EBVF4103 (Chapter 3) Fluid Mechanics for Civil Engineering

7/14/2019 EBVF4103 (Chapter 3) Fluid Mechanics for Civil Engineering

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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)

Faculty of Engineering andTechnical Studies

Copyright © ODL Jan 2005 Open University Malaysia

1

FLUID MECHANICS FORCIVIL ENGINEERING

TUTORIAL 2 – UNIT 2: Fluid Dynamics and Behaviour of Real Fluid

Chapter 3: Basic of Fluid Flow

Assoc. Prof. Dr Othman A. KarimEBVF4103 Fluid Mechanics for Civil EngineeringJan 2005





2Copyright © ODL Jan 2005 Open University Malaysia

SEQUENCE OF CHAPTER 3

Introduction

Objectives

3.1 Uniform Flow, Steady Flow

3.1.1 Laminar, Turbulent Flow

3.1.2 Relative Motion

3.1.3 Compressible or Incompressible

3.1.4 One, Two or Three-dimensional Flow

3.1.5 Streamlines

3.1.6 Streamtubes

3.2.1 Mass Flow Rate

3.2.2 Volume Flow Rate 3.3 The Fundamental Equations of Fluid Dynamics

3.3.1 Continuity (Principle of Conservation of Mass)

3.3.2 Work and Energy (Principle of Conservation of Energy) Summary






Introduction

Discusses the analysis of fluid in motion: fluid dynamics.

When a fluid flows through pipes and channel or around bodies such asaircraft and ships, the shape of the boundaries, the externally appliedforces and the fluid properties cause the velocities of the fluidparticles to vary from point to point throughout the flow field.

The motion of fluids can be predicted using the fundamental laws of physics together with the physical properties of the fluid.

The geometry of the motion of fluid particles in space and time isknown as the kinematics of the fluid motion.

A fluid motion may be specified by either tracing the motion of a

particle through the field of flow or examining the motion of allparticles as they pass a fixed point in space.

This course will use the second method where the emphasis is on thespatial position rather than on the particle, or known as EulerianApproach.






Objectives

1. comprehend the concepts necessary to analyse fluids inmotion.

2. identify differences between steady/unsteady,uniform/non-uniform and compressible/incompressible

flow.3. construct streamlines and stream tubes.

4. appreciate the Continuity principle through Conservation of Mass and Control Volumes.

5. derive the Bernoulli (energy) equation.

6. familiarise with the momentum equation for a fluid flow.






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3.1 Uniform Flow, Steady Flow

uniform flow: flow velocity is the same magnitude and direction at everypoint in the fluid.

non-uniform: If at a given instant, the velocity is not the same at everypoint the flow. (In practice, by this definition, every fluidthat flows near a solid boundary will be non-uniform - as thefluid at the boundary must take the speed of the boundary,usually zero. However if the size and shape of the of thecross-section of the stream of fluid is constant the flow isconsidered uniform.)

steady: A steady flow is one in which the conditions (velocity,pressure and cross-section) may differ from point to pointbut DO NOT change with time.

unsteady: If at any point in the fluid, the conditions change with time,the flow is described as unsteady . (In practice there isalways slight variations in velocity and pressure, but if theaverage values are constant, the flow is considered steady .)






Steady uniform flow: Conditions: do not change with position in the stream or with time.

Example: the flow of water in a pipe of constant diameter at constantvelocity.

Steady non-uniform flow:

Conditions: change from point to point in the stream but do not change withtime.

Example: flow in a tapering pipe with constant velocity at the inlet-velocitywill change as you move along the length of the pipe toward the exit.

Unsteady uniform flow:

At a given instant in time the conditions at every point are the same, but willchange with time.

Example: a pipe of constant diameter connected to a pump pumping at aconstant rate which is then switched off.

Unsteady non-uniform flow:

Every condition of the flow may change from point to point and with time atevery point.

Example: waves in a channel.

3.1 Uniform Flow, Steady Flow (cont.)


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3.1.1 Laminar and Turbulent Flow

Laminar flow

all the particles proceed along smooth parallel pathsand all particles on any path will follow it withoutdeviation.

Hence all particles have a velocity only in thedirection of flow.

Typical

particles

path

Figure 3.1a: Laminar flow





Turbulent Flow the particles move in an irregular manner through the flow field.

Each particle has superimposed on its mean velocity fluctuating velocitycomponents both transverse to and in the direction of the net flow.

Transition Flow

exists between laminar and turbulent flow.

