JNTUH Fluid Mechanics Hydraulic Machinery Study Materials

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

What make fluid mechanics different to solid mechanics?

LECTURE CONTENTS

om

x The nature of a fluid is different to that of a solid

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x In fluids we deal with continuous streams of fluid.

In solids we only consider individual elements.

or

ld

fo

In this section we will consider how we can classify the differences in nature of fluids and solids.

What do we mean by nature of a fluid?

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w

.jn

tu

w

Section 0: Introduction Section 1: Fluid Properties Fluids vs. Solids Viscosity Newtonian Fluids Properties of Fluids Section 2: Statics Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces Section 3: Dynamics The continuity equation.

The Bernoulli Equation.

Application of Bernoulli equation.

The momentum equation.

Application of momentum equation.

Section 4: Real Fluids Boundary layer.

Laminar flow in pipes.

Section 5: Dimensional Analysis An Intro to Dimensional analysis Similarity

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

1

Fluids are clearly different to solids.

But we must be specific.

We need some definable basic physical difference.

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

2

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

CIVE1400: Fluid Mechanics

We know that fluids flow under the action of a force, and the solids don’t but solids do deform.

Section 1: Fluid Properties

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What use can we make of these ideas?

In the analysis of fluids we often take small volumes (elements) and examine the forces on these.

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So we can say that

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x fluids lack the ability of solids to resist deformation.

ld or

w

Take the rectangular element below.

What forces cause it to deform?

A

B

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tu

(These definitions include both gasses and liquids as fluids.)

fo

x fluids change shape as long as a force acts.

D

w w

w

C

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

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CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

4

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

A’

B’

CIVE1400: Fluid Mechanics

F

Section 1: Fluid Properties

Fluids in motion

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Consider a fluid flowing near a wall.

- in a pipe for example -

F D

Fluid next to the wall will have zero velocity.

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C

Forces acting along edges (faces), such as F, are know as shearing forces.

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The fluid “sticks” to the wall.

Moving away from the wall velocity increases to a maximum.

fo

From this we arrive at the definition:

.jn

tu

This has the following implications for fluids at rest:

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or

ld

A Fluid is a substance which deforms continuously, or flows, when subjected to shearing forces.

Plotting the velocity across the section gives “velocity profile”

w w

w

If a fluid is at rest there are NO shearing forces acting on it, and any force must be acting perpendicular to the fluid

v

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

Change in velocity with distance is “velocity gradient” =

5

CIVE1400: Fluid Mechanics

du dy

Section 1: Fluid Properties

6

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

CIVE1400: Fluid Mechanics

As fluids are usually near surfaces there is usually a velocity gradient.

What use is this observation?

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Under normal conditions one fluid particle has a velocity different to its neighbour.

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It would be useful if we could quantify this shearing force.

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Particles next to each other with different velocities exert forces on each other (due to intermolecular action ) ‌‌

This may give us an understanding of what parameters govern the forces different fluid exert on flow.

fo

w

or

ld

i.e. shear forces exist in a fluid moving close to a wall.

.jn

tu

What if not near a wall?

Section 1: Fluid Properties

We will examine the force required to deform an element.

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w

Consider this 3-d rectangular element, under the action of the force F.

v

No velocity gradient, no shear forces.

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

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CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

8

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

CIVE1400: Fluid Mechanics

δx

A 2-d view may be clearer‌ b

a

δz

Section 1: Fluid Properties

A’

B

F

E x

φ

δy

F

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B

A

F

E’

y

C

D

.c

F

B’

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The shearing force acts on the area C

D

under the action of the force F

A Gz u Gx

b

a’

ld

a

fo

Shear stress, W is the force per unit area:

b’

B

B’

w

A’

or

F A

tu

E

D

w

C

.jn

F

Section 1: Fluid Properties

F A

The deformation which shear stress causes is measured by the angle I, and is know as shear strain.

Using these definitions we can amend our definition of a fluid: In a fluid I increases for as long as W is applied the fluid flows In a solid shear strain, I, is constant for a fixed shear stress W.

w w CIVE1400: Fluid Mechanics

W

9

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties 10

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

CIVE1400: Fluid Mechanics

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

u/y is the rate of change of velocity with distance,

om

I

in differential form this is

time Constant u

I t

ld

tu .jn w

w w

u

W

P

du dy

which is know as Newton’s law of viscosity

w

or

x y

rate of shear strain

x t

= velocity gradient.

giving

fo

If a particle at point E moves to point E’ in time t then: for small deformations

(note that

du dy

The constant of proportionality is known as the dynamic viscosity, P

We can express this in terms of the cuboid.

shear strain I

u

y

.c

W

Constant u

ru m

Wv

W

Section 1: Fluid Properties

is the velocity of the particle at E)

A fluid which obeys this rule is know as a Newtonian Fluid (sometimes also called real fluids) Newtonian fluids have constant values of P

Non-Newtonian Fluids

So CIVE1400: Fluid Mechanics

Section 1: Fluid Properties 11

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties 12

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

This graph shows how P changes for different fluids.

Some fluids do not have constant P. They do not obey Newton’s Law of viscosity.

Bingham plastic

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W

n

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The general relationship is:

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Shear stress, Ď„

They do obey a similar relationship and can be placed into several clear categories

§ Gu ¡ A B¨ ¸ Š Gy š

Pseudo plastic

plastic

fo

where A, B and n are constants.

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w

.jn

tu

w

or

ld

For Newtonian fluids A = 0, B = P and n = 1

Newtonian

Dilatant

Ideal, (τ=0) Rate of shear, δu/δy

x Plastic: Shear stress must reach a certain minimum before flow commences. x Bingham plastic: As with the plastic above a minimum shear stress must be achieved. With this classification n = 1. An example is sewage sludge. x Pseudo-plastic: No minimum shear stress necessary and the viscosity decreases with rate of shear, e.g. colloidial substances like clay, milk and cement. x Dilatant substances; Viscosity increases with rate of shear e.g. quicksand. x Thixotropic substances: Viscosity decreases with length of time shear force is applied e.g. thixotropic jelly paints. x Rheopectic substances: Viscosity increases with length of time shear force is applied

x Viscoelastic materials: Similar to Newtonian but if there is a sudden large change in shear they behave like plastic.

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties 13

CIVE1400: Fluid Mechanics

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CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

Properties of Fluids:

Liquids vs. Gasses Density Liquids and gasses behave in much the same way Some specific differences are 1. A liquid is difficult to compress and often regarded as being incompressible.

A gas is easily to compress and usually treated as such - it changes volume with pressure.

2. A given mass of liquid occupies a given volume and will form a free surface A gas has no fixed volume, it changes volume to expand to fill the containing vessel. No free surface is formed.

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There are three ways of expressing density:

U

mass per unit volume

U

mass of fluid volume of fluid

fo

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1. Mass density:

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Units: kg/m3

or

Causes of Viscosity in Fluids

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Viscosity in Gasses

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x Mainly due to molecular exchange between layers Mathematical considerations of this momentum exchange can lead to Newton law of viscosity.

ML 3

Typical values: Water = 1000 kg m Air = 1.23 kg m

3

3

, Mercury = 13546 kg m

, Paraffin Oil = 800 kg m

3

3 .

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x If temperature increases the momentum exchange between layers will increase thus increasing viscosity.

Dimensions:

Viscosity in Liquids

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x There is some molecular interchange between layers in liquids - but the cohesive forces are also important. x Increasing temperature of a fluid reduces the cohesive forces and increases the molecular interchange.

Resulting in a complex relationship between temperature and viscosity.

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties 15

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties 16

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

CIVE1400: Fluid Mechanics

2. Specific Weight:

Section 1: Fluid Properties

Viscosity

(sometimes known as specific gravity)

Ug 2 2

Dimensions: ML Typical values: Water =9814

1. Coefficient of Dynamic Viscosity

/ m3 (or kg/m2/s2)

P = the shear stress, W, required to drag one layer of fluid with unit velocity past another layer a unit distance away.

.

N m 3 , Mercury = 132943 N m 3 ,

fo

N m 3 , Paraffin Oil =7851 N m 3

ld

Air =12.07

T

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Units: Newton’s per cubic metre, N

There are two ways of expressing viscosity

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weight per unit volume

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Z Z

V

or

3. Relative Density:

ratio of mass density to

tu

U subs tan ce U $

.jn

V

w

a standard mass density

H 2O ( at 4 c )

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For solids and liquids this standard mass density is the

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maximum mass density for water (which occurs at at atmospheric pressure.

Units: none, as it is a ratio Dimensions: 1.

Typical values: Water = 1, Mercury = 13.5, Paraffin Oil =0.8.

CIVE1400: Fluid Mechanics

4$ c)

Section 1: Fluid Properties 17

du dy Force Velocity Area Distance Force u Time = Area Mass Length u Time

P W

Units: N s/m2 or kg/m s (kg m-1 s-1) (Note that P is often expressed in Poise, P, where 10 P = 1 N s/m2.)

Dimensions: ML-1T-1 Typical values: Water =1.14 u 10-3 Ns/m2, Air =1.78 u 10-5 Ns/m2, Mercury =1.552 Ns/m2, Paraffin Oil =1.9 Ns/m2.

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties 18

CIVE1400: Fluid Mechanics

CIVE1400: Fluid Mechanics

Section 1: Fluid Properties

2. Kinematic Viscosity

Q = the ratio of dynamic viscosity to mass density.

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P U

ru m

Units: m2s-1 Dimension: L2T-1

Water =1.14 u 10-6 m2/s, , Air =1.46 u 10-5 m2/s m s , Mercury =1.145 u 10-4 m2/s, Paraffin Oil =2.375 u 10-3 m2/s.

ld

fo

Typical values:

w w

w

.jn

tu

w

or

2 1

CIVE1400: Fluid Mechanics

LECTURE CONTENTS Section 0: Introduction Section 1: Fluid Properties Fluids vs. Solids Viscosity Newtonian Fluids Properties of Fluids Section 2: Statics Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces Section 3: Dynamics The continuity equation.

The Bernoulli Equation.

Application of Bernoulli equation.

The momentum equation.

Application of momentum equation.

Section 4: Real Fluids Boundary layer.

Laminar flow in pipes.

Section 5: Dimensional Analysis An Intro to Dimensional analysis Similarity

.c

Q

Section 2: Statics

Section 1: Fluid Properties 19

CIVE1400: Fluid Mechanics

Section 2: Statics 20

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

STATICS

What do we know about the forces involved with static fluids?

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4 Lectures Objectives

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From earlier we know that:

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x Introduce idea of pressure.

1. A static fluid can have no shearing force acting on it. x Prove unique value at any particular elevation.

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2. Any force between the fluid and the boundary must be acting at right angles to the boundary.

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x Show hydrostatic relationship.

F2

.jn

w w w Section 2: Statics 21

R2

Fn

x Determine the magnitude and direction of forces on submerged surfaces

CIVE1400: Fluid Mechanics

R1

F

tu

w

x Pressure measurement using manometers.

F1

or

x Express pressure in terms of head of fluid.

R

Rn

Pressure force normal to the boundary

CIVE1400: Fluid Mechanics

Section 2: Statics 22

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

Pressure

x curved surfaces

Convenient to work in terms of pressure, p, which is the force per unit area.

x any imaginary plane

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An element of fluid at rest is in equilibrium:

p

Units: Newton’s per square metre, N/m2, kg/m s2 (kg m-1s-2).

(Also known as a Pascal, Pa, i.e. 1 Pa = 1 N/m2) (Also frequently used is the alternative SI unit the bar, where 1bar = 105 N/m2)

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w

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tu

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or

ld

fo

3. The sum of forces in any direction is zero.

4. The sum of the moments of forces about any point is zero.

Force Area over which the force is applied F A

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pressure

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This is also true for:

CIVE1400: Fluid Mechanics

Section 2: Statics 23

Uniform Pressure: If the force on each unit area of a surface is equal then uniform pressure

CIVE1400: Fluid Mechanics

Section 2: Statics 24

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Summing forces in the x-direction:

Pascal’s Law

Force in the x-direction due to px,

Proof that pressure acts equally in all directions.

p x u Area ABFE

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Fx x ps

px

δy

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F

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δs

A

C

fo

θ D δx

ld

E

( sin T

Gy

Fx s

Force in x-direction due to py,

Fx y

Fx x Fx s Fx y

All forces at right angles to the surfaces

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0

To be at rest (in equilibrium)

w

.jn

Remember:

ps u Area ABCD u sin T Gy psGs Gz Gs psGy Gz

Gs )

tu

w

or

py

No shearing forces

0

p xGxGy psGyGz 0 px

CIVE1400: Fluid Mechanics

p x Gx Gy

Force in the x-direction due to ps,

B

δz

Section 2: Statics

Section 2: Statics 25

CIVE1400: Fluid Mechanics

ps Section 2: Statics 26

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

To be at rest (in equilibrium)

Summing forces in the y-direction.

