by looking over a bridge)
Downstream the boundary layer reattaches and
normal flow occurs.
Boundary layer separation is only local.                                                        Causes whistling in power cables.
Nevertheless downstream of a                                                            Caused Tacoma narrows bridge to collapse.
junction / bend /valve etc.                                                        Frequency of detachment was equal to the bridge
fluid will have lost energy.                                                                     natural frequency.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     212    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     213
Unit 4                                                                                        Unit 4
Aerofoil
Normal flow over a aerofoil or a wing cross-section.

(boundary layers greatly exaggerated)

The velocity increases as air flows over the wing. The
Fluid accelerates to get round the cylinder                                                pressure distribution is as below
Velocity maximum at Y.                                                             so transverse lift force occurs.
Pressure dropped.

Adverse pressure between here and downstream.
Separation occurs

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     214    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     215

Unit 4                                                                                        Unit 4
At too great an angle                                                   Examples:
boundary layer separation occurs on the top                                        Exam questions involving boundary layer theory are
Pressure changes dramatically.                                               typically descriptive. They ask you to explain the
This phenomenon is known as stalling.                                            mechanisms of growth of the boundary layers including
how, why and where separation occurs. You should also be
able to suggest what might be done to prevent separation.

All, or most, of the ‘suction’ pressure is lost.
The plane will suddenly drop from the sky!

Solution:
Prevent separation.
1 Engine intakes draws slow air from the boundary
layer at the rear of the wing though small holes
2 Move fast air from below to top via a slot.

3 Put a flap on the end of the wing and tilt it.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     216    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     217
Unit 4                                                                                                           Unit 4
Uses principle of dimensional homogeneity
Lectures 18 & 19: Dimensional Analysis                                           It gives qualitative results which only become quantitative
Unit 4: The Effect of the Boundary on Flow                                                         from experimental analysis.

Dimensions and units
Application of fluid mechanics in design makes use of
experiments results.                                                                                Any physical situation
Results often difficult to interpret.                                                  can be described by familiar properties.
Dimensional analysis provides a strategy for choosing
relevant data.                                                                e.g. length, velocity, area, volume, acceleration etc.
Used to help analyse fluid flow
Especially when fluid flow is too complex for                                                          These are all known as dimensions.
mathematical analysis.
Dimensions are of no use without a magnitude.
Specific uses:                                                                      i.e. a standardised unit
help design experiments                                                 e.g metre, kilometre, Kilogram, a yard etc.
Informs which measurements are important
Allows most to be obtained from experiment:                                                                Dimensions can be measured.
e.g. What runs to do. How to interpret.                                                     Units used to quantify these dimensions.

It depends on the correct identification of variables                                 In dimensional analysis we are concerned with the nature
of the dimension
Relates these variables together
i.e. its quality not its quantity.
Experiments necessary to complete solution
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     218    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                      Lectures 16-19     219

Unit 4                                                                                                           Unit 4

The following common abbreviations are used:                                      This table lists dimensions of some common physical
quantities:

length                               =L                                                         Quantity                          SI Unit                     Dimension
mass                                 =M                                                          velocity                           m/s           ms-1           LT-1
acceleration                           m/s2          ms-2           LT-2
time                                 =T
force                              N
force                                =F                                                                                          kg m/s2        kg ms-2         M LT-2

temperature                          =                                                    energy (or work)                       Joule J
N m,
kg m2s-2        ML2T-2
kg m2/s2
Here we will use L, M, T and F (not              ).                                            power                          Watt W
N m/s          Nms-1
kg m2/s3       kg m2s-3        ML2T-3

We can represent all the physical properties we are                                                      pressure ( or stress)                    Pascal P,
interested in with three:                                                                                                                            N/m2,         Nm-2
kg/m/s2        kg m-1s-2      ML-1T-2
3             -3
density                           kg/m          kg m            ML-3
L, T                                             specific weight                         N/m    3

kg/m2/s2       kg m-2s-2      ML-2T-2
and one of M or F
relative density                       a ratio                         1
no units                    no dimension
As either mass (M) of force (F) can be used to represent                                                         viscosity                         N s/m2         N sm-2
the other, i.e.                                                                                                               kg/m s        kg m-1s-1      M L-1T-1
-1
surface tension                           N/m          Nm
F = MLT-2
kg /s2        kg s-2         MT-2
M = FT2L-1

