lecture 1 (fluid mechanics)

Monash University Sunway campus is jointly owned by Monash University and the Jeffrey Cheah Foundation

MEC 3451

Fluid Mechanics 2

School of Engineering

Week 1

Semester 1, 2014


2

Course Objectives

MEC 3451 Fluid Mechanics 2

o Derive the conservation equations governing fluid flows using finite control volume

and differential analysis

o Apply these governing equations to solve simple potential flow and viscous flow

problems

o Analyse internal flow, external flow and open channel flow problems

o Examine how boundary layers affect the behaviour of a fluid close to a surface

o Calculate lift and drag effects on a body

o Understand the concept of turbulence

o Appreciate compressibility effects in fluids and apply simple techniques to analyse

such flows


Preliminaries

3


o Course delivery

• 33 Lectures (3 X 50 minute lectures per week)

• Practice Classes (1 X 3 hour session per week:

Check Allocate)

o Assessment

• Terminal examination (3 hours, 70%)

• Tests (30%)

o Course Text

• Munson, Young, and Okiishi, Fundamendal of

Fluid Mechanics, 6th Edition, John Wiley and

Sons

o Extra Help

• Moodle

• Tutors

• Consultation Hours

• Wednesday: 3PM – 5PM

• Thursday: 11AM – 12PM


4

Lecture Topics


Topic Textbook

(Munson et. al.)

Fluid Kinematics Chapter 4

Finite Control Volume Analysis Chapter 5

Differential Analysis Chapter 6

Similitude Chapter 7

Internal Flows Chapter 8

External Flows Chapter 9

Open Channel Flow Chapter 10

Compressible Flows Chapter 11


5

Tutorials (Tutors: Mr. Ang & Mr. Ooi)


o Practice classes (tutorials) held weekly starting Week 2

• Attendance at practice classes is compulsory

Test 1 (Assessment Task 1, 10%) o Approximately 1.5 to 2 hours

o 25th March, Tuesday (Week 4) • Materials covered in Week 1 to Week 3

• Approximately 1/3 of the marks on problems from tutorial sheets

• Approximately 1/10 of the marks on self-study topics


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Test 2 (Assessment Task 2, 10%)

Test 3 (Assessment Task 3, 10%)

o Approximately 1.5 to 2 hours

o 15th April, Tuesday (Week 7) • Materials covered in Week 4 to Week 6



o Approximately 1.5 to 2 hours

o 14th May, Wednesday (Week 10) • Materials covered in Week 7 to Week 9




Introduction:

Fluid Mechanics

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8

Fluid Mechanics


o Study of the behaviour of fluids when

subject to applies forces

o Two subcategories

• Fluid statics: Behaviour of fluids at rest

• Fluid dynamics: Behaviour of fluids in motion

o Why study fluid mechanics?

• Fluids everywhere

✴ Everyday phenomenon

✴ Environmental flows

✴ Biological flows

✴ Medical devices

✴ Aerodynamics


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o What is a fluid?

• Substance which continuously deform

(strained) when subject to a shear stress

• Solids, although deforming initially, do not

do so continuously

o Generally consists of liquids and

gases

• A liquid takes the shape of

the container it is in and

forms a free surface in the

presence of gravity

• Liquid is difficult to

compress

• A gas expands until

it encounters the walls

of the container and

fills the entire

available space

• Gases cannot form a

free surface


10

Fluid Properties


o Different fluids flow differently • This is because different fluids have different characteristics (for example

water, oil, honey, tar, air)

o Quantification of these fluids therefore requires the definition of

fluid properties • Density, specific volume, specific gravity

• Bulk modulus of compression

• Vapour pressure

• Surface tension


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Density, Specific Volume, Specific Gravity


o Density • Mass 𝑚 per unit volume ∀

𝜌 ≡𝑚

∀

o Specific volume • Volume ∀ per unit mass 𝑚

∀ ≡∀

𝑚=1

𝜌

o Specific gravity (relative density) • Density relative to density of water at 4℃

SG ≡𝜌

𝜌H2O,40C

o Specific weight o Weight per unit volume

𝛾 ≡ 𝜌𝑔 To measure specific

gravity


12

Viscosity


o Recall definition of a fluid

• Substance which continuously deforms when

subject to a shear (tangential stress)

• Introduce concept of viscosity to describe the

‘fluidity’ of a fluid, i.e., how easily it flows

o Shear stress (force 𝐹 applied tangentially

to area 𝐴)

𝜏 =𝐹

𝐴∝𝑑𝑢

𝑑𝑦= 𝜇

𝑑𝑢

𝑑𝑦

(Shear Stress) = (Dynamic Viscosity) X (Rate of Strain)

• Constant of proportionality is the dynamic (or

absolute) viscosity


13


o Measure of a fluid’s resistance to deformation and hence flow

o Acts like friction between layers of fluid when they are forced to

move relative to each other

o Determine from slope of shear stress vs strain rate (deformation rate

or velocity gradient 𝑑𝑢 𝑑𝑦 )

• Linear for most common fluids (Newtonian)

• Non-Newtonian flows deal with deviations from linearity


14

Hydrostatics


o Pressure

• When fluid is at rest, the shear (tangential) stress is zero

• The only stress acting on the fluid is the pressure (force per unit area

acting normal to a surface)

• Scalar field

Pressure is the same at all points on a horizontal plane in a given fluid.