Fy Fy Fy weight y s x 1 · § p yGxGy psGxGz ¨ Ug GxGyGz¸ ¹ © 2

Force due to py,

p y u Area ABCD

y

p y GxGz

om

Fy

ps u Area ABCD u cosT

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psGsGz

Gx Gs

Gs )

or

Gx

ld

psGxGz ( cos T

w

Component of force due to px,

Force due to gravity,

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w

weight = - specific weight u volume of element 1 = Ug u GxGyGz 2

CIVE1400: Fluid Mechanics

py

Section 2: Statics 27

ps

We showed above

px

ps

thus

tu

0

.jn

Fy x

0

The element is small i.e. Gx, Gx, and Gz, are small, so Gx u Gy u Gz, is very small and considered negligible, hence

fo

s

0

.c

Component of force due to ps,

Fy

Section 2: Statics

px

py

ps

Pressure at any point is the same in all directions.

This is Pascal’s Law and applies to fluids at rest.

CIVE1400: Fluid Mechanics

Section 2: Statics 28

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

The forces involved are: Vertical Variation Of Pressure In A Fluid Under Gravity

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Force due to p1 on A (upward) = p1A Force due to p2 on A (downward) = p2A

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p2, A Area A

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Fluid density Ď

Force due to weight of element (downward) = mg = mass density u volume u g = U g A(z2 - z1)

z1

.jn

tu

w

p1, A

or

ld

fo

z2

w w

w

Vertical cylindrical element of fluid cross sectional area = A mass density = U

CIVE1400: Fluid Mechanics

Taking upward as positive, we have

p1 A p2 A UgA z2 z1 = 0 p2 p1

UgA z2 z1

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

p2 p1

UgA z2 z1

This is the hydrostatic pressure change.

Section 2: Statics 29

CIVE1400: Fluid Mechanics

Section 2: Statics 30

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

Equality Of Pressure At The Same Level In A Static Fluid P

Q

om

Fluid density Ď

Area A

z

Face L

L

Face R

Horizontal cylindrical element

We have shown

cross sectional area = A

pl = pr

fo

mass density = U

For a vertical pressure change we have

ld

left end pressure = pl

pr

pq Ugz

p p Ugz pp

w

w w

Pressure in the horizontal direction is constant.

p p Ugz

so

.jn

tu

For equilibrium the sum of the forces in the x direction is zero.

pl and

w

or

right end pressure = pr

pl = pr

R

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weight, mg

pl A = pr A

z

.c

pr, A

pl, A

pq Ugz pq

Pressure at the two equal levels are the same.

This true for any continuous fluid.

CIVE1400: Fluid Mechanics

Section 2: Statics 31

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Section 2: Statics 32

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

General Equation For Variation Of Pressure In A Static Fluid

Horizontal If T

om

horizontal),so

§ dp · ¨ ¸ © ds ¹ T 90$

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Fluid density ρ

ru m

δs

fo

θ

z + δz

ld

mg pA

w

or

z

90$ then s is in the x or y directions, (i.e.

(p + δp)A

Area A

Section 2: Statics

.jn

tu

A cylindrical element of fluid at an arbitrary orientation.

dp dx

dp dy

0

Confirming that pressure change on any horizontal plane is zero.

Vertical If T

0$ then s is in the z direction (vertical) so § dp · ¨ ¸ © ds ¹ T 0$

dp dz

Ug

dp ds

CIVE1400: Fluid Mechanics

w w

w

From considering incremental forces (see detailed notes on WWW or other texts) we get

Ug cos T

Section 2: Statics 33

Confirming the hydrostatic result

p2 p1 z2 z1

Ug

p2 p1

Ug z2 z1

CIVE1400: Fluid Mechanics

Section 2: Statics 34

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

Pressure And Head It is convenient to take atmospheric pressure as the datum

We have the vertical pressure relationship

dp dz

om

Ug ,

Pressure quoted in this way is known as gauge pressure i.e.

.c

integrating gives p = -Ugz + constant

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Gauge pressure is

z

fo

measuring z from the free surface so that z = -h

ld

h

or

y

tu

Ugh constant

.jn

p

w

x

pgauge = U g h

The lower limit of any pressure is the pressure in a perfect vacuum.

Pressure measured above a perfect vacuum (zero) is known as absolute pressure

patmospheric

constant so

p CIVE1400: Fluid Mechanics

w w

w

surface pressure is atmospheric, patmospheric .

Absolute pressure is pabsolute = U g h + patmospheric

Ugh patmospheric Section 2: Statics 35

Absolute pressure = Gauge pressure + Atmospheric

CIVE1400: Fluid Mechanics

Section 2: Statics 36

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

A gauge pressure can be given

Section 2: Statics

Pressure Measurement By Manometer

using height of any fluid.

p

Ugh

Manometers use the relationship between pressure and head to measure pressure

om

This vertical height is the head.

The Piezometer Tube Manometer

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If pressure is quoted in head, the density of the fluid must also be given.

Example:

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The simplest manometer is an open tube.

This is attached to the top of a container with liquid at pressure. containing liquid at a pressure.

500 u 10 3

ld

B

3.75m of Mercury

.jn

h

h2

A

tu

In head of Mercury density U = 13.6u103 kgm-3.

3

or

50.95m of water

h1

w

h

500 u 103 1000 u 9.81

p Ug

fo

What is a pressure of 500 kNm-2 in head of water of density, U = 1000 kgm-3 Use p = Ugh,

13.6 u 10 u 9.81

h

w w

remember U = J u Uwater) 3

w

In head of a fluid with relative density J = 8.7.

500 u 10 586 . m of fluid J = 8.7 8.7 u 1000 u 9.81

CIVE1400: Fluid Mechanics

Section 2: Statics 37

The tube is open to the atmosphere,

The pressure measured is relative to atmospheric so it measures gauge pressure.

CIVE1400: Fluid Mechanics

Section 2: Statics 38

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

An Example of a Piezometer.

What is the maximum gauge pressure of water that can be measured by a Piezometer of height 1.5m?

And if the liquid had a relative density of 8.5 what would the maximum measurable gauge pressure?

Pressure at A = pressure due to column of liquid h1

om

pA = U g h 1

.c

Pressure at B = pressure due to column of liquid h2

fo

ru m

pB = U g h 2

or

ld

Problems with the Piezometer:

w

1. Can only be used for liquids

.jn

tu

2. Pressure must above atmospheric

w w

w

3. Liquid height must be convenient i.e. not be too small or too large.

CIVE1400: Fluid Mechanics

Section 2: Statics 39

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Section 2: Statics 40

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

We know:

“U”-Tube enables the pressure of both liquids and gases to be measured “U” is connected as shown and filled with manometric fluid.

Pressure in a continuous static fluid is the same at any horizontal level.

om

The “U”-Tube Manometer

.c

pressure at B = pressure at C

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Important points: 1. The manometric fluid density should be greater than of the fluid measured.

Uman > U

For the left hand arm

fo

pressure at B = pressure at A + pressure of height of liquid being measured

or

ld

2. The two fluids should not be able to mix they must be immiscible.

w

Fluid density ρ

tu

D

pB = pC

pB = pA + Ugh1

For the right hand arm pressure at C = pressure at D + pressure of height of manometric liquid

pC = patmospheric + Uman gh2

.jn

h2

A

C

w w

B

w

h1

Manometric fluid density ρ man

CIVE1400: Fluid Mechanics

Section 2: Statics 41

We are measuring gauge pressure we can subtract patmospheric giving

pB = pC pA = Uman gh2 - Ugh1

CIVE1400: Fluid Mechanics

Section 2: Statics 42

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

An example of the U-Tube manometer.

Using a u-tube manometer to measure gauge pressure of fluid density U = 700 kg/m3, and the manometric fluid is mercury, with a relative density of 13.6.

What is the gauge pressure if: a) h1 = 0.4m and h2 = 0.9m? b) h1 stayed the same but h2 = -0.1m?

What if the fluid is a gas?

om

Nothing changes.

.c

The manometer work exactly the same.

ru m

BUT:

ld or w

And Liquid density is much greater than gas,

fo

As the manometric fluid is liquid (usually mercury , oil or water)

.jn

tu

Uman >> U

w

Ugh1 can be neglected,

w w

and the gauge pressure given by pA = Uman gh2

CIVE1400: Fluid Mechanics

Section 2: Statics 43

CIVE1400: Fluid Mechanics

Section 2: Statics 44

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Pressure difference measurement Using a “U”-Tube Manometer.

Section 2: Statics

pressure at C = pressure at D pC = pD

om

The “U”-tube manometer can be connected at both ends to measure pressure difference between these two points

.c

pC = pA + U g ha

pD = pB + U g (hb + h) + Uman g h

ru m

B

pA + U g ha = pB + U g (hb + h) + Uman g h

fo

Fluid density ρ

ld

hb

or

E

pA - pB = U g (hb - ha) + (Uman - U)g h

w

h

Giving the pressure difference

ha D

Again if the fluid is a gas Uman >> U, then the terms involving U can be neglected,

pA - pB = Uman g h

w

.jn

C

tu

A

w w

Manometric fluid density ρman

CIVE1400: Fluid Mechanics

Section 2: Statics 45

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Section 2: Statics 46

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

An example using the u-tube for pressure difference measuring

Advances to the “U” tube manometer

In the figure below two pipes containing the same fluid of density U = 990 kg/m3 are connected using a u-tube manometer.

What is the pressure between the two pipes if the manometer contains fluid of relative density 13.6?

om

Problem: Two reading are required.

Solution: Increase cross-sectional area of one side.

Result: One level moves much more than the other.

.c

Fluid density ρ

ru m

Fluid density ρ

A

B ha = 1.5m

p

hb = 0.75m

p2

1

fo

E

diameter D diameter d

ld

h = 0.5m

Section 2: Statics

or

Datum line

D

z1

tu

w

C

z2

If the manometer is measuring the pressure difference of a gas of (p1 - p2) as shown,

we know p1 - p2 = Uman g h

w w

w

.jn

Manometric fluid density ρman = 13.6 ρ

CIVE1400: Fluid Mechanics

Section 2: Statics 47

CIVE1400: Fluid Mechanics

Section 2: Statics 48

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

volume of liquid moved from the left side to the right = z2 u ( Sd2 / 4)

om

Problem: Small pressure difference, movement cannot be read.

The fall in level of the left side is

.c

Volume moved Area of left side z 2 Sd 2 / 4

Solution 1: Reduce density of manometric fluid.

ru m

z1

SD 2 / 4

fo

Result:

ld

2

w

Putting this in the equation,

tu

2 ª §d¡ º Ug  z 2 z 2 ¨ ¸  Š Dš Ÿ 

.jn

p1 p2

Greater height change easier to read.

or

§d¡ z2 ¨ ¸ Š Dš

Section 2: Statics

w w

w

ª § d ¡2º Ugz 2 1 ¨ ¸   Š Dš Ÿ

If D >> d then (d/D)2 is very small so p1 p2 Ugz2

Solution 2: Tilt one arm of the manometer.

Result:

Same height change - but larger movement along the manometer arm - easier to read.

Inclined manometer CIVE1400: Fluid Mechanics

Section 2: Statics 49

CIVE1400: Fluid Mechanics

Section 2: Statics 50

CIVE1400: Fluid Mechanics

Section 2: Statics

diameter D

e ad

le ca

r

Example of an inclined manometer.

An inclined manometer is required to measure an air pressure of 3mm of water to an accuracy of +/- 3%.

The inclined arm is 8mm in diameter and the larger arm has a diameter of 24mm. The manometric fluid has density Uman = 740 kg/m3 and the scale may be read to +/- 0.5mm.

What is the angle required to ensure the desired accuracy may be achieved?

p2

diameter d

x

Re

z

2

S

Section 2: Statics

om

p1

CIVE1400: Fluid Mechanics

.c

Datum line

ru m

z1

Ugz2

Ugx sin T

.jn

x sin T

w w

w

p1 p2

tu

but, z2

ld

w

p1 p2

or

The pressure difference is still given by the height change of the manometric fluid.

fo

θ

The sensitivity to pressure change can be increased further by a greater inclination.

CIVE1400: Fluid Mechanics

Section 2: Statics 51

CIVE1400: Fluid Mechanics

Section 2: Statics 52

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

Forces on Submerged Surfaces in Static Fluids

Take care when fixing the manometer to vessel Burrs cause local pressure variations.

We have seen these features of static fluids

om

Choice Of Manometer

x Hydrostatic vertical pressure distribution x Pressures at any equal depths in a continuous fluid are equal x Pressure at a point acts equally in all directions (Pascal’s law).

.c

Disadvantages:

ru m

x Slow response - only really useful for very slowly varying pressures - no use at all for fluctuating pressures; x For the “U” tube manometer two measurements must be taken simultaneously to get the h value.

fo

x Forces from a fluid on a boundary acts at right angles to that boundary.

or

w

x It cannot be used for very large pressures unless several manometers are connected in series;

ld

x It is often difficult to measure small variations in pressure.

.jn

tu

x For very accurate work the temperature and relationship between temperature and U must be known;

Fluid pressure on a surface Pressure is force per unit area.

Pressure p acting on a small area GA exerted force will be

w w

x They are very simple.

w

Advantages of manometers:

F = puGA

x No calibration is required - the pressure can be calculated from first principles.

CIVE1400: Fluid Mechanics

Since the fluid is at rest the force will act at right-angles to the surface.

