An Introduction to Fluid Mechanics

Transcription

1 0. Contents of the Course Notes For the First Year Lecture Course: An Introduction to Fluid Mechanics School of Civil Engineering, University of Leeds. CIVE1400 FLUID MECHANICS Dr Andrew Sleigh January Objectives: The course will introduce fluid mechanics and establish its relevance in civil engineering. Develop the fundamental principles underlying the subject. Demonstrate how these are used for the design of simple hydraulic components. 0. Consists of: Lectures: 0 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 hours, worth 80% of the module credits. This consists of 6 questions of which you choose 4. Multiple choice question (MCQ) papers, worth 10% of the module credits. These will be for 30mins and set during the lectures. The timetable for these MCQs and lectures is shown in the table at the end of this section. 1 Marked problem sheet, worth 10% of the module credits. Laboratories: x 3 hours These two laboratory sessions examine how well the theoretical analysis of fluid dynamics describes what we observe in practice. During the laboratory you will take measurements and draw various graphs according to the details on the laboratory sheets. These graphs can be compared with those obtained from theoretical analysis. You will be expected to draw conclusions as to the validity of the theory based on the results you have obtained and the experimental procedure. After you have completed the two laboratories you should have obtained a greater understanding as to how the theory relates to practice, what parameters are important in analysis of fluid and where theoretical predictions and experimental measurements may differ. The two laboratories sessions are: 1. Impact of jets on various shaped surfaces - a jet of water is fired at a target and is deflected in various directions. This is an example of the application of the momentum equation.. The rectangular weir - the weir is used as a flow measuring device. Its accuracy is investigated. This is an example of how the Bernoulli (energy) equation is applied to analyses fluid flow. CIVE 1400: Fluid Mechanics Introduction 1 [As you know, these laboratory sessions are compulsory course-work. You must attend them. Should you fail to attend either one you will be asked to complete some extra work. This will involve a detailed report and further questions. The simplest strategy is to do the lab.] Homework: Example sheets: These will be given for each section of the course. Doing these will greatly improve your exam mark. They are course work but do not have credits toward the module. Lecture notes: Theses should be studied but explain only the basic outline of the necessary concepts and ideas. Books: It is very important do some extra reading in this subject. To do the examples you will definitely need a textbook. Any one of those identified below is adequate and will also be useful for the fluids (and other) modules in higher years - and in work. Example classes: There will be example classes each week. You may bring any problems/questions you have about the course and example sheets to these classes. Fluids Lecture and Test Schedule Week Date Subject Lecture Test Month day 0 January 17 Tue Fluid properties 1 18 Wed 1 4 Tue Statics 3 5 Wed 4 31 Tue 5 MCQ February 1 Wed Tue 7 8 Wed Dynamics Tue 9 15 Wed Tue 11 Wed 1 Problem Sheet 6 8 Tue Surveying March 1 Wed Surveying 7 7 Tue Surveying 8 Wed Surveying 8 14 Tue Real fluids Wed Tue 15 Wed 16 Easter Vacation 10 April 19 Tue Dimensional analysis 17 0 Wed Tue 19 MQC 7 Wed Revision Lectures 0 1 May 3 Tue 1 4 Wed CIVE 1400: Fluid Mechanics Introduction

2 0.3 Specific Elements: Introduction Fluid Properties Fluids vs. Solids Viscosity Newtonian Fluids Properties of Fluids Statics Hydrostatic pressure Manometry / pressure measurement Hydrostatic forces on submerged surfaces Dynamics The continuity equation. The Bernoulli Equation. Applications of the Bernoulli equation. The momentum equation. Application of the momentum equation. Real Fluids Boundary layer. Laminar flow in pipes. Introduction to dimensional analysis Dimensions Similarity These notes give more information than is found in the lectures. They do not replace textbooks. You must also read at least one of the recommended fluid mechanics books. The notes may be read online or printed off for personal use. 0.4 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. 0.5 Other Teaching Resources. There are some extra teaching/learning resources available for you to use that are computer based. Online Lecture Notes A more detailed set of lecture notes can be found on the WWW at he following address: You get to this using Netscape from any of the computers in the university. If you forget this address you can also get to the web pages via Dr Sleigh's web pages linked from the department's main page. CIVE 1400: Fluid Mechanics Introduction 3 CIVE 1400: Fluid Mechanics Introduction 4

3 0.6 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 are essential for the development of industrial society. It is these services which civil engineers provide. Fluid mechanics is involved in nearly all areas of Civil Engineering either directly or indirectly. Some examples of direct involvement are those where we are concerned with manipulating the fluid: Sea and river (flood) defences; Water distribution / sewerage (sanitation) networks; Hydraulic design of water/sewage treatment works; Dams; Irrigation; Pumps and Turbines; Water retaining structures. And some examples where the primary object is construction - yet analysis of the fluid mechanics is essential: Flow of air in / around buildings; Bridge piers in rivers; Ground-water flow. Notice how nearly all of these involve water. The following course, although introducing general fluid flow ideas and principles, will demonstrate many of these principles through examples where the fluid is water. 0.7 System of units As any quantity can be expressed in whatever way you like it is sometimes easy to become confused as to what exactly or how much is being referred to. This is particularly true in the field of fluid mechanics. Over the years many different ways have been used to express the various quantities involved. Even today different countries use different terminology as well as different units for the same thing - they even use the same name for different things e.g. an American pint is 4/5 of a British pint! To avoid any confusion on this course we will always use the SI (metric) system - which you will already be familiar with. It is essential that all quantities are expressed in the same system or the wrong solutions will results. Despite this warning you will still find that this is the most common mistake when you attempt example questions. 0.8 The SI System of units The SI system consists of six primary units, from which all quantities may be described. For convenience secondary units are used in general practise which are made from combinations of these primary units. Primary Units The six primary units of the SI system are shown in the table below: Quantity SI Unit Dimension length metre, m L mass kilogram, kg M time second, s T temperature Kelvin, K current ampere, A I luminosity candela Cd In fluid mechanics we are generally only interested in the top four units from this table. Notice how the term 'Dimension' of a unit has been introduced in this table. This is not a property of the individual units, rather it tells what the unit represents. For example a metre is a length which has a dimension L but also, an inch, a mile or a kilometre are all lengths so have dimension of L. (The above notation uses the MLT system of dimensions, there are other ways of writing dimensions - we will see more about this in the section of the course on dimensional analysis.) CIVE 1400: Fluid Mechanics Introduction 5 CIVE 1400: Fluid Mechanics Introduction 6

4 Derived Units There are many derived units all obtained from combination of the above primary units. Those most used are shown in the table below: Quantity SI Unit Dimension velocity m/s ms -1 LT -1 acceleration m/s ms - LT - force N kg m/s kg ms - M LT - energy (or work) Joule J N m, kg m /s kg m s - ML T - power Watt W N m/s kg m /s 3 Nms -1 kg m s -3 ML T -3 pressure ( or stress) Pascal P, N/m, kg/m/s Nm - kg m -1 s - ML -1 T - density kg/m 3 kg m -3 ML -3 specific weight N/m 3 kg/m /s kg m - s - ML - T - relative density a ratio no units 1 no dimension viscosity N s/m kg/m s N sm - kg m -1 s -1 M L -1 T -1 surface tension N/m kg /s Nm -1 kg s - MT Example: Units 1. A water company wants to check that it will have sufficient water if there is a prolonged drought in the area. The region it covers is 500 square miles and various different offices have sent in the following consumption figures. There is sufficient information to calculate the amount of water available, but unfortunately it is in several different units. Of the total area 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 00mm 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 m 3 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 (8 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? The above units should be used at all times. Values in other units should NOT be used without first converting them into the appropriate SI unit. If you do not know what a particular unit means find out, else your guess will probably be wrong. One very useful tip is to write down the units of any equation you are using. If at the end the units do not match you know you have made a mistake. For example is you have at the end of a calculation, 30 kg/m s = 30 m you have certainly made a mistake - checking the units can often help find the mistake. More on this subject will be seen later in the section on dimensional analysis and similarity. CIVE 1400: Fluid Mechanics Introduction 7 CIVE 1400: Fluid Mechanics Introduction 8

5 CIVE1400: Fluid Mechanics Section 1: Fluid Properties LECTURE CONTENTS Section 0: Introduction Section 1: Fluid Properties Fluids vs. Solids Viscosity Newtonian Fluids Properties of Fluids Section : Statics Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces Section 3: Dynamics The continuity equation. The Bernoulli Equation. Application of Bernoulli equation. The momentum equation. Application of momentum equation. Section 4: Real Fluids Boundary layer. Laminar flow in pipes. Section 5: Dimensional Analysis An Intro to Dimensional analysis Similarity CIVE1400: Fluid Mechanics Section 1: Fluid Properties What make fluid mechanics different to solid mechanics? The nature of a fluid is different to that of a solid In fluids we deal with continuous streams of fluid. In solids we only consider individual elements. In this section we will consider how we can classify the differences in nature of fluids and solids. What do we mean by nature of a fluid? Fluids are clearly different to solids. But we must be specific. We need some definable basic physical difference. CIVE1400: Fluid Mechanics Section 1: Fluid Properties 1 CIVE1400: Fluid Mechanics Section 1: Fluid Properties

6 CIVE1400: Fluid Mechanics Section 1: Fluid Properties We know that fluids flow under the action of a force, and the solids don t - but solids do deform. So we can say that fluids lack the ability of solids to resist deformation. fluids change shape as long as a force acts. CIVE1400: Fluid Mechanics Section 1: Fluid Properties What use can we make of these ideas? In the analysis of fluids we often take small volumes (elements) and examine the forces on these. Take the rectangular element below. What forces cause it to deform? (These definitions include both gasses and liquids as fluids.) A B C D CIVE1400: Fluid Mechanics Section 1: Fluid Properties 3 CIVE1400: Fluid Mechanics Section 1: Fluid Properties 4

7 CIVE1400: Fluid Mechanics Section 1: Fluid Properties CIVE1400: Fluid Mechanics Section 1: Fluid Properties A B F Fluids in motion F C Forces acting along edges (faces), such as F, are know as shearing forces. From this we arrive at the definition: D 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. 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 Plotting the velocity across the section gives velocity profile Change in velocity with distance is velocity gradient = du dy v CIVE1400: Fluid Mechanics Section 1: Fluid Properties 5 CIVE1400: Fluid Mechanics Section 1: Fluid Properties 6

8 CIVE1400: Fluid Mechanics Section 1: Fluid Properties 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? CIVE1400: Fluid Mechanics Section 1: Fluid Properties 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. No velocity gradient, no shear forces. v Consider this 3-d rectangular element, under the action of the force F. CIVE1400: Fluid Mechanics Section 1: Fluid Properties 7 CIVE1400: Fluid Mechanics Section 1: Fluid Properties 8

