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 T< 3000K is a good electrical insulator because there are almost no free electrons in it. For pressure >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 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 <3000K is a good electrical insulator, because there are almost no ∼ free electrons in it. For pressure > ∼ 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