mcmaster mech eng 3o04 - fluid mechanics notes by dr. mohamed s. hamed

9/3/2008

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ME3O04 – Chapter 1http://mech.mcmaster.ca/~hamedm/me3o04

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Outline and AnnouncementsOutline:-

Introduction – Course instructor and website.Assessment.Lectures, Class Notes, and DownloadsAssignments, Tutorials, and Exams.Role of the Textbook.Teaching Style.Course objectives.Interesting Flows.The “First Question”.

Important announcements:-1. Tutorials start the week of Sept 15.

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Assessment

Two-term tests (October and November) = 30%.

Assignments = 15% .

Final examination = 55% .

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LecturesClass discussions might include points that are not necessarily included in the textbook.

All exams will include questions on theory and concepts covered in lectures and class discussions.

Assignments might include questions on theory and concepts covered in lectures and class discussions.

Attending lectures is very important!!

Laptops and Cell Phones are not allowed during lectures, tutorials, and exams.

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Class Notes and Downloads

Class notes are print out of my PowerPoint presentations.

Notes taken by students during lectures are very important.

Lecture notes and other material will be posted on the course website in a password protected section.

All material is copyright protected and should not be shared with and/or distributed to others.

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Assignments

Assignments – will be assigned regularly, roughly every week.

Assignments might include problems from the textbook and from lectures and class discussions.

Due dates are posted on the web.

A late penalty of 10% per day will be applied.

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Tutorials and Exams

Tutorials are provided to:-Address any unclear points.Help you solve assignments.

All exams will include questions on theory and concepts covered in lectures and class discussions.

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Role of the Textbook

The textbook will be used to assign problems.

It supplements class discussions.

It is not a substitute for lectures!!

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Teaching StyleI love and enjoy teaching. So, my teaching style is dynamic and interactive.

Everybody is encouraged to interact (ask questions, inquire, answer questions, etc.)

Bonus cards will awarded to individuals based on their level of interaction and understanding of the material.

Each card is worth 0.25%. There is no limit on how many cards you can get!!

This bonus is added to your total. So, your total could be more than 100%!!

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Important Announcements

Please check the ““Important AnnouncementsImportant Announcements”section on the web on a regular basis.regular basis.

Tutorials will start the week of Sept 15Sept 15.

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Course Objectives

Introduce special “vocabularyvocabulary” and “basic concepts” used in fluid mechanics.

Develop a good understanding of these concepts.

Use them to analyze and understand fluid flows in “real (practical)real (practical)” problems.

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Why Fluid Mechanics is A Core Course?

Please try to think of or name:

An industry,A piece of machinery, orAny engineering system,

where fluid mechanics does not play an important role!

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Industries where Concepts of Fluid Mechanics Play a Vital Role

••Automobile EngineeringAutomobile Engineering••Aerospace EngineeringAerospace Engineering••Oil & Gas EngineeringOil & Gas Engineering••Power GenerationPower Generation••Thermal ManagementThermal Management••Environmental ControlEnvironmental Control••BiotechnologyBiotechnology••Energy ConversionEnergy Conversion••Process Engineering Process Engineering ▪▪

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Some Interesting Flows

Airfoil zero angle Airfoil 25° angle Wing Vortex

Personal PlumeAir flow over a CarReference: Multimedia Fluid Mechanics CD-ROM

Cambridge University Press ISBN-10: 0521604761

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OK, So what would be the first question we should address in this class?

What is Fluid Mechanics ?

Forms of Matter:

• Solid.• Liquid.• Gas. Motion and its “Cause”

Force

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Scope of Fluid Mechanics

Fluid Mechanics is the study of the behavior of fluids at restrest and in motionin motion.

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What is the Definition of a ““FluidFluid”” ??

areaforce tangential

Definition of A Fluid:

Fluid is a substance that deforms continuouslyunder the application (or the effect) of a shear stress.

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When a shear stress is applied:Solids deform or bend, then stop!Fluids continuously deform ⇒ i.e., they Flow

Definition of a Fluid – cont’d

Fig. 1.1

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Q. What makes a fluid deform?

A. The “no-slip condition”.

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Two Famous Flows in Fluid Mechanics

1. Couette flow.

2. Poiseuille flow.

Poiseuille Flow

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Two Famous Flows in Fluid Mechanics

1. Couette Flow2. Poiseuille Flow

Poiseuille Flow

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Basic Equations

In the analysis of any fluid mechanics problem, we need to use a set of equations called ““Governing EquationsGoverning Equations””.

These equations can be classified into:1. Basic or General Laws.2. Particular or Special Laws.

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Basic or General Laws

Conservation of mass.Newton’s second law of motion.The principle of angular momentum.

The first law of thermodynamics.The second law of thermodynamics.

From Mechanics

From Thermodynamics

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Basic or General Laws – Cont’d

Note that these laws do not depend on the type of fluid involved in the problem. That’s why they are called “General Laws”.

Not all these laws are required every time.

In many problems, it is necessary to bring into the analysis additional relations ⇒ “Particular Laws”

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Particular or Special LawsThese laws describe the behavior of physical properties of specific types of fluids.

Example: Ideal Gas Law ⇒

This law can be used only for ideal gases.It relates absolute pressure and temperature to gas density.P in Pa = N/m2, ρ in kg/m3, and T in K.R = gas constant, for air = 286.9 J/kg.K ■

TRP ρ=

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Methods of Description

1. The system approach (the Lagrangianapproach), which follows one specific particle or system (e.g., following the motion of a falling object).

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Methods of Description - Continue

2. The control volume approach (the Eurlerianapproach), which focuses on a certain region not on a certain particle (e.g., flow in a pipeline).

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Methods of Analysis

System(or “Closed System”)

Control Volume(or “Open System”)

in Thermodynamics

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Methods of Analysis in Fluid Mechanics

1. Infinitesimal or finite control volume.2. Large control volume.

Differential control volume Large Control volume

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Methods of Analysis in Fluid Mechanics

1. Infinitesimal or finite system.2. Control volume.

In each case the equations will look different.

In (1), the resulting equations are differential equations, which provide details of the flow.

In (2), the resulting equations are integral, which give a global or overall behavior of the flow. ■

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Why Use Infinitesimal Control Volume (Differential) Approach?

Differential approach allows us to determine flow details, e.g., velocity distribution:-

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Why Use Control Volume (Integral) Approach?

Integral approach allows the determination of global or overall values, such as, average velocities, forces, etc..

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Units and Dimensions

A unit is a specific quantitative measure of a physical quantity. Example, the foot and the meter, which are used to measure the physical quantity ‘length’.

A physical quantity, such as length, is called a dimension.

Dimensions can classified as basic or primarydimensions and secondary dimensions:-

1. Length and time are basic dimensions.2. Velocity is a secondary dimension because it can be

represented using primary dimensions (velocity = length / time).

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Systems of Basic Dimensions

1. [M], [L], [t], and [T].

2. [F], [L], [t], and [T].

3. [F],[M], [L], [t], and [T].

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Systems of Units

1. MLtT - SI (kg, m, s, K).

2. FLtT - British Gravitational (lbf, ft, s, oR).

3. FMLtT - English Engineering (lbf, lbm, ft, s, oR).

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Preferred Systems of Units

SI (kg, m, s, K)

British Gravitational (lbf, ft, s, oR)

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Dimensional Consistency

All equations and formulas must have consistent dimensions.

I.e., all terms in any equation must have the same dimension.

What is the dimension of each term of Bernoulli’s equation?

ρρ2

2

221

1

21

22pzgVpzgV

++=++rr

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Dimensional Consistency – Cont’dIn your engineering studies and practice, you might deal with two types of equations having inconsistent dimensions:-1. Semi-empirical equations, e.g., the Manning equation,

used to calculate the velocity of flow in an open channel (such as a canal):-

Where V is the velocity in m/s, R is the hydraulic radius of the channel in m, and S is the channel slope (ratio). For unfinished concrete, n = 0.014.

What is the unit of V? Does n have a unit?

If we use n =0.014, and R in ft, can we get the correct value of V in ft/s?

nSRV n

2/10

3/2

=

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Dimensional Consistency – Cont’d2. In the second type the dimensions of an equation are

consistent but the use of units is not.

For example, the commonly used Energy Efficiency ratio (EER) of an air conditioner is:-

The equation is dimensionally consistent, with EER being dimensionless (ratio). However, it is used in an inconsistent way.

A good A/C has EER = 10, which means 10 Btu/hr for each 1 W of electrical power.

One must say 10 (Btu/hr)/W because it is not dimensionless.

timeenergytimeenergyEER

//

input electricalrate cooling

==

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Exercise

A sky diver with a mass of 75 kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be FD = kV2, where k = 0.228 N.s2/m2.

Determine the maximum speed of free fall for the sky diver and the speed reached after 100 m of fall.

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Vocabulary List1. A Fluid.2. Mechanics.3. Shear force.4. Normal force (pressure).5. Flow = continuous deformation.6. Scope of Fluid Mechanics.7. Basic Equations.8. Methods of Description.9. Methods of Analysis.10. No-slip condition.11. Lagrangian approach.12. Eulerian approach.13. Infinitesimal or finite system = infinitesimal control volume

(C.V.).14. Infinitesimal C.V. = Differential approach.15. Control volume or integral approach.16. Basic dimensions.

9/22/2008

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ME3O04 Chapter 2http://mech.mcmaster.ca/~hamedm/me3o04/

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A flow is described by the velocity field,

Where a = 1 1/s and b = 2 m/s2. t in seconds.

(a) Plot the pathline of the particle that passed point (1,2) at t = 2.

(b) Streakeline at t = 3 of the particles that passed point (1,2).

jtbiyaV ˆˆ+=Problem

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Note the difference between the meaning of t and to

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Note the difference between the meaning of t and to

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Problem – time permitFluids of viscosities μ1 = 0.1 N.s/m2 and μ2 = 0.15 N. s/m2 are contained between two plates (each plate is 1 m2 in area). The thicknesses are h1 = 0.5 mm and h2 = 0.3 mm. respectively. Find the force F to make the upper plate move at a speed of 1 m/s. What is the fluid velocity at the interface between the two fluids?

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Chapter 2 - Fundamental ConceptsOutline:-

Fluid as a Continuum.Scalar quantities.Vector quantities - Velocity Field.Steady Vs Unsteady Flow Fields.Uniform Flow Fields.One-, Two-, and Three-Dimensional Flows.Visual Representation of flow fields:

Timeline.Streamlines. Pathlines. Streaklines.

Types of Forces acting on a fluid element.Stresses (name and sign convention).Stress Field (stress at a point).Shear Stress and rate of deformation of a fluid element.Viscosity.

1. Newtonian and 2. Non-Newtonian Fluids.Surface Tension.Classification of Fluid Motions.Compressibility effect.

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Nature of Fluids (Matter)

Fluids (gases or liquids) are forms of matter and they consist of molecules with atoms and space in between.

So, generally speaking, mass of a fluid is not continuously distributed in space:

(mass – space – mass – space – mass – etc…)

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Fluid as a Continuum Continuum Hypothesis: We can assume that fluids are continuous

medium.

1. What does it mean?It means that a fluid regardless of its molecular nature, can be treated as a continuous medium.

2. What is the result or benefitbenefit of this assumption?a) Each fluid property is assumed to have a definite value at every point

in space, thus

b) Fluid properties are considered to be continuous function of position and time.

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3. When it is not possible to make such an assumption?We can not use this assumption if the mean free path of the molecules is of the same order of magnitude as the smallest significant characteristic dimension of the problem.

Molecules always vibrate in space.

Mean free path is the distance the molecules travel as it vibrate.

Continuum Hypothesis – Cont’d

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Examples, for air:-

1. At STP (15º C and 101.3 kPa) ⇒ Lm = 6 × 10-8 m = 0.06 μm = 60 nm.

2. At P = 1.33 × 10-7 kPa (Rarefied Gas) ⇒ Lm = 0.9 m = 90 cm.

Applications = air flow between parallel plates with a = 10 cm at atmospheric pressure i.e., P = 101.3 kPa.

Mean Free Path, Lm

Can we assume that air is a continuous mediumin this case ?

?⇐

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Quantities of interest

In fluid mechanics we are interested in quantities that are:

1. Scalar quantities, such as: density, temperature, pressure, etc.

2. Vector quantities, such as: velocity, acceleration, stress, ect.

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Scalar Quantities – e.g., Density ρ

Since ρ is a scalar quantity, we need only the specification of its magnitude.

The complete or field representation of ρ is given by:

Equation (2.1) represents a scalar field.

(2.1)

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Density ρ - Cont’dDensity is sometimes expressed in terms of specificgravity, SG for liquids:-

where, ρwmax = max density of water = 1000 kg/m3 at 4 ºC.

For gases, where ρa= air density.

Or, in terms of specific weight, γ:

maxw

SGρρ

=

ggm ργ =∀

==volumeweight

a

SGρρ

=

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Similar to ρ, we can represent the velocity as:-

Eqn. 2.2 represents the velocity field, which is a vector field.

A Vector has magnitude and direction.

Vector Quantities – e.g., Velocity Field,

(2.2)

V

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Velocity can be written in terms of its 3 scalar components (u, v, w) in the x, y, and z directions:

where = unit vectors in x, y, and z directions, respectively.

In general each component u, v, and w is a function of x,y, z, and t, e.g., u = u(x,y,z,t)

Velocity Field, - Cont’dV

kji ˆ and,ˆ,ˆ

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Types of Flows - Steady Flow

If properties at every point in a flow field do not change with time (t), the flow is called steady.