9 CIVE1400: Fluid Mechanics Section 1: Fluid Properties CIVE1400: Fluid Mechanics Section 1: Fluid Properties δz a δx b A -d view may be clearer A B B F A B F φ E x E F δy C under the action of the force F D F y C The shearing force acts on the area A z x Shear stress, is the force per unit area: D a a b b F A A A B B E F The deformation which shear stress causes is measured by the angle, and is know as shear strain. F C D 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. CIVE1400: Fluid Mechanics Section 1: Fluid Properties 9 CIVE1400: Fluid Mechanics Section 1: Fluid Properties10

10 CIVE1400: Fluid Mechanics Section 1: Fluid Properties It has been shown experimentally that the rate of shear strain is directly proportional to shear stress time 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 So (note that x t shear strain x y rate of shear strain u is the velocity of the particle at E) CIVE1400: Fluid Mechanics Section 1: Fluid Properties Constant u y u/y is the rate of change of velocity with distance, du in differential form this is = velocity gradient. dy The constant of proportionality is known as the dynamic viscosity, giving 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 Non-Newtonian Fluids CIVE1400: Fluid Mechanics Section 1: Fluid Properties11 CIVE1400: Fluid Mechanics Section 1: Fluid Properties1

11 CIVE1400: Fluid Mechanics Section 1: Fluid Properties 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: A B u y where A, B and n are constants. n CIVE1400: Fluid Mechanics Section 1: Fluid Properties This graph shows how changes for different fluids. Shear stress, τ Bingham plastic Pseudo plastic plastic Newtonian Dilatant Ideal, (τ=0) For Newtonian fluids A = 0, B = and n = 1 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. colloidial substances like clay, milk and cement. Dilatant substances; Viscosity increases with rate of shear e.g. quicksand. Thixotropic substances: Viscosity decreases with length of time shear force is applied e.g. thixotropic jelly paints. Rheopectic substances: Viscosity increases with length of time shear force is applied Viscoelastic materials: Similar to Newtonian but if there is a sudden large change in shear they behave like plastic. CIVE1400: Fluid Mechanics Section 1: Fluid Properties13 CIVE1400: Fluid Mechanics Section 1: Fluid Properties14

12 CIVE1400: Fluid Mechanics Section 1: Fluid Properties Liquids vs. Gasses Liquids and gasses behave in much the same way Some specific differences are 1. A liquid is difficult to compress and often regarded as being incompressible. A gas is easily to compress and usually treated as such - it changes volume with pressure.. A given mass of liquid occupies a given volume and will form a free surface A gas has no fixed volume, it changes volume to expand to fill the containing vessel. No free surface is formed. Causes of Viscosity in Fluids Viscosity in Gasses Mainly due to molecular exchange between layers Mathematical considerations of this momentum exchange can lead to Newton law of viscosity. If temperature increases the momentum exchange between layers will increase thus increasing viscosity. Viscosity in Liquids There is some molecular interchange between layers in liquids - but the cohesive forces are also important. Increasing temperature of a fluid reduces the cohesive forces and increases the molecular interchange. Resulting in a complex relationship between temperature and viscosity. CIVE1400: Fluid Mechanics Section 1: Fluid Properties15 CIVE1400: Fluid Mechanics Section 1: Fluid Properties Density Properties of Fluids: There are three ways of expressing density: 1. Mass density: Units: kg/m 3 Dimensions: ML 3 Typical values: mass per unit volume mass of fluid volume of fluid Water = 1000kg m 3, Mercury = 13546kg m 3 Air = 1.3kg m 3, Paraffin Oil = 800kg m 3. CIVE1400: Fluid Mechanics Section 1: Fluid Properties16

13 CIVE1400: Fluid Mechanics Section 1: Fluid Properties. Specific Weight: (sometimes known as specific gravity) weight per unit volume g / m T. Units: Newton s per cubic metre, N Dimensions: ML Typical values: 3 (or kg/m /s ) Water =9814 Nm 3, Mercury = Nm 3, Air =1.07 Nm 3, Paraffin Oil =7851 Nm 3 3. Relative Density: ratio of mass density to a standard mass density subs tan ce HO( at4 c) For solids and liquids this standard mass density is the maximum mass density for water (which occurs at 4 c) at atmospheric pressure. Units: none, as it is a ratio Dimensions: 1. Typical values: Water = 1, Mercury = 13.5, Paraffin Oil =0.8. CIVE1400: Fluid Mechanics Section 1: Fluid Properties Viscosity There are two ways of expressing viscosity 1. Coefficient of Dynamic Viscosity = the shear stress,, required to drag one layer of fluid with unit velocity past another layer a unit distance away. du dy Force Velocity Area Distance = Force Time Area Mass Length Time Units: N s/m or kg/m s (kg m -1 s -1 ) (Note that is often expressed in Poise, P, where 10 P = 1 N s/m.) Dimensions: ML -1 T -1 Typical values: Water = Ns/m, Air = Ns/m, Mercury =1.55 Ns/m, Paraffin Oil =1.9 Ns/m. CIVE1400: Fluid Mechanics Section 1: Fluid Properties17 CIVE1400: Fluid Mechanics Section 1: Fluid Properties18

14 CIVE1400: Fluid Mechanics Section 1: Fluid Properties. Kinematic Viscosity = the ratio of dynamic viscosity to mass density. Units: m s -1 Dimension: L T -1 Typical values: Water = m /s,, Air = m 1 /s ms, Mercury = m /s, Paraffin Oil = m /s. CIVE1400: Fluid Mechanics Section : Statics LECTURE CONTENTS Section 0: Introduction Section 1: Fluid Properties Fluids vs. Solids Viscosity Newtonian Fluids Properties of Fluids Section : Statics Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces Section 3: Dynamics The continuity equation. The Bernoulli Equation. Application of Bernoulli equation. The momentum equation. Application of momentum equation. Section 4: Real Fluids Boundary layer. Laminar flow in pipes. Section 5: Dimensional Analysis An Intro to Dimensional analysis Similarity CIVE1400: Fluid Mechanics Section 1: Fluid Properties19 CIVE1400: Fluid Mechanics Section : Statics 0

15 CIVE1400: Fluid Mechanics Section : Statics STATICS 4 Lectures CIVE1400: Fluid Mechanics Section : Statics What do we know about the forces involved with static fluids? Objectives Introduce idea of pressure. Prove unique value at any particular elevation. Show hydrostatic relationship. Express pressure in terms of head of fluid. From earlier we know that: 1. A static fluid can have no shearing force acting on it.. Any force between the fluid and the boundary must be acting at right angles to the boundary. F 1 Pressure measurement using manometers. F F R 1 Determine the magnitude and direction of forces on submerged surfaces R F n R Pressure force normal to the boundary R n CIVE1400: Fluid Mechanics Section : Statics 1 CIVE1400: Fluid Mechanics Section : Statics

16 CIVE1400: Fluid Mechanics Section : Statics This is also true for: CIVE1400: Fluid Mechanics Section : Statics Pressure curved surfaces any imaginary plane An element of fluid at rest is in equilibrium: 3. The sum of forces in any direction is zero. 4. The sum of the moments of forces about any point is zero. Convenient to work in terms of pressure, p, which is the force per unit area. pressure Force Area over which the force is applied p F A Units: Newton s per square metre, N/m, kg/m s (kg m -1 s - ). (Also known as a Pascal, Pa, i.e. 1 Pa = 1 N/m ) (Also frequently used is the alternative SI unit the bar, where 1bar = 10 5 N/m ) Uniform Pressure: If the force on each unit area of a surface is equal then uniform pressure CIVE1400: Fluid Mechanics Section : Statics 3 CIVE1400: Fluid Mechanics Section : Statics 4

17 CIVE1400: Fluid Mechanics Section : Statics Pascal s Law CIVE1400: Fluid Mechanics Section : Statics Summing forces in the x-direction: Proof that pressure acts equally in all directions. Force in the x-direction due to p x, F p Area p x y x x ABFE x x A δz B δs p s Force in the x-direction due to p s, Fx ps Area s ABCD sin p x δy F C p s z y s s E δx θ D (sin y s ) p yz s p y Remember: Force in x-direction due to p y, F x y 0 No shearing forces All forces at right angles to the surfaces To be at rest (in equilibrium) F F F 0 x x x s x y p x y p y z x s p x 0 p s CIVE1400: Fluid Mechanics Section : Statics 5 CIVE1400: Fluid Mechanics Section : Statics 6

18 CIVE1400: Fluid Mechanics Section : Statics Summing forces in the y-direction. CIVE1400: Fluid Mechanics Section : Statics To be at rest (in equilibrium) Force due to p y, Fy p Area p x z y y ABCD y Component of force due to p s, (cos x s ) F p Area cos y s s ABCD p s z x s s p x z Component of force due to p x, F yx 0 Force due to gravity, weight = - specific weight 1 = g xyz s volume of element F F F weight 0 y y y s y x py xy ps xz 1 g xyz 0 The element is small i.e. x, x, and z, are small, so x y z, is very small and considered negligible, hence py ps thus We showed above p x p s p p p x y s Pressure at any point is the same in all directions. This is Pascal s Law and applies to fluids at rest. CIVE1400: Fluid Mechanics Section : Statics 7 CIVE1400: Fluid Mechanics Section : Statics 8

19 CIVE1400: Fluid Mechanics Section : Statics Vertical Variation Of Pressure In A Fluid Under Gravity CIVE1400: Fluid Mechanics Section : Statics The forces involved are: Force due to p 1 on A (upward) = p 1 A Force due to p on A (downward) = p A Area A p, A Fluid density ρ z Force due to weight of element (downward) = mg = mass density volume g = g A(z - z 1 ) Taking upward as positive, we have p 1, A z 1 pa 1 pa gaz z1 = 0 p p ga z z 1 1 Vertical cylindrical element of fluid cross sectional area = A mass density = Thus in a fluid under gravity, pressure decreases linearly with increase in height p p1 gaz z1 This is the hydrostatic pressure change. CIVE1400: Fluid Mechanics Section : Statics 9 CIVE1400: Fluid Mechanics Section : Statics 30

20 CIVE1400: Fluid Mechanics Section : Statics Equality Of Pressure At The Same Level In A Static Fluid CIVE1400: Fluid Mechanics Section : Statics P Q Area A Fluid density ρ p l, A p r, A z z Face L weight, mg Face R Horizontal cylindrical element cross sectional area = A mass density = left end pressure = p l right end pressure = p r For equilibrium the sum of the forces in the x direction is zero. p l A = p r A p l = p r L We have shown p l = p r For a vertical pressure change we have pl pp gz and so p p gz r q p gz p gz p p p p q q R Pressure in the horizontal direction is constant. Pressure at the two equal levels are the same. This true for any continuous fluid. CIVE1400: Fluid Mechanics Section : Statics 31 CIVE1400: Fluid Mechanics Section : Statics 3

