[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.]
Notes For the First Year Lecture Course:
An Introduction to Fluid Mechanics
School of Civil Engineering, University of Leeds.
x
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.
x
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.
CIVE1400 FLUID MECHANICS
Dr Andrew Sleigh
January 2006
0. Contents of the Course
0.1
Objectives:
x
The course will introduce fluid mechanics and establish its relevance in civil
engineering.
x
Develop the fundamental principles underlying the subject.
x
Demonstrate how these are used for the design of simple hydraulic components.
Fluids Lecture and Test Schedule
0.2
Consists of:
x
x
x
Week
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.
0
1
Assessment:
1 Exam of 2 hours, worth 80% of the module credits.
This consists of 6 questions of which you choose 4.
2
2 Multiple choice question (MCQ) papers, worth 10% of the module credits.
These will be for 30mins and set during the lectures. The timetable for these MCQs
and lectures is shown in the table at the end of this section.
3
1 Marked problem sheet, worth 10% of the module credits.
5
4
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.
CIVE 1400: Fluid Mechanics
Introduction
0.3
Specific Elements:
x Introduction
x Fluid Properties
x Fluids vs. Solids
x Viscosity
x Newtonian Fluids
x Properties of Fluids
x Statics
x Hydrostatic pressure
x Manometry / pressure measurement
x Hydrostatic forces on submerged surfaces
x Dynamics
x The continuity equation.
x The Bernoulli Equation.
x Applications of the Bernoulli equation.
x The momentum equation.
x Application of the momentum equation.
x Real Fluids
x Boundary layer.
x Laminar flow in pipes.
x Introduction to dimensional analysis
x Dimensions
x Similarity
0.4
Books:
6
7
8
9
10
day
17
18
24
25
31
February 1
7
8
14
15
21
22
28
March
1
7
8
14
15
21
22
April
11
12
1
Date
Month
January
May
CIVE 1400: Fluid Mechanics
19
20
26
27
3
4
Tue
Wed
Tue
Wed
Tue
Wed
Tue
Wed
Tue
Wed
Tue
Wed
Tue
Wed
Tue
Wed
Tue
Wed
Tue
Wed
Tue
Wed
Tue
Wed
Tue
Wed
Subject
Lecture
Fluid properties
1
2
3
4
5
6
7
8
9
10
11
12
Statics
Dynamics
Test
MCQ
Problem Sheet
Surveying
Surveying
Surveying
Surveying
Real fluids
Easter Vacation
Dimensional analysis
Revision Lectures
13
14
15
16
17
18
19
20
21
22
MQC
Introduction
2
These notes give more information than is found in the lectures. They do not replace textbooks.
You must also read at least one of the recommended fluid mechanics books. The notes
may be read online or printed off for personal use.
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.
0.5
Other Teaching Resources.
There are some extra teaching/learning resources available for you to use that are computer
based.
Online Lecture Notes
A more detailed set of lecture notes can be found on the WWW at he following address:
http://www.efm.leeds.ac.uk/cive
You get to this using Netscape from any of the computers in the university.
If you forget this address you can also get to the web pages via Dr Sleigh's web pages linked
from the department's main page.
CIVE 1400: Fluid Mechanics
Introduction
3
CIVE 1400: Fluid Mechanics
Introduction
4
0.6
Civil Engineering Fluid Mechanics
0.7
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 are 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:
x
Sea and river (flood) defences;
x
Water distribution / sewerage (sanitation) networks;
x
Hydraulic design of water/sewage treatment works;
x
Dams;
x
Irrigation;
x
Pumps and Turbines;
x
Water retaining structures.
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.
0.8
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 practise which are made from combinations
of these primary units.
And some examples where the primary object is construction - yet analysis of the fluid
mechanics is essential:
x
Flow of air in / around buildings;
x
Bridge piers in rivers;
x
Ground-water flow.
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!
Primary Units
The six primary units of the SI system are shown in the table below:
Notice how nearly all of these involve water. The following course, although introducing general
fluid flow ideas and principles, will demonstrate many of these principles through examples
where the fluid is water.
Quantity
SI Unit
Dimension
length
mass
time
temperature
current
luminosity
metre, m
kilogram, kg
second, s
Kelvin, K
ampere, A
candela
L
M
T
T
I
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
Introduction
5
Derived Units
CIVE 1400: Fluid Mechanics
0.9
There are many derived units all obtained from combination of the above primary units. Those
most used are shown in the table below:
Quantity
velocity
acceleration
force
energy (or work)
power
pressure ( or stress)
density
specific weight
relative density
viscosity
surface tension
SI Unit
m/s
m/s2
N
kg m/s2
Joule J
N m,
kg m2/s2
Watt W
N m/s
kg m2/s3
Pascal
P,
N/m2,
kg/m/s2
kg/m3
N/m3
kg/m2/s2
a ratio
no units
N s/m2
kg/m s
N/m
kg /s2
ms-1
ms-2
Dimension
LT-1
LT-2
kg ms-2
M LT-2
2 -2
2 -2
kg m s
ML T
Nms-1
kg m2s-3
ML2T-3
Nm-2
kg m-1s-2
ML-1T-2
kg m-3
ML-3
Introduction
6
Example: 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?
-2 -2
kg m s
N sm-2
kg m-1s-1
Nm-1
kg s-2
ML-2T-2
1
no dimension
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?
M L-1T-1
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.
One very useful tip is to write down the units of any equation you are using. If at the end the
units do not match you know you have made a mistake. For example is you have at the end of a
calculation,
30 kg/m s = 30 m
you have certainly made a mistake - checking the units can often help find the mistake.
More on this subject will be seen later in the section on dimensional analysis and similarity.
CIVE 1400: Fluid Mechanics
Introduction
7
CIVE 1400: Fluid Mechanics
Introduction
8
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
Section 0: Introduction
Section 1: Fluid Properties
Fluids vs. Solids
Viscosity
Newtonian Fluids
Properties of Fluids
Section 2: Statics
Hydrostatic pressure
Manometry/Pressure measurement
Hydrostatic forces on
submerged surfaces
Section 3: Dynamics
The continuity equation.
The Bernoulli Equation.
Application of Bernoulli equation.
The momentum equation.
Application of momentum equation.
Section 4: Real Fluids
Boundary layer.
Laminar flow in pipes.
Section 5: Dimensional Analysis
An Intro to Dimensional analysis
Similarity
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
xThe nature of a fluid is different to that of a solid
xIn fluids we deal with continuous
streams of fluid.
In solids we only consider individual elements.
In this section we will consider how we
can classify the differences in nature
of fluids and solids.
What do we mean by nature of a fluid?
Fluids are clearly different to solids.
But we must be specific.
We need some definable basic
physical difference.
1
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
CIVE1400: Fluid Mechanics
We know that fluids flow under the
action of a force, and the solids don’t but solids do deform.
xfluids lack the ability of solids to
resist deformation.
2
Section 1: Fluid Properties
What use can we make of these ideas?
Take the rectangular element below.
xfluids change shape as long as a
force acts.
What forces cause it to deform?
(These definitions include both
gasses and liquids as fluids.)
A
C
Section 1: Fluid Properties
Section 1: Fluid Properties
In the analysis of fluids
we often take small volumes (elements)
and examine the forces on these.
So we can say that
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
What make fluid mechanics different
to solid mechanics?
LECTURE CONTENTS
CIVE1400: Fluid Mechanics
CIVE1400: Fluid Mechanics
3
CIVE1400: Fluid Mechanics
B
D
Section 1: Fluid Properties
4
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
A’
B’
F
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
Fluids in motion
Consider a fluid flowing near a wall.
- in a pipe for example -
F
C
D
Fluid next to the wall will have zero velocity.
Forces acting along edges (faces), such as F,
are know as shearing forces.
The fluid “sticks” to the wall.
From this we arrive at the definition:
Moving away from the wall velocity increases
to a maximum.
A Fluid is a substance which deforms continuously,
or flows, when subjected to shearing forces.
This has the following implications
for fluids at rest:
v
Plotting the velocity across the section gives
“velocity profile”
If a fluid is at rest there are NO shearing forces acting
on it, and
any force must be acting perpendicular to the fluid
Change in velocity with distance is
“velocity gradient” =
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
CIVE1400: Fluid Mechanics
5
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
CIVE1400: Fluid Mechanics
As fluids are usually near surfaces
there is usually a velocity gradient.
du
dy
Section 1: Fluid Properties
6
Section 1: Fluid Properties
What use is this observation?
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.
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.
What if not near a wall?
Consider this 3-d rectangular element,
under the action of the force F.
v
No velocity gradient, no shear forces.
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
7
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
8
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
CIVE1400: Fluid Mechanics
δx
A 2-d view may be clearer…
b
a
δz
Section 1: Fluid Properties
A’
B
F
E x
φ
B
A
B’
F
E’
y
δy
F
C
F
D
The shearing force acts on the area
C
A Gz u Gx
D
under the action of the force F
Shear stress, W is the force per unit area:
b
a’
a
b’
F
A’
A
B
The deformation which shear stress causes is
measured by the angle I, and is know as
shear strain.
B’
E
F
C
F
A
W
Using these definitions we can amend our
definition of a fluid:
D
In a fluid I increases for as long as W is applied the fluid flows
In a solid shear strain, I, is constant for a fixed
shear stress W.
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
CIVE1400: Fluid Mechanics
9
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties 10
Section 1: Fluid Properties
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
It has been shown experimentally that the
rate of shear strain is directly
proportional to shear stress
Wv
W
I
W
Constant u
u/y is the rate of change of velocity with distance,
in differential form this is
time
Constant u
I
t
= velocity gradient.
giving
If a particle at point E moves to point E’ in
time t then:
for small deformations
x
y
du
dy
The constant of proportionality is known as
the dynamic viscosity, P
We can express this in terms of the cuboid.
shear strain I
u
y
W
P
du
dy
which is know as Newton’s law of viscosity
rate of shear strain
A fluid which obeys this rule is know as a
Newtonian Fluid
(sometimes also called real fluids)
Newtonian fluids have constant values of P
(note that
x
t
u
is the velocity of the particle at E)
Non-Newtonian Fluids
So
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties 11
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties 12
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
Some fluids do not have constant P.
They do not obey Newton’s Law of viscosity.
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
This graph shows how P changes for different fluids.
Bingham plastic
The general relationship is:
W
§ Gu ·
A B¨ ¸
© Gy ¹
n
Newtonian
Shear stress, τ
They do obey a similar relationship and can
be placed into several clear categories
Pseudo plastic
plastic
Dilatant
where A, B and n are constants.
Ideal, (τ=0)
For Newtonian fluids A = 0, B = P and n = 1
Rate of shear, δu/δy
x Plastic: Shear stress must reach a certain minimum before
flow commences.
x Bingham plastic: As with the plastic above a minimum shear
stress must be achieved. With this classification n = 1. An
example is sewage sludge.
x Pseudo-plastic: No minimum shear stress necessary and the
viscosity decreases with rate of shear, e.g. colloidial
substances like clay, milk and cement.
x Dilatant substances; Viscosity increases with rate of shear
e.g. quicksand.
x Thixotropic substances: Viscosity decreases with length of
time shear force is applied e.g. thixotropic jelly paints.
x Rheopectic substances: Viscosity increases with length of
time shear force is applied
xViscoelastic materials: Similar to Newtonian but if there is a
sudden large change in shear they behave like plastic.
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties 13
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties 14
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
Properties of Fluids:
Liquids vs. Gasses
Density
Liquids and gasses behave in much the same way
Some specific differences are
1. A liquid is difficult to compress and often regarded as
being incompressible.
A gas is easily to compress and usually treated as
such - it changes volume with pressure.
2. A given mass of liquid occupies a given volume and
will form a free surface
A gas has no fixed volume, it changes volume to
expand to fill the containing vessel. No free surface is
formed.
There are three ways of expressing density:
1. Mass density:
U
mass per unit volume
U
mass of fluid
volume of fluid
Units: kg/m3
Causes of Viscosity in Fluids
Dimensions:
ML3
Viscosity in Gasses
xMainly due to molecular exchange between layers
Mathematical considerations of this momentum
exchange can lead to Newton law of viscosity.
