1

Fluid kinematics

Joseph Lagrange (1736-

1813)

Leonhard Euler (1707-

1783)

2

Continuum hypothesis

• Cannot quantify motion of ﬂuid by tracking

motion of individual molecules (with a large

random component).

• Consider the ﬂuid to be made of lots of small

particles (with many molecules) that interact

with each other and surrounding

• The motion of the ﬂuid can is determined by the

velocity and acceleration of these ﬂuid particles

• Any ﬂuid property,

p,T,ρ,v

may be written as

a function of its spatial coordinates,

x,y,z

and

secondly the time

t

.

• So the temperature would be

T = T(x,y,z,t)

.

• The properties of the ﬂuid will generally be a

continuous function.

3

The velocity ﬁeld

The velocity ﬁeld is one of the most important ﬂuid

variable.It is a ﬁeld since velocity has direction and

well as magnitude.

v =

dr

A

dt

v = u(x,y,z,t)i +v(x,y,z,t)j +w(x,y,z,t)k

The velocity ﬁeld describes the motion of a piece of

the ﬂuid at

r

A

= (x,y,z)

at time

t

.

Sometimes

v = |v| =

√

u

2

+v

2

+w

2

called the

velocity.

4

Depiction of velocity ﬁeld

Draw the velocity ﬁeld for

v =

V

0

ℓ

(xi −yj)

and

identify the stagnation point.

• On

x

-axis

v =

V

0

x

ℓ

i

• On

y

-axis

v =

−V

0

y

ℓ

j

• Magnitude of velocity

v =

V

0

ℓ

p

x

2

+y

2

• Stagnation point at origin.

5

Euler and Lagrange descriptions

Euler approach The ﬂuid properties

p,ρ,v

are

written as functions of space and times.The

ﬂow is determined by the analyzing the behavior

of the functions.

Lagrange approach Pieces of the ﬂuid are

“tagged”.The ﬂuid ﬂow properties are

determined by tracking the motion and

properties of the particles as they move in time.

6

Euler vs Lagrange

Consider smoke going

up a chimney

Euler approach Attach thermometer to the top of

chimney,point

0

.Record

T

as a function of

time.As diﬀerent smoke particles pass through

O

,the temperature changes.Gives

T(x

0

,y

0

,z

0

,t)

.More thermometers to get

T(x,y,z,t)

.

Lagrange approach Thermometers are attached

to a particle,

A

.End up with

T

A

= T

A

(a)

.Can

have many particles and track

T

for all of them.

If we also know,position of each particle of

function of time,can translate Lagrange

information into Euler information.

7

Euler vs Lagrange

It is generally more common to use Eulerian

approach to ﬂuid ﬂows.Measuring water

temperature,or pressure at a point in a pipe.

Lagrangian methods sometimes used in experiments.

Throwing tracers into moving water bodies to

determine currents (see movie Twister).X-ray

opaque tracers in human blood.

Bird migration example.Ornithologists with

binoculars count migrating birds moving past a

(Euler) or scientists place radio transmitters on the

birds (Lagrange).

8

Streamlines,streaklines,pathlines

Streamlines,streaklines and

pathlines are used in the

visualization of ﬂuid ﬂow.

Streamlines mainly used in

analytic work,streaklines

and pathlines used in exper-

imental work.

x

y

v

u

v

streamlines are tangent to the velocity ﬁeld.For

steady ﬂow,the streamlines are ﬁxed in space.

Unsteady ﬂow,streamlines may change with

time.The slope of the streamline is equal to

tangent of velocity ﬁeld.

dy

dx

=

v

u

The streamlines can be determined from velocity

ﬁeld by integrating the lines deﬁne the tangents.

9

Streaklines

streak-lines Consist of all the particles in a ﬂow

that have passed through a common point.

Mainly a laboratory tool.

A streak-line can be made by injecting dye into

a moving ﬂuid at a speciﬁc point.For a steady

ﬂow,each particle follows the previous ones

precisely,and the streak-line is the same as the

streamline.

For unsteady ﬂows,particles injected at the

same point at diﬀerent times need not follow the

same path.An instantaneous photograph of the

marked ﬂuid would show the streak.The

streak-line would not be the same as the

streamline.

Pathline This is the trajectory followed by one

particle when it moves from one point to the

next.One injects dye at a point for an instant of

time,then does a time exposure photograph.

For steady ﬂow

streamline = streakline = pathline

10

Streaklines

Streaklines of dye moving past obstruction.They are

also the streamlines for the ﬂow.

