Unit 1
CIVE1400: An Introduction to Fluid Mechanics Dr P A Sleigh
[email protected] Dr CJ Noakes
C.J .Noak .Noakes@leed
[email protected] s.ac.uk J anuary 2008 Module web site: www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Unit 1: F luid Mechanics Basics Basics Flow Pressure P ropert roperties ies of Fluid F luids s F luids luids vs. Solid S olids s Viscosity
3 lectures lectures
Unit Unit 2: Statics tatics Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces
3 lectures
Unit Unit 3: Dynamics Dynamics The The con continu inuity ity equation ion. The The Ber Bernoulli Equa Equation. Application of Bernoulli equation. The The momentum equation. Application of momentum equation.
7 lectures
Unit Unit 4: Effect of the the boundary on flow Laminar and turbulent flow Boundary layer theory An Intro to Dimensional analysis Similarity
4 lectures
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/ w.efm.leeds.ac.uk/CIVE CIVE/FluidsLeve /FluidsLevel1
Lecture 1
1
Unit 1
Notes For the First Year Lecture Course:
An Introduction to Fluid Mechanics School of Civil Engineering, University of Leeds. CIVE1400 FLUID MECHANICS Dr Andrew Sleigh
J anuary ary 2008
Contents of the Course Objectives:
The The cou course will ill int introdu oduce fluid luid mechan chanic ics s and esta stablish lish its its rele elevance in civ civil engine ineerin ering g. Develop the fundamental principles underlying the subject. Demonstrate how these are used for the design of simple hydraulic components. Civil Engineering Fluid Mechanics
Why are we studying fluid mechanics on a Civil Engineering course? The provision of adequate water services such as the supply of potable water, drainage, sewerage is essential for the development of industrial society. It is these services which civil engineers provide. Fluid mechanics is involved in nearly all areas of Civil Engineering either directly or indirectly. Some examples of direct involvement are those where we are concerned with manipulating the fluid: Sea and river (flood) defences; Water distribution / sewerage (sanitation) networks; Hydraulic design of water/sewage treatment works; Dams; Irrigation; P umps umps and Turbines; Water retaining structures. And some examples where the primary object is construction - yet analysis of the fluid mechanics is essential: Flow of air in buildings; Flow of air around buildings; Bridge piers in rivers; Ground-water round-water flow – much larger scale in time time and space. space. Notice how nearly all of these involve water. The following course, although introducing general fluid flow ideas and principles, the course will demonstrate many of these principles through examples where the fluid is water.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/ w.efm.leeds.ac.uk/CIVE CIVE/FluidsLeve /FluidsLevel1
Lecture 1
2
Unit 1
Module Consists Consists of:
Lectures: 20 Classes presenting the concepts, theory and application. Worked examples will also be given to demonstrate how the theory is applied. You will be asked to do some calculations - so bring a calculator. Assessment: 1 Exam of 2 hours, worth 80% of the module credits. This This con consist sists s of 6 questio stion ns of which ich you choo choose se 4. 2 Multiple choice question (MCQ) papers, worth 10% of the module credits (5% each).
The These will ill be for 30mins ins and set set after the lect lectu ures. es. The The timetable able for these ese MCQs and and lectures is shown in the table at the end of this section. 1 Marked problem sheet, worth 10% of the module credits.
Laboratories: 2 x 3 hours Thes These e two lab laborat orato ory sessi session ons s exam examine ine how well the theor eoretical ical anal analy ysis sis of fluid luid dynamics ics 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 ill be exp expecte cted to draw con conclu clusio sions as to the valid alidit ity y of the theory ory based sed 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 The two lab laborator atorie ies s sess sessio ion ns are: 1. Impact of jets on various various shaped shaped surfaces - a jet jet of water water is fired at a target target and is deflected in various directions. This is an example of the application of the momentum equation. 2. The rectangular rectangular weir weir - the the weir is used as a flow measuring device. Its accuracy is investigated. This is an example of how the Bernoulli (energy) equation is applied to analyses fluid flow. [As you know, these laboratory sessions are compulsory course-work. You must attend them. Should you fail to attend either one you will be asked to complete some extra work. This will involve a detailed report and further questions. The simplest strategy is to do the lab.] Homework: Example sheets: These will be given for each section of the course. Doing these will greatly
improve your exam mark. They are course work but do not have credits toward the module. Lecture notes : Theses should be studied but explain only the basic outline of the necessary concepts and ideas. Books: It is very important do some extra reading in this subject. To do the examples you will definitely need a textbook. Any one of those identified below is adequate and will also be useful for the fluids (and other) modules in higher years - and in work. Example classes: Ther There e will ill be exam example clas classe ses s each each week eek. You may bring ing any problem lems/q s/questio stion ns you have ave about the course and example sheets to these classes.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/ w.efm.leeds.ac.uk/CIVE CIVE/FluidsLeve /FluidsLevel1
Lecture 1
3
Unit 1
Schedule: Lecture
Date
Week
15
0
2
16
0
Extra
22
1
3
23
1
4
29
2
5
30
2
5
3
7
6
3
8
12
4
9
13
4
10
19
5
1
Month
J anuary
6
February
Day Tue s
Time
Unit
3.00 pm
Unit 1: Fluid Mechanic Basics
Wed Tue s
9.00 am
Wed Tue s
9.00 am
Wed Tue s
9.00 am
Plane surfaces
3.00 pm
Curved surfaces
Wed Tue s
9.00 am
Design study 01 - Centre vale park
3.00 pm
Unit 3: Fluid Dynamics
Wed Tue s
9.00 am
Bernoulli
3.00 pm
Flow measurement
MCQ 20
5
26
6
27
6
4
7
13
5
7
14
11
8
15
12
8 Vacatio n
15
9
17
16
9
18
22
10
19
23
10
20
29
11
surveyin g
13 12
March
16
April
30
11
Viscosity, Flow Presentation of Case Studies
double lecture Flow calculations
Unit 2: Fluid Statics
Pressure
General
MCQ
Wed Tue s
9.00 am
Weir
3.00 pm
Momentum
Wed Tue s
9.00 am
Wed Tue s
9.00 am
Wed
9.00 am
problem sheet given out
Calculation
Tue s
3.00 pm
Unit 4: Effects of the Boundary on Flow
Boundary Layer
Design study 02 - Gaunless +Millwood
3.00 pm
Applications Design study 02 - Gaunless +Millwood
3.00 pm
Applications
Wed Tue s
9.00 am
Friction
3.00 pm
Dim. Analysis
Wed Tue s
9.00 am
problem sheet handed in
3.00 pm
Revision
4.00 pm
MCQ
MCQ 21
3.00 pm
4.00 pm
11 12
3.00 pm
Pressure, density
Wed
Dim. Analysis
9.00 am
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
4
Unit 1
Books:
Any of the books listed below are more than adequate for this module. (You will probably not need any more fluid mechanics books on the rest of the Civil Engineering course) Mechanics of Fluids, Massey B S., Van Nostrand Reinhold. Fluid Mechanics, Douglas J F, Gasiorek J M, and Swaffield J A, Longman. Civil Engineering Hydraulics, Featherstone R E and Nalluri C, Blackwell Science. Hydraulics in Civil and Environmental Engineering, Chadwick A, and Morfett J ., E & FN Spon Chapman & Hall.
Online Lecture Notes: http://www.efm.leeds.ac.uk/cive/FluidsLevel1
There is a lot of extra teaching material on this site: Example sheets, Solutions, Exams, Detailed lecture notes, Online video lectures, MCQ tests, Images etc. This site DOES NOT REP LACE LECTURES or BOOKS.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
5
Unit 1
Take care wit h the System of Unit s
As any quantity can be expressed in whatever way you like it is sometimes easy to become confused as to what exactly or how much is being referred to. This is particularly true in the field of fluid mechanics. Over the years many different ways have been used to express the various quantities involved. Even today different countries use different terminology as well as different units for the same thing - they even use the same name for different things e.g. an American pint is 4/5 of a British pint! To avoid any confusion on this course we will always use the SI (metric) system - which you will already be familiar with. It is essential that all quantities are expressed in the same system or the wrong solutions will results. Despite this warning you will still find that this is the most common mistake when you attempt example questions. The SI Syst em of un its
The SI system consists of six primary units, from which all quantities may be described. For convenience secondary units are used in general practice which are made from combinations of these primary units. Primary Units
The six primary units of the SI system are shown in the table below: Quantity
SI Unit
Dimension
Length Mass Time Temperature Current Luminosity
metre, m kilogram, kg second, s Kelvin, K ampere, A candela
L M T θ 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. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
6
Unit 1
Derived Units
There are many derived units all obtained from combination of the above primary units. Those most used are shown in the table below: Quantity 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 J oule 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 ms-2
Dimension LT-1 LT-2
kg ms-2
M LT-2
kg m2s-2
ML2 T-2
Nms-1 kg m2s-3
ML2 T-3
Nm-2 kg m-1s-2
ML-1 T-2
kg m-3
ML-3
kg m-2s-2
ML-2 T-2 1 no dimension
-1
N sm-2 kg m-1s-1 Nm-1 kg s-2
M L-1 T-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. More on this subject will be seen later in the section on dimensional analysis and similarity.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
7
Unit 1
Properties of Fluids: Density There are three ways of expressing density: 1. Mass density:
ρ = mass per unit volume ρ =
mass of fluid volume of fluid (units: kg/m3)
2. Specific Weight: (also known as specific gravity)
ω = weight per unit volume ω = ρ g 3
2
2
(units: N/m or kg/m /s ) 3. Relative Density:
σ = ratio of mass density to a standard mass density σ =
ρ subs tan ce ρ
o
H2 O ( at 4 c )
For solids and liquids this standard mass density is the maximum mass density for water (which occurs o
at 4 c) at atmospheric pressure. (units: none, as it is a ratio) CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
8
Unit 1
Pressure Convenient to work in terms of pressure, p, which is the force per unit area.
pressure = p =
Force Area over which the force is applied F A
Units: Newtons per square metre, 2
2
-1 -2
N/m , kg/m s (kg m s ). Also known as a Pascal, Pa, i.e. 1 Pa = 1 N/m
2
Also frequently used is the alternative SI unit the bar, 5 2 where 1bar = 10 N/m Standard atmosphere = 101325 Pa = 101.325 kPa 1 bar = 100 kPa (kilopascals) 1 mbar = 0.001 bar = 0.1 kPa = 100 Pa
Uniform Pressure: If the pressure is the same at all points on a surface uniform pressure CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
9
Unit 1
Pascal’s Law: pressure acts equally in all directions. ps B
δz δs
A
px
δy
F
C
θ E
D
δx
py
No shearing forces : All forces at right angles to the surfaces Summing forces in the x-direction: Force in the x-direction due to p x ,
F x
x
= p x × Area ABFE = p x δx δ y
Force in the x-direction due to p s ,
F x = − ps × Area ABCD × sinθ s
= − psδs δ z
δ y δ s
= − psδy δ z ( sinθ =
δ y δ s
)
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
10
Unit 1
Force in x-direction due to p y ,
F x
y
=0
To be at rest (in equilibrium) sum of forces is zero
F x x + Fx s + F x y = 0 p xδxδy + ( − psδyδz ) = 0 p x = ps Summing forces in the y-direction. Force due to p y ,
F y = p y × y
rea EFCD = p yδxδ z
Component of force due to p s ,
F y = − ps × Area ABCD × cosθ s
= − psδsδ z
δ x δ s
= − psδxδ z ( cosθ = δ x
δ s
)
Component of force due to p x ,
F y = 0 x
Force due to gravity, CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
11
Unit 1
weight = - specific weight × volume of element 1
= − ρg × δxδyδ z 2 To be at rest (in equilibrium)
F y + Fy + F y + weight = 0 y
s
x
1 ⎛ ⎞ p yδxδy + ( − psδxδz ) + ⎜ − ρg δxδyδz⎟ = 0 ⎝ ⎠ 2 The element is small i.e. x, x, and z, are small, so x y z, is very small and considered negligible, hence
p y = ps We showed above
p x = ps thus
p x = p y = ps Pressure at any point is the same in all directions. This is Pascal’s Law and applies to fluids at rest.
