Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Lecture 1:
Basic gas Properties – Fluid Dynamics
As gas atoms/molecules (or particles) move through out a volume they collide and
randomly distribute their energy. A basic tenant of statistics is that random processes result in
Normally distributed results. This is know as the central limit theorem, see for example Box
Hunter and Hunter, “Statistics for experimenters” (Yu Lei has the book right now so I don’t have
the title quite right.) The normal distribution is
y   2 
py   const
exp
.
2



 

2
Here  is the population variance,  is the population standard deviation, and  is the central
value. The distribution looks like
1
1.2
1
0.8
p(y)
Normal Distribution
0.6
0.4

0.2
0
-5
-4
-3
-2
-1
0
y
1
2
3
4
5
Concept of Temperature
Maxwell and Boltzman proposed that this same distribution can be used to model the velocity
distribution of particles in thermal equilibrium. (This is known as the Maxwell Distribution,
Boltzman Distribution and the Maxwell-Boltzman Distribution.) Using this assumption we have
that the velocity distribution is
mv 2 
f v   const exp 
,
 2kT 
Page 1
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
while the energy distribution is
 
.
f    const exp 

kT 

Here k is Boltzman’s constant and T is the temperature. Notice that the temperature is simply a
measure of the random energy of the system. NOTE that often ‘kT’ is simply written as ‘T’ and
hence ‘T’ is given units of energy.
Problem 1
Show that if in one dimension
 f v dv  n
then
1/ 2
m 
.
const  n
2kT
Also show that if in three dimensions
3
 f v dv  n
then
3/ 2
m 
.
const  n
2kT
Here n is the particle density.
Typically, the velocity distributions are well behaved and we have distributions as such:
The Maxwellian Distribution
Page 2
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
1.2
1
0.8
f(v)
Maxwellian
0.6
0.4
2Vth
0.2
0
-5
-4
-3
-2
-1
0
V/Vth
Where
1
2
3
4
5
mv 2 
m 1/ 2

f v  n
exp 
.
2kT 
 2kT 
This is what happens if there are enough collisions to evenly distribute the energy. In other
cases, however, we will be working in with a system in which the rate of collisions is not large
enough to evenly distribute the energy. Some typical examples are:
The bi-Maxwellian Distribution
Page 3
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
1.2
1
0.8
f(v)
Bi-Maxwellian
0.6
0.4
0.2
0
-5
-4
-3
-2
-1
0
V
Where
 m 
mv 2 
f v  1  n
exp



2kT1 
 2kT1 
.
1/ 2
2
 m 
m v  
  n
exp 

2kT2 

 2kT2 
1/ 2
The drifting Maxwellian Distribution
Page 4
1
2
3
4
5
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
1.2
1
0.8
f(v)
Drifting Maxwellian
0.6
0.4
0.2
0
-5
-4
-3
-2
-1
0
V
Where
1/ 2
mv  v0 
m 
f v  n
exp 
2kT 

 2kT
2

.


and the ‘bump-on-tail’
Page 5
1
2
3
4
5
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
1
0.9
0.8
0.7
0.6
f(v)
Bump-on-tail
0.5
0.4
0.3
0.2
0.1
0
-5
-4
-3
-2
-1
0
V
1
2
3
4
5
Where
f v  fMaxwel l  f drift .
While other distributions are certainly possible, these four comprise most of what is observed in
laboratory plasmas.
Normalization of the velocity distribution.
The velocity distribution is normalized so that if all particles of all velocities are accounted for
then one has the local particle density. This means that



f vdv  n . Thus in one dimension
mv 2 
m 1/ 2
f v  n
exp

 or in three-dimensions
2kT 
 2kT 
mv2 
m 3/ 2
f v  n
exp


2kT 
 2kT 
Concept of Thermal Velocity
We know that
m 1/ 2
mv 2 

f v  n
exp
 2kT .
2kT 


The central part of the distribution, ± sigma, will have ~69% of all the particles at or below a
given v. This velocity is known as the thermal velocity, vth. Thus we can ‘derive’
1/ 2
 v  v0 2 
m 

