Space Propulsi Prof. manuel martinez- sanchez Lecture 18: Hall Thruster Efficiency For a given mass flow m and thrust f, we would like to minimize the running power P. Define a thruster efficiency (1) ere is the minimum required power. The actual power is V s。,°。°°· B p。,°°·引 Where Va is the accelerating voltage and Ia the current through the power supply (or anode current or also cathode current). Of the Ia current of electrons injected by the cathode a fraction IB goes to neutralize the beam and the rest iBs back-streams into the thruster Since no net current is lost to the walls, 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 1 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 1 of 20 16.522, Space Propulsion Prof. Manuel Martinez-Sanchez Lecture 18: Hall Thruster Efficiency For a given mass flow m i and thrust F, we would like to minimize the running power P. Define a thruster efficiency 2 F = 2m P ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ η i (1) where 2 F 2m ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ i is the minimum required power. The actual power is P=IVa a (2) Where Va is the accelerating voltage and Ia the current through the power supply (or anode current, or also cathode current). Of the Ia current of electrons injected by the cathode, a fraction IB goes to neutralize the beam, and the rest, IBS back-streams into the thruster. Since no net current is lost to the walls, a B BS I =I +I (3) } Va (IB) e (IB) i IBS IBS Ia (IB) e
V The thrust is due to the accelerated ions only these are created at locations along the thruster which have different potentials v(x), and hence accelerate to different speeds. Then F=cdm (4) wnere Suppose the part dmi of mi is created in the region where v decreases by dv and define an"ionization distribution function"f(v) by dm=-fv.Va or, with dm=-f( o) do From the definition, f() satisfies f(o)do=1 (7) Then, from(4)and(5), 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 2 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 2 of 20 The thrust is due to the accelerated ions only. These are created at locations along the thruster which have different potentials V(x), and hence accelerate to different speeds. Then F = cdmi ∫ i (4) where i 2eV c = m (5) Suppose the part dmi i of mi i is created in the region where V decreases by dV, and define an “ionization distribution function” f(V) by i a a i dm V dV = -f V V m ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ i i (6) or, with a V = V ϕ , ( ) i i dm = -f d m ϕ ϕ. i i From the definition, f (ϕ) satisfies ( ) 1 0 f d =1 ϕ ϕ ∫ (7) Then, from (4) and (5), V Va 0 x
2ev and hence the efficiency is m √Gf(o)do 2m VI Notice that the beam current IB is related to mi by IB=mi. We can therefore re- m( o f() d there each of the factors is less than unity and can be assigned a separate meaning (11) is the utilization factor",i.e. it penalizes neutral gas flow (12) the backstreaming efficiency" penalizes electron backstreaming (13)o f(o)do=ne, the"nonuniformity factor"is less than unity because of the nonuniform ion velocity It is clear that, since f(o)do=1, we want to put most of f(o )where Vo is greatest, namely, we want to produce most of the ionization near the inlet. In that case f()=8(o-1), and ne =1. A somewhat pesimistic scenario would be f(9)=1 d mi proportional to dr. i.e., ionization rate proportional to field strength In that 「√@×1×do 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 3 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 3 of 20 ( ) 1 a i 0 i 2eV F = m f d m ϕϕϕ ∫ i (8) and hence the efficiency is ( ) 2 2 1 a i 0 a a i 2eV m f d m = 2mV I ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ϕϕϕ ⎝ ⎠ ⎝ ⎠ η ∫ i i (9) Notice that the beam current IB is related to mi i by B i i e I= m m i i . We can therefore rewrite (9) as ( ) 2 1 B a 0 mi I = f d I m ⎛ ⎞⎛ ⎞⎛ ⎞ η ϕϕϕ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ∫ i i (10) where each of the factors is less than unity and can be assigned a separate meaning: (11) u mi m ≡ η i i is the “utilization factor”, i.e., it penalizes neutral gas flow. (12) B a a I = I η , the “backstreaming efficiency” penalizes electron backstreaming. (13) ( ) 2 1 Ε 0 f d = ⎛ ⎞ ⎜ ⎟ ϕϕϕ η ⎝ ⎠ ∫ , the “nonuniformity factor” is less than unity because of the nonuniform ion velocity It is clear that, since ( ) 1 0 f d =1 ϕ ϕ ∫ , we want to put most of f (ϕ) where ϕ is greatest, namely, we want to produce most of the ionization near the inlet. In that case f = -1 () ( ) ϕ δϕ , and ηΕ = 1. A somewhat pesimistic scenario would be f =1 (ϕ) , namely dmi dx i proportional to dV - dx , i.e., ionization rate proportional to field strength. In that case 1 2 2 E 0 2 4 = ×1×d = = 3 9 ⎛ ⎞ ⎛ ⎞ η ϕϕ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∫
