5. 522, Space Propulsic Prof. manuel martinez-sanchez Lecture 15: Thrust Calculation(Single Grid, Single Potential) d xx 十 a=Pche x= so dx er dx xx0 ch 十 十 十 d d E 3 d A 9 8 Alternative F d2 Child-Langmuir 16.522, Space P pessan Lecture Prof. Manuel martinez Page 1 of
16.522, Space Propulsion Lecture 15 Prof. Manuel Martinez-Sanchez Page 1 of 11 16.522, Space Propulsion Prof. Manuel Martinez-Sanchez Lecture 15: Thrust Calculation (Single Grid, Single Potential) 2 x d 1 E dx 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ d d d 2 x x 0 x ch x 0 x 0 0 0 F dE E = E dx = E dx = A dx 2 ε ρ ε ∫ ∫ 4 3 a x = -V d ⎛ ⎞ φ ⎜ ⎟ ⎝ ⎠ 1 3 a x V 4 x E = d3 d ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ x 0 E =0 a x d 4 V E = 3 d 2 2 x 0a a 2 2 0 F 16 8 V V = = A 29 9 d d ⎡ ε ⎤ ⎢ ε ⎥ ⎣ ⎦ Alternative: 3 3 2 2 xi i aa a 0 0 2 2 i i Fm m 42 e 8 V 2eV V = jc = = A e e9 m m 9 d d ε ε m i j from Child-Langmuir
Bohm velocity: Why? Wall charged negatively, or facing extractor collisional"drag nmi=t =nE坛 small mall not near wall Add constant . my, dv d(+P)=-F ne vi =const =T amv+P。+P)=-Fn Pe+P=nk(Te+T)=nkTe dx/"、,/- d min at v.= 16.522, Space P pessan Lecture Prof. Manuel martinez Page 2 of
16.522, Space Propulsion Lecture 15 Prof. Manuel Martinez-Sanchez Page 2 of 11 Bohm velocity: Why? collisional “drag” i i i i i i x in dv dP nm v + = enE - F dx dx e e e e e e x en dv dP n m v + = -en E - F dx dx small small not near wall, n =n. e i Add: constant ( ) e i i e i i in dv d P +P n m v + - F dx dx and n v = const = e i Γi ( ) 2 e i i e in i d n m v + P + P - F dx P + P = n k T + T n kT e i e e i ee ( ) i i i i e in i d m v + kT = -F dx v ⎡ ⎤ Γ ⎢ ⎥ Γ ⎣ ⎦ min. at e i i kT v = m
The lines kT and mr v. must F ↓.1-↓ Supersonic cross at v kTe here their sum is minimum kTe Ii m So, no solution at Ions accelerate to vi = ve in the quasineutral plasma. Beyond that, becomes very strong, and ions just free-fall to wall (in the sheath)so, entering How big is the sheath? sheath Pre-sheath ne ni In sheath, say n =0 Child-Langmu 42 But also J=en 16.522, Space P pessan Lecture Prof. Manuel martinez Page 3 of 11
16.522, Space Propulsion Lecture 15 Prof. Manuel Martinez-Sanchez Page 3 of 11 The lines i e i kT v Γ and m v iii Γ must cross at e i i kT v m = , where their sum is minimum. So, no solution at e i B i kT v > =v m Ions accelerate to v =v i B in the quasineutral plasma. Beyond that, n << n , E e ix becomes very strong, and ions just free-fall to wall (in the sheath) so, entering sheath, v =v i B . How big is the sheath? In sheath, say n 0 e Child-Langmuir: 3 2 s i ii 0 2 i 42 e V j = en v = 9 m ε δ But also sh e i e i kT j en m sh 3 2 s e 0 e 2 i i 42 e V kT en 9m m ε δ
1 k。/ev en 1.018 deby =69 1m)~69,/3×110 =24×105m=24um 3×103 If wall not biased (insulator) evs -kTe In. / m 8=5d~3d For sheath in front of extractor grid, Vs-1000V18=78dD 6=1.9mm asma This approximately sizes the extractor holes 16.522, Space P pessan Lecture 15 Prof. Manuel martinez Page 4 of 11
