Download The Fundamental Equations of Gas Dynamics

The Fundamental Equations of Gas Dynamics
1 References for astrophysical gas (fluid) dynamics
1. L. D. Landau and E. M. Lifshitz Fluid mechanics
2. F.H. Shu The Physics of Astrophysics Volume II Gas Dynamics
3. L. Mestel Stellar Magnetism
Fundamental equations
© Geoffrey V. Bicknell
2 The equation of continuity
2.1 Derivation of the fundamental equation
Define
n
v
 = density
v i = velocity components
S
V
(1)
Mass within volume V is
M =
  dV
(2)
V
Fundamental equations
2/78
Rate of flow of mass out of V is
 vi ni dS
S
Therefore, the mass balance within V is given by:
d
-----  dV = – v i n i dS
dt

V

(3)

 v i  dV
 xi
(4)
S
Use the divergence theorem:

v i n i dS =
S
Fundamental equations

V
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and
d
-----  dV =
dt

V

V

dV
t
(5)
to give

V
 
+
 v i  dV = 0
 t  xi
(6)
Since the volume V is arbitrary, then
 
+
 v i  = 0
 t  xi
(7)
This is the equation of continuity.
Fundamental equations
4/78
Aside: Differentiation following the motion
Suppose we have a function f  x i t  which is a function of both
space and time. How does this function vary along the trajectory
of a fluid element described by x i = x i  t  ? We simply calculate
d
f f dx i
f
f
f  x i  t  t   =
+
------- =
+ vi
dt
 t  x i dt
t
 xi
Fundamental equations
(8)
5/78
Useful result from the equation of continuity
Write the above equation in the form:
v i


+ vi
+
= 0
t
 xi
 xi
(9)
Going back to the equation of continuity, then
v i
v i
d
1 d
------ = – 

= – --- -----dt
 xi  xi
 dt
Fundamental equations
(10)
6/78
Putting it another way:
v i
1 d
--- ------ = –
 dt
 xi
(11)
This tells us that in a diverging velocity field v i  x i  0 , the
density decreases and in a converging velocity field v i  x i  0
the density increases.
Fundamental equations
7/78
2.2 Examples of mass flux
Wind from a massive star
Massive stars such as O and B stars produce winds with velocities of around 1 000 km s – 1 and mass fluxes of around
10 – 6 solar masses per yr. We can use these facts to estimate the
density in the wind as follows.
Mass flux =
r
Fundamental equations
 vi ni dS
S
(12)
= 4r 2   r v  r 
8/78
where the integral is over a sphere of radius r . We assume that
the flow is steady so that the mass flux integrated over any 2 surfaces surrounding the star is the same. Hence,
·
M = 4r 2   r v  r  = constant
(13)
We shall show in later lectures that far from the star, the velocity
is constant, i.e.
v  r  = v  = constant
(14)
Therefore the density is given by:
M·
  r  = ------------------4r 2 v 
Fundamental equations
(15)
9/78
For typical parameters:
·
M = 10 – 6 solar masses per yr
15
r = 0.1 pc = 3.09 10 m
(16)
v  = 1 000 km s – 1 = 10 6 m s – 1
Units
30
1 solar mass = 2 10
16
1 pc = 3.09 10
kg
7
m
(17)
1year = 3.16 10 seconds
The density of the wind at 0.1 pc from the star is:
– 22
 = 5.2 10
Fundamental equations
kg m – 3
(18)
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This is not particularly informative by itself. We usually express
the density in terms of particles cm – 3 . Assuming the gas is totally ionised, then
 = n H m p + n He m He + ......
(19)
Now for solar cosmic abundances,
n He = 0.085n H
(20)
m He  4m
(21)
and
where m is an atomic mass unit
– 27
m = 1.66 10
Fundamental equations
kg
(22)
11/78
so that
  n H m + 4  0.085n H m = 1.34n H m
(23)
Therefore, for the O-star wind:
– 22
1.34n H m  5.2 10
kg m – 3
– 22
5.2 10
 n H = ------------------------- m – 3
1.34m
(24)
5
= 2.3 10 Hydrogen atoms m – 3
Fundamental equations
12/78
Often, instead of the Hydrogen density we use the total number
density of ions plus electrons. In this case we put
 = nm
 = mean molecular weight = 0.6156
(25)
Hence,
– 22
5
5.2 10
n = ------------------------- = 5.1 10 particles m – 3
0.6156m
Fundamental equations
(26)
13/78
3 Conservation of momentum
n
v i
V
Consider first the rate of change of
momentum within a volume as a rev i v j sult of the flux of momentum. Let
 i = total momentum , then
S
·
d
 i = ----- v i dV
(27)
dt

