Fluid Mechanics Notes

Fluid Mechanics I

Fluid Mechanics I EG-160/EG160c

Professor M.F. Webster Room 147

Credit 10 Core module for Civil, Mechanical and Aerospace BEng and MEng

Format: 3 Lectures / week

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Fluid Mechanics I

Content

Aims To create an interest in fluid flow To show that flow phenomena are amenable to analysis To show the relevance of fluid mechanics to Engineering To create confidence and ability in problem-solving in fluid mechanics

Contents Fluids properties Hydrostatics Conservation principles Viscous flow in pipes

Assessment 1 class test (Blackboard - online) 20% of final mark Week 4 - Week 5

Summer Exam - 2 hours closed book exam 80% 2

Fluid Mechanics I




Introduction

What is a fluid? What is the difference between a solid and a fluid? • A solid is “hard” and not easily deformed, whereas a fluid is “soft” and is easily deformed • A closer look at the molecular structure a solid (steel, concrete, etc.) » densely spaced molecules » large intermolecular cohesive forces

• However, for matter that we normally think of as a liquid (water, oil,etc.) » the molecules are spaced farther apart » the intermolecular forces are smaller than for solids,

• Gases (air, oxygen, etc.) have » greater molecular spacing and freedom of motion » negligible cohesive intermolecular forces

• A more specific distinction is based on how they deform under the action of an external load. Specifically, a fluid is defined as a substance that deforms continuously when acted

on by a shearing stress of any magnitude. • Common fluids such as water, oil, and air satisfy the definition of a fluid • Some materials, such as slurries, tar, putty, toothpaste, and so on, are not easily classified since they will behave as a solid if the applied shearing stress is small, but if the stress exceeds some critical value, the substance will flow. The study of such materials is called rheology

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Fluid Mechanics I

Introduction

• The study of fluid mechanics involves the same fundamental laws you have encountered in physics and other mechanics courses. These laws include Newton’s laws of motion, conservation of mass, and the first and second laws of thermodynamics. • The broad subject of fluid mechanics can be generally subdivided into fluid statics, in which the fluid is at rest, and fluid dynamics, in which the fluid is moving.

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Fluid Mechanics I

• • • • • • • • •

• • • • • • • •

Famous Names

ARCHIMEDES 287–212 B.C. Established elementary principles of buoyancy and flotation. SEXTUS JULIUS FRONTINUS A.D. 40–103. Wrote treatise on Roman methods of water distribution. LEONARDO da VINCI 1452–1519. Expressed elementary principle of continuity; observed and sketched many basic flow phenomena; suggested designs for hydraulic machinery. GALILEO GALILEI 1564–164. Indirectly stimulated experimental hydraulics; revised Aristotelian concept of vacuum. EVANGELISTA TORRICELLI 1608–164. Related barometric height to weight of atmosphere, and form of liquid jet to trajectory of free fall. BLAISE PASCAL 1623–1662. Finally clarified principles of barometer, hydraulic press, and pressure transmissibility. ISAAC NEWTON 1642–1727. Explored various aspects of fluid resistance–inertial, viscous, and wave; discovered jet contraction. HENRI de PITOT 1695–1771. Constructed double-tube device to indicate water velocity through differential head. DANIEL BERNOULLI 1700–1782. Experimented and wrote on many phases of fluid motion, coining name “hydrodynamics”; devised manometry technique and adapted primitive energy principle to explain velocity head indication; proposed jet propulsion. LEONHARD EULER 1707–1783. First explained role of pressure in fluid flow; formulated basic equations of motion and socalled Bernoulli theorem; introduced concept of cavitation and principle of centrifugal machinery. JEAN le ROND d’ALEMBERT 1717–1783. Originated notion of velocity and acceleration components, differential expression of continuity, and paradox of zero resistance to steady non-uniform motion. ANTOINE CHEZY 1718–1798. Formulated similarity parameter for predicting flow characteristics of one channel from measurements on another. GIOVANNI BATTISTA VENTURI 1746–1822. Performed tests on various forms of mouthpieces–in particular, conical contractions and expansions. LOUIS MARIE HENRI NAVIER 1785–1836. Extended equations of motion to include “molecular” forces. AUGUSTIN LOUIS de CAUCHY 1789–1857. Contributed to the general field of theoretical hydrodynamics and to the study of wave motion. GOTTHILF HEINRICH LUDWIG HAGEN 1797–1884. Conducted original studies of resistance in and transition between laminar and turbulent flow. JEAN LOUIS POISEUILLE 1799–1869. Performed meticulous tests on resistance of flow through capillary tubes.

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Fluid Mechanics I

• • • • • • • •

• • • • • • • •

Famous Names

HENRI PHILIBERT GASPARD DARCY 1803–1858. Performed extensive tests on filtration and pipe resistance; initiated open-channel studies carried out by Bazin. JULIUS WEISBACH 1806–1871. Incorporated hydraulics in treatise on engineering mechanics, based on original experiments; noteworthy for flow patterns, non-dimensional coefficients, weir, and resistance equations. WILLIAM FROUDE 1810–1879. Developed many towing-tank techniques, in particular the conversion of wave and boundary layer resistance from model to prototype scale. ROBERT MANNING 1816–1897. Proposed several formulas for open-channel resistance. GEORGE GABRIEL STOKES 1819–1903. Derived analytically various flow relationships ranging from wave mechanics to viscous resistance—particularly that for the settling of spheres. ERNST MACH 1838–1916. One of the pioneers in the field of supersonic aerodynamics. OSBORNE REYNOLDS 1842–1912. Described original experiments in many fields, cavitation, river model similarity, pipe resistance—and devised two parameters for viscous flow; adapted equations of motion of a viscous fluid to mean conditions of turbulent flow. JOHN WILLIAM STRUTT,LORD RAYLEIGH 1842–1919. Investigated hydrodynamics of bubble collapse, wave motion, jet instability, laminar flow analogies, and dynamic similarity. VINCENZ STROUHAL 1850–1922. Investigated the phenomenon of “singing wires.” EDGAR BUCKINGHAM 1867–1940. Stimulated interest in the United States in the use of dimensional analysis. MORITZ WEBER 1871–1951. Emphasized the use of the principles of similitude in fluid flow studies and formulated a capillarity similarity parameter. LUDWIG PRANDTL 1875–1953. Introduced concept of the boundary layer and is generally considered to be the father of present day fluid mechanics. LEWIS FERRY MOODY 1880–1953. Provided many innovations in the field of hydraulic machinery. Proposed a method of correlating pipe resistance data which is widely used. THEODOR VON KÁRMÁN 1881–1963. One of the recognized leaders of twentieth century fluid mechanics. Provided major contributions to our understanding of surface resistance, turbulence, and wake phenomena. PAUL RICHARD HEINRICH BLASIUS 1883–1970. One of Prandtl’s students who provided an analytical solution to the boundary layer equations. Also, demonstrated that pipe resistance was related to the Reynolds number.

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Fluid Mechanics I




Units

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

• 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

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Fluid Mechanics I




Units

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: • 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 8

Fluid Mechanics I




Units

Examples • During a study of a certain flow system the following equation relating the pressure p1 and p2 at two points was developed

p2 = p1 +

fLV Dg

In this equation V is a velocity, L the distance between the two points, D a diameter, g the acceleration of gravity, and f a dimensionless coefficient. Is the equation dimensionally consistent? • If V is a velocity, L a length, W a weight, and µ a fluid property having dimensions of FL-2T determine the dimensions of (a) VLW/µ, (b)WLµ, (c) Vµ /L and (d) VL2µ /W

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Fluid Mechanics I

Fluids Properties

• Before we can proceed, however, it will be necessary to define and discuss certain

fluid properties

Measures of Fluid Mass and Weight


Density • The density of a fluid, designated by the Greek symbol ρ, is defined as its mass per unit volume. Density is typically used to characterize the mass of a fluid system. In SI the units are kg/m3 • The value of density can vary widely between different fluids » For liquids, variations in pressure and temperature generally have only a small effect on the value of density » For gas, the density is strongly influenced by both pressure and temperature




Specific Volume

v=

1

ρ

• The specific volume, is the volume per unit mass and is therefore the reciprocal of the density

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Fluid Mechanics I




Fluids Properties

Specific Weight • The specific weight of a fluid, designated by the Greek symbol γ, is defined as its weight per unit volume. Thus, specific weight is related to density through the equation

γ = gρ

where g is the local acceleration of gravity

• Just as density is used to characterize the mass of a fluid system, the specific weight is used to characterize the weight of the system. In SI the units are N/m3


Specific Gravity • The specific gravity of a fluid, designated as SG, is defined as the ratio of the density of the fluid to the density of water at some specified temperature. Usually the specified temperature is taken as 4oC and at this temperature the density of water is 1000 kg/m3

SG = ρ / 1000 • and since it is the ratio of densities, the value of SG does not depend on the system of units used.