Change of Pressure in the Vertical Direction CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
12
Unit 1 p2, A Area A
Fluid density ρ
z2
p1, A
z1
Cylindrical element of fluid, area = A, density =
The forces involved are: Force due to p 1 on A (upward) = p 1 A Force due to p 2 on A (downward) = p 2 A Force due to weight of element (downward) = mg= density =
volume
g
g A(z2 - z1)
Taking upward as positive, we have p1 A − p2 A − gA( z2 − z1 ) = 0 p2 − p1 = − ρ g( z2 − z1 )
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
13
Unit 1
In a fluid pressure decreases linearly with increase in height p2 − p1 = − ρ g( z2 − z1 ) This is the hydrostatic pressure change. With liquids we normally measure from the surface. Measuring h down from the free surface so that h = -z z
h
y x
giving p 2 − p1 = ρ gh Surface pressure is atmospheric, patmospheric .
p = ρ gh + patmospheric
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
14
Unit 1
It is convenient to take atmospheric pressure as the datum Pressure quoted in this way is known as gauge pressure i.e. Gauge pressure is p gauge =
gh
The lower limit of any pressure is the pressure in a perfect vacuum. Pressure measured above a perfect vacuum (zero) is known as absolute pressure Absolute pressure is p absolute =
g h + p atmospheric
Absolute pressure = Gauge pressure + Atmospheric
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
15
Unit 1
Pressure density relationship Boyle’s Law
pV = constant Ideal gas law
pV = nRT where p is the absolute pressure, N/m 2, Pa
V is the volume of the vessel, m 3 n is the amount of substance of gas, moles R is the ideal gas constant,
T is the absolute temperature. K -1
-1
In SI units, R = 8.314472 J mol K 3 1 1 (or equivalently m Pa K mol ). −
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
−
Lecture 1
16
Unit 1
Lecture 2: Fluids vs Solids, Flow
What makes fluid mechanics different to solid mechanics?
Fluids are clearly different to solids. But we must be specific. Need definable basic physical difference. Fluids flow under the action of a force, and the solids don’t - but solids do deform.
• fluids lack the ability of solids to resist deformation. • fluids change shape as long as a force acts. Take a rectangular element CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 17
Unit 1
A
B
A’
F
B’
F C
D
C
D
Forces acting along edges (faces), such as F, are know as shearing forces. A Fluid is a substance which deforms continuously, or flows, when subjected to shearing forces.
This has the following implications for fluids at rest:
If a fluid is at rest there are NO shearing forces acting on it, and any force must be acting perpendicular to the fluid
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 18
Unit 1
Fluids in motion Consider a fluid flowing near a wall. - in a pipe for example Fluid next to the wall will have zero velocity. The fluid “sticks” to the wall. Moving away from the wall velocity increases to a maximum.
v
Plotting the velocity across the section gives “velocity profile”
Change in velocity with distance is “velocity gradient” =
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
du dy
Lecture 2 19
Unit 1
As fluids are usually near surfaces there is usually a velocity gradient. Under normal conditions one fluid particle has a velocity different to its neighbour. Particles next to each other with different velocities exert forces on each other (due to intermolecular action ) …… i.e. shear forces exist in a fluid moving close to a wall.
What if not near a wall?
v
No velocity gradient, no shear forces. CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 20
Unit 1
What use is this observation?
It would be useful if we could quantify this shearing force. This may give us an understanding of what parameters govern the forces different fluid exert on flow.
We will examine the force required to deform an element.
Consider this 3-d rectangular element, under the action of the force F.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 21
Unit 1
δx b
a
δz
F B
A
δy
F C
D
under the action of the force F
a
b
a’
b’
F A
A’
B
B’
E
F C
D
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 22
Unit 1
A 2-d view may be clearer… A’
B
E x
φ
B’
F
E’
y
F C
D
The shearing force acts on the area A = δz × δ x
Shear stress,
is the force per unit area: τ =
F A
The deformation which shear stress causes is measured by the angle , and is know as shear strain. Using these definitions we can amend our definition of a fluid: In a fluid
increases for as long as is applied the fluid flows In a solid shear strain, , is constant for a fixed shear stress .
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 23
Unit 1
It has been shown experimentally that the rate of shear strain is directly proportional to shear stress τ ∝
φ time
φ τ = Constant × t
We can express this in terms of the cuboid. If a particle at point E moves to point E’ in time t then: for small deformations shear strain φ =
x y
rate of shear strain =
=
=
= (note that
x t
=u
is the velocity of the particle at E)
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 24
Unit 1
So
τ = Constant ×
u y
u/y is the rate of change of velocity with distance, in differential form this is
du dy
= velocity gradient.
The constant of proportionality is known as the dynamic viscosity, giving
τ = μ
du dy
which is know as Newton’s law of viscosity
A fluid which obeys this rule is know as a Newtonian Fluid (sometimes also called real fluids) Newtonian fluids have constant values of
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 25
Unit 1
Non-Newtonian Fluids Some fluids do not have constant . They do not obey Newton’s Law of viscosity. They do obey a similar relationship and can be placed into several clear categories The general relationship is:
⎛ δ u ⎞ τ = A + B⎜ ⎟ ⎝ δ y ⎠
n
where A, B and n are constants. For Newtonian fluids A = 0, B =
and n = 1
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 26
Unit 1
This graph shows how
changes for different fluids.
Bingham plastic
Pseudo plastic
plastic Newtonian τ
, s s e r t s r a e h S
Dilatant
Ideal, (τ=0)
Rate of shear, δu/δy
• Plastic: Shear stress must reach a certain minimum before flow commences.
• Bingham plastic: As with the plastic above a minimum shear stress must be achieved. With this classification n = 1. An example is sewage sludge.
• Pseudo-plastic: No minimum shear stress necessary and the viscosity decreases with rate of shear, e.g. colloidal substances like clay, milk and cement.
• Dilatant sub stances; Viscosity increases with rate of shear e.g. quicksand.
• Thixotro pic substances: Viscosity decreases with length of time shear force is applied e.g. thixotropic jelly paints.
• Rheopectic sub stances: Viscosity increases with length of time shear force is applied
• Viscoelastic materials: Similar to Newtonian but if there is a sudden large change in shear they behave like plastic
Viscosity CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 27
Unit 1
There are two ways of expressing viscosity
Coefficient of Dynamic Viscosity τ μ = du dy 2
Units: N s/m or Pa s or kg/m s The unit Poise is also used where 10 P = 1 Pa·s 4
Water µ = 8.94 × 10 Pa s 3 Mercury µ = 1.526 × 10 Pa s Olive oil µ = .081 Pa s 8 Pitch µ = 2.3 × 10 Pa s Honey µ = 2000 – 10000 Pa s Ketchup µ = 50000 – 100000 Pa s (non-newtonian) −
−
Kinematic Viscosity = the ratio of dynamic viscosity to mass density
ν =
μ ρ
2
Units m /s 6
2
Water = 1.7 × 10− m /s. −5
2
Air = 1.5 × 10 m /s.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 28
Unit 1
Flow rate Mass flow rate
m &=
dm d t
=
mass time taken to accumulate this mass
A simple example: An empty bucket weighs 2.0kg. After 7 seconds of collecting water the bucket weighs 8.0kg, then:
mass flow rate = m & =
=
mass of fluid in bucket
time taken to collect the fluid 8.0 − 2.0
7 = 0857 . kg / s
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 29
Unit 1
Volume flow rate - Discharge. More commonly we use volume flow rate Also know as discharge. The symbol normally used for discharge is Q.
discharge, Q =
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 × 10 − 3 m3 25 sec 3
= 0.0008 m / s = 0.8 l / s
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 30
Unit 1
Discharge and mean velocity If we know the discharge and the diameter of a pipe, we can deduce the mean velocity um t
x
area A
Pipe
Cylinder of fluid
Cross sectional area of pipe is A Mean velocity is um. In time t, a cylinder of fluid will pass point X with a volume A× um × t. The discharge will thus be
Q=
volume
time Q = Aum
=
A × um × t t
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 31
Unit 1
A simple example: -3
2
If A = 1.2× 10 m
And discharge, Q is 24 l/s, mean velocity is
um =
=
Q A
2.4 × 10 − 3
1.2 × 10 − 3 = 2.0 m / s Note how we have called this the mean velocity. This is because the velocity in the pipe is not constant across the cross section. x
u um
uma x
This idea, that mean velocity multiplied by the area gives the discharge, applies to all situations - not just pipe flow. CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 32
Unit 1
Continuity This principle of conservation of mass says matter cannot be created or destroyed This is applied in fluids to fixed volumes, known as control volumes (or surfaces)
Mass flow in Control volume
Mass flow out
For any control volume the principle of conservation of mass says
Mass entering =
Mass leaving
per unit time
per unit time
+ Increase 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
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 33
Unit 1
In a real pipe (or any other vessel) we use the mean velocity and write 1 A1um1
= ρ 2
2 um2
= Constant = m &
For incompressible, fluid ρ 1 = ρ 2 = ρ (dropping the m subscript)
1u1
= A2 u2 = Q
This is the continuity equation most often used.
This equation is a very powerful tool. It will be used repeatedly throughout the rest of this course.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 2 34
Unit 1
Lecture 3: Examples from Unit 1: Fluid Mechanics Basics Units 1. A water company wants to check that it will have sufficient water if there is a prolonged drought in the area. The region it covers is 500 square miles and various different offices have sent in the following consumption figures. There is sufficient information to calculate the amount of water available, but unfortunately it is in several different units. Of the total area 100 000 acres are rural land and the rest urban. The density of the urban population is 50 per square kilometre. The average toilet cistern is sized 200mm by 15in by 0.3m and on average each person uses this 3 time per day. The density of the rural population is 5 per square mile. Baths are taken twice a week by each person with the average volume of water in the bath being 6 gallons. Local industry uses 1000 m3 per week. Other uses are estimated as 5 gallons per person per day. A US air base in the region has given water use figures of 50 US gallons per person per day. The average rain fall in 1in per month (28 days). In the urban area all of this goes to the river while in the rural area 10% goes to the river 85% is lost (to the aquifer) and the rest goes to the one reservoir which supplies the region. This reservoir has an average surface area of 500 acres and is at a depth of 10 fathoms. 10% of this volume can be used in a month. a) What is the total consumption of water per day? b) If the reservoir was empty and no water could be taken from the river, would there be enough water if available if rain fall was only 10% of average?
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
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Unit 1
Fluid Properties 1. The following is a table of measurement for a fluid at constant temperature. Determine the dynamic viscosity of the fluid. du/dy (s-1) 0.0 0.2 0.4 0.6 0.8 -2 0.0 1.0 1.9 3.1 4.0 τ (N m )
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
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Unit 1
2. The density of an oil is 850 kg/m3. Find its relative density and Kinematic viscosity if the dynamic viscosity is 5 × 10-3 kg/ms.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
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Unit 1 3.
The velocity distribution of a viscous liquid (dynamic viscosity μ = 0.9 Ns/m2) flowing over a fixed plate is given by u = 0.68y - y2 (u is velocity in m/s and y is the distance from the plate in m). What are the shear stresses at the plate surface and at y=0.34m?
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
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Unit 1
4. 5.6m3 of oil weighs 46 800 N. Find its mass density, ρ and relative density, γ.
5.
From table of fluid properties the viscosity of water is given as 0.01008 poises. What is this value in Ns/m2 and Pa s units?
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
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Unit 1
6.
In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125m/s. The fluid has absolute viscosity 0.048 Pa s and relative density 0.913. What is the velocity gradient and shear stress at the boundary assuming a linear velocity distribution.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
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Unit 1
Continuity
Section 1
Section 2
A liquid is flowing from left to right. By continuity
A1u1 ρ1 = A2 u2 ρ 2 As we are considering a liquid (incompressible), 1
=
2
=
Q1 = Q2 A1u1 = A2 u2 If the area A1=10 10-3 m 2 and A 2=3 10-3 m 2 And the upstream mean velocity u1=2.1 m/s. What is the downstream mean velocity?
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
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Unit 1
Now try this on a diffuser, a pipe which expands or diverges as in the figure below,
Section 1
Section 2
If d 1=30mm and d 2=40mm and the velocity u 2=3.0m/s.
What is the velocity entering the diffuser?
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
42
Unit 1
Velocities in pipes coming from a junction.
2
1
3
mass flow into the junction = mass flow out 1Q1
=
2Q 2
+
3Q 3
When incompressible
Q1 = Q 2 + Q 3 1u1
=
2u 2
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
+
3u 3
Lecture 1
43
Unit 1
If pipe 1 diameter = 50mm, mean velocity 2m/s, pipe 2 diameter 40mm takes 30% of total discharge and pipe 3 diameter 60mm. What are the values of discharge and mean velocity in each pipe?
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1
Lecture 1
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CIVE1400: Fluid Mechanics
Section 2: Statics
CIVE1400: Fluid Mechanics
Section 2: Statics
Pressure And Head It is convenient to take atmospheric pressure as the datum
We have the vertical pressure relationship
dp dz
U g ,
Pressure quoted in this way is known as gauge pressure i.e.
integrating gives p = - gz + constant
Gauge pressure is p gauge =
measuring z from the free surface so that z = -h z
gh
h
The lower limit of any pressure is y
the pressure in a perfect vacuum . x
Pressure measured above
p U gh constant
a perfect vacuum (zero) is known as absolute pressure
surface pressure is atmospheric, p atmospheric . Absolute pressure is
patmospheric constant
pabsolute =
so
Absolute pressure = Gauge pressure + Atmospheric
p U gh patmospheric CIVE1400: Fluid Mechanics
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g h + p atmospheric
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A gauge pressure can be given
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Pressure Measurement By Manometer
using height of any fluid.
p U gh This vertical height is the
Manometers use the relationship between pressure and head to measure pressure
head .