V
– p n i is the rate of change of momentum in
the volume, V .
Fundamental equations
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The flux of momentum out of the surface S is
 vi vj nj dS
(28)
S
3.1 Surface force
There are two complementary ways that we can look at the other
aspects of the conservation of momentum.
In a perfect fluid there is a force per unit area perpendicular to
the surface that exerts a force on the gas inside. This is the pessure p
Force on volume
= – pn i dS
within S

(29)
S
Fundamental equations
15/78
Therefore the total momentum balance for the volume is:
d
----- v i dV = – v i v j n j dS – pn i dS
dt


V

S
(30)
S
The surface integrals
Using the divergence we can write
 vi vj nj dS = 
S
V

 v i v j  dV
 xj
(31)
The surface integral of the pressure can be written
 pni dS =  pij nj dS = 
S
Fundamental equations
S
V

 p ij  dV
 xj
(32)
16/78
Entire momentum equation
Again, we take the time derivative inside and obtain:

V



 v i  +
 v i v j  +
 p ij  dV = 0
t
 xj
 xj
(33)
Since the volume is arbitrary, then


 v i  +
 v i v j + p ij  = 0
t
 xj
(34)
We often write this equation in the form:


p
 v i  +
 v i v j  = –
t
 xj
 xi
Fundamental equations
(35)
17/78
3.2 2nd way - momentum flux
Across any surface in a gas there is a flux of momentum due to
the random motions of atoms in the gas.
We write the flux per unit area of
n
 ij n j
momentum due to molecular mov i v j tions as  n
ij j
S
v i
V
–p ni
Fundamental equations
The flux of momentum out of the
volume due to molecular motions is
then:
 ij nj dS
(36)
V
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For a perfect fluid, the relation between  ij and p is
 ij = p ij
(37)
In viscous fluids (to be treated later) the tensor  ij is not diagonal.
Fundamental equations
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Momentum balance
When we adopt this approach, the momentum balance of the volume of gas is:
d
----- v i dV = – v i v j n j dS –  ij n j dS
dt


V


V
S

S


 v i  +
 v i v j +  ij  dV = 0
t
 xj
(38)
The corresponding partial differential equation is:


 v i  +
 v i v j +  ij  = 0
t
 xj
Fundamental equations
(39)
20/78
The point to remember from this analysis is that pressure in a fluid is the result of a flux of momentum resulting from the microscopic motions of the particles.
Particles of mass m crossing a surface within the fluid with random atomic velocity u i (relative to the bulk velocity) contribute
an amount mu i u j to the flux of momentum. A particle with an
equally opposite velocity contributes exactly the same amount to
the momentum flux. Hence, atomic motions in both directions
across the surface contribute equal amounts to the pressure.
Fundamental equations
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3.3 Alternate forms of the momentum equations
Consider, the isotropic case where  ij = p ij . Then,


p
 v i  +
 v i v j  = –
t
 xj
 xi
(40)
Expand the terms on the left:
v i
 v i 


p
vi + 
+ v i  v j  + v j   = –
t
t
 xj
 xi
  x j
(41)
Using the continuity equation to eliminate the blue terms:
v i
v i
p