Typical values:
xIf temperature increases the momentum exchange
between layers will increase thus increasing viscosity.
Air = 1.23 kg m
Water = 1000 kg m
3
3
, Mercury = 13546 kg m
, Paraffin Oil = 800 kg m
3
3
.
Viscosity in Liquids
xThere is some molecular interchange between layers
in liquids - but the cohesive forces are also important.
xIncreasing temperature of a fluid reduces the cohesive
forces and increases the molecular interchange.
Resulting in a complex relationship between
temperature and viscosity.
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties 15
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties 16
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
2. Specific Weight:
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
Viscosity
(sometimes known as specific gravity)
Z
Z
weight per unit volume
Ug
Units: Newton’s per cubic metre, N
2 2
Dimensions: ML
Typical values:
Water =9814
Air =12.07
T
/m
3
There are two ways of expressing viscosity
1. Coefficient of Dynamic Viscosity
(or kg/m2/s2)
.
P = the shear stress, W, required to drag one layer of
fluid with unit velocity past another layer a unit distance
away.
du
dy
Force Velocity
Area Distance
Force u Time
=
Area
Mass
Length u Time
N m 3 , Mercury = 132943 N m 3 ,
P W
N m 3 , Paraffin Oil =7851 N m 3
3. Relative Density:
V
ratio of mass density to
a standard mass density
V
U subs tan ce
U
$
H 2O ( at 4 c )
For solids and liquids this standard mass density is the
maximum mass density for water (which occurs at
at atmospheric pressure.
Units: none, as it is a ratio
Dimensions: 1.
Typical values:
Water = 1, Mercury = 13.5, Paraffin Oil =0.8.
4$ c)
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties 17
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties
2. Kinematic Viscosity
Q = the ratio of dynamic viscosity to mass density.
Q
P
U
Units: m2s-1
Dimension: L2T-1
Typical values:
Water =1.14 u 10-6 m2/s, , Air =1.46 u 10-5 m2/s m s ,
Mercury =1.145 u 10-4 m2/s, Paraffin Oil =2.375 u 10-3
m2/s.
2 1
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties 19
Units: N s/m2 or kg/m s (kg m-1 s-1)
(Note that P is often expressed in Poise, P,
where 10 P = 1 N s/m2.)
Dimensions: ML-1T-1
Typical values:
Water =1.14 u 10-3 Ns/m2, Air =1.78 u 10-5 Ns/m2,
Mercury =1.552 Ns/m2, Paraffin Oil =1.9 Ns/m2.
CIVE1400: Fluid Mechanics
CIVE1400: Fluid Mechanics
Section 1: Fluid Properties 18
Section 2: Statics
LECTURE CONTENTS
Section 0: Introduction
Section 1: Fluid Properties
Fluids vs. Solids
Viscosity
Newtonian Fluids
Properties of Fluids
Section 2: Statics
Hydrostatic pressure
Manometry/Pressure measurement
Hydrostatic forces on
submerged surfaces
Section 3: Dynamics
The continuity equation.
The Bernoulli Equation.
Application of
Bernoulli equation.
The momentum equation.
Application of momentum equation.
Section 4: Real Fluids
Boundary layer.
Laminar flow in pipes.
Section 5: Dimensional Analysis
An Intro to Dimensional analysis
Similarity
CIVE1400: Fluid Mechanics
Section 2: Statics 20
CIVE1400: Fluid Mechanics
Section 2: Statics
CIVE1400: Fluid Mechanics
Section 2: Statics
STATICS
What do we know about the forces
involved with static fluids?
4 Lectures
Objectives
From earlier we know that:
xIntroduce idea of pressure.
1. A static fluid can have no shearing force acting on it.
xProve unique value at any particular elevation.
2. Any force between the fluid and the boundary must
be acting at right angles to the boundary.
xShow hydrostatic relationship.
xExpress pressure in terms of head of fluid.
F1
F2
R1
xPressure measurement using manometers.
F
R2
Fn
x Determine the magnitude and
direction of forces on submerged surfaces
R
Rn
Pressure force normal to the boundary
CIVE1400: Fluid Mechanics
Section 2: Statics 21
CIVE1400: Fluid Mechanics
CIVE1400: Fluid Mechanics
Section 2: Statics
CIVE1400: Fluid Mechanics
Section 2: Statics 22
Section 2: Statics
This is also true for:
Pressure
x curved surfaces
Convenient to work in terms of pressure, p,
which is the force per unit area.
x any imaginary plane
pressure
An element of fluid at rest is in equilibrium:
p
3. The sum of forces in any
direction is zero.
4. The sum of the moments of forces
about any point is zero.
Force
Area over which the force is applied
F
A
Units: Newton’s 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)
Uniform Pressure:
If the force on each unit area
of a surface is equal then
uniform pressure
CIVE1400: Fluid Mechanics
Section 2: Statics 23
CIVE1400: Fluid Mechanics
Section 2: Statics 24
CIVE1400: Fluid Mechanics
Section 2: Statics
CIVE1400: Fluid Mechanics
Summing forces in the x-direction:
Pascal’s Law
Proof that pressure acts equally in all directions.
Force in the x-direction due to px,
p x u Area ABFE
Fx x
ps
px
Fx s
δs
A
δy
F
ps u Area ABCD u sin T
psGs Gz
C
θ
Gy
Gs
psGy Gz
D
E
p x Gx Gy
Force in the x-direction due to ps,
B
δz
Section 2: Statics
δx
( sin T
py
Gy
Gs )
Force in x-direction due to py,
Fx y
Remember:
No shearing forces
0
To be at rest (in equilibrium)
Fx x Fx s Fx y
All forces at right angles to the surfaces
0
p xGxGy psGyGz 0
px
ps
CIVE1400: Fluid Mechanics
Section 2: Statics 25
CIVE1400: Fluid Mechanics
Section 2: Statics 26
CIVE1400: Fluid Mechanics
Section 2: Statics
CIVE1400: Fluid Mechanics
Section 2: Statics
To be at rest (in equilibrium)
Summing forces in the y-direction.
Fy Fy Fy weight
y
s
x
0
1
§
·
p yGxGy psGxGz ¨ Ug GxGyGz¸
©
¹
2
0
Force due to py,
Fy
p y u Area ABCD
y
p y GxGz
Component of force due to ps,
Fy
s
ps u Area ABCD u cosT
psGsGz
Gx
Gs
The element is small i.e. Gx, Gx, and Gz, are small,
so Gx u Gy u Gz, is very small
and considered negligible, hence
py
psGxGz
( cos T
Gx
Gs )
ps
We showed above
px
Fy x
ps
thus
Component of force due to px,
0
px
py
ps
Force due to gravity,
weight = - specific weight u volume of element
1
= Ug u GxGyGz
2
CIVE1400: Fluid Mechanics
Pressure at any point is the same in all directions.
This is Pascal’s Law and applies to fluids at rest.
Section 2: Statics 27
CIVE1400: Fluid Mechanics
Section 2: Statics 28
CIVE1400: Fluid Mechanics
Section 2: Statics
CIVE1400: Fluid Mechanics
Section 2: Statics
The forces involved are:
Vertical Variation Of Pressure
In A Fluid Under Gravity
Force due to p1 on A (upward) = p1A
Force due to p2 on A (downward) = p2A
p2, A
Area A
Fluid density ρ
Force due to weight of element (downward)
= mg
= mass density u volume u g
= U g A(z2 - z1)
z2
Taking upward as positive, we have
p1, A
p1 A p2 A UgA z2 z1 = 0
z1
p2 p1
Vertical cylindrical element of fluid
cross sectional area = A
UgA z2 z1 Thus in a fluid under gravity, pressure
decreases linearly with increase in height
mass density = U
p2 p1
UgA z2 z1 This is the hydrostatic pressure change.
CIVE1400: Fluid Mechanics
Section 2: Statics 29
CIVE1400: Fluid Mechanics
Section 2: Statics 30
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
pr, A
pl, A
Face L
L
Face R
R
weight, mg
Horizontal cylindrical element
cross sectional area = A
We have shown
pl = pr
mass density = U
For a vertical pressure change we have
left end pressure = pl
right end pressure = pr
pl
p p Ugz
pr
pq Ugz
and
For equilibrium the sum of the
forces in the x direction is zero.
pl A = pr A
so
p p Ugz
pp
pl = pr
Pressure in the horizontal direction is constant.
pq Ugz
pq
Pressure at the two equal levels are the same.
This true for any continuous fluid.
CIVE1400: Fluid Mechanics
Section 2: Statics 31
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Section 2: Statics 32
CIVE1400: Fluid Mechanics
Section 2: Statics
General Equation For
Variation Of Pressure In A Static Fluid
CIVE1400: Fluid Mechanics
Horizontal
If T
$
90 then s is in the x or y directions, (i.e.
horizontal),so
(p + δp)A
Area A
Section 2: Statics
§ dp ·
¨ ¸
© ds ¹ T 90$
Fluid density ρ
δs
dp
dx
dp
dy
0
θ
Confirming that pressure change
on any horizontal plane is zero.
z + δz
mg
Vertical
pA
z
If T
A cylindrical element of fluid at an arbitrary
orientation.
$
0 then s is in the z direction (vertical) so
§ dp ·
¨ ¸
© ds ¹ T 0$
dp
dz
Ug
From considering incremental forces (see detailed
notes on WWW or other texts) we get
Confirming the hydrostatic result
dp
ds
Ug cos T
p2 p1
z2 z1
Ug
p2 p1
Ug z2 z1 CIVE1400: Fluid Mechanics
Section 2: Statics 33
CIVE1400: Fluid Mechanics
Section 2: Statics 34
CIVE1400: Fluid Mechanics
Section 2: Statics
CIVE1400: Fluid Mechanics
Section 2: Statics
Pressure And Head
We have the vertical pressure relationship
dp
dz
It is convenient to take atmospheric
pressure as the datum
Ug ,
Pressure quoted in this way is known as
gauge pressure i.e.
integrating gives
p = -Ugz + constant
Gauge pressure is
pgauge = U g h
measuring z from the free surface so that z = -h
z
h
The lower limit of any pressure is
the pressure in a perfect vacuum.
y
x
p
Pressure measured above
a perfect vacuum (zero)
is known as absolute pressure
Ugh constant
surface pressure is atmospheric, patmospheric .
Absolute pressure is
patmospheric
constant
pabsolute = U g h + patmospheric
so
p
CIVE1400: Fluid Mechanics
Ugh patmospheric
Section 2: Statics 35
Absolute pressure = Gauge pressure + Atmospheric
CIVE1400: Fluid Mechanics
Section 2: Statics 36
CIVE1400: Fluid Mechanics
Section 2: Statics
A gauge pressure can be given
CIVE1400: Fluid Mechanics
Section 2: Statics
Pressure Measurement By Manometer
using height of any fluid.
p
Ugh
Manometers use the relationship between pressure
and head to measure pressure
This vertical height is the head.
If pressure is quoted in head,
the density of the fluid must also be given.
Example:
What is a pressure of 500 kNm-2 in
head of water of density, U = 1000 kgm-3
Use p = Ugh,
h
500 u 103
1000 u 9.81
p
Ug
The Piezometer Tube Manometer
The simplest manometer is an open tube.
This is attached to the top of a container with liquid
at pressure. containing liquid at a pressure.
h1
A
In head of Mercury density U = 13.6u103 kgm-3.
3
h
500 u 10
3
h2
50.95m of water
B
3.75m of Mercury
13.6 u 10 u 9.81
In head of a fluid with relative density J = 8.7.
The tube is open to the atmosphere,
remember U = J u Uwater)
3
h
500 u 10
586
. m of fluid J = 8.7
8.7 u 1000 u 9.81
The pressure measured is relative to
atmospheric so it measures gauge pressure.
CIVE1400: Fluid Mechanics
Section 2: Statics 37
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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
pA = U g h 1
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
would the maximum measurable gauge pressure?
Pressure at B = pressure due to column of liquid h2
pB = U g h 2
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
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Section 2: Statics 40
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
and gases to be measured
“U” is connected as shown and filled with
manometric fluid.