In this region, the flow is very unpredictable and often changeable backand forth between laminar and turbulent states.

Modern experimentation has demonstrated that this type of flow maycomprise short ‘burst’ of turbulence embedded in a laminar flow.

Particlepaths

Figure 3.1b: Turbulent flow





3.1.2 Relative Motion

Observer

Observer

Boat moving

Flow pattern moves along

channel with boat

changes with time

UNSTEADY

Boat stationary

Fluid moving past boat

pattern stationary relative to

boat

does not change with time STEADY

Figure 3.2: Relative motion





3.1.3 Compressible or Incompressible

All fluids are compressible - even water - their density willchange as pressure changes.

Under steady conditions, and provided that the changes inpressure are small, it is usually possible to simplify analysis of

the flow by assuming it is incompressible and has constantdensity.

As you will appreciate, liquids are quite difficult to compress - so under most steady conditions they are treated asincompressible.





3.1.4 One, Two or Three-dimensional Flow

In general, all fluids flow three-dimensionally, withpressures and velocities and other flow propertiesvarying in all directions.

In many cases the greatest changes only occur in two

directions or even only in one.

In these cases changes in the other direction can beeffectively ignored making analysis much more simple.





Flow is one dimensional if the flow parameters (such asvelocity, pressure, depth etc.) at a given instant in time onlyvary in the direction of flow and not across the cross-section.The flow may be unsteady, in this case the parameter vary intime but still not across the cross-section.

An example of one-dimensional flow is the flow in a pipe.Note that since flow must be zero at the pipe wall - yet non-zero in the centre - there is a difference of parametersacross the cross-section.

Should this be treated as two-dimensional flow? Possibly -but it is only necessary if very high accuracy is required. Acorrection factor is then usually applied.





Flow is two-dimensional if it can be assumed that the flowparameters vary in the direction of flow and in one direction atright angles to this direction.

Streamlines in two-dimensional flow are curved lines on a

plane and are the same on all parallel planes. An example is flow over a weir for which typical streamlines

can be seen in the figure below. Over the majority of the lengthof the weir the flow is the same - only at the two ends does itchange slightly. Here correction factors may be applied.





3.1.5 Streamlines

• In analysing fluid flow it is useful to visualise the flowpattern.

• This can be done by drawing lines joining points of equalvelocity - velocity contours. These lines are known as

streamlines.

Figure 3.5: A streamline





Close to a solid boundary streamlines are parallel tothat boundary

At all points the direction of the streamline is the directionof the fluid velocity: this is how they are defined. Close to

the wall the velocity is parallel to the wall so the streamlineis also parallel to the wall.

It is also important to recognise that the position of streamlines can change with time - this is the case inunsteady flow. In steady flow, the position of streamlines

does not change.






Some things to know about streamlines

Because the fluid is moving in the same direction as thestreamlines, fluid can not cross a streamline.

Streamlines can not cross each other. If they were to cross

this would indicate two different velocities at the samepoint. This is not physically possible.

The above point implies that any particles of fluid starting onone streamline will stay on that same streamline throughoutthe fluid.






3.1.6 Streamtubes

A useful technique in fluid flow analysis is to consider only apart of the total fluid in isolation from the rest.

This can be done by imagining a tubular surface formed bystreamlines along which the fluid flows.

This tubular surface is known as a streamtube.

In a two-dimensional flow, we have a streamtube which isflat (in the plane of the paper).






• The "walls" of a streamtube are made of streamlines.

• As we have seen above, fluid cannot flow across a streamline, sofluid cannot cross a streamtube wall.

• The streamtube can often be viewed as a solid walled pipe.

• A streamtube is not a pipe - it differs in unsteady flow as the wallswill move with time. And it differs because the "wall" is movingwith the fluid.

Figure 3.7: (a)

A streamtube






3.2.1 Mass flow rate

mass flow rate = m =

time =

mass of fluid

time taken to collect the fluid

mass

mass flow rate






3.2.2 Volume flow rate - Discharge

discharge = Q =

=

=

volume of fluid

time

&

mass fluid rate

density

m

r =

mass of fluid density x time

massvolume

density =( )






3.2.3 Discharge and mean velocity

If the area of cross section of the pipe at point X is A, and the meanvelocity here is um, during a time t, a cylinder of fluid will pass point X with a volume A um t. The volume per unit time (the discharge) willthus be :

Q = Q=

or um=

Let um= V um = V =

Figure 3.8:

Discharge in pipe

t

t u A

time

volume m

mu A

A

Q

A

Q






• Note how carefully we have called this the mean velocity. This is becausethe velocity in the pipe is not constant across the cross section.