16.522, Space Propulsion Lecture 15 Prof. Manuel Martinez-Sanchez Page 4 of 11 sh sh 3 3 2 2 2 0s 0 e s 2 e e e e 42 1 42 V kT eV = 9 n 9 kT ekT e n ε ε ⎛ ⎞ δ ⎜ ⎟ ⎝ ⎠ e e sh -1 n = n exp 2 ∞ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 1 3 2 4 0e s -1 2 2 e e 4 2 kT eV e n kT 9e ∞ ε ⎛ ⎞ δ ⎜ ⎟ ⎝ ⎠ 1.018 dDebye ( ) ( ) e -5 Debye -3 17 e T K 3×11600 d 69 69 = 2.4×10 m = 24 m n m 3×10 = ∼ µ If wall not biased (insulator), i 3 4 se DD e m eV kT ln 5 d 3d m ⎛ ⎞ ∼ →δ ∼ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ For sheath in front of extractor grid, V 1000V s ∼ 78dD δ kTe 3V e ∼ δ 1.9 mm This approximately sizes the extractor holes
If holes much bigger than 8 If much smaller, ions lost plasma would escape to grid Space charge effects in the accel-decel gap For o< x<d d 2 dx Note: Slope here not necessarily zero Change integration variable 2 Xdv V=C m ceo 2 16.522, Space Propulsion Lecture 15 Prof. Manuel martinez-s Page 5 of 11
16.522, Space Propulsion Lecture 15 Prof. Manuel Martinez-Sanchez Page 5 of 11 If holes much bigger than δ, If much smaller, ions lost plasma would escape to grid Space charge effects in the accel-decel gap For 0 x<dd < , 2 i 2 0 0i 2 0 0 i d jj en =- =- =- dx v 2e v - m φ ε ε φ ε 2 0 0 2 0 i 1d j d =- +c 2 dx 2e v - m φ ⎛ ⎞ φ φ ⎜ ⎟ ⎝ ⎠ ε φ ∫ Note: Slope here not necessarily zero. Change integration variable : 2 0 i 2e v= v - m φ ( ) i 2 2 0 m = v -v 2e φ mi d = - vdv e φ ( ) 0 i 2 v i i 2 0 00 0 00 v m- vdv 1 d j j j 2e e m m =c- =c- v -v =c- v - v - 2 dx v e e m φ φ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ εεε ⎝ ⎠ ∫
here 2ev 2j m T 2c-2jm 2ev. 2e(V-o) d C, then all profiles follows 2jm|2ev2e(v-ψ Ege[y m, In the limit when the second gap becomes choked as well,(as in the case with no decel grid, in which case dd is the downstream sheath thickness) 0→c= jm,/2ev 2eVN Then d 2m|2e(W-9) spev m Change again variable 16.522, Space P pessan Lecture Prof. Manuel martinez Page 6 of 11
16.522, Space Propulsion Lecture 15 Prof. Manuel Martinez-Sanchez Page 6 of 11 where 2 T 0 i 2eV v = m ( ) 2 i T T 0i i 1 d 2j m 2eV 2e V - = 2c - - 2 dx e m m ⎛ ⎞ φ ⎛ ⎞ φ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ε ⎝ ⎠ ( ) i T T 0i i d dx = 2jm 2eV 2e V - 2c - - em m φ ⎛ ⎞ φ ⎜ ⎟ ε ⎝ ⎠ ( ) =V -V T N d 0 i T T 0i v d d = c, 2jm 2eV 2e V - 2c - - em m φ φ ⇒ ⎛ ⎞ φ ⎜ ⎟ ε ⎝ ⎠ ∫ then all profiles follows. In the limit when the second gap becomes choked as well, (as in the case with no decel grid, in which case dd is the downstream sheath thickness) d iT N x=d 0i i d jm 2eV 2eV =0 c= - dx e m m ⎛ ⎞ ⎛ ⎞ φ ⇒ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ε ⎝ ⎠ ( ) =V -V T N φ Then ( ) V -V T N d 0 i N T 0ii d d = 2jm 2eV 2e V - - em m φ ⎛ ⎞ φ ⎜ ⎟ ε ⎝ ⎠ ∫ Change again variable: mi d = - vdv e φ ( ) T i 2e V - = v m φ
。=2my-Nm m se da= aim [2v"-j2V-Vndv N 2VT-(Vr-VN) 2 +2 2×2 Now i_4 2 =2×26/) 9 define r V d,(e-wp(w2+22) dn=1-R为 1+2R 16.522, Space P pessan Lecture Prof. Manuel martinez Page 7 of
