Transcript Class 6 - Eqns of Motion

AOE 5104 Class 6
• Online presentations for next class:
– Equations of Motion 2
• Homework 2
• Homework 3 (revised this morning) due
9/18
• d’Alembert
Last class
…and their limitations
Integral theorems…
 d   ndS
 .Ad   A.ndS
   Ad   A  ndS
   A.nd S   A.ds
R
S
R
R
C
S
S
S
2D flow over airfoil with =0
C
Convective operator…
V. = change in density in direction of V, multiplied by magnitude of V
Irrotational and
Solenoidal Fields…
    0
.  A  0
Class Exercise
1.
Make up the most complex irrotational 3D velocity
field you can.
V  (esin x cos x  2 xy3 )i  3x 2 y 2 j  k / z 2 ?
We can generate an irrotational field by taking the gradient of any
scalar field, since     0
I got this one by randomly choosing
  esin x  x 2 y 3  1 / z
And computing



V
i
j
k
x
y
z
Acceleration??
2nd Order Integral Theorems
• Green’s theorem (1st form)

R    d  S  n dS
Volume R
with Surface S
2
• Green’s theorem (2nd form)
2
2






 d  

R
S
ndS
d
 
-
dS
n
n
These are both re-expressions of the divergence theorem.
The Equations of Motion
“Phrase of the Day”
Mutationem motus proportionalem esse vi
motrici impressae, & fieri secundum lineam
rectam qua vis illa imprimitur.
Go Hokies?
Supersonic Turbulent Jet Flow and
Near Acoustic Field
Freund at al. (1997)
Stanford Univ.
DNS
Conservation Laws
• Conservation of mass
R. O. C. of mass  0
• Conservation of momentum
R. O. C. of momentum  FBody  FPressure  FViscous
• Conservation of energy
R. O. C. of energy  WBody  WPressure  WViscous  Q
Supersonic Turbulent Jet Flow and
Near Acoustic Field
Freund at al. (1997)
Stanford Univ.
DNS
Conservation Laws
• Conservation of mass
R. O. C. of mass  0
• Conservation of momentum
R. O. C. of momentum  FBody  FPressure  FViscous
• Conservation of energy
R. O. C. of energy  WBody  WPressure  WViscous  Q
Apply to the fluid material (not the space)
Experimental observations
Assumption: Fluid is a homogeneous continuum
DTM
Kinematics of Continua
z
flow
1) Lagrangian Method
r  xi  y j zk
Position :
x  f1 ( xo ,yo, zo , t )
y  f 2 ( xo ,yo, zo , t )
ro  xoi  yo j  zok
z  f 3 ( xo ,yo, zo , t )
y
Velocity :
vx 
Df1 ( xo ,yo, zo , t )
vy 
Df 2 ( xo ,yo, zo , t )
Dt
where
x
Dt
,
D
 partial derivative wrt time
Dt
holding ( xo ,yo, zo ) constant
vz 
Df 3 ( xo ,yo, zo , t )
Dt
DTM
Acceleration :
ax 
D 2 f1 ( xo ,yo, zo , t )
Dt
2
,
ay 
D 2 f 2 ( xo ,yo, zo , t )
Dt
2
,
az 
D 2 f 3 ( xo ,yo, zo , t )
Dt 2
Concept is straightforward, but difficult to implement, often would produce more
information than we need or want, and doesn't fit the situation usually encountered
in fluid mechanics.
The Lagrangian Method is always used in solid mechanics :
P
xo
y
x  xo  f1 ( xo , yo , zo , t ),
y
P t 
x
6 EI
3
o
P
xo3  3lxo2 