Streaklines of smoke moving past obstruction.

11

The material derivative

In the Eulerian method,the fundamental property is

the velocity ﬁeld.The velocity ﬁeld does not track

the behaviour of individual partials,it describes the

velocity of whatever happens to be at a given

location.To do dynamics,need to apply

F = ma

.

Getting the acceleration is not trivial

For particle

A

,

x

A

(t),y

A

(t),z

A

(t)

describe the

motion of the particle.So

v

A

= v

A

(x

A

(t),y

A

(t),z

A

(t),t)

12

The material derivative

a

A

=

dv

A

dt

=

∂v

A

∂t

+

∂v

A

∂x

dx

A

dt

+

∂v

A

∂y

dy

A

dt

+

∂v

A

∂z

dz

A

dt

=

∂v

A

∂t

+

∂v

A

∂x

u

A

(t) +

∂v

A

∂y

v

A

(t) +

∂v

A

∂z

w

A

(t)

Since

A

is any particle,the acceleration ﬁeld is

a =

∂v

∂t

+u

∂v

∂x

+v

∂v

∂y

+w

∂v

∂z

The

a

is a vector with components

a

x

=

∂u

∂t

+u

∂u

∂x

+v

∂u

∂y

+w

∂u

∂z

a

y

=

∂v

∂t

+u

∂v

∂x

+v

∂v

∂y

+w

∂v

∂z

a

z

=

∂w

∂t

+u

∂w

∂x

+v

∂w

∂y

+w

∂w

∂z

13

The material derivative

The material derivative is often written

a =

Dv

Dt

with

D()

Dt

=

∂()

∂t

+u

∂()

∂x

+v

∂()

∂y

+w

∂()

∂z

One can deﬁne the material derivative for other

properties for a ﬂuid,e.g.temperature or pressure.

DT

Dt

=

∂T

∂t

+u

∂T

∂x

+v

∂T

∂y

+w

∂T

∂z

The material derivative allows for two types of

contribution.Unsteady eﬀects when

∂

∂t

6= 0

and

convective when

∂

∂xyz

6= 0

.

14

Unsteady eﬀects

Consider water from

a header tank ﬂowing

down a uniform cross

section pipe.

V

0

(t)

V

0

(t)

x

The water velocity at all points will be the same.

However the water velocity will gradually decrease as

the header tank empties.

a =

∂v

∂t

+u

∂v

∂x

+v

∂v

∂y

+w

∂v

∂z

a =

∂v

∂t

+0 +0 +0

The only term to survive is the local acceleration,

namely

∂v

∂t

.The

∂

∂t

part of the material derivative

is called the local derivative.

15

Convective derivative

Consider water going

through a water heater

under steady state ﬂow

conditions.

Cold

Hot

Pathline

Water

heater

T

out

> T

in

= 0

T

___

t

∂

∂

≠ 0

DT

___

Dt

T

in

x

The water temperature at any ﬁxed location is ﬁxed,

i.e.

∂T

∂t

= 0

.

However,the water temperature for a given piece of

water will increase as it progresses through the

heater.The rate of change is

dT

dt

=

Rate at which

T changes

with position

×

How quickly

water changes

position

=

∂T

∂s

u

s

16

Convective derivative

The convective part of the material derivative

D()

Dt

= u

∂()

∂x

+v

∂()

∂y

+w

∂()

∂z

represents changes in the ﬂow properties associated

with the movement of a particle from one point in

space to another.So movement to another location

can also aﬀect the net time rate of change of small

pieces of the ﬂuid.

17

Acceleration,example

An inviscid,incom-

pressible ﬂows past a

sphere of radius

R

.

The velocity along the

AB

stream line is

v = v

0

1 +

R

3

x

3

i

Determine the accelera-

tion

Along the

AB

streamline there is only the

x

velocity component.So

a =

∂v

∂t

+u

∂v

∂x

⇒a

x

= 0 +v

0

1 +

R

3

x

3

d

dx

v

0

1 +

R

3

x

3

⇒a

x

= v

0

1 +

R

3

x

3

v

0

−3R

3

x

4

⇒a

x

= −3

v

2

0

R

1 +

R

3

x

3

x

4

/R

4

!

The velocity is zero at

x = −R

.

18

Acceleration,example

u = v

0

1 +

R

3

x

3

a

x

= −3v

2

0

1 +

R

3

x

3

x

4

/R

4

The ﬂuid decelerates from

u = v

0

far away from the

obstruction to

u = 0

at stagnation point.