The Piezometer Tube Manometer
If pressure is quoted in head , the density of the fluid must also be given.
The simplest manometer is an open tube.
Example:
This is attached to the top of a container with liquid at pressure. containing liquid at a pressure.
-2
What is a pressure of 500 kNm in head of water of density, = 1000 kgm 3 Use p = gh,
h
p
500 u 103
U g 1000 u 9.81
In head of Mercury density
h
500 u 103 13.6 u 103 u 9.81
3.75m of Mercury
h
8.7 u 1000 u 9.81
CIVE1400: Fluid Mechanics
B
= 8.7. The tube is open to the atmosphere,
water )
500 u 103
h2
A
= 13.6 103 kgm-3.
In head of a fluid with relative density remember =
h1
5095 . m of water
The pressure measured is relative to atmospheric so it measures gauge pressure.
5.86m of fluid J = 8.7
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An Example of a Piezometer. What is the maximum gauge pressure of water that can be measured by a Piezometer of height 1.5m?
Pressure at A = pressure due to column of liquid h 1
p A =
And if the liquid had a relative density of 8.5 what would the maximum measurable gauge pressure?
g h1
Pressure at B = pressure due to column of liquid h 2
p B =
g h2
Problems with the Piezometer: 1. Can only be used for liquids 2. Pressure must above atmospheric 3. Liquid height must be convenient i.e. not be too small or too large.
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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
We have shown
cross sectional area = A
pl = pr
mass density =
For a vertical pressure change we have
left end pressure = pl
pl p p U gz
right end pressure = pr and
pr pq U gz
For equilibrium the sum of the forces in the x direction is zero.
so
pl A = pr A
p p Ugz pq U gz
pl = pr
p p pq
Pressure in the horizontal direction is constant.
Pressure at the two equal levels are the same.
This true for any continuous fluid. CIVE1400: Fluid Mechanics
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The “U”-Tube Manometer
We know:
“U”-Tube enables the pressure of both liquids and gases to be measured
Pressure in a continuous static fluid is the same at any horizontal level.
“U” is connected as shown and filled with manometric fluid .
pressure at B = pressure at C
pB = pC Important points: 1. The manometric fluid density should be greater than of the fluid measured. man >
For the left hand arm pressure at B = pressure at A + pressure of height of liquid being measured
pB = p A + gh1
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
Fluid density ρ
manometric liquid
D
pC = patmospheric +
man gh2
h2 A
We are measuring gauge pressure we can subtract patmospheric giving
h1 B
C
pB = pC Manometric fluid density ρ
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p A =
man
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man gh2 -
gh1
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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 = 700 kg/m 3, and the manometric fluid is mercury, with a relative density of 13.6. What is the gauge pressure if: a) h1 = 0.4m and h 2 = 0.9m? b) h1 stayed the same but h 2 = -0.1m?
BUT:
As the manometric fluid is liquid (usually mercury , oil or water) And Liquid density is much greater than gas, man
>>
gh1 can be neglected, and the gauge pressure given by p A =
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man gh2
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Pressure difference measurement Using a “U”-Tube Manometer.
CIVE1400: Fluid Mechanics
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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 = p A + pD = pB +
B
p A + Fluid density ρ
g ha
g (hb + h) +
g ha = pB +
man g
g (hb + h) +
h
man g
h
Giving the pressure difference E
hb
p A - pB =
g (hb - ha ) + (
man
- )g h
h A
Again if the fluid is a gas man >> , then the terms involving can be neglected,
ha D
C
p A - pB =
man
gh
Manometric fluid density ρman
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An example using the u-tube for pressure difference measuring
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Advances to the “U” tube manometer
In the figure below two pipes containing the same 3 fluid of density = 990 kg/m 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?
Problem: Two reading are required. Solution: Increase cross-sectional area of one side. Result: One level moves much more than the other.
Fluid density ρ
Fluid density ρ
A
ha = 1.5m
p2
p 1
B E
diameter D
hb = 0.75m
h = 0.5m
diameter d z2 Datum line
C
D
z1
Manometric fluid density ρ man = 13.6 ρ
If the manometer is measuring the pressure difference of a gas of (p1 - p2 ) as shown,
we know p1 - p2 =
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man
gh
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volume of liquid moved from the left side to the right 2
Problem: Small pressure difference,
= z 2 ( d / 4)
movement cannot be read.
The fall in level of the left side is z 1
Volume moved
Solution 1: Reduce density of manometric
Area of left side
fluid.
z 2 S d / 4 2
S D 2 / 4
d · z 2 § ¨ ¸ © D ¹
Result:
2
Greater height change easier to read.
Putting this in the equation, 2 ª d · º § p1 p 2 U g «z 2 z 2 ¨ ¸ » © D ¹ ¼ ¬
Solution 2: Tilt one arm of the manometer.
ª § d · 2 º U gz 2 «1 ¨ ¸ » ¬ © D ¹ ¼ 2 If D >> d then (d/D) is very small so p1 p2 U gz 2
Result:
Same height change - but larger movement along the manometer arm - easier to read.
Inclined manometer CIVE1400: Fluid Mechanics
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Example of an inclined manometer. p1
p
diameter d
diameter D
e r a d e R l e a S c
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 3 has density man = 740 kg/m and the scale may be read to +/- 0.5mm.
2
x z 2 Datum line
What is the angle required to ensure the desired accuracy may be achieved?
z1 θ
The pressure difference is still given by the height change of the manometric fluid. p1 p2 U gz 2
but, z2 x sin T p1 p2 Ugx sin T The sensitivity to pressure change can be increased further by a greater inclination.
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Forces on Submerged Surfaces in Static Fluids
Choice Of Manometer
We have seen these features of static fluids
Take care when fixing the manometer to vessel Burrs cause local pressure variations.
x Hydrostatic vertical pressure distribution x Pressures at any equal depths in a continuous
Disadvantages:
x Slow response - only really useful for very slowly
fluid are equal
varying pressures - no use at all for fluctuating pressures;
x Pressure at a point acts equally in all directions (Pascal’s law).
x For the “U” tube m anometer two measurements
xForces from a fluid on a boundary acts at right
must be taken simultaneously to get the h value.
angles to that boundary.
x It is often difficult to measure small variations in pressure.
x It cannot be used for very large pressures unless
Fluid pressure on a surface
several manometers are connected in series;
x For very accurate work the temperature and relationship between temperature and known;
Pressure is force per unit area.
must be
Pressure p acting on a small area A exerted force will be
Advantages of manometers:
x They are very simple. x No calibration is required - the pressure can be
F = p A
calculated from first principles.
Since the fluid is at rest the force will act at right-angles to the surface. CIVE1400: Fluid Mechanics
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Horizontal submerged plane
General submerged plane F =p δA 1 1 1 F2=p2δA2
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
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 GA n ¦ p GA
R = pA
Curved submerged surface
and This resultant force will act through the centre of pressure.
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.
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.
The sum of all the forces on each element will always be less than the sum of the pG A . individual forces,
¦
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¦ zG A is known as
Resultant Force and Centre of Pressure on a general plane surface in a liquid. O Fluid density ρ
z
z
Q
st
the 1 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
And it is known that ¦ zG A Az
d
area A
Sc
C 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 A, depth z ,
p = gz
In terms of distance from point O
¦ zGA Ax sin T
So force on element
= 1st moment of area
F = gz A (as z
R Ug
sin
about a line through O
Resultant force on plane
(assuming
Section 2: Statics
¦ zG A
x sinT )
The resultant force on a plane
and g as constant).
R U gAz
U gAx sin T CIVE1400: Fluid Mechanics
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This resultant force acts at right angles through the centre of pressure, C, at a depth D.
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Section 2: Statics
Sum of moments U g sin T
Moment of R about O = R
How do we find this position?
58
¦ s G A 2
u Sc = UgAx sin T S c
Equating
Take moments of the forces.
U gAx sin T S c Ug sin T
¦ s G A 2
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 A about the same point.
The position of the centre of pressure along the plane measure from the point O is:
¦ s G A 2
S c
It is convenient to take moment about O
Ax
How do we work out the summation term?
The force on each elemental area:
Force on G A UgzG A
U g s sin T G A the moment of this force is:
Moment of Force on G A about O Ug s sin T GA u s
U g sin T G As
2
This term is known as the nd 2 Moment of Area , I o, of the plane (about the axis through O)
, g and are the same for each element, giving the total moment as
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2nd moment of area about O I o ¦ s 2G A
How do you calculate the 2 area?
It can be easily calculated for many common shapes.
2
nd
1
moment of
It can be found from tables BUT only for moments about an axis through its centroid = I GG .
2 nd Moment of area about a line through O st
nd
moment of area is a geometric property.
The position of the centre of pressure along the plane measure from the point O is:
S c
Section 2: Statics
nd
Usually we want the 2 moment of area about a different axis.
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 sinT
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to give us what we want.
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nd
The 2 moment of area about a line through the centroid of some common shapes.
The parallel axis theorem can be written I o I GG Ax 2
Shape
Area A
an axis through the centroid
We then get the following equation for the position of the centre of pressure
Rectangle b
bd
S c
Ax
bd 3 12
h
I GG
2nd moment of area, I GG , about
G
G
x Triangle
§ I · D sinT ¨ GG x ¸ © Ax ¹
h G
h/3
G
b
bd 2
bd 3 36
S R 2
S R 4
Circle (In the examination the parallel axis theorem
R G
G
and the I GG will be given)
4 Semicircle G
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S R 2
R (4R)/(3π)
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2
01102 . R4
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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
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C
We know the relationship between pressure and depth: p = gz So we can draw the diagram below: ρgz
z H
2H 3 R p
ρgH
This is know as a pressure diagram. CIVE1400: Fluid Mechanics
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Pressure increases from zero at the surface linearly by p = gz , to a maximum at the base of p = gH . The area of this triangle represents the resultant force per unit width on the vertical wall,
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For a triangle the centroid is at 2/3 its height i.e. the resultant force acts 2 horizontally through the point z H . 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
D
2 1 H U gH 2 1 U gH 2 2
2 3
H
Resultant force per unit width 1 R U gH 2 2
( N / m)
The force acts through the centroid of the pressure diagram. CIVE1400: Fluid Mechanics
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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: The resultant force is given by:
oil ρo
R UgAz UgAx sinT H U g H u 1 sinT 2 1 U gH 2 2
0.8m
1.2m
water ρ
D
R
ρg0.8
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 the depth to the centre of pressure by:
§ I · D sinT ¨ o ¸ © Ax ¹ and by the parallel axis theorem (with width of 1)
I o I GG Ax 2
1 u H 3 12
2
3
H · H 1 u H § ¨ ¸ © 2 ¹ 3
Depth to the centre of pressure
§ H 3 / 3 · 2 ¸ H D ¨ © H 2 / 2 ¹ 3
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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
horizontal and vertical components.
B G O
FAC
RH
A
Combine these to obtain the resultant force and direction.
Rv
R
The fluid is at rest – in equilibrium.
So any element of fluid such as ABC is also in equilibrium.
(Although this can be done for all three dimensions we will only look at one vertical plane)
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Consider the Horizontal forces The sum of the horizontal forces is zero. C
B
RH
FAC
The resultant horizontal force of a fluid above a curved surface is: R H = Resultant force on the projection of the curved surface onto a vertical plane. We know 1. The force on a vertical plane must act horizontally (as it acts normal to the plane).
A
2. That R H must act through the same point.
No horizontal force on CB as there are no shear forces in a static fluid
So:
R H acts horizontally through the centre of pressure of the projection of the curved surface onto an vertical plane.
Horizontal forces act only on the faces AC and AB as shown.
F AC , must be equal and opposite to R H . We have seen earlier how to calculate resultant forces and point of action.
AC is the projection of the curved surface AB onto a vertical plane.
Hence we can calculate the resultant horizontal force on a curved surface.
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Consider the Vertical forces
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Resultant force
The sum of the vertical forces is zero. E
D
C
The overall resultant force is found by combining the vertical and horizontal components vectorialy,
B G
Resultant force R
A
2 2 RV R H
Rv
There are no shear force on the vertical edges, so the vertical component can only be due to the weight of the fluid .
And acts through O at an angle of .