It is clear that density, specific weight, and specific gravity are all interrelated, and from a knowledge of any one of the three the others can be calculated. 11

Fluid Mechanics I

Fluids Properties




It is clear that the previous properties are not sufficient to uniquely characterize how fluids behave since two fluids such as water and oil can have approximately the same value of density but behave quite differently when flowing.




There is apparently some additional property that is needed to describe the “fluidity” of the fluid which we will see later in the course.




Examples • Find the density of mercury if its specific gravity is 13.55 • A reservoir of glycerine has a mass of 1200 kg and a volume of 0.952 m3. Find the glycerine's weight, mass density, specific weight and specific gravity (Ans: 11.77kN, 1261 kg/m3, 12.36 kN/m3, 1.26) • The specific gravity of ethyl alcohol is 0.79. Calculate its specific weight and mass density (Ans: 7.73 kN/m3, 790 kg/m3) • The specific weight of a substance is 8.2 kN/m3, what is its mass density (Ans: 836 kg/m3)

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Fluid Mechanics I




Fluids Properties

Pressure • Pressure in a fluid at rest is defined as the normal force per unit area exerted on a plane surface (real or imaginary) immersed in a fluid and is created by the bombardment of the surface with the fluid molecules. • From the definition, pressure has the dimension FL-2 and in SI units is expressed as N/m2. In SI, 1N/m2 is defined as a pascal, abbreviated as Pa, and pressures are commonly specified in pascals • The equations of motion (Newton’s second law)

∑ Fy = p yδxδz − psδxδs sin θ = ρ

F = ma

in the y and z directions are

δxδyδz

ay 2 δxδyδz δxδyδz =ρ az ∑ Fz = p zδxδy − psδxδs cos θ − γ 2 2 13

Fluid Mechanics I

• Note that • Hence,

Fluids Properties

δy = δs cos θ

δz = δs sin θ

p y − p s = ρa y

δy 2

p z − p s = ( ρa z + γ )

δz 2

• Since we are interested in the pressure at a point, we take the limit as δx,δy and δz tend to zero and it follows that

p z = p y = ps • we can conclude that the pressure at a point in a fluid at rest, or in motion, is independent of direction as long as there are no shearing stresses present. This important result is known as Pascal’s law named in honour of Blaise Pascal

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Fluid Mechanics I




Fluids Properties

Compressibility of Fluids • This measure how easily can the volume (and thus the density) of a given mass of the fluid be changed when there is a change in pressure . A property that is commonly used to characterize compressibility is the bulk modulus, defined as

Ev = −

dp dp = dV V dρ ρ

• The bulk modulus (or the bulk modulus of elasticity) has dimensions of pressure. • Large values for the bulk modulus indicate that the fluid is relatively incompressible


Examples • A liquid compressed in a cylinder has a volume of 1000 cm3 at 1MN/m2 and volume of 995 cm3 at 2MN/m2. What is its bulk modulus of elasticity? • If the bulk modulus of elasticity for water is 2.2 GPa, what pressure is required to reduce a volume by 0.6 precent? (Ans: 13.2 Mpa)

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Fluid Mechanics I




Hydrostatics

Basic Equation for Pressure Field • Consider a cylindrical element of fluid inclined at an angle θ to the vertical. The pressure at the end with height z is p and at the end of height z+δz is p+δp • Resolving the forces in the direction along the central axis gives

pA − ( p + δp ) A − ρgAδs cos θ = 0

δp = − ρgδs cos θ ⇒

• Or in differential form

δp = − ρg cos θ δs dp = − ρg cos θ ds

• If θ=90ο then s is in the x or y directions, (i.e. horizontal),so

dp dp dp = = =0 ds dx dy

or in homogenous domain, Pressure in the horizontal direction is constant.

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Fluid Mechanics I

Hydrostatics

• Integrating the above equation gives

dp dp = = − ρg ds dz p = − ρgz + constant

• Integrating between z1 znd z2 will lead to

p2 − p1 = − ρg ( z2 − z1 )

• If θ=0ο then s is in the z directions, (i.e. vertical),so

• Thus in a fluid under gravity, pressure decreases with increase in height


Pressure and Head • In a liquid with a free surface the pressure at any depth z is normally measured from the free surface so that z = -h. This gives:

p = ρgh + constant • At the surface the pressure is the atmospheric pressure, patmospheric

p = ρgh + patmospheric p = γh + patmospheric 17

Fluid Mechanics I

Hydrostatics

• The lower limit of any pressure is zero- that is pressure in Vacuum. Pressure measured above this datum is known as absolute pressure • Since everything is under this pressure, it is convenient to take the atmospheric pressure as datum, hence, pressure quotes in this condition is called the Gauge pressure

p guage = ρgh = γh • Since g is constant, the gauge pressure can be given by stating the vertical height of any fluid of density ρ which is equal to the this pressure, this vertical height is know as head of fluid.


Example What is the pressure of 500 KN/m2 in terms of the height of water, ρ=1000kg/m3 , and in terms of Mercury, ρ=13600 kg/m3 .

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Fluid Mechanics I




Hydrostatics

Examples • Because of a leak in a buried gasoline storage tank, water has seeped in to the depth shown. If the specific gravity of the gasoline is SG= 0.68, determine the pressure at the gasoline-water interface and at the bottom of the tank.

5m 1m

• For the open tank, with piezometers attached on the side, containing two different immiscible liquids. Find (a) the elevation of the liquid surface in piezometer A (b) the elevation of the liquid surface in piezometer B (c) Total pressure at te bottom of the tank (Ans:2.0m, 0.82m, 18.9kPa) • The reading of an automobile fuel gauge is proportional to the gauge pressure at the bottom of the tank. If the tank is 32cm deep and is contaminated with 3cm of water, how many cm of air remains at the top when the gauge indicates full? Use γgasoline = 6670 N/m3 and γair = 11.8 N/m3. (Ans: 1.4cm)

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Fluid Mechanics I




Hydrostatics

Pressure Measurement • The relation between pressure and head is used to measure pressure with manometer, liquid gauge.

The Piezometer Tube Manometer • A tube which is attached to the top of a vessel containing fluid at a pressure higher than atmospheric • The pressure measured is relative to atmospheric, hence it is gauge pressure

p A = p1 = ρgh1 = γh1 • This method can only be used for liquids. And must not be too small or too large.