21 CIVE1400: Fluid Mechanics Section : Statics CIVE1400: Fluid Mechanics Section : Statics General Equation For Variation Of Pressure In A Static Fluid Area A (p + δp)a Horizontal If 90 then s is in the x or y directions, (i.e. horizontal),so Fluid density ρ δs dp dp dp 0 ds dx dy 90 z pa θ mg z + δz Confirming that pressure change on any horizontal plane is zero. Vertical If 0 then s is in the z direction (vertical) so A cylindrical element of fluid at an arbitrary orientation. From considering incremental forces (see detailed notes on WWW or other texts) we get dp ds g cos dp dp g ds 0 dz Confirming the hydrostatic result p p1 g z z 1 p p g z z 1 1 CIVE1400: Fluid Mechanics Section : Statics 33 CIVE1400: Fluid Mechanics Section : Statics 34

22 CIVE1400: Fluid Mechanics Section : Statics Pressure And Head We have the vertical pressure relationship dp g, dz integrating gives p = -gz + constant measuring z from the free surface so that z = -h CIVE1400: Fluid Mechanics Section : Statics 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 = g h z h y x p gh constant surface pressure is atmospheric, p atmospheric. p atmospheric constant so p gh p atmospheric 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 CIVE1400: Fluid Mechanics Section : Statics 35 CIVE1400: Fluid Mechanics Section : Statics 36

23 CIVE1400: Fluid Mechanics Section : Statics A gauge pressure can be given using height of any fluid. p gh This vertical height is the head. If pressure is quoted in head, the density of the fluid must also be given. Example: What is a pressure of 500 knm - in head of water of density, = 1000 kgm -3 Use p = gh, 3 h p g m of water In head of Mercury density = kgm h m of Mercury In head of a fluid with relative density = 8.7. remember = water ) 3 h m of fluid = 8.7 CIVE1400: Fluid Mechanics Section : Statics Pressure Measurement By Manometer Manometers use the relationship between pressure and head to measure pressure The Piezometer Tube Manometer The simplest manometer is an open tube. This is attached to the top of a container with liquid at pressure. containing liquid at a pressure. h1 A B The tube is open to the atmosphere, The pressure measured is relative to atmospheric so it measures gauge pressure. h CIVE1400: Fluid Mechanics Section : Statics 37 CIVE1400: Fluid Mechanics Section : Statics 38

24 CIVE1400: Fluid Mechanics Section : Statics Pressure at A = pressure due to column of liquid h 1 p A = g h 1 CIVE1400: Fluid Mechanics Section : Statics An Example of a Piezometer. What is the maximum gauge pressure of water that can be measured by a Piezometer of height 1.5m? And if the liquid had a relative density of 8.5 what would the maximum measurable gauge pressure? Pressure at B = pressure due to column of liquid h p B = g h Problems with the Piezometer: 1. Can only be used for liquids. Pressure must above atmospheric 3. Liquid height must be convenient i.e. not be too small or too large. CIVE1400: Fluid Mechanics Section : Statics 39 CIVE1400: Fluid Mechanics Section : Statics 40

25 CIVE1400: Fluid Mechanics Section : Statics The U -Tube Manometer CIVE1400: Fluid Mechanics Section : Statics We know: U -Tube enables the pressure of both liquids and gases to be measured U is connected as shown and filled with manometric fluid. Important points: 1. The manometric fluid density should be greater than of the fluid measured. man >. The two fluids should not be able to mix they must be immiscible. Fluid density ρ D Pressure in a continuous static fluid is the same at any horizontal level. pressure at B = pressure at C p B = p C For the left hand arm pressure at B = pressure at A + pressure of height of liquid being measured p B = p A + gh 1 For the right hand arm pressure at C = pressure at D + pressure of height of manometric liquid p C = p atmospheric + man gh A B h 1 h C We are measuring gauge pressure we can subtract p atmospheric giving p B = p C Manometric fluid density ρ man p A = man gh - gh 1 CIVE1400: Fluid Mechanics Section : Statics 41 CIVE1400: Fluid Mechanics Section : Statics 4

26 CIVE1400: Fluid Mechanics Section : Statics What if the fluid is a gas? Nothing changes. The manometer work exactly the same. CIVE1400: Fluid Mechanics Section : Statics An example of the U-Tube manometer. 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 What is the gauge pressure if: a) h 1 = 0.4m and h = 0.9m? b) h 1 stayed the same but h = -0.1m? BUT: As the manometric fluid is liquid (usually mercury, oil or water) And Liquid density is much greater than gas, man >> gh 1 can be neglected, and the gauge pressure given by p A = man gh CIVE1400: Fluid Mechanics Section : Statics 43 CIVE1400: Fluid Mechanics Section : Statics 44

27 CIVE1400: Fluid Mechanics Section : Statics Pressure difference measurement Using a U -Tube Manometer. The U -tube manometer can be connected at both ends to measure pressure difference between these two points B CIVE1400: Fluid Mechanics Section : Statics pressure at C = pressure at D p C = p D p C = p A + g h a p D = p B + g (h b + h) + man g h p A + g h a = p B + g (h b + h) + man g h Fluid density ρ Giving the pressure difference E h b p A - p B = g (h b - h a ) + ( man - )g h h A h a C D Again if the fluid is a gas man >>, then the terms involving can be neglected, p A - p B = man g h Manometric fluid density ρ man CIVE1400: Fluid Mechanics Section : Statics 45 CIVE1400: Fluid Mechanics Section : Statics 46

28 CIVE1400: Fluid Mechanics Section : Statics An example using the u-tube for pressure difference measuring In the figure below two pipes containing the same fluid of density = 990 kg/m 3 are connected using a u-tube manometer. What is the pressure between the two pipes if the manometer contains fluid of relative density 13.6? Fluid density ρ A Fluid density ρ CIVE1400: Fluid Mechanics Section : Statics Advances to the U tube manometer Problem: Two reading are required. Solution: Increase cross-sectional area of one side. Result: One level moves much more than the other. B p 1 p h a = 1.5m E h = 0.5m h b = 0.75m diameter D diameter d z Datum line C D z 1 Manometric fluid density ρ man = 13.6 ρ If the manometer is measuring the pressure difference of a gas of (p 1 - p ) as shown, we know p 1 - p = man g h CIVE1400: Fluid Mechanics Section : Statics 47 CIVE1400: Fluid Mechanics Section : Statics 48

29 CIVE1400: Fluid Mechanics Section : Statics volume of liquid moved from the left side to the right = z ( d / 4) The fall in level of the left side is Volume moved z1 Area of left side z d / 4 D / 4 d z D Putting this in the equation, d p1 p g z z D d gz 1 D If D >> d then (d/d) is very small so p p gz 1 CIVE1400: Fluid Mechanics Section : Statics Problem: Small pressure difference, movement cannot be read. Solution 1: Reduce density of manometric fluid. Result: Greater height change - easier to read. Solution : Tilt one arm of the manometer. Result: Same height change - but larger movement along the manometer arm - easier to read. CIVE1400: Fluid Mechanics Section : Statics 49 Inclined manometer CIVE1400: Fluid Mechanics Section : Statics 50

30 CIVE1400: Fluid Mechanics Section : Statics p p 1 diameter d diameter D z Datum line z 1 θ Scale Reader x CIVE1400: Fluid Mechanics Section : Statics Example of an inclined manometer. An inclined manometer is required to measure an air pressure of 3mm of water to an accuracy of +/- 3%. The inclined arm is 8mm in diameter and the larger arm has a diameter of 4mm. The manometric fluid has density man = 740 kg/m 3 and the scale may be read to +/- 0.5mm. What is the angle required to ensure the desired accuracy may be achieved? The pressure difference is still given by the height change of the manometric fluid. but, p p gz 1 z x sin p p gxsin 1 The sensitivity to pressure change can be increased further by a greater inclination. CIVE1400: Fluid Mechanics Section : Statics 51 CIVE1400: Fluid Mechanics Section : Statics 5

31 CIVE1400: Fluid Mechanics Section : Statics Choice Of Manometer CIVE1400: Fluid Mechanics Section : Statics Forces on Submerged Surfaces in Static Fluids Take care when fixing the manometer to vessel Burrs cause local pressure variations. Disadvantages: Slow response - only really useful for very slowly varying pressures - no use at all for fluctuating pressures; For the U tube manometer two measurements must be taken simultaneously to get the h value. It is often difficult to measure small variations in pressure. It cannot be used for very large pressures unless several manometers are connected in series; For very accurate work the temperature and relationship between temperature and must be known; Advantages of manometers: They are very simple. No calibration is required - the pressure can be calculated from first principles. We have seen these features of static fluids Hydrostatic vertical pressure distribution Pressures at any equal depths in a continuous fluid are equal Pressure at a point acts equally in all directions (Pascal s law). Forces from a fluid on a boundary acts at right angles to that boundary. Fluid pressure on a surface Pressure is force per unit area. Pressure p acting on a small area A exerted force will be F = pa Since the fluid is at rest the force will act at right-angles to the surface. CIVE1400: Fluid Mechanics Section : Statics 53 CIVE1400: Fluid Mechanics Section : Statics 54

32 CIVE1400: Fluid Mechanics Section : Statics General submerged plane F n =p n δa n F =p δa F 1 =p 1 δa 1 The total or resultant force, R, on the plane is the sum of the forces on the small elements i.e. R p1a1 pa pnan pa and This resultant force will act through the centre of pressure. For a plane surface all forces acting can be represented by one single resultant force, acting at right-angles to the plane through the centre of pressure. CIVE1400: Fluid Mechanics Section : Statics Horizontal submerged plane The pressure, p, will be equal at all points of the surface. The resultant force will be given by R pressure area of plane R= pa Curved submerged surface Each elemental force is a different magnitude and in a different direction (but still normal to the surface.). It is, in general, not easy to calculate the resultant force for a curved surface by combining all elemental forces. The sum of all the forces on each element will always be less than the sum of the individual forces, pa. CIVE1400: Fluid Mechanics Section : Statics 55 CIVE1400: Fluid Mechanics Section : Statics 56

33 CIVE1400: Fluid Mechanics Section : Statics Resultant Force and Centre of Pressure on a general plane surface in a liquid. CIVE1400: Fluid Mechanics Section : Statics za is known as Fluid density ρ Resultant Force R P C D G z z area δa x S c O θ Q s area A d O G elemental area δa x the 1 st Moment of Area of the plane PQ about the free surface. And it is known that za Az Take pressure as zero at the surface. Measuring down from the surface, the pressure on an element A, depth z, p = gz So force on element F = gza Resultant force on plane R g za (assuming and g as constant). CIVE1400: Fluid Mechanics Section : Statics 57 A is the area of the plane z is the distance to the centre of gravity (centroid) In terms of distance from point O za Axsin (as z x sin ) = 1 st moment of area sin about a line through O The resultant force on a plane R gaz gax sin CIVE1400: Fluid Mechanics Section : Statics 58