6 EI
 3lxo2   f 2 ( xo , yo , zo , t ),
z  0  f 3 ( xo , yo , zo , t )
DTM
1) Lagrangian Method
Position :
 x  f1 ( xo ,yo, zo , t )

 y  f 2 ( xo ,yo, zo , t )
 z  f ( x ,y z , t )
3
o o, o

solve for position
as a function of
time and “name”
Velocity :
Df1 ( xo ,yo, zo , t )
vx 
vy 
vz 
Dt
Df 2 ( xo ,yo, zo , t )
Dt
Df3 ( xo ,yo, zo , t )
Dt
Acceleration :
2
ax 
ay 
az 
D f1 ( xo ,yo, zo , t )
2
Dt
D 2 f 2 ( xo ,yo, zo , t )
Dt 2
D 2 f 3 ( xo ,yo, zo , t )
Dt 2
2) Eulerian Method
Position :
skip this step and do not try to find
the positions of fluid particles
Velocity :
express the velocity
as a function of time
and spatial position
 v x ( x, y , z , t )

v y ( x , y , z , t )
 v ( x, y , z , t )
 z
WOW! big, big difference: velocity as
a function of time and spatial position,
not velocity as a function of time and
particle name
complication: laws governing motion
apply to particles (Lagrange), not to
positions in space
Acceleration :
a


t 
Dv v

 Grad v v
Dt
t
denotes the derivative wrt time
holding the spatial position fixed,
often called the “local” derivative
DTM
Giuseppe Lodovico Lagrangia
(Joseph-Louis Lagrange)
born 25 January 1736 in Turin, Italy
died 10 April 1813 in Paris, France
Leonhard Paul Euler
born 15 April 1707 in Basel, Switzerland
died 18 September 1783 in St. Petersburg, Russia
Acceleration in the Eulerian Method:
A fluid particle, represented as a blue dot in the figure,
moves from position r to r  dr during the time interval dt.
z
dr
y
r
Its velocity changes from v(r, t ) to v(r  dr, t  dt )
where dr MUST be chosen  vdt
x
a
Dv v (r  dr, t  dt )  v (r, t )

Dt
dt
Dv v (r  dr, t  dt )  v (r, t ) v (r  dr, t  dt )  v (r  dr, t )  v (r  dr, t )  v (r, t )


Dt
dt
dt
v (r  dr, t  dt )  v (r  dr, t ) v
=
the derivative wrt time at a fixed location
dt
t
v (r  dr, t )  v (r, t ) Grad v dr change in v between two points in space at a fixed time


dt
dt
dt
a
t and dr are independent variables; so we are free to chose them anyway we want. In order
to follow a particle, we must chose dr  vdt : a 
v
 Grad v v
t
These equations define a line in terms of the
parameter, t , when xo , yo , zo are constant.
Such a line is called a pathline.
 x  f1 ( xo ,yo, zo , t )

 y  f 2 ( xo ,yo, zo , t )
 z  f ( x ,y z , t )
3
o o, o

.
If at a given instant we draw a line with the property that every point on
the line passed through the same reference point at some earlier time, the result
is known as a streakline.
It is called a streakline because, if the particles are dyed as they pass through
the common reference point, the result will be a line of dyed particles (i.e., a
streak) through the flowfield.
If at a given instant the velocity is calculated at all points in the flowfield
and then a line is drawn with the property that the velocities of all of the particles
lying on that line are tangent to it, the result is known as a streamline.
Streamlines are the velocity field lines. They provide a snapshot of the flowfield,
a picture at an instant. The surface formed by all the streamlines that pass
through a closed curve in space forms a stream tube.
Perspectives
Eulerian Perspective – the flow as as seen at fixed locations in space, or
over fixed volumes of space. (The perspective of most analysis.)
Lagrangian Perspective – the flow as seen by the fluid material. (The
perspective of the laws of motion.)
Control volume: finite fixed
region of space (Eulerian)
Coordinate: fixed point in space
(Eulerian)
Fluid system: finite piece of the fluid material
(Lagrangian)
Fluid particle: differentially small finite piece
of the fluid material (Lagrangian)
DTM
III
The Transport Theorem:
flow
A system moving along in the flow occupies volumes
I and II at time t. During the next interval dt some of
the system moves out of II into three and some
moves out of I into II. The rate of change of an
arbitrary property of the system, N, is given by the
following:
II
I
 N in II at t  dt    N in II at t    N in III at t  dt    N in I at t 
DN  N in II & III at t  dt    N in I & II at t 