The angle the resultant force makes to the horizontal is
So we can say The resultant vertical force of a fluid above a curved surface is:
§ R · T tan 1 ¨ V ¸ © R H ¹
R V = Weight of fluid directly above the curved surface. It will act vertically down through the centre of gravity of the mass of fluid.
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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 .
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A typical example application of this is the determination of the forces on dam walls or curved sluice gates.
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What are the forces if the fluid is curved surface?
Find the magnitude and direction of the resultant force of water on a quadrant gate as shown below.
below the
This situation may occur or a curved sluice gate. C
Gate width 3.0m
B G
1.0m
O
FAC
RH
3
Water ρ = 1000 kg/m
A R
Rv
The force calculation is very similar to when the fluid is above.
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Horizontal force
Section 2: Statics
Section 2: Statics
Vertical force C
B G
B
O
FAC
78
RH A
A
A’
Rv
What vertical force would keep this in equilibrium?
The two horizontal on the element are: The horizontal reaction force R H The force on the vertical plane A’B.
If the region above the curve were all water there would be equilibrium.
The resultant horizontal force, R H acts as shown in the diagram. Thus we can say:
The resultant horizontal force of a fluid below a curved surface is: R H = Resultant force on the projection of the curved surface onto a vertical plane.
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: R v =Weight of the imaginary volume of fluid vertically above the curved surface.
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The resultant force and direction of application are calculated in the same way as for fluids above the surface:
CIVE1400: Fluid Mechanics
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
Resultant force
Section 2: Statics
b) the reaction at B E
R
2 R H
C
2 RV
A
D
B
And acts through O at an angle of . The angle the resultant force makes to the horizontal is
§ R · T tan 1 ¨ V ¸ © R H ¹
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Unit 3: Fluid Dynamics
Unit 3: Fluid Dynamics
CIVE1400: An Introduction to Fluid Mechanics
Fluid Dynamics
Unit 3: Fluid Dynamics
Objectives
Dr P A Sleigh:
[email protected]
1.Identify differences between:
Dr CJ Noakes:
[email protected]
x steady/unsteady x uniform/non-uniform x compressible/incompressible flow
January 2008 Module web site: www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Unit 1: Fluid Mechanics Basics Flow Pressure Properties of Fluids Fluids vs. Solids Viscosity
3 lectures
Unit 2: Statics Hydrostatic pressure Manometry / Pressure measurement Hydrostatic forces on submerged surfaces
3 lectures
Unit 3: Dynamics The continuity equation. The Bernoulli Equation. Application of Bernoulli equation. The momentum equation. Application of momentum equation.
7 lectures
Unit 4: Effect of the boundary on flow Laminar and turbulent flow Boundary layer theory An Intro to Dimensional analysis Similarity
4 lectures
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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
Lecture 8
7.Demonstrate use of the momentum equation to predict forces induced by flowing fluids
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Unit 3: Fluid Dynamics
Fluid dynamics:
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Unit 3: Fluid Dynamics
Flow Classification Fluid flow may be
The analysis of fluid in motion
classified under the following headings
Fluid motion can be predicted in the same way as the motion of solids
uniform: Flow conditions (velocity, pressure, cross-section or depth) are the same at every point in the fluid.
By use of the fundamental laws of physics and the physical properties of the fluid
non-uniform: Flow conditions are not the same at every point.
Some fluid flow is very complex: e.g.
steady Flow conditions may differ from point to point but DO NOT change with time.
x Spray behind a car x waves on beaches;
unsteady Flow conditions change with time at any point.
x hurricanes and tornadoes x any other atmospheric phenomenon
Fluid flowing under normal circumstances - a river for example conditions vary from point to point we have non-uniform flow.
All can be analysed with varying degrees of success (in some cases hardly at all!). There are many common situations
If the conditions at one point vary as time passes then we have unsteady flow.
which analysis gives very accurate predictions CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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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. CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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Unit 3: Fluid Dynamics
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
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.
one direction at right angles to this.
Below shows flow pattern over a weir.
But: Since flow must be zero at the pipe wall - yet non-zero in the centre there is a difference of parameters across the cross-section.
Pipe
Ideal flow
Real flow
In this course we will be considering: x steady x incompressible x one and two-dimensional flow
Should this be treated as two-dimensional flow? Possibly - but it is only necessary if very high accuracy is required.
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Streamlines
Some points about streamlines :
It is useful to visualise the flow pattern.
x Close to a solid boundary, streamlines are parallel to that boundary
Lines joining points of equal velocity - velocity contours - can be drawn.
x The direction of the streamline is the direction of the fluid velocity
These lines are know as streamlines Here are 2-D streamlines around a cross-section of an aircraft wing shaped body:
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.
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Streamtubes
Some points about streamtubes
A circle of points in a flowing fluid each has a streamline passing through it.
x The “walls” of a streamtube are streamlines. x Fluid cannot flow across a streamline, so fluid
These streamlines make a tube-like shape known as a streamtube
cannot cross a streamtube “wall”.
x A 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):
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Flow rate
Discharge and mean velocity
Mass flow rate
Cross sectional area of a pipe is A Mean velocity is u m.
m
mass dm dt time taken to accumulate this mass
Q = Au m We usually drop the “m” and imply mean velocity.
Continuity
Volume flow rate - Discharge.
Mass entering = Mass leaving
More commonly we use volume flow rate
per unit time
Also know as discharge.
+
per unit time
Mass flow in Control volume
Mass flow out
Increase of mass in control vol per unit time
The symbol normally used for discharge is Q.
discharge, Q
For steady flow there is no increase in the mass within the control volume, so
volume of fluid
For steady flow
time
Mass entering = Mass leaving per unit time
per unit time
Q1 = Q2 = A1u1 = A2u2
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Applying to a streamtube:
Unit 3: Fluid Dynamics
In a real pipe (or any other vessel) we use the mean velocity and write
Mass enters and leaves only through the two ends (it cannot cross the streamtube wall). ρ2
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U1 A1u m1
U 2 A2 um2
Constant
m
u2 A2
For incompressible, fluid U 1 = U 2 = U (dropping the m subscript)
ρ1 u1 A1
Mass entering = per unit time
A1u1 A2 u2
Mass leaving per unit time
Q
This is the continuity equation most often used.
U1G A1u1 U2G A2u2 Or for steady flow,
U1G A1u1
U2G A2 u2
Constant
m
This equation is a very powerful tool. It will be used repeatedly throughout the rest of this course.
dm dt
This is the continuity equation. CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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Some example applications of Continuity Water flows in a circular pipe which increases in diameter from 400mm at point A to 500mm at point B. Then pipe then splits into two branches of diameters 0.3m and 0.2m discharging at C and D respectively.
1. What is the outflow?
3
1.5 m /s
If the velocity at A is 1.0m/s and at D is 0.8m/s, what are the discharges at C and D and the velocities at B and C? Solution: Draw diagram:
Qin = Qout
C
1.5 + 1.5 = 3 3 Qout = 3.0 m /s 2. What is the inflow?
dC=0.3m
dB=0.5m
A
B
dA=0.4m vA=1.0m/s
D
u = 1.5 m/s 2 A = 0.5 m
dD=0.2m
u 3. = 0.2 m/s 2 4. A = 1.3 m
u = 1.0 m/s 2 A = 0.7 m
Make a table and fill in the missing values
5.
3
Q
Q = 2.8 m /s Q = Area
vD=0.8m/s
u Mean Velocity = Au
Point
Velocity m/s
Diameter m
Area m²
Q m³/s
A
1.00
0.4
0.126
0.126
B
0.64
0.5
0.196
0.126
C
1.42
0.3
0.071
0.101
D
0.80
0.2
0.031
0.025
Q + 1.5u0.5 + 1u0.7 = 0.2u1.3 + 2.8 Q = 3.72 m3/s CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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Unit 3: Fluid Dynamics
potential head = z
total head = H
Restrictions in application of Bernoulli’s equation:
Lecture 9: The Bernoulli Equation Unit 3: Fluid Dynamics
x Flow is steady The Bernoulli equation is a statement of the principle of conservation of energy along a streamline
x Density is constant (incompressible) x Friction losses are negligible
It can be written:
p1
u12
z 1
U g 2 g
x It relates the states at two points along a single
H = Constant
streamline, (not conditions on two different streamlines)
These terms represent:
Pressure
Kinetic
All these conditions are impossible to satisfy at any instant in time!
Potential
Total
energy per energy per energy per energy per unit weight
unit weight
unit weight
unit weight
Fortunately, for many real situations where the conditions are approximately satisfied, the equation gives very good results.
These term all have units of length, they are often referred to as the following:
p pressure head = U g CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
velocity head =
u2 2 g Lecture 8
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Unit 3: Fluid Dynamics
Unit 3: Fluid Dynamics
The derivation of Bernoulli’s Equation: distance AA’ =
Cross sectional area a
B B’
work done = force u distance AA’
A z
A’
= pa
mg
An element of fluid, fluid, as that in the figure figure above, has potential 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
1 2
u
m Ua
pm U
p U g
work done per unit weight =
Summing all of these energy terms gives
mu 2
Pressure
Kinetic
Potential
Total
energy per energy per energy per energy per
kinetic energy per unit weight =
u
unit weight
2
unit weight
unit weight
p
u2
z H
U g 2 g
By the principle of conservation of energy, the total energy in energy in the system does not change, thus the total head does head does not change. So the Bernoulli equation can be written
force on AB = pa when the mass mg of mg of fluid has passed AB, cross-section AB will have moved to A’B’
u2
mg m
p
U g
U g 2 g
unit weight
or
2 g
At any cross-section the the pressure generates 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 is p and the area of the crosssection is a then
volume passing AB =
This term is know as the pressure energy of the flowing stream.
potential energy per unit weight = z kinetic energy =
m U a
U
z H Constant
therefore
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Unit 3: Fluid Dynamics
The Bernoulli equation is applied along _______________ _______________ like that joining points 1 and 2 below.
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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.
2
Example: Finding pressures and velocities within a contracting and expanding pipe.
1
total head at 1 = total head at 2 or
p1
u12
z 1
U g 2 g
p2
u22
z 2
U g 2 g
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: Total energy per
Total
unit weight at 1
p1
u12
Loss
Work done
energy per unit per unit weight at 2
z 1
U g 2 g
p2
per unit
weight
u22
weight
supplied
p1
p2
section 1
section 2
3
A fluid, density U = 960 kg/m is flowing steadily through the above tube. The section diameters are d 1=100mm and d 2 =80mm . 2=80mm. The velocity at 1 is u 1=5m/s. =5m/s.
per unit weight
The tube is horizontal ( z 1=z 2 2)
z2 h w q
U g 2 g
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u2
The gauge pressure at 1 is p1=200kN/m2
Energy
u1
Lecture 8
What is the gauge pressure at section 2?
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Apply the Bernoulli Bernoulli equation along a streamline streamline joining section 1 with section 2. p1 u12 p2 u22 z 1 z 2
U g 2 g
p2
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.
U g 2 g
p1
U 2 (u1 2
u22 )
Use the continuity equation to find u 2 2
Applications of the Bernoulli Equation
A1u1 A2u2 u2
A1u1 A2
2
The Bernoulli equation is applicable to many situations not just the pipe flow.
§ d · ¨ 1 ¸ u1 © d 2 ¹
8125m / s 7.8125
Here we will see its application to flow measurement from tanks, within pipes as well as in open channels.
So pressure at section 2
p2
Unit 3: Fluid Dynamics
200000 1729687 . 182703 N / m2 182 182.7 kN / m2
Note how the velocity has increased the pressure has decreased
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Apply Bernoulli Bernoulli along the streamline streamline joining point point 1 on the surface to point 2 at the centre of the orifice.
Applications of Bernoulli: Flow from Tanks Flow Through A Small Orifice
At the surface velocity is negligible (u (u 1 = 0 ) and the pressure atmospheric ( p p1 = 0 ). ).
Flow from a tank through a hole in the side.
At the orifice the jet is open to the air so 1
again the pressure is atmospheric p (p2 = 0 ). ).
Aactual
h
If we take the datum line through the orifice then z 1 = h and z 2 2 =0 , leaving 2
Vena contractor
h The edges of the hole are sharp to minimise frictional losses by minimising the contact between the hole and the liquid.
u2
u22 2 g
2 gh
This theoretical value of velocity is an overestimate as friction losses have not been taken into account.
The streamlines at the orifice contract reducing the area of flow.
A coefficient of velocity is used to correct the theoretical theoretical velocity,
This contraction is called the vena contracta
uactua ctuall
The amount of contraction must
Cv uthe theoret oretic ical al
be known to calculate the flow
Each orifice has its own coefficient of velocity , velocity , they usually lie in the range( 0.97 - 0.99)
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Unit 3: Fluid Dynamics
The discharge through the orifice
Time for the tank to empty
is jet area u jet velocity
We have an expression for the discharge from the tank
Q Cd Ao 2 gh
The area of the jet is the area of the vena contracta not the area of the orifice.