If η is any fluid property in a steady flow field, mathematically:

Remember : if f = f(x), then rate of change of f w.r.t. x

If f=f(x,y), then rate of change of f w.r.t. x = ■

),,(or,0 zyxt

ηηη==

∂∂

dxdf

=

xf∂∂

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Types of Flows - One-, Two-, and Three-Dimensional Flows

Depending on the number of space coordinates required to specify the velocity field, a flow can be described as 1D, 2D or 3D.

⇐ 3D, unsteady

⇐ 2D, unsteady

⇐ 1D, steday

),,,( tzyxVV =

),,( tyxVV =

)(xVV =

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Examples – Steady Flow

1D 2D

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Uniform Flow

Sometimes we can neglect the no-slip condition at the wall and assume that the velocity is uniform across the whole cross section, as shown below:

u = u(x,y) u=u(x)

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Visual Representation of Flow Fields

Sometime we want to have a visual representation of flow fields. Such a representation is provided by:

Timelines.

Streamlines.

Path lines.

Streak lines.

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TimelinesA timeline is a line connecting the positions of a set of fluid particles at a given instant.

This is a timeline

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StreamlinesA streamline is the line drawn in the flow field so that at any instant in time it is tangent to the direction of the flow, i.e., tangent to the velocity vector.

,tan.,.dxdy

uvei == α 0=−⇒ dxvdyu

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Streamlines – Cont’d

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Pathlines

A path line is the line traced out by a given particle as it flows from one point to another.

Path lines are useful in studying , for example, the trajectory of a contaminant leaving a smokestack.

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Pathlines – Cont’d

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Streaklines

A streakline is the locus of all particles that at an earlier instant in time, passed through a prescribed point in space.

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Streaklines – Cont’d

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What type of Lines are these?

The redlines, andThe white lines?

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Important points to note

A streamline and a timeline are instantaneous lines, e.g., snap shots.

While streaklines and pathlines are generated by the passage of time, e.g., video recording.

In steady flow, the velocity at each point in the flow field remains constant with time and consequently, in a steady flow, pathlinespathlines, streaklinesstreaklines, and streamlinesstreamlines are identical.

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Example Problem

A two-dimensional unsteady velocity field is given by u = x (1 + 2 t), v = y. Find:-

1. The time-varying streamlines which pass through some reference point (xo,yo). Sketch some for the case of xo=1, yo = 1.

2. Find the equation of the pathline which passes through the point (xo, yo) at t = 0. Sketch this pathlinefor the case of xo=1, yo = 1.

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Streamlines at different times

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Pathline for t < 0 to t > 0

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Types of Forces Acting on a Fluid Element

1. Surface Forces.2. Body Forces.

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Types of Forces Acting on a Fluid ElementA. Surfaces Forces:- e.g., Pressure and Friction

These are forces generated on the surface of the fluid element by contact with other fluid particles or a solid surface.

B. Body Forces:- e.g., Gravity and Electromagnetic Field

These forces are not concentrated at the surface of the fluid element; they are rather experienced throughout the particle.

Example: gravitational body force acting on a fluid element of volume, =

gdgmd ∀=×= ρforce nalgravitatio

∀d

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Surface Force on a Fluid Element

δAyδAx

δAz

z

x

y

Δx

Δy

Δz

F

Fluid Element in Cartesian Coordinates

xAδ

Surface Forces cause StressesStresses on the surfaces of the fluid element

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Stress Caused by Surface Forces

σ (sigma) is used to denote normal stresses.τ (Tao) is used to denote tangential stresses.

Fluid element

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Stress Caused by Surface Forces –Cont’d

Normal stress at point C =

Tangential stress at point C = Or,

x

x

xAxx A

Fδδσ

δlim

0→=

x

y

xAxy A

Fδδ

τδlim

0→=

x

z

xAxz A

Fδδτ

δlim

0→=

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Name Convection - Stress needs two directions to be define (force and area).

First letter on the left refers to the plan on which the stress acts.

Second letter refers to the direction in which the stress acts.

Note that the subscript inrefers to the direction

normal to the surface of interest.

xAδ

xAδ

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xAδ

Sign Convention

Velocity or force is positive if it is in the positive direction of the axis.

The plan is considered positive if the normal to it is in the positive direction of the axis.

Sign of stress is determined by the sign of its plan and force.

What is the sign of What is the sign of σσxx xx ??

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Stress Field

Note: all stresses shown here are positive.Note: all stresses shown here are positive.

Stress at a point = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

zzzyzx

yzyyyx

xzxyxx

σττ

τστ

ττσ

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Shear Stress in a Fluid Element Exposed to a Shear Force

Fluid continuously deforms (i.e., flows) under the effect of a shear force.

Shear Stress =y

x

y

x

yAyx dAdF

AFlim ==

→ δδτ

δ 0

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Rate of Deformation of a Fluid Element Exposed to a Shear Force

Rate of deformation = (1)dtd

tlimt

αδδα

δ=

→ 0

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Because δα is very small we can say that: δl = δα. δy (2)

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Due to the no slip condition, point M’ will be moving at δu, thus : δl = δu. δt (3)

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From (2) and (3), Rate of deformation =

dydu

dtd

yu

t==

αδδ

δδα ,or ,

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Relation Between Shear Stress and Rate of Deformation

Because every fluid element exposed to shear stress will deform, there must be a relationship between the shear stressshear stress and the rate of rate of deformationdeformation.

This relationship depends on the type of fluid:-1. Newtonian, or2. Non-Newtonian.

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Fluid ViscosityNewtonian Fluids:-

Most of the common fluids (water, air, oil, etc.)“Linear” fluids

For Newtonian fluids, the rate of deformation is in direct proportiondirect proportion with the shear stress, i.e.,

The constant of proportionality is the fluid viscosity, μ, i.e.,

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Non-Newtonian FluidsSpecial fluids (e.g., most biological fluids, toothpaste, some paints, etc.)“Non-linear” fluids

In this case, In this case, ηη is called the apparent viscosityis called the apparent viscosity

In this case the relation between Shear stress and rate of deformationTakes the form:

Which can also be written as :

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Non-Newtonian Fluids – Cont’d

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Dynamic and Kinematic Viscosity

This equation is called Newton’s law of viscosity for a one-dimensional flow (note note only y appears in the equationonly y appears in the equation).

μ (mu) is called the Dynamic Viscosity of the fluid and it is a physical property of the fluid.

ν (nu) =μ/ρ is called the Kinematic viscosityand it is another physical property.

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Dynamic and Kinematic Viscosity

Units of μ : Pa.s = N.S/m2 = kg/(m.s),

1 Poise = 1 gm/(cm.s).

Units of ν: m2/s,

1 Stoke = 1 cm2/s.

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ProblemA block of mass M slides on a thin film of oil. The film thickness is h and the area of the block is A. When released, mass m exerts tension on the cord, causing the block to accelerate. Neglect friction in the pulley and air resistance. Develop an expression for the viscous force that acts on the block when it moves at speed V. Obtain an expression for the block speed as a function of time. If mass M = 5 kg, m = 1 kg, A = 25 cm2 , and h = 0.5 mm. If it takes 1 s for the speed of the block to reach 1 m/s, find the oil viscosity μ.

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Surface Tension, σSurface tension is a force that appears along any common surface (interface) between two fluids.

Units of σ is force per unit length (length of the interface), e.g., N/m or lbf/ft.

interface

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Surface Tension – Cont’d

As shown below, if the two fluids are in contact with a solid surface, a contact angle, θ, develops.

Both of values of σ and θ depend on the type of fluids in contact.

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Capillary Rise (CR) and Capillary Depression (CD).

CR occurs when θ < 90°.

CD occurs when θ > 90°.

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Calculation of Δh

Note the sign of Δh when θ > 90º and θ < 90º.

0=∑ zF

0cos =∀Δ−=∑ gDFz ρθπσ (1)

hDΔ=∀Δ

4

2πQ (2)

From (2) in (1):-Dg

hρ

θσ cos4=Δ∴

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Description and Classification of Fluid Motions

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Viscous and Inviscid FlowsUnder some special circumstances, the effect of fluid viscosity can be ignored (neglected).

Example:- in region of flow away from solid surfaces.

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Laminar and Turbulent FlowsA laminar flow is a flow in which the fluid particles move in smooth layers, laminas.laminas.

A Turbulent Flow is a flow in which the fluid particles rapidly mix as they move due to random velocity fluctuations.

The flow in a pipe is considered laminar if Re < 2300, where,

Where, ρ,μ, V, & D are the density, viscosity, velocity, and diameter, respectively.

number ReynoldsRe ==μ

ρ DV

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Compressible and Incompressible Flows

Incompressible flows are those in which variations in density are negligible.

When variations in density are not negligible, the flow is called compressible.

Variations in density are due to changes in pressure and/or temperature.

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Compressible and Incompressible Flows

Mostly, liquids can be regarded as incompressible fluids.

Pressure and density changes in liquids are reflected by the bulk compressibility modulus, or modulus of elasticitymodulus of elasticity:

For water at 15°C, Ev = 2010 kPa = 2.92 x 105 psi.

ρρ /ddpE v =

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591 2 3

Can we treat a gas flow as an Incompressible flow?

For Mach number M < 0.3, the maximum density variation is less than 5%.

Thus, gas flows with M < 0.3 can be treated as incompressible.

Mach number = V/c, where V is the flow velosity and c is the speed of sound.

The speed of sound in an ideal gas is given by:

Where k = Cp/Cv, R = gas constant, and T is the absolute temperature.

TRkc=

8/28/2008

60


601 2 3

Problem

At what minimum speed (in mph) would an automobile have to travel for compressibility effects to be important? Assume the local air temperature is 60°F.

8/28/2008

61


611 2 3

Problem – time permitFluids of viscosities μ1 = 0.1 N.s/m2 and μ2 = 0.15 N. s/m2 are contained between two plates (each plate is 1 m2 in area). The thicknesses are h1 = 0.5 mm and h2 = 0.3 mm. respectively. Find the force F to make the upper plate move at a speed of 1 m/s. What is the fluid velocity at the interface between the two fluids?

8/28/2008

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621 2 3

Vocabulary List

1. Continuum hypothesis.2. Mean free path.3. Scalar quantity.4. Vector quantity.5. Uniform flow.6. Multi-dimension flow.7. Flow Visualization.

8/28/2008

63


631 2 3

Vocabulary List

1. Timeline.2. Streamline.3. Pathline.4. Streakline.

8/28/2008

64


641 2 3

Vocabulary List1. Body force.2. Surface force.3. Stress field.4. Shear stress.5. Normal stress.6. Rate of deformation.7. Viscid and Inviscid flows. 8. Laminar and Turbulent flows.9. Compressible and Incompressible flow.

9/13/2008

1

ME3O04 – Chapter 3http://mech.mcmaster.ca/~hamedm/me3o04/

11 3

Outline:- Fluid Statics – Chapter 3

The Basic Equation of Fluid StaticsTypes of Pressures.Pressure Variation in a Static Fluid.Example Problem.Hydrostatic Force on Submerged Surfaces.Example Problems.

9/13/2008

2


21 3

Statics means that the fluid is not moving, i.e., its velocity =0; and its acceleration = 0.

Fluid velocity = 0 means that it does not flowit does not flow.

If a static fluid does not flow, how much shear stress the fluid is exposed to?

no flow = no deformation = no shear

What does Static Fluid mean?

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31 3

What does Static Fluid mean?

In this case, fluid can be exposed to only normal forces and behaves as

““a rigid bodya rigid body”” –– no deformationno deformation

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41 3

The Basic Equation of Fluid StaticsConsider the following fluid element:-

9/13/2008

5


51 3

The Basic Equation of Fluid StaticsConsider the following fluid element:-

Newton’s 2nd law:-

Divide both sides by gives:0. ==∑ admFd rr

∀d∑ ==

∀0.a

dFd rr

ρ

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61 3

The Basic Equation of Fluid Statics – Cont’d

Forces affecting on the fluid element = surface + body forces, i.e.,

Body Force =

or,

∑ += Bs FdFdFdrrr

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7


71 3

Surface Force in y- direction, dFys

dzdydxyp

dzdxdyyppdy

yppFd yS

..

.22

∂∂

−=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−=r

Surface force in y-direction =

9/13/2008

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81 3

Surface Force, dFs

Combining the other two directions, we get:

Where, ∇p = gradient of p.

dzdydxp

dzdydxkzpj

ypi

xpFd S

...

..ˆˆˆ

−∇=

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

−=r

9/13/2008

9


91 3

Basic Equation of Fluid Statics

∑ ==∀

0.adFd rr

ρ

∑ += Bs FdFdFdrrr

dzdydxpFd S ...−∇=r

(2)

(1)

From (2) in (1):


Recall – Newton’s 2nd Law:-

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101 3


Note that equation above is not really one equation, it is rather three equationsthree equations in the three directions x, y, and z.

0,0,0 =+∂∂

−=+∂∂

−=+∂∂

− zyx gzpg

ypg

xp ρρρ

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11


111 3

Knowing that

Assuming that z is the vertical direction, we can say that gx = gy = 0, gz = - g, the three equations:


kgjgigg zyxˆˆˆ ++=

r

0,0,0 =+∂∂

−=+∂∂

−=+∂∂

− zyx gzpg

ypg

xp ρρρ

9/13/2008

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121 3

Can be written as :-

i.e.,


gzp

yp

xp ρ−=

∂∂

=∂∂

=∂∂ ,0,0

γρ −=−= gdzdp

i.e., pressure is not function of x i.e., pressure is not

function of y

9/13/2008

13


131 3

Important Restrictions:-

In order to use the following equation, three three conditions must be satisfiedconditions must be satisfied.

