Prof. dr. A. Achterberg, Astronomical Dept. , IMAPP ...achterb/Gasdynamica_2016/Gas Dynamics...F ad...
Transcript of Prof. dr. A. Achterberg, Astronomical Dept. , IMAPP ...achterb/Gasdynamica_2016/Gas Dynamics...F ad...
Prof. dr. A. Achterberg, Astronomical Dept. , IMAPP, Radboud Universiteit
( )2
29, 1 sin2 4UP a P ρθ θ∞
= + −
PARADOX OF D’ALAMBERT
surfacesphere
d ( , ) 0P a θ= − =∫F O
Viscosity = internal friction due to molecular diffusion, viscosity coefficient h:
Viscous force density: 2
visc visc η= − • ⇒ ∇Tf V
(incompressible flow!)
Equation of motion:
( ) 2Pt
ρ η∂ + • = − + ∇ ∂ V V V V
All flow quantities :
( )
3 3
2
0
3 1 3 1cos 1 , sin 12 2 4 4
3 cos,2
ra a a aV U V Ur r r r
U aP r Pa r
θθ θ
η θθ
= − + = − − −
= −
2
1 cos as sin
1 sin as sin
rV U rr
V U rr rθ
ψ θθ θ
ψ θθ
∂= + = →∞
∂
∂= − = − →∞
∂
( )11 1 21 2 31 3force ˆ ˆ ˆ ˆarea
momentum flux tensor
(Cartesian coordinates!)
ij ji
jiij
j i
T T T
T T
VVPx x
δ η
= = − • − + +
= =
∂∂= − + ∂ ∂
Tt n = x x x
Each particle transports local momentum to wall and “sticks”:
( cos )
d cosd
yp mV x
Vmx
θ
θ
= =
Momentum transferred per particle:
Each particle transports local momentum to wall and “sticks”:
2
d d =
d d dd cosd
y yx y
p Fn p
t O OVnmx
σ
σ θ
=
y-force per unit area:
Each particle transports local momentum to wall and “sticks”:
Average over all angles of incidence: 2 1cos
3θ =
d d dd 3 d d
yy
F nm V VtO x x
σ η ≡ ≡
( )
( )2 visc visc
surface element0
cos sin
= 2 sin cos sin |
D rr r
rr r r a
F dO T T
a d P T T
θ
π
θ
θ θ
π θ θ θ θ =
= − −
− + −
∫
∫
( )( )visc
†
-
P
P η
= +T I T
I = V + V
032 = = cos2
3 sin2
rrr
rr
u UT P P Pr a
u u UT rr r r a
θθ
ηη θ
η ηη θθ
∂= + −
∂
∂∂ = − − = ∂ ∂
( )2
0
20
0
2 sin cos sin
3 2 sin cos 62
D rr rF a d T T
Ua d P aUa
π
θ
π
π θ θ θ θ
ηπ θ θ θ πη
= − −
= − − =
∫
∫
For this particular flow at r=a:
Typical “ram pressure” : Projected area perpendicular to flow: Typical force = pressure x area = Drag coefficient =
21ram 2
2
21ram 2
DD 21
2
actual drag forcetypical drag force
P U
a
F U
FCU
ρ
π
ρ
ρ
=
=
=
= =
Very crude calculation for a sphere:
D 1/C Re∝ D constantC
In the stagnation point on symmetry axis flow comes to a standstill!
Bernoulli: 2 2st1 1st2 2constant PPU P P Uρ
ρ ρ+ = = ⇔ = +
In the stagnation point on symmetry axis flow comes to a standstill!
Net force: ( )
( ) 2 2 21st 2
Area Pressure difference front-back
= 8
K
P P U U Dπρ ρ
×
× − = =
Conclusion: 2 2Drag force constantK U Dρ= ×
• High Reynolds number UD/ν, low viscosity:
• Small Reynolds number UD/ν, large viscosity:
2 2
D 2 2
Drag force constant ,
Drag coefficient = constant
K U DKC
U D
ρ
ρ
= ×
=
D 2 2
Drag force constant , constantDrag Coefficient constant =
K UDKC
U D UD Re
ρνν
ρ
×
= ×
0t∂=
∂
Divergence product rule!
Combine mass cons + energy cons.:
Divergence chain rule!
Divergence product rule!
Bernoulli’s Law:
( )21
2 1is constant along stream lines
PV γγ ρ
≡ + +Φ−
Bernoulli’s Law:
( )21
2 1is constant along stream lines
PV γγ ρ
≡ + +Φ−
Adiabatic flow:
is constant along stream linesP γρ −
Astrophysical Application: Stellar and Solar Winds
Why is there a Solar Wind?