34 CIVE1400: Fluid Mechanics Section : Statics This resultant force acts at right angles through the centre of pressure, C, at a depth D. How do we find this position? Take moments of the forces. 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. It is convenient to take moment about O The force on each elemental area: Force on A gza gssin A the moment of this force is: Moment of Force on A about O g s sin A s gsinas, g and are the same for each element, giving the total moment as CIVE1400: Fluid Mechanics Section : Statics Sum of moments gsin s A Moment of R about O = R S = gax sin S Equating gax sin S g sin s A c The position of the centre of pressure along the plane measure from the point O is: s A Sc Ax How do we work out the summation term? This term is known as the nd Moment of Area, I o, of the plane (about the axis through O) c c CIVE1400: Fluid Mechanics Section : Statics 59 CIVE1400: Fluid Mechanics Section : Statics 60

35 CIVE1400: Fluid Mechanics Section : Statics nd moment of area about O It can be easily calculated for many common shapes. I s A o CIVE1400: Fluid Mechanics Section : Statics How do you calculate the nd moment of area? 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 nd Moment of area about a line through O st 1 Moment of area about a line through O It can be found from tables - BUT only for moments about an axis through its centroid = I GG. Usually we want the nd moment of area about a different axis. and Depth to the centre of pressure is D S c sin Through O in the above examples. We can use the parallel axis theorem to give us what we want. CIVE1400: Fluid Mechanics Section : Statics 61 CIVE1400: Fluid Mechanics Section : Statics 6

36 CIVE1400: Fluid Mechanics Section : Statics The parallel axis theorem can be written I I Ax o GG We then get the following equation for the position of the centre of pressure I Sc GG x Ax I D sin Ax GG x (In the examination the parallel axis theorem and the I GG will be given) CIVE1400: Fluid Mechanics Section : Statics The nd moment of area about a line through the centroid of some common shapes. Shape Area A nd moment of area, I GG, about an axis through the centroid Rectangle h G b Triangle G Circle G b Semicircle G R h h/3 G G bd bd 3 bd 1 bd 3 36 R G R R 4 4 R (4R)/(3π) R 4 CIVE1400: Fluid Mechanics Section : Statics 63 CIVE1400: Fluid Mechanics Section : Statics 64

37 CIVE1400: Fluid Mechanics Section : Statics An example: Find the moment required to keep this triangular gate closed on a tank which holds water. CIVE1400: Fluid Mechanics Section : Statics Submerged vertical surface - Pressure diagrams D G C.0m 1.m 1.5m For vertical walls of constant width it is possible to find the resultant force and centre of pressure graphically using a pressure diagram. We know the relationship between pressure and depth: p = gz So we can draw the diagram below: z ρgz H H 3 p R ρgh This is know as a pressure diagram. CIVE1400: Fluid Mechanics Section : Statics 65 CIVE1400: Fluid Mechanics Section : Statics 66

38 CIVE1400: Fluid Mechanics Section : Statics 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, Units of this are Newtons per metre. 1 Area AB BC 1 HgH 1 gh Resultant force per unit width R 1 gh ( N / m) CIVE1400: Fluid Mechanics Section : Statics For a triangle the centroid is at /3 its height i.e. the resultant force acts horizontally through the point z H. 3 For a vertical plane the depth to the centre of pressure is given by D H 3 The force acts through the centroid of the pressure diagram. CIVE1400: Fluid Mechanics Section : Statics 67 CIVE1400: Fluid Mechanics Section : Statics 68

39 CIVE1400: Fluid Mechanics Section : Statics Check this against the moment method: CIVE1400: Fluid Mechanics Section : Statics The same technique can be used with combinations of liquids are held in tanks (e.g. oil floating on water). For example: The resultant force is given by: R gaz gax sin H gh 1 sin oil ρ o water ρ 0.8m 1.m R D 1 gh and the depth to the centre of pressure by: I D o sin Ax and by the parallel axis theorem (with width of 1) I I Ax o GG H 1H H H 1 3 Depth to the centre of pressure ρg0.8 ρg1. Find the position and magnitude of the resultant force on this vertical wall of a tank which has oil floating on water as shown. H D H 3 / 3 / 3 H CIVE1400: Fluid Mechanics Section : Statics 69 CIVE1400: Fluid Mechanics Section : Statics 70

40 CIVE1400: Fluid Mechanics Section : Statics Submerged Curved Surface If the surface is curved the resultant force must be found by combining the elemental forces using some vectorial method. CIVE1400: Fluid Mechanics Section : Statics In the diagram below liquid is resting on top of a curved base. E D Calculate the horizontal and vertical components. F AC C G O B R H Combine these to obtain the resultant force and direction. A R R v The fluid is at rest in equilibrium. (Although this can be done for all three dimensions we will only look at one vertical plane) So any element of fluid such as ABC is also in equilibrium. CIVE1400: Fluid Mechanics Section : Statics 71 CIVE1400: Fluid Mechanics Section : Statics 7

41 CIVE1400: Fluid Mechanics Section : Statics Consider the Horizontal forces The sum of the horizontal forces is zero. C B CIVE1400: Fluid Mechanics Section : Statics 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. F AC A No horizontal force on CB as there are no shear forces in a static fluid Horizontal forces act only on the faces AC and AB as shown. F AC, must be equal and opposite to R H. AC is the projection of the curved surface AB onto a vertical plane. R H We know 1. The force on a vertical plane must act horizontally (as it acts normal to the plane).. That R H must act through the same point. So: R H acts horizontally through the centre of pressure of the projection of the curved surface onto an vertical plane. We have seen earlier how to calculate resultant forces and point of action. Hence we can calculate the resultant horizontal force on a curved surface. CIVE1400: Fluid Mechanics Section : Statics 73 CIVE1400: Fluid Mechanics Section : Statics 74

42 CIVE1400: Fluid Mechanics Section : Statics Consider the Vertical forces The sum of the vertical forces is zero. E C A G R v There are no shear force on the vertical edges, so the vertical component can only be due to the weight of the fluid. So we can say The resultant vertical force of a fluid above a curved surface is: D B CIVE1400: Fluid Mechanics Section : Statics Resultant force The overall resultant force is found by combining the vertical and horizontal components vectorialy, Resultant force H V R R R And acts through O at an angle of. The angle the resultant force makes to the horizontal is tan 1 R R V H 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. 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 R V. CIVE1400: Fluid Mechanics Section : Statics 75 CIVE1400: Fluid Mechanics Section : Statics 76

43 CIVE1400: Fluid Mechanics Section : Statics A typical example application of this is the determination of the forces on dam walls or curved sluice gates. Find the magnitude and direction of the resultant force of water on a quadrant gate as shown below. 1.0m Gate width 3.0m CIVE1400: Fluid Mechanics Section : Statics What are the forces if the fluid is below the curved surface? This situation may occur or a curved sluice gate. F AC C G O B R H Water ρ = 1000 kg/m 3 A R v R The force calculation is very similar to when the fluid is above. CIVE1400: Fluid Mechanics Section : Statics 77 CIVE1400: Fluid Mechanics Section : Statics 78

44 CIVE1400: Fluid Mechanics Section : Statics CIVE1400: Fluid Mechanics Section : Statics Horizontal force Vertical force B C G B F AC O R H A A The two horizontal on the element are: The horizontal reaction force R H The force on the vertical plane A B. The resultant horizontal force, R H acts as shown in the diagram. Thus we can say: A R v What vertical force would keep this in equilibrium? If the region above the curve were all water there would be equilibrium. 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. CIVE1400: Fluid Mechanics Section : Statics 79 CIVE1400: Fluid Mechanics Section : Statics 80

45 CIVE1400: Fluid Mechanics Section : Statics CIVE1400: Fluid Mechanics Section : Statics The resultant force and direction of application are calculated in the same way as for fluids above the surface: Resultant force H V R R R 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 50 kg find a) the reaction at A b) the reaction at B E D C A And acts through O at an angle of. The angle the resultant force makes to the horizontal is tan 1 R R V H B CIVE1400: Fluid Mechanics Section : Statics 81 CIVE1400: Fluid Mechanics Section : Statics 8

46 LECTURE CONTENTS Section 0: Introduction Section 1: Fluid Properties Fluids vs. Solids Viscosity Newtonian Fluids Properties of Fluids Section : Statics Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces Section 3: Dynamics The continuity equation. The Bernoulli Equation. Application of Bernoulli equation. The momentum equation. Application of momentum equation. Section 4: Real Fluids Boundary layer. Laminar flow in pipes. Section 5: Dimensional Analysis An Intro to Dimensional analysis Similarity Fluid Dynamics Objectives 1.Identify differences between: steady/unsteady uniform/non-uniform compressible/incompressible flow.demonstrate streamlines and stream tubes 3.Introduce the Continuity principle 4.Derive the Bernoulli (energy) equation 5.Use the continuity equations to predict pressure and velocity in flowing fluids 6.Introduce the momentum equation for a fluid 7.Demonstrate use of the momentum equation to predict forces induced by flowing fluids CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 83 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 84

47 Fluid dynamics: Flow Classification The analysis of fluid in motion Fluid motion can be predicted in the same way as the motion of solids By use of the fundamental laws of physics and the physical properties of the fluid Some fluid flow is very complex: e.g. Spray behind a car waves on beaches; hurricanes and tornadoes any other atmospheric phenomenon All can be analysed with varying degrees of success (in some cases hardly at all!). There are many common situations which analysis gives very accurate predictions Fluid flow may be classified under the following headings uniform: Flow conditions (velocity, pressure, crosssection or depth) are the same at every point in the fluid. non-uniform: Flow conditions are not the same at every point. steady: Flow conditions may differ from point to point but DO NOT change with time. unsteady: Flow conditions change with time at any point. Fluid flowing under normal circumstances - a river for example - conditions vary from point to point we have non-uniform flow. If the conditions at one point vary as time passes then we have unsteady flow. CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 85 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 86

48 Combining these four gives. 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. 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. 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. Compressible or Incompressible Flow? All fluids are compressible - even water. Density will change as pressure changes. 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. Pressures and velocities change in all directions. In many cases the greatest changes only occur in two directions or even only in one. Changes in the other direction can be effectively ignored making analysis much more simple. This course is restricted to Steady uniform flow - the most simple of the four. CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 87 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 88

49 One dimensional flow: Two-dimensional flow Conditions vary only in the direction of flow not across the cross-section. Conditions vary in the direction of flow and in one direction at right angles to this. The flow may be unsteady with the parameters varying in time but not across the cross-section. E.g. Flow in a pipe. 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. Flow patterns in two-dimensional flow can be shown by curved lines on a plane. Below shows flow pattern over a weir. Pipe Ideal flow Real flow Should this be treated as two-dimensional flow? Possibly - but it is only necessary if very high accuracy is required. In this course we will be considering: steady incompressible one and two-dimensional flow CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 89 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 90

50 Streamlines It is useful to visualise the flow pattern. Lines joining points of equal velocity - velocity contours - can be drawn. These lines are know as streamlines. Here are -D streamlines around a cross-section of an aircraft wing shaped body: Some points about streamlines: Close to a solid boundary, streamlines are parallel to that boundary The direction of the streamline is the direction of the fluid velocity Fluid can not cross a streamline Streamlines can not cross each other Any particles starting on one streamline will stay on that same streamline In unsteady flow streamlines can change position with time Fluid flowing past a solid boundary does not flow into or out of the solid surface. In steady flow, the position of streamlines does not change. Very close to a boundary wall the flow direction must be along the boundary. CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 91 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 9