1. Fluid must be static, i.e., velocity = acceleration = 0.

2. Gravity is the only body force.3. The Z axis is vertical and upward.

γρ −=−= gdzdp

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141 3

Types of Pressures

Fig. 3.2Pabs = Patm + Pgage

9/13/2008

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151 3

Pressure Variation in a Static Fluid

gdzdp ρ−=

∫ ∫−=p

op

z

ozdzgdp ρ

For Incompressible Fluid: Manometers

9/13/2008

16


161 3

Pressure Variation in a Static Fluid

∫ ∫−=p

op

z

ozdzgdp ρ

( ) ( )hgpp

zzgzzgpp

o

ooo

ρρρ

+=−=−−=− or,

For Incompressible Fluid: Manometers

9/13/2008

17


171 3

Simple Rules to Analyze Multiple-Liquid Manometer Problems

1. Any two points at the same elevation in a continuous in a continuous volume of the same liquidvolume of the same liquid are at the same pressure.

2. Pressure increases as one goes downincreases as one goes down a liquid column.

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181 3

Example- ProblemDetermine the gage pressure in psig at point ““aa””, if liquid A has SG = 0.75 and liquid B has SG = 1.20. The liquid surrounding point ““aa”” is water and the tank on the left is open to the atmosphere.

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191 3

Hydrostatic Force on Submerged Surfaces.

This equation allows us to determine how pressure varies in a static fluid.

We would like to determine the force due to that pressure on a surface submerged in a liquid.

hgpp o ρ+=

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201 3

Hydrostatic Force on Submerged Surfaces – Cont’d

In order to fully determine the force on a surface submerged in a liquid, we must determine the following:-

1. The magnitude of the force;

2. The direction of the force; and

3. The line of action of the force.

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211 3

1- Direction of the Force on a Plane Submerged SurfaceSince fluid is not moving (static), there is no shear, i.e., only normal forces might exist.

Since this force is caused by pressure of fluid, it will always be normal to the surfacenormal to the surface.

This determines the direction of the force.

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221 3

2- Magnitude of the Force on a Plane SubmergedSurface

dAPdF .=

hgPP o ρ+=

( )∫ ∫ +==∴A A

oR dAhgPdFF ρ

(1)

(2)

From (2) in (1)

(3)

9/13/2008

23


231 3

2- Magnitude of the Force on a Plane Submerged Surface

( )∫ ∫ +==∴A A

oR dAhgPdFF ρ

∫+=A

oR dAygAPF .sinθρ

θsinyh =

(3)

but,

(4)

From (4) in (3)

Where, Po is the pressure at the fluid surface, A is the surface area.

Note how Note how θθ is measuredis measured

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241 3

3- Line of Action of the Force on a Plane Submerged Surface

∫ ∫==∴A A

R ydAPydFyF .... '

( ) dAygyPyFA

oR .sin.. 2' ∫ +=∴ θρ

hgPP o ρ+=

θsinyh =

but,

and

9/13/2008

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251 3

3- Line of Action of the Force on a Plane Submerged Surface

dAygF

dAyPF

yA AR

oR

.sin.1...1 2' ∫ ∫+=∴ θρ

Solving for y’

( ) dAygyPyFA

oR .sin.. 2' ∫ +=∴ θρ

9/13/2008

26


261 3

Hydrostatic Force on Submerged Surfaces – Cont’d

1. The magnitude of the force;

2. The direction of the force = normal to the surface.normal to the surface.3. The line of action of the force.

∫+=A

oR dAygAPF .sinθρ

dAygF

dAyPF

yA AR

oR

.sin.1...1 2' ∫ ∫+=∴ θρ

9/13/2008

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271 3

Example ProblemThe pressure in the air gap is 8000 Pa gage. The tank is cylindrical. Calculate the net hydrostatic force

(a) On the bottom of the tank; (b) On the cylindrical sidewall CC; (c) On the annular plane panel BB.

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281 3

Example ProblemGate AB is a homogeneous mass of 180 kg, 1.2 m wide into the paper, resting on smooth bottom B. All fluids are at 20°C. For what water depth h will the force at point B be zero? Assume specific gravity of Glycerin = 1.26.

9/13/2008

29


291 3

Vocabulary List

1. Static fluid 2. Manometer.3. Hydrostatic pressure.4. Gauge pressure.5. Vacuum.6. Hydrostatic force on a submerged surface.

10/1/2008

1


11 2 3

Outline Chapter 4- Basic Equations in Integral Form for a Control Volume.

Definition of a control volume.Dot product of Two Vectors.Volume and Mass Rate of Flow through a C.V.

Volume Flux.Mass Flux.Sign Convention.

Conservation of Mass.Special cases.

Example Problem.Extensive and Intensive Fluid Properties.Reynolds Transport Theorem.Momentum Equation.Sign Convention of Terms in the Momentum Equation.Types of forces.Example problem.

10/1/2008

2


21 2 3

Basic EquationsBasic Equations in Integral Formfor a Control Volume

Chapter 4

Conservation of Mass.Conservation of Mass.Conservation of Momentum Conservation of Momentum

10/1/2008

3


31 2 3

Why do we need Basic Equations in integral Form ?

In some applications we do not need do not need detailsdetails of the flow field, we rather need need some global valuessome global values, such as:-

1. Average velocity at a certain section.2. Force due to fluid flow.3. Mass or Volume flow rates.

10/1/2008

4


41 2 3

Examples:-

Given velocity at the exit section, fine mass flow rate and average velocity at the inlet section.

10/1/2008

5


51 2 3

Examples:-Find velocity V3 and force on the scale due to fluid flow.

10/1/2008

6


61 2 3

Examples:-

Find force on the 90° elbow.

10/1/2008

7


71 2 3

Examples:-

Find flow rate and force on the gate in the open and closed positions.

gate

10/1/2008

8


81 2 3

Q1 How do we obtain these equations?

A1 By applying conservation lawsconservation laws on a controlvolume.

Q2 Which conservation laws?A2 1- Conservation of mass.

2- Conservation of momentum.

Q3 What is a control volume?

Basic Equations in Integral Form

10/1/2008

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91 2 3

Definition of a Control Volume1. A control volume is an arbitrary volume in space

through which fluid flows. The geometric boundary of the control volume is called the control surfacecontrol surface.

2. The control surface may be real or imaginary.

Fig 3.1

Real surface

Imaginary surface

10/1/2008

10


101 2 3

Background - Dot Product of Two Vectors

Dot product of V and dA = V dA cos α = projection of V on dA.

Fig 4.3

α

α

10/1/2008

11


111 2 3

Sign Convention

dAVdAVAdV +== 0cos.rr

Fig 4.3

dAVdAVAdV −== 180cos.rr

10/1/2008

12


121 2 3

Volume Flow Rate through a C.V.

Volume flow rate = volume flux through dA= dQ = V cosα . dA = V dA cos α

Total volume flow rate through control surface (CS)=

AdVrr

.=

∫∫ ==CSCS

AdVQdQrr

&& .

10/1/2008

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131 2 3

Mass Flow Rate through a C.V.

Mass flow rate = Mass flux through dA= ρ dQ = dm

Total mass flow rate through control surface (CS)=

AdVrr

.ρ=

∫∫ ==CSCS

AdVmdmrr

&& .ρ

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141 2 3

Sign Convention

dAVAdV +==rr

.flux volume

Fig 4.3

dAVAdV −==rr

.flux volume

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151 2 3

Conservation of Mass of any System

M = mass of the system = constant.

Or, the time rate of change of the mass of the system = 0

0 i.e.,system

=⎟⎠⎞

dtdM

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161 2 3

Conservation of Mass of a Control Volume

Time rate of decreasedecrease of mass within the control volume = net mass outflowoutflow rate from the control volume.

Time rate of decreasedecrease of mass within the control volume =

net mass outflowoutflow rate from the control volume =

∫ ∀∂∂

−CV

dt

ρ

∫CS

AdVrr

.ρ

0. =+∀∂∂

=⎟⎠⎞

∫∫CSCVsystem

AdVdtdt

dM rrρρ

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171 2 3

Special Cases 1. Unsteady Incompressible Flow, i.e., ρ = c

0. integrals, outside take =+∀∂∂

∫∫CSCV

AdVdt

rrρρρ

0. ,by divide =+∀∂∂

∫∫CSCV

AdVdt

rrρ

C.V. theof volumethe but, =∀=∀∫CV

d

0.thus, =+∂∂∀

∫CS

AdVt

rr

0. =+∀∂∂

=⎟⎠⎞

∫∫CSCVsystem

AdVdtdt

dM rrρρ

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181 2 3

Special Cases 1. Unsteady Incompressible Flow, i.e., ρ = c

0.thus, =+∂∂∀

∫CS

AdVt

rr

0. =∫CS

AdVrr

thus,0 C.V., (fixed) deformable-non afor =∂∂∀

t

10/1/2008

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191 2 3

Special Cases

2. Unsteady Incompressible flow through a fixed C.V.

0. =+∀∂∂

∫∫CSCV

AdVdt

rrρρ

0. =+∀∂∂

=⎟⎠⎞

∫∫CSCVsystem

AdVdtdt

dM rrρρ

0. ,by devide =+∀∂∂

∫∫CSCV

AdVdt

rrρ

0or c, c, =∂∂∀

=∀=t

ρ

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201 2 3

Special Cases

2. Unsteady Incompressible flow through a fixed C.V.

0. =+∀∂∂

=⎟⎠⎞

∫∫CSCVsystem

AdVdtdt

dM rrρρ

0.thus, =+∂∂∀

∫CS

AdVt

rr

0. =∫CS

AdVrr thus,0but =

∂∂∀

t

10/1/2008

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211 2 3

Special Cases

3. Steady Compressible flow

(1) 0. =∫CS

AdVrr

ρ

0. =+∀∂∂

=⎟⎠⎞

∫∫CSCVsystem

AdVdtdt

dM rrρρ

∑∫ == mAdVAdVmdCS

&rrrr

& . thus,,.Recall, ρρ

c leCompressib ,0 Steady ≠⇒=∂∂∀

⇒ ρt

outinCS

mmAdVm &&rr

& === ∫∑ i.e., ,0. (1),in ρ

10/1/2008

22


221 2 3

Example Problem

Water flows steadily through a pipe of length L and radius R = 3 in. Calculate the uniform inlet velocity, U, if the velocity distribution across the outlet is given by:

ft/s. 10 and , 1 max2

2

max =⎥⎦

⎤⎢⎣

⎡−= u

Rruu

10/1/2008

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231 2 3

Solution

1. Water = incompressible fluid → ρ = constant.

2. Flowing steadily = steady flow, i.e.,

Conservation of mass equation:-

Under conditions 1 and 2, equation (1) can be

written as:

.0=∂∂t

(1) 0. =+∀∂∂

∫∫CSCV

AdVdt

rrρρ

0. =∫CS

AdVrr

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241 2 3

Extensive Vs Intensive Fluid Properties

An extensiveextensive property is one that depends on the size of the C.V. Example:- volume or mass.

An intensiveintensive property is one that does not depend on the size of the C.V.

Example:- Specific volume, specific enthalpy.

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251 2 3

Reynolds Transport TheoremLet N = any extensive property.

the intensive property corresponding to N.

For any C.V., Reynolds Transport Theorem is:

==mNη

(I) .∫∫ +∀∂∂

=⎟⎠⎞

CSCVsystemAdVd

tdtdN rr

ρηρη

10/1/2008

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261 2 3

Reynolds Transport Theorem

If N = mass, M ⇒ in (I),

Conservation of mass equation:

0. =+∀∂∂

=⎟⎠⎞

∫∫CSCVsystem

AdVdtdt

dM rrρρ

1==mNη

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271 2 3

Interpretation of each term in R.T.T.

time rate of change of any extensive property of the system.

time rate of change of N within the C.V.

mass of an element contained in the C.V.

amount of N in that element.

=∀∂∂∫

CVd

tρη

=⎟⎠⎞

systemdtdN

=∀dρη

=∀dρ

(I) .∫∫ +∀∂∂

=⎟⎠⎞

CSCVsystemAdVd

tdtdN rr

ρηρη

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281 2 3

the net rate of flux of N out through the C.S.

the rate of mass flux exiting dA = mass flowrate.

=AdVrr

.ρ

=∫CS

AdVrr

.ρη

Interpretation of each term in R.T.T.

(I) .∫∫ +∀∂∂

=⎟⎠⎞

CSCVsystemAdVd

tdtdN rr

ρηρη

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291 2 3

Momentum Equation for an Inertial (not accelerating) C.V.

VMPrr

momentumlinear ==

VMPrr

NLet == VMN r

==∴η

∫∫ +∀∂∂

=⎟⎠⎞

CSCVsystemAdVd

tdtdN rr

.ρηρη

∫∫ +∀∂∂

=⎟⎟⎠

⎞

CSCVsystem

AdVVdVtdt

Pd rrrrr

.ρρ

In R.T.T.

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301 2 3

Momentum Equation “Newton’s Second Law” for an Inertial C.V.

∑ ==dtPdFr

r momentum of change of rate time

∫∫∑ +∀∂∂

=∴CSCV

AdVVdVt

Frrrrr

.ρρ

∫∫ +∀∂∂

=⎟⎟⎠

⎞

CSCVsystem

AdVVdVtdt

Pd rrrrr

. R.T.T., from ρρ

10/1/2008

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311 2 3

Momentum Equation – meaning of each term

∫∫∑ +∀∂∂

=∴CSCV

AdVVdVt

Frrrrr

.ρρ

(A) = sum of all forces acting on the fixed C.V.

