16.522, Space Propulsion Prof. Manuel Martinez-Sanchez Lecture 16: Ion Engine Performance. Brophy’s Theory DEFINITIONS: J B =Beam ion (and neutralizer electron current) J E =cathode emitted current J c =ion curent to cathode potential surfaces J D =current through disch. power supply J P =total ion production rate J ia =Ion current to anode J acc =Ion current intercepted by accel. grid Current Balances: J D = J E + J c + J B + J acc and also J D = J P + J E ? J ia J P = J B + J C + J ia + J acc (ion balance) B (Useful Power = V J + V D ) B ' Total Power = V J + V J accB B D D + V J B + P Heaters Energy cost ' B )P. Total - P. Useful (J D ? V J D + J acc V B + P H per beam ion ε B = = J B J B P H = V J cE J D ? J B = J E + J c + J acc ' ε B = ? ? D E + P H ? ? ? J P V J D + J acc (V + V D )? V J + c B ? J P ? J B J B ε p ε ' ε B = p + f c V D + f acc (V + V D ) (plasma ion cost) f B f B f B B 16.522, Space Propulsion Lecture 16 Prof. Manuel Martinez-Sanchez Page 1 of 8 V J + P H where f B = J B ,f c = J c , f acc = J acc and ε p = D E J J J J p p p p More definitions: U + = ionization energy per ion U j = excitation energy of level j J j = excitation rate (total) J LP = loss rate of primary electrons ε m = mean energy of Maxwellian electron group Discharge Energy Balance U + j + ∑ j J P j J U + P DLp J V J m + ε ( p J + E J P ia J J? ? ) Lp J = p D E J V J = p ε ? p H J P Define o ε = U + j + ∑ j J p j J U . Then ? P H J P p + ε = o ε m + ε + ( D V ) m ? ε E Lp J J p E J J ? p ia J J m ε p m ? ? ? ? + ε ε D H V P? / p J ? ? ? ? Use p ε D H V P? P J/ p ? ? ? ? ε 1 ? D mD V V ? ε E Lp J J ? D m V ε ? ? ? ? = o ε m + ε ? ( D V ) m ? ε E Lp J J D p H V J P ? p ia J J m ε m ? ε D p H V J P + D p H V J P + p H J P 1 ? E Lp J J ? D m V ε ? ? ? ? 1 ? E Lp J J ? ? ? ? 1 ? ? ? ? ? 1 ? E Lp J J ? ? ? ? ? ? ? ? 1 ? D m V ε ? ? ? ? 16.522, Space Propulsion Lecture 16 Prof. Manuel Martinez-Sanchez Page 2 of 8 1 J Lp ε m ? J Lp ε m + ? ? ? ? ? ? ? ? ? ? ? ? ? ε m ? ? ? ? ? ? ? ? P H V D J ia ε p = o m ε+ε ? J m ε? 1 ? 1 ? J E V J V J V D? ? ? ? D D Ep p ? ? ? ε m ?J 1 + ? ? ? ? ? ? P H Lp 1 J ε ? ? ?? ? - V J ? ? p p J ia 1 o m J p P H JJ Lp 1 m p 1 J E V D + ? ? ? ? ? ? ? ? ε ? ? ? ? ? ? ? ? ? ? ? ? ? ? ε+ ? ε ? ? ? ? = ε p NOTE: If we write P H = V J the equation for ε p becomes c E , ? ? ? ? ? ? J ia o m ε+ε ?1 J ? ? ? ? V C? ? p ? ? ? ? ? ? ? ? ? =ε p ? 1 ? 1 + VJ ? ? ? V ε m ? ? ? ?DLp J 1 ? ? ??E D Survival Equation for Primary Electrons J Lp )σσ tot exc primaries . primary electon before it would be captured by the anode, if it did not collide with a neutral before before that. This path lenght is that of the electron’s helical path around one of the magnetic lines of force created by the confinement magnets. The neutral density is related to the flow rate by σ( ?σ A tot n e n , where Also, Ae is the path length for a += e = + J E Γ n c n 4 η 4 u  m ?(1 ) (φ = grid system transparency forn n = = φc n A m g i φ ηneutrals; u = utilization efficiency) φ η(m  1 u c A m i n g σ) ? ? J 4 ? A LP J e ?? ?= exp ? ? ? ? E 16.522, Space Propulsion Lecture 16 Prof. Manuel Martinez-Sanchez Page 3 of 8 ε * p ε = p  1 ? exp[? m c (1 ?η )] o n [ε + ε m (1 ? f ia )]? ? 1 + V c ? ? o ? ? V D ? ? C o 4 σ T Ae ε * = =where , and p c A m φε m n g i 1 ? V D The quantity C o is a measure of the confinement effectivenss for primary electrons (better for long electron path Ae , small grid open area A g φ ). If C o ∞ → , the energy cost per ion, ε p , tends to the limit ε * , which then represents the cost per ion no p primary losses. Calculation of Primary/Secondary Population Ratio Primaries are endowed initially with an energy V D , and, if they did not escape, would all thermalize eventually, to an energy ε m . The rate at which they disappear in that case is simply the rate of ionization or excitation by primaries (a primary is assumed to become a secondary-Maxwellian - after one ionization or one excitation). So, the net energy input rate per unit volume due to injection of primaries is (without escape) υ σ T (V )(V ? ε m ) (σ = σ + +σ exc )n n p p D D Tn This energy is used by the primaries and their secondary “progenie” to (a) Produce ionization by primaries. Per