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Fluid Mechanics I

Hydrostatics

The “U” Tube Manometer • The “U” tube measure the pressure of both liquids and gases • The “U” tube is filled with a fluid called the Manometric fluid • The density of the fluid whose pressure to be measured must be less than that of the manomatric fluid • We know that,

p A = p1

• However,

p2 = p A + γh1

p2 = p3

and

• Since we are measuring gauge pressure • Hence,

p3 = γ man h2

p A = γ man h2 − γh1

• If the fluid to be measure is gas, the

ρ man >> ρ

, and the gauge pressure

p A = γ man h2 21

Fluid Mechanics I

Hydrostatics

Measurement of pressure difference using the “U” Tube Manometer • The “U” tube is connected at the two points where the pressure difference is to be measured • Using the figure • And

p A = p1

and

pB = p5

p2 = p3 p A + γ 1h1 = pB + γ man h2 + γ 3h3 p A − pB = γ man h2 − γ 1h1 + γ 3h3

• If the fluid is the same at the two points then • Again if the fluid is gas then

p A − pB = γ man h2 + γ 1 ( h3 − h1 )

p A − pB = γ man h2

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Fluid Mechanics I

Hydrostatics

Inclined Tube Manometer • If the pressure difference to be measured is small, one leg of the tube is inclined at an angle θ

• In this case

p A − pB = γ manl2 sin θ − γ 1h1 + γ 3h3

• If the fluid is the same at the two points then • Again if the fluid is gas then

p A − pB = γ manl2 sin θ + γ 1 ( h3 − h1 )

p A − pB = γ manl2 sin θ

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Fluid Mechanics I




Hydrostatics

Examples • A closed tank contains compressed air and oil (SG. 0.9). A U-tube manometer using mercury is connected to the tank. For column heights h1=90cm, h2= 15cm and h3=22.5cm determine the pressure reading of the gage. • the volume rate of flow, Q, through a pipe can be determined by means of a flow nozzle located in the pipe as illustrated. The nozzle creates a pressure drop, along the pipe which is related to the flow through the equation

Q = K p A − pB

where K is a constant depending on the pipe and nozzle size. The pressure drop is frequently measured with a differential U-tube manometer. Determine an equation for pA – pB in terms of the specific weight of the flowing fluid, γ1 the specific weight of the gage fluid, γ2 and the various heights indicated. For γ1 = 9.8 kN/m3, γ2 = 15.6 kN/m3 ,h1 = 1.0m and h2 = 0.5m what is the value of the pressure drop, pA – pB ?

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Fluid Mechanics I




Hydrostatics

Problems • A Differential manometer is attached to two tanks. Calculate the pressure difference between chambers A and B. Take SGMercury = 13.6, SGOil = 0.89 and SGTetrachloride = 1.59. (Ans:-37kN/m2)

• Calculate the level h of the oil in the right hand tube (Ans: 0.18m)

• The liquid at A and B is water and the manometer liquid is oil with SG = 0.8, h1 = 300mm, h2 = 200mm and h3 = 600mm. (a) determine pA-pB. (b) If pB = 50 kPa and the barometer reading is 730 mmHg, find the absolute pressure at A in meters of waters. (Ans:-1.37kPa, 14.9m)

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Fluid Mechanics I




Hydrostatics

Forces on Submerged Surfaces in Static Fluids We have seen that: • Hydrostatic vertical pressure distribution, the pressure varies linearly with depth • Pressures at any equal depth 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 • Pressure is defined as force per unit area

Fluid Pressure on a Surface • The determination of the forces developed on the surface due to the fluid is important in the design of storage tanks, ships, dams, and other hydraulic structures. • For a horizontal surface, the magnitude of the resultant force is FR=p A • where p = γh is the uniform pressure on the bottom • If atmospheric pressure acts on both sides of the surface, the resultant force is due to the liquid in the tank. • Since the pressure is constant and uniformly distributed over the bottom, the resultant acts through the centroid of the area 26

Fluid Mechanics I

Hydrostatics

• For a general case, assuming that the fluid surface is open to the atmosphere and using the x–y coordinate system shown. • We wish to determine the direction, location, and magnitude of the resultant force acting on one side of this area due to the liquid. • At any given depth, h, the force acting on dA is dF = γhdA and is perpendicular to the surface. • Thus, the magnitude of the resultant

FR = ∫A γhdA = ∫A γy sin θdA = γ sin θ ∫A ydA

∫A ydA = yc A

• The integral is the first moment of the area • yc is the y coordinate of the centroid, thus

FR = γAyc sin θ = γAhc

• The resultant force is equal to the pressure at the centroid multiplied by the total area

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Fluid Mechanics I

Hydrostatics

• Since all differential forces are perpendicular to the surface , the resultant force must also be perpendicular to the surface • To find the location of the resultant force we consider the moment around the x axis

FR y R = ∫A ydF = ∫A γy 2 sin θdA • Since

FR = γAyc sin θ

y 2 dA ∫ A yR = yc A

• The numerator is the second moment of the area, Ix. Using the parallel axis theorem

I x = I xc + Ayc2 • Ixc is the second moment of the area with respect to an axis passing through the centroid and parallel to the x axis. Thus,

yR =

I xc + yc yc A 28

Fluid Mechanics I

Hydrostatics

• It is clear that the resultant force does not pass through the centroid but always below it • Similarly the x coordinate of the resultant force can be obtained by summing moment about the y axis

xR =

∫A xydA = I xy = I xyc + x c yc A

yc A

yc A

• Ixy is the product of inertia with respect to the x and y axes. Ixyc is the product of inertia with respect to an orthogonal system passing through the centroid

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Fluid Mechanics I




Hydrostatics

Examples • The 4m-diameter circular gate is located in the inclined wall of a large reservoir containing water γ = 9.80 kN/m3The gate is mounted on a shaft along its horizontal diameter. For a water depth of 10 m above the shaft determine: (a) the magnitude and location of the resultant force exerted on the gate by the water, and (b) the moment that would have to be applied to the shaft to open the gate.

• A pressurized tank contains oil (S.G=0.9) and has a square, 0.6m by 0.6m plate bolted to its side. When the pressure gage on the top of the tank reads 50 kPa, what is the magnitude and location of the resultant force on the attached plate?

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Fluid Mechanics I

Hydrostatics

Hydrostatic Forces on a Curved Surfaces • Integration method can be used, however, it can be tedious • The easiest way is to consider the fluid volume enclosed by the curved surface

• The magnitude and location of the forces on the horizontal and vertical surfaces can be determined from the relationship of planer surfaces • In order for force system to be in equilibrium the horizontal components must be equal and collinear and the vertical components must be equal and collinear

FH = F2

,

FV = F1 + W

⇒

FR = ( FH ) 2 + ( FV ) 2

• The location of the resultant force is found by Summing moment about an appropriate axis

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Fluid Mechanics I




Hydrostatics

Examples • The 3m long cylinder floats in oil and rest against a wall Determine the horizontal force the cylinder exerts on the wall at the point of contact. 1m

960N/m3

• The 2m-diameter drainage conduit is half full of water at rest. Determine the magnitude and line of action of the resultant force that the water exerts on a 1m length of the curved section BC of the conduit wall.

1m

C

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Fluid Mechanics I




Hydrostatics

Problems • Isosceles triangular gate AB is hinged at A. Compute the horizontal force P at point B for equilibrium, neglecting the weight of the gate. (Ans: 22.47 kN)

• The triangular channel is hinged at A and held together by cable BC at the top. If cable spacing is 1m into the paper, what is the cable tension. (Ans: 88.5 kN)

• Determine the pivot locaion y of the square gate so that it will rotate open when the liquid surface is shown. (Ans: 0.833m)

• Compute the air pressure required to keep the gate closed. The gate is a circular plate of diameter 0.8m and weight 2.0kN.(Ans: 43 kPa)

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Fluid Mechanics I




Hydrostatics

Problems • Compute the horizontal and vertical components of the hydrostatic force on the quarter-circle face of the tank. (Ans: 308kN, 289kN)

• Compute the hydrostatic force and its line of action on semicylindrical indentation ABC per meter of width into the paper. (Ans: 115.1 kN, φ = 10.6o)

• The face of the dam retaining water to depth 10m is shaped as shown. Determine the magnitude of the hydrostatic force acting on the curve portion AB per meter width of the dam and the moment of this force about A. (Ans: 658 kN, 2470kNm)

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Fluid Mechanics I




Fluid Dynamics

Fluid Dynamics This section discusses the analysis of fluid in motion - fluid dynamics. The motion of fluids can be predicted in the same way as the motion of solids are predicted using the fundamental laws of physics together with the physical properties of the fluid Objectives • Introduce concepts necessary to analyse fluids in motion • Identify differences between Steady/unsteady uniform/non-uniform compressible/incompressible flow • Demonstrate streamlines and stream tubes • Introduce the Continuity principle through conservation of mass and control volumes • Derive the Bernoulli (energy) equation • Demonstrate practical uses of the Bernoulli and continuity equation in the analysis of flow • Introduce the momentum equation for a fluid • Demonstrate how the momentum equation and principle of conservation of momentum is used to predict forces induced by flowing fluids.