(B) = time rate of change of momentum inside the C.V.

(C) = net rate of flux of momentum out through the C.S.

(A) = (B) + (C)

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321 2 3

Types of Forces ∑∑ ∑ += BS FFFrrr

∫∑ −=⇒=CS

S dAPF sF pressure e.g., forces, surfacerr

∫∑ ∀=⇒=CV

B dgF ρBF gravity e.g., forces,body rr

∫∫∑ +∀∂∂

=CSCV

AdVVdVt

Frrrrr

.in sub ρρ

(I) .∫∫ +∀∂∂

=+CSCV

BS AdVVdVt

FFrrrrrr

ρρ

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331 2 3

Scalar Equations of the Momentum Equation

(I) .∫∫ +∀∂∂

=+CSCV

BS AdVVdVt

FFrrrrrr

ρρ

Equation (I) is a vector equation. Therefore, it may be written as three scalar component equations.

In Cartesian coordinates (x, y, z), equation (I) can be written as:

. :equationdiretion -z

. :equationdiretion -y

. :equationdiretion -x

∫∫

∫∫

∫∫

+∀∂∂

=+

+∀∂∂

=+

+∀∂∂

=+

CSCVBzsz

CSCVBysy

CSCVBxsx

AdVwdwt

FF

AdVvdvt

FF

AdVudut

FF

rr

rr

rr

ρρ

ρρ

ρρ

NoteNote NoteNote

NoteNote

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341 2 3

Sign Convention of Terms in the Momentum Equation.

(I) .∫∫ +∀∂∂

=+CSCV

BS AdVVdVt

FFrrrrrr

ρρ

F is positive if it is in the positive direction of the coordinate.

V is positive if it is in the positive direction of the coordinate.

Sign of depends on the relative directions ofand .

AdVrr

. Vr

Adr

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351 2 3

ProblemA jet of water issuing from a stationary nozzle at 15 m/s (Aj = 0.05 m2) strikes a turning vane mounted on a cart as shown. The vane turns the jet through angle θ = 50º. Determine the value of mass, M, required to hold the cart stationary. If the vane angle θ is adjustable, plot the mass, M, needed to hold the cart stationary versus θfor 0 ≤ θ ≤ 180°.

10/1/2008

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361 2 3

Solution Procedure for This Type of Problems

1. Identify your control volume, C.V.

2. Identify its C.S., and number important sections to consider.

3. Identify your coordinates and draw them.

4. Identify which equation (s) will be used to solve the problem.

5. Identify which assumptions you can make to simplify your equations.

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371 2 3

Solution Procedure for This Type of Problems

6. Common assumptions are:-a. Steady.b. Incompressible, i.e., density, ρ = constant .c. Uniform flow.d. Body force = 0.e. Effect of atmospheric pressure is negligible.

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381 2 3

Solution

1. Identify your control volume, C.V.2. Identify its C.S., and number important sections to

consider.

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391 2 3

Solution

1. Identify your coordinates and draw them.

10/1/2008

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401 2 3

Solution – Cont’d

4. Identify which equation (s) will be used to solve the problem.

. :equationdiretion -z

. :equationdiretion -y

. :equationdiretion -x

-:Equation Momentum

∫∫

∫∫

∫∫

+∀∂∂

=+

+∀∂∂

=+

+∀∂∂

=+

CSCVBzsz

CSCVBysy

CSCVBxsx

AdVwdwt

FF

AdVvdvt

FF

AdVudut

FF

rr

rr

rr

ρρ

ρρ

ρρ

Rx

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411 2 3


5. Identify which assumptions you can make to simplify your equations.

6. Common assumptions are:-a. Steady.b. Incompressible, i.e., density, ρ = constant .c. Uniform flow.d. Body force = 0.e. Effect of atmospheric pressure is negligible.

Rx

.∫∫ +∀∂∂

=+CSCV

Bxsx AdVudut

FFrr

ρρ

10/1/2008

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421 2 3

Rx


( )θρ cos1M2

−=g

AV

10/1/2008

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431 2 3

Vocabulary List

1. Basic equations (conservation laws).2. Control volume.3. Dot product of two vectors.4. Volume flux.5. Extensive Property.6. Intensive Property.7. Reynolds Transport Theorem

10/7/2008

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251 2 3

Exercise

Water flows steadily through a pipe of length L and radius R = 3 in. Calculate the uniform inlet velocity, U, if the velocity distribution across the outlet is given by:

ft/s. 10 and , 1 max2

2

max =⎥⎦

⎤⎢⎣

⎡−= u

Rruu

10/7/2008

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261 2 3

Solution

1. Water = incompressible fluid → ρ = constant.

2. Flowing steadily = steady flow, i.e.,

Conservation of mass equation:-

Under conditions 1 and 2, equation (1) can be

written as:

.0=∂∂t

(1) 0. =+∀∂∂

∫∫CSCV

AdVdt

rrρρ

0. =∫CS

AdVrr

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271 2 3

Solution – cont’d

(a) 0.. 0.21

=+⇒= ∫∫∫ AdVAdVAdVCS

rrrrrr

02)1(2. (a),in 0

2

2

max0

=−++− ∫∫RR

drrRrurdrU ππ

drrπdARr- u V 2 ,1 (2)at constant,UV (1)at 2

2

max =⎥⎦

⎤⎢⎣

⎡===

rr

Note the sign of the first and the second integrals.

10/7/2008

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281 2 3

Solution – cont’d

(b) 042

2.0

2

42

max2 =⎥

⎦

⎤⎢⎣

⎡−+−

R

RrruRU ππ

ft/s.052

:for U (b) Solving max .uU ==

02)1(2. (a),in 0

2

2

max0

=−+− ∫∫RR

drrRrurdrU ππ

10/7/2008

1


11 2 3

Outline Chapter 5- Differential Analysis of Fluid Motion.

Why differential.Region of interest.Conservation of Mass for an infinitesimal C.V. (fluid particle) -Special cases.Lagrangian and Eulerian descriptions of the motion of a fluid particle.Material, substantial, or particle derivative.Example problem.Particle acceleration.Types of motion of a fluid particle. Rotation and vorticity vectors.Fluid deformation:1. Angular deformation.2. Linear deformation.Differential momentum equation.

10/7/2008

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21 2 3

Differential Analysis of Fluid Motion

Why differential?

To obtain detailed knowledge, we must apply the basic equations of fluid motion in differential form.

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31 2 3

Region of interestSince we are interested in formulating differential equations, our analysis will be in terms of infinitesimal systems or infinitesimal control volumes (i.e., differential C.V.)

Fig 5.1

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41 2 3

Conservation of Mass for an infinitesimal C.V. (fluid particle)

(I) =Time rate of decreasedecrease of mass inside C.V. = (II) = net rate of mass flux (mass flow rate) outout through the C.S.

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51 2 3


(I) = Time rate of decreasedecrease of mass inside C.V. =

∀∂∂

−= dtρ dzdydx

t∂∂

−=ρ

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61 2 3

(II) = net rate of mass flux (mass flow rate) outout through the C.S. in the x-direction =

( ) dzdyudzdydxxuu ρρρ −⎟

⎠⎞

⎜⎝⎛

∂∂

+=)(

dzdydxxu

∂∂

+=)(ρ

10/7/2008

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71 2 3


(II) = net rate of mass flux (mass flow rate) outout through the C.S. in the three directions =

dzdydxzw

yv

xuII ⎥

⎦

⎤⎢⎣

⎡∂

∂+

∂∂

+∂

∂+=

)()()()( ρρρ

0)()()(=

∂∂

+∂

∂+

∂∂

+∂∂

∴zw

yv

xu

tρρρρ

dzdydxt

dt

I∂∂

−=∀∂∂

−=ρρ)(

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81 2 3

Conservation of Mass - Special Cases

1. Incompressible Flow:- ρ = constant

2. Steady Flow:-

0)()()(=

∂∂

+∂

∂+

∂∂

+∂∂

zw

yv

xu

tρρρρ

0=∂∂t

0 =∂∂

⇒tρ

0. .,.

0 0)()()(

=∇

=∂∂

+∂∂

+∂∂

⇒=∂

∂+

∂∂

+∂

∂

Vei

zw

yv

xu

zw

yv

xu

r

ρρρ

0 =∂∂

⇒tρ

0)(. .,. 0)()()(=∇⇒=

∂∂

+∂

∂+

∂∂ Vei

zw

yv

xu r

ρρρρ

⇐ Compressible , UnsteadyCompressible , Unsteady

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9


91 2 3

Example Problem

Which of the following sets of equations represent possible two-dimensional incompressible flow cases?

a)

b)

)2(2

23

222

yyxxvyxyxu

−+=

−+=

22

2

22

xyxyvyxxyu

+−=

+−=

10/7/2008

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101 2 3

Lagrangian and Eulerian Descriptions of the Motion of a Fluid Particle

1. Lagrangian Description:-In the Lagrangian description, any fluid property, “F”, is function of the position vector and time (t), i.e., X

r

),( tXFFr

=

dtXd Vr

r=∴ velocity,particle

2

2

on,accelerati particle anddt

XddtVd a

rrr

==

Xr

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111 2 3

Lagrangian and Eulerian Descriptions of the Motion of a Fluid Particle

2. Eulerian Description:-In the Elurian description, any fluid property, “F”, is function of the space coordinates x, y, z, and time (t), i.e., ),,,( tzyxFF =

). velocityparticlecertain a(not

direction - xin the velocity local=∂∂

∴tx

location) z y,certain x, a(at F of change local =∂∂

tF

EulerianEulerian description is used in most fluid mechanics problems.description is used in most fluid mechanics problems.

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121 2 3

ProblemThe temperature, T, in a long tunnel is known to vary approximately as:

where To , α, L, and τ are constants, and x is measured from the entrance of the tunnel.

A particle moves into the tunnel with a constant speed, U. Obtain an expression for the rate of change of temperature experienced by the particle.

)/2sin( τπα teTT Lx

o−

−=

? toequal change of rate required theIstT∂∂

10/7/2008

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131 2 3

Substantial Derivative

If we want to find the time rate of change of any fluid property,”F”, following a certain particlecertain particle and still use Eulerian descriction, we have to use what is called a “substantialsubstantial” or “particleparticle” or “materialmaterial””derivativederivative.

10/7/2008

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141 2 3


In the Eulerian description:-

If we are following a certain particle:-

),,,( tzyxFF =

dzzFdy

yFdx

xFdt

tFdF

∂∂

+∂∂

+∂∂

+∂∂

=

(1) dt

dt,by devidedtdz

zF

dtdy

yF

dtdx

xF

tFdF

∂∂

+∂∂

+∂∂

+∂∂

=

wdtdzv

dtdyu

dtdx

=== , ,

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151 2 3


(1) dt

dtdz

zF

dtdy

yF

dtdx

xF

tFdF

∂∂

+∂∂

+∂∂

+∂∂

=

wdtdzv

dtdyu

dtdx

=== , ,

FVtF

zFw

yFv

xFu

tFDF

∇+∂∂

=

∂∂

+∂∂

+∂∂

+∂∂

=

.DtDF or,

Dt (1)in

r

10/7/2008

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161 2 3

Substantial Derivative – meaning of each term

FVtF

zFw

yFv

xFu

tFDF

∇+∂∂

=

∂∂

+∂∂

+∂∂

+∂∂

=

.DtDF or,

Dt

r

particle.certain aby

dexperienceor seen F of change of rate time total=DtDF

F. of change of rate timelocal=∂∂

tF

particle. theofmotion the todue change of rate change of rate convective.

==∇FV

r

10/7/2008

17


171 2 3

ProblemThe temperature, T, in a long tunnel is known to vary approximately as:

where To , α, L, and τ are constants, and x is measured from the entrance of the tunnel.

A particle moves into the tunnel with a constant speed, U. Obtain an expression for the rate of change of temperature experienced by the particle.

)/2sin( τπα teTT Lx

o−

−=

10/7/2008

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181 2 3

Particle Acceleration

In the Eulerian description:- ),,,( tzyxVVrr

=

paDt

VD onaccelerati particle ==r

onaccelerati local=∂∂

tVr

VVtVa

zVw

yVv

xVu

tVVDa

p

p

rrr

rrrrr

∇+∂∂

=

∂∂

+∂∂

+∂∂

+∂∂

==∴

. or,

(I) Dt

Note:Note:-- Equation (I) is a vector equation, so three equations can be written in the three coordinates.

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191 2 3

Particle Acceleration

VVtVa

zVw

yVv

xVu

tVVDa

p

p

rrr

rrrrr

∇+∂∂

=

∂∂

+∂∂

+∂∂

+∂∂

==∴

. or,

(I) Dt

zuw

yuv

xuu

tu

DtDuapx ∂

∂+

∂∂

+∂∂

+∂∂

==∴ :direction-in x

zvw

yvv

xvu

tv

DtDvapy ∂

∂+

∂∂

+∂∂

+∂∂

==∴ :direction-yin

zww

ywv

xwu

tw

DtDwapz ∂

∂+

∂∂

+∂∂

+∂∂

==∴ :direction-zin

10/7/2008

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201 2 3

Types of Motion of A Fluid Particle (Kinematics)

1. Translational.2. Rotation.3. Linear Deformation.4. Angular Deformation.

Fig 5.5

10/7/2008

21


211 2 3

Deformation and Rotation of a Fluid Element

Rate of Deformation =

Rate of rotation = average rotational speed =

( )tΔΔ+Δ βα

( )tΔΔ−Δ βα

21

10/7/2008

22


221 2 3

Rotation of a Fluid Element

(1) ,yx Δ

Δ=Δ

ΔΔ

=Δξβηα

tyyututy

yuu ΔΔ

∂∂

=Δ−Δ⎟⎟⎠

⎞⎜⎜⎝

⎛Δ

∂∂

+=Δ ...ξ

txxvtvtx

xvv ΔΔ

∂∂

=Δ−Δ⎟⎠⎞

⎜⎝⎛ Δ

∂∂

+=Δ ...η

(2) , (1)in yu

xv

∂∂

=Δ∂∂

=Δ∴ βα

10/7/2008

23


231 2 3

Rotation Vector

tΔΔ−Δ

==βαω

21 velocity rotational Average z

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

=∴yu

xv

21 (2) from zω

ωz = component of the rotation vector about the z-axis.