+ v j
= –
t
 xj
 xi
Fundamental equations
(42)
22/78
In terms of the derivative following the motion:
dv i
p
 ------- = –
dt
 xi
(43)
3.4 Additional forces
We can in general have additional forces acting on the fluid. In
particular, we can have a gravitational force derived from a gravitational potential. This is not a surface force like the pressure
but a body force which is proportional to the volume of the region. We write:

Gravitational force per unit mass = –
 xi
Fundamental equations
(44)
23/78
where  is the gravitational potential. Therefore the gravitational
force acting on volume V is

F i = –  dV
 xi

(45)
V
We add this term to the momentum equations to obtain:

V



 v i  +
 v i v j +  ij  dV = –  dV
t
 xj
 xi

(46)
V
implying that:



 v i  +
 v i v j +  ij  = – 
t
 xj
 xi
Fundamental equations
(47)
24/78
or the alternative form:
v i
v i
 ij

= –
–

+ v j
 xj
 xj
 xi
t
(48)
When the momentum flux is diagonal
 ij = p ij


p

 v i  +
 v i v j  = –
–
t
 xj
 xi
 xi
v i
v i
(49)
(50)
p

= –
–

+ v j
t
 xj
 xi
 xi
Fundamental equations
25/78
3.5 Hydrostatic application of the momentum equations
X-ray emitting atmosphere in an elliptical galaxy
Elliptical galaxies have extended hot atmospheres extending for
10 – 100 kpc from the centre of the galaxy.
Our first approach to understanding the distribution of gas in
such an atmosphere is to consider an hydrostatic model.
In this case v i = 0 and the momentum equations reduce to
1 p

--= –
  xi
 xi
Fundamental equations
(51)
26/78
Let us restrict ourselves to the case of spherical symmetry:
1 p

GM  r 
--= –
= – ----------------r
r
r2
(52)
G = Newtons constant of gravitation
– 11
= 6.67 10 SI units
M  r  = Mass within r
(53)
This can be used to estimate the mass of the galaxy. Put
kT
p = nkT = ---------m
Fundamental equations
(54)
27/78
then
1 d kT
GM  r 
--- ----- ---------- = – ---------------- dr m
r2
1 kT d k dT
GM  r 
--- -------- ------ + -------- ------ = – ---------------- m dr m dr
r2
(55)
Rearrangement of terms gives:
r 2 kT 1 d 1 dT
M  r  = – ----- -------- --- ------ + --- -----G m  dr T dr
Fundamental equations
(56)
28/78
Usually, at large radii, the temperature of the atmosphere is isothermal and the density of the X-ray emitting gas may be approximated by a power-law:
T  constant
  r –
r d
 --- ------ = – 
 dr
  0.7
(57)
Therefore,
r 2 kT 1 d 1 dT
kT
M  r  = – ----- ----------- --- ------ + --- ------ = ---------------r
G m p  dr T dr
m p G
11  T   r 
= 3.1 10  --------- ---------------- solar masses
 10 7  10 kpc
Fundamental equations
(58)
29/78
The amount of matter implied by such observations is larger than
can be accounted for by the stellar light from elliptical galaxies
and implies that elliptical galaxies, like spiral galaxies, have
large amounts of dark matter.
4 Entropy
In any dynamical system, the conservation of mass, momentum
and energy are the fundamental principles to consider.
However, before proceeding with the conservation of energy it
is necessary to make an excursion into the domain of entropy.
Fundamental equations
30/78
4.1 Entropy of a fluid
Consider an element of fluid with:

p

s
m
=
=
=
=
=
Internal energy density (per unit volume)
pressure
density
entropy per unit mass
mass of the element
Fundamental equations
(59)
31/78
V
m

s
m
Volume = ------
Entropy = sm
m
Internal energy = ---------
(60)
The relationship between internal energy  U  , pressure, volume
 V  and entropy  S  of a gas is
kTdS = dU + pdV
Fundamental equations
(61)
32/78
For the infinitesimal volume, above
m
m