We will mostly use LTM:

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1   Lectures 16-19     220    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                      Lectures 16-19     221
Unit 4                                                                                                       Unit 4
Dimensional Homogeneity                                                                                    What exactly do we get
from Dimensional Analysis?
Any equation is only true if both sides
have the same dimensions.
A single equation,
It must be dimensionally homogenous.
Which relates all the physical factors
of a problem to each other.
What are the dimensions of X?
2                                                                                                            An example:
B 2 gH 3/ 2        X
3                                                                       Problem: What is the force, F, on a propeller?
L (LT-2)1/2 L3/2 = X                                                                      What might influence the force?
L (L1/2T-1) L3/2 = X
L3 T-1 = X                                                 It would be reasonable to assume that the force, F,
depends on the following physical properties?
The powers of the individual dimensions must be equal
on both sides.                                                                                       diameter,                                                        d
(for L they are both 3, for T both -1).                                                             forward velocity of the propeller
(velocity of the plane),                             u
Dimensional homogeneity can be useful for:                                                                   fluid density,
1. Checking units of equations;                                                                                    revolutions per second,                                          N
2. Converting between two sets of units;                                                                           fluid viscosity,
3. Defining dimensionless relationships

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1               Lectures 16-19     222    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                  Lectures 16-19     223

Unit 4                                                                                                       Unit 4
How do we get the dimensionless groups?
From this list we can write this equation:
There are several methods.

F=           ( d, u, , N,    )                                                 We will use the strategic method based on:
or                                                                                                                         Buckingham’s                     theorems.
0=             ( F, d, u, , N,       )
There are two                        theorems:
and                                                                        st
1 are       unknown functions.                               1          theorem:
A relationship between m variables (physical properties
Dimensional Analysis produces:                                                such as velocity, density etc.) can be expressed as a
relationship between m-n non-dimensional groups of
variables (called groups), where n is the number of
F     Nd
,    ,                0                             fundamental dimensions (such as mass, length and time)
u2d 2 u     ud                                           required to express the variables.

These groups are dimensionless.
So if a problem is expressed:
will be determined by experiment.
( Q1 , Q2 , Q3 ,………, Qm ) = 0

These dimensionless groups help
to decide what experimental measurements to take.                                                                              Then this can also be expressed
(     1   ,   2   ,    3   ,………,    m-n   )=0

In fluids, we can normally take n = 3
(corresponding to M, L, T)

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1               Lectures 16-19     224    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                  Lectures 16-19     225
Unit 4                                                                                                                        Unit 4

2nd         theorem
An example
Each group is a function of n governing or repeating
variables plus one of the remaining variables.
Taking the example discussed above of force F induced
Choice of repeating variables                                                           on a propeller blade, we have the equation

Repeating variables appear in most of the                                          groups.                                                        0=            ( F, d, u, , N,                       )
They have a large influence on the problem.                                                                                                    n = 3 and m = 6
There is great freedom in choosing these.
There are m - n = 3                                           groups, so
Some rules which should be followed are                                                                                                        (         ,            ,       )=0
1            2       3
There are n ( = 3) repeating variables.
In combination they must contain                                                                              The choice of                        , u, d satisfies the criteria above.
all of dimensions (M, L, T)
The repeating variables must not form a
dimensionless group.                                                                                                                                      They are:

They do not have to appear in all                                     groups.                                                                             measurable,

The should be measurable in an experiment.                                                                                                 good design parameters

They should be of major interest to the designer.                                                                            contain all the dimensions M,L and T.