Dt
dt
dt
dt

   dV    v ndS
t II
II
the unit vector
normal
to ABC,
the same two material elements,
but now approximating the
surface of volume III at time =
n
C’
A’
B’
vdt
S
C
A
B
material that flowed through the
surface of volume II during the
interval and now fills volume III:
dN    vdt n  S
two triangular elements from the
family approximating the surface of
volume II at time = t
Strategy
• Write down equations of motion for
Lagrangian rates of change seen by fluid
particle or system
• Derive relationship between Lagrangian
and Eulerian rates of change
• Substitute to get Eulerian equations of
motion
Conservation of Mass
From a Lagrangian Perspective
Law: Rate of Change of Mass of Fluid Material = 0
For a Fluid Particle:
Volume d
Density 
d
t
d


 d
t part
t
1 d



d t part t

.V 
t
.V 
For a Fluid System:
Volume R
d
Density
=(x,y,z,t)

t
‘Seen by the
particle’
0
part
0
part
0
part
D 

is referred to as
Dt t part
the SUBSTANTIAL DERIVATIVE
(or total, or material, or Lagrangian…)
where
0
part
D
0
Dt
 d  0
sys R
D
d  0
Dt R
AXIOMATA SIVE LEGES MOTUS
• Lex I.
– Corpus omne perseverare in statuo suo quiescendi vel movendi
uniformiter in directum, nisi quatenus a viribus impressis cogitur
statum illum mutare.
• Lex II.
– Mutationem motus proportionalem esse vi motrici impressae, &
fieri secundum lineam rectam qua vis illa imprimitur.
• Lex III.
– Actioni contrariam semper & æqualem esse reactionem: sive
corporum duorum actiones in se mutuo semper esse æquales &
in partes contrarias dirigi.
• Corol. I.
– Corpus viribus conjunctis diagonalem parallelogrammi eodem
tempore describere, quo latera separatis.
Conservation of Momentum
From a Lagrangian Perspective (Fluid Particle)
Law: Rate of Change of Momentum = Fbody+Fpressure+Fviscous
dV
ROC of Momentum 
t
V
 d
t
part
part
DV
 d
Dt
k
P
i
Fbody:  fd
dz
j
dx
dy
Net …
density 
volume d
velocity V
body force per unit mass f
Elemental Volume, Surface Forces
Sides of volume have lengths dx,
dy, dz
z, k
 zy 
 xy 
 xy 
 xy dx
y 2
 xy dx
 zy dz
z 2
 yy dy
y 2
p dy
p
y 2
 dy
 yy  yy
y 2
p dy
p
y 2
 yy 
y, j
y 2
On front
and rear
faces
y-component
x, i
 zy dz
 zy 
z 2
• Volume d = dxdydz
• Density 
• Velocity V
Conservation of Momentum
From a Lagrangian Perspective (Fluid Particle)
Law: Rate of Change of Momentum = Fbody+Fpressure+Fviscous
dV
ROC of Momentum 
t
V
 d
t
part
part
k
DV
 d
Dt
P
i
dz
j
Fbody:  fd
dx




p
p
p
Fpressure: y component   p  12 dy dxdzj   p  12 dy dxdzj   dj
y
y
y




so Fpressure  p d
dy
z, k
p dy
p
y 2
P
p
p dy
y 2
y, j
x, i
Conservation of Momentum
z, k
 zy 
 zy dz
z 2
From a Lagrangian Perspective (Fluid Particle)
Law: Rate of Change of Momentum = Fbody+Fpressure+Fviscous
ROC of Momentum 
dV
t
 d
part
 dy
V yy  yy
yd2
t
part
DV
Dt
P
k
P
i
Fviscous:
 yy dy
y dz
2
j
y, j
dx
Fbody:  fd