The deceleration is

largest at

x = 1.205R

.

The decelerations can be very large (many times

g

)

for fast ﬂuid ﬂows.Note,

a

y

,

and

a

z

are nonzero oﬀ

the

AB

Streamline.

19

Streamline coordinates

It is convenient to use a coordinate system deﬁned in

terms of the ﬂow streamlines.The coordinate along

the streamline is

s

and the coordinate normal to the

streamline is

n

.The unit vectors for the streamline

coordinates are

ˆ

s

and

ˆ

n

.

The direction of

ˆs

will be chosen to be in the same

direction as the velocity.So

v = v

ˆ

s

.

s

n

^

s

^

V

s = 0

s = s

1

s = s

2

n = n

2

n = n

1

n = 0

Streamlines

y

x

The ﬂow plane is covered with an orthogonal curved

net of coordinate lines and

v = v(s,n)

ˆ

s

and

ˆs =ˆs(s,n)

for steady ﬂow.

20

Streamline coordinates

In the streamline coordinates

a = a

s

ˆs +a

n

ˆn =

Dv

Dt

Note,the acceleration has a component parallel and

perpendicular to the direction of travel.It can be

equated to the material derivative.For steady-ﬂow

a =

D(v(s,n)ˆs(s,n))

Dt

=

Dv

Dt

ˆ

s(s,n) +v

Dˆs

Dt

Dv

Dt

ˆs =

∂v

∂t

+

∂v

∂s

ds

dt

+

∂v

∂n

dn

dt

ˆs

v

Dˆs

Dt

= v

∂ˆs

∂t

+

∂ˆs

∂s

ds

dt

+

∂ˆs

∂n

dn

dt

Simpliﬁcations

•

v(s,n)

no explicit

t

dependence

⇒

∂v

∂t

= 0

•

ˆs(s,n)

no explicit

t

dependence

⇒

∂ˆs

∂t

= 0

• Along streamline

v =

ds

dt

• Particles follow streamline

n = Constant

21

Streamline acceleration

With simpliﬁcations

a = v

∂v

∂s

ˆs +v

v

∂ˆs

∂s

= v

∂v

∂s

ˆs +v

2

∂

ˆ

s

∂s

= v

∂v

∂s

ˆ

s +

v

2

R

ˆ

n

The manner in which

ˆs

changes direction with

s

is

simply a matter of geometry.

∂ˆs

∂s

= lim

δs→0

δˆs

δs

=

ˆn

R

2

O

O

O

δθ

δθ

δθ

s

s

A

A

A'

B

B

B'

δ

s

δ

δ

n

^

s

^

s(s)

^

s(s)

^

s(

s

+

s

)

^

δ

s(

s

+

s

)

^

δ

The acceleration that does occur is the convective

acceleration.The direction of

ˆ

n

is towards center of

curvature.

22

Example of material derivative

Water ﬂows steadily through a

5.0 m

long pipe at a

velocity of

2.0 m/s

.At the inlet the water

temperature is

85

o

C

while at the outlet it is

75

o

C

.Determine the time rate of change of water

temperature as it ﬂows along the pipe assuming the

temperature gradient is constant.

Want

DT

Dt

Along a streamline we would have

DT

Dt

=

∂T

∂t

+

∂T

∂s

ds

dt

= 0 +

75 −85

5

2 = −4.0

o

C/s

23

Control volume and system representation

When describing a ﬂuid,we can look at a region of

space (Euler) or look at what happens to speciﬁc

pieces of the ﬂuid (Lagrange).When applying the

laws of motion,can use either the system or control

volume approach.

A system consists of a speciﬁc,identiﬁable chunk of

the ﬂuid.It can be a large of small chunk,but the all

the particles that make the chunk are identiﬁable

and the mass of the system does not change.It can

change,shape or speed as forces act on it.In

dynamics,want to keep track of given chunk of

matter by isolating in and drawing a free body

diagram.Can be problematic in a ﬂuid since harder

to identify and keep of a particular chunk of matter.

Sometimes interested in the forces on a fan or

automobile resulting from the air ﬂowing past (as

opposed to what happens to air).A speciﬁc volume

of space is identiﬁed (the control volume) and the

ﬂow within,through or around that volume is

investigated.

24

The control volume

The control surface is just the surface that encloses

the control volume.

Fixed control volume

The control volume con-

sists of the inside of the

pipe between

(1)

and

(2)

Part of the control surface consists of the physical

surface of the pipe.Fluid can ﬂow across the ends of

the control surface.