kTd  sm  = d ---------- + pd ------  
 
(62)
The mass m is constant, therefore

1


kTds = d --- + pd -- 
 
Fundamental equations
(63)
33/78
Other ways of expressing the entropy relation
Expanding the differentials:
1
 + p

kTds = --- d – ------------ d
 2 

 + p

kTds = d – ------------ d
  
(64)
Specific enthalpy
+p
h = ------------ = Enthalpy per unit mass

(65)
= Specific enthalpy
Fundamental equations
34/78
Expression using the enthalpy
1
 + p

kTds = --- d – ------------ d
 2 

dp
1
 + p

= --- d   + p  – ------------ d – ----- 2 


(66)
dp
= dh – -----
To be symmetric with the previous expression
kTds = dh – dp
Fundamental equations
(67)
35/78
Relation between thermodynamic variables throughout the
fluid
These relationships have been derived for a given fluid element.
However, the equation of state of a gas can be expressed in the
form
p = p   s 
(68)
so that any relationship derived between the thermodynamic variables is valid everywhere. Therefore the differential expressions
 + p

kTds = d – ------------ d = d – hd
  
(69)
kTds = dh – dp
Fundamental equations
36/78
are valid relationships between the differentials of s   p
throughout the fluid.
In particular when considering the energy equation, we look at
the changes in these quantities resulting from temporal or spatial
changes. We often use the first form for temporal or spatial
changes and the second form for spatial changes.
Thus consider the change in thermodynamic variables due to
temporal changes alone:
The first differential form tells us that:
s
   + p  


kT =
– ---------------=
–h
t
t
 t
t
t
Fundamental equations
(70)
37/78
and for spatial changes
s
   + p  
kT
=
– --------------- xi
 xi
  xi
(71)
For spatial changes we often use:
s
h p
kT
= 
–
 xi
 xi  xi
(72)
Derivation along a fluid trajectory
Another use for the entropy equation is to consider the variation
of the entropy along the trajectory of a fluid element. Take
 + p

kTds = d – ------------ d
  
Fundamental equations
(73)
38/78
then
ds
d   + p d
d
d
kT ----- = ----- – ------------ ------ = ----- – h -----dt
dt
dt
dt    dt
(74)
4.2 Adiabatic gas
If the gas is adiabatic, then the change of entropy along the trajectory of an element of fluid is zero, i.e.
ds
----- = 0
dt
(75)
d
d
----- – h ------ = 0
dt
dt
(76)
and
Fundamental equations
39/78
This is often re-expressed in a different form using:
v i
1 d
--- ------ = –
 dt
 xi
(77)
v i
d
----- = –   + p 
dt
 xi
(78)
to give
This equation describes the change in internal energy of the fluid
resulting from expansion or compression.
Fundamental equations
40/78
4.3 Equation of state
The above equations can be used to derive the equation of state
for a gas, given relationships between  and p . One important
case is the  -law equation of state where the pressure and internal energy density are related by:
1
p =   – 1    = ----------- p
–1
(79)
cp
 = ----- = Constant ratio of specific heats
cv
(80)
where
Fundamental equations
41/78
We use this in conjunction with the perfect gas law
kT
p = nkT = ---------m
 kT = mp
(81)
Substitute this into:
ds
d   + p d
kT ----- = ----- – ------------ -----dt
dt    dt
ds
1 dp
 p d
 mp ----- = ----------- ------ – ----------- --- -----dt
 – 1 dt  – 1  dt
Fundamental equations
(82)
42/78
–1
Multiply by ----------- :
p
1 dp
1 d
ds
--- ------ –  --- ------ = m   – 1  ---- dt
dt
p dt
Fundamental equations
(83)
43/78
so that
d
d
ds
-----  ln p  –  ----- ln  = m   – 1  ----dt
dt
dt
d  p
ds
----- ln ----- = m   – 1  ----dt   
dt
p