It is usually possible to take
, u and d
This freedom of choice means:
many different groups - all are valid.
There is not really a wrong choice.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                     Lectures 16-19     226    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                   Lectures 16-19     227

Unit 4                                                                                                                        Unit 4
a1
For the first                  group,             1                 ub1 d c1 F
We can now form the three groups                                                       In terms of dimensions
according to the 2nd theorem,                                                                                                                        a1               1 b1        c1
M 0 L0 T 0               ML3                      LT              L        M LT   2

a1
1
u b1 d c1 F
a2                                                    The powers for each dimension (M, L or T), the powers
u b2 d c2 N
2
must be equal on each side.
a3
3
u b3 d c3
for M:                     0 = a1 + 1
The            groups are all dimensionless,                                                                       a1 = -1
i.e. they have dimensions M0L0T0
for L:                     0 = -3a1 + b1 + c1 + 1
We use the principle of dimensional homogeneity to                                                                                   0 = 4 + b1 + c1
equate the dimensions for each group.

for T:                     0 = -b1 - 2
b1 = -2
c1 = -4 - b1 = -2

Giving            1   as
1
1
u 2d 2 F
F
1
u2d 2

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                     Lectures 16-19     228    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                   Lectures 16-19     229
Unit 4                                                                                                                                         Unit 4
a3
And a similar procedure is followed for the other                                                                            And for the third,                                     3                ub3 d c3
groups.                                                                                                                                                     0       0       0                           a3             1 b3        c3

a2    b2    c2
M LT                                 ML3              LT               L         ML 1T       1

Group             2
u d N
a1        1 b1       c
M 0 L0T 0                    ML3              LT          L 1T    1

for M:                     0 = a3 + 1
a3 = -1
for M:                     0 = a2
for L:                     0 = -3a3 + b3 + c3 -1
for L:                     0 = -3a2 + b2 + c2                                                                                                           b3 + c3 = -2
0 = b2 + c2
for T:                     0 = -b3 - 1
for T:                     0 = -b2 - 1                                                                                                                  b3 = -1
b2 = -1                                                                                                                      c3 = -1
c2 = 1

Giving            3   as
Giving            2 as                                                                                                                                                              u 1d
1           1
3
0    1       1
2              u d N
Nd                                                                                                                      3
ud
2
u

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                  Lectures 16-19     230    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                                    Lectures 16-19     231

Unit 4                                                                                                                                         Unit 4

Thus the problem may be described by                                                                                                                                Manipulation of the                                       groups

(           ,       ,        )=0                                                        Once identified the                                            groups can be changed.
1       2        3
The number of groups does not change.
Their appearance may change drastically.
F     Nd
,    ,                          0
u2d 2 u     ud
Taking the defining equation as:
This may also be written:
(     1      ,       2   ,           3   ………              m-n   )=0
The following changes are permitted:
F                      Nd
,                                                   i. Combination of exiting groups by multiplication or division
u2d 2                    u   ud
to form a new group to replaces one of the existing.

E.g. 1 and 2 may be combined to form                                                     1a   =    1   /   2   so the defining
Wrong choice of physical properties.                                                                      equation becomes

If, extra, unimportant variables are chosen :                                                                                             (    1a      ,       2   ,       3   ………              m-n    )=0
*        Extra groups will be formed                                                                                         ii. Reciprocal of any group is valid.
*        Will have little effect on physical performance                                                                                           (    1    ,1/            2   ,       3   ……… 1/              m-n    )=0
*        Should be identified during experiments                                                                             iii. A group may be raised to any power.
If an important variable is missed:                                                                                              ( ( 1 )2, ( 2 )1/2, ( 3 )3………                                                 m-n   )=0
A       group would be missing.                                                                        iv. Multiplied by a constant.