⎟⎠⎞

⎜⎝⎛

∂∂

−∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

=xw

zu

zv

yw

21 ,

21

yx ωω

wvuzyx

kji

kωjωiω wyx ∂∂

∂∂

∂∂

==×∇=++==

ˆˆˆ

21V curl

21 V

21ˆˆectorrotation v

rr)rω

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24


241 2 3

Vorticity Vector

wvuzyx

kji

ω∂∂

∂∂

∂∂

=×∇===

ˆˆˆ

V2vectorvorticity rrr

ζ

wvuzyx

kji

kωjωiω wyx

∂∂

∂∂

∂∂

==

=×∇=++==

ˆˆˆ

21V curl

21

V21ˆˆectorrotation v

r

r)rω

10/7/2008

25


251 2 3

Fluid Deformation: (1) Angular Deformation

tΔΔ+Δ

=βα planey -in xn deformatioangular of Rate

, but,yu

xv

∂∂

=Δ∂∂

=Δ βα

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=∴yu

xvplaney -in xn deformatioangular of Rate

⎟⎠⎞

⎜⎝⎛

∂∂

+∂∂

=∴zu

xwplane z-in xn deformatioangular of Rate

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=∴zv

ywplane z-yin n deformatioangular of Rate

10/7/2008

26


261 2 3

Types of Motion of A Fluid Particle (Kinematics)

1. Translational.2. Rotation.3. Linear Deformation.4. Angular Deformation.

Fig 5.5

10/7/2008

27


271 2 3

Fluid Deformation: (2) Linear Deformation

tx

tutxΔ

∂∂

=Δ

Δ−Δ⎥⎦⎤

⎢⎣⎡ Δ

∂∂

+=

xu x

uudirection-in xn deformatioLinear

xudirection-in xn deformatiolinear of Rate∂∂

=

10/7/2008

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281 2 3

Fluid Deformation: (2) Linear Deformation

xudirection-in xn deformatiolinear of Rate∂∂

=

yv∂∂

=direction-yin n deformatiolinear of rate Similarly,

zw∂∂

=direction-zin n deformatiolinear of rate and

10/7/2008

29


291 2 3

Rate of Volume Deformation or Dilation

V

zw

yv

r.

xu

dilation or n deformatio volumeof Rate

∇=

∂∂

+∂∂

+∂∂

=

=

10/7/2008

30


301 2 3

0. .,.

0 0)()()(

=∇

=∂∂

+∂∂

+∂∂

⇒=∂

∂+

∂∂

+∂

∂

Vei

zw

yv

xu

zw

yv

xu

r

ρρρ

Recall - Conservation of Mass - Special Cases

Incompressible Flow:- ρ = constant

0)()()(=

∂∂

+∂

∂+

∂∂

+∂∂

zw

yv

xu

tρρρρ

0 =∂∂

⇒tρ

⇐⇐Compressible Compressible ⇐⇐ , Unsteady, Unsteady

i.e., for an incompressible fluid, rate of volume deformation = 0

10/7/2008

31


311 2 3

Differential Momentum Equation.

For an incompressible fluids with constant density and viscosity, momentum equations are:-

1) x–direction,

2) y-direction,

3) z-direction,

10/20/2008

1


11 2

Outline Chapter 6- Incompressible Inviscid Flow

Momentum equation - special cases:Incompressible flow with constant viscosity.Inviscid (frictionless flow), i.e., μ = 0 - Euler’s equation.

Euler’s equations along a streamline.Bernoulli’s Equation. Hydrostatic, Static, Dynamic, and Stagnation Pressures.Applications of Bernoulli’s equation.Energy Grade Line (EGL) and Hydraulic Grade Line (HGL).Example problem.

10/20/2008

2


21 2

Differential Momentum Equation.

For an incompressible fluids with constant density and viscosity, momentum equations are:-

1) x–direction,

2) y-direction,

3) z-direction,

10/20/2008

3


31 2

Incompressible Inviscid Flow

Incompressible means ρ = constant.

Inviscid means that viscosity μ = 0.

10/20/2008

4


41 2

Differential Momentum Equation For an incompressible fluid (incompressible fluid (ρρ = c)= c) with constant viscosity, momentum equation is:-

1) x–direction,

2) y-direction,

3) z-direction,

VpgDtVD rr

2∇+∇−= μρρOr,

Inviscid ⇒ μ = 0

10/20/2008

5


51 2

Euler’s EquationFor an incompressible (incompressible (ρρ = c= c)) , inviscid flow inviscid flow ((μμ = 0)= 0), momentum equation is:-

1) x–direction,

2) y-direction,

3) z-direction,

Or,pg

DtVD

∇−= ρρr

10/20/2008

6


61 2

Euler’s Equation in Streamline Coordinates

Euler’s equations shown in the previous slide are written using x-y-z coordinates.

In steady flow a fluid particle will move along a streamline because, for a steady flow, pathlinesand streamlines coincide.

Thus, in describing the motion of a fluid particle in a steady flow, the distance along a streamline is a logical coordinate to use in writing the equations of motion.

10/20/2008

7


71 2

Fluid Particle Moving along a Streamline

Fig 6.1

10/20/2008

8


81 2

Euler’s Equation in Streamline Coordinates1. Apply Newton’s 2nd Law in the s-direction

∑ =+= sBS amFFF

saddgdxdndssppdxdnds

spp ∀=∀−⎥⎦

⎤⎢⎣⎡

∂∂

+−⎥⎦⎤

⎢⎣⎡

∂∂

− ρβρ sin.2

.2

10/20/2008

9


91 2

Euler’s Equation in S-direction

saddgdxdndssppdxdnds

spp ∀=∀−⎥⎦

⎤⎢⎣⎡

∂∂

+−⎥⎦⎤

⎢⎣⎡

∂∂

− ρβρ sin.2

.2

sVV

tV

DtVDa

szg

sp

s ∂∂

+∂∂

===∂∂

−∂∂

−r

rrr

ρ1 (1)in

dsdz

=βsin but,

(1)

If we neglect the body force and consider only steady flow:-

1 sVV

sp

∂∂

=∂∂

−r

r

ρ

10/20/2008

10


101 2

10/20/2008

11


111 2

Euler’s Equation in Streamline Coordinates2. Apply Newton’s 2nd Law in the n-direction

∑ =+= nBS amFFF

naddgdxdsdnnppdxdsdn

npp ∀=∀−⎥⎦

⎤⎢⎣⎡

∂∂

+−⎥⎦⎤

⎢⎣⎡

∂∂

− ρβρ cos.2

.2

10/20/2008

12


121 2

Euler’s Equation in n-direction

RVa

nzg

np

n

21 (2)in r

−==∂∂

−∂∂

−ρ

dndz

=βcos but,

(2)

an = centripetal acceleration, R = radius of curvature.If we neglect the body force and consider only steady flow:-

1 2

RV

np

r

=∂∂

ρ

naddgdxdsdnnppdxdsdn

npp ∀=∀−⎥⎦

⎤⎢⎣⎡

∂∂

+−⎥⎦⎤

⎢⎣⎡

∂∂

− ρβρ cos.2

.2

10/20/2008

13


131 2

Euler’s Equation in n-direction

This equation indicates that pressure increases in the direction outwards from the center of curvature of the streamline.

In case of flow in a straight line:-

i.e., there is no variation in pressure in the n-direction.

1 2

RV

np

r

=∂∂

ρ

0 =∂∂

∴∞=npR

10/20/2008

14


141 2

Measurement of Static Pressure

1 2

RV

np

r

=∂∂

ρ

0 =∂∂

∴∞=npR

10/20/2008

15


151 2

Measurement of Static Pressure

Can we put the pressure gage at the elbow?Can we put the pressure gage at the elbow?

1 2

RV

np

r

=∂∂

ρ

10/20/2008

16


161 2

Bernoulli’s Equation

Bernoulli’s equation results from the integration of Euler’s equation along a streamline for a steady flow.

Euler’s equation in s-direction (along a streamline):-

For steady flow:-

sVV

tV

szg

sp

∂∂

+∂∂

=∂∂

−∂∂

−r

rr

ρ1

(3) 1sVV

szg

sp

∂∂

=∂∂

−∂∂

−r

r

ρ

10/20/2008

17


171 2

Bernoulli’s Equation – Cont’dSince we are moving a long a streamline, thus,

p=p(s,t)ds

spdt

tpds

spdp

∂∂

=∂∂

+∂∂

=∴

dsszdt

tzds

szdz

∂∂

=∂∂

+∂∂

=and,

dssVdt

tVds

sVVd

∂∂

=∂∂

+∂∂

=rrr

rand,

(3) 1 (3) RecallsVV

szg

sp

∂∂

=∂∂

−∂∂

−r

r

ρ

10/20/2008

18


181 2

Integration of (4) along S:-

For an incompressible fluid ρ = c, in (5):

( )4 0 or,

1 (3)in

=++

=−−

ρ

ρdpdzgVdV

VdVdzgdp

rr

rr

( )5 constant 2

2=++ ∫ ρ

dpzgVr

constant2

2=++

ρpzgV

r⇐ Bernoulli’s equation

10/20/2008

19


191 2

Restrictions of The Application of Bernoulli’s Equation

Bernoulli’s equation is a very powerful tool, but is has to be

used very carefullyvery carefully. Because it has very strict applicability

limitations.

1. Incompressible flow, i.e., ρ = c, Mach number, Ma < 0.3.

2. Inviscid flow, i.e., μ = 0.

3. Steady flow, i.e.,

4. Along a streamline.

constant2

2=++

ρpzgV

r

0=∂∂t

10/20/2008

20


201 2

Hydrostatic, Static, Dynamic, and Stagnation Pressures1. Hydrostatic pressure is pressure resulting from weight

of a fluid column. P = ρ g h.

2. Ps=Static pressure is pressure due to the thermodynamic state of the fluid.

3. Pd=Dynamic pressure is pressure due to the velocity of the fluid.

4. Po=Stagnation pressure is pressure exerted by a moving fluid when brought from motion to rest.

Po=Ps + Pd

2

2VPd

rρ

=

10/20/2008

21


211 2

Measurement of Static Pressure, Ps

Fig 6.2

10/20/2008

22


221 2

Measurement of Stagnation Pressure

Po = Ps + Pd

10/20/2008

23


231 2

Simultaneous Measurement of Static and Stagnation Pressures

gzVP

gzVPo

oo ++=++22

sBernoulli’ From22rr

ρρ

2 z,z and 0

2

o

VPPV oo

rr

ρ+=∴==

10/20/2008

24


241 2

Applications of Bernoulli’s Equation

Bernoulli’s equation can be written for any two points along a same stream line as:

ρρ2

2

221

1

21

22pzgVpzgV

++=++rr

constant2

2=++

ρpzgV

r

10/20/2008

25


251 2

Measurement of Fluid Velocity at a Point

Apply Bernoulli’s equation between points A and B:

(1) 22

22

BBB

AAA zgVPzgVP

++=++rr

ρρ

o

B

Pzz

===

BB

A

P thus,0V and :Note

2

(1)in 2

ρρoAA PVP

=+r

( )ρ

AoA

PPV −=∴

2

10/20/2008

26


261 2

Applications – Nozzle Flow

ρρ2

2

221

1

21

22pzgVpzgV

++=++rr

One can relate info at section 2 to those at section 1 using:

10/20/2008

27


271 2

Flow through a Siphon One can relate info at sections 1, A, and 2 using:

ρρρ2

2

22

21

1

21

222pzgVpzgVpzgV A

AA ++=++=++

rrr

.21 :Note atmPPP ==

0 then,

areaarea Since

1

pipereservoir

≈

>>

VIf one uses Patm =0, this means that pressures are gage.If one uses Patm = 101.3 kPa, this means that pressures are absolute.

10/20/2008

28


281 2

Hzg

Vg

p=++

2 :gby devide

2r

ρ

Meaning of Each Term in Bernoulli’s Equation

constant2

2=++ zgVp

r

ρ

=∀

=gm

pg

pρ flow energy per unit weight of the flowing fluid=

head due to local static pressure.

2V

2V

22

==mg

m

g

rr

kinetic energy per unit weight = head due to localdynamic pressure.

==gmzgmz Potential energy per unit weight = head

due to elevation.H = total mechanical energy per unit weight

= total head of the flow

10/20/2008

29


291 2

Example ProblemA tank with a reentrant orifice called a Bordamouthpiece is shown. The fluid is inviscid and incompressible. The reentrant orifice essentially eliminates flow along the tank walls, so the pressure there is nearly hydrostatic. Calculate the contraction coefficient, Cc = Aj/Ao.

10/20/2008

30


301 2

Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)

EGL is a line representing the total head.

HGL is a line representing the sum of elevation heads and static pressure head.

The difference EGL-HGL = dynamic head.