Pressure in a continuous static fluid
is the same at any horizontal level.
pressure at B = pressure at C
pB = pC
Important points:
1. The manometric fluid density should be
greater than of the fluid measured.
Uman > U
For the left hand arm
pressure at B = pressure at A + pressure of height of
liquid being measured
pB = pA + Ugh1
2. The two fluids should not be able to mix
they must be immiscible.
For the right hand arm
pressure at C = pressure at D + pressure of height of
manometric liquid
Fluid density ρ
D
pC = patmospheric + Uman gh2
h2
A
h1
B
C
We are measuring gauge pressure we can subtract
patmospheric giving
pB = pC
Manometric fluid density ρ
pA = Uman gh2 - Ugh1
man
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?
Nothing changes.
The manometer work exactly the same.
Using a u-tube manometer to measure gauge
pressure of fluid density U = 700 kg/m3, and the
manometric fluid is mercury, with a relative density
of 13.6.
What is the gauge pressure if:
a) h1 = 0.4m and h2 = 0.9m?
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,
Uman >> U
Ugh1 can be neglected,
and the gauge pressure given by
pA = Uman gh2
CIVE1400: Fluid Mechanics
Section 2: Statics 43
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Section 2: Statics 44
CIVE1400: Fluid Mechanics
Section 2: Statics
Pressure difference measurement
Using a “U”-Tube Manometer.
CIVE1400: Fluid Mechanics
Section 2: Statics
pressure at C = pressure at D
pC = pD
The “U”-tube manometer can be connected
at both ends to measure pressure difference between
these two points
pC = pA + U g ha
pD = pB + U g (hb + h) + Uman g h
B
pA + U g ha = pB + U g (hb + h) + Uman g h
Fluid density ρ
Giving the pressure difference
hb
E
pA - pB = U g (hb - ha) + (Uman - U)g h
h
A
Again if the fluid is a gas Uman >> U, then the terms
involving U can be neglected,
ha
D
C
pA - pB = Uman g h
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
difference measuring
In the figure below two pipes containing the same
fluid of density U = 990 kg/m3 are connected using a
u-tube manometer.
What is the pressure between the two pipes if the
manometer contains fluid of relative density 13.6?
Fluid density ρ
Fluid density ρ
A
B
ha = 1.5m
Advances to the “U” tube manometer
Problem: Two reading are required.
Solution: Increase cross-sectional area
of one side.
Result:
One level moves
much more than the other.
p2
p1
E
diameter D
hb = 0.75m
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 = Uman g h
CIVE1400: Fluid Mechanics
Section 2: Statics 47
CIVE1400: Fluid Mechanics
Section 2: Statics 48
CIVE1400: Fluid Mechanics
Section 2: Statics
volume of liquid moved from
the left side to the right
= z2 u ( Sd2 / 4)
CIVE1400: Fluid Mechanics
Section 2: Statics
Problem: Small pressure difference,
movement cannot be read.
The fall in level of the left side is
Volume moved
Area of left side
z 2 Sd 2 / 4
z1
Solution 1: Reduce density of manometric
fluid.
SD 2 / 4
§d·
z2 ¨ ¸
© D¹
Result:
2
Greater height change easier to read.
Putting this in the equation,
2
ª
d º
Ug « z 2 z 2 §¨ ·¸ »
© ¹
p1 p2
D
¬
Solution 2: Tilt one arm of the manometer.
¼
2
ª
d º
Ugz 2 «1 §¨ ·¸ »
© ¹
D
¬
Result:
¼
If D >> d then (d/D)2 is very small so
p1 p2 Ugz2
Same height change - but larger
movement along the
manometer arm - easier to read.
Inclined manometer
CIVE1400: Fluid Mechanics
Section 2: Statics 49
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Section 2: Statics 50
CIVE1400: Fluid Mechanics
Section 2: Statics
CIVE1400: Fluid Mechanics
Section 2: Statics
p1
p2
diameter d
diameter D
er
x
ad
e
eR
al
Sc
z2
Datum line
z1
Example of an inclined manometer.
An inclined manometer is required to measure an air
pressure of 3mm of water to an accuracy of +/- 3%.
The inclined arm is 8mm in diameter and the larger
arm has a diameter of 24mm. The manometric fluid
has density Uman = 740 kg/m3 and the scale may be
read to +/- 0.5mm.
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
Ugz2
but,
z2
p1 p2
x sin T
Ugx sin T
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
Burrs cause local pressure variations.
We have seen these features of static fluids
Disadvantages:
x Slow response - only really useful for very slowly
varying pressures - no use at all for fluctuating
pressures;
x For the “U” tube manometer two measurements
must be taken simultaneously to get the h value.
x It is often difficult to measure small variations in
pressure.
x It cannot be used for very large pressures unless
several manometers are connected in series;
x For very accurate work the temperature and
relationship between temperature and U must be
known;
x Hydrostatic vertical pressure distribution
x Pressures at any equal depths in a continuous
fluid are equal
x Pressure at a point acts equally in all
directions (Pascal’s law).
xForces from a fluid on a boundary acts at right
angles to that boundary.
Fluid pressure on a surface
Pressure is force per unit area.
Pressure p acting on a small area GA exerted
force will be
Advantages of manometers:
x They are very simple.
F = puGA
x 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
F =p δA
n n n
The total or resultant force, R, on the
plane is the sum of the forces on the
small elements i.e.
R p1GA1 p 2 GA2 p n GAn ¦ pGA
and
This resultant force will act through the
centre of pressure.
For a plane surface all forces acting
can be represented by one single
resultant force,
acting at right-angles to the plane
through the centre of pressure.
CIVE1400: Fluid Mechanics
Section 2: Statics 55
The pressure, p, will be equal at all points of
the surface.
The resultant force will be given by
R pressure u area of plane
R = pA
Curved submerged surface
Each elemental force is a different
magnitude and in a different direction (but
still normal to the surface.).
It is, in general, not easy to calculate the
resultant force for a curved surface by
combining all elemental forces.
The sum of all the forces on each element
will always be less than the sum of the
individual forces, ¦ pGA .
CIVE1400: Fluid Mechanics
Section 2: Statics 56
CIVE1400: Fluid Mechanics
Section 2: Statics
CIVE1400: Fluid Mechanics
¦ zGA is known as
Resultant Force and Centre of Pressure on a
general plane surface in a liquid.
O
O
Fluid
density ρ
Q
z
z
the 1st Moment of Area of the
plane PQ about the free surface.
θ
elemental
area δA
s
Resultant
Force R D
G
area δA
G
x
C
And it is known that
¦ zGA Az
d
area A
Sc
P
x
Take pressure as zero at the surface.
A is the area of the plane
z is the distance to the centre of gravity
(centroid)
Measuring down from the surface, the pressure on
an element GA, depth z,
p = Ugz
In terms of distance from point O
¦ zGA
So force on element
Ax sin T
about a line through O
Resultant force on plane
(as
Ug ¦ zGA
z
x sin T )
The resultant force on a plane
(assuming U and g as constant).
R
CIVE1400: Fluid Mechanics
Section 2: Statics 57
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Section 2: Statics
CIVE1400: Fluid Mechanics
This resultant force acts at right angles
through the centre of pressure, C, at a depth D.
How do we find this position?
Take moments of the forces.
UgAz
UgAx sin T
Section 2: Statics 58
Section 2: Statics
Sum of moments
Moment of R about O =
As the plane is in equilibrium:
The moment of R will be equal to the sum of the
moments of the forces on all the elements GA
about the same point.
R u S c = UgAx sin T S c
Ug sin T ¦ s 2 GA
The position of the centre of pressure along the
plane measure from the point O is:
¦ s GA
2
Sc
It is convenient to take moment about O
Ax
How do we work out
the summation term?
The force on each elemental area:
UgzGA
Ug s sin T GA
the moment of this force is:
Moment of Force on GA about O
Ug sin T ¦ s 2 GA
Equating
UgAx sin T S c
Force on GA
u sinT
= 1st moment of area
F = UgzGA
R
Section 2: Statics
Ug s sin T GA u s
Ug sin T GAs 2
This term is known as the
2nd Moment of Area , Io,
of the plane
(about the axis through O)
U , g and T are the same for each element, giving the
total moment as
CIVE1400: Fluid Mechanics
Section 2: Statics 59
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Section 2: Statics 60
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Section 2: Statics
2nd moment of area about O
Io
CIVE1400: Fluid Mechanics
¦ s 2GA
It can be easily calculated
for many common shapes.
How do you calculate the 2nd moment of
area?
2nd moment of area is a geometric property.
It can be found from tables BUT only for moments about
an axis through its centroid = IGG.
The position of the centre of pressure
along the plane measure from the point O is:
Sc
Section 2: Statics
2 nd Moment of area about a line through O
Usually we want the 2nd moment of area
about a different axis.
1st Moment of area about a line through O
Through O in the above examples.
and
We can use the
Depth to the centre of pressure is
parallel axis theorem
D S c sin T
to give us what we want.
CIVE1400: Fluid Mechanics
Section 2: Statics 61
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Section 2: Statics 62
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Section 2: Statics
CIVE1400: Fluid Mechanics
Section 2: Statics
nd
The 2 moment of area about a line
through the centroid of some common
shapes.
The parallel axis theorem can be written
Io
I GG Ax 2
Shape
We then get the following
equation for the
position of the centre of pressure
Area A
2nd moment of area, I GG ,
about
an axis through the centroid
bd
bd 3
12
bd
2
bd 3
36
SR 2
SR 4
Rectangle
b
h
Sc
D
G
G
I GG
x
Ax
§I
·
sin T ¨ GG x ¸
© Ax
¹
Triangle
h
G
h/3
G
b
Circle
(In the examination the parallel axis theorem
R
G
G
and the I GG will be given)
4
Semicircle
G
CIVE1400: Fluid Mechanics
Section 2: Statics 63
SR
R
(4R)/(3π)
CIVE1400: Fluid Mechanics
2
2
01102
.
R4
Section 2: Statics 64
CIVE1400: Fluid Mechanics
Section 2: Statics
An example:
Find the moment required to keep this triangular
gate closed on a tank which holds water.
1.2m
Section 2: Statics
Submerged vertical surface Pressure diagrams
For vertical walls of constant width
it is possible to find the resultant force and
centre of pressure graphically using a
D
2.0m
pressure diagram.
1.5m
G
CIVE1400: Fluid Mechanics
C
We know the relationship between
pressure and depth:
p = Ugz
So we can draw the diagram below:
ρgz
z
2H
3
H
R
p
ρgH
This is know as a pressure diagram.
CIVE1400: Fluid Mechanics
Section 2: Statics 65
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Section 2: Statics 66
CIVE1400: Fluid Mechanics
Section 2: Statics
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Section 2: Statics
Pressure increases from zero at the
surface linearly by p = Ugz, to a
maximum at the base of p = UgH.
The area of this triangle represents the
resultant force per unit width on the
vertical wall,
For a triangle the centroid is at 2/3 its height
i.e. the resultant force acts
2
H.
horizontally through the point z
3
For a vertical plane the
depth to the centre of pressure is given by
Units of this are Newtons per metre.
Area
1
u AB u BC
2
1
HUgH
2
1
UgH 2
2
D
2
H
3
Resultant force per unit width
R
1
UgH 2
2
( N / m)
The force acts through the centroid of
the pressure diagram.
CIVE1400: Fluid Mechanics
Section 2: Statics 67
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Section 2: Statics 68
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Section 2: Statics
The resultant force is given by:
UgAz
oil ρo
UgAx sinT
Ug H u 1
H
sinT
2
0.8m
1.2m
water ρ
1
UgH 2
2
D
R
ρg0.8
and the depth to the centre of pressure by:
D
Section 2: Statics
The same technique can be used with combinations
of liquids are held in tanks (e.g. oil floating on water).