51 Streamtubes Some points about streamtubes A circle of points in a flowing fluid each has a streamline passing through it. These streamlines make a tube-like shape known as a streamtube The walls of a streamtube are streamlines. Fluid cannot flow across a streamline, so fluid cannot cross a streamtube wall. A streamtube is not like a pipe. Its walls move with the fluid. In unsteady flow streamtubes can change position with time In steady flow, the position of streamtubes does not change. In a two-dimensional flow the streamtube is flat (in the plane of the paper): CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 93 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 94

52 m dm dt Flow rate Mass flow rate mass time taken to accumulate this mass A simple example: An empty bucket weighs.0kg. After 7 seconds of collecting water the bucket weighs 8.0kg, then: mass of fluid in bucket mass flow rate m = time taken to collect the fluid kg / s 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.0 litres in 5 seconds, what is the discharge? Q sec m m / s 08. l / s CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 95 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 96

53 Discharge and mean velocity If we know the discharge and the diameter of a pipe, we can deduce the mean velocity x Pipe u m t area A Cylinder of fluid Cross sectional area of pipe is A Mean velocity is u m. In time t, a cylinder of fluid will pass point X with a volume A u m t. The discharge will thus be If A = m And discharge, Q is 4 l/s, mean velocity is A simple example: u m Q A m/ s 3 3 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 Q = Q volume time Au m = A um t t u m u max This idea, that mean velocity multiplied by the area gives the discharge, applies to all situations - not just pipe flow. u CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 97 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 98

54 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) Applying to a streamtube: Mass enters and leaves only through the two ends (it cannot cross the streamtube wall). ρ u A Mass flow in Control volume Mass flow out ρ 1 u 1 A 1 For any control volume the principle of conservation of mass says Mass entering = Mass leaving + Increase per unit time per unit time of mass in control vol per unit time For steady flow there is no increase in the mass within the control volume, so For steady flow Mass entering = Mass leaving per unit time per unit time Mass entering = Mass leaving per unit time per unit time Or for steady flow, A u A u Au Au Constant m dm 1 1 dt This is the continuity equation. CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 99 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 100

55 In a real pipe (or any other vessel) we use the mean velocity and write Some example applications of Continuity 1A1u 1 Au Constant m m m For incompressible, fluid 1 = = (dropping the m subscript) A u A u Q 1 1 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. Section 1 Section A liquid is flowing from left to right. By the continuity A1u 11 Au As we are considering a liquid, 1 = = Q1 Q Au 11 Au An example: If the area A 1 = m and A =310-3 m And the upstream mean velocity u 1 =.1 m/s. The downstream mean velocity is A u 1 A u m/ s CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 101 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 10

56 Now try this on a diffuser, a pipe which expands or diverges as in the figure below, Velocities in pipes coming from a junction. 1 Section 1 Section 3 If d 1 =30mm and d =40mm and the velocity u =3.0m/s. The velocity entering the diffuser is given by, A u A u d d 1 u d d u 1 / 4 1 / 4 1 d d u m/ s mass flow into the junction = mass flow out 1 Q 1 = Q + 3 Q 3 When incompressible Q 1 = Q + Q 3 1 u 1 = u + 3 u 3 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 103 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 104

57 If pipe 1 diameter = 50mm, mean velocity m/s, pipe diameter 40mm takes 30% of total discharge and pipe 3 diameter 60mm. What are the values of discharge and mean velocity in each pipe? Discharge in: Discharges out: Velocities out: Q d Au u m / s Q 0. 3Q m / s Q1 Q Q3 Q Q 03. Q 07. Q m / s Q Au u m/ s Q3 A3u3 u m/ s 3 The Bernoulli equation The Bernoulli equation is a statement of the principle of conservation of energy along a streamline It can be written: p1 u1 g g z 1 H = Constant These terms represent: Pressure Kinetic Potential Total energy per energy per energy per energy per unit weight unit weight unit weight unit weight These term all have units of length, they are often referred to as the following: p pressure head = velocity head = u g g potential head = z total head = H CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 105 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 106

58 Restrictions in application of Bernoulli s equation: The derivation of Bernoulli s Equation: Cross sectional area a B B Flow is steady z mg A A Density is constant (incompressible) Friction losses are negligible It relates the states at two points along a single streamline, (not conditions on two different streamlines) All these conditions are impossible to satisfy at any instant in time! Fortunately, for many real situations where the conditions are approximately satisfied, the equation gives very good results. An element of fluid, as that in the figure above, has potential energy due to its height z above a datum and kinetic energy due to its velocity u. If the element has weight mg then potential energy = mgz potential energy per unit weight = z kinetic energy = 1 mu u kinetic energy per unit weight = g At any cross-section the pressure generates a force, the fluid will flow, moving the cross-section, so work will be done. If the pressure at cross section AB is p and the area of the cross-section is a then force on AB = pa when the mass mg of fluid has passed AB, cross-section AB will have moved to A B mg volume passing AB = g therefore m CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 107 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 108

59 distance AA = m a work done = force distance AA = pa m a pm work done per unit weight = p g This term is know as the pressure energy of the flowing stream. Summing all of these energy terms gives Pressure Kinetic Potential Total energy per energy per energy per energy per unit weight unit weight unit weight unit weight or The Bernoulli equation is applied along streamlines like that joining points 1 and below. 1 total head at 1 = total head at p1 1 g ug z p u 1 g g z or p u g g z H 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: By the principle of conservation of energy, the total energy in the system does not change, thus the total head does not change. So the Bernoulli equation can be written p u g g z H Constant Total Total Loss Work done Energy energy per energy per unit per unit per unit supplied unit weight at 1 weight at weight weight per unit weight p1 1 g ug z p u 1 g g z hwq CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 109 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 110

60 Practical use of the Bernoulli Equation The Bernoulli equation is often combined with the continuity equation to find velocities and pressures at points in the flow connected by a streamline. An example: Finding pressures and velocities within a contracting and expanding pipe. u 1 u p 1 p section 1 section A fluid, density = 960 kg/m 3 is flowing steadily through the above tube. The section diameters are d 1 =100mm and d =80mm. The gauge pressure at 1 is p 1 =00kN/m The velocity at 1 is u 1 =5m/s. The tube is horizontal (z 1 =z ) What is the gauge pressure at section? Apply the Bernoulli equation along a streamline joining section 1 with section. p1 1 g ug z p u 1 g g z p p1 ( u1 u ) Use the continuity equation to find u A11 u Au Au d u 11 1 u1 A d m/ s So pressure at section p N / m kn / m Note how the velocity has increased the pressure has decreased CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 111 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 11

61 We have used both the Bernoulli equation and the Continuity principle together to solve the problem. Use of this combination is very common. We will be seeing this again frequently throughout the rest of the course. Pitot Tube The Pitot tube is a simple velocity measuring device. Uniform velocity flow hitting a solid blunt body, has streamlines similar to this: 1 Applications of the Bernoulli Equation The Bernoulli equation is applicable to many situations not just the pipe flow. Here we will see its application to flow measurement from tanks, within pipes as well as in open channels. Some move to the left and some to the right. The centre one hits the blunt body and stops. At this point () velocity is zero The fluid does not move at this one point. This point is known as the stagnation point. Using the Bernoulli equation we can calculate the pressure at this point. Along the central streamline at 1: velocity u 1, pressure p 1 at the stagnation point of: u = 0. (Also z 1 = z ) p1 u1 p 1 p p u 1 1 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 113 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 114

62 How can we use this? The blunt body does not have to be a solid. I could be a static column of fluid. Two piezometers, one as normal and one as a Pitot tube within the pipe can be used as shown below to measure velocity of flow. Pitot Static Tube The necessity of two piezometers makes this arrangement awkward. The Pitot static tube combines the tubes and they can then be easily connected to a manometer. h1 h h X We have the equation for p, 1 p p1 u1 1 gh gh1 u1 u g( h h ) 1 We now have an expression for velocity from two pressure measurements and the application of the Bernoulli equation. CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 115 [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 are considered to be at the same level] The holes on the side connect to one side of a manometer, while the central hole connects to the other side of the manometer CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 116 A B

63 Using the theory of the manometer, p p g X h gh p p gx p A B A p 1 B p gx p g X h gh 1 1 We know that p p1 u1, giving man man u p1hgman p1 gh( u m ) 1 1 Venturi Meter The Venturi meter is a device for measuring discharge in a pipe. It is a rapidly converging section which increases the velocity of flow and hence reduces the pressure. It then returns to the original dimensions of the pipe by a gently diverging diffuser section. about 0 about 6 The Pitot/Pitot-static is: Simple to use (and analyse) 1 Gives velocities (not discharge) May block easily as the holes are small. z 1 h z datum CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 117 Apply Bernoulli along the streamline from point 1 to point CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 118

64 By continuity p1 1 g ug z p u 1 g g z Qu A u A u ua A Substituting and rearranging gives p p g 1 u A z1 z g A u 1 1 u1 A1 A g A A 1 1 g p 1 p z g A A 1 z 1 The theoretical (ideal) discharge is ua. Actual discharge takes into account the losses due to friction, we include a coefficient of discharge (C d 0.9) p Giving Qideal u1a1 Q C Q C u A Q actual d ideal d actual p g 1 Q C A A d 1 g p 1 1 p z g A A 1 1 In terms of the manometer readings z 1 p gz p gh g( z h) 1 1 man z z h man 1 1 C A A gh man 1 A A actual d 1 1 This expression does not include any elevation terms. (z 1 or z ) When used with a manometer The Venturimeter can be used without knowing its angle. CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 119 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 10

65 Venturimeter design: The diffuser assures a gradual and steady deceleration after the throat. So that pressure rises to something near that before the meter. Flow Through A Small Orifice Flow from a tank through a hole in the side. 1 A actual The angle of the diffuser is usually between 6 and 8 degrees. h Wider and the flow might separate from the walls increasing energy loss. Vena contractor If the angle is less the meter becomes very long and pressure losses again become significant. The efficiency of the diffuser of increasing pressure back to the original is rarely greater than 80%. Care must be taken when connecting the manometer so that no burrs are present. The edges of the hole are sharp to minimise frictional losses by minimising the contact between the hole and the liquid. The streamlines at the orifice contract reducing the area of flow. This contraction is called the vena contracta. The amount of contraction must be known to calculate the flow. CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 11 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 1