• Crossing the centre line of the pipe, the velocity is zero at the walls,increasing to a maximum at the centre then decreasing symmetrically tothe other wall.

• This variation across the section is known as the velocity profile ordistribution. A typical one is shown in the figure

• This idea, that mean velocity multiplied by the area gives the discharge,applies to all situations - not just pipe flow.

Figure 3.9: A typicalvelocity profileacross a pipe






Example 3.1

An empty bucket weighs 2.0 kg. After 7 seconds of collecting

water the bucket weighs 8.0 kg, then:

mass flow rate = ṁ =

= = 0.857 kg/s (kg s-1)

mass of fluid in bucket

time taken to collect the fluid

8.0 -2.0

7






Example 3.2

If we know the mass flow is 1.7 kg/s, how long will it take to

fill a container with 8 kg of fluid?

time =

=

massmass flow rate

8

1.7

= 4.7s






Example 3.3

If the density of the fluid in the above example is 850 kg/m3

what is the volume per unit time (the discharge)?

Q =

=

= 0.00108 m3/s (m3s-1)

= 1.008 10-3 m3/s but 1 litre = 1.0 10 -3m3,

so Q = 1.008 l/s

0.857

850

mass fluid rate

density

m

r =






Example 3.4

If the cross-section area, A, is 1.2 x 10-3 m2 and the discharge,

Q is 24 l/s, what is the mean velocity, of the fluid?

Let mean velocity, um = V

um = V =

=

= 2.0 m/s

A

Q

2.4 x 10-3 m3 /s

1.2 x 10-3






3.3 The Fundamental Equations of FluidDynamics

1. The law of conservation of matter

stipulates that matter can be neither created nordestroyed, though it may be transformed (e.g. by a

chemical process). Since this study of the mechanics of fluids excludes

chemical activity from consideration, the law reduces tothe principle of conservation of mass.

F l f E i i d






2. The law of conservation of energy

states that energy may be neither created nor destroyed.

Energy can be transformed from one guise to another (e.g.potential energy can be transformed into kinetic energy),but none is actually lost.

Engineers sometimes loosely refer to ‘energy losses’ due tofriction, but in fact the friction transforms some energy intoheat, so none is really ‘lost’.

F lt f E i i d






3. The law of conservation of momentum

states that a body in motion cannot gain or losemomentum unless some external force is applied.

The classical statement of this law is Newton's SecondLaw of Motion, i.e.

force = rate of change of momentum

F lt f E i i d






3.3.1 Continuity (Principle of Conservation of Mass)

• Matter cannot be created nor destroyed - (it is simplychanged in to a different form of matter).

• This principle is known as the conservation of mass and we

use it in the analysis of flowing fluids. • The principle is applied to fixed volumes, known as control

volumes or surfaces

CONTROL

VOLUME

Control surface

OutflowInflow

Figure 3.10: A control volume

F lt f E i i d






Figure 3.11: A streamtube section

Mass entering per unittime at end 1 = Massleaving per unit time atend 2

For any control volume the principle of conservation of mass saysMass entering = Mass leaving + Increase of mass in the control

per unit time per unit time volume per unit time

For steady flow:(there is no increase in the mass within the control volume)

Mass entering per unit time = Mass leaving per unit time

Fac lt of Enginee ing and






flow is incompressible, the density of the fluid is constantthroughout the fluid continum. Mass flow, m, entering maybe calculated by taking the product

(density of fluid, r ) (volume of fluid entering per second Q )

Mass flow is therefore represented by the product r Q , hence

r Q (entering) = r Q (leaving)

But since flow is incompressible, the density is constant, so

Q (entering) = Q (leaving) (3.5a)

This is the ‘continuity equation’ for steady incompressibleflow.

Faculty of Engineering and






If the velocity of flow across the entry to the controlvolume is measured, and that the velocity is constant atV 1 m/s. Then, if the cross-sectional area of the streamtubeat entry is A1,

Q (entering) = V 1

A1

Thus, if the velocity of flow leaving the volume is V 2 andthe area of the streamtube at exit is A2, then

Q (leaving) = V 2 A2

Therefore, the continuity equation may also be written as V 1 A1 = V 2 A2 (3.5b)







Application of Continuity Equation

We can apply the principle of continuity to pipes with cross sections whichchange along their length.