For example:
Check this against
the moment method:
R
CIVE1400: Fluid Mechanics
§I ·
sin T ¨ o ¸
© Ax ¹
ρg1.2
Find the position and magnitude of the resultant
force on this vertical wall of a tank which has oil
floating on water as shown.
and by the parallel axis theorem (with width of 1)
Io
I GG Ax 2
1u H3
H
1 u H §¨ ·¸
©
12
2¹
2
H3
3
Depth to the centre of pressure
§ H 3 / 3·
D ¨ 2 ¸
© H / 2¹
2
H
3
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Section 2: Statics 69
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Section 2: Statics 70
CIVE1400: Fluid Mechanics
Section 2: Statics
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Section 2: Statics
Submerged Curved Surface
If the surface is curved the resultant force
must be found by combining the elemental
forces using some vectorial method.
In the diagram below liquid is resting on
top of a curved base.
E
Calculate the
D
C
B
horizontal and vertical
components.
G
O
FAC
RH
A
Combine these to obtain the resultant
force and direction.
Rv
R
The fluid is at rest – in equilibrium.
(Although this can be done for all three
dimensions we will only look at one vertical
plane)
CIVE1400: Fluid Mechanics
Section 2: Statics 71
So any element of fluid
such as ABC is also in equilibrium.
CIVE1400: Fluid Mechanics
Section 2: Statics 72
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Section 2: Statics
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Section 2: Statics
Consider the Horizontal forces
The sum of the horizontal forces is zero.
C
B
The resultant horizontal force of a fluid
above a curved surface is:
RH = Resultant force on the projection of the
curved surface onto a vertical plane.
RH
FAC
We know
1. The force on a vertical plane must act
horizontally (as it acts normal to the plane).
A
No horizontal force on CB as there are
no shear forces in a static fluid
2. That RH must act through the same point.
So:
Horizontal forces act only on the faces
AC and AB as shown.
RH acts horizontally through the centre of
pressure of the projection of
the curved surface onto an vertical plane.
FAC, must be equal and opposite to RH.
AC is the projection of the curved surface
AB onto a vertical plane.
We have seen earlier how to calculate
resultant forces and point of action.
Hence we can calculate the resultant
horizontal force on a curved surface.
CIVE1400: Fluid Mechanics
Section 2: Statics 73
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Section 2: Statics 74
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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
components vectorialy,
C
B
G
Resultant force
R
A
2
RH
RV2
Rv
There are no shear force on the vertical edges,
so the vertical component can only be due to
the weight of the fluid.
So we can say
The resultant vertical force of a fluid above a
curved surface is:
RV = Weight of fluid directly above the curved
surface.
It will act vertically down through the centre of
gravity of the mass of fluid.
CIVE1400: Fluid Mechanics
Section 2: Statics 75
And acts through O at an angle of T.
The angle the resultant force makes to the
horizontal is
§R ·
T tan 1 ¨ V ¸
© RH ¹
The position of O is the point of interaction of
the horizontal line of action of R H and the
vertical line of action of RV .
CIVE1400: Fluid Mechanics
Section 2: Statics 76
CIVE1400: Fluid Mechanics
Section 2: Statics
A typical example application of this is the
determination of the forces on dam walls or curved
sluice gates.
Find the magnitude and direction of the
resultant force of water on a quadrant gate as
shown below.
CIVE1400: Fluid Mechanics
Section 2: Statics
What are the forces if the fluid is below the
curved surface?
This situation may occur or a curved sluice gate.
C
Gate width 3.0m
B
G
1.0m
O
FAC
RH
Water ρ = 1000 kg/m3
A
R
Rv
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
O
FAC
RH
A
A
A’
Rv
The two horizontal on the element are:
The horizontal reaction force RH
The force on the vertical plane A’B.
The resultant horizontal force, RH acts as shown in
the diagram. Thus we can say:
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.
CIVE1400: Fluid Mechanics
Section 2: Statics 79
What vertical force would
keep this in equilibrium?
If the region above the curve were all
water there would be equilibrium.
Hence: the force exerted by this amount of fluid
must equal he resultant force.
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 80
CIVE1400: Fluid Mechanics
Section 2: Statics
The resultant force and direction of application
are calculated in the same way as for fluids
above the surface:
Resultant force
CIVE1400: Fluid Mechanics
Section 2: Statics
An example of a curved sluice gate which
experiences force from fluid below.
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
E
R
2
RH
C
RV2
A
D
B
And acts through O at an angle of T.
The angle the resultant force makes to the horizontal
is
T
§R ·
tan 1 ¨ V ¸
© RH ¹
CIVE1400: Fluid Mechanics
Section 2: Statics 81
CIVE1400: Fluid Mechanics
LECTURE CONTENTS
Fluid Dynamics
Section 0: Introduction
Section 1: Fluid Properties
Fluids vs. Solids
Viscosity
Newtonian Fluids
Properties of Fluids
Section 2: Statics
Hydrostatic pressure
Manometry/Pressure measurement
Hydrostatic forces on
submerged surfaces
Section 3: Dynamics
The continuity equation.
The Bernoulli Equation.
Application of Bernoulli equation.
The momentum equation.
Application of momentum equation.
Section 4: Real Fluids
Boundary layer.
Laminar flow in pipes.
Section 5: Dimensional Analysis
An Intro to Dimensional analysis
Similarity
CIVE 1400: Fluid Mechanics
Section 2: Statics 82
Section 3: Fluid dynamics 83
Objectives
1.Identify differences between:
xsteady/unsteady
xuniform/non-uniform
xcompressible/incompressible flow
2.Demonstrate streamlines and stream tubes
3.Introduce the Continuity principle
4.Derive the Bernoulli (energy) equation
5.Use the continuity equations to predict
pressure and velocity in flowing fluids
6.Introduce the momentum equation for a fluid
7.Demonstrate use of the momentum equation
to predict forces induced by flowing fluids
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 84
Fluid dynamics:
Flow Classification
The analysis of fluid in motion
Fluid flow may be
classified under the following headings
Fluid motion can be predicted in the
same way as the motion of solids
By use of the fundamental laws of physics and
the physical properties of the fluid
Some fluid flow is very complex:
e.g.
xSpray behind a car
xwaves on beaches;
xhurricanes and tornadoes
xany other atmospheric phenomenon
All can be analysed
with varying degrees of success
(in some cases hardly at all!).
There are many common situations
which analysis gives very accurate predictions
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 85
uniform:
Flow conditions (velocity, pressure, crosssection or depth) are the same at every point in
the fluid.
non-uniform:
Flow conditions are not the same at every point.
steady:
Flow conditions may differ from point to point
but DO NOT change with time.
unsteady:
Flow conditions change with time at any point.
Fluid flowing under normal circumstances
- a river for example conditions vary from point to point
we have non-uniform flow.
If the conditions at one point vary as time
passes then we have unsteady flow.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 86
Combining these four gives.
Compressible or Incompressible Flow?
Steady uniform flow.
Conditions do not change with position
in the stream or with time.
E.g. flow of water in a pipe of constant diameter at
constant velocity.
All fluids are compressible - even water.
Density will change as pressure changes.
Steady non-uniform flow.
Conditions change from point to point in the stream
but do not change with time.
E.g. Flow in a tapering pipe with constant velocity at
the inlet.
Under steady conditions
- provided that changes in pressure are small we usually say the fluid is incompressible
- it has constant density.
Three-dimensional flow
In general fluid flow is three-dimensional.
Unsteady uniform flow.
At a given instant in time the conditions at every
point are 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.
Unsteady non-uniform flow.
Every condition of the flow may change from point to
point and with time at every point.
E.g. Waves in a channel.
Pressures and velocities change in all
directions.
In many cases the greatest changes only occur
in two directions or even only in one.
Changes in the other direction can be effectively
ignored making analysis much more simple.
This course is restricted to Steady uniform flow
- the most simple of the four.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 87
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 88
One dimensional flow:
Two-dimensional flow
Conditions vary only in the direction of flow
not across the cross-section.
Conditions vary in the direction of flow and in
one direction at right angles to this.
The flow may be unsteady with the parameters
varying in time but not across the cross-section.
E.g. Flow in a pipe.
Flow patterns in two-dimensional flow can be
shown by curved lines on a plane.
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
Should this be treated as two-dimensional flow?
Possibly - but it is only necessary if very high
accuracy is required.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 89
In this course we will be considering:
xsteady
xincompressible
xone and two-dimensional flow
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 90
Streamlines
Some points about streamlines:
It is useful to visualise the flow pattern.
Lines joining points of equal velocity - velocity
contours - can be drawn.
x Close to a solid boundary, streamlines are
parallel to that boundary
These lines are know as streamlines.
Here are 2-D streamlines around a cross-section
of an aircraft wing shaped body:
x The direction of the streamline is the direction
of the fluid velocity
x Fluid can not cross a streamline
x Streamlines can not cross each other
x Any particles starting on one streamline will
stay on that same streamline
x 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.
x In steady flow, the position of streamlines
does not change.
Very close to a boundary wall the flow direction
must be along the boundary.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 91
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 92
Streamtubes
Some points about streamtubes
A circle of points in a flowing fluid each
has a streamline passing through it.
xThe “walls” of a streamtube are streamlines.
These streamlines make a tube-like shape
known as a streamtube
xFluid cannot flow across a streamline, so fluid
cannot cross a streamtube “wall”.
xA streamtube is not like a pipe.
Its “walls” move with the fluid.
x In unsteady flow streamtubes can change
position with time
x 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):
CIVE 1400: Fluid Mechanics
m
dm
dt
Section 3: Fluid dynamics 93
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 94
Flow rate
Volume flow rate - Discharge.
Mass flow rate
More commonly we use volume flow rate
Also know as discharge.
mass
time taken to accumulate this mass
The symbol normally used for discharge is Q.
discharge, Q
A simple example:
An empty bucket weighs 2.0kg. After 7 seconds of
collecting water the bucket weighs 8.0kg, then:
mass flow rate
mass of fluid in bucket
time taken to collect the fluid
8.0 2.0
7
0.857kg / s
m
=
volume of fluid
time
A simple example:
If the bucket above fills with 2.0 litres in 25 seconds,
what is the discharge?
Q
2.0 u 10 3 m3
25 sec
0.0008 m3 / s
0.8 l / s
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 95
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 96
A simple example:
Discharge and mean velocity
-3
If we know the discharge and the diameter of a
pipe, we can deduce the mean velocity
2
If A = 1.2u10 m
And discharge, Q is 24 l/s,
mean velocity is
Q
A
um
um t
2.4 u 10 3
x
12
. u 10 3
2.0 m / s
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 Au um u t.
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
The discharge will thus be
Q=
Q
volume A u um u t
=
t
time
Aum
CIVE 1400: Fluid Mechanics
u
um
umax
This idea, that mean velocity multiplied by the
area gives the discharge, applies to all
situations - not just pipe flow.
Section 3: Fluid dynamics 97
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 98
Applying to a streamtube:
Continuity
This principle of conservation of mass says matter
cannot be created or destroyed
Mass enters and leaves only through the two
ends (it cannot cross the streamtube wall).
ρ2
u2
This is applied in fluids to fixed volumes, known as
control volumes (or surfaces)
A2
Mass flow in
ρ1
Control
volume
u1
A1
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
For steady flow
Mass entering = Mass leaving
per unit time
per unit time
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 99
Mass entering = Mass leaving
per unit time
per unit time
U1GA1u1
U2GA2u2
Or for steady flow,
U1GA1u1
U2GA2 u2
Constant
m
dm
dt
This is the continuity equation.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 100
In a real pipe (or any other vessel) we use the
mean velocity and write
U1 A1um1
U2 A2 um2
Constant
Some example applications of Continuity
m
Section 1
For incompressible, fluid U1 = U2 = U
(dropping the m subscript)
Section 2
A liquid is flowing from left to right.
By the continuity
A1u1U1
A1u1
A2 u2
A2 u2 U2
As we are considering a liquid,
U1 = U 2 = U
Q
Q1
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.
A1u1
Q2
A2u2
An example:
If the area A1=10u10-3 m2 and A2=3u10-3 m2
And the upstream mean velocity u1=2.1 m/s.
The downstream mean velocity is
u2
A1
u1
A2
7.0 m / s
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 101
Now try this on a diffuser, a pipe which expands
or diverges as in the figure below,
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 102
Velocities in pipes coming from a junction.