ionization event, this uses U + + E m , since the new electron created has energy ε m . Total p.u. volume n n σ υ T (V )(U + + ε m ) n p p D (b) Excite atom, by primaries Total energy rate p.u. volume n n υ σ exc () p p n U v exco (a shorthand for υ ∑ σ j U j )n p p j (c) Produce ionization by secondaries (Mawellian). The rate p.u. volume is ∞ 2 () cn n c c f σ + ()4π c dc ≡ n c n m σ n e + ∫ m o 16.522, Space Propulsion Lecture 16 Prof. Manuel Martinez-Sanchez Page 4 of 8 ∞ 2 c c σ ()4π dc c where σ + = 1 f m () + c c n e ∫ m o ∞ 8 kTe 2 and c e = , n m = ∫ f m 4π dc c π m e o The ionization cross-section σ t (c) in zero below c + = 2eU + . Using a Maxwellian m e form for f m (), we find easily c ∞ σ t = ∫ e ?u uσ + () ? E ? du u ? ? u = ? ? kT e ? ? + u and the energy spent by secondaries in ionization (p.u. time and volume) is then c n n e σ + (U + + ε m ) n m Similarly, the energy spent in excitation is c n n σ exc U excn e m The energy balance is therefore (dividing by n n throughout) σ υ (V σ υ (V )(U + + ε m )+σ ( U V D ) exc ]+ c n e [σ (U + + ε m )+σ exc U exc ]n p p + D ? ε m ) = n p p [ + D exc m + n p This can be solved for : n m U + + ε m + U exc σ exc n p = c e σ + n m (V ) ? σ (V ) σ exc (V ) D + D D υ p (V ? ε m ) σ T K (U + + ε m )? D σ + σ + σ + * which is a function of T e for a fixed V D . Hence ε p is also a function of T e . This is because, given ε * p ε = p  1 ? exp[? m C (1 ?η )] o u 16.522, Space Propulsion Lecture 16 Prof. Manuel Martinez-Sanchez Page 5 of 8 ε * is seen to be the energy per ion created if no primary electrons were to escape p * (ε o ∞ → ) . The expression for ε (neglecting ion capture by the screen, f ia = o , and p heating power, V c = o) was p ε * = D m mo V ? + ε εε 1 ; o ε = U + + p exc J J exc U or ε o = + U + exc U () () + + + σ υ σ υ Dp p Dexc p p Vn Vn + σ σ e m exce m c n c n and this does depend on T e and () e m p T n n . Substituting the expression for m p n n found above, and simplifying, we obtain * σ ( V σ + ( U + + ε m )+σ exc U exc ε p = V T D D ) [σ σ ( V )?σ ( V )σ ] U + σ σ T ( V )( V ? ε m ) exc + B exc D + exc + D D n * p NOTE: An intermediate expression for ε p (still containing ), which will be useful n m later, is ?σ T ? ? ? * V D ? ? σ + ? ? ε p = n m c e σ + 1 + n υ σ ( V D ) p p + Calculation of utilization efficiency  J p = ( n Vd σ υ ( V )+ c n e σ + ) n J B = J f = e η u m)( p p + D m n p B e m i  ↑ m( 1 ?η ) 4 u c n φ A g m i ( η u V( n σ υ ( V )+ c n e σ + ) 1 4 ?η u ) = p p + D m c n φ Ag f B But also n + = n m + n p and J B = n e + 61.0 υ A φ i g B 16.522, Space Propulsion Lecture 16 Prof. Manuel Martinez-Sanchez Page 6 of 8 J B so n m + n p = e υ B Ag φ i n p υ p (V D ) σ + (V D ) 1 + 1 4 ?η ) = e 6.0 A g φ i ( n Divide: c V e σ n m c e σ + c n φAg η u+ B 1 + n p f J B n m 2 e 61.0 υ A g φφ c n ? ? 1 + n p ? ? ? 11 ?η u = B i ? n m ? = 1 + yη u + D 4 c V f J e σ + ? ? ? 1 + n p ? υ p σ (V ) ? ? = y → η u B B ? n m c e σ + ? ? V D σ T (V ) n p υ p D ? ε P σ + n m c e ? n ? * 2 15. 0 e υ ε A g φφ i c n ? ? 1 + p ? p B ? n m ? ? y = V f J D υ p V σ (V ) ? ? n p ? ? B B T D ? ? ? n m ? So: Given V D , J B ,T walls → c n Geometry: f B ,φ ,φ ,V , Ag i Gas ( U + , U exc ,σ + (),σ ( E ), M )E exc n * Can conclude p ()→ ε ()→η () NOTE: m = m i J B so real parameter is mT e p T e u T e   , n m e η n not J B .  Then given also A e (magnetic geometry) and m → C o → ε p → ε B Summary Inputs: Geometry and Gas Operating Aditonal Magentic Field Properties Parameters Parameter f B , f c ,... U + , U exc V D , J B ,φ φ i , V , Ag E Eσ + (),σ exc () T e → ε m T w → c n Ae M 16.522, Space Propulsion Lecture 16 Prof. Manuel Martinez-Sanchez Page 7 of 8 U + +ε m + U exc σ exc n p c e σ = n m p + V D υ ( V ?ε m ) σ () U exc D σ + ↓ n e υ p σ ( V D ) + ε T = V D ? ? ?σ T ? ? ? n m c e σ +* ? σ + ? V D 1 + n p υ p σ + ( V D ) n m c e σ + ↓ Y = ( ) n n VVV f J n n cAe m p DTp D B B m e n i gB p ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? +115.0 2* συ φφυ ε ; u η = Y+1 1 → m =  u Bi J e m η o C = ( ) φ σ n g i DT cA m eV A4 → p ε ( )[ ] uo p mC η ε ??? = 1exp1 *  ↓ B ε = B p f ε + B c f f D V 16.522, Space Propulsion Lecture 16 Prof. Manuel Martinez-Sanchez Page 8 of 8