7/28/2019 Goebel_AppC_ Electric Propulsion EnergyLoss
1/3
467
Appendix C
Energy Loss by Electrons
The energy lost from the plasma due to electrons being lost to
an anode that ismore negative than the plasma is derived. Figure
C-1 shows the plasmapotential distribution in the negative-going
sheath towards the anode wall. TheMaxwellian electrons are
decelerated and repelled by the sheath potential. Todetermine the
average energy removed from the plasma by each electron,moments of
the Maxwellian distribution are taken. The electron current
densityreaching the wall is given by
Je = en dvx
dvy
vz
m
2kTe
3/2
2e/m
exp
m vx2+ vy
2+ vz
2( )2kTe
dvz
=
1
4en
8kTe
mexp
e
kTe
. (C-1)
The electrons must overcome the sheath potential to reach the
wall so theminimum electron speed toward the wall (assumed to be in
the z-direction) is
2e m . The plasma electrons lose kinetic energy as they traverse
the sheath,
so the power flux from plasma is
Pe = n dvx dvy vz2e/m
me vx2+ vy
2+ vz
2( )2
m
2kTe
3/2
expm vx
2+ vy
2+ vz
2( )2kTe
dvz
=1
4ne
8kTe
m2kTe
e+
exp
e
kTe
,
(C-2)
7/28/2019 Goebel_AppC_ Electric Propulsion EnergyLoss
2/3
468 Appendix C
Plasma
Sheath
SheathEdge
Pre-Sheath
WallPotential
Fig. C-1. Schematic of plasma in contact with theanode wall.
whereis expressed in electron volts (eV). The average energy
that an electron
removes from the plasma (in eV) is then the ratio of the power
per electron to
the flux of electrons:
Eave =Pe
Je
= 2kTe
e+= 2TeV + , (C-3)
where TeV is in electron volts (eV). This is the energy removed
from the
plasma per electron striking the wall through a negative-going
sheath.
It should be noted that this energy loss from the plasma per
electron is differentthan the average energy that each electron has
when it hits the wall. The flux ofelectrons hitting the anode wall
is the same as analyzed above. The plasmaelectrons lose kinetic
energy as they traverse the sheath; hence, a e term
must be included in the particle energy expression for each
electron. The powerflux to the insert from plasma electrons is
then
Pe = n dvx dvy vz2e/m
me vx2+ vy
2+ vz
2( )2
e
m
2kTe
3/2
expm vx
2+ vy
2+ vz
2( )2kTe
dvz
=1
4ne
8kTe
m2kTe
e
exp
e
kTe
.
(C-4)
7/28/2019 Goebel_AppC_ Electric Propulsion EnergyLoss
3/3
Energy Loss by Electrons 469
The average energy of each electron is then the ratio of the
power to the flux:
Eave =Pe
Je
= 2kTe
e= 2TeV [energy per electron that strikes the wall] (C-5)