2
1
Section 1
Section 2
3
If d1=30mm and d2=40mm and the velocity u2=3.0m/s.
mass flow into the junction = mass flow out
The velocity entering the diffuser is given by,
U 1 Q1 = U 2 Q 2 + U 3 Q3
u1
A2
u2
A1
S d 22 / 4
S d12 / 4
u2
d 22
d12
u2
When incompressible
2
§ d2 ·
¨ ¸ u2
© d1 ¹
Q1 = Q 2 + Q3
2
§ 40·
¨ ¸ 3.0 5.3 m / s
© 30¹
$ 1 u1 = $ 2 u2 + $ 3 u3
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 103
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 104
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?
Discharge in:
Q1
The Bernoulli equation
The Bernoulli equation is a statement of the
principle of conservation of energy along a
streamline
§ Sd12 ·
¨
¸ u1
© 4 ¹
A1u1
It can be written:
p1 u12
z
Ug 2 g 1
0.00392 m3 / s
Discharges out:
0.001178m3 / s
Q2
0.3Q1
Q1
Q2 Q3
Q1 0.3Q1
Q3
Velocities out:
A2 u2
u2
0.936 m / s
Q3
A3u3
u3
Potential
Total
energy per
unit weight
These term all have units of length,
they are often referred to as the following:
p
Ug
potential head = z
velocity head =
pressure head =
0.972 m / s
CIVE 1400: Fluid Mechanics
Kinetic
energy per energy per energy per
unit weight unit weight unit weight
0.00275 m3 / s
Q2
These terms represent:
Pressure
0.7Q1
H = Constant
Section 3: Fluid dynamics 105
u2
2g
total head = H
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 106
The derivation of Bernoulli’s Equation:
Restrictions in application
of Bernoulli’s equation:
Cross sectional area a
B
B’
A
xFlow is steady
z
A’
mg
xDensity is constant (incompressible)
xFriction losses are negligible
xIt relates the states at two points along a single
streamline, (not conditions on two different
streamlines)
All these conditions are impossible to satisfy at
any instant in time!
Fortunately, for many real situations where the
conditions are approximately satisfied, the
equation gives very good results.
An element of fluid, as that in the figure above, has
potential energy due to its height z above a datum and
kinetic energy due to its velocity u. If the element has
weight mg then
potential energy = mgz
potential energy per unit weight =
kinetic energy =
z
1 2
mu
2
kinetic energy per unit weight =
u2
2g
At any cross-section the pressure generates a force, the
fluid will flow, moving the cross-section, so work will be
done. If the pressure at cross section AB is p and the area
of the cross-section is a then
force on AB = pa
when the mass mg of fluid has passed AB, cross-section
AB will have moved to A’B’
volume passing AB =
mg
Ug
m
U
therefore
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 107
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 108
distance AA’ =
m
The Bernoulli equation is applied along
streamlines
like that joining points 1 and 2 below.
Ua
work done = force u distance AA’
=
m
pa u
Ua
2
pm
U
work done per unit weight =
p
Ug
1
This term is know as the pressure energy of the flowing
stream.
Summing all of these energy terms gives
Pressure
Kinetic
Potential
energy per energy per energy per
unit weight unit weight unit weight
This equation assumes no energy losses (e.g. from friction)
or energy gains (e.g. from a pump) along the streamline. It
can be expanded to include these simply, by adding the
appropriate energy terms:
H
Total
By the principle of conservation of energy, the total energy
in the system does not change, thus the total head does
not change. So the Bernoulli equation can be written
p u2
z
Ug 2 g
CIVE 1400: Fluid Mechanics
p2 u22
z
Ug 2 g 2
p1 u12
z
Ug 2 g 1
Total
energy per
unit weight
or
p u2
z
Ug 2 g
total head at 1 = total head at 2
or
Total
Section 3: Fluid dynamics 109
Practical use of the Bernoulli Equation
The Bernoulli equation is often combined with
the continuity equation to find velocities and
pressures at points in the flow connected by a
streamline.
An example:
Finding pressures and velocities within a
contracting and expanding pipe.
u1
u2
p1
p2
Work done
Energy
p2 u22
z h wq
Ug 2 g 2
p1 u12
z
Ug 2 g 1
H Constant
Loss
energy per
energy per unit per unit per unit supplied
unit weight at 1
weight at 2
weight
weight
per unit weight
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 110
Apply the Bernoulli equation along a streamline
joining section 1 with section 2.
p1 u12
p2 u22
z1
z2
Ug 2 g
p2
Ug 2 g
p1 U
2
(u12 u22 )
Use the continuity equation to find u2
A1u1
A2u2
u2
A1u1
A2
2
§ d1 ·
¨ ¸ u1
© d2 ¹
7.8125 m / s
So pressure at section 2
section 1
section 2
A fluid, density U = 960 kg/m3 is flowing steadily
through the above tube.
The section diameters are d1=100mm and d2=80mm.
The gauge pressure at 1 is p1=200kN/m2
The velocity at 1 is u1=5m/s.
The tube is horizontal (z1=z2)
What is the gauge pressure at section 2?
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 111
p2
200000 17296.87
182703 N / m2
182.7 kN / m2
Note how
the velocity has increased
the pressure has decreased
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 112
We have used both the Bernoulli equation and
the Continuity principle together to solve the
problem.
Use of this combination is very common. We will
be seeing this again frequently throughout the
rest of the course.
Pitot Tube
The Pitot tube is a simple velocity measuring device.
Uniform velocity flow hitting a solid blunt body, has
streamlines similar to this:
2
1
Applications of the Bernoulli Equation
The Bernoulli equation is applicable to many
situations not just the pipe flow.
Here we will see its application to flow
measurement from tanks, within pipes as well as
in open channels.
Some move to the left and some to the right.
The centre one hits the blunt body and stops.
At this point (2) velocity is zero
The fluid does not move at this one point.
This point is known as the stagnation point.
Using the Bernoulli equation we can calculate the
pressure at this point.
Along the central streamline at 1: velocity u1 , pressure p1
at the stagnation point of: u2 = 0. (Also z1 = z2)
p1 u12
U 2
p2
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 113
How can we use this?
The blunt body does not have to be a solid.
I could be a static column of fluid.
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.
h1
1
p2
U
1
p1 Uu12
2
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 114
Pitot Static Tube
The necessity of two piezometers makes this
arrangement awkward.
The Pitot static tube combines the tubes and
they can then be easily connected to a
manometer.
h2
2
1
2
1
X
h
A
We have the equation for p2 ,
p2
Ugh2
u
1
p1 Uu12
2
1
Ugh1 Uu12
2
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.
CIVE 1400: Fluid Mechanics
B
Section 3: Fluid dynamics 115
The holes on the side connect to one side of a
manometer, while the central hole connects to the
other side of the manometer
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 116
Using the theory of the manometer,
p1 Ug X h Uman gh
pB
p2 UgX
pA
pB
p2
The Venturi meter is a device for measuring
discharge in a pipe.
p1 Ug X h Uman gh
p2 UgX
We know that
Venturi Meter
pA
1
p1 Uu12 , giving
2
p1 hg Uman U p1 It is a rapidly converging section which increases the
velocity of flow and hence reduces the pressure.
It then returns to the original dimensions of the pipe
by a gently diverging ‘diffuser’ section.
Uu12
2
2 gh( Um U )
u1
about 6°
U
about 20°
The Pitot/Pitot-static is:
2
xSimple to use (and analyse)
1
xGives velocities (not discharge)
z2
xMay block easily as the holes are small.
z1
h
datum
Apply Bernoulli along the streamline from point 1 to point 2
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 117
p1 u12
z
Ug 2 g 1
p2 u22
z
Ug 2 g 2
By continuity
Q
u2
u1 A1
u1 A1
Qactual
2
º
u12 ª§ A1 ·
«¨ ¸ 1»
2 g «© A2 ¹
»¼
¬
Cd Qideal
Cd u1 A1
ª p p2
º
z1 z2 »
2g« 1
¬ Ug
¼
Cd A1 A2
A12 A22
In terms of the manometer readings
p1 Ugz1
u12 ª A12 A22 º
»
«
2 g ¬ A22 ¼
u1
Actual discharge takes into account the losses due to
friction, we include a coefficient of discharge (Cd |0.9)
Qactual
u1 A1
A2
Section 3: Fluid dynamics 118
The theoretical (ideal) discharge is uuA.
Qideal
u2 A2
Substituting and rearranging gives
p1 p2
z1 z2
Ug
CIVE 1400: Fluid Mechanics
p1 p2
z1 z2
Ug
p2 Uman gh Ug ( z2 h)
§U
·
h¨ man 1¸
© U
¹
Giving
º
ª p p2
z1 z2 »
2g« 1
¼
¬ Ug
A2
2
2
A1 A2
Qactual
Cd A1 A2
§U
·
2 gh¨ man 1¸
© U
¹
A12 A22
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.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 119
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 120
Flow Through A Small Orifice
Venturimeter design:
Flow from a tank through a hole in the side.
xThe diffuser assures a gradual and steady deceleration
after the throat. So that pressure rises to something near
that before the meter.
xThe angle of the diffuser is usually between 6 and 8
degrees.
1
Aactual
h
xWider and the flow might separate from the walls
increasing energy loss.
xIf the angle is less the meter becomes very long and
pressure losses again become significant.
xThe efficiency of the diffuser of increasing pressure back
to the original is rarely greater than 80%.
xCare must be taken when connecting the manometer so
that no burrs are present.
2
Vena contractor
The edges of the hole are sharp to minimise frictional
losses by minimising the contact between the hole and the
liquid.
The streamlines at the orifice
contract reducing the area of flow.
This contraction is called the vena contracta.
The amount of contraction must
be known to calculate the flow.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 121
Apply Bernoulli along the streamline joining point 1
on the surface to point 2 at the centre of the orifice.
At the surface velocity is negligible (u1 = 0) and the
pressure atmospheric (p1 = 0).
At the orifice the jet is open to the air so
again the pressure is atmospheric (p2 = 0).
If we take the datum line through the orifice
then z1 = h and z2 =0, leaving
h
u2
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 122
The discharge through the orifice
is
jet area u jet velocity
The area of the jet is the area of the vena contracta
not the area of the orifice.
We use a coefficient of contraction
to get the area of the jet
Aactual
u22
2g
Cc Aorifice
Giving discharge through the orifice:
2 gh
This theoretical value of velocity is an overestimate
as friction losses have not been taken into account.
Q
Qactual
Au
Aactual uactual
Cc Cv Aorifice utheoretical
A coefficient of velocity is used to correct the
theoretical velocity,
uactual
Cv utheoretical
Cd Aorifice 2 gh
Each orifice has its own coefficient of velocity, they
usually lie in the range( 0.97 - 0.99)
CIVE 1400: Fluid Mechanics
Cd Aorifice utheoretical
Section 3: Fluid dynamics 123
Cd is the coefficient of discharge,
C d = C c u Cv
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 124
Time for the tank to empty
This Q is the same as the flow out of the orifice so
We have an expression for the discharge from the tank
Q
Cd Ao 2 gh
A
Cd Ao 2 gh
We can use this to calculate how long
it will take for level in the to fall
Gh
Gt
A
Gh
Cd Ao 2 g h
Gt
As the tank empties the level of water falls.
The discharge will also drop.
Integrating between the initial level, h1, and final
level, h2, gives the time it takes to fall this height
h1
t
h2
§ 1
¨³
© h
The tank has a cross sectional area of A.
In a time dt the level falls by dh
The flow out of the tank is
Q
t
³ h 1/2
Gh
Gt
·
2 h¸
¹
2h1/ 2
A
>2 h @hh12
Cd Ao 2 g
2A
Cd Ao 2 g
Au
Q A
A
h2 Gh
³
h
Cd Ao 2 g 1 h
>
h2 h1
@
(-ve sign as dh is falling)
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 125
Submerged Orifice
What if the tank is feeding into another?