16.522, Space Propulsion Lecture 15 Prof. Manuel Martinez-Sanchez Page 7 of 11 N i T T N i 2eV i m v 0 i d 2eV v N i N m 0 i m- vdv e m v dv d= = 2jm 2ev 2je v-v v - e m ε ⎛ ⎞ ⎜ ⎟ ε ⎝ ⎠ ∫ ∫ N 2d v - v ( ) T T N N v v 0i 0i d N N TN T TN v v m m 4 d = 2v v - v - 2 v - v dv = v - v 2v - v - v 2je 2je 3 ε ε ⎧ ⎫ ⎪ ⎪ ⎛ ⎞ ⎡ ⎤ ⎨ ⎬ ⎜ ⎟ ⎣ ⎦ ⎪ ⎪ ⎝ ⎠ ⎩ ⎭ ∫ ( ) T N v 3 2 N v 4 v-v 3 T N 2 4 v+ v 3 3 ( ) 0 i d T NT N 2 m d = v - v v + 2v 3 ej ε ( ) 1 1 4 4 1 11 1 0 2 22 2 d T NT N i 2×2 e d = V - V V + 2V 3 jm ε ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Now 3 2 T 0 2 i a 42 e V j = 9 m d ε , so 1 3 4 4 1 1 4 2 T a 0 1 2 i 2 e V d = ×2 3 m j ⎛ ⎞ ε ⎜ ⎟ ⎝ ⎠ define N T V R = V ( ) ( ) 1 12 12 12 12 2 TN T T d 3 4 a T d V - V V + 2V = d V ( ) ( ) 1 1 1 2 d 2 2 a d = 1 - R 1+ 2R d
Appendix B ELECTRON DIFFUSION IN A MAGNETIC FIELD 1) No B Field In electron momentum balance, main forces are pressure gradient and collisional retardation(no inertia ) -nme ve Also P =n. kT. VP =kT Vn Solve for flux In, v. =m. -Vr This is Ficks law of diffusion n Ve =-D. Vn, with a diffusivity (v。=∑ncQ, collision frequency) 2)WthB(⊥tovP) Add magnetic force neme Vev- en. ve×B (4) To solve for n ve form VPe xB=-mevene Ve XB-ene ve XB xB, ×xB=(.6,B Eliminate ve xb between these two equations, simplify 16.522, Space P pessan Lecture Prof. Manuel martinez Page 8 of
16.522, Space Propulsion Lecture 15 Prof. Manuel Martinez-Sanchez Page 8 of 11 Appendix B ELECTRON DIFFUSION IN A MAGNETIC FIELD 1) No B JG Field In electron momentum balance, main forces are pressure gradient and collisional retardation (no inertia): e ∇ ν P -n m v e ee e JG (1) Also P = n kT e ee , ∇ ∇ P kT n . e ee Solve for flux: e e e e e e kT n v =- n m ∇ ν JG (2) This is Fick’s law of diffusion e n v = -D n e ee ∇ JG , with a diffusivity e e e e kT D = m ν (3) ( e je e j ν = ncQ ∑ , collision frequency) 2) With B to P ( ) ⊥ ∇ e JG Add magnetic force: e e ∇ ν P = -n m v - en v ×B e ee e e JG JG JG (4) To solve for e n ve JG , form ∇ ν P ×B = -m n v ×B - en v ×B ×B e eee e e e ( ) JG JG JG JG JG JG , and use 0 ( ) () 2 v ×B ×B = B v . B - B v . e ee JG JG JG JG JG JG JG Eliminate ( ) v ×B e JG JG between these two equations, simplify:
n ve=- P (5) NOTE:This leaves the E field out. To include it, just replace Vne by Vn.+ ene - E Define the nondimensional factor ÷、eBc (Hall parameter (6) here o=eB is the cyclotron frequency for electrons Then n ve= 1+ (-D。vn-阝xDn Of these two terms, the second is perpendicular to both b and Vn,, and is called the"VP. xb drift". The main interest is on the first term which is along -Vn, as a regular diffusion We see that this"cross-field diffusion"is governed by n Ve =-D,Vne, witl D So, a high Hall parameter B can greatly reduce diffusion, compared to that in the absence of a magnetic field. High B means both, high B and/or low collision frequency In an ion engine, with T. 4ev =46400 K, the e-n and e-i cross-section are roughly Q。=10m2 Q。=4×1048m2, and 8 kTe e=1.34×10°m/ If also n=28×103m3, 16.522, Space P pessan Lecture Prof. Manuel martinez Page 9 of