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Fluid Mechanics I




Fluid Dynamics

Flow Classification It is possible - and useful - to classify the type of flow which is being examined into small number of groups. The following terms describe the states which are used to classify fluid flow: • uniform flow: If the flow velocity is the same magnitude and direction at every point in the fluid it is said to be uniform • non-uniform: If at a given instant, the velocity is not the same at every point the flow is non-uniform. (In practice, by this definition, every fluid that flows near a solid boundary will be non-uniform – as the fluid at the boundary must take the speed of the boundary, usually zero. However if the size and shape of the of the cross-section of the stream of fluid is constant the flow is considered uniform) • steady: A steady flow is one in which the conditions (velocity, pressure and cross-section) may differ from point to point but DO NOT change with time • unsteady: If at any point in the fluid, the conditions change with time, the flow is described as unsteady. (In practise there is always slight variations in velocity and pressure, but if the average values are constant, the flow is considered steady

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Fluid Mechanics I




Fluid Dynamics

Combining the above we can classify any flow in to one of four type: • Steady uniform flow. Conditions do not change with position in the stream or with time. An example is the 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. An example is flow in a tapering pipe with constant velocity at the inlet velocity will change as you move along the length of the pipe toward the exit. • Unsteady uniform flow. At a given instant in time the conditions at every point are the same, but will change with time. An example is 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. For example waves in a channel.

steady flow is by far the most simple of the four. You will then be pleased to hear that this course is restricted to only this class of flow

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Fluid Mechanics I




Fluid Dynamics

Three-dimensional flow • Although in general all fluids flow three-dimensionally, in many cases the greatest changes only occur in two directions or even only in one. • Flow is one dimensional if the flow parameters (such as velocity, pressure, depth etc.) at a given instant in time only vary in the direction of flow and not across the cross-section. An example of one-dimensional flow is the ideal flow in a pipe.

• Note that 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. Which is only necessary if very high accuracy is required. • Flow is two-dimensional if it can be assumed that the flow parameters vary in the direction of flow and in one direction at right angles to this direction.

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Fluid Mechanics I




Fluid Dynamics

Streamlines and Streamtubes • The motion of each fluid particle is described in terms of its velocity vector, V • If it is steady flow, each successive particle that passes through a given point will follow the same path. For such cases the path is a fixed line in the x–z plane.

• For steady flows each particle slides along its path, and its velocity vector is everywhere tangent to the path. The lines that are tangent to the velocity vectors throughout the flow field are called streamlines • Close to a solid boundary streamlines are parallel to that boundary • The fluid is moving in the same direction as the streamlines, hence, fluid can not cross it • Streamlines can not cross each other • Any particle starting on one streamline will stay on the same streamline • Streamlines are two dimension while streamtubes are three dimensions 39

Fluid Mechanics I




Fluid Dynamics

Flow rate • Mass flow rate: is the mass of fluid flowing per unit time • Volume flow rate – Discharge: is the volume of fluid flowing per unit time. (It is also commonly, but inaccurately, simply called flow rate). The symbol normally used for discharge is Q. Multiplying this by the density of the fluid gives us the mass flow rate




Mean Velocity: • This is the discharge divided by the area cross section. • This does not imply that the velocity is constant across the cross section




Inviscid Flow: • That is the fluid is assumed to have zero viscosity. • In practice there are no inviscid fluid. However, for many flow situation the viscous effect is small compared to other forces such as pressure gradient and gravitation.

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Fluid Mechanics I




Conservation Principles

Continuity • Matter cannot be created or destroyed. This principle is know as the conservation of mass.The principle is applied to fixed volumes, known as control volumes (or surfaces)

Mass entering / unit time = Mass leaving / unit time + Increase of mass in the control volume/unit time • For steady state Mass entering / unit time = Mass leaving / unit time

ρ1 A1Vm1 = ρ 2 A2Vm 2 = constant • For incompressible flow ρ = constant, hence

A1Vm1 = A2Vm 2 = Q 41

Fluid Mechanics I




Conservation Principles

Examples • For the pipe contraction shown, determine the velocity of water at section 2 if the velocity at section 1 is 2.1 m/s and the surface area of section 1 and section 2 are 0.01 m2 and 0.003 m2 respectively • For the pipe expansion shown, determine the velocity of water at section 1 if the velocity at section 2 is 3.0 m/s and the diameters of section 1 and section 2 are 30 mm and 40 mm respectively • If the mean velocity in pipe 1 is 2 m/s and its diameter is 50mm and pipe 2 diameter is 40 mm and takes 30% of the total discharge and pipe 3 diameter is 60mm, determine the values of discharge and the mean velocity in each pipe

42

Fluid Mechanics I




Conservation Principles

Energy – The Bernoulli’s Equation • Mass passing in 1 sec = ρAV • Weight passing in 1 sec =

V

ρgAV

• Kinetic Energy passing in 1 sec = • KE per unit weight =

V2 2g

A

mV 2 1 = (ρAV )V 2 2 2

h

Datum

mgh = ( ρAV )gh

• Potential Energy passing in 1 sec = • PE per unit weight = h

• Work done by pressure in 1 sec = force * Distance = • Work done per unit weight = • Energy per unit weight =

( pA)V

p ρg

p V2 +h+ ρg 2g

p V2 +h+ = Constant 2g ρg

• Along a streamtube If there is no energy dissipation

43

Fluid Mechanics I




Conservation Principles

Energy – The Bernoulli’s Equation • In the direction of the streamtube we have the following forces » Pressure force at upstream end =

pA

» Pressure force at downstream end =

− ( p + δp ) A

» Pressure forces around the circumference = zero » Weight force =

− mg cos θ = − ρAδLg cos θ = − ρAδhg

• Using Newton 2nd Law » deviding by

» Noting that

F = ma ⇒

ρgAδL ⇒

pA − ( p + δp ) A − ρAδhg = ρAδL( dV / dt )

1 dp 1 dV dh + + =0 ρg dL g dt dL

dV dV dL dV d (V 2 / 2 ) = =V = dt dL dt dL dL

gives

1 dp 1 dV 2 dh + + =0 ρg dL 2 g dL dL

» This is known as the Euler’s Equation, for incompressible fluid this can be integrated to yield

p V2 + + h = Constant ρg 2g 44

Fluid Mechanics I




Conservation Principles

Energy – The Bernoulli’s Equation • Note that all the individual terms in the Bernoulli’s equation have units of length • The term h is know as potential head

V2 • The term is know as velocity head 2g p • The term is know as pressure head ρg • The term H =




p V2 +h+ is know as total head ρg 2g

Example • A fluid of constant density of 960 kg/m3 is flowing steadily through the tube shown. The diameter at section 1 is 100 mm and at section 2 is 80 mm. The pressure gauge at section 1 indicated a pressure of 200 kN/m2 and the velocity was 5 m/s. Determine the pressure at section 2.