Area A1
Area A2
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 126
Time for Equalisation of Levels in Two Tanks
The time for the levels to equalise can be found by
integrating this equation as before.
h1
By the continuity equation
h2
Gh1
Gh
A2 2
Gt
Gt
A1Gh1 A2Gh2
A1
Q
QGt
Orifice area Ao
Apply Bernoulli from point 1 on the surface of the
deeper tank to point 2 at the centre of the orifice,
p1 u12
p2 u22
z1
z2
Ug
Ug
2g
0 0 h1
u2
2g
Ugh2 u22
0
Ug 2 g
Gh Gh1 Gh2 so
A1Gh1 A2Gh1 A2Gh
A2Gh
Gh1
A1 A2
and
QGt
2 g (h1 h2 )
And the discharge is given by
Q
writing
Cd Ao u
Cd Ao 2 g (h1 h2 ) Gt
Cd Ao 2 g (h1 h2 )
but
Gt
So the discharge of the jet through the submerged orifice
depends on the difference in head across the orifice.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 127
h
CIVE 1400: Fluid Mechanics
A1Gh1
A1 A2
Gh
A1 A2
h1 h2
A1 A2
Gh
A1 A2 Cd Ao 2 g h
Section 3: Fluid dynamics 128
Integrating between the two levels
t
Flow Over Notches and Weirs
A1 A2
Gh
h
³ final
( A1 A2 )Cd Ao 2 g hinitial h
xA notch is an opening in the side of a tank or reservoir.
xIt is a device for measuring discharge.
h final
2 A1 A2
> h @hinitial
( A1 A2 )Cd Ao 2 g
2 A1 A2
( A1 A2 )Cd Ao 2 g
>
xA weir is a notch on a larger scale - usually found in
rivers.
h final hinitial
xIt is used as both a discharge measuring device and a
device to raise water levels.
@
xThere are many different designs of weir.
h is the difference in height between
the two levels (h2 - h1)
xWe will look at sharp crested weirs.
Weir Assumptions
The time for the levels to equal use
hinitial = (h1 - h2) and hfinal = 0
xvelocity of the fluid approaching the weir is small so we
can ignore kinetic energy.
xThe 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
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 129
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 130
A General Weir Equation
Rectangular Weir
Consider a horizontal strip of
width b, depth h below the free surface
The width does not change with depth so
b
b
constant
B
h
H
B
δh
H
velocity through the strip, u
discharge through the strip, GQ
2 gh
Au bGh 2 gh
Integrating from the free surface, h=0, to the weir
crest, h=H, gives the total theoretical discharge
H
Qtheoretical
2 g ³ bh1/ 2 dh
0
This is different for every differently
shaped weir or notch.
Substituting this into the general weir equation gives
H
Qtheoretical B 2 g ³ h1/ 2 dh
0
2
B 2 gH 3/ 2
3
To get the actual discharge we introduce a coefficient
of discharge, Cd, to account for
losses at the edges of the weir
and contractions in the area of flow,
Qactual
We need an expression relating the width of flow
across the weir to the depth below the free surface.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 131
CIVE 1400: Fluid Mechanics
Cd
2
B 2 gH 3 / 2
3
Section 3: Fluid dynamics 132
‘V’ Notch Weir
The Momentum Equation
And Its Applications
The relationship between width and depth is
dependent on the angle of the “V”.
b
We have all seen moving
fluids exerting forces.
h
H
θ
x The lift force on an aircraft is exerted by the air
moving over the wing.
The width, b, a depth h from the free surface is
b
x A jet of water from a hose exerts a force on
whatever it hits.
§T ·
2 H h tan¨ ¸
© 2¹
So the discharge is
H
Qtheoretical
§T ·
2 2 g tan¨ ¸ ³ H hh1/ 2 dh
© 2¹
The analysis of motion is as in solid mechanics:
by use of Newton’s laws of motion.
0
H
2
º
§ T · ª2
2 2 g tan¨ ¸ « Hh3/ 2 h5/ 2 »
© 2¹ ¬ 3
5
¼0
8
§T ·
2 g tan¨ ¸ H 5/ 2
© 2¹
15
The actual discharge is obtained by introducing a
coefficient of discharge
Qactual
Cd
CIVE 1400: Fluid Mechanics
8
§T ·
2 g tan¨ ¸ H 5 / 2
© 2¹
15
Section 3: Fluid dynamics 133
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.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 134
From solid mechanics you will recognise
F = ma
In time Gt a volume of the fluid moves
from the inlet a distance u1Gt, so
What mass of moving fluid we should use?
volume entering the stream tube = area u distance
= A 1u1 Gt
We use a different form of the equation.
Consider a streamtube:
And assume steady non-uniform flow
The mass entering,
mass entering stream tube = volume u density
= U1 A1 u1 Gt
A2
u2
A1
u1
ρ2
And momentum
momentum entering stream tube = mass u velocity
= U1 A1 u1 Gt u1
ρ1
u1 δt
Similarly, at the exit, we get the expression:
momentum leaving stream tube = U2 A 2 u 2 Gt u 2
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 135
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 136
By Newton’s 2nd Law.
An alternative derivation
From conservation of mass
Force = rate of change of momentum
F=
( U2 A2u2Gt u2 U1 A1u1Gt u1 )
Gt
mass into face 1 = mass out of face 2
we can write
rate of change of mass
dm
dt
U1 A1u1 U2 A2u2
m
We know from continuity that
The rate at which momentum enters face 1 is
U1 A1u1u1 mu
1
Q A1u1 A2 u2
And if we have a fluid of constant density,
i.e. U1 U2 U , then
The rate at which momentum leaves face 2 is
U2 A2 u2 u2 mu
2
F QU (u2 u1 )
Thus the rate at which momentum changes
across the stream tube is
U2 A2 u2 u2 U1 A1u1u1 mu
2 mu
1
So
Force = rate of change of momentum
F m ( u2 u1 )
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 137
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 138
So we have these two expressions,
either one is known as the momentum equation
The previous analysis assumed the inlet and
outlet velocities in the same direction
i.e. a one dimensional system.
F m ( u2 u1 )
What happens when this is not the case?
F
u2
θ2
QU ( u2 u1 )
The Momentum equation.
This force acts on the fluid
in the direction of the flow of the fluid.
θ1
u1
We consider the forces by resolving in the
directions of the co-ordinate axes.
The force in the x-direction
Fx
m u2 cosT2 u1 cosT1 m u2 x u1 x or
Fx
UQu2 cosT2 u1 cosT1 UQu2 x u1 x CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 139
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 140
And the force in the y-direction
Fy
m u2 sin T2 u1 sin T1 In summary we can say:
Total force
on the fluid =
m §¨© u2 y u1 y ·¸¹
or
Fy
UQu2 sin T2 u1 sin T1 F
UQ¨§© u2 y u1 y ¸·¹
F
rate of change of
momentum through
the control volume
m uout uin or
UQuout uin The resultant force can be found by combining
these components
Fy
Remember that we are working with vectors so
F is in the direction of the velocity.
FResultant
φ
Fx
Fresultant
Fx2 Fy2
And the angle of this force
I
§ Fy ·
tan 1 ¨ ¸
© Fx ¹
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 141
This force is made up of three components:
FR = Force exerted on the fluid by any solid body
touching the control volume
FB = Force exerted on the fluid body (e.g. gravity)
FP = Force exerted on the fluid by fluid pressure
outside the control volume
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 142
Application of the Momentum Equation
Force due the flow around a pipe bend
A converging pipe bend
lying in the horizontal plane
turning through an angle of T.
So we say that the total force, FT,
is given by the sum of these forces:
ρ2
y
u2
A2
x
F T = FR + FB + F P
ρ1
u1
θ
A1
The force exerted
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
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 143
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 144
Why do we want to know the forces here?
Control Volume
As the fluid changes direction
a force will act on the bend.
y
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.
We need to know how much force a support
(thrust block) must withstand.
x
2 Co-ordinate axis system
Any co-ordinate axis can be chosen.
Choose a convenient one, as above.
Step in Analysis:
1.Draw a control volume
2.Decide on co-ordinate axis system
3.Calculate the total force
4.Calculate the pressure force
5.Calculate the body force
6.Calculate the resultant force
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 145
In the y-direction:
UQ u2 y u1 y
FT y
u1 y
u1 sin 0
u2 y
u2 sin T
FT y
UQu2 sin T
0
FT x
p2 u22
z hf
Ug 2 g 2
where hf is the friction loss
(this can often be ignored, hf=0)
As the pipe is in the horizontal plane, z1=z2
And with continuity, Q= u1A1 = u2A2
u1 x
u1
u2 x
u2 cosT
FT x
UQu2 cosT u1 Section 3: Fluid dynamics 146
pressure force at 1 - pressure force at 2
FP x
p1 A1 cos 0 p2 A2 cosT
FP y
p1 A1 sin 0 p2 A2 sin T
CIVE 1400: Fluid Mechanics
p2 The body force due to gravity is acting in the
z-direction so need not be considered.
6 Calculate the resultant force
2§
FT y
Section 3: Fluid dynamics 147
p2 A2 sin T
There are no body forces in the x or y directions.
FRx = FRy = 0
1
1·
¨
2¸
2
2 © A2 A1 ¹
UQ
p1 A1 p2 A2 cosT
5 Calculate the body force
FT x
p1
UQu2 x u1 x CIVE 1400: Fluid Mechanics
FP
Use Bernoulli.
Ug 2 g
In the x-direction:
Knowing the pressures at each end the pressure
force can be calculated,
4 Calculate the pressure force
If we know pressure at the inlet we can relate
this to the pressure at the outlet.
p1 u12
z1
3 Calculate the total force
CIVE 1400: Fluid Mechanics
FR x FP x FB x
FR y FP y FB y
Section 3: Fluid dynamics 148
FR x
FT x FP x 0
Impact of a Jet on a Plane
UQu2 cosT u1 p1 A1 p2 A2 cosT
FR y
A jet hitting a flat plate (a plane)
at an angle of 90q
FT y FP y 0
UQu2 sin T p2 A2 sin T
And the resultant force on the fluid is given by
FRy
We want to find the reaction force of the plate.
i.e. the force the plate will have to apply to stay
in the same position.
FResultant
1 & 2 Control volume and Co-ordinate axis are
shown in the figure below.
φ
FRx
y
u2
FR2 x FR2 y
FR
x
u1
And the direction of application is
I
§ FR y ·
¸
tan 1 ¨
© FR x ¹
u2
the force on the bend is the same magnitude but
in the opposite direction
R
FR
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 149
CIVE 1400: Fluid Mechanics
3 Calculate the total force
In the x-direction
FT x
Section 3: Fluid dynamics 150
FT x
FR x
UQu2 x u1 x FR x FP x FB x
FT x 0 0
UQu1 x
UQu1 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
FR x
R
FT y
0
4 Calculate the pressure force.
The pressures at both the inlet and the outlets
to the control volume are atmospheric.
The pressure force is zero
FP x
FP y
If the plane were at an angle
the analysis is the same.
But it is usually most convenient to choose the
axis system normal to the plate.
y
0
u1
5 Calculate the body force
As the control volume is small
we can ignore the body force due to gravity.
FB x
FB y
u2
x
θ
u3
0
6 Calculate the resultant force
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 151
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 152
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.
UQu2 u1 cosT FT x
Pressures at ends are equal - atmospheric
by continuity u1
Q
, so
A
u2
Both the cross-section and velocities
(in the direction of flow) remain constant.
U
FT x
Q2
1 cosT A
u2
y
and in the y-direction
FT y
x
UQu2 sin T 0
U
u1
θ
Q2
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.
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 153
5 Calculate the body force
FP x
And the direction of application is
I
6 Calculate the resultant force
FT x
FR x FP x FB x
FR x
FT x
FR y FP y FB y
FR y
FT y
U
§ FR y ·
¸
tan 1 ¨
© FR x ¹
exerted on the fluid.
The force on the vane is the same magnitude
but in the opposite direction
R
Q2
1 cosT A
FT y
FR2 x FR2 y
FR
UgV
(This is often small as the jet volume is small
and sometimes ignored in analysis.)
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 154
And the resultant force on the fluid is given by
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,
U
0
CIVE 1400: Fluid Mechanics
No body forces in the x-direction, FB x = 0.