Deforming control vol-

ume

The control volume con-

sists of the interior sur-

face of the collapsing bal-

loon.The balloon may

even be moving.

25

The control volume

The control volume is designed to surround a jet

engine.

Air is continually passing through the engine.The

system that was in the engine at

t = t

1

is well past

the engine at

t = t

2

.

The control volume is stationary if the jet itself is

stationary.If the jet is moving then the control

volume itself is moving.

Laws of physics are framed in a systems approach.

To use control volume ideas requires a way to

translate between systems and control volume

approaches,the Reynolds transport theorem.

26

The Reynolds transport theorem

When the describing a system,there are physical

properties like mass,energy,momentum that need

consideration.Let

B

be the property of interest.

We can write

B = mb

m

is system mass,

B

is the total amount of

whatever is in the system.

b

is the amount of

B

per

unit mass.Note,

B

can be a vector (if interested in

momentum).The present interest is for mass,energy

and momentum.

B

is called an extensive variable.It is proportional

to the mass of the system.

b

is called an intensive variable

One writes

B =

X

δB

i

=

X

i

b

i

ρ

i

δV

i

=

ZZZ

sys

ρb dV

The amount of

B

in the system is determined by

adding up

δB

i

for each piece of the system.

27

System versus control volume

Typically interested in the rate of change of the

system

B

sys

and the control volume contents

B

cv

.

˙

B

sys

=

dB

sys

dt

=

d

RRR

sys

ρb dV

dt

˙

B

cv

=

dB

cv

dt

=

d

RRR

cv

ρb dV

dt

The limits of integration for

˙

B

sys

and

˙

B

cv

are

diﬀerent.Consider a system consisting of the ﬂuid in

a ﬁre extinguisher at

t = 0

.Once it is turned on

˙

M

sys

= 0

˙

M

cv

< 0

(a) (b)

System

Control

surface

t = 0 t > 0

Although

B

sys

and

B

cv

look superﬁcially similar,

they keep track of diﬀerent entities.

28

The Reynolds transport theorem

The Reynolds transport

theorem provides a way

to relate what is hap-

pening to the system

and what is happening

in the control volume.

It is

CV–I

II

I

Inflow

Outflow

Fixed control surface and system

boundary at time t

System boundary at time t + t

δ

DB

sys

Dt

=

∂

∂t

ZZZ

cv

ρbdV +

I I

cv

ρbv

ˆ

ndA

for a ﬁxed,non-deforming control volume.The unit

vector normal to the closed surface

ˆn

points out.

The system volume and control volume coincide at

at the instant the equation is evaluated.

The ﬁrst term is the rate of change of the property

B

for a speciﬁc part of the ﬂuid.System is moving,and

control volume is ﬁxed,then rate of change of

B

sys

not necessarily the same as

B

cv

.

29

Reynolds transport theorem:Interpretation

DB

sys

Dt

=

∂

∂t

ZZZ

cv

ρbdV +

I I

cv

ρbv ˆndA

The ﬁrst term on the right hand side is the rate of

change of

B

within the control volume.All we are

doing is adding up all the little pieces of

B

i

= ρ

i

b

i

within the control volume.

control surface

V

i

dV

i

b

i

30

Reynolds transport theorem:Interpretation

DB

sys

Dt

=

∂

∂t

ZZZ

cv

bρdV +

I I

cv

bρv ˆndA

The last term represents of ﬂow of

B

into/outof the

control volume.

V

•

n < 0

V = 0

^

V

•

n = 0

^

V

•

n > 0

^

n

^

n

^

n

^

CS

in

CS

out

CS

(a) (b) (c)

When

v

ˆ

n > 0

,there is a net outﬂow of

B

from the

control volume.When

v ˆn < 0

,there is a net inﬂow

of

B

into the control volume.When

v ˆn = 0

,there

is no ﬂow of

B

into or out of the control volume.

31

Reynolds transport theorem:Steady ﬂow

In a steady-state situation,the amount of

B

in any

control volume,

B

cv

will not change.So

DB

sys

Dt

=

I I

cv

ρbv ˆndA

In order for

B

sys

to change there must be diﬀerences

in the inﬂows and outﬂows of

B

into the control

volume.

F

V

out

V

in

Control volume

Consider the ﬂow of ﬂuid through the black box.