Hg

Vg

pEGL

HGL

=++

4484476 r

3212

z 2

ρ

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At point (1) p1 = patm.=0 (gage) and V1 = 0, thus H1 = z1.

At point 4, p4=patm. = 0, thus the height of HGL = z4.

Hg

Vg

pEGL

HGL

=++

4484476 r

3212

z 2

ρ

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321 2


Flow through a constant cross section will have a horizontal HGL (bec. V = c).

Hg

Vg

pEGL

HGL

=++

4484476 r

3212

z 2

ρ

11/1/2008

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381 2

Example 1A tank with a reentrant orifice called a Bordamouthpiece is shown. The fluid is inviscid and incompressible. The reentrant orifice essentially eliminates flow along the tank walls, so the pressure there is nearly hydrostatic. Calculate the contraction coefficient, Cc = Aj/Ao.

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391 2

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3


401 2

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4


411 2

+

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5


421 2

11/1/2008

6


431 2

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441 2

21 ==∴

o

jc A

AC

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451 2

Heavy loads can be moved with relative ease on air cushions by using a load pallet as shown. Air is supplied from the plenum through porous surface AB. It enters the gap vertically at uniform speed, q. Once in the gap, all air flows in the positive x direction (there is no flow in across the plane at x = 0) Assume air flow in the gap isincompressible and uniform at each cross section, with speed u(x) as shown in the enlarged view. Although the gap is narrow (h<<L), neglect frictional effects as a first approximation. Use a suitably chosen control volume to show that u(x) = qx/h in the gap. Calculate the acceleration of a fluid particle in the gap. Evaluate the pressure gradient dp/dx, and sketch the pressure distribution within the gap. Be sure to indicate the pressure at x=L.

Example 2

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461 2

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471 2

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481 2

11/17/2008

1

7ME3O04 –Reviewhttp://mech.mcmaster.ca/~hamedm/me3o04/

1 2

Water is admitted at the center pipe of the platform shown below at a rate of 1 m3/s and discharged into the air around the periphery. The upper circular plate in the figure is horizontal and is fixed in position to the ceiling. The lower annular plate is free to movevertically and is not supported by the pipe. The annular plate weighs 30 N, and the weight of water on it should be considered. If the distance, d, between the two plates is to be maintained at 3.5 cm, what is the total weight W that this platform can support?

11/17/2008

2

18ME3O04 –Reviewhttp://mech.mcmaster.ca/~hamedm/me3o04/

1 2

Water is discharged from a narrow slot in a 150 mm diameter pipe. The resulting horizontal two-dimensional jet is 1 m long and 15 mm thick, but of non-uniform velocity. The pressure at the inlet section is 30 kPag. Calculate (a) the volume flow rate at the inlet section and (b) the forces required at the coupling to hold the spray pipe in place. Neglect the mass of the pipe and the mass of water it contains.

Coupling

10/30/2008

1

ME3O04 - Chapter 7http://mech.mcmaster.ca/~hamedm/me3o04/

1

1 2

Outline Chapter 7- Dimensional Analysis and Similitude

Meaning of Similitude.Dimensionless numbers.Methods of dimensional analysis:-

1. Nondimensionalizing the basic differential equations.2. Using Buckingham PI Theorem.

Buckingham PI Theorem. Procedure to determine the PI groups (illustrative example - Drag force on a sphere).Significant Dimensionless Groups in Fluid Mechanics.Significant Dimensionless Groups in Fluid Mechanics.Flow Similarity and Model Studies.Scaling with Multiple Dependent Parameters.Scaling with Multiple Dependent Parameters.Example Problem

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2

1 2

Similitude

????

ghV 2

One of the important dimensionlessdimensionless

numbers in this problem =

mms

sm

ghV 12

2

22=

= dimensionless

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3

1 2

Similitude

????V2

2

22

1

21

ghV

ghV

=

21

1

222 V

hhV =

VVV 22 i.e., 12 ==2

12

2 2 VV =

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4

1 2

Roasting Time of a TurkeyWhat are the parameters affecting roasting time of a turkey (t) ?

1. Mass, m2. Density, ρ3. Thermal conductivity, k4. Specific heat, cp

Using dimensional analysis we can show that:-Or,

kmc

t p3 2

constant ρ

=

3 2mckt

p ρ=constant = dimensionless number

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5

1 2

Roasting Time of a Turkey

2 no. bird3 2

1 no. bird3 2 ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛=⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛

mckt

mckt

pp ρρ

So, if we know the cooking time of bird no.1, we can calculatethe cooking time of bird no. 2 from this equation.

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6

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Drag Force on a Sphere Drag force on a sphere, Fdepends on:-1. Diameter, D2. Fluid density, ρ3. Fluid viscosity, μ4. Fluid velocity, U

Dimensional analysis shows that this problem is governed by two dimensionless numbers :

f that And 22 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

μρVDρ

DVF

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7

1 2

Does the size of sphere (i.e., D) matter in using this figure?Does the type of fluid (i.e., ρ and μ) matter in using this figure?

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8

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Similitude

So, if we know which dimensionless parameters are important in a problem, all other similar problems can be dealt with easily, or we should say, similarlysimilarly.

The “QUESTION” now is how can we determine important “Dimensionless NumbersDimensionless Numbers” in any problem of interest?

Methods to do that:1. Nondimensionalizing basic differential equations.2. Using Buckingham PI Theorem.

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9

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Nondimensionalizing the Basic Differential EquationsExample: Steady, incompressible, two-dimensional

flow, of a Newtonian fluid, with constant viscosity.

Assumptions:

1. Steady

2. Incompressible

3. Two-dimensional

4. Newtonian fluid with constant μ.

0=∂∂

⇒t

c=⇒ ρ

0==∂∂

⇒ wz

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10

1 2

For an incompressible fluid with constant viscosity, momentum equations are:-

1) x–direction,

2) y-direction,

3) z-direction,

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11

1 2

Momentum Equations after Applying all Other Assumptions

(3) direction y

(2) direction x

2

2

2

2

2

2

2

2

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

+−∂∂

−=⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

−

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

+∂∂

−=⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

−

yv

xvg

yp

yvv

xvu

yu

xu

xp

yuv

xuu

μρρ

μρ

1) x–direction,

2) y-direction,

3) z-direction,

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12

1 2

Non- Dimensionalizing Equations 1 and 2

Non-dimensionalizing these equations means :

1. To divide all lengths by a reference length, L;

2. To divide all velocities by a reference velocity, V∞;

Note: dimensionless quantities are denoted with asterisks:

Lyy

Lxx == ** ,

∞∞

==Vvv

Vuu ** ,

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13

1 2

Non- Dimensionalizing Equations 1 and 2

3. To divide pressure by twice the dynamic head = ρV∞

2.

Note: dimensionless quantities are denoted with asterisks:

2*

∞

=Vpp

ρ

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14

1 2

(3)

:directiony Momentum,

(2)

:direction xMomentum,

2*

*2

2*

*2

2*

*

*

**

*

**

2*

*2

2*

*2

*

*

*

**

*

**

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

+−∂∂

−=∂∂

+∂∂

−

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

−

∞∞

∞

yv

xv

LVVgL

yp

yvv

xvu

yu

xu

LVxp

yuv

xuu

ρμ

ρμ

Equations in Dimensionless Form

number ReynoldsRe == ∞

μρ LV

number Froud2== ∞

LgVFr

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15

1 2

RecallRecall - Similitude

So, if we know which dimensionless parameters are important in a problem, all other similar problems can be dealt with easily, or we should say, similarly.

The “QUESTION” now is how can we determine these important “Dimensionless NumbersDimensionless Numbers” in any problem of interest?

Methods to do that:1. Nondimensionalizing the basic differential equations.2. Using Buckingham PI Theorem.

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16

1 2

Methods to find the Dimensionless Numbers Relevant to a Certain Problem of Interest

The method of non-dimensionalizing differential equations depends on knowing which equations to use, which is not always the case.

If we do not know the equations, we use the second method known as the ““Buckingham PI TheoremBuckingham PI Theorem”” .

Before we discuss the Buckingham PI Theorem, we need first to discuss two important concepts:-

1. Fundamental, or Independent, or Primary Dimensions.2. Dimensional homogeneity.

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17

1 2

Fundamental, or Independent, or Primary Dimensions

Dimensions of all physical quantities can be expressed in terms of a group of ““independent independent or primary primary dimensionsdimensions””.

These primary dimensions are: 1. Mass, M;2. Length, L;3. Time, t;4. Temperature, T.Examples:-

1. Velocity −=== tLtLV

,. 22

−=== tLtL

tVonAccelerati

22 ... −=== tLM

tLMmaForce

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18

1 2

Dimensional HomogeneityAny valid equation that relates physical quantities must be dimensionally homogeneousdimensionally homogeneous, i.e., each term in the equation must have the same dimensions.

Example: Newton’s Law of Viscosity

(1) yu∂∂

= μτ

LtM

LtLM

.1..

areaforce (1) of L.H.S. 222 ===τ

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19

1 2

Dimensional Homogeneity

(1) yu∂∂

= μτ

LtM

LtL

Lt

tLM

mN.sμ

.1... (1) of R.H.S. 2222 ==∴=Q

0y)u,,,(f i.e.,

,01

yu :as written becan (1)Equation

=

=−

∂∂

μτ

μ

τ

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20

1 2

Buckingham PI TheoremThis theorem states that: Given a relation among nn

parameters, q1, q2, …, qn, of the form:

The nn parameters may be grouped into (nn--mm) independent dimensionless groups or ratios (πparameters), expressible in functional form by:

where m m = rr = the minimum number of independent dimensions required to specify the dimensions of the n n parameters.

0),.....,,( 21 =nqqqg

0),.....,,( 21 =−mnG πππ

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21

1 2

Illustrative example: Drag force on a sphere.Step 1

List all dimensional parameters involved in the problem and determine n. Where n is the number of dimensional parameters.Drag force on a sphere, FD depends on:-1. Diameter, D2. Fluid density, ρ3. Fluid viscosity, μ4. Fluid velocity, U

Procedure to Determine the π Groups

0),,,,( =ρμUDFg D ∴ n = 5

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22

1 2

Procedure to Determine the π Groups –Cont’d

Step 2 Select a set of primary dimensions = M, L, t, TM, L, t, TStep 3 Construct the ““Dimensional MatrixDimensional Matrix”” by listing all

parameters in terms of the primary dimensions.

000000110231111

11001

−−−−−

TtLM

FD D U μ ρ

““Dimensional MatrixDimensional Matrix””33

11

1

212

,

,

,

−

−−

−

−−

==

==

==

=

==

LMLM

tLMtL

M

tLtLU

LD

tLMtL

MF

ρ

μ

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23

1 2


Step 4 Determine the rank of the dimensional matrix, r.where the rank, r, is the order of the largest non-zero determinant in the matrix.

000000110231111

11001

−−−−−

TtLM

FD D U μ ρ

∴ r = 3

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24

1 2

Procedure to Determine the π Groups –Cont’dStep 5 Select a set of r dimensional parameters that

includes all the primary dimensions used in step 3 (i.e., M, L, and t).

Since r = 3, then select [ U, D, ρ ]

These parameters are called the repeatingrepeating parameters.Note: It is common to select a velocitya velocity, a dimensiona dimension, and a a

fluid propertyfluid property.

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25

1 2

Procedure to Determine the π Groups –Cont’dStep 6 Set dimensional equations using the repeating

parameters and one of the other parameters, one-at-a-time.

Solving for the exponents:of t: -a -2 = 0 ⇒ a = -2of M: c + 1 = 0 ⇒ c = -1of L: a + b -3c + 1 = 0 ⇒ b = -2

...1 == Dcba FDU ρπ

221 DUFD

ρπ =∴ = drag coefficient

000 .. tLM=23 ... t

LMLML

tL c

ba

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

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26

1 2

Procedure to Determine the π Groups –Cont’dStep 6 Set dimensional equations using the repeating

parameters and one of the other parameters, one-at-a-time.

Solving for the exponents:of t: -a -1 = 0 ⇒ a = -1of M: c + 1 = 0 ⇒ c = -1of L: a + b -3c -1 = 0 ⇒ b = -1

...2 == μρπ cba DU

Re1

2 ==∴DUρ

μπ

000 .. tLM=tLM

LML

tL c

ba

.... 3 ⎟⎠

⎞⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

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27

1 2


Step 7 Check that each π group is dimensionless.

The use of Buckingham theorem in the problem of drag force on a sphere shows there are two two important dimensionless numbers in this problem, which are:

But how these two groups relate, needs to be determined experimentally.

22 DUFD

ρdrag coefficient =

μρ DUand, Reynolds number =

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28

1 2

Relationship between Drag Coefficient and Reynolds number

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29

1 2

Buckingham PI Theorem

Important points to note:-

Buckingham theorem allows us to determine:-1. The number of π groups (dimensionless numbers)

involved in the problem.2. The form of each one of these dimensionless

numbers (π’s).

But, it does not allow us to determine the form of the function, , which has to be determined experimentallyexperimentally.

0),.....,,( 21 =−mnG πππ

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1 2

Forces encountered in Fluid MechanicsForces encountered in flowing fluids include forces

due to:1. Inertia2. Viscosity3. Pressure4. Gravity5. Surface tension6. Compressibility.The ratio of any two forces will be dimensionless, and defines a significant dimensionless number in Fluid Mechanics.

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31

1 2

Dimensions of These Forces

1. Inertia force =

2. Viscous force =

3. Pressure force =

( ) , VLbut t ,.. 3 =⎟

⎠⎞

⎜⎝⎛=

tVLam ρ

LVLLVA

yuA μμμτ ==∂∂

= 2...