A liquid is flowing from left to right and the pipe is narrowing in the samedirection. By the continuity principle, the mass flow rate must be thesame at each section - the mass going into the pipe is equal to the massgoing out of the pipe. So we can write:

r 1 A1V 1= r 2 A2V 2

Figure 3.12: Pipe with a contraction

As we are considering a liquid, usually water, which is not very compressible,the density changes very little so we cansay r 1 = r 2 = r . This also says that thevolume flow rate is constant or that

Discharge at section 1 = Discharge atsection 2

Q 1 = Q 2

A1V 1 = A2V 2 or V 2 = A1V 1

A2







As the area of the circular pipe is a function of thediameter we can reduce the calculation further,

V 2 =

V 2 = (3.6)

A1

A2

V 1=p d 1

2/4

p d 22

/4

V 1=d 1

2

d 22

V 1

d 12

d 22

V 1( ) 2







Another example is a diffuser , a pipe which expands ordiverges as in the figure below







The continuity principle can also be used to determine thevelocities in pipes coming from a junction.

Total mass flow into the junction = Total mass flow out of thejunction

r 1Q1 = r 2Q2 + r 3Q3 When the flow is incompressible (e.g. water) r 1 = r 2 = r

Q1 = Q2 + Q3

A1V 1 = A2V 2 + A3V 3 (3.7)







Example 3.5 If the area in Figure 3.12 A1 = 10 10-3 m2 and A1 = 10 10-3

m2 and and the upstream mean velocity, V 1 = 2.1 m/s, what

is the downstream mean velocity?

V 2 =

= 7.0 m/s

A1V 1

A2

10 x 10-3 x 2.1

3 x 10-3=







Example 3.6

If the diameter of a diffuser (Figure 3.13) at section 1 is d 1 =30 mm and at section 2 d 2 = 40 mm and the mean velocity atsection 2 is V 2 = 3.0 m/s. Calculate the velocity entering the

diffuser.

V 2 = ( ) 40

303.0 = 5.3m/s

2







Example 3.7 For a junction (Figure 3.14), if pipe 1 diameter = 50 mm,

mean velocity 2 m/s, pipe 2 diameter 40 mm takes 30% of total discharge and pipe 3 diameter 60 mm. What are thevalues of discharge and mean velocity in each pipe?

Q 1 = A1V 1 = = 0.00392 m3/s

But Q 2 = 0.3Q 1 = 0.001178 m3/s

Also Q 1 = Q 2 + Q 3

Q 3 = Q 1 – 0.3Q 1 = 0.7Q 1 = 0.00275 m3/s

V 2 = Q 2 / V 2 = 0.936 m/s

V 3 = Q 3 / V 3 = 0.972 m/s







3.3.2 Work and Energy (Principle Of Conservation Of Energy)

friction: negligible

sum of kinetic energy and gravitational potential

energy is constant. Recall :

Kinetic energy = ½ mV 2

Gravitational potential energy = mgh

(m: mass, V : velocity, h: height above the datum).






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To apply this to a falling body we have an initial velocity of zero, and it falls through a height of h.

Initial kinetic energy = 0

Initial potential energy = mgh

Final kinetic energy = ½ mV 2

Final potential energy = 0We know that,

kinetic energy + potential energy = constant

Initial

kineticEnergy

Initial

potentialEnergy

Final

KineticEnergy

Final

PotentialEnergy}+{

mgh = ½ mV 2 or ghV 2

{ }={ }+{ }






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continuous jet of liquid

a continuous jet of water coming from a pipe with velocity V 1. One particle of the liquid with mass m travels with the jet and falls from

height z 1 to z 2. The velocity also changes from V 1 to V 2. The jet is traveling in air where

the pressure is everywhere atmospheric so there is no force due to pressureacting on the fluid.

The only force which is acting is that due to gravity. The sum of the kineticand potential energies remains constant (as we neglect energy losses dueto friction) so :

mgz 1 + mV 12 = mgz 2 + mV 2

2

As m is constant this becomes : V 1

2 + gz 1 = V 22 + gz 2

Figure 3.15 : The trajectory of a jet of water






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Flow from a reservoir

• The level of the water in the reservoir is z 1.Considering the energy situation - there is nomovement of water so kinetic energy is zero butthe gravitational potential energy is mgz 1.