f v  n
exp 

2
2kT 

 2vth 

Page 6
Class notes for EE6318/Phys 6383 – Spring 2001
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thus
vth2 
kT
.
m
Note that in some texts vth2 
2kT
. We will not use this definition.
m
Average velocity

v 






vf vdv
f v dv
1 
vf v dv
n 
v 2 
1  m 1/ 2 
 n
v
exp

dv
n 2kT   odd function 2vth2 
even function
 0 (Odd functions int egrated from   to  is zero.)
This is very different than the average speed, which is
v 
1 
v f vdv
n 
1/ 2
m 
 
2kT 
v 2 
v
exp
 2 dv

2vth 

v 2 
1  1 1/2 
   2 0 vexp  2 dv
vth 2
2vth 
 v2   v 
1 1/2  v

 vth   2 0
exp 2 d 
2
vth
2v th  vth 

1 
u 2  u 2 

 vth   2 0 exp
d 
2
 2   2 
1/2
1/2

1 
 vth 
2
expy dy

2 
0
1/ 2
2kT 
 m 
in one-dimension. In three-dimension the average speed is
1 
v   v f v4v2 dv
n 
8kT1/ 2

 
m 

Page 7
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Average energy in one-dimension
E 
1
1 
m v 2   v 2 f vdv
2
n 
v 2 
1  m 1/ 2  2
 m
v exp  2 dv
2
2kT  
2vth 
v 2 
1  1 1/ 2  2

m
2 v exp 2 dv
2vth 2  
2v th 
2
1/2
 v 
v 2   v 
exp
 vth  2vth2 d vth 
1/2
y 2 
2
y
exp
 dy

 2 
 1 
 kT 
8
1
 kT 
8 



1/2

1  
y 2 
 y 2  

y exp     exp 
 kT
dy 
8  

2
2



 



1/2
 1 
 kT
8 
2  kT2
1/2
In three dimensions the average energy is
3kT
E 
. The derivation of this is left homework.
2
Concept of Mean-free path
As atoms/molecules pass through a volume, they collide with other atoms/molecules. A prime
example of this is Brownian motion – the motion of a dust particle in air that moves in disjointed
fashion.
also kno wna s
a random walk
We can calculate the mean-free path, mfp, if we consider the following picture.
Page 8
Class notes for EE6318/Phys 6383 – Spring 2001
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If we were to repeatedly fire test particles at the target, a fraction of the test particles will be
scattered by collisions with atoms in the target. The fraction scattered is simply the area ratio.
N
Fraction scattered 
A
where N is the number of atoms in the target. Assuming that our target is a part of a larger piece
of material, in which we know the density, n, of the material then
Fraction scattered 
nAdx 
 ndx .
A
Now let us send a continuous flux, , of test particle – or a current density J, at the target then
the change in J across the target is
J  Jafter  Jbefore
 1  ndx J before  Jbefore
 Jndx
or
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dJ
 Jn
dx

J  J 0 expx / mf p 
1
where  mf p 
.
n
We can make approximations of the cross section of the target atom, based on the radius of the
target atom. This is known as the hard-sphere approximation, which is at best imprecise. The
reality is that the interaction, and hence collisions, is related to the target atom electron orbitals
and particle energies, i.e. it is quantum mechanical in nature. From our above discussion, we can
calculate the
Mean frequency of collisions
v

 vn
mf p
And the mean period of collisions
1


Of course the average collision frequency for a distribution of velocities is
  n v noting    v .
Gas pressure
Lets assume that we have a particle in a box of sides l . Assume that the particle is reflected with
M  mvx 
mvx
init
ial
momentum
fi
nal
momentum
no loss of energy (why?) when it hits wall A1. Thus E  0 but
 2mvx
Where we have assumed that v y  0  v z . Now let us assume that the particle travels across the
2l
volume without striking anything else and reflects off side A2. The round trip time is t 
vx
The force applied to side A1 is
M 2mvx mvx2
F


t
2l vx
l
and the pressure is
F mv 2
P  2  3x
l
l
Putting in more particles we find
Page 10
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
P
all
mv2x
N
m 3