ln ----- = m   – 1   s – s 0 
  
(84)
p
----- = exp  m   – 1   s – s 0  

We write
(85)
Fundamental equations
44/78
K  s  = exp  m   – 1   s – s 0   = Pseudo-entropy
The equation of state is therefore written:
p = K  s  
(86)
ds
----- = 0  s = constant along a streamline
dt
(87)
Adiabatic flow:
In adiabatic flow
and K  s  is a streamline constant, but may differ from streamline to streamline.
Fundamental equations
45/78
Special cases
5
Monatomic gas (e.g. completely ionised gas)   = --3
7
Diatomic gas   = --- = 1.4
5
Fundamental equations
(88)
46/78
4.4 Radiating gas
In astrophysics we often have to take account of the fact that the
gas radiates energy at a sufficient rate, that we have to take into
account the effect on the internal energy. Let the energy radiated
per unit volume per unit time per steradian be j then the internal
energy equation
ds
d   + p d
d
d
kT ----- = ----- – ------------ ------ = ----- – h -----dt
dt
dt
dt    dt
(89)
d   + p d
----- – ------------ ------ = – 4j
dt    dt
(90)
becomes
Fundamental equations
47/78
The quantity j represents the total emissivity in units of energy
per unit time per unit volume per steradian. The factor of 4 results from integrarting over 4 steradians.
Thermal gas
When we are dealing with a thermal gas, that is one in which the
ions and electrons are more or less in thermal equilibrium, then
the total emissivity may be expressed in the form:
4j = n e n p   T 
Fundamental equations
(91)
48/78
where
n e = electron density
n p = proton density
(92)
  T  = cooling function
This has been the time-honoured way of expressing thermal
cooling. A more modern approach is to write
4j = n 2   T 
(93)
where n is the total number density.
The cooling function is calculated using complex atomic physics
calculations. Ralph Sutherland has published cooling functions
for different plasma conditions.
Fundamental equations
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5 The energy equations
5.1 Conservation of energy in classical mechanics
It’s a good idea to look at how we derive the expression for energy in classical mechanics. Suppose we have a particle moving
in a time invariant potential field,  , with its equation of motion:
dv i

m ------- = – m
dt
 xi
(94)
Take the scalar product of this equation with v i .
dv i

mv i ------- = – mv i
dt
 xi
Fundamental equations
(95)
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Now
dv i
d  1 2
1d
v i ------- = --- -----  v i v i  = ----- --- v
dt  2 
dt
2 dt
(96)
Also, the differentiation following the particle motion of 
d



------ =
+ vi
= vi
dt
t
 xi
 xi
(97)

since
= 0 . Hence
t
d 1 2
----- --- mv + m = 0

dt  2
Fundamental equations
(98)
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As a consequence, the total energy,
1 2
E = --- mv + m
2
(99)
is a constant of the motion, i.e. it is conserved.
5.2 Conservation of energy in gas dynamics
The consideration of energy in gas dynamics follows a similar
line – We start by taking the scalar product of the momentum
equations with the velocity. The end result is not as simple but
has some interesting and useful consequences.
Fundamental equations
52/78
We start with the momentum equations in the form
v i
v i
p

= –
–

+ v j
 xj
 xi
 xi
t
(100)
and take the scalar product with the velocity:
v i
v i
p

v i
+ v j v i
= – vi
– v i
t
 xj
 xi
 xi
Fundamental equations
(101)
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Now use:
v i
 1
  1 2

vi
=
--- v i v i =
--- v

t
t 2
t 2 
v i
(102)
 1
  1 2

vi
=
--- v i v i =
--- v



 xj
 xj 2
 xj 2 
then:
  1 2
  1 2
p

 --- v + v j
--- v = – v i
– v i
t 2 
 xj  2 
 xi
 xi
Fundamental equations
(103)
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We can now take the density and momentum inside the derivatives on the left hand side, using the continuity equation:
  1 2
 1 2 
p