The length or the cross-sectional area of the tube has no effect on the height of the fluid column of a barometer.

𝑝𝐴1 = 𝑝𝐴2 = 𝑝𝐴3


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Hydrostatics


o Variation with depth

• Newton’s Second Law

𝛿𝐹𝑧 = 𝜌𝑔 𝛿𝑧 𝐴 = 𝑝𝐴 − 𝑝 +𝑑𝑝

𝑑𝑧𝛿𝑧 𝐴

𝑑𝑝

𝑑𝑧= −𝜌𝑔

• Negative sign: pressure increases with depth

• Integrate between two elevations 𝑧1 and 𝑧2 to get

𝑝 = 𝑝0 + 𝜌𝑔ℎ


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Spatial Flow Field Variation (1, 2, 3-D Flow)


o Flow can be exceedingly complex • A flow is either one-, two-, or three-dimensional depending on the number of

spatial components (𝑢𝑥 , 𝑢𝑦, 𝑢𝑧) in the velocity vector

𝑢 𝑥, 𝑦, 𝑧 = 𝑢𝑥𝐢 + 𝑢𝑦𝐣 + 𝑢𝑧𝐤

o However, it is possible to simplify the flow analysis • Flow between two flat plates

o For wide plates, negligible variation in the 𝑧 −direction

o Thin gaps, vertical velocity component 𝑢𝑦 is negligible

o Only one velocity component, i.e., 𝑢 = 𝑢(𝑦) needs to be considered

o Flow is one dimensional (although 𝑢𝑥 is a function of 𝑦)


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o Flow over an infinitely long cylinder (into the plane) • Negligible variation in the 𝑧 −direction

• Both 𝑢𝑥 and 𝑢𝑦 important since flow circumnavigates cylinder

• Flow is two-dimensional


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Spatial Flow Field Variation (Uniform/Non-uniform Flow)


o Uniform flow o A flow is uniform if the velocity does not vary along a streamline

• Flow between two plates

o An example of non-uniform flow is the flow over an aerofoil • The fluid accelerates on streamlines over the aerofoil and decelerates on

streamlines under the aerofoil to main flow conservation

• The velocity along a given streamline is therefore not constant


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Temporal Flow Field Variation (Steady/Unsteady Flow)


o Steady flow • Flow invariant in time • All fluid properties at any spatial position in the flow do

not change with time 𝑑 𝑑𝑡 = 0 • Example of unsteady flow (𝑑 𝑑𝑡 ≠ 0)

o Flow over an oscillating plate

o Flow through a diffuser channel with moving walls

Steady flow

Unsteady flow


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Laminar and Turbulent Flow


o Laminar flow • Flow is regular and highly ordered

o Each fluid layer moves smoothly and steadily with respect to the other layers (laminae) adjacent to it

o Deterministic system o Usually occur in viscous fluids where the velocity is low

o Turbulent flow • Flow is random and highly disordered

o Irregular and unsteady – characterised by velocity fluctuations o Chaotic movements of part of liquid in different directions superimposed on

main flow direction o All fluid properties at any spatial position in the flow field do not change with

time o System no longer deterministic

o Can only be described in term of statistical averages o Usually occurs in high velocity inviscid fluids


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Conservation Laws


o Mass conservation (continuity equation)

o Mass cannot be created or destroyed

o Steady flow: 𝜕 𝜕𝑡 = 0

𝑚 in = 𝑚 out

𝜌𝑉in𝐴in = 𝜌𝑉out𝐴out


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Conservation Laws


• Momentum conservation flow • Newton’s Second Law

Time rate of change of linear

momentum of a system

Sum of external forces

acting on the system =

𝑚 𝑉

out

− 𝑚 𝑉

in

= 𝐹sys


Control Volume Analysis (Review):

Reynolds Transport Theorem

25

Osborne Reynolds

(1842-1912)



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o Difficult to identify a fluid mass and track this for all times (Lagrangian description)

o Moreover, often not interested in a particular mass of fluid but rather the effect of

a flow in a structure or device

o Thus, helpful to formulate the fundamental equations of fluid flow for a finite

spatial region (geometric identity independent of mass), i.e., the control volume

(Eulerian description)

o Equations developed will be expressed in integral form

• Volume integrals are a convenient way to capture spatial variations in the fluid

properties

• These are related by the Reynolds Transport Theorem


CV is arbitrarily chosen, however, selection of CV can either simplify or

complicate analysis.

