5. 522, Space Propulsic Prof. Manuel martinez-Sanchez Lecture 10: Electric Propulsion- Some Generalities on Plasma(and Arcjet Engines) Ionization and Conduction in a High-pressure plasma A normal gas at T0.1 atm collision among molecule and other particles are frequent enough that we can assume local Thermodynamic equilibrium and in particular, ionization-recombination reactions are governed by the Law of Mass Action. Consider neutral atoms(n) which ionize singly to ions (i and electrons (e): n ee+ (1) One form of the Law of Mass Action(in terms of number densities n kT, where T is the same for all species )is nn =s(T) no Where the"Saha function"S is given(according to Statistical Mechanics)as Ground state degeneracy of ion(= 1 for H) Ground state degeneracy of neutral(=2 for H) me mass of electron = 0.91 x 10-30 Kg k= Boltzmann constant =1.38x10-J/K (Note: k= R/Avogadro's number) h= Plank's constant=6.62x10-34 ]s Vi= Ionization potential of the atom(volts) (Vi=13.6 V for H Except for very narrow" sheaths"near walls, plasmas are quasi-neutral n2 nn can be used 16.522, Space P pessan Lecture 10 Prof. Manuel martinez Page 1 of 12
16.522, Space Propulsion Prof. Manuel Martinez-Sanchez Lecture 10: Electric Propulsion - Some Generalities on Plasma (and Arcjet Engines) Ionization and Conduction in a High-pressure Plasma A normal gas at T ∼ 0.1 atm, collision among molecule and other particles are frequent enough that we can assume local Thermodynamic Equilibrium, and in particular, ionization-recombination reactions are governed by the Law of Mass Action. Consider neutral atoms (n) which ionize singly to ions (i) and electrons (e): n R e+i (1) One form of the Law of Mass Action (in terms of number densities n = Pj kT j , where T is the same for all species) is n nei = S T ( ) (2) nn Where the “Saha function” S is given (according to Statistical Mechanics) as 3 ST ( ) i qn = 2 q ⎛ π e ⎜ h2 ⎟ 2 m kT ⎞ ⎝ ⎠ eVi e kT - (3) qi = Ground state degeneracy of ion (= 1 for H+) qn = Ground state degeneracy of neutral (= 2 for H) me = mass of electron = 0.91 × 10-30 Kg k = Boltzmann constant = 1.38 ×10-23 J/K (Note: k = R/Avogadro’s number) h = Plank’s constant = 6.62x10-34 J.s. Vi = Ionization potential of the atom (volts) (Vi = 13.6 V for H) Except for very narrow “sheaths” near walls, plasmas are quasi-neutral: n = n e i (4) So that e n n n = S T ( ) (3’) can be used. 2 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 1 of 12 2
Given T, this relates n, to n,. A second relation is needed and very often it is a specification of the overall pressure P=(ne+n,+nn)kT =(2ne +n)kT Combining(3)and(5) =s( s(T)(n-2n2) Where nP kT is the total member density of all particles We then have n2+2sn。-Sn=0 Since s increases very rapidly with t, the limits of (6)are n sn→0 (Weak ionization) T→0 heT→2 (Full ionization) Once an electron population exists, an electric field E will drive a current density j through the plasma. To understand this quantitatively consider the momentum balance of a" typical"electron It sees an electrostatic force FE=-eE [t also sees a"frictional" force due to transfer of momentum each time it collides with some other particle(neutral or ion). Collisions with other electrons are not counted, because the momentum transfer is in that case internal to the electron species. The ions and neutrals are almost at rest compared to the fast-moving electrons and we define an effective collision as one in which the electrons directed momentum is fully given up. Suppose there are ve of these collisions per second (ve=collision frequency per electron). The electron loses momentum at a rate m, Vev, where V. =mean directed velocity of electrons, and so 16.522, Space P pessan Lecture 10 Prof. Manuel martinez