Experimental analysis may miss significant                                                             v. Expressed as a function of the other groups
behavioural changes.                                                                                                         2   =            (       1   ,       3   ………              m-n   )

Initial choice of variables                                                                  In general the defining equation could look like
should be done with great care.
(     1      , 1/            2   ,(      3   )i……… 0.5                 m-n     )=0

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                  Lectures 16-19     232    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                                    Lectures 16-19     233
Unit 4                                                                                                                Unit 4

An Example                                                                                                                     Common                   groups
Q. If we have a function describing a problem:
Several groups will appear again and again.
Q, d , , , p                             0
d 2 p 1/ 2                 d   1/ 2
p 1/ 2                                                                                                    These often have names.
Show that Q                            1/ 2

They can be related to physical forces.
Ans.
Other common non-dimensional numbers
Dimensional analysis using Q, , d will result in:
or (           groups):
d d4p
, 2                               0                                             Reynolds number:
Q  Q
ud
Re            inertial, viscous force ratio

The reciprocal of square root of                                                          2:                         Euler number:
1/ 2
1           Q                                                                                                       p
2 1/ 2  2a ,                                                                                             En                                       pressure, inertial force ratio
2  d p                                                                                                            u2
Multiply                     1   by this new group:                                                            Froude number:
1/ 2
d           Q                                                                         u2
1a               1    2a             2 1/ 2                                                                      Fn                                       inertial, gravitational force ratio
Q d p         d 1/ 2 p1/ 2                                                            gd
then we can say                                                                                                                               Weber number:
1/ 2       1/ 2           2   1/ 2
d             p              d p                                                ud
1/       1a   ,         2a                                        ,                  0                             We                                        inertial, surface tension force ratio
Q 1/ 2
or                                                                                                                  Mach number:
2        1/ 2             1/ 2       1/ 2                                                                      u
d p                     d           p                                                                        Mn                                        Local velocity, local velocity of sound ratio
Q                1/ 2                                                                                                       c

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                                   Lectures 16-19     234    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                           Lectures 16-19     235

Unit 4                                                                                                                Unit 4

Similarity                                                                                                                   Kinematic similarity

The similarity of time as well as geometry.
Similarity is concerned with how to transfer
It exists if:
measurements from models to the full scale.
i. the paths of particles are geometrically similar
ii. the ratios of the velocities of are similar
Three types of similarity
which exist between a model and prototype:
Some useful ratios are:
Vm L m / Tm
Geometric similarity:                                                                                                                   Velocity                                                     L
u
Vp L p / Tp
The ratio of all corresponding dimensions                                                                                                                                                                       T

in the model and prototype are equal.
am      Lm / Tm2        L
Acceleration                                                                      a
For lengths                                                                                                                                         ap      L p / Tp2       2
T

Lmodel    Lm
L
Lprototype Lp                                                                                                                                       Qm        L3 / Tm       3
m            L
Discharge                                                              Q
L     is the scale factor for length.                                                                                                                                     Qp        L3p / Tp      T

For areas                                                                                                     A consequence is that streamline
Amodel    L2
m                             2
patterns are the same.
L
Aprototype L2
p

All corresponding angles are the same.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                                   Lectures 16-19     236    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                           Lectures 16-19     237
Unit 4                                                                                                                  Unit 4
Dynamic similarity                                                                                               Modelling and Scaling Laws
Measurements taken from a model needs a scaling law
If geometrically and kinematically similar and
applied to predict the values in the prototype.
the ratios of all forces are the same.

An example:
Force ratio
2
3
Fm          M m am              m Lm    L                            2   L               2    2
3   2                             L                   L    u                                               For resistance R, of a body
Fp          M pa p              p Lp   T                                 T
moving through a fluid.
R, is dependent on the following:
This occurs when
the controlling group                                                                                              ML-3                      u:        LT-1                  l:(length) L     :     ML-1T-1
is the same for model and prototype.
So
(R, , u, l,                  )=0
The controlling group is usually Re.
So Re is the same for
model and prototype:                                                                                              Taking , u, l as repeating variables gives:
R                         ul
m um d m            pupd p                                                                                               u2l 2
m                      p                                                                                                                                ul
R            u2l 2