We can use this to calculate how long it will take for level in the to fall
We use a coefficient of contraction
As the tank empties empties the level of water water falls.
to get the area of the jet
The discharge will also drop.
Cc Aorifice
Aactual
h1
Giving discharge through the orifice:
h2
Q Au Qactua ctuall
Aactu ctual uactu actual al Cc Cv Aorif orific icee u theo theore reti tica cal l Cd Aorific orificee u theore theoretic tical al Cd Aorif orific icee
The tank has a cross sectional area of A of A.. In a time G t the t the level falls by G h The flow out of the tank is Q Au
2 g h
Q A
C d d is the coefficient of discharge, C d d = C c c u C v v
G h G t
(-ve sign as G h is falling)
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This Q is the same as the flow out of the orifice so
Cd Ao 2 gh A
G t
Unit 3: Fluid Dynamics
Submerged Orifice What if the tank is feeding into another?
G h
Area A1
G t
A
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Area A2
h1 h2
G h
Cd Ao 2 g h Orifice area Ao
Integrating between the initial level, h1, and final level, h2 , gives the time it takes to fall this height
t
A Cd Ao
h G h
³ 2 2 g h1
Apply Bernoulli Bernoulli from point 1 on the surface of the the deeper tank to point 2 at the centre of the orifice, p1 u12 p2 u22 z 1 z 2
U g 2 g
h
1 § ¨© ³ ³ h 1/2 2h1/2 2 h · ¹¸ h
U g 2 g
0 0 h1
u2
U gh2 U g
u22 2 g
0
2 g (h1 h2 )
And the discharge discharge is given by
t
A Cd Ao
Q Cd Ao u
>2 h @hh12 2 g
2 A
>
Cd Ao 2 g
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h2
Cd Ao
h1 @
2 g (h1 h2 )
So the discharge of the jet through the submerged orifice depends on the difference in head across the orifice.
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Using the Bernoulli equation we can calculate the pressure at this point.
Lecture 10: Flow Measurement Devices Unit 3: Fluid Dynamics
Along the central streamline at 1: velocity u 1 , pressure p1 At the stagnation point (2): u 2 = 0 . (Also z 1 = z 2 )
p1 u12 U 2
Pitot Tube The Pitot tube is a simple velocity measuring device.
p2
Uniform velocity flow hitting a solid blunt body, has streamlines similar to this:
p2 U
p1
1 2 U u1 2
How can we use this? 2
1
The blunt body does not have to be a solid. It could be a static column of fluid. Some move to the left and some to the right. 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.
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 .
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h1
1
h2
2
The Pitot static tube combines the tubes and they can then be easily connected to a manometer.
1 2
1 2 U u1 2 1 U gh2 Ugh1 U u12 2 u 2 g (h2 h1 )
p1
1 X
h A
B
[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.
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Unit 3: Fluid Dynamics
Pitot Static Tube The necessity of two piezometers makes this arrangement awkward.
We have the equation for p2 ,
p2
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The holes on the side connect to one side of a manometer, while the central hole connects to the other side of the manometer
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Using the theory of the manometer,
Pitot-Static Tube Example
p1 Ug X h U man gh p B p2 U gX p A p B p2 UgX p1 Ug X h U man gh p A
We know that p2
p1
p1 hg Uman
1 2
A pitot-static tube is used to measure the air flow at the centre of a 400mm diameter building ventilation duct. If the height measured on the attached manometer is 10 mm and the density of the manometer fluid is 1000 kg/m3, determine the volume flow rate in the duct. 3 Assume that the density of air is 1.2 kg/m .
U u12 , giving U u12
U p1
2
2 gh( Um U )
u1
U
The Pitot/Pitot-static is:
x Simple to use (and analyse) x Gives velocities (not discharge)
x May block easily as the holes are small.
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Apply Bernoulli along the streamline from point 1 to point 2
Venturi Meter
p1 The Venturi meter is a device for measuring
u12
z 1
U g 2 g
discharge in a pipe.
p2
u22
z 2
U g 2 g
By continuity
Q u1 A1
It is a rapidly converging section which increases the velocity of flow and hence reduces the pressure.
u2
It then returns to the original dimensions of the pipe by a gently diverging ‘diffuser’ section.
p1 p2 z1 z 2 U g
2
u1 A1 A2
Substituting and rearranging gives
about 6°
about 20°
u2 A2
u12 ª§ A1 · «¨ ¸ 2 g «© A2 ¹
º 1» »¼ ¬ u12 ª A12 A22 º « » 2 g ¬ A22 ¼ 2
1
z2 z1
u1 A2
h
ª p p2 º 2 g « 1 z1 z 2 » ¬ U g ¼ A12 A22
datum
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The theoretical (ideal) discharge is uu A.
Venturimeter design:
Actual discharge takes into account the losses due to friction, we include a coefficient of discharge (C d |0.9)
x The diffuser assures a gradual and steady deceleration after
Qideal u1 A1 Qactual
Qactual
Cd Qideal Cd u 1 A1 Cd A1 A2
x The angle of the diffuser is usually between 6 and 8 degrees.
ª p p2 º 2 g « 1 z1 z 2 » ¬ U g ¼ A12
the throat. So that pressure rises to something near that before the meter.
x Wider and the flow might separate from the walls increasing energy loss.
A22
x If the angle is less the meter becomes very long and pressure losses again become significant.
In terms of the manometer readings
p1 Ugz1
p2 Uman gh U g ( z2 h ) p1 p2 § U · z1 z2 h¨ man 1¸ © U ¹ U g
x The efficiency of the diffuser of increasing pressure back to the original is rarely greater than 80%.
x Care must be taken when connecting the manometer so that no burrs are present.
Giving
Qactual
Cd A1 A2
§ U 2 gh¨ man © U A12
· 1¸ ¹
A22
This expression does not include any elevation terms. (z 1 or z 2) When used with a manometer The Venturimeter can be used without knowing its angle. CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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Venturimeter Example A venturimeter is used to measure the flow of water in a 150 mm diameter pipe. The throat diameter of the venturimeter is 60 mm and the discharge coefficient is 0.9. If the pressure difference measured by a manometer is 10 cm mercury, what is the average velocity in the pipe? Assume water has a density of 1000 kg/m 3 and mercury has a relative density of 13.6.
Lecture 11: Notches and Weirs Unit 3: Fluid Dynamics
x A notch is an opening in the side of a tank or reservoir. x It is a device for measuring discharge x A weir is a notch on a larger scale - usually found in rivers. x It is used as both a discharge measuring device and a device to raise water levels.
x There are many different designs of weir. x We will look at sharp crested weirs. Weir Assumptions
x velocity of the fluid approaching the weir is small so we can ignore kinetic energy.
x The velocity in the flow depends only on the depth below the free surface.
u
2 g h
These assumptions are fine for tanks with notches or reservoirs with weirs, in rivers with high velocity approaching the weir is substantial the kinetic energy must be taken into account CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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A General Weir Equation
Rectangular Weir
Consider a horizontal strip of
The width does not change with depth so
width b, depth h below the free surface
b constant b
B
h
H
B
δh
H
velocity through the strip, u 2 gh discharge through the strip, GQ Au bG h 2 gh
Substituting this into the general weir equation gives
Integrating from the free surface, h=0 , to the weir crest, h=H , gives the total theoretical discharge H Qtheoretical 2 g bh1/ 2 dh 0
Qtheoretical
shaped weir or notch. We need an expression relating the width of flow across the weir to the depth below the free surface. CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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2 g ³ h1/ 2 dh
B
0
2
2 gH 3/ 2
B
³
This is different for every differently
H
3
To get the actual discharge we introduce a coefficient of discharge, C d, to account for losses at the edges of the weir and contractions in the area of flow,
Qactual
2
Cd B 3
2 gH 3 / 2
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Rectangular Weir Example
‘V’ Notch Weir The relationship between width and depth is dependent on the angle of the “V”.
Water enters the Millwood flood storage area via a rectangular weir when the river height exceeds the weir crest. For design purposes a flow rate of 162 litres/s over the weir can be assumed
b
h
H θ
1. Assuming a height over the crest of 20cm and Cd=0.2, what is the necessary width, B, of the weir? The width, b, a depth h from the free surface is
b
T 2 H h tan§ ¨© · ¸ 2 ¹
So the discharge is H
Qtheoretical
§ T · 2 g tan¨ ¸ ³ H hh1/ 2 dh © 2 ¹
2
0
2. What will be the velocity over the weir at this design?
H T · ª 2 § 3/ 2 2 5 / 2 º 2 g tan¨ ¸ « Hh h » © 2 ¹ ¬ 3 5 ¼0
2
8 15
§ T ·¸ 5/ 2 © 2 ¹ H
2 g tan¨
The actual discharge is obtained by introducing a coefficient of discharge
Qactual CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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C d
8 15
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§ T · H 5 / 2 ¸ © 2 ¹
2 g tan¨
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‘V’ Notch Weir Example Water is flowing over a 90 o ‘V’ Notch weir into a tank 2 with a cross-sectional area of 0.6m . After 30s the depth of the water in the tank is 1.5m.
Lecture 12: The Momentum Equation Unit 3: Fluid Dynamics
If the discharge coefficient for the weir is 0.8, what is the height of the water above the weir?
We have all seen moving fluids exerting forces.
x The lift force on an aircraft is exerted by the air moving over the wing.
x A jet of water from a hose exerts a force on whatever it hits. The analysis of motion is as in solid mechanics: by use of Newton’s laws of motion.
The Momentum equation is a statement of Newton’s Second Law It relates the sum of the forces to the acceleration or rate of change of momentum.
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From solid mechanics you will recognise F = ma
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In time G t a volume of the fluid moves from the inlet a distance u 1G t, so
What mass of moving fluid we should use?
volume entering the stream tube = area
u distance
= A 1u1 G t
We use a different form of the equation. The mass entering, Consider a streamtube:
mass entering stream tube = volume
density
= U1 A1 u1 G t
And assume steady non-uniform flow
And momentum momentum entering stream tube = mass
A2 u2
A1 u1
u
u velocity
= U1 A1 u1 G t u1
ρ2
ρ1
Similarly, at the exit, we get the expression:
u1 δt
momentum leaving stream tube = U2 A 2 u 2 G t u 2
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By Newton’s 2
nd
Unit 3: Fluid Dynamics
Law.
An alternative derivation From conservation of mass
Force = rate of change of momentum
F =
mass into face 1 = mass out of face 2
( U2 A2u2Gt u2 U1 A1u1G t u1 )
we can write
G t
rate of change of mass
dm dt U1 A1u1 U 2 A2u2
m
We know from continuity that The rate at which momentum enters face 1 is
U 1 A1u1u1 mu 1
Q A1u1 A2 u2
The rate at which momentum leaves face 2 is
And if we have a fluid of constant density, i.e. U1
F
U 2 A2 u2 u2 mu 2
U2 U , then
Q U (u2 u1 )
Thus the rate at which momentum changes across the stream tube is
U2 A2 u2 u2
U 1 A1u1u1 mu 2 mu 1
So
Force = rate of change of momentum
F m ( u2 u1 ) CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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The previous analysis assumed the inlet and outlet velocities in the same direction i.e. a one dimensional system.
So we have these two expressions, either one is known as the momentum equation
What happens when this is not the case? u2
F m ( u2 u1 ) F
θ2
Q U ( u2 u1 ) θ1
The Momentum equation.
u1
This force acts on the fluid in the direction of the flow of the fluid.
We consider the forces by resolving in the directions of the co-ordinate axes. The force in the x-direction
F x
m u 2 cosT2 u1 cosT 1 m u 2 x u1 x or
F x CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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UQu2 cosT2 u1 cosT 1 U Qu2 x u1x
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And the force in the y-direction
F y
m u2 sin T2 u1 sinT 1
In summary we can say:
¨ u2 u1 ·¸ m § © y y ¹
Total force on the fluid
or
F y
rate of change of =
the control volume
UQu2 sinT2 u1 sinT 1 ¨ u2 u1 ·¸ U Q§ © y y ¹
F
m u out uin or
F
The resultant force can be found by combining these components Fy
momentum through
U Quout uin
FResultant
Remember that we are working with vectors so F i s in the direction of the velocity.