– Clearly define all boundaries. Analysis is often simplified if control surface CS

is normal to flow direction.

– Clearly identify forces and torques of interest acting on the CV and CS.

– Clearly identify all fluxes crossing the CS.

Choosing a Control Volume


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Let 𝐵 represents fluid parameters (mass, momentum, acceleration…..)

Let 𝑏 represents the amount of 𝐵 per unit mass (𝑚)

𝑏 =𝐵

𝑚

𝐵 = 𝑚 𝑏

Extensive Property (𝑩) Intensive Property (𝒃)

𝑚 1

1

2𝑚𝑉2

1

2𝑉2

Extensive and Intensive Property


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RTT - Physical Interpretation

The purpose of Reynolds transport theorem is to provide a link between control volume

ideas and system ideas.

A physical understanding of the concepts involved will show that it is a straightforward,

easy-to-use tool.

𝑑𝐵sys

𝑑𝑡 =

𝑑

𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝑏 𝜌 𝐕 ∙ 𝐧 𝑑𝐴

CSCV


29


𝑑𝐵sys

𝑑𝑡 =

𝑑

𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝜌 𝑏 𝐕 ∙ 𝐧

CSCV

𝐴

• The time rate of change of an arbitrary extensive parameter of a system

• This may represent the rate of change of mass, momentum, energy, or

angular momentum of the system, depending on the choice of the

parameter 𝐵.


30


𝑑𝐵sys

𝑑𝑡 =

𝑑

𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝜌 𝑏 𝐕 ∙ 𝐧

CS

𝐴CV

This term represents the rate of change of 𝐵 within the control volume as the

fluid flow through it.

Because the system is moving and the

control volume is stationary, the time rate

of change of the amount of 𝐵 within the

control volume is not necessarily equal to

that of the system.


31


𝑑𝐵sys

𝑑𝑡 =

𝑑

𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝜌 𝑏 𝐕 ∙ 𝐧 𝐴

CSCV

This term represents

the net flowrate of the

parameter 𝐵 across the

entire control surface.

Over this portion of the control surface this

property is being carried out of the control

volume (𝑉 ∙ 𝑛 > 0)

Over this portion of the control surface

this property is being carried into the

control volume (𝑉 ∙ 𝑛 < 0)

Over the remainder of the control

surface there is no transport of 𝐵

across the surface since 𝑏𝑉 ∙ 𝑛 = 0,

because either 𝑉 = 0 or 𝑉 is parallel

to the surface at those locations.


Reynolds Transport Theorem

Linear

Momentum

Equation

Moment (angular)

Momentum Equation

Continuity

Equation

The Energy Equation


33


The general form of the continuity equation (conservation of mass) is obtained by

substituting the properties for mass into the Reynolds transport theorem

𝑑𝐵sys

𝑑𝑡 =

𝑑

𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝑏 𝜌 𝐕 ∙ 𝐧 𝑑A

CSCV

Let 𝐵sys = 𝑚sys and 𝑏 = 𝑚sys/𝑚sys = 1 , resulting in

𝑑𝑚sys

𝑑𝑡=

𝑑

𝑑𝑡 𝜌 𝑑∀ + 𝜌 𝐕 ∙ 𝐧 𝑑A

CSCV

Control Volume Analysis:

Conservation of mass


34


𝑑𝑚sys

𝑑𝑡=

𝑑

𝑑𝑡 𝜌 𝑑∀ + 𝜌 𝐕 ∙ 𝐧 𝑑A

CSCV

However, conservation of mass

𝑑𝑚sys

𝑑𝑡= 0

so the general, or integral form of the continuity equation is

𝑑

𝑑𝑡 𝜌 𝑑∀ + 𝜌 𝐕 ∙ 𝐧 𝑑A = 0

CSCV

This equation can be expressed in words as

The accumulation rateof mass in thecontrol volume

+The net flowrateof mass throughthe control surface

= 0


35


For steady flow, the total amount of mass contained in CV is constant, that is,

total amount of mass entering must be equal to total amount of mass leaving,

𝑑

𝑑𝑡𝑚CV + 𝑚 𝑜 − 𝑚 𝑖

CSCS

= 0

For single-stream steady-flow systems,

𝑚 = 𝑚

outin

𝜌𝑖𝐴𝑖𝑉𝑖 in = 𝜌𝑖𝐴𝑖𝑉𝑖 out

𝑖𝑖

Conservation of mass:

Steady flow processes (𝝏 𝝏𝒕 → 𝟎 )


36


For incompressible flows (𝜌 = constant),

The 𝑄𝑖 = 𝑉𝑖𝐴𝑖 is called the volume flow passing through the given cross

section. The volume flow 𝑄 = 𝑉𝐴 will have units of cubic meters per

second (m3/s).