p
p
p
Fpressure: y component   p  12 dy dxdzj   p  12 dy dxdzj   zy ddzj
 zy  y
y
y




z 2
x, i
so Fpressure  p d
 yy 
dy
 yy 1 
 yy 1 





y component...  yy 
dy
dxdz
j



yy
2
2 dy  dxdzj 


y
y




Likewise for
 xy 1 
 xy 1 


x and z
 xy 



2 dx  dydzj   xy 
2 dx  dydzj 
x
x







 




 
 zy  zy 12 dz dxdyj   zy  zy 12 dz dxdyj   xy  yy  zy  jd
z
z
y
z 




 x
Conservation of Momentum
From a Lagrangian Perspective (Fluid Particle)
Law: Rate of Change of Momentum = Fbody+Fpressure+Fviscous
dV
ROC of Momentum 
t
V
 d
t
part
part
k
DV
 d
Dt
P
i
Fbody:  fd
So,
j
dx




p
p
p
Fpressure: y component   p  12 dy dxdzj   p  12 dy dxdzj   dj
y
y
y




so Fpressure  p d
Fviscous:
dz
dy
 yy 1 
 yy 1 





y component...  yy 
dy
dxdz
j



yy
2
2 dy  dxdzj 


y
y




Likewise for
 xy 1 
 xy 1 


x and z
 xy 



2 dx  dydzj   xy 
2 dx  dydzj 
x
x







 




 
 zy  zy 12 dz dxdyj   zy  zy 12 dz dxdyj   xy  yy  zy  jd
z
z
y
z 




 x
DV

 f  p  (.τ x )i  (.τ y ) j  (.τ z )k
Dt
τ x   xx i   yx j   zxk
where
τ y   xy i   yy j   zyk
τ z   xz i   yz j   zzk
AXIOMS CONCERNING MOTION
• Law 1.
– Every body continues in its state of rest or of uniform motion in a
straight line, unless it is compelled to change that state by forces
impressed upon it.
• Law 2.
– Change of motion is proportional to the motive force impressed;
and is in the same direction as the line of the impressed force.
• Law 3.
– For every action there is always an opposed equal reaction; or,
the mutual actions of two bodies on each other are always equal
and directed to opposite parts.
• Corollary 1.
– A body, acted on by two forces simultaneously, will describe the
diagonal of a parallelogram in the same time as it would describe
the sides by those forces separately.
Isaac Newton
1642-1727
Conservation of Energy
From a Lagrangian Perspective (Fluid Particle)
Law: Rate of Change of Energy = Wbody+Wpressure+Wviscous+Q
ROC of Energy
Wbody  f.Vd
d (e  12 V 2 )

t
D (e  12 V 2 )
 d
Dt
part
k
P
i
dz
j
dx
dy
• Total energy is internal energy + kinetic energy
= e + V2/2 per unit mass
• Rate of work (power) = force x velocity in
direction of force
• Fourier’s law to gives rate of heat added by
conduction
Elemental Volume, Surface Force
Work and Heat Transfer
Sides of volume have lengths dx,
z, k
y-contributions
 T 
 k

y  dy
T

k

y
y
2
p dy
p
y 2
v dy
v
y 2

T 
  k


y
T
 dy
k
 
y
y
2
p dy
p
y 2
v dy
v
y 2
Viscous work
requires expansion
of v velocity on all
six sides
dy, dz
y, j
x, i
Velocity
components
u, v, w
• Volume d = dxdydz
• Density 
• Velocity V
Conservation of Energy
From a Lagrangian Perspective (Fluid Particle)
Law: Rate of Change of Energy = Wbody+Wpressure+Wviscous+Q
ROC of Energy
d (e  12 V 2 )

t
D (e  12 V 2 )
 d
Dt
part
Wbody  f.Vd
Wpressure  .( pV ) d
Wviscous  (.(uτ x )  .(vτ y )  .( wτ z ) )d
Q:
 T   T  1 
 T   T  1 
  T 
 2 dy dxdz   k
 2 dy dxdz   k
d
y contribution   k
  k
  k

y

y

y

y

y

y

y

y










so Q  .(kT ) d
So,
D(e  12 V 2 )

 f .V  .( pV)  .(uτ x )  .(vτ y )  .( wτ z )  .(kT )
Dt
Equations for Changes Seen From
a Lagrangian Perspective
Differential Form (for a particle)
D
  .V
Dt