FB x
FP y
FR
Q2
A
Section 3: Fluid dynamics 155
CIVE 1400: Fluid Mechanics
Section 3: Fluid dynamics 156
LECTURE CONTENTS
Real fluids
x Introduction to Laminar and Turbulent flow
Section 0: Introduction
Section 1: Fluid Properties
Fluids vs. Solids
Viscosity
Newtonian Fluids
Properties of Fluids
Section 2: Statics
Hydrostatic pressure
Manometry/Pressure measurement
Hydrostatic forces on
submerged surfaces
Section 3: Dynamics
The continuity equation.
The Bernoulli Equation.
Application of Bernoulli equation.
The momentum equation.
Application of momentum equation.
Section 4: Real Fluids
Laminar and turbulent flow
Boundary layer theory
Section 5: Dimensional Analysis
An Intro to Dimensional analysis
Similarity
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 157
x Head loss in pipes
x Hagen-Poiseuille equation
x Boundary layer theory
x Boundary layer separation
x Losses at bends and junctions
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 158
Flowing real fluids exhibit
viscous effects, they:
Laminar and turbulent flow
x “stick” to solid surfaces
x have stresses within their body.
Injecting a dye into the middle of flow in a pipe,
what would we expect to happen?
This
From earlier we saw this relationship between
shear stress and velocity gradient:
W v
du
dy
The shear stress, W, in a fluid
is proportional to the velocity gradient
- the rate of change of velocity across the flow.
this
For a “Newtonian” fluid we can write:
W
P
du
dy
where P is coefficient of viscosity
(or simply viscosity).
or this
Here we look at the influence of forces due to
momentum changes and viscosity
in a moving fluid.
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 159
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 160
All three would happen but for different flow rates.
The was first investigated in the 1880s
by Osbourne Reynolds
in a classic experiment in fluid mechanics.
Top: Slow flow
Middle: Medium flow
Bottom: Fast flow
Top:
Middle:
Bottom:
A tank arranged as below:
Laminar flow
Transitional flow
Turbulent flow
Laminar flow:
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
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
Section 4: Real Fluids 161
After many experiments he found this
expression
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 162
What are the units of Reynolds number?
We can fill in the equation with SI units:
Uud
P
U = density,
d = diameter
U kg / m3 , u m / s,
P Ns / m2 kg / m s
u = mean velocity,
P = viscosity
Re
CIVE 1400: Fluid Mechanics
kg m m m s
1
m3 s 1 kg
A quantity with no units is known as a
non-dimensional (or dimensionless) quantity.
This value is known as the
Reynolds number, Re:
Laminar flow:
Transitional flow:
Turbulent flow:
m
It has no units!
This could be used to predict the change in
flow type for any fluid.
Re
Uud
P
d
(We will see more of these in the section on
dimensional analysis.)
Uud
P
Re < 2000
2000 < Re < 4000
Re > 4000
Section 4: Real Fluids 163
The Reynolds number, Re,
is a non-dimensional number.
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 164
At what speed does the flow pattern change?
What is the MINIMUM velocity when flow is
turbulent i.e. Re = 4000
We use the Reynolds number in an example:
Re
A pipe and the fluid flowing
have the following properties:
water density
pipe diameter
(dynamic) viscosity,
U = 1000 kg/m3
d = 0.5m
P = 0.55x103 Ns/m2
u
Uud
P
4000
0.0044 m / s
In a house central heating system,
typical pipe diameter = 0.015m,
limiting velocities would be,
0.0733 and 0.147m/s.
What is the MAXIMUM velocity when flow is
laminar i.e. Re = 2000
Re
u
u
Uud
P
In practice laminar flow rarely occurs
in a piped water system.
2000
2000P 2000 u 0.55 u 10 3
1000 u 0.5
Ud
0.0022 m / s
CIVE 1400: Fluid Mechanics
Both of these are very slow.
Section 4: Real Fluids 165
Laminar flow does occur in
fluids of greater viscosity
e.g. in bearing with oil as the lubricant.
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 166
Laminar flow
What does this abstract number mean?
xRe < 2000
We can give the Re number a physical meaning.
x‘low’ velocity
xDye does not mix with water
This may help to understand some of the
reasons for the changes from laminar to
turbulent flow.
Re
Uud
P
xFluid particles move in straight lines
xSimple mathematical analysis possible
xRare in practice in water systems.
Transitional flow
x2000 > Re < 4000
x‘medium’ velocity
inertial forces
viscous forces
xDye stream wavers - mixes slightly.
When inertial forces dominate
(when the fluid is flowing faster and Re is larger)
the flow is turbulent.
Turbulent flow
xRe > 4000
x‘high’ velocity
xDye mixes rapidly and completely
When the viscous forces are dominant
(slow flow, low Re)
they keep the fluid particles in line,
the flow is laminar.
xParticle paths completely irregular
xAverage motion is in flow direction
xCannot be seen by the naked eye
xChanges/fluctuations are very difficult to
detect. Must use laser.
xMathematical analysis very difficult - so
experimental measures are used
xMost common type of flow.
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 167
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 168
Pressure loss due to friction in a pipeline
Up to now we have considered ideal fluids:
no energy losses due to friction
Attaching a manometer gives
pressure (head) loss due to the energy lost by
the fluid overcoming the shear stress.
L
Because fluids are viscous,
energy is lost by flowing fluids due to friction.
This must be taken into account.
Δp
The effect of the friction shows itself as a
pressure (or head) loss.
In a real flowing fluid shear stress
slows the flow.
How can we quantify this pressure loss
in terms of the forces acting on the fluid?
To give a velocity profile:
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 169
Consider a cylindrical element of
incompressible fluid flowing in the pipe,
τw
το
The pressure at 1 (upstream)
is higher than the pressure at 2.
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 170
As the flow is in equilibrium,
driving force = retarding force
το
τw
'p
area A
Ww is the mean shear stress on the boundary
Sd 2
4
'p
Upstream pressure is p,
Downstream pressure falls by 'p to (p-'p)
W wSdL
Ww 4 L
d
Giving pressure loss in a pipe in terms of:
The driving force due to pressure
xpipe diameter
xshear stress at the wall
driving force = Pressure force at 1 - pressure force at 2
pA p 'p A
'p A
'p
Sd 2
4
The retarding force is due to the shear stress
shear stress u area over which it acts
= W w u area of pipe wall
= W wSdL
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 171
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 172
What is the variation of shear stress in the flow?
Shear stress and hence pressure loss varies
with velocity of flow and hence with Re.
τw
Many experiments have been done
with various fluids measuring
the pressure loss at various Reynolds numbers.
R
r
τw
A graph of pressure loss and Re look like:
At the wall
Ww
R 'p
2 L
At a radius r
W
W
r 'p
2 L
r
Ww
R
A linear variation in shear stress.
This graph shows that the relationship between
pressure loss and Re can be expressed as
This is valid for:
xsteady flow
xlaminar flow
xturbulent flow
CIVE 1400: Fluid Mechanics
'p v u
laminar
turbulent 'p v u1.7 ( or
Section 4: Real Fluids 173
Pressure loss during laminar flow in a pipe
CIVE 1400: Fluid Mechanics
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.
Section 4: Real Fluids 174
The fluid is in equilibrium,
shearing forces equal the pressure forces.
In general the shear stress Ww. is almost
impossible to measure.
For laminar flow we can calculate
a theoretical value for
a given velocity, fluid and pipe dimension.
W 2Sr L
'p A
W
'p r
L 2
P
du
,
dy
We are measuring from the pipe centre, so
W
P
du
dr
Giving:
'p r
du
P
L 2
dr
'p r
du
dr
L 2P
In an integral form this gives an
expression for velocity,
δr
r
u
R
CIVE 1400: Fluid Mechanics
'pSr 2
Newtons law of viscosity saysW
And the stress on the fluid in laminar flow is
entirely due to viscose forces.
As before, consider a cylinder of fluid, length L,
radius r, flowing steadily in the centre of a pipe.
r
2 .0 )
Section 4: Real Fluids 175
CIVE 1400: Fluid Mechanics
'p 1
r dr
L 2P ³
Section 4: Real Fluids 176
The value of velocity at a
point distance r from the centre
ur
'p r 2
C
L 4P
What is the discharge in the pipe?
The flow in an annulus of thickness Gr
GQ ur Aannulus
At r = 0, (the centre of the pipe), u = umax, at
r = R (the pipe wall) u = 0;
Aannulus
2
C
'p R
L 4P
At a point r from the pipe centre when the flow is
laminar:
ur
GQ
'p 1
R 2 r 2 2SrGr
L 4P
Q
'p S R 2
³ R r r 3 dr
L 2P 0
'p Sd 4
L128P
So the discharge can be written
Q
'p Sd 4
L 128P
This is the Hagen-Poiseuille Equation
for laminar flow in a pipe
v
CIVE 1400: Fluid Mechanics
'p SR 4
L 8P
'p 1
R2 r 2 L 4P
This is a parabolic profile
(of the form y = ax2 + b )
so the velocity profile in the pipe looks similar to
S (r Gr )2 Sr 2 | 2SrGr
Section 4: Real Fluids 177
To get pressure loss (head loss)
in terms of the velocity of the flow, write
pressure in terms of head loss hf, i.e. p = Ughf
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Section 4: Real Fluids 178
Boundary Layers
Recommended reading: Fluid Mechanics
by Douglas J F, Gasiorek J M, and Swaffield J A.
Longman publishers. Pages 327-332.
Mean velocity:
u
Q/ A
u
Ugh f d 2
32 PL
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
This gives a, now familiar, velocity profile:
umax
Head loss in a pipe with laminar flow by the
Hagen-Poiseuille equation:
τo
zero velocity
hf
32 PLu
Ugd
Wall
2
Pressure loss is directly proportional to the
velocity when flow is laminar.
It has been validated many time by experiment.
It justifies two assumptions:
1.fluid does not slip past a solid boundary
2.Newtons hypothesis.
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 179
Zero at the wall
A maximum at the centre of the flow.
The profile doesn’t just exit.
It is build up gradually.
Starting when it first flows past the surface
e.g. when it enters a pipe.
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Section 4: Real Fluids 180
Considering a flat plate in a fluid.
Understand this Boundary layer growth diagram.
Upstream the velocity profile is uniform,
This is known as free stream flow.
Downstream a velocity profile exists.
This is known as fully developed flow.
Free stream flow
Fully developed flow
Some question we might ask:
How do we get to the fully developed state?
Are there any changes in flow as we get there?
Are the changes significant / important?
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Section 4: Real Fluids 181
Boundary layer thickness:
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Section 4: Real Fluids 182
First: viscous forces
(the forces which hold the fluid together)
G = distance from wall to where u = 0.99 umainstream
G increases as fluid moves along the plate.
It reaches a maximum in fully developed flow.
The G increase corresponds to a
drag force increase on the fluid.
When the boundary layer is thin:
velocity gradient du/dy, is large
by Newton’s law of viscosity
shear stress, W = P (du/dy), is large.
The force may be large enough to
drag the fluid close to the surface.
As the boundary layer thickens
velocity gradient reduces and
shear stress decreases.
As fluid is passes over a greater length:
xmore fluid is slowed
xby friction between the fluid layers
xthe 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:
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Section 4: Real Fluids 183
Eventually it is too small
to drag the slow fluid along.
Up to this point the flow has been laminar.
Newton’s law of viscosity has applied.
This part of the boundary layer is the
laminar boundary layer
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Section 4: Real Fluids 184
Second: momentum transfer
If the viscous forces were the only action
the fluid would come to a rest.
Viscous shear stresses have held the fluid
particles in a constant motion within layers.
Eventually they become too small to
hold the flow in layers;
the fluid starts to rotate.
Close to boundary velocity gradients are very large.
Viscous shear forces are large.
Possibly large enough to cause laminar flow.
This region is known as the laminar sub-layer.
This layer occurs within the turbulent zone
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.
In laminar flow:
Roughness has very little effect
The fluid motion rapidly becomes turbulent.
Momentum transfer occurs between fast moving
main flow and slow moving near wall flow.
Thus the fluid by the wall is kept in motion.
The net effect is an increase in momentum in the
boundary layer.
This is the turbulent boundary layer.
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Use Reynolds number to determine which state.
Uud
Re
P
Laminar flow:
Re < 2000
Transitional flow: 2000 <
Turbulent flow:
Boundary layers in pipes
Initially of the laminar form.