45

Fluid Mechanics I




Conservation Principles

Application of the Energy Equation

Pitot Tube • If a stream of uniform velocity flows into a blunt body, some move to the left and some to the right. But one, in the centre, goes to the tip of the blunt body and stops. This point is known as the stagnation point • Applying the Bernoulli’s equation between point 1 and 2 gives

1 p2 = p1 + ρV12 2 • The term

1 ρV12 is called dynamic pressure 2

• Knowledge of the static and stagnation pressure will enable the calculation of the velocity of the fluid • This is the principle on which the Pitot-static tube is based • Two concentric tubes are attached to two pressure gauges

46

Fluid Mechanics I




Conservation Principles

Example • For the Pitot tube shown, show that the relation between the fluid velocity and the manometer reading h

V1 =

2 gh(ρ man − ρ )

h

ρ

Venturi Meter • The Venturi meter is a device for measuring discharge in a pipe. • It consists of a rapidly converging section which increases the velocity of flow and hence reduces the pressure. • By measuring the pressure differences the discharge can be calculated.

h

47

Fluid Mechanics I

Conservation Principles

• Using the Bernoulli’s equation between 1 and 2

p1 V12 p V2 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g • Using the continuity equation

Q = V1 A1 = V2 A2 ⇒ V2 = • However,

V1 A1 A2

p1 + ρgz1 = p2 + ρg ( z2 − h) + ρ man gh • Hence,

ρ −ρ 2 gh man  ρ   V1 = 2  A1    − 1  A2 

• The theoretical discharge is

A

h B

Qideal = V1 A1

• To get the actual discharge we account for losses due to friction, we include a coefficient of discharge

Qactual = Cd Qideal

48

Fluid Mechanics I

Conservation Principles

Flow Through a Small orifice • At the surface velocity is negligible and the pressure atmospheric. • At the orifice the jet is open to the air so again the pressure is atmospheric • If we take the datum line through the orifice then z1 = h and z2 =0. • Hence,

V22 h= ⇒ V2 = 2 gh 2g

• This is the theoretical value of velocity. • To incorporate friction we use the coefficient of velocity

Vactual = CvVtheoretical

• Each orifice has its own coefficient of velocity, they usually lie in the range (0.97 - 0.99) • The actual area of the jet is the area of the vena contracta not the area of the orifice. We obtain this area by using a coefficient of contraction for the orifice

Aactual = Cc Atheoretical

49

Fluid Mechanics I

Conservation Principles

• So the discharge through the orifice is given by

Qactual = AactualVactual = CcCv AorificeVtheoretical Qactual = Cd Aorifice gh

50

Fluid Mechanics I

Conservation Principles

Flow Over Notches and Weirs • A notch is an opening in the side of a tank or reservoir which extends above the surface of the liquid. • It is usually a device for measuring discharge. • A weir is a notch on a larger scale - usually found in rivers. • Weir can be sharp crested but also may have a substantial width in the direction of flow - it is used as both a flow measuring device and a device to raise water levels. • We will assume that the velocity of the fluid approaching the weir is small so that kinetic energy can be neglected. • We will also assume that the velocity through any elemental strip depends only on the depth below the free surface. • These are acceptable assumptions for tanks with notches or reservoirs with weirs, but for flows where the velocity approaching the weir is substantial the kinetic energy must be taken into account

51

Fluid Mechanics I

Conservation Principles

• To determine the theoretical discharge over a weir we will consider a strip of thickness δh, width l and at depth h • Velocity through the strip

V = 2 gh

• Discharge through the strip

δQ = AV = lδh 2 gh

• Integrating from the free surface to the weir crest H

Q = 2 g ∫ lh1/ 2 dh 0

Rectangular Weir

H

• The width is constant, b, Hence

2 Q = b 2 g ∫ h1/ 2 dh = b 2 g H 3 / 2 3 0

• To obtain the actual discharge we introduce the coefficient of discharge

2 Qactual = Cd b 2 g H 3 / 2 3 Triangular Weir • The width b at depth h is

b = 2( H − h) tan (θ / 2 ) , hence Qactual = Cd

8 tan (θ / 2 ) 2 g H 5 / 2 15 52

Fluid Mechanics I




Conservation Principles

Examples • Water flows along a circular duct from A to B where conditions are those shown. Assuming no losses, estimate the pressure at B.

• If the velocity of the water jet at point A is 20 m/s, what is the pressure at point B? Neglect all losses due to viscous effects and assume that the nozzle outlet is at the same height as point B.

• A horizontal venturi tube, for measuring the flow of water, tapers from 300 mm diameter at the inlet to 100 mm diameter at the throat and has a discharge coefficient of 0.98. If the differential U-tube manometer, containing water over mercury (specific gravity 13.6), connecting the inlet and the throat, shows a difference in mercury levels of 55 mm, determine the volume flow.

53

Fluid Mechanics I




Conservation Principles

Problems • A pitot-static tube used to measure the air speed in a wind tunnel is coupled to a water manometer. If the dynamic pressure is h mm of water, obtain an expression for the air speed in m/s. (Ans: 3.99h0.5)

• Water collects in the bottom of a rectangular oil tank as shown. How long will it take for the water to drain from the tank through a 0.02- m-diameter drain hole in the bottom of the tank? Assume quasisteady flow. (Ans: 2.45 hr)

• A triangular orifice is cut in the vertical side of a tank containing a liquid. The base of the orifice is horizontal and of breadth b. The apex of the orifice is at a height d above the base and is located at a depth d below the liquid surface level. If the coefficient of • discharge is unity, derive an expression for the volume flow rate. (Ans: 0.91b(gd3)0.5)

54

Fluid Mechanics I




Conservation Principles

Problems • Water flows through the pipe contraction shown. For the given 0.2-m difference in manometer level, determine the flowrate as a function of the diameter of the small pipe, D. (Ans: 0.0156 m3/s)

• Water flows steadily through the pipe shown such that the pressures at sections (1) and (2) are 300 kPa and 100 kPa, respectively. Determine the diameter of the pipe at section (2), D2, if the velocity at section 1 is 20 m/s and viscous effects are negligible. (Ans: 0.0688 m)

55

Fluid Mechanics I




Conservation Principles

The Momentum Equation • Moving fluids exerting forces on whatever it hits • In fluid mechanics the analysis of motion is performed in the same way as in solid mechanics by the use of Newton’ s Second Law of motion • The momentum equation is a statement of Newton’s Second Law and relates the sum of the forces acting on an element of fluid to its acceleration or rate of change of momentum. • Newton’s 2nd Law can be written:

The Rate of change of momentum of a body is equal to the resultant force acting on the body, and takes place in the direction of the force

F=

d ( mV ) dt

• We start by assuming that we have steady flow • In time δt » momentum of fluid entering stream tube » momentum of fluid leaving stream tube » Applying Newton’s Second Law

F=

ρ1 A1V1δtV1 ρ 2 A2V2δtV2

Note 1 represent inflow 2 represent outflow

ρ 2 A2V2δtV2 − ρ1 A1V1δtV1 δt 56

Fluid Mechanics I

Conservation Principles

» Assuming fluid with a constant density and using the continuity equation

Q1 = A1V1

Q2 = A2V2

F = ρ (Q2 V2 − Q1V1 ) » Since the velocity have components in the x, y, and z direction, it is more convenient to consider each direction separately

Fx = ρ (Q2Vx 2 − Q1Vx1 )

Fy = ρ (Q2V y 2 − Q1V y1 )

Fz = ρ (Q2Vz 2 − Q1Vz1 )

• The force F is made up of following components: » FR = Force exerted on the fluid by any solid body touching the control volume » FB = Force exerted on the fluid body (e.g. gravity) » FP = Force exerted on the fluid by fluid pressure outside the control volume

• When using the momentum equation, the following steps need to be considered: » » » » » »

Draw a control volume Decide on co-ordinate axis system Calculate the total force Calculate the pressure force Calculate the body force Calculate the resultant force

57

Fluid Mechanics I




Conservation Principles

Application of the Momentum Equation The Force Due to the Flow Around a Pipe Bend • Consider a pipe bend with a constant cross section lying in the horizontal plane and turning through an angle of θ • Control Volume : The control volume include the faces at the inlet and outlet of the bend and the pipe walls • Co-ordinate system It is convenient to choose the co-ordinate axis so that one is pointing in the direction of the inlet velocity. • Calculate the total force:

FTx = ρQ (Vx 2 − Vx1 ) = ρQ (V2 cosθ − V1 ) FTy = ρQ (Vy 2 − Vy1 ) = ρQ (V2 sin θ − 0)

• Calculate the pressure force:

FPx = p1 A1 − p2 A2 cos θ FPy = 0 − p2 A2 sin θ

• Calculate the body force: The only body force is that exerted by gravity and have not component in the x and y directions

58

Fluid Mechanics I

Conservation Principles

• Calculate the resultant force:

FTx = FRx + FPx + FBx FTy = FRy + FPy + FBy FRx = ρQ (V2 cosθ − V1 ) − p1 A1 + p2 A2 cosθ FRy = ρQV2 sin θ + p2 A2 sin θ FRx

» Hence

FR

FR = FRx2 + FRy2  FRy    F  Rx 

φ = tan −1 

Φ

FRx

» the force on the bend is the same magnitude but in the opposite direction

V5_5.mov

59

Fluid Mechanics I

Conservation Principles

Force on a Pipe Nozzle • Because the fluid is contracted at the nozzle forces are induced in the nozzle • Control Volume and Co-ordinate system are shown • Calculate the total force:

FT = FTx = ρQ (Vx 2 − Vx1 )

» Using the continuity equation

Q = A1V1 = A2V2

• Calculate the pressure force:

⇒

FPx = p1 A1 − p2 A2

 1 1 FT = ρQ 2  −   A2 A1 

» We use the Bernoulli equation and noting that the pressure outside is atmospheric

p1 =

ρQ 2  1

1   2 − 2  2  A2 A1 

• Calculate the body force: gravity have no component in the x direction • Calculate the resultant force:

FTx = FRx + FPx + FBx 1 1 1 ρQ 2 1 FRx = ρQ ( − ) − ( 2 − 2 ) A1 A2 A1 2 A2 A1 2

60

Fluid Mechanics I

Conservation Principles

Force Due to an Inclined Jet Hitting a Plate V2

• Control Volume and Co-ordinate system are shown • Calculate the total force:

FTx = ρ ((Q2Vx 2 + Q3Vx 3 ) − Q1Vx1 ) = − ρQ1V1 cosθ FTy = ρ ((Q2V y 2 + Q3Vy 3 ) − Q1V y1 ) = ρ ((Q2V2 − Q3V3 ) − Q1V1 sin θ )

V1

V3

» Using the energy equation and noting that z1=z2=z3 and the pressure is all atmospheric we can prove that v1 = v2 = v3 = v • Calculate the pressure force: all zero as the pressure is everywhere atmospheric • Calculate the body force: gravity have no component in the x and y directions • Calculate the resultant force:

FTx = FRx + FPx + FBx

⇒ FRx = − ρQ1V cosθ

FTy = FRy + FPy + FBy

⇒ FRy = − ρV (Q3 − Q2 + Q1 sin θ ) 61

Fluid Mechanics I

Conservation Principles

Force on a Pelton Wheel Blade • The above analysis of impact of jets be extended and applied to analysis of turbine blades • One clear demonstration of this is with the blade of a turbine called the Pelton wheel • A narrow jet is fired at blades which stick out around the periphery of a large metal disk • The jet is deflected by the blade and the change of its momentum transfer a force to the blade and hence a torque to the drive shaft • Calculate the total force:

Q Q FTx = ρ (( Vx 2 + Vx 2 ) − Q Vx1 ) 2 2 = ρQ (V2 cos β + V1 ) • Calculate the pressure force: all zero as the pressure is everywhere atmospheric • Calculate the body force: gravity have no component in the x and y directions • Calculate the resultant force: FTx = FRx + FPx + FBx

⇒

FRx = ρQ (V2 cos β + V1 )

62

Fluid Mechanics I




Conservation Principles

Examples • Air flows from a 600 mm diameter pipe, through a nozzle which is bolted on to the end of the pipe, and discharges into the atmosphere. The outlet diameter of the nozzle is 300 mm. A U-tube manometer, connected to the pipe, shows a pressure difference of 250mm of water. Assuming that there are no losses, estimate the speed of the air at the outlet of the nozzle and the force in the bolts required to hold the nozzle in position. (Ans: 65.2 m/s; 415 N) • A horizontal pipeline has a bend which changes the direction of the water flowing through it by 45° and at the same time changes in diameter from 0.5 m upstream to 0.25 m downstream. The gauge pressure upstream is 2x1O5N/m2 and the volume flow is O.4m3/s. Neglecting losses, determine the force required to hold the bend in position. (Ans:R=33kN)

63

Fluid Mechanics I




Conservation Principles

Problems • A nozzle providing a horizontal jet of water 25 rnm diameter at a speed of 10 m/s is supplied by a pipe from an open reservoir whose free surface is 7 m above the nozzle. The jet powers a simple turbine made up of flat plate blades which the jet strikes at 90°. The blades are connected to a shaft so that the point of impact between the jet and the blades is 300 mm from the centre of the shaft Mate an estimate of: (i) the loss of head in the pipe. (ii) the force exerted by the jet if the blades are stationary. If the shaft is allowed to rotate at 200 rev/min, calculate: (iii) the force now exerted on the blades;(iv) the power available at the shaft; and (v) the overall efficiency, taking the free surface of the reservoir as the input (Ans: (i) 1.9 m; (ii) 49.1 N (iii) 18.3 N; (iv) 115 W; (v) 34%) • A horizontal streamlined nozzle issues a jet of fluid of density ρ, cross sectional area a and velocity U. Ignoring any viscous losses, derive an expression for the force required to hold the nozzle in position at the end of a pipeline of cross sectional area A.

64

Fluid Mechanics I




Conservation Principles

Problems • Air of density 1.22 kg/m3 flows in a duct of internal diameter 600 mm and is discharged to the atmosphere. At the outlet end of the duct, and co-axial with it, is a cone with a base diameter greater than 600 mm and a vertex angle of 90°. The flow through the duct is controlled by moving the cone into the duct, the air then escaping along the sloping sides of the cone. The mean velocity in the duct upstream of the cone is 15 m/s and the air leaves the cone with a mean velocity of 60 m/s parallel to the sides. Neglecting viscous effects, calculate the force exerted by the air on the cone. (Ans: 441N) • In the system shown, air is drawn from the atmosphere into a 250 mm diameter duct by a fan and flows out past a 200 mm diameter obstacle with a speed of 30 m/s. If the air is stationary in the atmosphere and there are no losses in the duct, calculate: (i) the speed of the air in the duct (ii) the pressure in the duct upstream of the fan (iii) the force F required to hold the obstacle in position (iv) the power delivered to the air by the fan (Ans: (i)10.8 m/s; (ii) -71.7 N/m2 gauge; (iii) 11.1 N; (iv) 293 W) 65

Fluid Mechanics I

Real Fluid

Viscosity • It is clear that the previous properties are not sufficient to uniquely characterize how fluids behave since two fluids such as water and oil can have approximately the same value of density but behave quite differently when flowing. • There is apparently some additional property that is needed to describe the “fluidity” of the fluid. • From the definition of fluid, deforms continuously when subjected to shear forces, if a fluid is at rest there are no shearing forces. • Shear stresses develop if the particles of moving fluid move relative to one another. • At all solid boundaries the flow particles have zero relative velocity to the boundaries and it will increase as we move toward the centre • Since we are concerned with flow past solid boundaries; cars, aeroplanes, pipes and channels, shear forces will be present

66

Fluid Mechanics I

Fluids Properties

Viscosity in Gasses • The molecules of gasses are only weakly kept in position by molecular cohesion (as they are so far apart). As adjacent layers move by each other there is a continuous exchange of molecules. Molecules of a slower layer move to faster layers causing a drag, while molecules moving the other way exert an acceleration force. Mathematical considerations of this momentum exchange can lead to Newton law of viscosity. • If temperature of a gas increases the momentum exchange between layers will increase thus increasing viscosity. • Viscosity will also change with pressure - but under normal conditions this change is negligible in gasses.