66 Apply Bernoulli along the streamline joining point 1 on the surface to point at the centre of the orifice. At the surface velocity is negligible (u 1 = 0) and the pressure atmospheric (p 1 = 0). At the orifice the jet is open to the air so again the pressure is atmospheric (p = 0). If we take the datum line through the orifice then z 1 = h and z =0, leaving u h g u gh This theoretical value of velocity is an overestimate as friction losses have not been taken into account. A coefficient of velocity is used to correct the theoretical velocity, u actual C u v theoretical Each orifice has its own coefficient of velocity, they usually lie in the range( ) The discharge through the orifice is jet area jet velocity The area of the jet is the area of the vena contracta not the area of the orifice. We use a coefficient of contraction to get the area of the jet A C A actual c orifice Giving discharge through the orifice: Q Au Q A u CC A actual actual actual C A u u c v orifice theoretical d orifice theoretical C A gh d orifice C d is the coefficient of discharge, C d = C c C v CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 13 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 14

67 Time for the tank to empty We have an expression for the discharge from the tank Q C A gh d We can use this to calculate how long it will take for level in the to fall As the tank empties the level of water falls. The discharge will also drop. o This Q is the same as the flow out of the orifice so C A gh A h d o t A h t C A g h Integrating between the initial level, h 1, and final level, h, gives the time it takes to fall this height d o h 1 h t A C A g d o h h1 h h The tank has a cross sectional area of A. 1 h 1 / 1 / h h h In a time dt the level falls by dh The flow out of the tank is Q Au QA h t t A C A g d o A C A g d o h h h h1 h 1 (-ve sign as dh is falling) CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 15 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 16

68 Submerged Orifice What if the tank is feeding into another? Area A 1 Area A Time for Equalisation of Levels in Two Tanks The time for the levels to equalise can be found by integrating this equation as before. h 1 h Orifice area A o Apply Bernoulli from point 1 on the surface of the deeper tank to point at the centre of the orifice, p1 1 g ug z p u 1 g g z gh u 00h1 0 g g u g( h h ) 1 And the discharge is given by Q Cd Aou C A g( h h ) d o 1 So the discharge of the jet through the submerged orifice depends on the difference in head across the orifice. writing hh h By the continuity equation QA h 1 A h 1 t t Qt AhAh 1 so 1 1 A h A h A h h 1 Ah A A 1 and Qt A h AA CdAo g( h h t 1 ) 1 A A h but hh h t AA 1 A A C A g 1 d o h h CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 17 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 18

69 Integrating between the two levels Flow Over Notches and Weirs t A1A ( A A ) C A g 1 d o h final hinitial h h A notch is an opening in the side of a tank or reservoir. It is a device for measuring discharge. AA 1 ( A A ) C A g 1 d o h h final hinitial A weir is a notch on a larger scale - usually found in rivers. AA 1 ( A A ) C A g 1 d o h final h initial It is used as both a discharge measuring device and a device to raise water levels. h is the difference in height between the two levels (h - h 1 ) The time for the levels to equal use h initial = (h 1 - h ) and h final = 0 There are many different designs of weir. We will look at sharp crested weirs. Weir Assumptions velocity of the fluid approaching the weir is small so we can ignore kinetic energy. The velocity in the flow depends only on the depth below the free surface. u gh These assumptions are fine for tanks with notches or reservoirs with weirs, in rivers with high velocity approaching the weir is substantial the kinetic energy must be taken into account CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 19 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 130

70 A General Weir Equation Rectangular Weir Consider a horizontal strip of width b, depth h below the free surface The width does not change with depth so b constant B H b δh h B H velocity through the strip, u gh discharge through the strip, Q Aubh gh Integrating from the free surface, h=0, to the weir crest, h=h, gives the total theoretical discharge H 1 / Qtheoretical g bh dh 0 This is different for every differently shaped weir or notch. We need an expression relating the width of flow across the weir to the depth below the free surface. Substituting this into the general weir equation gives H 1 / Qtheoretical B g h dh 0 B 3 gh 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 Cd B gh 3 3 / CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 131 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 13

71 V Notch Weir The relationship between width and depth is dependent on the angle of the V. H θ b The width, b, a depth h from the free surface is So the discharge is b H htan Qtheoretical gtan H hh dh 1 / gtan Hh h gtan H 15 5 / H 0 h 3 / 5 / The actual discharge is obtained by introducing a coefficient of discharge 8 Q Cd g / H actual tan 15 5 H 0 The Momentum Equation And Its Applications We have all seen moving fluids exerting forces. The lift force on an aircraft is exerted by the air moving over the wing. A jet of water from a hose exerts a force on whatever it hits. The analysis of motion is as in solid mechanics: by use of Newton s laws of motion. The Momentum equation is a statement of Newton s Second Law It relates the sum of the forces to the acceleration or rate of change of momentum. CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 133 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 134

72 From solid mechanics you will recognise F = ma What mass of moving fluid we should use? We use a different form of the equation. In time t a volume of the fluid moves from the inlet a distance u 1 t, so volume entering the stream tube = area distance = A u 11 t Consider a streamtube: And assume steady non-uniform flow The mass entering, mass entering stream tube = volume density = A u t A A 1 u 1 ρ 1 u 1 δt u ρ And momentum momentum entering stream tube = mass velocity = A u tu Similarly, at the exit, we get the expression: momentum leaving stream tube = Aut u CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 135 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 136

73 By Newton s nd Law. Force = rate of change of momentum F= ( Autu 1Au 1 1tu1) t We know from continuity that QA u A u 1 1 And if we have a fluid of constant density,, then i.e. 1 F Q( u u ) 1 An alternative derivation From conservation of mass mass into face 1 = mass out of face we can write rate of change of mass m dm dt Au A u The rate at which momentum enters face 1 is 1 A 1 uu 1 1 mu 1 The rate at which momentum leaves face is A uu mu Thus the rate at which momentum changes across the stream tube is Auu 1A1uu 1 1mu mu 1 So Force = rate of change of momentum Fm ( u u ) 1 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 137 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 138

74 So we have these two expressions, either one is known as the momentum equation F m ( u u ) 1 F Q ( u u ) 1 The previous analysis assumed the inlet and outlet velocities in the same direction i.e. a one dimensional system. What happens when this is not the case? u θ The Momentum equation. This force acts on the fluid in the direction of the flow of the fluid. u 1 θ 1 We consider the forces by resolving in the directions of the co-ordinate axes. The force in the x-direction Fx m ucos u1cos1 mu u F Q u cos u cos x or Qu x 1x u 1 1 x 1x CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 139 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 140

75 And the force in the y-direction Fy m usin u1sin1 mu u y 1y or F Q u sin u sin y Qu 1 1 u y 1y The resultant force can be found by combining these components F y F Resultant In summary we can say: Total force on the fluid = F m u u or rate of change of momentum through the control volume out in F Q u u out Remember that we are working with vectors so F is in the direction of the velocity. in φ F x x y F F F resultant And the angle of this force tan 1 F y Fx CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 141 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 14

76 This force is made up of three components: F R = Force exerted on the fluid by any solid body touching the control volume F B = Force exerted on the fluid body (e.g. gravity) F P = Force exerted on the fluid by fluid pressure outside the control volume Application of the Momentum Equation Force due the flow around a pipe bend A converging pipe bend lying in the horizontal plane turning through an angle of. So we say that the total force, F T, is given by the sum of these forces: y ρ u A x F T = F R + F B + F P The force exerted ρ 1 u 1 A 1 θ 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 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 143 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 144

77 Why do we want to know the forces here? Control Volume 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 to prevent breakage at the joints. y x We need to know how much force a support (thrust block) must withstand. Step in Analysis: 1.Draw a control volume.decide on co-ordinate axis system 3.Calculate the total force 4.Calculate the pressure force 5.Calculate the body force 6.Calculate the resultant force Co-ordinate axis system Any co-ordinate axis can be chosen. Choose a convenient one, as above. 3 Calculate the total force In the x-direction: FT x Qux u1x u1x u1 ux ucos FT x Qucos u1 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 145 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 146

78 In the y-direction: F Q u u u u F u u T y y 1y 1y 1 y T y Qu sin0 0 sin sin 4 Calculate the pressure force If we know pressure at the inlet we can relate this to the pressure at the outlet. Use Bernoulli. p1 u1 p u z 1 z h f g g g g where h f is the friction loss (this can often be ignored, h f =0) As the pipe is in the horizontal plane, z 1 =z And with continuity, Q= u 1 A 1 = u A p 1 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 147 Q 1 1 p A A 1 Knowing the pressures at each end the pressure force can be calculated, F P F p A cos0 p A cos p A p A cos Px F p A sin0 p A sin p A sin P y pressure force at 1 - pressure force at Calculate the body force There are no body forces in the x or y directions. F Rx = F Ry = 0 The body force due to gravity is acting in the z-direction so need not be considered. 6 Calculate the resultant force F F F F F F F F T x R x P x B x T y R y P y B y CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 148

79 F F F 0 Rx T x Px Qu cos u pa p A cos F F F 0 R y T y P y Qu sin p A sin And the resultant force on the fluid is given by F Ry F Resultant Impact of a Jet on a Plane A jet hitting a flat plate (a plane) at an angle of 90 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. φ 1 & Control volume and Co-ordinate axis are shown in the figure below. F Rx R Rx R y F F F y x u And the direction of application is tan 1 F R y FR x u 1 u the force on the bend is the same magnitude but in the opposite direction R F R CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 149 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 150

80 3 Calculate the total force In the x-direction F Q u u T x x 1x Qu 1x The system is symmetrical the forces in the y-direction cancel. F T y 0 4 Calculate the pressure force. The pressures at both the inlet and the outlets to the control volume are atmospheric. The pressure force is zero FPx FP y 0 Exerted on the fluid. F F F F FR x FT x 00 Qu T x R x P x B x 1x The force on the plane is the same magnitude but in the opposite direction R F R x If the plane were at an angle the analysis is the same. But it is usually most convenient to choose the axis system normal to the plate. y x u 5 Calculate the body force As the control volume is small we can ignore the body force due to gravity. F F Bx B 0 y u 1 u 3 θ 6 Calculate the resultant force CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 151 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 15

81 Force on a curved vane This case is similar to that of a pipe, but the analysis is simpler. Pressures at ends are equal - atmospheric Both the cross-section and velocities (in the direction of flow) remain constant. 3 Calculate the total force in the x direction FT x Q u u1 cos Q by continuity u1 u, so A F T x Q A 1 cos y x u and in the y-direction FT y Qu sin 0 u 1 θ Q A 1 & Control volume and Co-ordinate axis are shown in the figure above. 4 Calculate the pressure force. The pressure at both the inlet and the outlets to the control volume is atmospheric. F Px F P y 0 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 153 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 154

82 5 Calculate the body force And the resultant force on the fluid is given by 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, FB x gv (This is often small as the jet volume is small and sometimes ignored in analysis.) R Rx R y F F F And the direction of application is exerted on the fluid. tan 1 F R y FR x 6 Calculate the resultant force F F F F F T x Rx Px Bx Rx Q FT x 1cos A The force on the vane is the same magnitude but in the opposite direction R F R F F F F F T y R y P y B y R y F T y Q A CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 155 CIVE 1400: Fluid Mechanics Section 3: Fluid dynamics 156