Escape velocity ~ Thermal Velocity in Solar Corona (T ~ 1 MK)
‘Something’ bends comet tails
2620 km/s
3 200 km/s
esc
bth
p
GMV
R
k TVm
= ≈
≈
Aurora: “something” acts as a medium supporting perturbations
which propagate from Sun to Earth
Solar wind velocity as measured by Ulysses satellite
The Parker Model
Assumptions: 1. The wind is steady and adiabatic
2. The flow is spherically symmetric
3. Neglect effect of magnetic fields and rotation star
b
p
sound speed: k TPmρ
b
p
sound speed: k TPmρ
There must be a sonic radius where flow speed = sound speed
2
212
4 ( ) ( ) constant
( ) constant( 1)
( ) ( ) constant
( )
r r V r M
PV r
P r r
GMrr
γ
π ρ
γγ ρ
ρ−
∗
= ≡
+ +Φ = ≡−
=
Φ = −
Conservation of mass in steady flow
Bernouilli: conservation of energy
Entropy is constant: Adiabatic Flow
Gravitational potential of a single star
2
area mass flux
4 constant= r V Mπ ρ× =
Steady flow in radial direction:
( ) ( ) ( )22
This is in spherical coordinates!
1 1 1sin 0sin sin
0 , 0
rr V V Vt r r r r
V Vt
θ φ
ρ
θ φ
ρ ρ θρ ρθ θ θ φ•
∂ ∂ ∂ ∂+ + + =
∂ ∂ ∂ ∂
∂= = =
∂⇔
V
( )22 2
1 0 r rAr V V
r r rρ ρ∂
= ⇔ =∂
dd = ( ) ( ) 0d
r r r rr
+ ∆ − = ∆ =
( )
( ) ( )
( ) ( )
212
2
1
d =d ( ) 01
d d 4 0
d d d 0
PV r
M r V
P P P γγ γ
γγ ρ
π ρ
ρ ρ γ ρ ρ− +− −
+ +Φ = −
= =
= − =
( ) ( ) ( )2 2 2d 4 ( ) ( ) 0 4 d d 2 d
d d d2
r r V r r V r V Vr r
V rV r
π ρ π ρ ρ ρ
ρρ
= = + + ⇔
= − +
Step 1: calculate density change
212 2
dd ( ) 0 d d( 1)
GMP PV r V V rr
γ γ ργ ρ ρ ρ
∗ + +Φ = ≡ + + −
( ) ( ) constantP r rγρ − = ( ) GMrr
∗Φ = −
d d d2V rV r
ρρ
= − +
2
dd d 0GMPV V rr
γ ρρ ρ
∗ + + =
( )2 2 2d d2 , s s sGMV r PV C C C
V r rγρ
∗ − = − ≡
Adiabatic sound speed
( )2 2 2d d2 , s s sGMV r PV C C C
V r rγρ
∗ − = − ≡
( )d d lnd ln d lnd ln
V VV rV r
= =
( )d d lnr rr=
( )2 2 2d ln 2d lns s
GMVV C Cr r
∗ − = −
sV C= ± 22cs
GMr rC
∗= =
Special velocity: sound speed (“Mach One”)
Special radius: critical radius
( )2 2 2d ln 2d lns s
GMVV C Cr r
∗ − = −
( )2 2 2d ln 2d lns s
GMVV C Cr r
∗ − = −
Accelerating wind solution: V > 0 and dV/dr > 0! Solution should remain regular at all radii!
Solution space for Parker’s Equation
( )2 2 2d ln 2d lns s
GMVV C Cr r
∗ − = −
Critical Point Condition:
2 at 2s c
s
GMV C r rC
∗= = =
Wind and Breeze Solutions
Special case: Isothermal Wind with constant temperature
constant
(case with 1 ; )
siP TC
P
ρ µ
γ ρ
= = =
= ∝
Accretion Solution
Bondi Accretion
( )2 2 2ln 2lns s
GMd VV c cd r r
∗ − = −
Critical Point Condition:
2 at 2s c
s
GMV c r rc
∗= = =
( ) ( ) rV r V r e= −
Isothermal Bondi Accretion
2 at 2s c
s
GMV c r rc
∗= = =
constant
(case with 1 ; )
siP TC
P
ρ µ
γ ρ
= = =
= ∝
1. Laval Nozzle (jet engines)
2
( ) ( ) ( ) constant
constant
= constant2 ( 1)
M z V z z
P
V P
γ
ρ
ρ
γγ ρ
−
= =
=
+ =−
Basic equations:
2. Astrophysical jets:
Stellar Winds and Jets: similarities and differences
• Steady flow Steady flow • Large opening angle Small opening angle
• Parker-equation Parker-type equation
• Flow geometry known Pressure known