--- v +
--- v v j = – v i
– v i
 x  2

t 2
 xi
 xi
j
(104)
Bring in entropy equations
We introduce the entropy equations in order to eliminate the
term v i p  x i . The first one to use is:
s
h p
= 
–
kT
 xi
 xi  xi
p
s
h
 –vi
= kTv i
– v i
 xi
 xi
 xi
Fundamental equations
(105)
55/78
We then manipulate the enthalpy term as follows:
h


v i
=
 hv i  – h  v i 
 xi
 xi
 xi


=
 hv i  + h
 xi
t
(106)
And now use the other form of the entropy equation
s


kT =
–h
t
t
t
 
s
h =
– kT
t t
t
Fundamental equations
(107)
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Putting all of these equations together with appropriate signs:
p
s
h
= kTv i
– v i
–vi
 xi
 xi
 xi
h


– v i
= –
 hv i  – h
 xi
 xi
t
(108)


s
–h
= – + kT
t
t
t
Fundamental equations
57/78
Add all of these up:
p
s
s
 
= kT
 hv i 
–vi
+ vi
– –
 xi
t
 xi  t  xi
(109)
ds  
= kT ----- – –
 hv i 
dt  t  x i
Substitute in the intermediate form of the energy equation:
  1 2
 1 2 
p

--- v +
--- v v j = – v i
– v i
 x  2

t 2
 xi
 xi
j
Fundamental equations
(110)
58/78
gives:
 1 2
 1 2
--- v +  +
--- v v j + hv j
t 2
 xj 2

ds
= – v i
+ kT ---- xi
dt
Fundamental equations
(111)
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Gravitational potential term
Tha last term to deal with is – v i   x i :



– v i
= –
 v i   +   v i 
 xi
 xi
 xi


= –  v i   – 
t
 xi
(112)


= –
 v i  –   
 xi
t
Fundamental equations
60/78
Final form of energy equation
Substitute the last expression into
 1 2
 1 2
--- v +  +
--- v v j + hv j
t 2
 xj 2

ds
= – v i
+ kT ---- xi
dt
 1 2
 1 2
--- v +  +  +
--- v v j + hv j + v j
t 2
 xj 2
ds
= kT ----dt
Fundamental equations
(113)
(114)
61/78
and since j is a dummy repeated subscript, we can write
 1 2

--- v +  +  +
t 2
 xi
 1--- v 2 + h +  v
2
 i
ds
= kT ----dt
(115)
ds
When the flow is adiabatic then kT ----- = 0 and
dt
 1 2

--- v +  +  +
t 2
 xi
Fundamental equations
 1--- v 2 + h +  v = 0
2
 i
(116)
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When there is radiation
ds
kT ----- = – 4j
dt
(117)
and
 1 2

--- v +  +  +
t 2
 xi
Fundamental equations
 1--- v 2 + h +  v = – 4j (118)
2
 i
63/78
Terms in the total energy
1 2
--- v = Kinetic energy density
2
 = Internal energy density
 = Gravitational energy density
(119)
1 2
--- v +  +  = E tot  Total energy density 
2
1 2
Note analogy with E = --- mv + m for a single particle.
2
Fundamental equations
64/78
The energy flux
1 2

Energy flux = F Ei = --- v + h +  v i
2

1 2
--- v v i = Flux of kinetic energy
2
hv i = Enthalpy flux
(120)
v i = Flux of gravitational potential energy
An interesting point is that the flux associated with the internal
energy is not v i as one might expect but the enthalpy flux
hv i =   + p v i .
Fundamental equations
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5.3 Integral form of the energy equation
n
We can integrate the energy equaF E i
tion over volume giving:
S
E tot

