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 ⎝ ⎠ ∫