The velocity of ﬂuid entering is diﬀerent from ﬂuid

exiting.Let

B

be the momentum,

p

of the system,

Dp

sys

Dt

=

I I

cv

ρv(v ˆn)dA

The system momentum is changing because

p

ﬂow

out is diﬀerent from

p

ﬂow in.

32

Reynolds transport theorem:Steady ﬂow

F

V

out

V

in

Control volume

The net change of momentum of the system is equal

to the force that the device inside the control volume

exerts on the system.

F =

Dp

sys

Dt

=

I I

cv

ρv(v ˆn)dA

33

Reynolds transport theorem:Unsteady ﬂow

First,consider a situation where the ﬂows across the

control surface sum to zero.An example is ﬂow down

a uniform pipe with an increasing water velocity (the

height of the header tank may be increasing).

(1)

Control surface

(2)

n = –j

^

V

0

i

V

2

= V

0

(t)

V

1

= V

0

(t)

x

y

^

^

n = –i

^ ^

n = i

^ ^

D B

s y s

D t

=

∂

∂ t

Z Z Z

c v

ρ b d V

L e t

v = v

0

(t)i

.O n c e a g a i n,l o o k a t t h e m o m e n t u m,

p

o f t h e s y s t e m.

Dp

s y s

D t

=

∂

∂ t

Z Z Z

c v

ρ v d V

=

∂

∂ t

Z Z Z

c v

ρ v

0

(t)id V

=

∂

∂ t

ρ v

0

(t)i

Z Z Z

c v

d V = ρ V

c v

d v

0

d t

i

34

Reynolds transport theorem:Unsteady ﬂow

(1)

Control surface

(2)

n = –j

^

V

0

i

V

2

= V

0

(t)

V

1

= V

0

(t)

x

y

^

^

n = –i

^ ^

n = i

^ ^

E v a l u a t i o n o f ﬂ o w t e r m

Dp

s y s

D t

=

I I

ρv(t) (v(t) ˆn)d A

Dp

s y s

D t

=

I I

ρ v

0

(t)i [v

0

(t)ii +v

0

(t)i (−i) ]d A

= 0

T h e c o n v e c t i v e t e r m i n t h i s c a s e h a p p e n s t o b e z e r o.

T h i s i s n o t a l w a y s t h e c a s e.

35

Reynolds transport theorem:Example

In a region of downstream of a sluice gate,the water

can develop a reverse ﬂow region.One region had

v

a

= 10.0 ft/s

while the other has

v

b

= 3.0 ft/s

.

Sluice gate

Control surface

V

b

= 3 ft/s

V

a

= 10 ft/s

1.8 ft

1.2 ft

(1) (2)

Determine the mass ﬂowrate of water across the

portion of the control surface at

(2)

if the channel is

20 ft

wide?

First convert to SI units.

3.0ft/s →0.9144 m/s

,

10.0ft/s →3.048 m/s

,

1.2ft →0.3658 m

,

1.8ft →0.5486 m

,

20.0ft →6.096 m

36

Reynolds transport theorem:Example

3.0ft/s →0.9144 m/s

10.0ft/s →3.048 m/s

1.2ft →0.3658 m

1.8ft →0.5486 m

20.0ft →6.096 m

Need to evaluate

Flow =

RR

(2)

ρ(v ˆn)dA v

a

= 3.048i

and so

v

b

= −0.9144i

Flow

(2)v

a

=

ZZ

v

a

ρ(v

a

ˆn)dA

= 10

3

×3.048ii ×(0.3658 ×6.096)

= 6.797 ×10

3

kg/s

Flow

(2)v

b

=

ZZ

v

b

ρ(v

b

ˆn)dA

= 10

3

×(−0.9144i)i ×(0.5486 ×6.096)

= −3.058 ×10

3

kg/s

Net ﬂowrate out of surface is

(6.797 −3.058) ×10

3

= 3.739 ×10

3

kg/s

37

Summary:ﬂuid kinematics

Working out the rate of change of any ﬂuid system

with time is non-trivial.One has a jet-engine

occupying a ﬁxed region of space,acting on a mass

of ﬂuid that moves past it.

The material derivative and Reynolds transport

theorem provide the mechanism to relate what is

happening to the ﬁxed region of space to the ﬁxed

system mass that goes through that space.

The control volume concept provides a useful

visualization tool like the free-body diagram does in

solid mechanics.Choosing the ”best” control-volume

is like choosing the ”best” free-body diagram.A

control-volume is a useful mathematical and

visualization asset,and a good choice can simplify

the analysis.

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