Given T, this relates ne to nn . A second relation is needed and very often it is a specification of the overall pressure P = (n + ni + n ) kT = (2n + n ) kT (5) e n e n Combining (3’) and (5), 2 ( ) ⎛ ⎜ P n = S T - 2ne ⎟ = S T ⎝ kT ⎞ e ( )(n - 2ne ) ⎠ P Where n= is the total member density of all particles. kT We then have 2 n + 2Sn - Sn = 0 e e S2 (6) n = -S + + Sn = n e 1 + 1 + n S T( ) Since S increases very rapidly with T, the limits of (6) are n ⎯⎯⎯⎯⎯→ Sn (→ 0) (Weak ionization) e T0 → T 0 → ⎯⎯⎯⎯⎯ n ne T→∞ → 2 (Full ionization) G Once an electron population exists, an electric field E will drive a current density G j through the plasma. To understand this quantitatively, consider the momentum balance of a “typical” electron. It sees an electrostatic force G G F = -eE E (7) It also sees a “frictional” force due to transfer of momentum each time it collides with some other particle (neutral or ion). Collisions with other electrons are not counted, because the momentum transfer is in that case internal to the electron species. The ions and neutrals are almost at rest compared to the fast-moving electrons, and we define an effective collision as one in which the electron’s directed momentum is fully given up. Suppose there are νe of these collisions per second ( νe =collision frequency per electron). The electron loses momentum at a rate JG JJG -m Ve ν , where Ve =mean directed velocity of electrons, and so e e G JG FFriction = -m Ve ν (8) e e On average, 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 2 of 12
FE +FRiction =0 m -e (9) The group Hem is the electron "mobility"((m/s)/(volt/m)). The current density is the flux of charge due to motion of all charges. If only the electron motion is counted (it dominates in this case from(9), The group is the conductivity of the plasma (si, Let us consider the collision frequency. Suppose a neutral is regarded as a sphere with a cross-section area Qen. Electrons moving at random with (thermal) velocity Ce intercept the area Qen at a rate equal to their flux n c Qen. Since a whole range of speeds ce exists, we use the average value Ce for all electrons but this is for all electrons colliding with one en neutral. We are interested in the reverse (all neutrals, one electron), the part of v due to neutrals uld be n ceQen. Adding the part due to ions, 16.522, Space P pessan Lecture 10 Prof. Manuel martinez Page 3 of 12
G G F + F E Friction = 0, or JG G m Veν = -e E e e JG ⎛ e ⎞ G V =- e ⎜ ⎟E (9) ν ⎝mee ⎠ The group µe = e is the electron “mobility” ((m/s)/ (volt/m)). The current density m νee is the flux of charge due to motion of all charges. If only the electron motion is counted (it dominates in this case) G JG j -en V ≡ e e (10) and from (9), G 2 ⎛ en ⎞ G e j= ⎜ ⎜ m ν ⎟ ⎟E (11) ⎝ ee ⎠ The group 2 e νe σ = (12) m νee is the conductivity of the plasma (Si/m). Let us consider the collision frequency. Suppose a neutral is regarded as a sphere with a cross-section area Qen. Electrons moving at random with (thermal) velocity ce intercept the area Qen at a rate equal to their flux n c Qen . Since a whole range of ee speeds ce exists, we use the average value ce for all electrons. But this is for all electrons colliding with one neutral. We are interested in the reverse (all neutrals, one electron), so the part of νe due to neutrals should be n c Qen e . Adding the part due to ions, n ν = n c Qen + n c Qei e (13) e n e i 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 3 of 12
NOTE c is very different(usually much larger) than v. Most of the thermal motion is fast, but in random directions, so that on average it nearly cancels out. The non- cancelling remainder is v. Think of a swarm of bees moving furiously to and fro, but oving(as a swarm) slowly The number of electrons per unit volume that have a velocity vector c ending in a dc。 dcdc。=d3 c. in velocity space is defined as d . d Where f( x is the Distribution function of the electrons which depends(for a given location x and tir t on the three components of C. In an equilibrium situation all directions are equally likely, so f=fe(Gel=fe(ce)only, and one can show that the form is maxwellian fe c.+C+ With the normalization 「fd3 The mean velocity is then ∫∫ and direct calculation gives k (15) For Hydrogen atoms, Ce=6210千。 (m/s, with T. in k) NOTE If there is current, the distribution cannot be strictly Maxwellian(or even isotropic) But since Ve < Ce, the mean thermal velocity is very close to Equation(15)anyway 16.522, Space Propulsion Lecture 10 Prof. Manuel martinez-Sanchez