Section 2: Statics 53

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Section 2: Statics 54

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

Horizontal submerged plane

General submerged plane F =p δA1 1 1

F =p δA 2 2 2

The pressure, p, will be equal at all points of the surface.

The resultant force will be given by R pressure u area of plane R = pA

.c

om

F =p δA n n n

ru m

The total or resultant force, R, on the plane is the sum of the forces on the small elements i.e.

R p1GA1 p 2 GA2 p n GAn ÂŚ pGA

ld

Each elemental force is a different magnitude and in a different direction (but still normal to the surface.).

tu

w

or

and This resultant force will act through the centre of pressure.

fo

Curved submerged surface

It is, in general, not easy to calculate the resultant force for a curved surface by combining all elemental forces.

w w

w

.jn

For a plane surface all forces acting can be represented by one single resultant force, acting at right-angles to the plane through the centre of pressure.

CIVE1400: Fluid Mechanics

Section 2: Statics 55

The sum of all the forces on each element will always be less than the sum of the individual forces, ÂŚ pGA .

CIVE1400: Fluid Mechanics

Section 2: Statics 56

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

¦ zGA is known as

Resultant Force and Centre of Pressure on a general plane surface in a liquid.

O

O Q

z

z

the 1st Moment of Area of the plane PQ about the free surface.

θ elemental area δA

s

om

Fluid density ρ Resultant Force R D

area δA

C

Sc

.c

G

x

d

area A

ru m

G

P

x

Ug ¦ zGA

In terms of distance from point O

¦ zGA

Ax sin T

u sinT

= 1st moment of area about a line through O (as

z

x sin T )

The resultant force on a plane

(assuming U and g as constant).

CIVE1400: Fluid Mechanics

A is the area of the plane

.jn

w w

w

F = UgzGA Resultant force on plane

And it is known that ¦ zGA Az

z is the distance to the centre of gravity (centroid)

tu

w

p = Ugz

R

ld

or

Measuring down from the surface, the pressure on an element GA, depth z,

fo

Take pressure as zero at the surface.

So force on element

Section 2: Statics

R Section 2: Statics 57

CIVE1400: Fluid Mechanics

UgAz UgAx sin T Section 2: Statics 58

Section 2: Statics

CIVE1400: Fluid Mechanics

This resultant force acts at right angles through the centre of pressure, C, at a depth D.

Section 2: Statics

Sum of moments

Moment of R about O =

How do we find this position?

om

CIVE1400: Fluid Mechanics

Equating

.c

Take moments of the forces.

The force on each elemental area:

Ug s sin T GA u s

w w

Moment of Force on GA about O

.jn

the moment of this force is:

tu

UgzGA Ug s sin T GA

w

Force on GA

Ug sin T GAs 2

fo

¦ s GA 2

Sc

or

w

It is convenient to take moment about O

Ug sin T ¦ s 2 GA

The position of the centre of pressure along the plane measure from the point O is:

ld

As the plane is in equilibrium: The moment of R will be equal to the sum of the moments of the forces on all the elements GA about the same point.

R u S c = UgAx sin T S c

ru m

UgAx sin T S c

Ug sin T ¦ s 2 GA

Ax

How do we work out the summation term?

This term is known as the 2nd Moment of Area , Io, of the plane (about the axis through O)

U , g and T are the same for each element, giving the total moment as

CIVE1400: Fluid Mechanics

Section 2: Statics 59

CIVE1400: Fluid Mechanics

Section 2: Statics 60

CIVE1400: Fluid Mechanics

Section 2: Statics

2nd moment of area about O

Io

CIVE1400: Fluid Mechanics

ÂŚ s 2GA

How do you calculate the 2nd moment of area?

om

It can be easily calculated for many common shapes.

.c

2nd moment of area is a geometric property.

It can be found from tables BUT only for moments about an axis through its centroid = IGG.

ru m

The position of the centre of pressure along the plane measure from the point O is:

fo

2 nd Moment of area about a line through O

ld

st

1 Moment of area about a line through O

w

or

Sc

Usually we want the 2nd moment of area about a different axis.

Through O in the above examples.

.jn

tu

and

Section 2: Statics

We can use the

w w

D S c sin T

w

Depth to the centre of pressure is

CIVE1400: Fluid Mechanics

Section 2: Statics 61

parallel axis theorem to give us what we want.

CIVE1400: Fluid Mechanics

Section 2: Statics 62

Section 2: Statics

CIVE1400: Fluid Mechanics

The 2nd moment of area about a line through the centroid of some common shapes.

The parallel axis theorem can be written

Io

Section 2: Statics

I GG Ax 2

om

CIVE1400: Fluid Mechanics

Area A

2nd moment of area, I GG , about an axis through the centroid

bd

bd 3 12

bd 2

bd 3 36

SR 2

SR 4

.c

Shape

We then get the following equation for the position of the centre of pressure

ru m

Rectangle b

ld or .jn

tu

D

I GG x Ax §I ¡ sin T ¨ GG x ¸ Š Ax š

w

Sc

fo

h

G

G

Triangle h G

h/3 b

Circle

w w

w

(In the examination the parallel axis theorem and the I GG will be given)

CIVE1400: Fluid Mechanics

G

R G

G

4

SR 2

Semicircle G

Section 2: Statics 63

R (4R)/(3Ď€)

CIVE1400: Fluid Mechanics

2

01102 .

R4

Section 2: Statics 64

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

An example: Find the moment required to keep this triangular gate closed on a tank which holds water.

om

Submerged vertical surface Pressure diagrams

1.2m

For vertical walls of constant width it is possible to find the resultant force and centre of pressure graphically using a

D

.c

2.0m

ru m

1.5m

G

Section 2: Statics

We know the relationship between pressure and depth: p = Ugz

So we can draw the diagram below: Ď gz

z H

2H 3 R

w

.jn

tu

w

or

ld

fo

C

pressure diagram.

w w

p Ď gH

This is know as a pressure diagram.

CIVE1400: Fluid Mechanics

Section 2: Statics 65

CIVE1400: Fluid Mechanics

Section 2: Statics 66

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Pressure increases from zero at the surface linearly by p = Ugz, to a maximum at the base of p = UgH.

om

For a triangle the centroid is at 2/3 its height i.e. the resultant force acts 2 horizontally through the point z H. 3

.c

The area of this triangle represents the resultant force per unit width on the vertical wall,

ru m

For a vertical plane the depth to the centre of pressure is given by 2 H 3

w

.jn

tu

w

or

D

( N / m)

w w

1 UgH 2 2

ld

1 u AB u BC 2 1 HUgH 2 1 UgH 2 2

Resultant force per unit width R

fo

Units of this are Newtons per metre.

Area

Section 2: Statics

The force acts through the centroid of the pressure diagram.

CIVE1400: Fluid Mechanics

Section 2: Statics 67

CIVE1400: Fluid Mechanics

Section 2: Statics 68

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

The same technique can be used with combinations of liquids are held in tanks (e.g. oil floating on water).

For example:

Check this against the moment method: The resultant force is given by:

UgAz

oil Ď o

om

R

UgAx sinT

H Ug H u 1 sinT 2 1 UgH 2 2

1.2m

D

R

ru m

Ď g0.8

Ď g1.2

Find the position and magnitude of the resultant force on this vertical wall of a tank which has oil floating on water as shown.

fo or

ld

§I ¡ sin T ¨ o ¸ Š Ax š

and by the parallel axis theorem (with width of 1) 2

w

Depth to the centre of pressure

2 H 3

w w

§ H 3 / 3¡ D ¨ 2 ¸ Š H / 2š

H3 3

tu

H 1u H3 1 u H §¨ ¡¸ Š 2š 12

w

I GG Ax 2

.jn

Io

0.8m

.c

water Ď

and the depth to the centre of pressure by:

D

Section 2: Statics

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Section 2: Statics 69

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Section 2: Statics 70

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

Submerged Curved Surface In the diagram below liquid is resting on top of a curved base.

om

If the surface is curved the resultant force must be found by combining the elemental forces using some vectorial method.

.c

E

ru m

Calculate the

O

FAC

RH

A

ld or

B

Rv

R

The fluid is at rest – in equilibrium.

w

.jn

tu

w

Combine these to obtain the resultant force and direction.

C G

fo

horizontal and vertical components.

D

w w

(Although this can be done for all three dimensions we will only look at one vertical plane)

CIVE1400: Fluid Mechanics

Section 2: Statics 71

So any element of fluid such as ABC is also in equilibrium.

CIVE1400: Fluid Mechanics

Section 2: Statics 72

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

Consider the Horizontal forces The resultant horizontal force of a fluid above a curved surface is: RH = Resultant force on the projection of the curved surface onto a vertical plane.

om

The sum of the horizontal forces is zero.

C B

.c

RH

We know 1. The force on a vertical plane must act horizontally (as it acts normal to the plane).

ru m

FAC

A

2. That RH must act through the same point.

w

tu

Horizontal forces act only on the faces AC and AB as shown.

or

ld

fo

No horizontal force on CB as there are no shear forces in a static fluid

So:

RH acts horizontally through the centre of pressure of the projection of the curved surface onto an vertical plane.

.jn

FAC, must be equal and opposite to RH.

w w

w

AC is the projection of the curved surface AB onto a vertical plane.

CIVE1400: Fluid Mechanics

We have seen earlier how to calculate resultant forces and point of action.

Hence we can calculate the resultant horizontal force on a curved surface.

Section 2: Statics 73

CIVE1400: Fluid Mechanics

Section 2: Statics 74

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Consider the Vertical forces

Section 2: Statics

Resultant force

The sum of the vertical forces is zero.

E

D

om

The overall resultant force is found by combining the vertical and horizontal components vectorialy,

C B

.c

G

ru m

Resultant force A Rv

w

fo ld

The angle the resultant force makes to the horizontal is §R ¡ T tan 1 ¨ V ¸ Š RH š

.jn

tu

So we can say The resultant vertical force of a fluid above a curved surface is:

2 RH RV2

And acts through O at an angle of T.

or

There are no shear force on the vertical edges, so the vertical component can only be due to the weight of the fluid.

R

w w

w

RV = Weight of fluid directly above the curved surface.

It will act vertically down through the centre of gravity of the mass of fluid.

CIVE1400: Fluid Mechanics

Section 2: Statics 75

The position of O is the point of interaction of the horizontal line of action of R H and the vertical line of action of RV .

CIVE1400: Fluid Mechanics

Section 2: Statics 76

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

A typical example application of this is the determination of the forces on dam walls or curved sluice gates.

Section 2: Statics

What are the forces if the fluid is below the curved surface?

Find the magnitude and direction of the resultant force of water on a quadrant gate as shown below.

om

This situation may occur or a curved sluice gate.

C

.c

Gate width 3.0m

ru m

1.0m

O

FAC

RH

A Rv

R

The force calculation is very similar to when the fluid is above.

w w

w

.jn

tu

w

or

ld

fo

Water Ď = 1000 kg/m3

B G

CIVE1400: Fluid Mechanics

Section 2: Statics 77

CIVE1400: Fluid Mechanics

Section 2: Statics 78

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Horizontal force

Section 2: Statics

Vertical force C B G

O

.c

A

RH

A’

ru m

FAC

om

B

Rv

What vertical force would keep this in equilibrium?

If the region above the curve were all water there would be equilibrium.

.jn

tu

w

The resultant horizontal force, RH acts as shown in the diagram. Thus we can say:

or

ld

fo

The two horizontal on the element are: The horizontal reaction force RH The force on the vertical plane A’B.

A

w w

w

The resultant horizontal force of a fluid below a curved surface is: RH = Resultant force on the projection of the curved surface onto a vertical plane.

CIVE1400: Fluid Mechanics

Section 2: Statics 79

Hence: the force exerted by this amount of fluid must equal he resultant force.

The resultant vertical force of a fluid below a curved surface is: Rv =Weight of the imaginary volume of fluid vertically above the curved surface.

CIVE1400: Fluid Mechanics

Section 2: Statics 80

CIVE1400: Fluid Mechanics

Section 2: Statics

CIVE1400: Fluid Mechanics

Section 2: Statics

An example of a curved sluice gate which experiences force from fluid below.

A 1.5m long cylinder lies as shown in the figure, holding back oil of relative density 0.8. If the cylinder has a mass of 2250 kg find a) the reaction at A b) the reaction at B

om

The resultant force and direction of application are calculated in the same way as for fluids above the surface:

A

D

B

tu

w

§R ¡ tan 1 ¨ V ¸ Š RH š

w w

w

.jn

T

C

or

And acts through O at an angle of T. The angle the resultant force makes to the horizontal is

fo

ru m

2 RH RV2

E

ld

R

.c

Resultant force

CIVE1400: Fluid Mechanics

Section 2: Statics 81

CIVE1400: Fluid Mechanics

Section 2: Statics 82

LECTURE CONTENTS

Fluid Dynamics

om

Objectives

ru m

.c

1.Identify differences between: x steady/unsteady x uniform/non-uniform x compressible/incompressible flow

tu

w

or

ld

fo

2.Demonstrate streamlines and stream tubes

w w

w

.jn

Section 0: Introduction Section 1: Fluid Properties Fluids vs. Solids Viscosity Newtonian Fluids Properties of Fluids Section 2: Statics Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces Section 3: Dynamics The continuity equation.