83 LECTURE CONTENTS Section 0: Introduction Section 1: Fluid Properties Fluids vs. Solids Viscosity Newtonian Fluids Properties of Fluids Section : Statics Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces Section 3: Dynamics The continuity equation. The Bernoulli Equation. Application of Bernoulli equation. The momentum equation. Application of momentum equation. Section 4: Real Fluids Laminar and turbulent flow Boundary layer theory Section 5: Dimensional Analysis An Intro to Dimensional analysis Similarity Real fluids Introduction to Laminar and Turbulent flow Head loss in pipes Hagen-Poiseuille equation Boundary layer theory Boundary layer separation Losses at bends and junctions CIVE 1400: Fluid Mechanics Section 4: Real Fluids 157 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 158

84 Flowing real fluids exhibit viscous effects, they: stick to solid surfaces have stresses within their body. Laminar and turbulent flow Injecting a dye into the middle of flow in a pipe, what would we expect to happen? This From earlier we saw this relationship between shear stress and velocity gradient: du dy The shear stress,, in a fluid is proportional to the velocity gradient - the rate of change of velocity across the flow. this For a Newtonian fluid we can write: du dy where is coefficient of viscosity (or simply viscosity). or this Here we look at the influence of forces due to momentum changes and viscosity in a moving fluid. CIVE 1400: Fluid Mechanics Section 4: Real Fluids 159 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 160

85 All three would happen - but for different flow rates. Top: Slow flow Middle: Medium flow Bottom: Fast flow The was first investigated in the 1880s by Osbourne Reynolds in a classic experiment in fluid mechanics. A tank arranged as below: Top: Laminar flow Middle: Transitional flow Bottom: Turbulent flow Laminar flow: Motion of the fluid particles is very orderly all particles moving in straight lines parallel to the pipe walls. Turbulent flow: Motion is, locally, completely random but the overall direction of flow is one way. But what is fast or slow? At what speed does the flow pattern change? And why might we want to know this? CIVE 1400: Fluid Mechanics Section 4: Real Fluids 161 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 16

86 After many experiments he found this expression ud = density, u = mean velocity, d = diameter = viscosity What are the units of Reynolds number? We can fill in the equation with SI units: 3 kg / m, u m / s, d m Ns / m kg / ms ud kg m m ms Re 1 3 m s 1 kg This could be used to predict the change in flow type for any fluid. This value is known as the Reynolds number, Re: Re ud It has no units! A quantity with no units is known as a non-dimensional (or dimensionless) quantity. (We will see more of these in the section on dimensional analysis.) Laminar flow: Re < 000 Transitional flow: 000 < Re < 4000 Turbulent flow: Re > 4000 The Reynolds number, Re, is a non-dimensional number. CIVE 1400: Fluid Mechanics Section 4: Real Fluids 163 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 164

87 At what speed does the flow pattern change? We use the Reynolds number in an example: A pipe and the fluid flowing have the following properties: water density = 1000 kg/m 3 pipe diameter d = 0.5m (dynamic) viscosity, = 0.55x10 3 Ns/m What is the MAXIMUM velocity when flow is laminar i.e. Re = 000 What is the MINIMUM velocity when flow is turbulent i.e. Re = 4000 ud Re 4000 u m/ s In a house central heating system, typical pipe diameter = 0.015m, limiting velocities would be, and 0.147m/s. Both of these are very slow. ud Re u d u m/ s 3 In practice laminar flow rarely occurs in a piped water system. Laminar flow does occur in fluids of greater viscosity e.g. in bearing with oil as the lubricant. CIVE 1400: Fluid Mechanics Section 4: Real Fluids 165 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 166

88 What does this abstract number mean? We can give the Re number a physical meaning. This may help to understand some of the reasons for the changes from laminar to turbulent flow. ud Re inertial forces viscous forces When inertial forces dominate (when the fluid is flowing faster and Re is larger) the flow is turbulent. When the viscous forces are dominant (slow flow, low Re) they keep the fluid particles in line, the flow is laminar. CIVE 1400: Fluid Mechanics Section 4: Real Fluids 167 Laminar flow Re < 000 low velocity Dye does not mix with water Fluid particles move in straight lines Simple mathematical analysis possible Rare in practice in water systems. Transitional flow 000 > Re < 4000 medium velocity Dye stream wavers - mixes slightly. Turbulent flow Re > 4000 high velocity Dye mixes rapidly and completely Particle paths completely irregular Average motion is in flow direction Cannot be seen by the naked eye Changes/fluctuations are very difficult to detect. Must use laser. Mathematical analysis very difficult - so experimental measures are used Most common type of flow. CIVE 1400: Fluid Mechanics Section 4: Real Fluids 168

89 Pressure loss due to friction in a pipeline Up to now we have considered ideal fluids: no energy losses due to friction Attaching a manometer gives pressure (head) loss due to the energy lost by the fluid overcoming the shear stress. L Because fluids are viscous, energy is lost by flowing fluids due to friction. This must be taken into account. The effect of the friction shows itself as a pressure (or head) loss. Δp In a real flowing fluid shear stress slows the flow. To give a velocity profile: The pressure at 1 (upstream) is higher than the pressure at. How can we quantify this pressure loss in terms of the forces acting on the fluid? CIVE 1400: Fluid Mechanics Section 4: Real Fluids 169 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 170

90 Consider a cylindrical element of incompressible fluid flowing in the pipe, τw το το τw area A w is the mean shear stress on the boundary Upstream pressure is p, Downstream pressure falls by p to (p-p) The driving force due to pressure driving force = Pressure force at 1 - pressure force at pa p p A p A p d 4 As the flow is in equilibrium, driving force = retarding force p d wdl 4 L p w4 d Giving pressure loss in a pipe in terms of: pipe diameter shear stress at the wall The retarding force is due to the shear stress shear stress area over which it acts = area of pipe wall w = dl w CIVE 1400: Fluid Mechanics Section 4: Real Fluids 171 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 17

91 What is the variation of shear stress in the flow? Shear stress and hence pressure loss varies with velocity of flow and hence with Re. R r τw Many experiments have been done with various fluids measuring the pressure loss at various Reynolds numbers. At the wall R p w L τw A graph of pressure loss and Re look like: At a radius r r p L r w R A linear variation in shear stress. This is valid for: steady flow laminar flow turbulent flow This graph shows that the relationship between pressure loss and Re can be expressed as laminar turbulent p u p u 17. ( or 0. ) CIVE 1400: Fluid Mechanics Section 4: Real Fluids 173 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 174

92 Pressure loss during laminar flow in a pipe 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. The fluid is in equilibrium, shearing forces equal the pressure forces. rl pa pr p r L Newtons law of viscosity says du dy, We are measuring from the pipe centre, so du dr Giving: p r du L dr du p r dr L r r δr R In an integral form this gives an expression for velocity, p u 1 L rdr CIVE 1400: Fluid Mechanics Section 4: Real Fluids 175 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 176

93 The value of velocity at a point distance r from the centre u r p r C L 4 At r = 0, (the centre of the pipe), u = u max, at r = R (the pipe wall) u = 0; p R C L 4 At a point r from the pipe centre when the flow is laminar: u p 1 r R r L 4 This is a parabolic profile (of the form y = ax + b ) so the velocity profile in the pipe looks similar to What is the discharge in the pipe? The flow in an annulus of thickness r Q ur Aannulus Aannulus ( rr) r rr p 1 Q R r r r L 4 Q p L p L R 3 Rrr dr R pd 8 L18 So the discharge can be written 4 p d Q L 18 v This is the Hagen-Poiseuille Equation for laminar flow in a pipe CIVE 1400: Fluid Mechanics Section 4: Real Fluids 177 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 178

94 To get pressure loss (head loss) in terms of the velocity of the flow, write pressure in terms of head loss h f, i.e. p = gh f Mean velocity: u Q/ A ghf d u 3L Head loss in a pipe with laminar flow by the Hagen-Poiseuille equation: Lu h f 3 gd Pressure loss is directly proportional to the velocity when flow is laminar. It has been validated many time by experiment. It justifies two assumptions: 1.fluid does not slip past a solid boundary.newtons hypothesis. Boundary Layers Recommended reading: Fluid Mechanics by Douglas J F, Gasiorek J M, and Swaffield J A. Longman publishers. Pages 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 This gives a, now familiar, velocity profile: zero velocity Wall τ o u max Zero at the wall A maximum at the centre of the flow. The profile doesn t just exit. It is build up gradually. Starting when it first flows past the surface e.g. when it enters a pipe. CIVE 1400: Fluid Mechanics Section 4: Real Fluids 179 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 180

95 Considering a flat plate in a fluid. Understand this Boundary layer growth diagram. Upstream the velocity profile is uniform, This is known as free stream flow. Downstream a velocity profile exists. This is known as fully developed flow. Free stream flow Fully developed flow Some question we might ask: How do we get to the fully developed state? Are there any changes in flow as we get there? Are the changes significant / important? CIVE 1400: Fluid Mechanics Section 4: Real Fluids 181 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 18

96 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: First: viscous forces (the forces which hold the fluid together) When the boundary layer is thin: velocity gradient du/dy, is large by Newton s law of viscosity shear stress, = (du/dy), is large. The force may be large enough to drag the fluid close to the surface. As the boundary layer thickens velocity gradient reduces and shear stress decreases. Eventually it is too small to drag the slow fluid along. Up to this point the flow has been laminar. Newton s law of viscosity has applied. This part of the boundary layer is the laminar boundary layer CIVE 1400: Fluid Mechanics Section 4: Real Fluids 183 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 184

97 Second: momentum transfer If the viscous forces were the only action the fluid would come to a rest. Viscous shear stresses have held the fluid particles in a constant motion within layers. Eventually they become too small to hold the flow in layers; the fluid starts to rotate. Close to boundary velocity gradients are very large. Viscous shear forces are large. Possibly large enough to cause laminar flow. This region is known as the laminar sub-layer. This layer occurs within the turbulent zone it is next to the wall. It is very thin a few hundredths of a mm. Surface roughness effect Despite its thinness, the laminar sub-layer has vital role in the friction characteristics of the surface. In turbulent flow: Roughness higher than laminar sub-layer: increases turbulence and energy losses. In laminar flow: Roughness has very little effect The fluid motion rapidly becomes turbulent. Momentum transfer occurs between fast moving main flow and slow moving near wall flow. Thus the fluid by the wall is kept in motion. The net effect is an increase in momentum in the boundary layer. This is the turbulent boundary layer. Boundary layers in pipes Initially of the laminar form. It changes depending on the ratio of inertial and viscous forces; i.e. whether we have laminar (viscous forces high) or turbulent flow (inertial forces high). CIVE 1400: Fluid Mechanics Section 4: Real Fluids 185 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 186

