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).Xray
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
streaklines Consist of all the particles in a ﬂow
that have passed through a common point.
Mainly a laboratory tool.
A streakline 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 streakline 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
streakline 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 steadyﬂow
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,nondeforming 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 steadystate 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 nontrivial.One has a jetengine
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 freebody diagram does in
solid mechanics.Choosing the ”best” controlvolume
is like choosing the ”best” freebody diagram.A
controlvolume is a useful mathematical and
visualization asset,and a good choice can simplify
the analysis.
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