Viscosity in Liquids • There is some molecular interchange between adjacent layers. The molecules are much closer than in gasses, hence, the cohesive forces hold the molecules in place more rigidly. • Increasing the temperature of a fluid reduces the cohesive forces and increases the molecular interchange. • Reducing cohesive forces reduces shear stress, while increasing molecular interchange increases shear stress. • High pressure can also change the viscosity of a liquid. As pressure increases the relative movement of molecules requires more energy hence viscosity increases.

67

Fluid Mechanics I

Fluids Properties

Newton’s Law of Viscosity • For a 3d rectangular element of fluid • The shear stress, τ, is

τ=

F A

d

• The deformation caused by τ is

d

the shear strain and it is the angle φ • For a particle at point E which moves under the shear stress to E’ in time t • The shear strain is

φ=

dx dy

and the rate of shear strain is

φ t

=

dx dV = tdy dy

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

τ = Constant *

dV dy

• The proportionality constant is known as the dynamic viscosity,

Newton’s law of Viscosity is

τ =µ

dV dy

µ, Hence

68

Fluid Mechanics I

Fluids Properties

•

For a solid the strain is a function of the applied stress (providing that the elastic limit has not been reached). For a fluid, the rate of strain is proportional to the applied stress.

•

The strain in a solid is independent of the time over which the force is applied and (if the elastic limit is not reached) the deformation disappears when the force is removed. A fluid continues to flow for as long as the force is applied and will not recover its original form when the force is removed.

• Newtonian Fluids: Fluids obeying Newton’s law where the value of µ is constant • Non-Newtonian Fluids: Fluids in which the value of µ is not constant

69

Fluid Mechanics I

Fluids Properties

• Viscosity, µ, is the property of a fluid, due to cohesion and interaction between molecules, which offers resistance to sheer deformation. • Different fluids deform at different rates under the same shear stress. Fluid with a high viscosity such as syrup, deforms more slowly than fluid with a low viscosity such as water. • All fluids are viscous, “Newtonian Fluids” obey the linear relationship given by Newton’s law of viscosity.

τ =µ

dV dy

• Where

τ, is the shear stress, its dimensions ML-1 T -2

and its Units N m

-2

or kg m-1 s

-2

dV is the velocity gradient or rate of shear strain, its Dimension t –1 and Units radians s-1 dy

µ is the “coefficient of dynamic viscosity”

70

Fluid Mechanics I

Fluids Properties

Coefficient of Dynamic Viscosity • The Coefficient of Dynamic Viscosity, µ, is defined as the shear force, per unit area, (or shear stress τ), required to drag one layer of fluid with unit velocity past another layer a unit distance away.

µ =τ /

dV Force Velocity Force × Time Mass = / = = dy Area Distance Area Length × Time

• Units: N s m-2 or kg m-1 s-1 . (Although kg m-1 s-1 )

µ is often expressed in Poise, P, where 10 P = 1

• Typical values: Water =1.14 x 10-3 kg m-1 s-1 1 , Air =1.78 x 10-5 kg m-1 s-1 , Mercury =1.552 kg m-1 s-1 , Paraffin Oil =1.9 kg m-1 s-1 .

71

Fluid Mechanics I

Fluids Properties

Kinematic Viscosity • Kinematic Viscosity, ν, is defined as the ratio of dynamic viscosity to mass density.

ν=

• Dimensions: L2 T -1 and Units: m2 s 104 St = 1 m2 s –1 .)

–1

µ ρ

(Although

ν is often expressed in Stokes, St, where

• Typical values: Water =1.14 x 10-6 m2 s –1 - , Air =1.46 x 10-5 m2 s 10-4 m2 s –1 , Paraffin Oil =2.375 x 10-3 m2 s –1 .

–1

, Mercury =1.145 x

72

Fluid Mechanics I




Fluids Properties

Examples • A dimensionless combination of variables that is important in the study of viscous flow through pipes is called the Reynolds number, Re, defined as ρVD/µ where ρ is the fluid density, V the mean fluid velocity, D the pipe diameter, and µ the fluid viscosity. A Newtonian fluid having a viscosity 0.38 N.s/m2 of and a specific gravity of 0.91 flows through a 25-mm-diameter pipe with a velocity of 2.6 m/s. Determine the value of the Reynolds number • The velocity distribution for the flow of a Newtonian fluid between two wide, parallel plates is given by the equation 2 3Vm   y   V= 1 −    2  h 

where Vm is the mean velocity. The fluid has a viscosity of 0.2 N.s/m2. When Vm=0.6 m/s and h=0.5 cm. determine: (a) the shearing stress acting on the bottom wall, and (b) the shearing stress acting on a plane parallel to the walls and passing through the centerline (midplane).

73

Fluid Mechanics I




Fluids Properties

Examples • A large plate moves with speed vo over a stationary plate on a layer of oil of thickness d and viscosity µ. If the velocity profile is that of a parabola, with the oil at the plates having the same velocity as the plates, what is the shear stress on the moving plate from the oil? If a linear profile is assumed, what is the shear stress on the moving plate? (Ans: µvo/(2d), µvo/d )

• A 250 mm square block weighing 1.1.kN slide down an incline on a film of oil 0.006 mm thick. Assuming a linear velocity profile in the oil, what is the terminal speed of the block? The viscosity of the oil is 7 mPa.s. (Ans: 5.16 m/s)

74

Fluid Mechanics I




Viscous Flow in Pipes

Characteristic of Pipe Flow • Most of conduits used to transport fluid are round in cross section • They are designed to withstand a considerable pressure difference across their walls • Most of the basic principles involved are independent of the cross-sectional shape • For all flows involved in this section, we assume that the pipe is completely filled with the fluid being transported

• The difference between open-channel flow and the pipe flow is in the fundamental mechanism that drives the flow » For open-channel flow, gravity alone is the driving force » For pipe flow, gravity may be important, but the main driving force is likely to be a pressure gradient along the pipe

• If the pipe is not full, it is not possible to maintain this pressure difference

75

Fluid Mechanics I

Viscous Flow in Pipes

• The flow of real fluids exhibits viscous effect, that is they tend to .stick. to solid surfaces and have stresses within their body • You might remember from earlier in the course Newtons law of viscosity:

τ∝

dV dy

• This tells us that the shear stress, τ, in a fluid is proportional to the velocity gradient - the rate of change of velocity across the fluid path. For a Newtonian fluid we can write:

τ =µ

dV dy

• The constant of proportionality, µ, is known as the viscosity. • In this part we shall look at how the forces due to momentum changes on the fluid and viscous forces compare and what changes take place.