NOTE: ce is very different (usually much larger) than Ve . Most of the thermal motion is fast, but in random directions, so that on average it nearly cancels out. The noncancelling remainder is Ve . Think of a swarm of bees moving furiously to and fro, but moving (as a swarm) slowly. G The number of electrons per unit volume that have a velocity vector ce ending in a “box” 3 dc dc dc ≡ d c in velocity space is defined as ex y e ez e G 3 fc ( ) d c e e e ( G JG Where f c , x ) is the Distribution function of the e e JG electrons which depends (for a given location x and time G t) on the three components of ce . In an equilibrium JJG situation all directions are equally likely, so f = fe ( c ) = f ( ) c only, and one can e e e e show that the form is Maxwellian. 3 2 2 m c 2 2 2 2 - ee f = n ⎛ m ⎞ ⎟ e 2kTe ; (c = c + cey + cez ) (14) e e ⎜ e e ex ⎝2 kTe ⎠ π With the normalization 3 f d c = n . ∫∫∫ e e e The mean velocity is then 3 ce ≡ ∫∫∫ c f d c ee e and direct calculation gives 8 kTe c =e (15) π me For Hydrogen atoms, ce = 6210 T (m/s, with Te in K) (16) e NOTE: If there is current, the distribution cannot be strictly Maxwellian (or even isotropic). But since Ve << ce, the mean thermal velocity is very close to Equation (15) anyway. 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 4 of 12
Regarding the cross sections Qen, Qeir they depend on the collision velocity c especially Qej. This is because the e-i coulombic interaction is soft", so a very fast electron can pass nearly undeflected near an ion, whereas a slow one will be strongly deflected. The complete theory yields an expression Q。=2.95×10 (T in Kelvin) here In A=-11.35+ 2In T(K)-=In P(atm) so that In A is usually around 6-12, and can even be taken as a constant(8) rough calculations For the neutral hydrogen atoms, the collisions are fairly hard and one can use the approximation Q=2×10°m Numerical Example Consider Hydrogen at P=l atm. For T >4000K, diatomic H2 is not present anymore (H,->2H). So ionization is from atomic hydrogen, H, for which V=13.6 volts, s=2.42×1021Te (m3) We also find 7.34×1027 6210√T(m/s) σ=2821×10na/v 16.522, Space P pessan Lecture 10 Prof. Manuel martinez Page 5 of 12
Regarding the cross sections Qen, Qei, they depend on the collision velocity c , e especially Qei. This is because the e-i coulombic interaction is “soft”, so a very fast electron can pass nearly undeflected near an ion, whereas a slow one will be strongly deflected. The complete theory yields an expression Qei 2.95 ×10-10 ln Λ (m2 ) (17) T (T in Kelvin) where 1 ln Λ -11.35 + 2ln T(K) - ln P(atm) (18) 2 so that ln Λ is usually around 6-12, and can even be taken as a constant (~8) in rough calculations. For the neutral hydrogen atoms, the collisions are fairly “hard”, and one can use the approximation 2 Q 2 ×10-19m (19) eH Numerical Example Consider Hydrogen at P=1 atm. For T ≥ 4000K , diatomic H2 is not present anymore (H2 → 2H) . So ionization is from atomic hydrogen, H, for which V = 13.6 volts, i q = 1, i q = 2, n so that 3 2 157,800 - e S = 2.42 ×1021 T T (m-3). We also find 7.34 ×1027 (m-3 n= ), T c = e 6210 T (m/s), σ = 2.821×10-8 n ν e e (ν = νei + ν ). e en 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 5 of 12
The results are shown below T(K) 5,000 000 7,000 8,000 10,000 12,000 14,000 16,000 nm 147×1024122X1 1.048×1024918×10237.34×10236.12×1035.24×10234.59×1023 s(m3)1.68×1034.26x10152.30×1027470×10183.393x1 5104×102|2.55×1 )4.97×1017216×104906×102072×1021545×10215565×011.203×1021.716×102 34×10659x10547×10-400023 0.0215 0.1000 0.298 c(m/s)4.39×105481×105520×1055.56×105621×1056.80×105.35×1057.86×10 Ina 5.68 05 6.36 662 7.07 744 7.74 8.01 a(s+)|1.46x10°1.72×1099.76x1093.51x10102.00×1031|5.77×1011.030×102|124×1012 van(s1)1.29×1011.15×1011.09×1011.02x101873x1010681x1010416×10101.82x100 v。(s1)1.29×101117×1011.19×1011.37x101287x1016.45×1011.072×102|1.262×1012 (S/m)1.09 17.4 116.3 426.7 1519 2434 3166 3836 Notice (a)Coulomb-dominated (v >>v)for T>8000 K (b)Rapid rise of ionization fraction a for 70007000 K (c) Conductivity above 1,000Si/m T>9000K Ohmic Dissipation -Stability, constriction The conductivity o increases rapidly with T in the fully Coulomb-dominated range 0.0153 (Te in K, o in Si/m) Notice also how, in this limit (which occurs at high temperature as a approaches the conductivity becomes independent of the kind of gas in question, except for 1) small influences hidden in In a 16.522, Space P pessan Lecture 10 Prof. Manuel martinez Page 6 of 12