( ) 222

3 .m.a Thus, VLL

VL ρρ =⎟⎟⎠

⎞⎜⎜⎝

⎛=

2.. LpAp Δ=Δ

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32

1 2

Dimensions of These Forces –Cont’d4. Gravity force =

5. Surface tension force = , where, σ = surface tension = force per unit length.

6. Compressibility force = , where Ev is the compressibility modulus, or modulus of elasticity.

2.. LEAE vv =

L.σ

stressareaforce

==⎟⎠⎞⎜

⎝⎛

=

ρρddpEv

( ) gLgm .. 3ρ=

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33

1 2

Significant Dimensionless Groups in Fluid Mechanics

1.

2.

3.

4.

number Reynolds Reforce viscousforce inertia 22

====μ

ρμρ LV

LVVL

numberEuler Eu

21force inertia

force pressure2

22

2

==Δ

=Δ

=V

pVLLp

ρρ

( )222

3

22

number Froudforcegravity force inertia

==== FrgLV

gLVL

ρρ

numberWeber Weforce tension surface

force inertia 222

====σ

ρσ

ρ VLLVL

1.

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34

1 2

Significant Dimensionless Groups in Fluid Mechanics – Cont’d

5.

and C = speed of sound =

Note:- in case of incompressible flow, ρ = constant, i.e., dρ = 0. Thus, C = ∞ ⇒ Ma = 0.

222

2

22

forceility compressibforce inertia Ma

EV

EV

LEVL

vvv

====ρ

ρρ

numberMach Ma where, ====CV

ddpV

EV

v

ρρ

ρddp

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35

1 2

Significant Dimensionless Groups in Fluid Mechanics – Cont’d

6.

where, p = pressure in liquid stream.pv = vapor pressure of liquid.

Note:- As Ca ↓, the more likely cavitation to occur.

number Cavitation

21force inertia

force pressure2

22

2

==−

=Δ

= CaV

ppVLLp v

ρρ

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36

1 2

Flow Similarity and Model Studies

A model test must yield data that can be scaled to obtain information of interest on the full-scale prototype.

To be able to do that, there are conditions that have to be met to ensure similarity of model and prototype flow.

Model and prototype must have:-1. Geometrical similarity.2. Kinematic similarity.3. Dynamic similarity.

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37

1 2

Geometrical Similarity

Geometrical similarity requires that:-

1. The model and prototype be of the same shape.

2. All linear dimensions of the model related to corresponding dimensions of the prototype by a constant scale factorconstant scale factor.

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38

1 2

Kinematic SimilarityTwo flows are kinematically similar when the velocities at corresponding points are in the same direction and differ by a constant scale factor.

Thus two kinematically similar flows have streamlines related by a constant scale factor.

Since the boundaries form the boundary streamlines of the flow, flows that are kinematically similar must be geometrically similar.

i.e., geometrical similarity is a prerequisite for kinematic similarity.

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39

1 2

Dynamic Similarity

Dynamic similarity requires that Identical types of forces to be :-1. Parallel, and2. Related in magnitude by a constant scale.

For condition 1, dynamic similarity requires kinematic similarity, hence geometric similarity too.

For condition 2, each independent dimensionless group (force ratio) must have the same value in the model and the prototype.

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40

1 2

Example – Drag Force on a Sphere

),,,( ρμUDfF =

⎟⎟⎠

⎞⎜⎜⎝

⎛==

μρ

ρDVff

DUF

2222 (Re)Drag coefficient =

Dynamic similarity is achieved if we use a model sphere(could be smaller or bigger) and keep:-

prototypemodelprototypemodel i.e., ,ReRe ⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

μDVρ

μDVρ

prototype22

model22 and, ⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

DUF

DUF

ρρ

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41

1 2

Example – Drag Force on a Sphere –Cont’d

Equations (1) and (2) state that:-

1. We do not need to use the same fluid to test the model.

2. The resulted drag force in the model will not be equal to the drag force in the prototype. However, the drag coefficients are the same.

(1) prototypemodel⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛μ

DVρμ

DVρ

(2) prototype

22model

22 ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

DUF

DUF

ρρ

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42

1 2

Scaling with Multiple Dependent Parameters

In the example of drag force on a sphere, we were interested in one dependent parameterdependent parameter, which is the drag force.

In some practical applications there might be more than one dependent parameter of interest.

In such cases, dimensionless groups must be formed separately for each one of those dependent parameters.

Example – performance of a typical centrifugal pump.

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43

1 2

Centrifugal Pump or Fan

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44


44

1 2

Example for Scaling with Multiple Dependent Parameters

Performance of a typical centrifugal pump.

In this case, dependent parameters of interest are:-1. Pressure rise or head developed by the pump, h.2. Power input required to derive the pump, P.

We are interested to know how h and P depend on:-1. Volume flow rate, Q.2. Angular speed, ω.3. Impeller diameter, D.4. Fluid properties ρ and μ.

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45

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Performance of a Typical Centrifugal Pump

If we apply Buckingham Theorem, we can find out that:-

),,,,( ),,,,(

2

1

DQf PandDQfh

μρωμρω

==∴

⎟⎟⎠

⎞⎜⎜⎝

⎛=

μωρ

ωω

2

3122 , DDQf

Dh

⎟⎟⎠

⎞⎜⎜⎝

⎛=

μωρ

ωωρ

2

3253 , and, DDQf

DP

22 where,Dh

ω= head coefficient

=53 ,D

Pωρ

power coefficient

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46

1 2

Performance of a Typical Centrifugal Pump

If we apply Buckingham Theorem, we can find out that:-

),,,,( ),,,,(

2

1

DQf PandDQfh

μρωμρω

==∴

⎟⎟⎠

⎞⎜⎜⎝

⎛=

μωρ

ωω

2

3122 , DDQf

Dh

⎟⎟⎠

⎞⎜⎜⎝

⎛=

μωρ

ωωρ

2

3253 , and, DDQf

DP

=3DQ

ωflow coefficient,

( )μ

ρμωρ

μωρ DVDDD

≡=2

and is a form of Re number

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47

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Performance of a Typical Centrifugal Pump – Cont’d

From practice it has been found that viscous effects are relatively unimportant w.r.t. inertial effects.

So, we can exclude Re number, and thus:

⎟⎟⎠

⎞⎜⎜⎝

⎛= 3

'122 D

QfDh

ωω ⎟⎟⎠

⎞⎜⎜⎝

⎛= 3

'253 and,

DQf

DP

ωωρ

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48

1 2

Performance Curves of a Typical Centrifugal Pump

⎟⎟⎠

⎞⎜⎜⎝

⎛= 3

'122 D

QfDh

ωω

⎟⎟⎠

⎞⎜⎜⎝

⎛= 3

'253 D

QfD

Pωωρ

HorsepowerShaft Efficiency P

=Fig. 7.5

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49

1 2

Similarity in Pump Performance

Complete similarity in pump performance test would require:-

222

22

11

1

DQ

DQ

ωω=

52

322

25

13

11

1

DP

DP

ωρωρ=

22

22

22

12

1

1

Dh

Dh

ωω=

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50

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Example ProblemThe drag of an airfoil at zero angle of attack is a function of density, viscosity, and velocity, in addition to a length parameter.

A 1/10-scale model of an airfoil was tested in a wind tunnel at a Reynolds number of 5.5 X 106, based on chord length. Test conditions in the wind tunnel air stream were 15°C and 10 atmospheres absolute pressure. The prototype airfoil has a chord length of 2 m, and it is to be flown in air at standard conditions.

Determine the speed at which the wind tunnel model was tested, and the corresponding prototype speed.

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α = angle of attack.

α = 0

11/8/2008

1

1ME3O04 – Chapter 8http://mech.mcmaster.ca/~hamedm/me3o04/

1 2

Outline Chapter 8- Internal Incompressible Viscous FlowClassification of Continuum Fluid Mechanics.Internal Incompressible Viscous Flow.Flow Regimes – Laminar and Turbulent.Boundary Layer and meaning of fully developed flow. Fully developed, Laminar Flow between Infinite Parallel Plates.Flow in Pipes and Ducts:-1. Velocity Profiles in fully developed pipe flow.2. Turbulent velocity profiles in fully developed pipe flow -

“Power Law” .3. Calculation of head loss:-

a. Major losses.b. Minor losses.

11/8/2008

2


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Classification of Continuum Fluid Mechanics

To be coveredin 4th year

Chapters 8 and 9

444 3444 21 Chapter 6

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Internal Incompressible Viscous Flow

Internal means that the flow is completely bounded by solid surfaces.

Examples:- flow in nozzles, ducts, diffusers, pipes, etc.

Flow Regimes:- internal flows can be classified, based on the flow regime, into:-

1. Laminar flow: fluid flows in layers or laminas.

2. Turbulent flow: flow is characterized by high-frequency velocity fluctuations.

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a) Laminar Flow (Re < 2300). b) Turbulent Flow (Re > 2300).

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Osborne Reynolds Experiments

a) Laminar Flow (Re < 2300). b) Turbulent Flow (Re > 2300).

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Boundary LayerViscosity is responsible for what we call the no-slip condition.

Therefore, as fluid approaches a solid surface and due to the no-slip condition, a region of significant deformation, i.e., significant velocity gradient is formed.

This region is called the boundary layer.

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Entrance Regionthickness of boundary layer

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Fully Developed FlowWhen the boundary layer reaches its maximum thickness the velocity distribution in the direction of the velocity distribution in the direction of flow does not change anymoreflow does not change anymore, at which case, the flow is said to have been fully developedfully developed.

thickness of boundary layer

u(r)u i.e., ,0.,. ==∂∂

xuei x)u(r,u i.e., ,0.,. =≠

∂∂

xuei

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Entrance length = Development length, Lfor Flow in a Pipe.

1. For Laminar Flow:

e.g., at Re = 2300, L = 138 D. where D = pipe diameter

2. For Turbulent Flow: L = 80 D

Re06.0=DL

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Fully-Developed Laminar Flows

Common cases are:-

1. Fully developed laminar flow between infinite parallel plates –two cases:-

a. Both plates stationary.b. Upper plate moving with constant speed , U.

2. Fully developed laminar flow in a pipe.

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Fully Developed Laminar Flow Between Infinite Parallel Plates –upper plate moving with constant speed U.

Assumptions:-

1. Steady flow, i.e.,

2. Incompressible flow, i.e., ρ = constant.

3. Constant viscosity, i.e., μ = constant.

4. If L>> a, then flow is fully developed, i.e.,

where, L = length of plates and a = height of gap between plates.

5. Infinite plates means , i.e., flow is two-dimensional.

6. Neglect body forces in x and y directions.

0=∂∂t

0 u(y)uxu

=≡=∂∂

0=∂∂z

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Fully Developed Laminar Flow Between Infinite Parallel Plates – upper plate moving with constant speed U.

Boundary conditions:-

1. at y = 0, u = v = 0.

2. at y = a, u = U, v = 0.

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Analysis

For this type of flow, we would like to determine the following :-

1. Velocity distribution, 2. Pressure distribution.3. Shear stress distribution.4. Volume flow rate.5. Average velocity.6. Point of maximum velocity.

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Analysis – Differential Equations

1. Conservation of mass:-

Applying assumptions number 1, 5 , and 2:-

0)()()(=

∂∂

+∂

∂+

∂∂

+∂∂

zw

yv

xu

tρρρρ

0=∂∂

+∂∂

yv

xu 0 =

∂∂

⇒yv

constant

)(

=

=

⇒or

xf

v

)( any x at 0 Since xfvv ≠⇒=.everywhere 0constant ==∴ v

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Analysis – Differential Equations – cont’dFor an incompressible fluid (incompressible fluid (ρρ = c)= c) with constant viscosity, momentum equation is:-

Momentum equation in y - direction:-

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

+∂∂

++∂∂

−=⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

2

zv

yv

xvg

yp

zvw

yvv

xvu

tv

y μρρ

VpgDt

VD rr

2∇+∇−= μρρ

constant

)(0

=

=⇒=

∂∂

∴ orxp

pyp

Can not be constant. If it is constant, there will be no flow.

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Analysis – Differential Equations – cont’dMomentum equation in x-direction

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

+∂∂

++∂∂

−=⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

2

zu

yu

xug

xp

zuw

yuv

xuu

tu

x μρρ

(1) 2

2

yu

xp

∂∂

=∂∂

∴ μ

)( and )( Recall yuuxpp ==⇒

constant (1)equation of R.H.S. and or

f(y)

constant (1)equation of L.H.S. or

f(x)∴

)()( yfxf ≠

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(2) say constant, 12

2c

yu

xp

==∂∂

=∂∂

∴ μ

(3) . 211 cxcpcxp

+=⇒=∂∂

∴

2

1

at 0at

p L p xp px

====

Lp

Lpppc Δ

−=−

−==∴ 21112 c and

1 (3)in pxLpp +Δ

−=∴ ⇐ Pressure distribution

Pressure Distribution

Note that p is function of x.

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Velocity Distribution

Lpc

xp Δ

−==∂∂

1 but

(1) -Recall 2

2

yu

xp

∂∂

=∂∂ μ

L p- (1)in 2

2

μΔ

=∂∂yu

3CyL p- +

Δ=

∂∂

∴μy

u

U a u u

====

yat 00yat

( ) ( )22

21

2yya

xpu

aUyay

Lpy

aUu −

∂∂

−=−Δ

+=μμ

(4) CyL 2

p- u and 432 Cy ++

Δ=

μ

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Shear Stress Distribution

yu

yx ∂∂

= μτQ

( )yaLp

aU 2

2−

Δ+=∴ μτ

( )ya−Δ= 2y

L 2p-y

aU u and

μ

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Volume Flow Rate, Q

WdyyuAdVQa

A∫∫ ==0

.)(.rr

( ) dyWyayL

pyaUQ

a.