The Bernoulli Equation.

Application of Bernoulli equation.

The momentum equation.

Application of momentum equation.

Section 4: Real Fluids Boundary layer.

Laminar flow in pipes.

Section 5: Dimensional Analysis An Intro to Dimensional analysis Similarity CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 83

3.Introduce the Continuity principle 4.Derive the Bernoulli (energy) equation 5.Use the continuity equations to predict pressure and velocity in flowing fluids 6.Introduce the momentum equation for a fluid 7.Demonstrate use of the momentum equation to predict forces induced by flowing fluids

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 84

Flow Classification

The analysis of fluid in motion

Fluid flow may be classified under the following headings

om

Fluid dynamics:

.jn

tu

w

or

ld

fo

ru m

By use of the fundamental laws of physics and the physical properties of the fluid Some fluid flow is very complex: e.g. x Spray behind a car x waves on beaches; x hurricanes and tornadoes x any other atmospheric phenomenon

w w

w

All can be analysed with varying degrees of success (in some cases hardly at all!).

There are many common situations which analysis gives very accurate predictions CIVE 1400: Fluid Mechanics

uniform: Flow conditions (velocity, pressure, crosssection or depth) are the same at every point in the fluid. non-uniform: Flow conditions are not the same at every point. steady: Flow conditions may differ from point to point but DO NOT change with time.

.c

Fluid motion can be predicted in the same way as the motion of solids

Section 3: Fluid dynamics 85

unsteady: Flow conditions change with time at any point.

Fluid flowing under normal circumstances - a river for example conditions vary from point to point we have non-uniform flow.

If the conditions at one point vary as time passes then we have unsteady flow.

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 86

Compressible or Incompressible Flow?

Steady uniform flow.

Conditions do not change with position in the stream or with time.

E.g. flow of water in a pipe of constant diameter at constant velocity.

All fluids are compressible - even water.

Density will change as pressure changes.

om

Combining these four gives.

.c

Under steady conditions - provided that changes in pressure are small we usually say the fluid is incompressible - it has constant density.

fo ld or

Three-dimensional flow

In general fluid flow is three-dimensional.

Pressures and velocities change in all directions.

.jn

tu

w

Unsteady uniform flow.

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

E.g. A pipe of constant diameter connected to a pump pumping at a constant rate which is then switched off.

ru m

Steady non-uniform flow.

Conditions change from point to point in the stream but do not change with time.

E.g. Flow in a tapering pipe with constant velocity at the inlet.

w w

w

Unsteady non-uniform flow.

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

E.g. Waves in a channel.

In many cases the greatest changes only occur in two directions or even only in one.

Changes in the other direction can be effectively ignored making analysis much more simple.

This course is restricted to Steady uniform flow - the most simple of the four.

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 87

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 88

Two-dimensional flow

Conditions vary only in the direction of flow not across the cross-section.

Conditions vary in the direction of flow and in one direction at right angles to this.

The flow may be unsteady with the parameters varying in time but not across the cross-section.

E.g. Flow in a pipe.

Flow patterns in two-dimensional flow can be shown by curved lines on a plane.

ru m

.c

om

One dimensional flow:

Below shows flow pattern over a weir.

Ideal flow

Real flow

w

Pipe

.jn

ld or

tu

w

Since flow must be zero at the pipe wall - yet non-zero in the centre there is a difference of parameters across the cross-section.

fo

But:

w w

Should this be treated as two-dimensional flow?

Possibly - but it is only necessary if very high accuracy is required.

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 89

In this course we will be considering: x steady x incompressible x one and two-dimensional flow CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 90

Some points about streamlines:

It is useful to visualise the flow pattern.

Lines joining points of equal velocity - velocity contours - can be drawn.

x Close to a solid boundary, streamlines are parallel to that boundary

om

Streamlines

.c

x The direction of the streamline is the direction of the fluid velocity

ru m

These lines are know as streamlines.

x Fluid can not cross a streamline

Here are 2-D streamlines around a cross-section of an aircraft wing shaped body:

tu

w

or

ld

fo

x Streamlines can not cross each other

x In unsteady flow streamlines can change position with time x In steady flow, the position of streamlines does not change.

w

.jn

Fluid flowing past a solid boundary does not flow into or out of the solid surface.

x Any particles starting on one streamline will stay on that same streamline

w w

Very close to a boundary wall the flow direction must be along the boundary.

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 91

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 92

Some points about streamtubes

A circle of points in a flowing fluid each has a streamline passing through it.

x The “walls” of a streamtube are streamlines.

om

Streamtubes

x Fluid cannot flow across a streamline, so fluid cannot cross a streamtube “wall”.

.c

These streamlines make a tube-like shape known as a streamtube

x In steady flow, the position of streamtubes does not change.

tu

w

or

ld

fo

ru m

x A streamtube is not like a pipe.

Its “walls” move with the fluid.

x In unsteady flow streamtubes can change position with time

w w

w

.jn

In a two-dimensional flow the streamtube is flat (in the plane of the paper):

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 93

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 94

More commonly we use volume flow rate Also know as discharge.

om

Mass flow rate

The symbol normally used for discharge is Q.

mass time taken to accumulate this mass

.c

dm dt

Volume flow rate - Discharge.

ru m

m

Flow rate

ld or

mass of fluid in bucket time taken to collect the fluid 8.0 2.0 7 0.857kg / s m =

A simple example: If the bucket above fills with 2.0 litres in 25 seconds, what is the discharge?

w

mass flow rate

volume of fluid time

fo

A simple example: An empty bucket weighs 2.0kg. After 7 seconds of collecting water the bucket weighs 8.0kg, then:

discharge, Q

tu

Q

2.0 u 10 3 m3 25 sec

.jn

0.0008 m3 / s

w w

w

0.8 l / s

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 95

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 96

A simple example:

Discharge and mean velocity

If A = 1.2u10-3m2 And discharge, Q is 24 l/s, mean velocity is

om

If we know the discharge and the diameter of a pipe, we can deduce the mean velocity

um

.c

um t

area A Cylinder of fluid

Cross sectional area of pipe is A Mean velocity is um.

This is because the velocity in the pipe is not constant across the cross section.

w

or

ld

Note how we have called this the mean velocity.

In time t, a cylinder of fluid will pass point X with a volume Au um u t.

Q

CIVE 1400: Fluid Mechanics

tu

w

volume A u um u t = t time Aum

w w

Q=

x

.jn

The discharge will thus be

12 . u 10 3 2.0 m / s

fo

Pipe

2.4 u 10 3

ru m

x

Q A

u um

umax

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

Section 3: Fluid dynamics 97

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 98

Applying to a streamtube:

Continuity This principle of conservation of mass says matter cannot be created or destroyed

om

Mass enters and leaves only through the two ends (it cannot cross the streamtube wall).

ru m

Mass flow in Control volume

A2

Ď 1 u1 A1

Mass entering = Mass leaving per unit time per unit time

w

Mass entering = Mass leaving + Increase per unit time per unit time of mass in control vol per unit time

or

For any control volume the principle of conservation of mass says

ld

fo

Mass flow out

U1GA1u1

tu

.jn

w w

w

For steady flow there is no increase in the mass within the control volume, so

For steady flow Mass entering = Mass leaving per unit time per unit time CIVE 1400: Fluid Mechanics

Ď 2 u2

.c

This is applied in fluids to fixed volumes, known as control volumes (or surfaces)

Section 3: Fluid dynamics 99

U2GA2u2

Or for steady flow, U1GA1u1

U2GA2 u2

Constant

m

dm dt

This is the continuity equation.

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 100

In a real pipe (or any other vessel) we use the mean velocity and write

U2 A2 um2

Constant

m

om

U1 A1um1

Some example applications of Continuity

For incompressible, fluid U1 = U2 = U

A liquid is flowing from left to right.

By the continuity

A2 u2 U2

As we are considering a liquid, U1 = U 2 = U

Q1 A1u1

.jn

tu

w

This is the continuity equation most often used.

Q2 A2u2

An example: If the area A1=10u10-3 m2 and A2=3u10-3 m2 And the upstream mean velocity u1=2.1 m/s.

The downstream mean velocity is

w w

w

This equation is a very powerful tool.

It will be used repeatedly throughout the rest of this course.

CIVE 1400: Fluid Mechanics

A1u1U1

fo ld

Q

or

A2 u2

ru m

(dropping the m subscript)

A1u1

Section 2

.c

Section 1

u2

A1 u1 A2 7.0 m / s

Section 3: Fluid dynamics 101

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 102

Now try this on a diffuser, a pipe which expands or diverges as in the figure below,

Velocities in pipes coming from a junction.

om

2

Section 2

If d1=30mm and d2=40mm and the velocity u2=3.0m/s.

mass flow into the junction = mass flow out

/4 /4

u2

ld

A2 u2 A1

d 22 u 2 2 d1

or

u1

fo

The velocity entering the diffuser is given by,

S d 22 S d12

2

U 1 Q1 = U 2 Q 2 + U 3 Q3

When incompressible

w

§ d2 · ¨ ¸ u2 © d1 ¹

3

ru m

Section 1

.c

1

tu

Q1 = Q 2 + Q 3

2

w

.jn

§ 40· ¨ ¸ 3.0 5.3 m / s © 30¹

w w

$ 1 u1 = $ 2 u2 + $ 3 u3

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 103

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 104

If pipe 1 diameter = 50mm, mean velocity 2m/s, pipe 2 diameter 40mm takes 30% of total discharge and pipe 3 diameter 60mm.

What are the values of discharge and mean velocity in each pipe?

Discharge in:

om

The Bernoulli equation is a statement of the principle of conservation of energy along a streamline

A1u1

.c

§ Sd12 ¡ ¸ u1 ¨ 4 š Š

ru m

Q1

The Bernoulli equation

0.00392 m3 / s Discharges out:

Q2 Q3 Q1 0.3Q1

Q3

0.7Q1

w

0.00275 m3 / s

CIVE 1400: Fluid Mechanics

0.936 m / s

Q3

A3u3

u3

0.972 m / s

.jn

u2

w

A2 u2

w w

Q2

tu

Velocities out:

H = Constant

These terms represent:

ld

Q1

or

0.3Q1

p1 u12 z Ug 2 g 1

fo

0.001178m3 / s

Q2

It can be written:

Section 3: Fluid dynamics 105

Pressure

Kinetic

Potential

Total

energy per energy per energy per unit weight unit weight unit weight

energy per unit weight

These term all have units of length, they are often referred to as the following:

p pressure head = Ug potential head = z CIVE 1400: Fluid Mechanics

u2 velocity head = 2g total head = H Section 3: Fluid dynamics 106

The derivation of Bernoulli’s Equation:

Restrictions in application of Bernoulli’s equation:

Cross sectional area a

B B’ A

x Flow is steady

om

z

A’

mg

x Density is constant (incompressible)

.c

An element of fluid, as that in the figure above, has potential energy due to its height z above a datum and kinetic energy due to its velocity u. If the element has weight mg then potential energy = mgz

ru m

x Friction losses are negligible x It relates the states at two points along a single streamline, (not conditions on two different streamlines)

w w

w

.jn

fo

tu

Fortunately, for many real situations where the conditions are approximately satisfied, the equation gives very good results.

CIVE 1400: Fluid Mechanics

kinetic energy =

ld or

w

All these conditions are impossible to satisfy at any instant in time!

potential energy per unit weight =

z

1 2 mu 2

kinetic energy per unit weight =

u2 2g

At any cross-section the pressure generates a force, the fluid will flow, moving the cross-section, so work will be done. If the pressure at cross section AB is p and the area of the cross-section is a then force on AB = pa when the mass mg of fluid has passed AB, cross-section AB will have moved to A’B’ volume passing AB =

mg Ug

m

U

therefore

Section 3: Fluid dynamics 107

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 108

distance AA’ =

m Ua

The Bernoulli equation is applied along streamlines like that joining points 1 and 2 below.

work done = force u distance AA’

2

om

pm

U p Ug

.c

work done per unit weight =

ru m

=

m pa u Ua

This term is know as the pressure energy of the flowing stream.

Summing all of these energy terms gives

fo w

2

.jn

tu

H

CIVE 1400: Fluid Mechanics

w w

w

By the principle of conservation of energy, the total energy in the system does not change, thus the total head does not change. So the Bernoulli equation can be written

p u2

z Ug 2 g

p1 u12 z Ug 2 g 1

ld

Total energy per unit weight

or

p u

z Ug 2 g

total head at 1 = total head at 2

or

or

Pressure Kinetic Potential energy per energy per energy per unit weight unit weight unit weight

1

H Constant

Section 3: Fluid dynamics 109

p2 u22 z Ug 2 g 2

This equation assumes no energy losses (e.g. from friction) or energy gains (e.g. from a pump) along the streamline. It can be expanded to include these simply, by adding the appropriate energy terms: Total

Total

Loss

Work done

Energy

energy per energy per unit per unit per unit supplied unit weight at 1 weight at 2 weight weight per unit weight

p1 u12 z Ug 2 g 1

CIVE 1400: Fluid Mechanics

p2 u22 z h w q Ug 2 g 2

Section 3: Fluid dynamics 110

Apply the Bernoulli equation along a streamline joining section 1 with section 2. p1 u12 p2 u22 z1 z2

Practical use of the Bernoulli Equation The Bernoulli equation is often combined with the continuity equation to find velocities and pressures at points in the flow connected by a streamline.

om

Ug 2 g

.c

p2

p2

fo

p1

section 2

3

u2

A1u1 A2

p2

Section 3: Fluid dynamics 111

2

§ d1 ¡ ¨ ¸ u1 Š d2 š

200000 17296.87 182703 N / m2

.jn

w

w w

What is the gauge pressure at section 2?