DV
 f  p  (.τ x )i  (.τ y ) j  (.τ z )k
Dt
D(e  12 V 2 )

 f .V  .( pV)  .(uτ x )  .(vτ y )  .( wτ z )  .(kT )
Dt
D
Integral Form (for a system)

d

=
0
Dt R
D
V d =  f d -  pn dS +  ( τ x .n ) i + ( τ y .n )j + ( τ z .n )k dS

Dt R
R
S
S


2
D
V
(e + ) d =  V.f d +  - pn + ( τ x .n) i  ( τ y .n) j + ( τ z .n) k .V dS +  k( T).n dS

Dt R
2
R
S
S
Conversion from Lagrangian to
Eulerian rate of change - Derivative

t
D
The Substantial Derivative
Dt
part
  x  y  z 








u
v
w
t x t y t z t t
x
y
z


 V.
t

y
Time Derivative
Convective Derivative
x
(x(t),y(t),z(t),t)
z
Conversion from Lagrangian to
Eulerian rate of change - Integral

t
D
D d

d


R Dt
Dt R
D
Dd 

d  
Dt
Dt
R
D

d  .Vd
Dt
R
 


V
.





.
V

 d
  t

R 
 



.(

V
)

 d
  t

R 
α

d   V.ndS
t
R
S
  d =
sys R
D 

 V.
Dt
t
Apply
Divergence
Theorem
The Reynolds
Transport
Theorem
y
x
z
Volume R
Surface S
Equations for Changes Seen From
a Lagrangian Perspective
D
  .V
Dt
D 

Dt t

part
Differential Form (for a particle)
DV
 f  p  (.τ x )i  (.τ y ) j  (.τ z )k
Dt
D(e  12 V 2 )

 f .V  .( pV)  .(uτ x )  .(vτ y )  .( wτ z )  .(kT )
Dt
D
Integral Form (for a system)

d

=
0
Dt R
D
V d =  f d -  pn dS +  ( τ x .n ) i + ( τ y .n )j + ( τ z .n )k dS
Dt R
R
S
S


2
D
V
(e + ) d =  V.f d +  - pn + ( τ x .n) i  ( τ y .n) j + ( τ z .n) k .V dS +  k( T).n dS
Dt R
2
R
S
S
Equations for Changes Seen From
an Eulerian Perspective
D
  .V
Dt
D

  V.
Dt t

Differential Form (for a fixed volume element)
DV
 f  p  (.τ x )i  (.τ y ) j  (.τ z )k
Dt
D(e  12 V 2 )

 f .V  .( pV)  .(uτ x )  .(vτ y )  .( wτ z )  .(kT )
Dt

Integral Form (for a system)
d



V.
n
dS
=
0
R t
S
V
R t d   V(V.n)dS = R f d - S pn dS + S ( τ x .n ) i + ( τ y .n )j + ( τ z .n )k dS
 (e + 12 V 2)
) d    (e + 12 V 2)V.ndS =  V.f d +  - pn + ( τ x .n) i  ( τ y .n) j + ( τ z .n) k .V dS +  k( T).n dS
R t
S
R
S
S


Equivalence of Integral and
Differential Forms
Cons. of mass
(Integral form)

R t d  S V.ndS = 0
Divergence
Theorem
 V.ndS =  . V  d
R
S
Conservation of
mass for any
volume R
 





.

V
 d  0
R  t

Then we get

 . V   0
t
Cons. of mass
(Differential form)
D
  .V
Dt
or

 V.  .V  0
t