Measurements tend to indicate ne w 0.6-0.9, which means that ionization tends to occur early in the channel. This is to be expected, because that is where the backstreaming electrons have had the most chance to gain energy by falling"up the potential The factor n, = is related to the ionization fraction. Putting mi=neCA, m=m+mn=(nc +ncn)A, (11) 1+(%) Since ci /cn is large(cn w neutral speed of sound, i.e., a few hundred m/sec, while C:= gIsp-20,000m/sec), nu can be high even with ne /nn no more than a few percent. Datall show n, ranging from 40% to 90% The factor ne = IB/I requires some discussion. Most of the ionization is due to the backstreaming electrons, so that we are not really free to drive IB towards Ia (Ies =I-I). What we need to strive for is (a) Conditions which favor creation of as many ions as possible per backstreaming electron, and (b) Minimization of ion-electron losses to the walls, once they are created This can be quantified as follows: Let b be the number of secondary electrons(and of ions) produced per backstreaming electron and let a be the fraction of these new e-i pairs which is lost by recombination on walls. Then, per backstreaming electron, (1-a)B ions make it to the beam, and an equal number of cathode electrons are used to neutralize them therefore I_(1-a)阝 I31+(1-a)阝 Clearly, we want B>>1 and a<<1. The first(B>>1)implies lengthening the electron path by means of the applied radial magnetic field, and also using accelerating potentials which are not too far from 5/2 times the range of energies where ionization is most efficient(typically 30-80 Volts). This last condition creates some difficulties with heavy ions, which require higher accelerating potentials for a given exit speed The condition a < 1 implies minimization of insulation surfaces on which the recombination can take place and arrangement of the electric fields such that ions 16.522, Space P pessan Lecture 18 Prof. Manuel martinez
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 4 of 20 Measurements[1] tend to indicate ηΕ ~ 0.6 - 0.9 , which means that ionization tends to occur early in the channel. This is to be expected, because that is where the backstreaming electrons have had the most chance to gain energy by “falling” up the potential. The factor u mi = m η i i is related to the ionization fraction. Putting e i m = n c A , i i n ( ) ei nn m=m+m = n c +n c A i i ii , e i n n u e i n n n c n c = n c 1+ n c ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ η ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ (11) Since i n c c is large (cn ~ neutral speed of sound, i.e., a few hundred m/sec, while i sp c gI 20,000m/sec ∼ ), ηu can be high even with n n e n no more than a few percent. Data[1] show ηu ranging from 40% to 90%. The factor ηE Ba =I I requires some discussion. Most of the ionization is due to the backstreaming electrons, so that we are not really free to drive BI towards I I =I -I a BS a B ( ) . What we need to strive for is (a) Conditions which favor creation of as many ions as possible per backstreaming electron, and (b) Minimization of ion-electron losses to the walls, once they are created. This can be quantified as follows: Let β be the number of secondary electrons (and of ions) produced per backstreaming electron, and let α be the fraction of these new e-ipairs which is lost by recombination on walls. Then, per backstreaming electron, ( ) 1 - α β ions make it to the beam, and an equal number of cathode electrons are used to neutralize them. Therefore ( ) ( ) B a a I 1 - = = I 1+ 1- α β η α β (12) Clearly, we want β >> 1and α << 1. The first (β >> 1) implies lengthening the electron path by means of the applied radial magnetic field, and also using accelerating potentials which are not too far from 5/2 times the range of energies where ionization is most efficient (typically 30-80 Volts). This last condition creates some difficulties with heavy ions, which require higher accelerating potentials for a given exit speed. The condition α << 1implies minimization of insulation surfaces on which the recombination can take place, and arrangement of the electric fields such that ions
are not directly accelerated into walls. This is difficult to achieve without detailed surveys of equipotential surfaces Reference Komurasaki, K, Hirakowa, M. and Arakawa, Y, IEPC paper 91-078. 22 Electric Propulsion Conference, Viareggio, Italy Oct. 1991 1-D Model of hall thruster M=-1 Drift (not included) Anode pre-sheath ionization Define「e=n(v。<0,sor。<0) T=neVi (2) Tn=nvn v,=constant=SKT w (3) 1. Conservation of particles n=n vion dx (4) 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 5 of 20
16.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 5 of 20 are not directly accelerated into walls. This is difficult to achieve without detailed surveys of equipotential surfaces. Reference: Komurasaki, K, Hirakowa, M. and Arakawa, Y., IEPC paper 91-078. 22nd Electric Propulsion Conference, Viareggio, Italy, Oct. 1991. 1-D Model of Hall Thruster Define Γe ee =n v ( ) v < 0, so < 0 e e Γ (1) Γi ei =n v ( ) + or - (2) Γn nn =n v w n i 5 kT v constant 3 m (3) 1. Conservation of particles e i n e ion d d d = =- =n dx dx dx Γ Γ Γ ν (4) Drift Zone SHEATH (not included) Sonic point Ionization region M = -1 Tw x Anode Pre-sheath Diffusion