CIVE 1400: Fluid Mechanics

A2u2

So pressure at section 2

tu

A fluid, density U = 960 kg/m is flowing steadily through the above tube.

The section diameters are d1=100mm and d2=80mm.

The gauge pressure at 1 is p1=200kN/m2 The velocity at 1 is u1=5m/s.

The tube is horizontal (z1=z2)

2

(u12 u22 )

7.8125 m / s

w

section 1

U

A1u1

ld

u2

p1

Use the continuity equation to find u2

or

u1

ru m

An example: Finding pressures and velocities within a contracting and expanding pipe.

Ug 2 g

182.7 kN / m2 Note how the velocity has increased the pressure has decreased

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 112

We have used both the Bernoulli equation and the Continuity principle together to solve the problem.

Pitot Tube The Pitot tube is a simple velocity measuring device.

om

Uniform velocity flow hitting a solid blunt body, has streamlines similar to this:

.c

Use of this combination is very common. We will be seeing this again frequently throughout the rest of the course.

fo

Applications of the Bernoulli Equation

or

Some move to the left and some to the right.

The centre one hits the blunt body and stops.

At this point (2) velocity is zero The fluid does not move at this one point.

This point is known as the stagnation point.

Using the Bernoulli equation we can calculate the pressure at this point.

Along the central streamline at 1: velocity u1 , pressure p1 at the stagnation point of: u2 = 0. (Also z1 = z2)

p1 u12

U 2

w w

w

.jn

tu

w

Here we will see its application to flow measurement from tanks, within pipes as well as in open channels.

ld

The Bernoulli equation is applicable to many situations not just the pipe flow.

CIVE 1400: Fluid Mechanics

2

ru m

1

p2 Section 3: Fluid dynamics 113

CIVE 1400: Fluid Mechanics

p2

U 1 p1 Uu12 2 Section 3: Fluid dynamics 114

How can we use this?

Pitot Static Tube The necessity of two piezometers makes this arrangement awkward.

om

The blunt body does not have to be a solid.

I could be a static column of fluid.

The Pitot static tube combines the tubes and they can then be easily connected to a manometer.

1

ru m

h2

fo

h1

.c

Two piezometers, one as normal and one as a Pitot tube within the pipe can be used as shown below to measure velocity of flow.

ld

2

u

or tu

.jn

We now have an expression for velocity from two pressure measurements and the application of the Bernoulli equation.

CIVE 1400: Fluid Mechanics

X

h A

B

[Note: the diagram of the Pitot tube is not to scale. In reality its diameter is very small and can be ignored i.e. points 1 and 2 are considered to be at the same level]

w

Ugh2

1 p1 Uu12 2 1 Ugh1 Uu12 2 2 g (h2 h1 )

w w

p2

1

w

We have the equation for p2 ,

1 2

Section 3: Fluid dynamics 115

The holes on the side connect to one side of a manometer, while the central hole connects to the other side of the manometer

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 116

Using the theory of the manometer,

pA

p1 Ug X h Uman gh

pB

p2 UgX

pA

pB

om

p1 Ug X h Uman gh

.c

It is a rapidly converging section which increases the velocity of flow and hence reduces the pressure.

u1

p1

It then returns to the original dimensions of the pipe by a gently diverging ‘diffuser’ section.

Uu12

2 2 gh( Um U )

U

about 20°

2

w

The Pitot/Pitot-static is:

w

x Gives velocities (not discharge)

1

.jn

tu

x Simple to use (and analyse)

about 6°

fo

p1 hg Uman U

ru m

1 p1 Uu12 , giving 2

ld

p2

The Venturi meter is a device for measuring discharge in a pipe.

or

p2 UgX We know that

Venturi Meter

w w

x May block easily as the holes are small.

z2 z1

h

datum

Apply Bernoulli along the streamline from point 1 to point 2 CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 117

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 118

p2 u22

z Ug 2 g 2

The theoretical (ideal) discharge is uuA. Actual discharge takes into account the losses due to friction, we include a coefficient of discharge (Cd |0.9)

By continuity

Qideal

u2 A2

Qactual

u1 A1 A2

Qactual

Substituting and rearranging gives

fo

2 º u12 ª§ A1 · «¨ ¸ 1» 2 g «© A2 ¹ »¼ ¬

ld

p1 p2 z1 z2 Ug

u1 A1

Cd Qideal

Cd u1 A1

.c

u2

u1 A1

ru m

Q

om

p1 u12

z Ug 2 g 1

In terms of the manometer readings

p1 Ugz1 p1 p2 z1 z2 Ug

p2 Uman gh Ug ( z2 h) §U · h¨ man 1¸ © U ¹

Giving

tu

w

or

u12 ª A12 A22 º « » 2 g ¬ A22 ¼

ª p p2 º z1 z2 » 2g« 1 ¬ Ug ¼ Cd A1 A2 2 2 A1 A2

.jn

w

w w

u1

ª p p2 º z1 z2 » 2g« 1 ¬ Ug ¼ A2 2 2 A1 A2

CIVE 1400: Fluid Mechanics

Qactual

Cd A1 A2

· §U 2 gh¨ man 1¸ ¹ © U A12 A22

This expression does not include any elevation terms. (z1 or z2) When used with a manometer The Venturimeter can be used without knowing its angle.

Section 3: Fluid dynamics 119

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 120

Flow Through A Small Orifice Venturimeter design:

x The diffuser assures a gradual and steady deceleration after the throat. So that pressure rises to something near that before the meter. x The angle of the diffuser is usually between 6 and 8 degrees.

ru m

h

2

Vena contractor

fo

x Wider and the flow might separate from the walls increasing energy loss.

Aactual

.c

1

om

Flow from a tank through a hole in the side.

w

x The efficiency of the diffuser of increasing pressure back to the original is rarely greater than 80%.

or

ld

x If the angle is less the meter becomes very long and pressure losses again become significant.

The streamlines at the orifice contract reducing the area of flow.

This contraction is called the vena contracta.

The amount of contraction must be known to calculate the flow.

w w

w

.jn

tu

x Care must be taken when connecting the manometer so that no burrs are present.

The edges of the hole are sharp to minimise frictional losses by minimising the contact between the hole and the liquid.

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 121

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 122

Apply Bernoulli along the streamline joining point 1 on the surface to point 2 at the centre of the orifice.

The discharge through the orifice is jet area u jet velocity

om

At the surface velocity is negligible (u1 = 0) and the pressure atmospheric (p1 = 0).

The area of the jet is the area of the vena contracta not the area of the orifice.

.c

At the orifice the jet is open to the air so again the pressure is atmospheric (p2 = 0).

ru m

We use a coefficient of contraction to get the area of the jet

If we take the datum line through the orifice then z1 = h and z2 =0, leaving

2 gh

tu

w

This theoretical value of velocity is an overestimate as friction losses have not been taken into account.

Q Qactual

.jn

w w

Each orifice has its own coefficient of velocity, they usually lie in the range( 0.97 - 0.99)

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 123

Au Aactual uactual Cc Cv Aorifice utheoretical Cd Aorifice utheoretical Cd Aorifice 2 gh

w

Cv utheoretical

Cc Aorifice

Giving discharge through the orifice:

A coefficient of velocity is used to correct the theoretical velocity,

uactual

Aactual

fo ld

u2

or

h

u22 2g

Cd is the coefficient of discharge, C d = C c u Cv CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 124

This Q is the same as the flow out of the orifice so

Time for the tank to empty We have an expression for the discharge from the tank

Cd Ao 2 gh

A

Cd Ao 2 gh

om

Q

.c

We can use this to calculate how long it will take for level in the to fall

Gh A Cd Ao 2 g h

Gt

ru m

As the tank empties the level of water falls.

The discharge will also drop.

Gh Gt

fo

Integrating between the initial level, h1, and final level, h2, gives the time it takes to fall this height

ld

h1

t

w

or

h2

tu

§ 1 ¨³ © h

.jn

The tank has a cross sectional area of A.

Q

t

w w

w

In a time dt the level falls by dh The flow out of the tank is

A h2 Gh ³ Cd Ao 2 g h1 h

³ h 1/2

A >2 h @hh12 Cd Ao 2 g 2A Cd Ao 2 g

Au

Gh Q A Gt

· 2 h¸ ¹

2h1/ 2

>

h2 h1

@

(-ve sign as dh is falling) CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 125

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 126

Submerged Orifice What if the tank is feeding into another?

Time for Equalisation of Levels in Two Tanks The time for the levels to equalise can be found by integrating this equation as before.

Area A1

om

Area A2

h1

By the continuity equation

ru m

.c

h2

Orifice area Ao

Apply Bernoulli from point 1 on the surface of the deeper tank to point 2 at the centre of the orifice, p1 u12 p2 u22

z1

z2

u2

Ugh2 u22

0 Ug 2 g 2 g (h1 h2 )

Cd Ao u

fo ld

A1 A2

and

QGt Cd Ao 2 g (h1 h2 ) Gt

w

Q

.jn

And the discharge is given by

w

0 0 h1

2g

or

Ug

2g

Cd Ao 2 g (h1 h2 )

w w

but

Section 3: Fluid dynamics 127

h

Gt

So the discharge of the jet through the submerged orifice depends on the difference in head across the orifice.

CIVE 1400: Fluid Mechanics

QGt

Gh Gh1 Gh2 so A1Gh1 A2Gh1 A2Gh A2Gh Gh1

tu

Ug

writing

Gh1 Gh A2 2 Gt Gt A1Gh1 A2Gh2 A1

Q

CIVE 1400: Fluid Mechanics

A1Gh1 A1 A2 Gh A1 A2 h1 h2

Gh A1 A2 A1 A2 Cd Ao 2 g h

Section 3: Fluid dynamics 128

Integrating between the two levels

A1 A2 h final Gh Âł ( A1 A2 )Cd Ao 2 g hinitial h

x A notch is an opening in the side of a tank or reservoir.

om

t

Flow Over Notches and Weirs

x It is a device for measuring discharge.

h final 2 A1 A2 > h @hinitial ( A1 A2 )Cd Ao 2 g

.c

ru m

h final hinitial

@

fo

>

x It is used as both a discharge measuring device and a device to raise water levels.

ld

2 A1 A2 ( A1 A2 ) Cd Ao 2 g

x A weir is a notch on a larger scale - usually found in rivers.

CIVE 1400: Fluid Mechanics

x There are many different designs of weir. x We will look at sharp crested weirs.

Weir Assumptions

w tu

w w

w

.jn

The time for the levels to equal use hinitial = (h1 - h2) and hfinal = 0

or

h is the difference in height between the two levels (h2 - h1)

x velocity of the fluid approaching the weir is small so we can ignore kinetic energy. x The velocity in the flow depends only on the depth below 2 gh the free surface. u These assumptions are fine for tanks with notches or reservoirs with weirs, in rivers with high velocity approaching the weir is substantial the kinetic energy must be taken into account

Section 3: Fluid dynamics 129

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 130

Rectangular Weir

Consider a horizontal strip of width b, depth h below the free surface

The width does not change with depth so

om

A General Weir Equation

b

b

.c

h

H

Au bGh 2 gh

B

H

Substituting this into the general weir equation gives H Qtheoretical B 2 g ³ h1/ 2 dh 0

w

.jn

tu

w

Integrating from the free surface, h=0, to the weir crest, h=H, gives the total theoretical discharge H Qtheoretical 2 g ³ bh1/ 2 dh 0

w w

This is different for every differently shaped weir or notch.

2 B 2 gH 3/ 2 3 To get the actual discharge we introduce a coefficient of discharge, Cd, to account for losses at the edges of the weir and contractions in the area of flow,

Qactual

We need an expression relating the width of flow across the weir to the depth below the free surface.

CIVE 1400: Fluid Mechanics

B

fo ld

2 gh

or

discharge through the strip, GQ

ru m

δh

velocity through the strip, u

constant

Section 3: Fluid dynamics 131

CIVE 1400: Fluid Mechanics

Cd

2 B 2 gH 3 / 2 3 Section 3: Fluid dynamics 132

‘V’ Notch Weir

The Momentum Equation And Its Applications

The relationship between width and depth is dependent on the angle of the “V”.

om

b

We have all seen moving fluids exerting forces.

h

H

.c

θ

ru m

x The lift force on an aircraft is exerted by the air moving over the wing.