φ Fx
Fresultant
F x2 F y2
And the angle of this force
§ F y · I tan 1 ¨ ¸ © F x ¹ CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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This force is made up of three components: F R = Force exerted on the fluid by any solid body
Application of the Momentum Equation
touching the control volume
Forces on a Bend F B = Force exerted on the fluid body (e.g. gravity)
Consider a converging or diverging pipe bend lying in the vertical or horizontal plane
F P = Force exerted on the fluid by fluid pressure outside the control volume
turning through an angle of T . So we say that the total force, F T, Here is a diagram of a diverging pipe bend.
is given by the sum of these forces:
y
p2 u 2
F T = F R + F B + F P
A2
x
The force exerted
1m
p1 u1
45°
A1
by the fluid on the solid body touching the control volume is opposite to F R. So the reaction force, R , is given by R = -F R CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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Why do we want to know the forces here?
An Example of Forces on a Bend
As the fluid changes direction a force will act on the bend. This force can be very large in the case of water supply pipes. The bend must be held in place
The outlet pipe from a pump is a bend of 45q rising in the vertical plane (i.e. and internal angle of 135q). The bend is 150mm diameter at its inlet and 300mm diameter at its outlet. The pipe axis at the inlet is horizontal and at the outlet it is 1m higher. By neglecting friction, calculate the force and its direction if the inlet pressure is 100kN/m 2 and the flow of water through the pipe is 0.3m3 /s. The volume of the pipe is 0.075m3. [13.95kN at 67q 39’ to the horizontal]
1&2 Draw the control volume and the axis system
to prevent breakage at the joints.
y
We need to know how much force a support (thrust block) must withstand.
p2 u 2
A2
x
1m
p1
Step in Analysis:
45°
u1 A1
1.Draw a control volume 2.Decide on co-ordinate axis system 3.Calculate the total force
2
4.Calculate the pressure force
p1 = 100 kN/m ,
5.Calculate the body force
Q = 0.3 m /s
3
= 45
6.Calculate the resultant force
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d1 = 0.15 m
d2 = 0.3 m
A1 = 0.177 m 2
A2 = 0.0707 m 2
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3 Calculate the total force in the x direction
Unit 3: Fluid Dynamics
4 Calculate the pressure force. F P pressure force at 1 - pressure force at 2 T 1
FT x
by continuity A1u1
u1
u2
F T x
U Qu2 x u1x UQ u2 cosT u1
A2u 2
Q , so
0.3 S 0.152 / 4 0.3 0.0707
. m/s 1698
T 2
T
F P x
p1 A1 cos 0 p 2 A2 cos T p1 A1 p 2 A2 cos T
F P y
p1 A1 sin 0 p 2 A2 sin T p 2 A2 sin T
we can use the Bernoulli equation to calculate this unknown pressure.
4.24 m / s
p1 u12 p2 u22 z 1 z2 h f U g 2 g U g 2 g where hf is the friction loss
and in the y-direction
In the question it says this can be ignored, hf =0
U Qu2 y u1 y UQu2 sin T 0 1000 u 0.34.24 sin 45 899.44 N
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0,
We know pressure at the inlet but not at the outlet
1000 u 0.34.24 cos45 16.98 419368 . N
FT y
159
The height of the pipe at the outlet is 1m above the inlet. Taking the inlet level as the datum: z 1 = 0 Lecture 8
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z 2 = 1m Lecture 8
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6 Calculate the resultant force So the Bernoulli equation becomes:
FR x FP x F B x FT y FR y FP y F B y
4.24 2 p2 0 10 . 1000 u 9.81 2 u 9.81 1000 u 9.81 2 u 9.81 100000
1698 . 2
p2
2253614 . N / m2
F P x
100000 u 0.0177 225361.4 cos45 u 0.0707 1770 11266.34 9496.37 kN
F P y
225361.4 sin 45 u 0.0707 11266.37
5 Calculate the body force The only body force is the force due to gravity. That is the weight acting in the -ve y direction.
F B y
FT x
U g u volume 1000 u 9.81 u 0.075 1290156 . N
FT x FP x F B x 4193.6 9496.37 5302.7 N
F R y
FT y FP y F B y 899.44 11266.37 735.75 1290156 . N
And the resultant force on the fluid is given by FRy
FResultant
φ FRx
F R
There are no body forces in the x direction,
F B x
F R x
F F 2 R x
2 Ry
. 2 5302.7 2 1290156 1395 . kN
0
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And the direction of application is
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Lecture 14: Momentum Equation Examples Unit 3: Fluid Dynamics
§ F R y · 1¨ ¸ I tan ¨ © F R x ¹¸ 1290156 . · ¨© tan 1§ ¹¸
Impact of a Jet on a Plane
5302.7
$
67.66 67
$
A jet hitting a flat plate (a plane) at an angle of 90q
39 '
The force on the bend is the s ame magnitude but in the opposite direction
R F R
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.
. kN 1395 1 & 2 Control volume and Co-ordinate axis are shown in the figure below. y
u2
x
Lecture 13: Design Study 2
u1
See Separate Handout u2
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3 Calculate the total force
Unit 3: Fluid Dynamics
6 Calculate the resultant force
In the x-direction
FT x
FR x FP x F B x F T x 0 0 U Qu1 x
FT x
U Qu2 x u1 x U Qu1 x
F R x
Exerted on the fluid. The system is symmetrical the forces in the y-direction cancel.
F T y
The force on the plane is the s ame magnitude but in the opposite direction
R F R x
0
If the plane were at an angle the analysis is the same.
4 Calculate the pressure force.
But it is usually most convenient to choose the axis system normal to the plate.
The pressures at both the inlet and the outlets to the control volume are atmospheric.
y
The pressure force is zero
F P x
u2
x
F P y 0
5 Calculate the body force As the control volume is small
u1 θ
we can ignore the body force due to gravity.
F B x
u3
F B y 0
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Force on a curved vane
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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 at atmospheric by continuity u1
u2
Both the cross-section and velocities (in the direction of flow) remain constant.
F T x
Q A
U
, so
Q2 A
1 cosT
u2
y
and in the y-direction x
FT y
UQu2 sin T 0
u1
U
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.
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F P x 168
F P y 0
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5 Calculate the body force
F R
No body forces in the x-direction, F B x = 0. 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,
F B x
FR2 x F R2 y
And the direction of application is
§ F R y · ¸ I tan 1 ¨ © F R x ¹
U gV
exerted on the fluid. (This is often small as the jet volume is small and sometimes ignored in analysis.)
The force on the vane is the same magnitude but in the opposite direction
6 Calculate the resultant force
FT x F R x
R F R
FR x FP x F B x
FT y F R y
Q2 F T x U 1 cosT A
FR y FP y F B y
Q2 F T y U A
And the resultant force on the fluid is given by CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1
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We work with components of the force:
SUMMARY
u2 θ2
The Momentum equation is a statement of Newton’s Second Law For a fluid of constant density, θ1
Total force on the fluid
=
rate of change of momentum through
u1
the control volume
F
m uout uin U Quout uin
F x
UQu2 x u1x UQu2 cosT2 u1 cosT 1
F y
UQu2 y u1 y UQu2 sin T2 u1 sinT 1
The resultant force can be found by combining these components
This force acts on the fluid in the direction of the velocity of fluid.
Fy
FResultant
This is the total force F T where: F T = F R + F B + F P
φ
F R = External force on the fluid from any solid body
Fx
touching the control volume
§ F y · I tan 1 ¨ ¸ © F x ¹
F P = Pressure force on the fluid by fluid pressure outside the control volume Lecture 8
F x2 F y2
And the angle this force acts:
F B = Body force on the fluid body (e.g. gravity)
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Fresultant
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2. A 600mm diameter pipeline carries water under a head of 30m with a velocity of 3m/s. This water main is fitted with a horizontal bend which turns the axis of the pipeline through 75q (i.e. the internal angle at the bend is 105q). Calculate the resultant force on the bend and its angle to the horizontal.
Lecture 15: Calculations Unit 3: Fluid Dynamics
1. The figure below shows a smooth curved vane attached to a rigid foundation. The jet of water, rectangular in section, 75mm wide and 25mm thick, strike the vane with a velocity of 25m/s. Calculate the vertical and horizontal components of the force exerted on the vane and indicate in which direction these components act.
45q 25q
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3. A 75mm diameter jet of water having a velocity of 25m/s strikes a flat plate, the normal of which is inclined at 30q to the jet. Find the force normal to the surface of the plate.
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Unit 3: Fluid Dynamics
4. In an experiment a jet of water of diameter 20mm is fired vertically upwards at a sprung target that deflects the water at an angle of 120° to the horizontal in all directions. If a 500g mass placed on the target balances the force of the jet, was is the discharge of the jet in litres/s?
5. Water is being fired at 10 m/s from a hose of 50mm diameter into the atmosphere. The water leaves the hose through a nozzle with a diameter of 30mm at its exit. Find the pressure just upstream of the nozzle and the force on the nozzle.
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Unit 4
Unit 4
CIVE1400: An Introduction to Fluid Mechanics
Real fluids
Dr P A Sleigh
[email protected]
Flowing real fluids exhibit viscous effects, they:
Dr CJ Noakes
x “stick” to solid surfaces
[email protected]
x have stresses within their body.
January 2008
From earlier we saw this relationship between shear stress and velocity gradient:
Module web site: www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Unit 1: Fluid Mechanics Basics Flow Pressure Properties of Fluids Fluids vs. Solids Viscosity
3 lectures
Unit 2: Statics Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces
3 lectures
Unit 3: Dynamics The continuity equation. The Bernoulli Equation. Application of Bernoulli equation. The momentum equation. Application of momentum equation.
7 lectures
Unit 4: Effect of the boundary on flow Laminar and turbulent flow Boundary layer theory An Intro to Dimensional analysis Similarity
4 lectures
CIVE 1400: Fluid Mechanics. www.efm.leeds. ac.uk/CIVE /FluidsLevel1
W
v
du dy
The shear stress, , in a fluid is proportional to the velocity gradient - the rate of change of velocity across the flow. For a “Newtonian” fluid we can write:
W where
Lectures 16-19
P
du dy
is coefficient of viscosity (or simply viscosity).
Here we look at the influence of forces due to momentum changes and viscosity in a moving fluid. 178
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Lectures 16-19
Unit 4
179
Unit 4
Laminar and turbulent flow
All three would happen but for different flow rates.
Injecting a dye into the middle of flow in a pipe, what would we expect to happen? This
Top: Slow flow Middle: Medium flow Bottom: Fast flow Top: Middle: Bottom:
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.
this
Turbulent flow: Motion is, locally, completely random but the overall direction of flow is one way.
or this
But what is fast or slow? At what speed does the flow pattern change? And why might we want to know this?
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Unit 4
The was first investigated in the 1880s by Osbourne Reynolds in a classic experiment in fluid mechanics.
Unit 4
After many experiments he found this expression
U ud P
A tank arranged as below:
= density,
u = mean velocity,
d = diameter
= viscosity
This could be used to predict the change in flow type for any fluid. This value is known as the Reynolds number, Re:
Re
U ud P
Laminar flow: Transitional flow: Turbulent flow:
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Re < 2000 2000 < Re < 4000 Re > 4000
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Unit 4
Unit 4
What are the units of Reynolds number?
At what speed does the flow pattern change?
We can fill in the equation with SI units:
We use the Reynolds number in an example:
U kg / m3 , u m / s, P Ns / m
Re
U ud P
2
kg / m s
kg m m ms 1 3 m s 1 kg
d
m
A pipe and the fluid flowing have the following properties: water density
= 1000 kg/m
pipe diameter
3
d = 0.5m
(dynamic) viscosity,
3
2
= 0.55x10 Ns/m
It has no units! A quantity with no units is known as a non-dimensional (or dimensionless) quantity.
What is the MAXIMUM velocity when flow is laminar i.e. Re = 2000
(We will see more of these in the section on dimensional analysis.)
Re
The Reynolds number, Re,
u
is a non-dimensional number.
U ud 2000 P 2000P U d
2000 u 0.55 u 10 3 1000 u 0.5
u 0.0022 m / s
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Unit 4
What is the MINIMUM velocity when flow is turbulent i.e. Re = 4000
Unit 4
What does this abstract number mean? We can give the Re number a physical meaning.
Re
U ud 4000 P
This may help to understand some of the reasons for the changes from laminar to turbulent flow.
u 0.0044 m / s In a house central heating system, typical pipe diameter = 0.015m ,
Re
limiting velocities would be, 0.0733
and 0.147m/s .
Both of these are very slow.
viscous forces
When the viscous forces are dominant (slow flow, low Re) they keep the fluid particles in line, the flow is laminar.
Laminar flow does occur in fluids of greater viscosity e.g. in bearing with oil as the lubricant.
Lectures 16-19
inertial forces
When inertial forces dominate (when the fluid is flowing faster and Re is larger) the flow is turbulent.
In practice laminar flow rarely occurs in a piped water system.
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Laminar flow
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Unit 4
Pressure loss due to friction in a pipeline
x Re < 2000 x ‘low’ velocity x Dye does not mix with water x Fluid particles move in straight lines x Simple mathematical analysis possible x Rare in practice in water systems.