16.522, Space Propulsion Lecture 15 Prof. Manuel Martinez-Sanchez Page 9 of 11 e e e e 2 2 e e e e e e 2 e e kT ekT - n + n ×B m m nv = eB 1+ m ∇ ∇ ν ν ⎛ ⎞ ⎜ ⎟ ν ⎝ ⎠ JG JG (5) NOTE: This leaves the E G field out. To include it, just replace ∇ne by e e e en n+ E kT ∇ G Define the nondimensional factor c ee e eB β m ω ≡ ≡ ν ν (Hall parameter) (6) where c e eB = m ω is the cyclotron frequency for electrons. Then e ee ee e 2 ( ) 1 n v = -D n - ×D n 1+ ∇β ∇ β JG G (7) Of these two terms, the second is perpendicular to both, B JG and ∇ne , and is called the “ ∇P ×B e JG drift”. The main interest is on the first term, which is along e - n∇ , as a regular diffusion. We see that this “cross-field diffusion” is governed by e n v = -D n e e ⊥∇ JG , with e 2 D D = 1+ ⊥ β (8) So, a high Hall parameter β can greatly reduce diffusion, compared to that in the absence of a magnetic field. High β means both, high B and/or low collision frequency. In an ion engine, with T = 4eV = 46400 K, e the e-n and e-i cross-section are roughly -19 2 Q 10 m , en -18 2 Q 4×10 m , ei and e 6 e e 8 kT c = 1.34×10 m/s. π m If also 17 -3 n 2.8×10 m , e
n.=7.4×1018m3 then v=7.4×1018×1.34×105×1019=99×105s V=28X0×134X10×4X10=1.50×195=249×0 At a point in the engine where b=100 gauss=0.01 Tesla, 16×1019x1y=1.76×105→B0=706>>1 ec=0.91×1030 Under these conditions,( 8)reduces to kT kT a2m。o2 (10) This last form shows that collisions favor diffusion. In contrast, recall Equation (3) (no magnetic field or B<< 1), which shows that in that case, collisions impede diffusion, Equation(10) also shows that D, scales as e2 and so, increasing B should provide very strong confinement of electrons With the given numbers, we find 0=n:38×1023×46400=2.83×105m2/s X 3×105 =0.57m2/s 706 a diffusing substance spreads(in 1-D)roughly as x-2 Dt. So, to spread by 1 cm electrons would require a time 16.522, Space P pessan Lecture 15 Prof. Manuel martinez Page 10 of 11
16.522, Space Propulsion Lecture 15 Prof. Manuel Martinez-Sanchez Page 10 of 11 18 -3 n 7.4×10 m , n then 18 6 -19 5 -1 en 6 -1 17 6 -18 6 -1 e ei = 7.4×10 ×1.34×10 ×10 = 9.9×10 s = 2.49×10 s = 2.8×10 ×1.34×10 × 4×10 = 1.50×10 s ν ⎫⎪ ⎬ ν ν ⎪⎭ At a point in the engine where B = 100 gauss = 0.01 Tesla, -19 -2 9 -1 c c -30 e 1.6 ×10 ×10 = = 1.76×10 s = = 706 >> 1 0.91×10 ω ω ⇒β ν Under these conditions, (8) reduces to e 2 D D⊥ β (9) or e e ee 22 2 e e ec c kT kT D m m + ⊥ ⎡ ⎤ ν ν ⎢ ⎥ ⎣ ⎦ νω ω = (10) This last form shows that collisions favor diffusion. In contrast, recall Equation (3), (no magnetic field, or β << 1), which shows that in that case, collisions impede diffusion, Equation (10) also shows that D⊥ scales as 2 1 B : e e eee 22 2 2 e 2 e kT m kT D = m e B e B m ⊥ ν ν and so, increasing B should provide very strong confinement of electrons. With the given numbers, we find -23 e 5 2 e -30 6 e e kT 1.38×10 × 46400 D = = = 2.83×10 m /s m ν 0.91×10 ×2.49×10 and 5 2 2 2.83×10 D = = 0.57 m /s 706 ⊥ A diffusing substance spreads (in 1-D) roughly as x 2 Dt ∼ . So, to spread by 1 cm, electrons would require a time