The width, b, a depth h from the free surface is

§T · 2 H h tan¨ ¸ © 2¹

fo

b

H §T · 0

w

8 §T · 2 g tan¨ ¸ H 5/ 2 © 2¹ 15

.jn

tu

H § T · ª 2 3/ 2 2 5/ 2 º 2 2 g tan¨ ¸ « Hh h » © 2¹ ¬ 3 5 ¼0

or

2 2 g tan¨ ¸ ³ H h h1/ 2 dh © 2¹

CIVE 1400: Fluid Mechanics

Cd

w w

The actual discharge is obtained by introducing a coefficient of discharge

Qactual

The analysis of motion is as in solid mechanics: by use of Newton’s laws of motion.

w

Qtheoretical

ld

So the discharge is

x A jet of water from a hose exerts a force on whatever it hits.

8 §T · 2 g tan¨ ¸ H 5 / 2 © 2¹ 15 Section 3: Fluid dynamics 133

The Momentum equation is a statement of Newton’s Second Law

It relates the sum of the forces to the acceleration or rate of change of momentum.

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 134

In time Gt a volume of the fluid moves from the inlet a distance u1Gt, so

What mass of moving fluid we should use?

volume entering the stream tube = area u distance

om

From solid mechanics you will recognise F = ma

ru m

.c

We use a different form of the equation.

= A 1u1 Gt

Consider a streamtube: And assume steady non-uniform flow

The mass entering,

fo

mass entering stream tube = volume u density

ld

A2

or

u2

A1 u1

w

ρ2

ρ1

And momentum

momentum entering stream tube = mass u velocity = U1 A1 u1 Gt u1

Similarly, at the exit, we get the expression:

momentum leaving stream tube = U2 A 2 u 2 Gt u 2

w w

w

.jn

tu

u1 δt

= U1 A1 u1 Gt

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 135

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 136

By Newton’s 2nd Law.

An alternative derivation From conservation of mass

Force = rate of change of momentum

om

we can write

( U2 A2u2Gt u2 U1 A1u1Gt u1 ) Gt

.c

rate of change of mass

ru m

F=

mass into face 1 = mass out of face 2

We know from continuity that

dm dt U1 A1u1 U2 A2u2 m

fo

The rate at which momentum enters face 1 is

or w

U2 A2 u2 u2 mu 2 Thus the rate at which momentum changes across the stream tube is

w w

w

.jn

F QU (u2 u1 )

CIVE 1400: Fluid Mechanics

U1 A1u1u1 mu 1

The rate at which momentum leaves face 2 is

tu

And if we have a fluid of constant density, i.e. U1 U2 U , then

ld

Q A1u1 A2 u2

U2 A2 u2 u2 U1 A1u1u1 mu 2 mu 1 So

Force = rate of change of momentum F m ( u2 u1 ) Section 3: Fluid dynamics 137

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 138

The previous analysis assumed the inlet and outlet velocities in the same direction i.e. a one dimensional system.

F m ( u2 u1 )

What happens when this is not the case?

.c

QU ( u2 u1 )

fo

The Momentum equation.

θ2

θ1 u1

We consider the forces by resolving in the directions of the co-ordinate axes.

tu

w

or

ld

This force acts on the fluid in the direction of the flow of the fluid.

.jn

The force in the x-direction

Fx

m u2 cosT2 u1 cosT1

w

m u2 x u1 x

or

w w CIVE 1400: Fluid Mechanics

u2

ru m

F

om

So we have these two expressions, either one is known as the momentum equation

Fx

UQ u2 cosT2 u1 cosT1

UQ u2 x u1 x

Section 3: Fluid dynamics 139

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 140

And the force in the y-direction

Fy

In summary we can say:

m u2 sin T2 u1 sin T1

Total force on the fluid =

om

m §¨Š u2 y u1 y ¡¸š or

.c

UQ u2 sin T2 u1 sin T1

ru m

Fy

ld

The resultant force can be found by combining these components Fy

UQ uout uin

Remember that we are working with vectors so F is in the direction of the velocity.

w

or

FResultant

φ

.jn

w

Fx2 Fy2

tu

Fx

Fresultant

F

m uout uin

or

fo

UQ§¨Š u2 y u1 y ¡¸š

F

rate of change of momentum through the control volume

I

CIVE 1400: Fluid Mechanics

w w

And the angle of this force

§ Fy ¡ tan 1 ¨ ¸ Š Fx š

Section 3: Fluid dynamics 141

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 142

This force is made up of three components:

Application of the Momentum Equation

FR = Force exerted on the fluid by any solid body touching the control volume

om

Force due the flow around a pipe bend

FB = Force exerted on the fluid body (e.g. gravity)

A converging pipe bend lying in the horizontal plane turning through an angle of T.

ru m

.c

FP = Force exerted on the fluid by fluid pressure outside the control volume

fo

So we say that the total force, FT, is given by the sum of these forces:

or

ld

F T = FR + FB + F P

A2

x

ρ1 u1

θ

A1

tu .jn

w

by the fluid on the solid body

u2

w

The force exerted

ρ2

y

w w

touching the control volume is opposite to FR. So the reaction force, R, is given by R = -FR

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 143

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 144

Why do we want to know the forces here?

Control Volume

om

As the fluid changes direction a force will act on the bend.

y

fo

2 Co-ordinate axis system

or

ld

We need to know how much force a support (thrust block) must withstand.

Any co-ordinate axis can be chosen.

Choose a convenient one, as above.

3 Calculate the total force In the x-direction:

FT x

w w

w

.jn

tu

w

Step in Analysis: 1.Draw a control volume 2.Decide on co-ordinate axis system 3.Calculate the total force 4.Calculate the pressure force 5.Calculate the body force 6.Calculate the resultant force

CIVE 1400: Fluid Mechanics

x

ru m

.c

This force can be very large in the case of water supply pipes.

The bend must be held in place to prevent breakage at the joints.

Section 3: Fluid dynamics 145

CIVE 1400: Fluid Mechanics

UQ u2 x u1 x

u1 x

u1

u2 x

u2 cosT

FT x

UQ u2 cosT u1

Section 3: Fluid dynamics 146

u1 y

u1 sin 0

u2 y

u2 sin T

FT y

UQu2 sin T

0

FP

FP y

ld or

.jn

where hf is the friction loss (this can often be ignored, hf=0)

w

Ug 2 g

tu

Ug 2 g

p1 CIVE 1400: Fluid Mechanics

p2

There are no body forces in the x or y directions.

FRx = FRy = 0 The body force due to gravity is acting in the z-direction so need not be considered.

6 Calculate the resultant force

w w

w

As the pipe is in the horizontal plane, z1=z2 And with continuity, Q= u1A1 = u2A2

FT x

UQ 2 § 1

1¡ ¨ ¸

2 Š A22 A12 š

FT y Section 3: Fluid dynamics 147

p2 A2 sin T

5 Calculate the body force

Use Bernoulli.

p2 u22

z2 h f

p1 A1 sin 0 p2 A2 sin T

p1 A1 p2 A2 cosT

fo

4 Calculate the pressure force If we know pressure at the inlet we can relate this to the pressure at the outlet.

p1 u12

z1

p1 A1 cos 0 p2 A2 cosT

ru m

FP x

pressure force at 1 - pressure force at 2

om

UQ u2 y u1 y

FT y

Knowing the pressures at each end the pressure force can be calculated,

.c

In the y-direction:

CIVE 1400: Fluid Mechanics

FR x FP x FB x FR y FP y FB y Section 3: Fluid dynamics 148

FR x

FT x FP x 0

Impact of a Jet on a Plane

UQ u2 cosT u1 p1 A1 p2 A2 cosT

om

FR y

A jet hitting a flat plate (a plane) at an angle of 90q

FT y FP y 0

We want to find the reaction force of the plate. i.e. the force the plate will have to apply to stay in the same position.

ru m

.c

UQu2 sin T p2 A2 sin T And the resultant force on the fluid is given by FRy

fo

FResultant

ld

φ

or

FRx

tu

u1

w

¨ ¸ Š FR x š

x

u2

w w

tan

F 1 § R y ¡

u2

.jn

And the direction of application is

I

y

w

FR2 x FR2 y

FR

1 & 2 Control volume and Co-ordinate axis are shown in the figure below.

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

R CIVE 1400: Fluid Mechanics

FR Section 3: Fluid dynamics 149

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 150

3 Calculate the total force In the x-direction

FR x

UQ u2 x u1 x

UQu1 x

UQu1 x

.c

Exerted on the fluid.

The system is symmetrical the forces in the y-direction cancel.

ru m

The force on the plane is the same magnitude but in the opposite direction

y

ld

.jn

0

u1

5 Calculate the body force As the control volume is small we can ignore the body force due to gravity.

θ

w w

FB x

FB y

u2

x

w

FP y

If the plane were at an angle the analysis is the same.

But it is usually most convenient to choose the axis system normal to the plate.

tu

w

4 Calculate the pressure force.

The pressures at both the inlet and the outlets to the control volume are atmospheric.

The pressure force is zero

FP x

FR x

R

fo

0

or

FT y

FR x FP x FB x FT x 0 0

om

FT x

FT x

u3

0

6 Calculate the resultant force CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 151

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 152

Force on a curved vane

3 Calculate the total force in the x direction

This case is similar to that of a pipe, but the analysis is simpler.

om

FT x

.c

Pressures at ends are equal - atmospheric

by continuity u1

fo

ru m

Both the cross-section and velocities (in the direction of flow) remain constant.

w w

w

.jn

θ

Section 3: Fluid dynamics 153

UQ u2 sin T 0

4 Calculate the pressure force.

The pressure at both the inlet and the outlets to the control volume is atmospheric.

1 & 2 Control volume and Co-ordinate axis are shown in the figure above.

CIVE 1400: Fluid Mechanics

FT x

Q2 U A

tu

u1

Q , so A

Q2 U 1 cosT

A

FT y

w

x

u2

and in the y-direction

or

ld

u2

y

UQ u2 u1 cosT

FP x CIVE 1400: Fluid Mechanics

FP y

0

Section 3: Fluid dynamics 154

5 Calculate the body force

And the resultant force on the fluid is given by

No body forces in the x-direction, FB x = 0.

om

In the y-direction the body force acting is the weight of the fluid.

If V is the volume of the fluid on the vane then,

.c

And the direction of application is

ru m

UgV

FB x

(This is often small as the jet volume is small and sometimes ignored in analysis.)

CIVE 1400: Fluid Mechanics

fo ld

w

The force on the vane is the same magnitude but in the opposite direction

tu

R

FR

.jn

FT x

Q2 U 1 cosT

A

§ FR y ¡ ¸ tan 1 ¨ F Š Rx š

w

FR x

FR x FP x FB x

I

exerted on the fluid.

or

6 Calculate the resultant force

FT x

FR2 x FR2 y

FR

FR y FP y FB y

FR y

Q2 U A

FT y

w w

FT y

Section 3: Fluid dynamics 155

CIVE 1400: Fluid Mechanics

Section 3: Fluid dynamics 156

LECTURE CONTENTS

Real fluids x Introduction to Laminar and Turbulent flow

om

x Head loss in pipes

.c

x Hagen-Poiseuille equation

ru m

x Boundary layer theory

x Boundary layer separation

tu

w

or

ld

fo

x Losses at bends and junctions

w w

w

.jn

Section 0: Introduction Section 1: Fluid Properties Fluids vs. Solids Viscosity Newtonian Fluids Properties of Fluids Section 2: Statics Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces Section 3: Dynamics The continuity equation.

The Bernoulli Equation.

Application of Bernoulli equation.

The momentum equation.

Application of momentum equation.

Section 4: Real Fluids Laminar and turbulent flow Boundary layer theory Section 5: Dimensional Analysis An Intro to Dimensional analysis Similarity CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 157

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 158

Laminar and turbulent flow

x “stick� to solid surfaces x have stresses within their body.

Injecting a dye into the middle of flow in a pipe, what would we expect to happen?

This

om

Flowing real fluids exhibit viscous effects, they:

ru m this

P

du dy

.jn

W

tu

For a “Newtonian� fluid we can write:

w

or

The shear stress, W, in a fluid is proportional to the velocity gradient - the rate of change of velocity across the flow.

fo

du dy

ld

W v

.c

From earlier we saw this relationship between shear stress and velocity gradient:

or this

w w

w

where P is coefficient of viscosity (or simply viscosity).

Here we look at the influence of forces due to momentum changes and viscosity in a moving fluid.

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 159

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 160

All three would happen but for different flow rates.

The was first investigated in the 1880s

by Osbourne Reynolds in a classic experiment in fluid mechanics.

Top: Slow flow Middle: Medium flow Bottom: Fast flow

om

.c ld

tu

w

Laminar flow: Motion of the fluid particles is very orderly all particles moving in straight lines parallel to the pipe walls.

fo

ru m

Laminar flow Transitional flow Turbulent flow

or

Top: Middle: Bottom:

A tank arranged as below:

w w

w

.jn

Turbulent flow: Motion is, locally, completely random but the overall direction of flow is one way.

But what is fast or slow?

At what speed does the flow pattern change?

And why might we want to know this?

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 161

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 162

After many experiments he found this expression

What are the units of Reynolds number?

