Joseph Lagrange (1736- 1813) Leonhard Euler (1707- 1783)

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Nov 13, 2013 (3 years and 7 months ago)

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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

(xi −yj)
and
identify the stagnation point.
• On
x
-axis
v =
V
0
x

i
• On
y
-axis
v =
−V
0
y

j
• Magnitude of velocity
v =
V
0

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.