The results are shown below. T(K) 5,000 6,000 7,000 8,000 10,000 12,000 14,000 16,000 -3 ) 1.47x1024 1.22x1024 1.048x1024 9.18x1023 7.34x1023 6.12x1023 5.24x1023 4.59x1023 nm S(m-3) ( 1.68x1013 4.26x1015 2.30x1017 4.70x1018 3.393x1020 6.189x1021 5.104x1022 2.55x1023 -3 ) 4.97x1018 7.216x1019 4.906x1020 2.072x1021 1.545x1022 5.565x1022 1.203x1023 1.716x1023 n me ( α 3.4x10-6 5.9x10-5 4.7x10-4 0.0023 0.0215 0.1000 0.298 0.597 ce (m/s) 4.39x105 4.81x105 5.20x105 5.56x105 6.21x105 6.80x105 7.35x105 7.86x105 ln Λ 5.68 6.05 6.36 6.62 7.07 7.44 7.74 ν 8.01 ei (s-1) 1.46x108 1.72x109 9.76x109 3.51x1010 2.00x1011 5.77x1011 1.030x1012 1.244x1012 ν (s-1) 1.29x1011 1.15x1011 1.09x1011 1.02x1011 8.73x1010 6.81x1010 4.16x1010 1.82x1010 en ν (s-1) 1.29x1011 1.17x1011 1.19x1011 1.37x1011 2.87x1011 6.45x1011 1.072x1012 1.262x1012 e σ (Si/m) 1.09 17.4 116.3 426.7 1519 2434 3166 3836 Notice (a) Coulomb-dominated ( ν >> νen ) for T > 8000 K ei ∼ (b) Rapid rise of ionization fraction α for 7000 7000 K ∼ (c) Conductivity above 1, 000 Si/m T > 9000 K ∼ Ohmic Dissipation - Stability, constriction The conductivity σ increases rapidly with T in the fully Coulomb-dominated range, 0.0153 T 3 2 e σ = (Te in K, σ in Si/m). ln Λ Notice also how, in this limit (which occurs at high temperature, as α approaches 1), the conductivity becomes independent of the kind of gas in question, except for small influences hidden in ln Λ. 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 6 of 12
One important consequence of o=o(T) is the tendency for current to concentrate into"filaments"or arcs". To understand this, consider the amount of work done by electric forces to overcome the "friction"on the electrons due to collisions the force on the ne electrons in a unit volume is-en e, and these electrons reach a terminal velocity V as they slide against friction. Hence the power dissipated per unit volume s Dom=-enEV。=(enV)E DoH =jE(Ohmic dissipation) (20) Since ohm's law gives j= oE, we can put (21) The simplest situation is one with an initially uniform plasma subject to a constant applied field E, such E as would occur between the plates of a plane capacitor Regardless of the path taken by d the current if the plates are large and the gap is small, the field E=- remains unchanged If we now look at we see that the dissipation becomes large wherever the conductivity(hence the temperature)is large. Starting from uniform temperature if a small non-uniformity arises such that T is higher along a certain path that path becomes more conductive, heats up due to extra Ohmic dissipation and this reinforces the initia nonuniformity. The result is a constriction of the current into a filament or arc In principle the constriction process would continue indefinitely and lead to arcs of zero radius and infinite current density. But as the temperature profile steepens heat will increasingly diffuse away from the hot core to the cooler surroundings, and provided it can be removed efficiently from there an equilibrium is eventually reached at some finite arc radius and arc core temperature. Clearly the detailed end result will depend on the details of the thermal management of the gas: the more efficient the cooling of the background the more the constriction can progress, and the hotter the eventual arc. This counter-intuitive result(more cooling leads to hotter arcs) is one of several paradoxical properties of arcs, all of them related to their being the result of a statically unstable situation. We analyze this behavior 16.522, Space P pessan Lecture 10 Prof. Manuel martinez