20

2∫ ⎥⎦

⎤⎢⎣

⎡−

Δ−=

μ

WL

apUaQ ⎥⎦

⎤⎢⎣

⎡ Δ+=∴

μ122

3

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Average Velocity

aWW

LapUa

AQuav

1122

3×⎥

⎦

⎤⎢⎣

⎡ Δ+==

μ

LapUuav μ122

2Δ+=∴

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Point of Maximum VelocityPoint of maximum velocity is defined by:-

0 when i.e., ,0 yx ==∂∂ τ

yu

( )pa

LUayyaLp

aU

Δ+=⇒=−

Δ+=

μμτ2

022

Note: - Point of maximum velocity is not at y =a/2.

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Velocity Distribution

Velocity distribution will change as pressure gradient changes.

( )yayxp

−∂∂

+= 2

21y

aU u

μ

Pressure decreases Pressure decreases to the right. to the right.

Pressure increases Pressure increases to the right, i.e., decreasesto the right, i.e., decreasesto the left. to the left.

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Velocity distribution will change if the upper plate is not moving

Pressure decreases Pressure decreases to the right. to the right.

( )yayxp

−∂∂

+= 2

21y

aU u

μ

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For an incompressible, InviscidInviscid (i.e., μ = 0) flow, Bernoulli’s equation can be applied between points 1 and 2 as:-

Flow in Pipes and Ducts

(1) constant 22 2

222

1

211 =++=++ zgVpzgVp

rr

ρρ

inviscid flow

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Due to friction, i.e., due to shear stress at the wall, we do not expect the right hand side of Bernoulli’s equation to remain constant for an incompressible viscid flow.

Flow in Pipes and Ducts

(2) 22 2

222

1

211

lThzgVpzgVp+++=++

rr

ρρ

viscid flow

where hlT = total energy loss per unit mass.

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Velocity Profiles for Fully Developed Pipe Flow

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Velocity Profile in Fully Developed Pipe Flow – Laminar Flow

1. Laminar Flow:-

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

∂∂

−==22

14

)(Rr

xpRruu

μ

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Velocity Profile in Fully Developed Pipe Flow – Turbulent Flow

2. Turbulent Flow:- velocity vector in case of fully developed turbulent flow in a pipe can be represented by:

time-mean velocity.u’ and v’ are fluctuating velocity components in x- and y-directions, respectively

jviuuV ˆˆ)( ′+′+=r

=u where,

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Velocity Profile in Fully Developed Pipe Flow – Turbulent FlowThe velocity profile for turbulent flow through a smooth pipe may be approximated by the empirical power-law

where, U = maximum velocity = velocity at the centerline.

for ReU > 2x104, n = -1.7+1.8 log ReU

ReU = Reynolds number calculated using U =

(1) 1/1 n

Rr

Uu

⎟⎠⎞

⎜⎝⎛ −=

μρ DU

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Energy Equation for Viscid Flow in Pipes

Equation (I) is not the known Bernoulli’s equation. This equation has two major differences:-

1. hlT = total energy loss per unit mass, which takes into account energy loss due to friction between points 1 and 2.

2. Kinetic energy coefficients α1 and α2, which allow us to use average velocities V1 and V2.

α = 2 in case of laminar flow, and = 1 in case of turbulent flow.

(I) 22 2

22

22

1

21

11

lThzgVpzgVp+++=++

rr

αρ

αρ

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Solution of Pipe Flow Problems

The energy conditions relating conditions at any two points 1 and 2 for a single-path pipe system (i.e., no branching) is:

where hlT = total energy loss per unit mass.

Δhpump= head caused by a pump=

α = kinetic energy coefficient = 2.0 for laminar flow, Re < 2300.= 1.0 for turbulent flow Re ≥ 2300.

(II) 2

h2 2

22

22

pump1

21

11

lThzgVpzgVp+++=Δ+++

rr

αρ

αρ

mWp pumppump

&

&==

Δ

rate flow Masspower Pump

ρ

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Calculation of Head Loss, hlT

hlT = total energy loss per unit mass due to friction, is calculated from:

where∑hl = sum of major losses due to frictional effects in fully

developed flow in constantconstant--area sectionsarea sections.

∑hlm=sum of minor losses due to changes in flow direction flow direction and cross section areaand cross section area, e.g., entrances, elbows, contractions, etc.

∑ ∑+= lmllT hhh

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Calculation of Major Losses, ∑hl

For each straight part of the pipe system

Where, D = pipe diameter, L = pipe length (length of straight part), f = friction coefficient, which can be calculated from:-1. Moody chart, or2. The following equations:

2

2av

lV

DLfh =

Re64 2300 Re i.e., flow,laminar for =< f

⎥⎦

⎤⎢⎣

⎡+−=≥ 5.0Re

51.27.3

/log0.21 2300 Re i.e., flow,Turbulent for f

Def

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Moody Chart – Fig 8.12

e = surface roughness

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Surface Roughness, e

Depends on pipe material – Table 8.1

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Calculation of Major Losses, ∑hlm

These are additional losses encountered, primarily as a result of flow separation (due to change in flow direction and/or change in cross section area) in pipe fittings.Depending on the type of fitting, minor losses are computed in one of two ways:-1. Using k = the loss coefficient, which is determined

experimentally.

2. Or, using Le = the equivalent length of straight pipe, which is also determined experimentally.

2

2av

lmVkh =

2

2ave

lmV

DLfh =

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Inlets and Exits – Table 8.2

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Sudden Enlargements and Contractions –Fig 8.14

Note the difference in the velocity to be used in each case

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Gradual Contractions – Nozzles

2

22av

lmVkh =

Value of k given in Table 8.3

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Gradual Enlargements - Diffusers

Where, Cp is the pressure recovery coefficient – from Fig 8.15. Cpi is the ideal pressure recovery coefficient.

AR = area ratio, to calculated as shown below

( )2

21av

ilmVCpCph −=

211

ARCpi −=

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Fig 8.15

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Bends – Fig 8.16

2

2ave

lmV

DLfh =

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Valves and Other Fittings – Table 8.4

2

2ave

lmV

DLfh =

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Pumps, Fans, and Blowers

ρpump

pumpp

hΔ

=Δ

QW

p pumppump &

&=Δ

mW

QW

h pumppumppump &

&

&

&==Δ∴

ρ

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Non-Circular Ducts

Instead of using D, we use the hydraulic diameter, Dh, defined by:-

Where, A = cross-section area and P = wetted perimeter.

For a rectangle with two sides a and b, A = a x b, P = 2(a+b), thus:

For a square a = b, thus:

PADh

4=

)(2

)(24

baba

babaDh +

=+

=

aDh =

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Example ProblemWater is pumped at the rate of 2 ft3/S from a reservoir 20 ft above a pump to a free discharge 90 ft above the pump. The pressure on the intake side of the pump is 5 psig and the pressure on the discharge side is 50 psig. All pipes are commercial steel of 6 in. diameter. Determine (a) the head supplied by the pump and (b) the total head loss between the pump and point of free discharge.

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Outline Chapter 9- External Incompressible Viscous Flow

The Concept of Boundary LayerEffect of Boundary Layer on a Blunt BodyEffect of Boundary Layer on a Streamlined Body

Boundary Layer Thicknesses:1. Disturbance Thickness, δ992. Displacement Thickness, δ*Example Problem.Fluid Flow about Immersed BodiesDrag and Lift.Types of Drag.CD and CL.

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External Flows

External flows are flows over bodies immersed in an unbounded fluid.

Examples of external flows are the flow fields around such objects as airfoils, automobiles, and airplanes.

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Definition of Boundary Layer

The boundary layer is the region adjacent to a solid surface in which viscous stresses are present.

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Boundary Layer Thicknesses

1. Disturbance Thickness, δ99

2. Displacement Thickness, δ*

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Disturbance Thickness, δ99

δ99 is the boundary layer thickness at which u equals to 99% of the free stream velocity, U.

In other words, it is the distance from the surface at which the velocity is within 1 % of the free stream

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Displacement Thicknesses , δ*

What do we do to make both mass flow rates equal?

h h

∫=h

AdU0

flow massr

ρ ∫=h

Adu0

flow massr

ρ>

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Raise the plate a distance δ* so that mass flow rates are equal.The displacement thickness, δ*, is the distance the plate would be moved so that the loss of mass flux (due to reduction in uniform flow area) is equivalent to the loss the boundary layer causes.

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∫ ∫∞ ∞

==0 0

W.dy UU B.L noflux with mass ρρ Adr

∫ ∫∞ ∞

==0 0

W.dy B.Lflux with mass uAdu ρρr

( ) ( ) (1) W.dy flux massin difference0 0∫ ∫∞ ∞

−=−= uUAduU ρρr

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(2)W flux massin difference *δρ U=

∫∫ ⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −=∴=

∞ δ

δ00

* dy 1dy 1 (2)(1) fromUu

Uu

Note: W is depth normal to paper.

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The Use of Displacement Thicknesses , δ*, in Practical Applications

One can use Bernoulli’s equation to design or determine the pressure drop in a duct by reducing duct dimensions by 2δ* as shown above.

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Example Problem

Laboratory wind tunnels have test sections 1 ft square and 2 ft long. With nominal air speed U1 = 80 ft/s at the test section inlet, turbulent boundary layers form on the top, bottom, and side walls of the tunnel. The boundary-layer thickness is δ1 = 0.8 in. at the inlet and δ2 = 1.2 in. at the outlet from the test section. The boundary-layer velocity profiles are of power-law form, with:

a) Evaluate the freestream velocity, U2, at the exit from the wind-tunnel test section.b) Determine the change in static pressure along the test section.

7/1

⎟⎠⎞

⎜⎝⎛=δy

Uu

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Fluid Flow about Immersed Bodies

Source: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001.

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Forces from the Surrounding Fluid on a Two-Dimensional Object

(a) Pressure force (b) Viscous (shear or friction) force

Source: Fundamentals of Fluid Mechanics by B. R. Munson et. al., Wiley, 1994.

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Resultant Force

Source: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001.

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Drag and Lift

Source of fig: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001.

Net force F is resolved into:-1. The drag force, FD, defined as the component of the force parallel to

the direction of motion, and

2. The lift force, FL, defined as the component of the force perpendicular to the direction of motion

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Types of Drag1. Pressure or Form Drag:

This type of drag is due to pressure difference in front of and at the back of the object. The formation of a low pressure wake behind the object depends on the shape or “formform” of the object. This is why this type of drag is called “form” drag.

In case of a streamlined object, the total drag will be due to friction, as in (a). In (b), due to this large wake (region of low pressure), form drag will be significant.

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171 2

Types of Drag

2. Friction or Skin Friction Drag:

This type of drag results from viscous (shear) stresses at the surface of the object.

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Drag Coefficient

with

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191 2

Flow over a Flat Plate Parallel to the Flow: only friction drag, no pressure drag

Boundary Layer can be 100% laminar, partly laminar and partly turbulent, or essentially 100% turbulent; hence several different drag coefficients are available

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201 2

Flow over a Flat Plate Parallel to the Flow: Only Friction Drag

Turbulent BL:

Laminar BL:

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211 2

Drag Coefficient for a Smooth Flat Plate –Fig 9.8

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221 2

Flow over a Flat Plate Perpendicular to the Flow: Pressure Drag, No Friction Drag.

Drag coefficients are usually obtained empirically

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231 2

Flow over a Flat Plate Perpendicular to the Flow: Pressure Drag (Continued)

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241 2

Flow over a Sphere: Friction and Pressure Drag

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251 2

Drag Coefficient - Typical Values

Source: Principles of Fluid Mechanics by A. Alexandrou, Prentice Hall, 2001.

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261 2

Effect of flow Regime on Form or Pressure Drag - flow over a cylinder

Source: Fluid Mechanics by F.M. White, Wiley, 2003.

< 1.0

> 1.0

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271 2

StreamliningUsed to Reduce Wake and hence reduce pressure drag

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281 2

Important Note Regarding the Area, A

1. In case of pure friction drag, the area, A, is the total surface area in contact with the fluid (i.e., the wetted the wetted areaarea).

2. In case of pure pressure drag, the area A is the frontal frontal area or projected areaarea or projected area of the object.

3. For combined cases, drag coefficient for flow over an immersed object usually is based on the frontal area or frontal area or projected area of the objectprojected area of the object, except for airfoils and wings.

4. for airfoils and wings, use the planformplanform area.area.

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291 2

Airfoils and Wings Planform Area

Planform area is the maximum projected area of the wing.Source: Fluid Mechanics by F.M. White, Wiley, 2003.

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301 2

How de calculate Lift? Lift Coefficient, CL

Note: Ap is the planform area = maximum projected area.

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311 2

CL is Function of Re and Angle of Attack, αExamples: NACA 23015; NACA 662-215

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321 2

Minimum Flight Speed, Vmin

At steady-state flight conditions, lift force, FL, must be equal to aircraft weight, W, thus:

V = Vmin = Minimum flight speed when CL=CLmax

The question is how to maximize CL?

AVCWF LL2

21 ρ==

ACWV

Lmaxmin

2ρ

=

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Increase of CL using Winglets (Flaps)

Figure 9.23

mcmaster mech eng 3o04 - fluid mechanics notes by dr. mohamed s. hamed

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