• If a pipe is attached at the bottom water flowsalong this pipe out of the tank to a level z 2. Amass m has flowed from the top of the reservoir

to the nozzle and it has gained a velocity V 2. Thekinetic energy is now ½mV 22 and the potential

energy mgz 2. Summarising :

Initial kinetic energy = 0 Initial potential energy = mgz 1

Final kinetic energy = ½ mV 22

Final potential energy = mgz 2

So

mgz 1 = ½ mV 22 + mgz 2

mg ( z 1 - z 2 ) = ½ mV 22

V 2 = (3.8)

Figure 3.16 : Flow from a reservoir

)(2 21 z z g






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Example 3.8A reservoir of water has the surface at 310 m above the outlet nozzle of a pipe with

diameter 15mm. What is the velocity; the discharge out of the nozzle; and mass flow rate. (Neglect all friction in the nozzle and the pipe)

Solution:

a)

b) Volume flow rate is equal to the area of the nozzle multiplied by the velocity Q = AV

=

=

= 0.01378 m3/s c) The density of water is 1000 kg/m3 so the mass flow rate is

ṁ = density volume flow rate= r Q = 1000 0.01378

= 13.78 kg/s

)(2 212 z z g V

3102 g sm /0.78

V d

4

2

p

0.784

015.02

p





y g gTechnical Studies

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Bernoulli's Equation

We see that from applying equal pressure or zero velocities we getthe two equations from the section above. They are both just specialcases of Bernoulli's equation.

Bernoulli's equation has some restrictions in its applicability, they are:

Flow is steady;

Density is constant (which also means the fluid is incompressible);

Friction losses are negligible.

The equation relates the states at two points along a single streamline,(not conditions on two different streamlines).

2

2

221

2

11

22 z

g

V

g

p z

g

V

g

p

r r






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• A fluid of constant density = 960 kg/m3 is flowing steadily throughthe above tube. The diameters at the sections are d 1 = 100mm andd 2 = 80mm. The gauge pressure at 1 is P 1 = 200 kN/m2 and thevelocity here is V 1 = 5m/s. What is the gauge pressure at section 2.

• Bernoulli equation is applied along a streamline joining section 1

with section 2.

• The tube is horizontal, with z 1 = z 2 so Bernoulli gives us thefollowing equation for pressure at section 2:

P 2 = P 1 + (V 12 – V 2

2)

Figure 3.19 : A contractingexpanding pipe






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But we do not know the value of V 2. We can calculate this from thecontinuity equation: Discharge into the tube is equal to the discharge outi.e.

= 7.8125 m/s

So we can now calculate the pressure at section 2

1

2

2

1

2

2

112

2211

V d

d

V

A

V AV

V AV A

508.0

1.02

p2 = 200000 -17296.87

= 182703 N/m2

= 182.7 kN/m2

P 2 = P 1+ (V 12 – V 2

2) = 200000 + (52 – 7.81252) 9602

r

2






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Modifications of Bernoulli Equation

• In practice, the total energy of a streamline does not remain constant.Energy is ‘lost’ through friction, and external energy may be either :

added by means of a pump or

extracted by a turbine.

• Consider a streamline between two points 1 and 2. If the energy head lostthrough friction is denoted by H f and the external energy head added (sayby a pump) is or extracted (by a turbine) H E , then Bernoulli's equation may

be rewritten as :

± HE

= H2

+ Hf (3.11)

or

H E = energy head added/loss due to external source such as pump/turbines

This equation is really a restatement of the First Law of Thermodynamics for an incompressiblefluid.

f E H z g

V

g

p H z

g

V

g

p

2

2

221

2

11

22 r r (3.12)





Technical Studies

50


The Power Equation

In the case of work done over a fluid the power input into theflow is :

P = r gQH E (3.13)

where Q = discharge,

H E = head added / loss If p = efficiency of the pump, the power input required,

P in = (3.14)

p

E gQH

r

Faculty of Engineering andh i l di




Technical Studies

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Summary

This chapter has outlined and discussed on the fundamental of fluidin motion. Students are aspect to be able to discuss and visualise onthe following aspect:

Able to classify FOUR (4) types of flow- Steady uniform flow, Steady non-uniform flow, Unsteady uniform flow and Unsteady non-uniform

flow

The differences between Laminar Flow, Turbulent Flow andalso Transition Flow

The idea of using the streamline to visualise the flow pattern

The calculation of mass flow rate, volume flow rate and the

mean velocity of the flow

Able to explain and apply the THREE (3) laws- conservation of matter (conservation of mass); conservation of energy andconservation of momentum

The important of Bernoulli Equation and the derivation

Faculty of Engineering andT h i l S di



Technical Studies

52

Thank You

EBVF4103 (Chapter 3) Fluid Mechanics for Civil Engineering

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