It is possible another group is dominant.                                                                                     This applies whatever the size of the body
In open channel i.e. river Froude number is                                                                                        i.e. it is applicable to prototype and
often taken as dominant.                                                                                                   a geometrically similar model.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                    Lectures 16-19     238    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                             Lectures 16-19     239

Unit 4                                                                                                                  Unit 4

For the model                                                                                                                  Example 1
An underwater missile, diameter 2m and length 10m is tested in a
Rm                          m
um lm                                          water tunnel to determine the forces acting on the real prototype. A
2 2
1/20th scale model is to be used. If the maximum allowable speed of the
m um lm                           m
prototype missile is 10 m/s, what should be the speed of the water in
the tunnel to achieve dynamic similarity?

and for the prototype
Dynamic similarity so Reynolds numbers equal:
Rp                         p   uplp
2 2                                                                                             m um d m pupd p
p
uplp                         p
m                      p

Dividing these two equations gives                                                                                             The model velocity should be
2 2
Rm /        m um lm                     m um lm /        m                                                                                                     p   dp        m
2 2                                                                                                                        um          up
Rp /        puplp                       p
uplp /       p                                                                                                     m
dm        p

W can go no further without some assumptions.                                                                       Both the model and prototype are in water then,
Assuming dynamic similarity, so Reynolds number are                                                                                  m = p and m = p so
the same for both the model and prototype:
m   um d m          p   upd p                                                                                              dp                 1
um        up               10                     200 m / s
m                      p                                                                                              dm              1 / 20
so
2 2
Rm             mum lm                                                                                       This is a very high velocity.
Rp              u2l 2
p p p                                                       This is one reason why model tests are not always done
i.e. a scaling law for resistance force:                                                                           at exactly equal Reynolds numbers.
2       2                                               A wind tunnel could have been used so the values of the
R              u       L
and ratios would be used in the above.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                                    Lectures 16-19     240    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                             Lectures 16-19     241
Unit 4                                                                                                           Unit 4
Example 2                                                                                                           So the model velocity is found to be
A model aeroplane is built at 1/10 scale and is to be tested in a wind                                                                                                    1 1
tunnel operating at a pressure of 20 times atmospheric. The aeroplane                                                                                        um        up                        0.5u p
will fly at 500km/h. At what speed should the wind tunnel operate to give                                                                                                20 1 / 10
dynamic similarity between the model and prototype? If the drag                                                                                              um        250 km / h
measure on the model is 337.5 N what will be the drag on the plane?
Earlier we derived an equation for resistance on a body
moving through air:                                                                                                 And the ratio of forces is
ul                                                                                           Rm              u2l 2     m
R             u2l 2                         u 2 l 2 Re
Rp              u2l 2     p
2        2
Rm         20 0.5              .
01
0.05
For dynamic similarity Rem = Rep, so                                                                                                                      Rp         1 1                 1

p   dp    m
So the drag force on the prototype will be
um         up
m
dm    p                                                                               1
Rp              Rm             20 337.5 6750 N
0.05
The value of                     does not change much with pressure so
m= p

For an ideal gas is p = RT so the density of the air in the
model can be obtained from
pm            m RT             m

pp            p
RT             p

20 p p              m

pp              p

m
20       p

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                         Lectures 16-19     242    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                      Lectures 16-19     243

Unit 4                                                                                                           Unit 4
Geometric distortion in river models

For practical reasons it is difficult to build a geometrically
similar model.

A model with suitable depth of flow will often be far too
big - take up too much floor space.

Keeping Geometric Similarity result in:
depths and become very difficult to measure;
the bed roughness becomes impracticably small;
laminar flow may occur -
(turbulent flow is normal in rivers.)

Solution: Abandon geometric similarity.

Typical values are
1/100 in the vertical and 1/400 in the horizontal.

Resulting in:
Good overall flow patterns and discharge
local detail of flow is not well modelled.

The Froude number (Fn) is taken as dominant.
Fn can be the same even for distorted models.

CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                         Lectures 16-19     244    CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1                      Lectures 16-19     245

```
To top