V
V
E tot
t
+
F E i
 xi
dV = 0
and then using the divergence theorem

E tot dV + F E i n i dS = – 4j dV
t

Fundamental equations


(121)
66/78
This says that the total energy within a volume changes as a result of the energy flux out of that volume and the radiative losses
from the volume.
5.4 Examples of energy flux
Wind from O star
Suppose that the wind is spherically symmetric:
FE =

sphere
Fundamental equations
 p
 1--- v 2 + ---------- --- v i n i dS
2
 – 1 
(122)
67/78
Since all variables are constant over the surface of the sphere,
then
1 2
 p

F E = --- v + ----------- --- 
2
 – 1 

sphere
v i n i dS
(123)
1 2
 p
·

= M --- v + ----------- --2
 – 1 
We neglect the enthalpy term in comparison to v 2  2 so that
1 · 2
1
F E  --- M v = ---  10 – 6 solar masses / yr   10 3 km s – 1  2
2
2
(124)
28
= 3.2 10
W
Why do we neglect the enthalpy term?
Fundamental equations
68/78
Consider
2
c
2 p
1 2
2 s
1 2
 p
1 2
--- v + ----------- --- = --- v 1 + ----------- --------- = --- v 1 + ----------- ----2
 – 1 v 2
2
 – 1 v2
2
 – 1
(125)
2 1
1 2
= --- v 1 + ----------- ------- – 1 M2
2
where
c s = sound speed
M = Mach number (126)
In the asymptotic zone the Mach number is high, so that
2 1
----------- -------- « 1
 – 1 M2
(127)
We shall justify these statements later.
Fundamental equations
69/78
Velocity of a jet
20 cm image of the radio galaxy, IC4296.
This image shows the jets
(unresolved close to the core)
and the lobes of the radio
source.
Fundamental equations
70/78
Images of the jets in the radio galaxy IC4296 close to the core.
The resolutions are 1" and 3.2" on the left and right respectively.
Fundamental equations
71/78
In this case we shall reverse the process to estimate the jet velocity. For a radio jet such as we observe in IC4296, once the jet has
widened appreciably, then the Mach number is quite low probably of order unity. Hence the energy flux is dominated by the enthalpy flux.
This is the case when
2 1
2
4
2
----------- --------  1  M  ----------- = 6 for  = -- – 1 M2
–1
3
r
z
FE 

jet

----------- pv z dS
–1
(128)
(129)
2
= 4pv z R jet
Fundamental equations
72/78
Contour image of
the inner 150" of
IC4296. The diameters of jets can be
worked out from images such as this.
Fundamental equations
73/78
Using the radio data we can estimate the following parameters at
a distance 100 from the core:
Diameter = 15 = 2.57 kpc
Minimum pressure = 5 10
– 13
N m –2
(130)
We also know from an analysis from the radio emission from the
lobes of this radio galaxy that
F E  10 36 W
Fundamental equations
(131)
74/78
Since,
FE
10 36 W
v z = --------------- = --------------------------------------------------------------------2
– 13  2.57
4pr 2
4  5 10
 ---------- kpc
 2

(132)
7
= 2.5 10 m s – 1 = 25 000 km s – 1  0.08c
This is actually an upper estimate of the velocity since we are using a minimum estimate of the pressure. However, it is unlikely
that the pressure is too far from the minimum value, so that this
is a reasonable estimate of the velocity in the western jet of
IC4296 at this distance.
Fundamental equations
75/78
6 Summary of gas dynamics equations
6.1 Mass
 
+
 v i  = 0
 t  xi
(133)
6.2 Momentum


p

 v i  +
 v i v j  = –
–
t
 xj
 xi
 xi
v i
v i
(134)
p


+ v j
= –
–
t
 xj
 xi
 xi
Fundamental equations
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6.3 Internal energy
v i
d
----- = –   + p 
– 4j
dt
 xi
(135)
6.4 Total energy
 1 2

--- v +  +  +
t 2
 xi
 1--- v 2 + h +  v = – 4j (136)
2
 i
6.5 Thermal cooling
4j = n 2   T  
Fundamental equations
(137)
77/78
6.6 Equation of state
p =   – 1  = K  s  
Fundamental equations
(138)
78/78