Up to now we have considered ideal fluids: no energy losses due to friction Because fluids are viscous, energy is lost by flowing fluids due to friction.
Transitional flow
This must be taken into account.
x 2000 > Re < 4000 x ‘medium’ velocity x Dye stream wavers - mixes slightly.
The effect of the friction shows itself as a pressure (or head) loss.
Turbulent flow
x Re > 4000 x ‘high’ velocity x Dye mixes rapidly and completely x Particle paths completely irregular x Average motion is in flow direction x Cannot be seen by the naked eye x Changes/fluctuations are very difficult to
In a real flowing fluid shear stress slows the flow. To give a velocity profile:
detect. Must use laser.
x Mathematical analysis very difficult - so experimental measures are used
x Most common type of flow. CIVE 1400: Fluid Mechanics. www.efm.leeds. ac.uk/CIVE /FluidsLevel1
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Unit 4
Attaching a manometer gives pressure (head) loss due to the energy lost by the fluid overcoming the shear stress.
Unit 4
Consider a cylindrical element of incompressible fluid flowing in the pipe, τ
w
το L
το τ w is
area A
w
the mean shear stress on the boundary Upstream pressure is p,
Downstream pressure falls by p to ( p- p)
Δp
The driving force due to pressure driving force = Pressure force at 1 - pressure force at 2
The pressure at 1 (upstream) is higher than the pressure at 2.
pA p 'p A 'p A
How can we quantify this pressure loss in terms of the forces acting on the fluid?
S d 2 'p 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 wS dL
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As the flow is in equilibrium,
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What is the variation of shear stress in the flow? τ
driving force = retarding force
w
R
S d 2 ' p 4
r
W wS dL
' p
W w 4 L
τ
d
w
At the wall
W w
Giving pressure loss in a pipe in terms of:
x pipe diameter x shear stress at the wall
R 'p 2 L
At a radius r
r 'p 2 L r W W w R W
A linear variation in shear stress. This is valid for:
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Unit 4
Shear stress and hence pressure loss varies with velocity of flow and hence with Re.
Unit 4
Pressure loss during laminar flow in a pipe
Many experiments have been done with various fluids measuring the pressure loss at various Reynolds numbers. A graph of pressure loss and Re look like:
In general the shear stress w . is almost impossible to measure. For laminar flow we can calculate a theoretical value for a given velocity, fluid and pipe dimension. In laminar flow the paths of individual particles of fluid do not cross. Flow is like a series of concentric cylinders sliding over each other. And the stress on the fluid in laminar flow is entirely due to viscose forces. As before, consider a cylinder of fluid, length L, radius r, flowing steadily in the centre of a pipe.
This graph shows that the relationship between pressure loss and Re can be expressed as
laminar turbulent
' p v u ' p v u1.7 ( or 2.0)
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r
u
R
W 2Sr L 'p A 'pS r 2
³ r dr
' p r 2 L 4P
C
At r = 0 , (the centre of the pipe), u = umax , at r = R (the pipe wall) u = 0 ;
L 2
P
C
du , dy
' p R 2 L 4P
At a point r from the pipe centre when the flow is laminar:
We are measuring from the pipe centre, so
ur
du W P dr
' p L
1
R 4 P
2
r 2
This is a parabolic profile 2 (of the form y = ax + b ) so the velocity profile in the pipe looks similar to
Giving:
' p r
1
L 2P
ur
' p r
Newtons law of viscosity says W
' p
The value of velocity at a point distance r from the centre
The fluid is in equilibrium, shearing forces equal the pressure forces.
W
Unit 4
In an integral form this gives an expression for velocity,
δr r
195
du L 2 dr ' p r du dr L 2 P
P
v
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Unit 4
Unit 4
To get pressure loss (head loss) What is the discharge in the pipe?
in terms of the velocity of the flow, write pressure in terms of head loss hf , i.e. p = ghf
The flow in an annulus of thickness r
G Q ur Aannulus
Mean velocity:
u Q/ A
S (r Gr )2 Sr 2 | 2SrGr ' p 1 2 2 G Q R r 2SrG r
Aannulus
u
L 4 P
Q
' p L
S
U gh f d 2 32 P L
R
R 2r r 3 dr ³ 2 P 0 4
' p S R
L 8P
Head loss in a pipe with laminar flow by the Hagen-Poiseuille equation:
'p S d 4 L128P
h f
32 P Lu U gd 2
So the discharge can be written Pressure loss is directly proportional to the velocity when flow is laminar.
4
Q
' p S d
L 128P It has been validated many time by experiment. It justifies two assumptions:
This is the Hagen-Poiseuille Equation
1.fluid does not slip past a solid boundary 2.Newtons hypothesis.
for laminar flow in a pipe
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Unit 4
Considering a flat plate in a fluid.
Boundary Layers Recommended reading: Fluid Mechanics by Douglas J F, Gasiorek J M, and Swaffield J A. Longman publishers. Pages 327-332.
Upstream the velocity profile is uniform, This is known as free stream flow.
Fluid flowing over a stationary surface, e.g. the bed of a river, or the wall of a pipe, is brought to rest by the shear stress t o
Downstream a velocity profile exists.
This gives a, now familiar, velocity profile:
This is known as fully developed flow.
umax Free stream flow
τ
zero velocity
o Fully developed flow
Wall
Zero at the wall A maximum at the centre of the flow. The profile doesn’t just exit. It is build up gradually. Some question we might ask: Starting when it first flows past the surface e.g. when it enters a pipe.
How do we get to the fully developed state? Are there any changes in flow as we get there? Are the changes significant / important?
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Unit 4
Understand this Boundary layer growth diagram.
Boundary layer thickness:
= distance from wall to where u = 0.99 u mainstream
increases as fluid moves along the plate. It reaches a maximum in fully developed flow. The increase corresponds to a drag force increase on the fluid. As fluid is passes over a greater length:
*
* more fluid is slowed * by friction between the fluid layers the thickness of the slow layer increases.
Fluid near the top of the boundary layer drags the fluid nearer to the solid surface along. The mechanism for this dragging may be one of two types:
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First: viscous forces
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Second: momentum transfer
(the forces which hold the fluid together) If the viscous forces were the only action the fluid would come to a rest.
When the boundary layer is thin: velocity gradient du/dy, is large
Viscous shear stresses have held the fluid particles in a constant motion within layers.
by Newton’s law of viscosity shear stress, = (du/dy), is large.
Eventually they become too small to hold the flow in layers;
The force may be large enough to drag the fluid close to the surface.
the fluid starts to rotate.
As the boundary layer thickens velocity gradient reduces and shear stress decreases. Eventually it is too small to drag the slow fluid along. Up to this point the flow has been laminar. The fluid motion rapidly becomes turbulent. Newton’s law of viscosity has applied.
Momentum transfer occurs between fast moving main flow and slow moving near wall flow.
This part of the boundary layer is the
Thus the fluid by the wall is kept in motion.
laminar boundary layer
The net effect is an increase in momentum in the boundary layer. This is the turbulent boundary layer.
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Unit 4
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.
Unit 4
Use Reynolds number to determine which state. U ud Re P Laminar flow:
Re < 2000
Transitional flow: 2000 <
This layer occurs within the turbulent zone it is next to the wall. It is very thin – a few hundredths of a mm.
Turbulent flow:
Re < 4000
Re > 4000
Surface roughness effect Despite its thinness, the laminar sub-layer has vital role in the friction characteristics of the surface. In turbulent flow: Roughness higher than laminar sub-layer: increases turbulence and energy losses. Laminar flow: profile parabolic (proved in earlier lectures) The first part of the boundary layer growth diagram.
In laminar flow: Roughness has very little effect
Turbulent (or transitional), Laminar and the turbulent (transitional) zones of the boundary layer growth diagram.
Boundary layers in pipes Initially of the laminar form. It changes depending on the ratio of inertial and viscous forces;
Length of pipe for fully developed flow is the entry length.
i.e. whether we have laminar (viscous forces high) or turbulent flow (inertial forces high).
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Laminar flow
120
diameter
Turbulent flow
60
diameter
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Unit 4
Boundary layer separation
Boundary layer separation: *
Divergent flows: Positive pressure gradients. Pressure increases in the direction of flow.
*
increases the turbulence
increases the energy losses in the flow.
Separating / divergent flows are inherently unstable
The fluid in the boundary layer has so little momentum that it is brought to rest, and possibly reversed in direction.
Convergent flows:
Reversal lifts the boundary layer.
x Negative pressure gradients x Pressure decreases in the direction of flow.
u1
u2
p1
x Fluid accelerates and the boundary layer is thinner.
p2
p1 < p2
u1
u1 > u2
u2 p2
p1
p1 > p2
u1 < u2
x Flow remains stable x Turbulence reduces. This phenomenon is known as boundary layer separation. CIVE 1400: Fluid Mechanics. www.efm.leeds. ac.uk/CIVE /FluidsLevel1
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Unit 4
Examples of boundary layer separation
Unit 4
Tee-Junctions
A divergent duct or diffuser velocity drop (according to continuity) pressure increase (according to the Bernoulli equation).
Assuming equal sized pipes), Velocities at 2 and 3 are smaller than at 1. Pressure at 2 and 3 are higher than at 1. Causing the two separations shown Y-Junctions Y-Junctions Tee junctions are special cases of the Y-junction. Increasing the angle increases the probability of boundary layer separation. Venturi meter Diffuser angle of about 6 A balance between: *
length of meter
*
danger of boundary layer separation.
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Bends
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Flow past a cylinder Slow flow, Re < 0.5 no separation:
Moderate flow, Re < 70, separation vortices form.
Two separation zones occur in bends as shown above. Pb > Pa causing separation.
Fast flow Re > 70 vortices detach alternately.
Pd > Pc causing separation
Form a trail of down stream. Karman vortex vortex trail or street.
Localised effect
(Easily seen by looking over a bridge)
Downstream the boundary layer reattaches and normal flow occurs. Boundary layer separation is only local.
Causes whistling in power cables. Caused Tacoma narrows bridge to collapse.
Nevertheless downstream downstream of a junction / bend bend /valve etc. fluid will have lost energy. CIVE 1400: Fluid Mechanics. www.efm.leeds. ac.uk/CIVE /FluidsLevel1
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Unit 4
Aerofoil Normal flow over a aerofoil or a wing cross-section.
(boundary layers greatly exaggerated)
The velocity increases as air flows over the wing. The pressure distribution is as below so transverse lift force occurs.
Fluid accelerates to get round the cylinder Velocity maximum at Y. Pressure dropped. Adverse pressure between here and downstream. Separation occurs
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At too great an angle boundary layer separation occurs on the top Pressure changes dramatically. This phenomenon is known as stalling.
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Examples: Exam questions involving boundary layer theory are typically descriptive. 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.
All, or most, of the ‘suction’ pressure is lost. The plane will suddenly drop from the sky! Solution: Prevent separation. 1 Engine intakes draws slow air from the boundary layer at the rear of the wing though small holes 2 Move fast air from below to top via a slot.
3 Put a flap on the end of the wing and tilt it.
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Unit 4
Uses principle of dimensional homogeneity It gives qualitative results which only become quantitative from experimental analysis.
Lectures 18 & 19: Dimensional Analysis Unit 4: The Effect of the Boundary on Flow
Dimensions and units Application of fluid fluid mechanics in design makes use use of experiments results.
Any physical situation
Results often difficult to interpret.
can be described by familiar properties.
Dimensional analysis provides a strategy for choosing relevant data.
e.g. length, velocity, area, volume, acceleration etc.
Used to help analyse fluid flow These are all known as dimensions.
Especially when fluid flow is too complex for mathematical analysis.
Dimensions are of no use without a magnitude. Specific uses:
i.e. a standardised unit
x help design experiments
e.g metre, kilometre, Kilogram, a yard etc.
x Informs which measurements are important 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
i.e. its quality not its quantity.
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Unit 4
The following common abbreviations are used:
Unit 4
This table lists dimensions of some common physical quantities:
length
=L
Quantity
SI Unit
mass
=M
velocity
m/s
ms-1
LT-1
time
=T
acceleration
m/s 2
ms-2
LT-2
force
=F
kg ms-2
M LT-2
temperature
=
kg m2s-2
ML2T-2
force
N kg m/s2
4
energy (or work)
N m,
4).
power
Watt W Nms-1
N m/s 2
3
kg m /s
We can represent all the physical properties we are interested in with three:
pressure ( or stress)
Nm-2 2
density specific weight relative density
As either mass (M) (M) of force (F) can can be used to represent represent the other, i.e. F = MLT
viscosity surface tension
-2
-1 -2
ML-1T-2
kg m s
kg/m3
kg m-3
ML-3
kg m-2s-2
ML-2T-2
N/m
3
a ratio
1
no units
no dimension
N s/m2
N sm-2
kg/m s
kg m-1s-1
N/m
Nm-1
kg /s
M = FT 2L-1
ML2T-3
kg/m/s
kg/m2/s2
and one of M or F
kg m2s-3
Pascal P, N/m2,
L, T
Dimension
Joule J kg m2/s2
Here we will use L, M, T and F (not
219
2
kg s
-2
M L-1T-1
-2
MT
We will mostly use LTM:
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Unit 4
Dimensional Homogeneity
What exactly do we get from Dimensional Analysis?