We can fill in the equation with SI units:

Uud P

om

.c ru m

d = diameter

u = mean velocity, P = viscosity

fo

U = density,

U kg / m3 , u m / s, P Ns / m2 kg / m s

or tu

w

This value is known as the Reynolds number, Re:

CIVE 1400: Fluid Mechanics

w w

Laminar flow: Transitional flow: Turbulent flow:

Uud P

m

kg m m m s 1 m3 s 1 kg

It has no units!

A quantity with no units is known as a non-dimensional (or dimensionless) quantity.

(We will see more of these in the section on dimensional analysis.)

.jn

Uud P

w

Re

ld

This could be used to predict the change in flow type for any fluid.

Re

d

Re < 2000 2000 < Re < 4000 Re > 4000

Section 4: Real Fluids 163

The Reynolds number, Re, is a non-dimensional number.

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 164

At what speed does the flow pattern change?

What is the MINIMUM velocity when flow is turbulent i.e. Re = 4000

We use the Reynolds number in an example:

Re

U = 1000 kg/m3

CIVE 1400: Fluid Mechanics

ru m ld or

.jn

tu

w u

2000

w

u

Uud P

2000P 2000 u 0.55 u 10 3 1000 u 0.5 Ud 0.0022 m / s

w w

Re

0.0044 m / s

In a house central heating system, typical pipe diameter = 0.015m,

d = 0.5m P = 0.55x103 Ns/m2

What is the MAXIMUM velocity when flow is laminar i.e. Re = 2000

4000

.c

u

fo

water density pipe diameter (dynamic) viscosity,

om

A pipe and the fluid flowing have the following properties:

Uud P

Section 4: Real Fluids 165

limiting velocities would be, 0.0733 and 0.147m/s.

Both of these are very slow.

In practice laminar flow rarely occurs in a piped water system.

Laminar flow does occur in fluids of greater viscosity e.g. in bearing with oil as the lubricant.

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 166

Laminar flow

What does this abstract number mean? x Re < 2000

x ‘low’ velocity

We can give the Re number a physical meaning.

x Dye does not mix with water x Fluid particles move in straight lines x Simple mathematical analysis possible

.c

x Rare in practice in water systems.

ru m

Re

om

This may help to understand some of the reasons for the changes from laminar to turbulent flow.

Uud P

Transitional flow

x 2000 > Re < 4000 x ‘medium’ velocity x Dye stream wavers - mixes slightly.

or

ld

fo

inertial forces viscous forces

.jn

tu

w

When inertial forces dominate (when the fluid is flowing faster and Re is larger) the flow is turbulent.

w w

w

When the viscous forces are dominant (slow flow, low Re) they keep the fluid particles in line, the flow is laminar.

Turbulent flow x Re > 4000 x ‘high’ velocity x Dye mixes rapidly and completely x Particle paths completely irregular x Average motion is in flow direction x Cannot be seen by the naked eye x Changes/fluctuations are very difficult to detect. Must use laser. x Mathematical analysis very difficult - so experimental measures are used

x Most common type of flow.

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 167

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 168

Attaching a manometer gives pressure (head) loss due to the energy lost by the fluid overcoming the shear stress.

Pressure loss due to friction in a pipeline Up to now we have considered ideal fluids: no energy losses due to friction

om

L

ru m

.c

Because fluids are viscous, energy is lost by flowing fluids due to friction.

This must be taken into account.

w

.jn

tu

In a real flowing fluid shear stress slows the flow.

or

ld

fo

The effect of the friction shows itself as a pressure (or head) loss.

Î&#x201D;p

How can we quantify this pressure loss in terms of the forces acting on the fluid?

w w

w

To give a velocity profile:

The pressure at 1 (upstream) is higher than the pressure at 2.

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 169

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 170

Consider a cylindrical element of incompressible fluid flowing in the pipe,

As the flow is in equilibrium,

τw το

Sd 2

om

τw

'p

area A

.c

το

driving force = retarding force

ru m

Ww is the mean shear stress on the boundary Upstream pressure is p, Downstream pressure falls by 'p to (p-'p)

4 'p

W wSdL Ww 4 L d

fo

Giving pressure loss in a pipe in terms of:

4

w

'p

tu

'p A

.jn

pA p 'p A

Sd 2

x pipe diameter x shear stress at the wall

or

driving force = Pressure force at 1 - pressure force at 2

ld

The driving force due to pressure

w

The retarding force is due to the shear stress

= W wSdL

CIVE 1400: Fluid Mechanics

w w

shear stress u area over which it acts = W w u area of pipe wall

Section 4: Real Fluids 171

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 172

What is the variation of shear stress in the flow?

Shear stress and hence pressure loss varies with velocity of flow and hence with Re.

Ď&#x201E;w Many experiments have been done with various fluids measuring the pressure loss at various Reynolds numbers.

om

R

.c

r

Ď&#x201E;w

ru m

A graph of pressure loss and Re look like:

At the wall

fo

R 'p 2 L

ld

Ww

w tu

W

r 'p 2 L r Ww R

.jn

W

or

At a radius r

w

A linear variation in shear stress.

This graph shows that the relationship between pressure loss and Re can be expressed as

w w

This is valid for: x steady flow x laminar flow x turbulent flow CIVE 1400: Fluid Mechanics

laminar

'p v u

turbulent 'p v u1.7 ( or Section 4: Real Fluids 173

CIVE 1400: Fluid Mechanics

2 .0 )

Section 4: Real Fluids 174

Pressure loss during laminar flow in a pipe

The fluid is in equilibrium, shearing forces equal the pressure forces.

In general the shear stress Ww. is almost impossible to measure.

W 2Sr L

om

W

.c

For laminar flow we can calculate a theoretical value for a given velocity, fluid and pipe dimension.

fo ld

P

du dr

Giving:

'p r du P L 2 dr 'p r du

dr L 2P

w

or

W

tu

.jn

w w w CIVE 1400: Fluid Mechanics

du , dy

We are measuring from the pipe centre, so

And the stress on the fluid in laminar flow is entirely due to viscose forces.

As before, consider a cylinder of fluid, length L, radius r, flowing steadily in the centre of a pipe.

r

P

ru m

Newtons law of viscosity saysW

In laminar flow the paths of individual particles of fluid do not cross.

Flow is like a series of concentric cylinders sliding over each other.

'p A 'pSr 2 'p r L 2

In an integral form this gives an expression for velocity,

δr r

u

R

Section 4: Real Fluids 175

CIVE 1400: Fluid Mechanics

'p 1 r dr L 2P Âł Section 4: Real Fluids 176

The value of velocity at a point distance r from the centre

What is the discharge in the pipe?

'p r 2

C L 4P

The flow in an annulus of thickness Gr

om

GQ ur Aannulus Aannulus

At r = 0, (the centre of the pipe), u = umax, at r = R (the pipe wall) u = 0; 2

At a point r from the pipe centre when the flow is laminar:

Q

'p S R 2 R r r 3 dr Âł L 2P 0

ld

Q

w w w

'p Sd 4 L128P

'p Sd 4 L 128P

This is the Hagen-Poiseuille Equation for laminar flow in a pipe

v

CIVE 1400: Fluid Mechanics

So the discharge can be written

.jn

tu

w

This is a parabolic profile (of the form y = ax2 + b ) so the velocity profile in the pipe looks similar to

'p SR 4 L 8P

or

ur

'p 1 R2 r 2

L 4P

'p 1 R 2 r 2 2SrGr L 4P

ru m

'p R L 4P

GQ

fo

C

S (r Gr )2 Sr 2 | 2SrGr

.c

ur

Section 4: Real Fluids 177

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 178

Boundary Layers

To get pressure loss (head loss) in terms of the velocity of the flow, write pressure in terms of head loss hf, i.e. p = Ughf

Recommended reading: Fluid Mechanics

by Douglas J F, Gasiorek J M, and Swaffield J A. Longman publishers. Pages 327-332.

u

Ugh f d 2 32 PL

Fluid flowing over a stationary surface, e.g. the bed of a river, or the wall of a pipe, is brought to rest by the shear stress to This gives a, now familiar, velocity profile:

.c

Q/ A

ru m

u

om

Mean velocity:

ld

w

Ugd

2

Wall

Zero at the wall A maximum at the centre of the flow.

.jn

Pressure loss is directly proportional to the velocity when flow is laminar.

The profile doesnâ&#x20AC;&#x2122;t just exit.

It is build up gradually.

w w w

It has been validated many time by experiment.

It justifies two assumptions: 1.fluid does not slip past a solid boundary 2.Newtons hypothesis.

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 179

Ď&#x201E;o

zero velocity

or

32 PLu

tu

hf

fo

Head loss in a pipe with laminar flow by the Hagen-Poiseuille equation:

umax

Starting when it first flows past the surface e.g. when it enters a pipe.

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 180

Understand this Boundary layer growth diagram.

Considering a flat plate in a fluid.

.c

om

Upstream the velocity profile is uniform, This is known as free stream flow.

ru m

Downstream a velocity profile exists.

This is known as fully developed flow.

or

ld

fo

Free stream flow

w

.jn

tu

w

Fully developed flow

w w

Some question we might ask:

How do we get to the fully developed state?

Are there any changes in flow as we get there?

Are the changes significant / important?

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 181

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 182

Boundary layer thickness:

First: viscous forces (the forces which hold the fluid together)

G = distance from wall to where u = 0.99 umainstream

om

When the boundary layer is thin: velocity gradient du/dy, is large

by Newtonâ&#x20AC;&#x2122;s law of viscosity shear stress, W = P (du/dy), is large.

ru m

.c

G increases as fluid moves along the plate.

It reaches a maximum in fully developed flow.

The force may be large enough to drag the fluid close to the surface.

fo

The G increase corresponds to a drag force increase on the fluid.

As the boundary layer thickens velocity gradient reduces and shear stress decreases.

Eventually it is too small to drag the slow fluid along.

tu

w

xmore fluid is slowed xby friction between the fluid layers xthe thickness of the slow layer increases.

or

ld

As fluid is passes over a greater length:

w

.jn

Fluid near the top of the boundary layer drags the fluid nearer to the solid surface along.

w w

The mechanism for this dragging may be one of two types:

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 183

Up to this point the flow has been laminar.

Newtonâ&#x20AC;&#x2122;s law of viscosity has applied.

This part of the boundary layer is the laminar boundary layer

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 184

Close to boundary velocity gradients are very large.

Viscous shear forces are large.

Possibly large enough to cause laminar flow.

This region is known as the laminar sub-layer.

Second: momentum transfer

om

If the viscous forces were the only action the fluid would come to a rest.

This layer occurs within the turbulent zone it is next to the wall.

It is very thin â&#x20AC;&#x201C; a few hundredths of a mm.

ru m

.c

Viscous shear stresses have held the fluid particles in a constant motion within layers.

Eventually they become too small to hold the flow in layers; the fluid starts to rotate.

Surface roughness effect

In turbulent flow: Roughness higher than laminar sub-layer: increases turbulence and energy losses.

.jn

tu

w

or

ld

fo

Despite its thinness, the laminar sub-layer has vital role in the friction characteristics of the surface.

w w

w

The fluid motion rapidly becomes turbulent.

Momentum transfer occurs between fast moving main flow and slow moving near wall flow.

Thus the fluid by the wall is kept in motion.

The net effect is an increase in momentum in the boundary layer.

This is the turbulent boundary layer.

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 185

In laminar flow: Roughness has very little effect Boundary layers in pipes Initially of the laminar form.

It changes depending on the ratio of inertial and viscous forces; i.e. whether we have laminar (viscous forces high) or turbulent flow (inertial forces high).

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 186

Use Reynolds number to determine which state.

Uud Re

Boundary layer separation Divergent flows: Positive pressure gradients.

Pressure increases in the direction of flow.

P

Re < 2000

Transitional flow: 2000 <

Re > 4000

The fluid in the boundary layer has so little momentum that it is brought to rest, and possibly reversed in direction.

Reversal lifts the boundary layer.

u1

u2

p1

or

ld

fo

ru m

.c

Turbulent flow:

Re < 4000

om

Laminar flow:

p2

w

p1 < p2

u1 > u2

.jn

tu

Laminar flow: profile parabolic (proved in earlier lectures) The first part of the boundary layer growth diagram.

w w

w

Turbulent (or transitional), Laminar and the turbulent (transitional) zones of the boundary layer growth diagram.

Length of pipe for fully developed flow is the entry length.

Laminar flow

|120 u diameter

This phenomenon is known as boundary layer separation.

Turbulent flow | 60 u diameter CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 187

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 188

Boundary layer separation:

Examples of boundary layer separation

xincreases the turbulence A divergent duct or diffuser velocity drop (according to continuity) pressure increase (according to the Bernoulli equation).

xincreases the energy losses in the flow.

om

Separating / divergent flows are inherently unstable

ru m

.c

Convergent flows: x Negative pressure gradients

u1

w

u2

tu

p2

Increasing the angle increases the probability of boundary layer separation.

Venturi meter

.jn

p1

u1 < u 2

Diffuser angle of about 6q A balance between:

w w

w

p1 > p 2

ld

or

x Fluid accelerates and the boundary layer is thinner.

fo

x Pressure decreases in the direction of flow.

x Flow remains stable

xlength of meter xdanger of boundary layer separation.

x Turbulence reduces.

x Boundary layer separation does not occur.