V One important consequence of σ = σ (Τ ) is the tendency for current to concentrate into “filaments”, or “arcs”. To understand this, consider the amount of work done by electric forces to overcome the “friction” on the electrons due to collisions. The force G on the ne electrons in a unit volume is-en E, and these electrons reach a terminal JG e velocity Veas they slide against friction. Hence the power dissipated per unit volume is G JG JG G DOH = -en E.V = - (en V ). E e e e e G G D = j . E (Ohmic dissipation) (20) OH G G Since Ohm’s law gives j= σ E , we can put j 2 = σ Ε2 D or DOH = (21) OH σ The simplest situation is one with an initially uniform plasma subject G to a constant applied field E , such as would occur between the plates of a plane capacitor: Regardless of the path taken by the current, if the plates are large and the gap is small, the field E= remains unchanged. If we d now look at = σ Ε2 D , OH we see that the dissipation becomes large wherever the conductivity (hence the temperature) is large. Starting from uniform temperature, if a small non-uniformity arises such that T is higher along a certain path, that path becomes more conductive, heats up due to extra Ohmic dissipation, and this reinforces the initial nonuniformity. The result is a constriction of the current into a filament or “arc”. In principle, the constriction process would continue indefinitely and lead to arcs of zero radius and infinite current density. But as the temperature profile steepens, heat will increasingly diffuse away from the hot core to the cooler surroundings, and, provided it can be removed efficiently from there, an equilibrium is eventually reached at some finite arc radius and arc core temperature. Clearly, the detailed end result will depend on the details of the thermal management of the gas: the more efficient the cooling of the background, the more the constriction can progress, and the hotter the eventual arc. This counter-intuitive result (more cooling leads to hotter arcs) is one of several paradoxical properties of arcs, all of them related to their being the result of a statically unstable situation. We analyze this behavior next. 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 7 of 12
Sample Physical Properties of High-temperature Gases(Near Equilibrium) 05101520253035.103 Fig 1 Number of Particles per cm3 for Nitrogen at Atmospheric Presuure Source: Dr W. FinkeInburg and Dr H Maecker, Elektrische Bogen und Thermisches Plasma, Encyclopedia of Physics, Vol 22, Gas Discharges ll Springer-Verlag, Berlin, 1956 Figure 1: Equilibrium Composition of Nitrogen at p=l atm Note: N2 replaced by N at w 7,000K, then by N at w 14, 000K and then by N at 29, 000K. Electron density satisfies ne = n 2n For Hydrogen, similar, but all transitions at lower temperature(and, of course here is no HTt) 16.522, Space P pessan Lecture 10 Prof. Manuel martinez Page 8 of 12
Sample Physical Properties of High-temperature Gases (Near Equilibrium) Figure 1: Equilibrium Composition of Nitrogen at P=1 atm. Note: N2 replaced by N at ~ 7,000K, then by N+ at ~ 14,000K and then by N++ at ~ 29,000K. Electron density satisfies ne = n+ + 2n++ For Hydrogen, similar, but all transitions at lower temperature (and, of course, there is no H++) 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 8 of 12
p=l atm 1xt04 104 °K Fig 2 Electrical Conductivity of Nitrogen Versus Temperature g 3 Electrical Conductivity of Hydrogen Versus Temperature Source: J Yos, Transport Properties of Nitrogen, Hydrogen, Oxygen, and Air to 30,000 K, AVcO, Technical Memorandum RAD-TM-63-7, 1963, reprint-1966 Figure 2,3: Electrical Conductivity of Nitrogen and oxygen Note: Weak dependence on gas type Pressure Units are mho/cm= si/cm 1 Si/cm= 100 Si/m 16.522, Space P pessan Lecture 10 Prof. Manuel martinez Page 9 of 12
Figure 2, 3: Electrical Conductivity of Nitrogen and Oxygen gas type Note: Weak dependence on Pressure Units are mho/cm ≡ Si/cm. 1 Si/cm = 100 Si/m 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 9 of 12
10 Ke 2z9 10 3 TEMPERATURI Fig 4 Individual contributions to the heat conductivity of a nitrogen plasma( Burhorn, 1959) Contribution due to molecules Km, atoms Kat, electrons Ke, ions K, dissociation K ionization K 16.522, Space P pessan Prof. Manuel martinez
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