3
l
part icles l
all
mv 2x

part icles N
 mn v 2x
1
 2n n v 2x  2n Ex
2
 nkT  the ideal gas law!
Lecture 2 Gas Flow
Conservation of Flux, which we will prove later
If a gas passes through a series of pipes with various cross sectional areas, then the flux is
conserved. Physically this should be obvious. Think of this as the same as saying that we do not
have rarefaction/compression of the gas, nor do we have a source/sink for the gas.
1
2
3
P2
P1
P3
P4
A1
  A1 n1 v1  A2 n2 v2  A3 n3 v3 ...
Flow resistance/conductance
If a gas flow through a pipe, we expect some pressure differential to cause the flow. The ratio of
the pressure to the flow is known as the resistance, R.
P
R
.

The conductance is simply the inverse of the R,
1

# /s / area
C 
R P Force / area
These equations are very similar to the resistance in a circuit, where we replace voltage with
pressure. Thus for a series of pipes
Page 11
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Ptotal 

but
Ctotal
Ptotal   Pi and so
i
Ptotal 


1
   
Ctotal
i Ci
i Ci

1
1
Ctotal
i Ci
or in a similar vane,
Rtotal   Ri .

i
For Parallel pipes we find
Ctotal   Ci and
i
1
1

Rtotal
i Ri
Pumps
We can think of pumps as simply another conductance, S. After all we are trying to have a
certain  go through the pump and the pump acts as if the back side has a pressure of zero. Here
however, the pump conductance is given a special name, S, for ‘speed’.
If a pump has a pipe connecting it to the chamber, we get an effective pump speed from
1
1
1
 
Seff S Cpipe
Example:
Given a pipe with a conductance C= 100 L/s and a pump speed of S = 200 L/s we find that the
effective pump speed is 66.6 L/s. This means that we using only 33% of our possible pumping
ability. This can become critical in the design of a process system.
Types of gas flow.
There are three radically different types of gas flow. They are:
Laminar fluid (viscous) flow
Turbulent fluid flow
Molecular flow
Page 12
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The distinction between Laminar (‘layer’) and turbulent is often seen in wind tunnel experiments
on new cars.
Lamin ar
flow
Turbulent
flow
One can determine if one is operating in fluid flow or molecular flow by determining the
Knudsen number, Kn.
mean free path
for all colli sions
Kn 
mf p
d
charact eristic distan ce
across system
 1 Molecular

 ~ 1 Transition

Fluid
 1
We can calculate the conductance in molecular and fluid flow regimes from the following
equations. (Note these equations are at least partly experimental approximations.)
Fluid Flow through a round tube
P1
d
P2
L
with the requirements P1 > P2 and L >> d.
Laminar Flow (the most common for us)
 d 4 P1  P2
C
;   gas viscosity
128 L 2
Here P1 and P2 are in mbar and L and d are in cm, giving C in L/s.
Turbulent Flow
3/ 7
gas cons tan t 
1/ 7
2
2
2 4 /7
d  4   5 3 P1  P2   R

C
d
T
  


P1  P2    3.2 4
2L   M molar 
 Molar mass 
Here, Mmolar is in g/mole, R is 83.14 mBar L/Mole/K and T is in K.
Fluid Flow through an orifice
Page 13
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P1
P2
D
d
with the requirements P1 > P2.
1/ 2
1/ 
(1  )/ 

P1 d 2  2RT  P2     P2 
C

1


   


P1  P2 4  Mmolar   P1    1 P1 



C
where   P is the adiabatic cons tan t
CV
1/ 2
Molecular flow through a round tube
1/ 2
 RT 


d 2 2Mmolar 
C
; where 1    1.12 is a fudge factor
4 1  3 L 
 4 d 
Page 14