The Electron Energy Equation
Since ve is often very different from other species mean velocities,
it makes sense to formulate
o
i
the energy equation in terms of Te; , defined as ne 23 kTe; =
1
m c2
2 e e
, vce = w
v − ve . We found
e
before (with no inelastic effects)
�
�
∂
3 ;
ne kTe +
∂t
2
�
�
3 ;
;
· neve kTe + vqe + ⇒:
P
2
;
Ere
ve =
(1)
r
and, at least for Maxwellian collision (but generalizable to others),
;
Ere
�
= ne ver µre
;
Ere
3k(Tr; − Te; )
mr
+
(vvr − ve )2
me + mr
mr + me
�
�
�
3k(Tr; − Te; )
2
∼
+ (vvr − ve )
= ne ver me
mr
(2a)
(2b)
where (2b) results from me << mr .
One first observation is that the “heat transfer” portion of this, namely
El = ne ver
2me 3
k(Tr; − Te; )
mr 2
(3)
can be thought of as transferring the mean thermal energy difference per particle, 32 k(Tr; −Te; ),
per collision, but with a very poor efficiency
nel =
2me
<< 1
mr
(4)
In other words, while Maxwellian distributions about Te; and Tr; established in a few e.g. e-e
mr
and r-r collisions, it takes about 2m
collisions (tens to hundreds of thousands) to drive Te;
e
1
toward Tr; . In practice, a good approximation is that there are separate Maxwellian popula­
tions for electrons (at Te; ) vs. the heavy species (at close to the same Tr; , since the collisional
rs
efficiency among them m2µ
is the of order 1).
r +ms
We next examine the irreversible second term in (2b). The summation over r-species includes
ions and neutrals, and we assume a single kind of each. We then have a dissipation
D = ne me [vei (vvi − ve )2 + ven (vvn − ve )2 ]
(5)
The electron and ion current densities are
vje = −eneve
(6a)
vji = enev1
(6b)
�
and so D = ne me vei
�
vj
ene
�2
�
�
�
vj 2
+ ven vn +
ene
(7)
This expression simplifies several limits:
(a) vn << vi , ve
�
�
j2
je2
j2
j2
∼
�
+�
e �
D = ne me vei 2 2 + ven 2 2 = �
e ne
e ne
e2 ne
e 2 ne
me vei
me ven
The quantities in the denominator are the conductivities if only ei or en collisions occurred:
j2
j2
+ e
D∼
=
σei σen
(8)
and, in particular
(a.1) For a neutral-dominated gas (ven >> vei ) (Hall Thruster), D ∼
=
(a.2) For a Coulomb-dominated gas (vei >> ven )D ∼
=
2
j2
σ
je2
σ
(9)
(10),
(a.3) If ji << je , je ∼
=
= j and D ∼
j2
,
σ
with σ =
e2 ne
).
me (ven +vei
(b) When density is relatively high, ions and neutrals couple strongly and vn ∼
= vi (as in
MPD thrusters or MHD generators). In that case (5) yields
j2
D∼
= ne me (vei + ven )(vvi − ve )2 =
σ
with
σ =
e2 ne
me (vei + ven )
(11)
(12)
One approximation which is routinely made is to neglect the viscous dissipation of the elec­
tron gas, i.e., the contribution of the off-diagonal terms in ⇒: ∇ve :
P
⇒I · ve ∼
= Pe; ∇.vve
Pe
(13)
where Pe; = ne kTe; is the scalar pressure (the trace of ⇒I ). Breaking this into ∇ · (ne kTe;ve ) −
Pe
ve · ∇Pe; and substituting in (1), we get
�
�
�
�
∂
3 ;
5 ;
;
ne kTe + ∇ · ne ve kTe + vqe = D + El + ve · Pe;
∂t
2
2
(14)
with D given by (7) and El given by (3).
Finally,oalthoughiwe will not prove it here, it stands to reason that the heat flux vector
vqe; = ne 12 me ce2vce will be expressible in the form of a Fourier law
e
vqe; = −Ke (Te )∇Te;
(15)
where Ke is the electron thermal conductivity. Note that a simple form like this may not be
accurate when the alternative definition Te of temperature is used.
3
MIT OpenCourseWare
http://ocw.mit.edu
16.55 Ionized Gases
Fall 2014
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