It changes depending on the ratio of inertial and
viscous forces;
i.e. whether we have laminar (viscous forces high) or
turbulent flow (inertial forces high).
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Boundary layer separation
Divergent flows:
Positive pressure gradients.
Pressure increases in the direction of flow.
Re < 4000
Re > 4000
The fluid in the boundary layer has so little
momentum that it is brought to rest,
and possibly reversed in direction.
Reversal lifts the boundary layer.
u1
u2
p1
p2
p1 < p2
u1 > u2
Laminar flow: profile parabolic (proved in earlier lectures)
The first part of the boundary layer growth diagram.
Turbulent (or transitional),
Laminar and the turbulent (transitional) zones of the
boundary layer growth diagram.
Length of pipe for fully developed flow is
the entry length.
Laminar flow
|120 u diameter
This phenomenon is known as
boundary layer separation.
Turbulent flow | 60 u diameter
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Section 4: Real Fluids 187
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Boundary layer separation:
Examples of boundary layer separation
xincreases the turbulence
xincreases the energy losses in the flow.
Separating / divergent flows are inherently
unstable
A divergent duct or diffuser
velocity drop
(according to continuity)
pressure increase
(according to the Bernoulli equation).
Convergent flows:
xNegative pressure gradients
xPressure decreases in the direction of flow.
xFluid accelerates and the boundary layer is thinner.
u1
Increasing the angle increases the probability of
boundary layer separation.
u2
p2
p1
Venturi meter
p1 > p2
Diffuser angle of about 6q
A balance between:
u1 < u2
xFlow remains stable
xlength of meter
xdanger of boundary layer separation.
xTurbulence reduces.
xBoundary layer separation does not occur.
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Section 4: Real Fluids 189
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 190
Tee-Junctions
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.
Two separation zones occur in bends as shown
above.
Pb > Pa causing separation.
Pd > Pc causing separation
Localised effect
Downstream the boundary layer reattaches and
normal flow occurs.
Boundary layer separation is only local.
Nevertheless downstream of a
junction / bend /valve etc.
fluid will have lost energy.
Bends
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Flow past a cylinder
Slow flow, Re < 0.5 no separation:
Moderate flow, Re < 70, separation
vortices form.
Fast flow Re > 70
vortices detach alternately.
Form a trail of down stream.
Karman vortex trail or street.
(Easily seen by looking over a bridge)
Fluid accelerates to get round the cylinder
Velocity maximum at Y.
Pressure dropped.
Adverse pressure between here and downstream.
Separation occurs
Causes whistling in power cables.
Caused Tacoma narrows bridge to collapse.
Frequency of detachment was equal to the bridge
natural frequency.
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Section 4: Real Fluids 193
Aerofoil
Normal flow over a aerofoil or a wing cross-section.
(boundary layers greatly exaggerated)
CIVE 1400: Fluid Mechanics
Section 4: Real Fluids 194
At too great an angle
boundary layer separation occurs on the top
Pressure changes dramatically.
This phenomenon is known as stalling.
All, or most, of the ‘suction’ pressure is lost.
The plane will suddenly drop from the sky!
The velocity increases as air flows over the wing. The
pressure distribution is as below
so transverse lift force occurs.
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
Section 4: Real Fluids 195
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Section 4: Real Fluids 196
Examples:
LECTURE CONTENTS
Exam questions involving boundary layer theory are
typically descriptive. They ask you to explain the
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.
Section 5: Dimensional Analysis
An Intro to Dimensional analysis
Similarity
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Section 4: Real Fluids 197
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Section 5: Dimensional Analysis
198
Dimensions and units
Dimensional Analysis
Application of fluid mechanics in design makes use
of experiments results.
Any physical situation
can be described by familiar properties.
Results often difficult to interpret.
Dimensional analysis provides a strategy for
choosing relevant data.
Used to help analyse fluid flow
e.g. length, velocity, area, volume, acceleration etc.
Especially when fluid flow is too complex for
mathematical analysis.
These are all known as dimensions.
Specific uses:
Dimensions are of no use without a magnitude.
x help design experiments
i.e. a standardised unit
x Informs which measurements are important
e.g metre, kilometre, Kilogram, a yard etc.
x 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
Doesn’t give the complete answer
i.e. its quality not its quantity.
Experiments necessary to complete solution
Uses principle of dimensional homogeneity
Give qualitative results which only become
quantitative from experimental analysis.
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Section 5: Dimensional Analysis
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The following common abbreviations are used:
This table lists dimensions of some common
physical quantities:
length
=L
Quantity
SI Unit
mass
=M
velocity
m/s
ms-1
LT-1
=T
acceleration
m/s2
ms-2
LT-2
force
N
kg ms-2
M LT-2
kg m2s-2
ML2T-2
time
force
=F
temperature
=4
kg m/s2
energy (or work)
Joule J
N m,
kg m2/s2
Here we will use L, M, T and F (not 4).
power
We can represent all the physical properties we are
interested in with three:
Watt W
pressure ( or stress)
N m/s
Nms-1
kg m2/s3
kg m2s-3
N/m2,
Nm-2
kg/m/s2
kg m-1s-2
3
kg/m
specific weight
relative density
kg m
ML-3
kg m-2s-2
ML-2T-2
a ratio
1
no units
no dimension
viscosity
F = MLT
ML-1T-2
N/m
kg/m2/s2
-2
-3
3
and one of M or F
As either mass (M) of force (F) can be used to
represent the other, i.e.
ML2T-3
Pascal P,
density
L, T
Dimension
surface tension
2 -1
M = FT L
N s/m2
N sm-2
kg/m s
kg m-1s-1
M L-1T-1
-1
N/m
Nm
kg /s2
kg s-2
MT-2
We will always use LTM:
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Section 5: Dimensional Analysis
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Dimensional Homogeneity
Section 5: Dimensional Analysis
202
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
B 2 gH 3/ 2
3
An example:
X
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,
(for L they are both 3, for T both -1).
d
forward velocity of the propeller
(velocity of the plane), u
Dimensional homogeneity can be useful for:
fluid density,
U
1. Checking units of equations;
revolutions per second,
N
2. Converting between two sets of units;
fluid viscosity,
P
3. Defining dimensionless relationships
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Section 5: Dimensional Analysis
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CIVE 1400: Fluid Mechanics
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Common S groups
From this list we can write this equation:
Several groups will appear again and again.
F = I ( d, u, U, N, P )
These often have names.
0 = I ( F, d, u, U, N, P )
They can be related to physical forces.
or
Other common non-dimensional numbers
I and I1 are unknown functions.
or ( S groups):
Reynolds number:
Uud
Re
inertial, viscous force ratio
Dimensional Analysis produces:
§ F
Nd P ·
,
,
¸
© Uu 2 d 2 u Uud ¹
I¨
P
0
Euler number:
p
En
Uu 2
These groups are dimensionless.
I will be determined by experiment.
These dimensionless groups help
to decide what experimental measurements to take.
Froude number:
u2
Fn
gd
Weber number:
Uud
We
V
Mach number:
u
Mn
c
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pressure, inertial force ratio
inertial, gravitational force ratio
inertial, surface tension force ratio
Local velocity, local velocity of sound ratio
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Kinematic similarity
Similarity
The similarity of time as well as geometry.
It exists if:
i. the paths of particles are geometrically similar
ii. the ratios of the velocities of are similar
Similarity is concerned with how to transfer
measurements from models to the full scale.
Three types of similarity
which exist between a model and prototype:
Some useful ratios are:
Vm L m / Tm O L
Velocity
Vp L p / Tp OT
Geometric similarity:
The ratio of all corresponding dimensions
in the model and prototype are equal.
Acceleration
For lengths
Lmodel
Lm
OL
Lprototype
Lp
Discharge
OL is the scale factor for length.
For areas
Amodel
L2m
O2L
Aprototype L2p
am
ap
Qm
Qp
Lm / Tm2
L p / Tp2
L3m / Tm
L3p / Tp
Ou
OL
O2T
O3L
OT
Oa
OQ
A consequence is that streamline
patterns are the same.
All corresponding angles are the same.
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Dynamic similarity
Modelling and Scaling Laws
Measurements taken from a model needs a scaling
law applied to predict the values in the prototype.
If geometrically and kinematically similar and
the ratios of all forces are the same.
Fm
Fp
M mam
M pa p
Force ratio
2
Um L3m O L
2 § OL ·
O
O
u
¸
U L¨
U p L3p O2T
© OT ¹
An example:
O U O2L O2u
For resistance R, of a body
moving through a fluid.
R, is dependent on the following:
This occurs when
the controlling S group
is the same for model and prototype.
U
P:
ML-1T-1
Taking U, u, l as repeating variables gives:
R
Uu 2 l 2
R
It is possible another group is dominant.
In open channel i.e. river Froude number is
often taken as dominant.
Section 5: Dimensional Analysis
l:(length) L
So
U p upd p
Pp
CIVE 1400: Fluid Mechanics
LT-1
u:
I (R, U, u, l, P ) = 0
The controlling S group is usually Re.
So Re is the same for
model and prototype:
Um um dm
Pm
ML-3
§ Uul ·
¸
© P ¹
I¨
§ Uul ·
¸
© P ¹
Uu 2 l 2 I ¨
This applies whatever the size of the body
i.e. it is applicable to prototype and
a geometrically similar model.
219
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Example 1
For the model
Rm
Um um2 lm2
An underwater missile, diameter 2m and length 10m is tested in a
water tunnel to determine the forces acting on the real prototype.
A 1/20th scale model is to be used. If the maximum allowable speed
of the prototype missile is 10 m/s, what should be the speed of the
water in the tunnel to achieve dynamic similarity?
§ Um um lm ·
¸
© Pm ¹
I¨
and for the prototype
Rp
U p u p2 l p2
Dynamic similarity so Reynolds numbers equal:
Um um dm U p u p d p
§U u l ·
I¨ p p p ¸
© Pp ¹
Pm
Dividing these two equations gives
Rm / Um um2 lm2
R p / U p u p2 l p2
The model velocity should be
U d P
um u p p p m
Um d m P p
I Um um lm / P m IU puplp / P p W can go no further without some assumptions.
Assuming dynamic similarity, so Reynolds number
are the same for both the model and prototype:
Um um dm
Pm
Pp
Both the model and prototype are in water then,
Pm = Pp and Um = Up so
U p upd p
Pp
um
up
dp
dm
10
1
1 / 20
200 m / s
so
Rm
Rp
Um um2 lm2
U p u 2p l p2
This is a very high velocity.
This is one reason why model tests are not always
done at exactly equal Reynolds numbers.
i.e. a scaling law for resistance force:
OR
CIVE 1400: Fluid Mechanics
A wind tunnel could have been used so the values of
the U and P ratios would be used in the above.
OU O2u O2L
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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 tunnel operating at a pressure of 20 times atmospheric. The
aeroplane will fly at 500km/h. At what speed should the wind
tunnel operate to give dynamic similarity between the model and
prototype? If the drag measure on the model is 337.5 N what will
be the drag on the plane?
R
§ Uul ·
Uu l I ¨
¸
© P ¹
Rm
Rp
Uu l I Re
2 2
Rm
Rp
For dynamic similarity Rem = Rep, so
0.5u p
Uu l Uu l 2 2
m
2 2
20 0.5
1 1
p
2
2
.
01
1
0.05
So the drag force on the prototype will be
U d P
up p p m
Um d m P p
um
um
1 1
20 1 / 10
250 km / h
up
And the ratio of forces is
Earlier we derived an equation for resistance on a
body moving through air:
2 2
um
Rp
1
Rm
0.05
20 u 337.5 6750 N
The value of P does not change much with pressure
so Pm = Pp
For an ideal gas is p = URT so the density of the air in
the model can be obtained from
pm
pp
20 p p
pp
Um
Um RT
U p RT
Um
Up
Um
Up
20U p
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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:
xdepths and become very difficult to measure;
xthe bed roughness becomes impracticably
small;
xlaminar 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:
xGood overall flow patterns and discharge
xlocal detail of flow is not well modelled.
The Froude number (Fn) is taken as dominant.
Fn can be the same even for distorted models.
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Section 5: Dimensional Analysis
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