Conservation of mass:

Incompressible flows (𝝆 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭)

𝐴𝑖𝑉𝑖 in = 𝐴𝑖𝑉𝑖 out

𝑖𝑖

𝑄in = 𝑄out𝑖𝑖


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Control Volume Analysis:

Linear Momentum Equation

𝑑𝐵sys

𝑑𝑡 =

𝑑

𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝑏 𝜌 𝐕 ∙ 𝐧 𝑑𝐴

CSCV

Rate of changeof property 𝐵of system

=Rate of change of property 𝐵

in control volume

+Net outflowof property 𝐵

through control surface

The extensive property 𝐵𝑠𝑦𝑠 becomes the momentum of the system:

𝐵𝑠𝑦𝑠 = 𝑀𝑠𝑦𝑠

The intensive property of mass 𝑚 in the system is 𝑚𝑉, and so

𝑏 =𝑚𝑉

𝑚= 𝑉



𝑑 𝑀sys

𝑑𝑡=

𝑑

𝑑𝑡 𝑉𝜌 𝑑∀ + 𝑉𝜌 𝑉 ∙ 𝑛 𝑑𝐴

CSCV

Newton’s second law for a system of mass 𝑚 subjected to a force 𝐹 is expressed

as

𝐹 = 𝑚𝑎

𝐹 =𝑑 𝑚𝑉

𝑑𝑡

The law can also be formulated for a system composed of a group of particles.

𝐹 =𝑑 𝑀sys

𝑑𝑡

The 𝑀sys denotes the total momentum of all mass comprising the system.


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Forces acting on CV consist of body forces that act throughout the entire body of

the CV (such as gravity, electric, and magnetic forces) and surface forces that

act on the control surface (such as pressure and viscous forces, and reaction

forces at points of contact).

Forces Acting on a CV

Body forces act on each volumetric

portion 𝑑𝑉 of the CV

Surface forces act on each portion

𝑑𝐴 of the CS


40


The most common body force is

gravity, which exerts a downward

force on every differential element of

the CV

Total body force acting on CV

𝐹 𝑏𝑜𝑑𝑦 = 𝜌𝑔 𝑑∀ CV

= 𝑚CV𝑔

Body Force


41


A surface force is defined as a force that requires physical contact, meaning that

the surface forces act at the control surface.

For example, 𝑝1𝐴1 acts at the control surface and requires contact between the

fluid outside the control volume and the fluid inside the control volume.

In addition to pressure, surface forces can be caused by shear stress, for

example the force 𝜏𝐴

Surface

Force


42


If the flow crossing the control surface occurs through a series of inlet and outlet

ports, then:

𝐹 = 𝐹 S + 𝐹 B =𝑑

𝑑𝑡 𝑉𝜌 𝑑∀ + CV

𝑚 𝑜𝑉𝑜CS

− 𝑚 𝑖CS

𝑉𝑖

where the subscripts 𝑜 and 𝑖 refer to the outlet and inlet ports, respectively.

𝑑 𝑀sys

𝑑𝑡= 𝐹 = 𝐹 S + 𝐹 B =

𝑑

𝑑𝑡 𝑉𝜌 𝑑∀ + 𝑉𝜌 𝑉 ∙ 𝑛 𝑑𝐴

CSCV

Time rate of change of

momentum in control

volume

Net outflow rate of

momentum through

control surface

Sum of forces acting on

the matter in control

volume


43


The first term on the right hand side of the equation represent the momentum

accumulation term. This term is zero when the momentum in each differential

volume is constant with time, that is, steady flow.

𝐹 = 𝐹 S + 𝐹 B =𝑑


𝑚 𝑜𝑉𝑜CS

− 𝑚 𝑖CS

𝑉𝑖

Steady flow


44


It is important to know that the momentum equation is a vector equation

(there is a direction associated with each term).

The 𝑚 (scalar) is the rate at which mass is passing across the control surface,

and 𝑉 (vector) is velocity evaluated at the control surface.

𝐹 = 𝐹 S + 𝐹 B =𝑑


𝑚 𝑜𝑉𝑜CS

− 𝑚 𝑖CS

𝑉𝑖


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lecture 1 (fluid mechanics)

Documents

hours o

topics o

body o

surface o

fluid flows

fundamendal of fluid

jeffrey cheah foundation

differential analysis