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 189

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 190

ru m

.c

om

Tee-Junctions

or

Two separation zones occur in bends as shown above.

Pb > Pa causing separation.

Pd > Pc causing separation

w w

w

.jn

tu

w

Y-Junctions Tee junctions are special cases of the Y-junction.

ld

fo

Assuming equal sized pipes), Velocities at 2 and 3 are smaller than at 1.

Pressure at 2 and 3 are higher than at 1.

Causing the two separations shown

Localised effect Downstream the boundary layer reattaches and normal flow occurs.

Boundary layer separation is only local.

Nevertheless downstream of a junction / bend /valve etc. fluid will have lost energy.

Bends CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 191

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 192

.c

om

Flow past a cylinder Slow flow, Re < 0.5 no separation:

Fluid accelerates to get round the cylinder Velocity maximum at Y. Pressure dropped.

Adverse pressure between here and downstream.

Separation occurs

w w

w

.jn

tu

Fast flow Re > 70 vortices detach alternately.

Form a trail of down stream.

Karman vortex trail or street.

(Easily seen by looking over a bridge)

w

or

ld

fo

ru m

Moderate flow, Re < 70, separation vortices form.

Causes whistling in power cables.

Caused Tacoma narrows bridge to collapse.

Frequency of detachment was equal to the bridge natural frequency.

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 193

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 194

At too great an angle boundary layer separation occurs on the top Pressure changes dramatically.

This phenomenon is known as stalling.

ru m

.c

om

Aerofoil Normal flow over a aerofoil or a wing cross-section.

All, or most, of the â&#x20AC;&#x2DC;suctionâ&#x20AC;&#x2122; pressure is lost.

The plane will suddenly drop from the sky!

Solution: Prevent separation.

1 Engine intakes draws slow air from the boundary layer at the rear of the wing though small holes 2 Move fast air from below to top via a slot.

w w

w

.jn

tu

w

or

The velocity increases as air flows over the wing. The pressure distribution is as below so transverse lift force occurs.

ld

fo

(boundary layers greatly exaggerated)

CIVE 1400: Fluid Mechanics

3 Put a flap on the end of the wing and tilt it.

Section 4: Real Fluids 195

CIVE 1400: Fluid Mechanics

Section 4: Real Fluids 196

Examples:

LECTURE CONTENTS

w w

w

.jn

tu

w

or

ld

fo

ru m

.c

om

Exam questions involving boundary layer theory are typically descriptive. They ask you to explain the mechanisms of growth of the boundary layers including how, why and where separation occurs. You should also be able to suggest what might be done to prevent separation.

CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis An Intro to Dimensional analysis Similarity Section 4: Real Fluids 197

CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

198

Dimensions and units

Dimensional Analysis Application of fluid mechanics in design makes use of experiments results.

Any physical situation can be described by familiar properties.

Results often difficult to interpret.

om

Dimensional analysis provides a strategy for choosing relevant data.

e.g. length, velocity, area, volume, acceleration etc.

.c

Used to help analyse fluid flow Especially when fluid flow is too complex for mathematical analysis.

ru m

These are all known as dimensions.

Specific uses:

fo

Dimensions are of no use without a magnitude.

ld

x help design experiments

tu

e.g. What runs to do. How to interpret.

w

x Allows most to be obtained from experiment:

e.g metre, kilometre, Kilogram, a yard etc.

or

x Informs which measurements are important

i.e. a standardised unit

Dimensions can be measured.

Units used to quantify these dimensions.

.jn

It depends on the correct identification of variables In dimensional analysis we are concerned with the nature of the dimension

Relates these variables together

w

Doesnâ&#x20AC;&#x2122;t give the complete answer

i.e. its quality not its quantity.

w w

Experiments necessary to complete solution Uses principle of dimensional homogeneity Give qualitative results which only become quantitative from experimental analysis.

CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

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CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

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The following common abbreviations are used:

This table lists dimensions of some common physical quantities:

length

=L

mass

=M

velocity

time

=T

force

=F

temperature

=4

Quantity

SI Unit

Dimension ms

LT-1

acceleration

m/s2

ms-2

LT-2

force

N kg ms-2

M LT-2

kg m2s-2

ML2T-2

om

m/s

-1

.c

kg m/s2

ru m

energy (or work)

Here we will use L, M, T and F (not 4).

fo

power

or w

L, T

.jn

tu

and one of M or F

As either mass (M) of force (F) can be used to represent the other, i.e.

2

kg m /s

2

Watt W N m/s

Nms-1

kg m2/s3

kg m2s-3

density

kg/m3

specific weight

N/m3

Nm-2 2

kg/m2/s2 relative density

kg m-1s-2

ML-1T-2

kg m-3

ML-3

kg m-2s-2

ML-2T-2

a ratio

1

no units

surface tension

N s/m

2

no dimension -2

N sm

kg/m s

kg m-1s-1

N/m

Nm-1

kg /s

M = FT2L-1

ML2T-3

Pascal P, kg/m/s

viscosity

w

N m,

N/m2,

w w

F = MLT-2

pressure ( or stress)

ld

We can represent all the physical properties we are interested in with three:

Joule J

2

kg s-2

M L-1T-1

MT-2

We will always use LTM:

CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

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CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

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Dimensional Homogeneity

What exactly do we get from Dimensional Analysis?

Any equation is only true if both sides have the same dimensions.

A single equation,

It must be dimensionally homogenous.

om

Which relates all the physical factors of a problem to each other.

X

ru m

2 B 2 gH 3/ 2 3

.c

What are the dimensions of X?

Problem: What is the force, F, on a propeller?

L (LT-2)1/2 L3/2 = X 3/2

L (L T ) L 3

What might influence the force?

=X

fo

1/2 -1

-1

.jn

tu

(for L they are both 3, for T both -1).

w

The powers of the individual dimensions must be equal on both sides.

or

ld

L T =X

w

Dimensional homogeneity can be useful for: 1. Checking units of equations;

An example:

w w

2. Converting between two sets of units;

It would be reasonable to assume that the force, F, depends on the following physical properties? diameter,

d

forward velocity of the propeller (velocity of the plane), u fluid density,

U

revolutions per second,

N

fluid viscosity,

P

3. Defining dimensionless relationships

CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

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CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

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Common S groups From this list we can write this equation:

Several groups will appear again and again.

These often have names.

0 = I ( F, d, u, U, N, P )

They can be related to physical forces.

om

F = I ( d, u, U, N, P )

.c

or

Other common non-dimensional numbers

ru m

I and I1 are unknown functions.

Reynolds number: Uud Re inertial, viscous force ratio

ld

fo

Dimensional Analysis produces:

or

0

w

§ F Nd P · I¨ 2 2 , , ¸ © Uu d u Uud ¹

tu

These groups are dimensionless.

.jn

I will be determined by experiment.

w w

w

These dimensionless groups help to decide what experimental measurements to take.

CIVE 1400: Fluid Mechanics

or ( S groups):

P

Euler number: p En Uu 2 Froude number: u2 Fn gd Weber number: Uud We

V

Mach number: u Mn c Section 5: Dimensional Analysis

205

CIVE 1400: Fluid Mechanics

pressure, inertial force ratio

inertial, gravitational force ratio

inertial, surface tension force ratio

Local velocity, local velocity of sound ratio

Section 5: Dimensional Analysis

216

Kinematic similarity

Similarity

The similarity of time as well as geometry.

It exists if: i. the paths of particles are geometrically similar ii. the ratios of the velocities of are similar

om

Similarity is concerned with how to transfer measurements from models to the full scale.

Three types of similarity which exist between a model and prototype:

.c

Some useful ratios are: Vm L m / Tm O L Velocity Vp L p / Tp OT

fo

ru m

Geometric similarity: The ratio of all corresponding dimensions in the model and prototype are equal.

or

ld

For lengths Lmodel Lm OL Lprototype Lp

Discharge

Qm Qp

Lm / Tm2 L p / Tp2 L3m / Tm L3p / Tp

OL O2T O3L OT

Oa

OQ

A consequence is that streamline patterns are the same.

w w

w

.jn

tu

w

OL is the scale factor for length.

For areas Amodel L2m O2L 2 Aprototype Lp

Acceleration

am ap

Ou

All corresponding angles are the same.

CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

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CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

218

Dynamic similarity

Modelling and Scaling Laws Measurements taken from a model needs a scaling law applied to predict the values in the prototype.

If geometrically and kinematically similar and the ratios of all forces are the same.

om

O U O2L O2u

For resistance R, of a body moving through a fluid.

R, is dependent on the following:

.c

M m am M pa p

An example:

ru m

Fm Fp

Force ratio 2 Um L3m O L 2 § OL · OU O L ¨ ¸ u U p L3p O2T © OT ¹

This occurs when the controlling S group is the same for model and prototype.

ld

R Uu 2 l 2

.jn w

Section 5: Dimensional Analysis

ML-1T-1

§ Uul · ¸ © P ¹

I¨

§ Uul · ¸ © P ¹

Uu 2 l 2 I ¨

This applies whatever the size of the body i.e. it is applicable to prototype and a geometrically similar model.

w w

CIVE 1400: Fluid Mechanics

P:

Taking U, u, l as repeating variables gives:

R

It is possible another group is dominant.

In open channel i.e. river Froude number is often taken as dominant.

l:(length) L

I (R, U, u, l, P ) = 0

tu

U p upd p Pp

LT-1

u:

So

w

or

The controlling S group is usually Re. So Re is the same for model and prototype:

Um um dm Pm

ML-3

fo

U

219

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Section 5: Dimensional Analysis

220

Example 1

For the model

An underwater missile, diameter 2m and length 10m is tested in a water tunnel to determine the forces acting on the real prototype.

A 1/20th scale model is to be used. If the maximum allowable speed of the prototype missile is 10 m/s, what should be the speed of the water in the tunnel to achieve dynamic similarity?

§U u l · I¨ m m m ¸ © Pm ¹

om

Rm Um um2 lm2 and for the prototype

.c

U p u 2p l p2

Dynamic similarity so Reynolds numbers equal: Um um dm U p u p d p

§U u l · I¨ p p p ¸ © Pp ¹

Dividing these two equations gives

I Um um lm / P m

I U puplp / P p

w

W can go no further without some assumptions.

CIVE 1400: Fluid Mechanics

up

dp dm

10

1 1 / 20

200 m / s

This is a very high velocity.

This is one reason why model tests are not always done at exactly equal Reynolds numbers.

i.e. a scaling law for resistance force:

OR

um

w

Um um2 lm2 U p u 2p l p2

w w

Rm Rp

Both the model and prototype are in water then, Pm = Pp and Um = Up so

.jn

U p upd p Pp

so

The model velocity should be U d P um u p p p m Um d m P p

tu

Assuming dynamic similarity, so Reynolds number are the same for both the model and prototype:

Um um dm Pm

Pp

fo

Rm / Um u l Rp / U p u l

ld

2 m 2 p

or

2 m 2 p

Pm

ru m

Rp

A wind tunnel could have been used so the values of the U and P ratios would be used in the above.

OU O2u O2L Section 5: Dimensional Analysis

221

CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

222

Example 2

So the model velocity is found to be

A model aeroplane is built at 1/10 scale and is to be tested in a wind tunnel operating at a pressure of 20 times atmospheric. The aeroplane will fly at 500km/h. At what speed should the wind tunnel operate to give dynamic similarity between the model and prototype? If the drag measure on the model is 337.5 N what will be the drag on the plane?

um

om

um

.c

Uu l I Re

2 2

Rm Rp

fo

For dynamic similarity Rem = Rep, so

Uu l

Uu l

2 2

m

2 2

20 0.5

1 1

p 2

2 .

01

1

0.05

So the drag force on the prototype will be

ld

U d P up p p m Um d m P p

Rp

1 Rm 0.05

20 u 337.5 6750 N

w

or

um

Rm Rp

ru m

R

§ Uul · Uu l I ¨ ¸ © P ¹

0.5u p

And the ratio of forces is

Earlier we derived an equation for resistance on a body moving through air: 2 2

1 1 20 1 / 10 250 km / h

up

.jn

tu

The value of P does not change much with pressure so Pm = Pp

20 p p pp

Um CIVE 1400: Fluid Mechanics

Um RT U p RT Um Up

Um Up

w w

pm pp

w

For an ideal gas is p = URT so the density of the air in the model can be obtained from

20U p Section 5: Dimensional Analysis

223

CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

224

Geometric distortion in river models

om

For practical reasons it is difficult to build a geometrically similar model.

.c

A model with suitable depth of flow will often be far too big - take up too much floor space.

ru m

Keeping Geometric Similarity result in: x depths and become very difficult to measure;

fo

x the bed roughness becomes impracticably small;

w

tu

Solution: Abandon geometric similarity.

or

ld

x laminar flow may occur (turbulent flow is normal in rivers.)

Resulting in:

w

.jn

Typical values are 1/100 in the vertical and 1/400 in the horizontal.

w w

x Good overall flow patterns and discharge x local detail of flow is not well modelled.

The Froude number (Fn) is taken as dominant.

Fn can be the same even for distorted models.

CIVE 1400: Fluid Mechanics

Section 5: Dimensional Analysis

225