Any equation is only true if both sides have the same dimensions.
A single equation,
It must be dimensionally homogenous.
Which relates all the physical factors of a problem to each other.
What are the dimensions of X? 2 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 1/2 -1
3/2
L (L T ) L 3
What might influence the force?
=X
-1
L T =X
It would be reasonable to assume that the force, F , depends on the following physical properties?
The powers of the individual dimensions must be equal on both sides.
diameter,
d
forward velocity of the propeller (velocity of the plane),
u
fluid density,
U
1. Checking units of equations;
revolutions per second,
N
2. Converting between two sets of units;
fluid viscosity,
P
(for L they are both 3, for T both -1). Dimensional homogeneity can be useful for:
3. Defining dimensionless relationships
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Unit 4
How do we get the dimensionless groups? From this list we can write this equation:
There are several methods. We will use the strategic method based on:
F = I ( d, u, U , N, P )
or
Buckingham’s S theorems. 0 = I ( F, d, u, U , N, P )
There are two st
I and I 1 are unknown functions.
1
S theorem:
A relationship between m variables (physical properties such as velocity, density etc.) can be expressed as a relationship between m-n non-dimensional groups of variables (called S groups), where n is the number of fundamental dimensions (such as mass, length and time) required to express the variables.
Dimensional Analysis produces:
§ F Nd P · I ¨ 2 2 , , © U u d u U ud ¹¸
S theorems:
0
These groups are dimensionless.
So if a problem is expressed:
I will be determined by experiment.
I ( Q1 , Q2 , Q3 ,………, Qm ) = 0
These dimensionless groups help Then this can also be expressed
to decide what experimental measurements to take.
I ( S 1 , S 2 , S 3 ,………, S m-n ) = 0
In fluids, we can normally take n = 3 (corresponding to M, L, T)
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Unit 4
nd
S theorem Each S group is a function of n governing or repeating 2
An example
variables plus one of the remaining variables.
Taking the example discussed above of force F induced on a propeller blade, we have the equation
Choice of repeating variables Repeating variables appear in most of the
S groups.
0 = I ( F, d, u, U , N, P )
They have a large influence on the problem.
n = 3 and m = 6
There is great freedom in choosing these. There are m - n = 3 Some rules which should be followed are
S groups, so
I ( S 1 , S 2 , S 3 ) = 0
x There are n ( = 3) repeating variables. x In combination they must contain
The choice of U , u, d satisfies the criteria above.
all of dimensions (M, L, T)
x The repeating variables must not form a
They are:
dimensionless group.
x measurable,
x They do not have to appear in all S groups. x The should be measurable in an experiment. x They should be of major interest to the designer.
x good design parameters x contain all the dimensions M,L and T.
It is usually possible to take U, u and d This freedom of choice means: many different S groups - all are valid. There is not really a wrong choice. CIVE 1400: Fluid Mechanics. www.efm.leeds. ac.uk/CIVE /FluidsLevel1
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For the first We can now form the three groups according to the 2nd theorem, S1
U a1 ub1 d c1 F
S2
U a2 u b2 d c2 N
S3
U a3 ub3 d c3 P
227
S group, S 1 U
a1
b1
c1
u d F
In terms of dimensions M 0 L0 T 0
a
b
M L 3 1 L T 1 1 L 1 M L T 2 c
The powers for each dimension (M, L or T), the powers must be equal on each side. for M:
The S groups are all dimensionless,
0 = a1 + 1 a1 = -1
0 0 0
i.e. they have dimensions M L T
for L: We use the principle of dimensional homogeneity to equate the dimensions for each S group.
0 = -3a 1 + b1 + c1 + 1 0 = 4 + b 1 + c1
for T:
0 = -b 1 - 2 b1 = -2 c1 = -4 - b 1 = -2
Giving S1 as
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S1
U 1u 2 d 2 F
S 1
F U u 2 d 2
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Unit 4
And a similar procedure is followed for the other S groups. Group S 2
Unit 4
And for the third, S 3 M 0 L0T 0
U u d N a2
b2
c2
a
b
M L 3 1 L T 1 1 L 1 T1
M 0 L0T 0
U u d P a3
b3
c3
a
b
M L3 3 L T 1 3 L 3 ML1T 1 c
c
for M:
0 = a3 + 1 a3 = -1
for M:
0 = a2 for L:
for L:
0 = -3a 3 + b3 + c3 -1
0 = -3a 2 + b2 + c2
b3 + c3 = -2
0 = b2 + c2 for T: for T:
0 = -b 3 - 1
0 = -b 2 - 1
b3 = -1
b2 = -1
c3 = -1
c2 = 1 Giving S3 as
Giving S2 as S2
U 0u 1d 1 N
Nd S 2 u
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S3
U 1u 1d 1P
S 3
P U ud
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Lectures 16-19
Unit 4
Unit 4
Manipulation of the
Thus the problem may be described by
Once identified the
I ( S 1 , S 2 , S 3 ) = 0
231
S groups
S groups can be changed.
The number of groups does not change.
§ F Nd P · I ¨ 2 2 , , ¸ © U u d u U ud ¹
Their appearance may change drastically.
0 Taking the defining equation as:
This may also be written:
I ( S 1 , S 2 , S 3 ……… S m-n ) = 0
F U u 2 d 2
The following changes are permitted:
§ Nd P · I ¨ , ¸ © u U ud ¹
i. Combination of exiting groups by multiplication or division to form a new group to replaces one of the existing.
Wrong choice of physical properties.
E.g. S1 and S2 may be combined to form equation becomes
If, extra, unimportant variables are chosen :
I ( S 1a , S 2 , S 3 ……… S m-n ) = 0
*
Extra S groups will be formed
*
Will have little effect on physical performance
*
Should be identified during experiments
ii. Reciprocal of any group is valid.
I ( S 1 ,1/ S 2 , S 3 ……… 1/ Sm -n ) = 0
iii. A group may be raised to any power. I ( ( S 1 )2 , ( S2 )1/2 , ( S3 )3……… S m-n ) = 0
If an important variable is missed:
iv. Multiplied by a constant.
x A S group would be missing. x Experimental analysis may miss significant
v. Expressed as a function of the other groups
S2 = I ( S 1 , S 3 ……… S m-n )
behavioural changes.
Initial choice of variables should be done with great care.
CIVE 1400: Fluid Mechanics. www.efm.leeds. ac.uk/CIVE /FluidsLevel1
S1a = S1 / S2 so the defining
In general the defining equation could look like I ( S 1 , 1/ S2 ,( S 3 )i……… 0.5S m-n ) = 0 Lectures 16-19
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CIVE 1400: Fluid Mechanics. www.efm.leeds .ac.uk/CI VE/FluidsLevel 1
Lectures 16-19
233
Unit 4
Unit 4
Common S groups
An Example Q. If we have a function describing a problem:
Several groups will appear again and again.
I Q, d , U , P , p 0
Show that Q
d 2 p 1/ 2 § dU 1/ 2 p 1/ 2 · I ¨ ¸ U 1/ 2 © P ¹
These often have names. They can be related to physical forces.
Ans. Other common non-dimensional numbers
Dimensional analysis using Q, U , d will result in:
or (
§ d P d 4 p · I ¨ , ¸ 0 © Q U U Q2 ¹
Reynolds number: U ud inertial, viscous force ratio Re P
The reciprocal of square root of S2: 1 S 2
U 1/ 2Q
d 2 p1/ 2
Euler number:
S 2 a ,
En
Multiply S1 by this new group: S 1a
S 1S 2 a
d P U 1/ 2Q Q U d 2 p 1/ 2
1/ 2
d U p
Fn
1/ 2
then we can say
§ dU p I 1 / S 1a , S 2 a I ¨ © P 1/ 2
1/ 2
d p · ¸ QU 1/ 2 ¹ 2
,
1/ 2
0
d 2 p1/ 2 § dU 1/ 2 p1/ 2 · I ¨ 1/ 2 © P ¹¸ U
u2 gd
Mn
CIVE 1400: Fluid Mechanics. www.efm.leeds. ac.uk/CIVE /FluidsLevel1
pressure, inertial force ratio
Lectures 16-19
inertial, gravitational force ratio
Weber number: U ud We V Mach number:
or Q
p 2 U u
Froude number:
P
S groups):
234
inertial, surface tension force ratio
u c
Local velocity, local velocity of sound ratio
CIVE 1400: Fluid Mechanics. www.efm.leeds .ac.uk/CI VE/FluidsLevel 1
Lectures 16-19
Unit 4
The similarity of time as well as geometry.
Similarity is concerned with how to transfer measurements from models to the full scale.
It exists if: i. the paths of particles are geometrically similar ii. the ratios of the velocities of are similar
Three types of similarity which exist between a model and prototype:
Some useful ratios are: V m L m / T m O L Velocity V p L p / T p O T
Geometric similarity: The ratio of all corresponding dimensions in the model and prototype are equal.
Acceleration
For lengths Lmodel L m O L L prototype L p
Discharge
O L is the scale factor for length.
For areas A prototype
L2m L p2
Unit 4
Kinematic similarity
Similarity
Amodel
235
am a p Qm Q p
Lm / T m2 2 p
L p / T
L3m / T m L p3 / T p
O L3 O T
O u
O L O 2T
O a
O Q
A consequence is that streamline patterns are the same.
O L2
All corresponding angles are the same.
CIVE 1400: Fluid Mechanics. www.efm.leeds. ac.uk/CIVE /FluidsLevel1
Lectures 16-19
236
CIVE 1400: Fluid Mechanics. www.efm.leeds .ac.uk/CI VE/FluidsLevel 1
Lectures 16-19
237
Unit 4
Unit 4
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.
An example:
Force ratio 2
§ O · u 2 O U O L ¨ L ¸ OU O L2 O 2u 3 F p M p a p U p L p O T © O T ¹
F m
M m am
3
U m Lm
O L
2
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
ML-3
LT -1
u:
l:(length) L
P :
ML-1T -1
So I (R, U , u, l, P ) = 0
The controlling S group is usually Re. So Re is the same for model and prototype:
Taking U , u, l as repeating variables gives:
§ U ul · I ¨ ¸ © P ¹ U u l § U ul · R Uu 2 l 2I ¨ ¸ © P ¹ R
U m um d m P m
U p u p d p
2 2
P p
It is possible another group is dominant. In open channel i.e. river Froude number is often taken as dominant.
CIVE 1400: Fluid Mechanics. www.efm.leeds. ac.uk/CIVE /FluidsLevel1
Lectures 16-19
This applies whatever the size of the body i.e. it is applicable to prototype and a geometrically similar model.
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CIVE 1400: Fluid Mechanics. www.efm.leeds .ac.uk/CI VE/FluidsLevel 1
Lectures 16-19
Unit 4
239
Unit 4
Example 1
For the model
An underwater missile, diameter 2m and length 10m is tested in a water tunnel to determine the forces acting on the real prototype. A th 1/20 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?
§ U u l · I ¨ m m m ¸ 2 2 U m um l m © P m ¹ Rm
and for the prototype Dynamic similarity so Reynolds numbers equal: U m um d m U p u p d p
§ U u l · I ¨ p p p ¸ 2 2 U p u p l p © P p ¹ R p
P m
Dividing these two equations gives 2 2 m m
Rm / U m u l
2 2
R p / U p u p l p
I U p u p l p / 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: U m um d m P m
P p
The model velocity should be U p d p P m um u p U m d m P p
I Um um l m / P m
U p u p d p
Both the model and prototype are in water then, P m = P p and U m = U p so
um
P p
up
d p d m
10
1 1 / 20
200 m / s
so Rm R p
U m um2 l m2
This is a very high velocity.
2 2 p p
U p u l
This is one reason why model tests are not always done at exactly equal Reynolds numbers.
i.e. a scaling law for resistance force: O R
A wind tunnel could have been used so the values of the U and P ratios would be used in the above.
O U O2u O2L
CIVE 1400: Fluid Mechanics. www.efm.leeds. ac.uk/CIVE /FluidsLevel1
Lectures 16-19
240
CIVE 1400: Fluid Mechanics. www.efm.leeds .ac.uk/CI VE/FluidsLevel 1
Lectures 16-19
241