98 Use Reynolds number to determine which state. Re ud Laminar flow: Re < 000 Transitional flow: 000 < Re < 4000 Turbulent flow: Re > 4000 Boundary layer separation Divergent flows: Positive pressure gradients. Pressure increases in the direction of flow. The fluid in the boundary layer has so little momentum that it is brought to rest, and possibly reversed in direction. Reversal lifts the boundary layer. u 1 p 1 u p p 1 <p u 1 > u Laminar flow: profile parabolic (proved in earlier lectures) The first part of the boundary layer growth diagram. Turbulent (or transitional), Laminar and the turbulent (transitional) zones of the boundary layer growth diagram. Length of pipe for fully developed flow is the entry length. Laminar flow 10 diameter Turbulent flow 60 diameter This phenomenon is known as boundary layer separation. CIVE 1400: Fluid Mechanics Section 4: Real Fluids 187 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 188

99 Boundary layer separation: increases the turbulence increases the energy losses in the flow. Separating / divergent flows are inherently unstable Examples of boundary layer separation A divergent duct or diffuser velocity drop (according to continuity) pressure increase (according to the Bernoulli equation). Convergent flows: Negative pressure gradients Pressure decreases in the direction of flow. Fluid accelerates and the boundary layer is thinner. u 1 p 1 p 1 >p u 1 < u u p Flow remains stable Turbulence reduces. 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. Boundary layer separation does not occur. CIVE 1400: Fluid Mechanics Section 4: Real Fluids 189 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 190

100 Tee-Junctions Assuming equal sized pipes), Velocities at and 3 are smaller than at 1. Pressure at and 3 are higher than at 1. Causing the two separations shown Y-Junctions Tee junctions are special cases of the Y-junction. Two separation zones occur in bends as shown above. P b > P a causing separation. P d > P c causing separation Localised effect Downstream the boundary layer reattaches and normal flow occurs. Boundary layer separation is only local. Nevertheless downstream of a junction / bend /valve etc. fluid will have lost energy. Bends CIVE 1400: Fluid Mechanics Section 4: Real Fluids 191 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 19

101 Flow past a cylinder Slow flow, Re < 0.5 no separation: Moderate flow, Re < 70, separation vortices form. Fast flow Re > 70 vortices detach alternately. Form a trail of down stream. Karman vortex trail or street. (Easily seen by looking over a bridge) Fluid accelerates to get round the cylinder Velocity maximum at Y. Pressure dropped. Adverse pressure between here and downstream. Separation occurs Causes whistling in power cables. Caused Tacoma narrows bridge to collapse. Frequency of detachment was equal to the bridge natural frequency. CIVE 1400: Fluid Mechanics Section 4: Real Fluids 193 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 194

102 Aerofoil Normal flow over a aerofoil or a wing cross-section. At too great an angle boundary layer separation occurs on the top Pressure changes dramatically. This phenomenon is known as stalling. (boundary layers greatly exaggerated) The velocity increases as air flows over the wing. The pressure distribution is as below so transverse lift force occurs. 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 Move fast air from below to top via a slot. 3 Put a flap on the end of the wing and tilt it. CIVE 1400: Fluid Mechanics Section 4: Real Fluids 195 CIVE 1400: Fluid Mechanics Section 4: Real Fluids 196

103 Examples: Exam questions involving boundary layer theory are typically descriptive. They ask you to explain the mechanisms of growth of the boundary layers including how, why and where separation occurs. You should also be able to suggest what might be done to prevent separation. LECTURE CONTENTS Section 5: Dimensional Analysis An Intro to Dimensional analysis Similarity CIVE 1400: Fluid Mechanics Section 4: Real Fluids 197 CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 198

104 Dimensional Analysis Application of fluid mechanics in design makes use of experiments results. Results often difficult to interpret. Dimensional analysis provides a strategy for choosing relevant data. Used to help analyse fluid flow Especially when fluid flow is too complex for mathematical analysis. Dimensions and units Any physical situation can be described by familiar properties. e.g. length, velocity, area, volume, acceleration etc. These are all known as dimensions. Specific uses: help design experiments Informs which measurements are important Allows most to be obtained from experiment: e.g. What runs to do. How to interpret. It depends on the correct identification of variables Relates these variables together Doesn t give the complete answer Experiments necessary to complete solution Uses principle of dimensional homogeneity Dimensions are of no use without a magnitude. i.e. a standardised unit e.g metre, kilometre, Kilogram, a yard etc. Dimensions can be measured. Units used to quantify these dimensions. In dimensional analysis we are concerned with the nature of the dimension i.e. its quality not its quantity. Give qualitative results which only become quantitative from experimental analysis. CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 199 CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 00

105 The following common abbreviations are used: length = L mass = M time = T force = F temperature = Here we will use L, M, T and F (not ). We can represent all the physical properties we are interested in with three: L, T and one of M or F As either mass (M) of force (F) can be used to represent the other, i.e. F = MLT - M = FT L -1 This table lists dimensions of some common physical quantities: Quantity SI Unit Dimension velocity m/s ms -1 LT -1 acceleration m/s ms - LT - force N kg m/s kg ms- M LT - energy (or work) Joule J N m, kg m /s kg m s - ML T - power Watt W N m/s kg m /s 3-1 Nms kg m s -3 ML T -3 pressure ( or stress) Pascal P, N/m, kg/m/s - Nm kg m -1 s - ML-1 T - density kg/m 3 kg m -3 ML -3 specific weight N/m 3 kg/m /s kg m- s - ML - T - relative density a ratio no units 1 no dimension viscosity N s/m N sm - surface tension kg/m s N/m kg /s kg m -1 s -1 M L-1 T -1-1 Nm kg s - MT - We will always use LTM: CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 01 CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 0

106 Dimensional Homogeneity Any equation is only true if both sides have the same dimensions. It must be dimensionally homogenous. What are the dimensions of X? 3 3 / B gh X L (LT - ) 1/ L 3/ = X L (L 1/ T -1 ) L 3/ = X L 3 T -1 = X The powers of the individual dimensions must be equal on both sides. (for L they are both 3, for T both -1). Dimensional homogeneity can be useful for: 1. Checking units of equations;. Converting between two sets of units; 3. Defining dimensionless relationships What exactly do we get from Dimensional Analysis? A single equation, Which relates all the physical factors of a problem to each other. An example: Problem: What is the force, F, on a propeller? What might influence the force? It would be reasonable to assume that the force, F, depends on the following physical properties? diameter, d forward velocity of the propeller (velocity of the plane), u fluid density, revolutions per second, N fluid viscosity, CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 03 CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 04

107 From this list we can write this equation: F = ( d, u,, N, ) or 0 = ( F, d, u,, N, ) and 1 are unknown functions. Dimensional Analysis produces: F ud Nd,, u ud 0 These groups are dimensionless. will be determined by experiment. These dimensionless groups help to decide what experimental measurements to take. Common groups Several groups will appear again and again. These often have names. They can be related to physical forces. Other common non-dimensional numbers or ( groups): Reynolds number: Re ud inertial, viscous force ratio Euler number: En p u Froude number: Fn u gd Weber number: We ud pressure, inertial force ratio inertial, gravitational force ratio inertial, surface tension force ratio Mach number: Mn u c Local velocity, local velocity of sound ratio CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 05 CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 16

108 Similarity Similarity is concerned with how to transfer measurements from models to the full scale. Three types of similarity which exist between a model and prototype: Geometric similarity: The ratio of all corresponding dimensions in the model and prototype are equal. For lengths Lmodel Lm L L L prototype p L is the scale factor for length. Kinematic similarity The similarity of time as well as geometry. It exists if: i. the paths of particles are geometrically similar ii. the ratios of the velocities of are similar Some useful ratios are: V T m m L Velocity u V L / m L T / Acceleration Discharge p Q Q a a m p m p p p Lm / Tm L a L / T p 3 3 L / T m m L 3 Q L / T p p p T T T A A For areas model prototype Lm L L p A consequence is that streamline patterns are the same. All corresponding angles are the same. CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 17 CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 18

109 Dynamic similarity If geometrically and kinematically similar and the ratios of all forces are the same. F F m p Force ratio 3 Ma m m mlm L L L 3 Ma L p p p p T T L u This occurs when the controlling group is the same for model and prototype. The controlling group is usually Re. So Re is the same for model and prototype: ud m m m m ud p p p It is possible another group is dominant. In open channel i.e. river Froude number is often taken as dominant. p Modelling and Scaling Laws Measurements taken from a model needs a scaling law applied to predict the values in the prototype. An example: For resistance R, of a body moving through a fluid. R, is dependent on the following: ML -3 u: LT -1 l:(length) L : ML -1 T -1 So (R,, u, l, ) = 0 Taking, u, l as repeating variables gives: R ul ul ul R u l This applies whatever the size of the body i.e. it is applicable to prototype and a geometrically similar model. CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 19 CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 0

110 For the model and for the prototype Rm mul m m ul m m m Rp pul p p ul p p p Dividing these two equations gives R R / u l / u l m m m m p p p p m p ul m m m m ul / / p p p p Example 1 An underwater missile, diameter m and length 10m is tested in a water tunnel to determine the forces acting on the real prototype. A 1/0 th 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? Dynamic similarity so Reynolds numbers equal: mud m m pud p p The model velocity should be p d u u m p d m m p m p m p W can go no further without some assumptions. Assuming dynamic similarity, so Reynolds number are the same for both the model and prototype: mud m m pud p p so m Rm mul m m R ul p p p p i.e. a scaling law for resistance force: p R u L Both the model and prototype are in water then, m = p and m = p so u m u d p p m/ s d 1/ 0 m This is a very high velocity. This is one reason why model tests are not always done at exactly equal Reynolds numbers. A wind tunnel could have been used so the values of the and ratios would be used in the above. CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 1 CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis

111 Example A model aeroplane is built at 1/10 scale and is to be tested in a wind tunnel operating at a pressure of 0 times atmospheric. The aeroplane will fly at 500km/h. At what speed should the wind tunnel operate to give dynamic similarity between the model and prototype? If the drag measure on the model is N what will be the drag on the plane? Earlier we derived an equation for resistance on a body moving through air: ul R u l ul Re For dynamic similarity Re m = Re p, so p d p m u u m p d m m p So the model velocity is found to be And the ratio of forces is R R R R 1 1 u u 05. u 0110 / u 50km/ h m p m p m p p m ul ul 0 1 m p So the drag force on the prototype will be 1 Rp Rm N 005. The value of does not change much with pressure so m = p For an ideal gas is p = RT so the density of the air in the model can be obtained from pm mrt m pp p RT p 0pp m p p p 0 m p CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 3 CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 4

112 Geometric distortion in river models For practical reasons it is difficult to build a geometrically similar model. A model with suitable depth of flow will often be far too big - take up too much floor space. Keeping Geometric Similarity result in: depths and become very difficult to measure; the bed roughness becomes impracticably small; laminar flow may occur - (turbulent flow is normal in rivers.) Solution: Abandon geometric similarity. Typical values are 1/100 in the vertical and 1/400 in the horizontal. Resulting in: Good overall flow patterns and discharge local detail of flow is not well modelled. The Froude number (Fn) is taken as dominant. Fn can be the same even for distorted models. CIVE 1400: Fluid Mechanics Section 5: Dimensional Analysis 5