76

Fluid Mechanics I




Viscous Flow in Pipes

Laminar and Turbulent Flow • If we were to take a pipe of free flowing water and inject a dye into the middle of the stream, what would we expect to happen? • This phenomenon was first investigated in the 1880s by Osbourne Reynolds in an experiment which has become a classic in fluid mechanics • Reynolds discovered that dependent on the speed of the flow the dye will flow smoothly, in a wavy manner or in a vigorous eddying motion where it mixed completely with the water

• In laminar flow the motion of the particles of fluid is very orderly with all particles moving in straight lines parallel to the pipe walls • In transitional the flow comprises short burst of turbulence embedded in a laminar flow • In turbulent the flow incorporate an eddying or mixing action. The motion of the fluid particle is complex and involve fluctuations in velocity and direction 77

Fluid Mechanics I

Viscous Flow in Pipes

• Reynolds’ experiments revealed that the onset of turbulence was a function of » Fluid velocity » Fluid viscosity » Typical dimension

• This led to the formation of the dimensionless Reynolds Number Where:

L is a representative length V is mean velocity ρ is density of fluid µ is absolute viscosity ν is kinematic viscosity

Re =

ρVL VL = µ ν

• It can be shown that

The Reynolds Number = Inertia force / viscous force • For commercial pipe flow » for Re < 2000 laminar flow » for 2000 < Re < 4000 flow is transitional » for Re > 4000 the flow is turbulent

78

Fluid Mechanics I




Viscous Flow in Pipes

Examples: Are the flows laminar or turbulent: • A flow of water trough a pipe of square cross section. The section is 500x500mm and the mean velocity of flow is 3m/s. take viscosity to be 1.2*10-3 kg/m s. • A flow of air through a pipe of diameter 35mm. The air velocity is 0.1 m/s and viscosity is 1.7*10-5 kg /m s

79

Fluid Mechanics I




Viscous Flow in Pipes

Energy Losses Due to Friction • In the derivation of the Conservation of energy equation we have assumed ideal fluid 1

• The friction which results from the the shear stress in real fluid will absorb some of the energy available, hence

hL

Energy Line

V22 2g

P1 ρg

V 1

Z1

Head loss Velocity head

P2

ρg

Piezometric head

Pressure head

» hf is often know as the

V12 2g

Piezometric head

p V2 p1 V12 + + h1 = 2 + 2 + h2 + h f ρg 2g ρg 2g

2

V

head loss due to friction

2 Z2

• To determine the hf we consider the Free-body diagram of a cylinder of fluid

Datum

• Using Newton’s second law and noting that the fluid is flowing at a constant velocity

p1πr 2 − ( p1 − ∆p1 )πr 2 − τ 2πrL = 0 ∆p 2τ = L r 80

Fluid Mechanics I

Viscous Flow in Pipes

• Assuming laminar flow and using Newton’s law of viscosity • Thus,

dV  ∆p  = − r ⇒ dr  2 µL 

• Integrating will give

V =−

τ = −µ

∆p

dV dr

∫ dV = − 2µL ∫ rdr

∆p 2 r + C1 4 µL

This is represent a parabolic velocity distribution

∆pD 2 • At r = D/2 we have V = 0 this will give C1 = 16 µL • The discharge is Q = ∫ V dA =

•

Using the Energy equation

r=D / 2

∫

r =0

∆pD 2   2r  1 −   16 µL   D 

2 ∆pD 2   2r   , hence V = 1 −    16µL   D  

2

 πD 4 ∆p π 2 rdr =  128µL 

p1 p2 − = h f ⇒ ∆p = ρgh f ρg ρg

πD 4 ρgh f Q= 128µL

⇒

hf =

128µLQ πD 4 ρg 81

Fluid Mechanics I

• Using the mean velocity Vmean=Q/A will give

• However, Re =

ρVd µ

therefore,

hf =

hf =

32 µLVmean ρgD 2

Viscous Flow in Pipes

. We will drop the mean.

64 L V 2 L V2 = f Re D 2 g D 2g

• For Turbulent flow, Newton’s law of viscosity does not apply, here semi imperical formula are used to determine the velocity profile and hence the turbulent shear stress • Following the experiments of Reynolds and Darcy and Wiesbach, the shear stress was proportional to V2. • Following the same analogy as above we get

L V2 hf = f D 2g

• The various experiments showed that the friction is dependent of the Re and the relative roughness of the pipe k/D or ε/D • Moody produced plots of the friction f as a function of Re and ε/D for commercial pipes

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Fluid Mechanics I

Viscous Flow in Pipes

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Fluid Mechanics I




Viscous Flow in Pipes

Energy Losses Due to Friction •

General theoretical treatment for local losses is not available.

•

It is usual to assume rough turbulence which lead to the simple equation

hL = K L •

Energy Line

hL

V2 2 g Here, K is the loss coefficient

The losses can be due to :

K L = (1 −

A1 2 ) A2

»

Pipe expansions

»

Pipe contractions: depend on A1/A2

»

Elbow and junctions

»

Pipe Entrance and exit

»

Valves

KL = 0.8

KL = 0.5

KL = 0.2

KL = 0.04

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Fluid Mechanics I

Viscous Flow in Pipes

85

Fluid Mechanics I




Viscous Flow in Pipes

Example • Determine the head lost to friction when water flows through 300m of 150mm diameter galvanised steel pipe at 50 litres/s. • Calculate the steady rate at which oil (ν =10-5 m2/s) will flow through a cast-iron pipe 100mm diameter and 120 m long under a head difference of 5 m. • Determine the size of galvanised steel pipe needed to carry water a distance of 180 m at 85 litres/s with a head loss of 9 m. • Determine the discharge for the flow situation shown in the figure. The loss coefficient for a fully opened valve, a standard elbow and a flush entrance can be taken as 10., 0.9 and 0.5 respectively. The pipe is 150mm diameter and made of cast-iron. The water temperature is 15oc.

10m Fully Open valve

12m

30m

60m

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Fluid Mechanics I




Viscous Flow in Pipes

Examples • Oil of viscosity 0.048 Pa s and density 930 kglm3 flows through a horizontal 25 mm diameter pipe with an average speed of 0.3 m/s. (i) Check that the flow is laminar (ii) Calculate the pressure drop in a 30 m length of pipe (iii) Find the speed of the fluid at a distance of 6 mm from the wall of the pipe. (Ans: Re = 145; (ii) 2.21 x 104 N/m2; (iii) 0.438 m/s) • Calculate the power required to pump 50,000 kg of oil per hour along a horizontal pipeline 100 mm diameter and 1.6 km long if the density and kinematic viscosity of the oil are 915 kglm3 and 1.86 1O-3 m2/s. (Ans: 257 kw) • Calculate the pressure drop and power required per 100 m length of horizontal 250 mm diameter cast iron pipe to pump water (ν = 1.14 X 10-6 m2/s) at the rate of 2.0 litres per second. (Ans: 11.1 N/m2; 2.22 x 10-2 W)

87

Fluid Mechanics I




Viscous Flow in Pipes

Problems • A pipeline connecting two water reservoirs having a difference of level of 6 m is 720 m long and rises to a height of 3 m above the upper reservoir, at a distance of 240 m from it, before descending to the lower reservoir. If the pipe diameter is 1.2 m and the friction factor is 0.04, estimate the volume flow and the gauge pressure at the highest point in the pipeline. Assume losses are those due to friction only. (Ans: 2.51 m3/s; -51.6x103 N/m2) • Water from a large reservoir is discharged to atmosphere through a pipe 450 m long. The outlet is 12 m below the surface level in the reservoir. Taking the friction factor as .04, calculate the diameter of pipe necessary if the discharge is to be 9 x 1O-3 m3/s. Assume that the loss at the inlet to the pipe and the kinetic energy in the water at the pipe outlet are negligible and then justify this assumption. (Ans: 100mm) • A pipe 50 mm diameter and 45 m long is connected to a large tank, the inlet to the pipe being 3 m below the surface. The lower end of the pipe, which is 6 m below the upper end, is connected to a 100 mm diameter horizontal pipe 75 m long which discharges to atmosphere. If the friction factor is 0.036 calculate the discharge, taking into account the entry loss and the sudden enlargement between the two pipes. (Ans: 4.66 x 10-3m3/s)

88

Fluid Mechanics I




Viscous Flow in Pipes

Problems • A 200 mm diameter commercial steel pipeline 500 m long is to convey 4 m3 of water per minute from a reservoir to a storage tank whose free surface level is 20 m above that of the reservoir. If there are four standard elbow bends (loss coefficient 0.9 per elbow) and two gate valves (loss coefficient 0.3 per valve) in the line, calculate the power required, assuming the efficiency of the pump is 80%. (Ans: 24.9 kW) • • In the water system shown in Fig 9, a turbine is situated in a 150 m length of 150 mm diameter galvanised steel pipe, containing four 90° elbow bends, which discharges to atmosphere. If the turbine absorbs 10 kW from the water, estimate the depth H required in the reservoir to give a flow rate of 0.1 m3/s. (Ans: 51.25 m)

89