Hall-effect thruster — cross-field discharge, annular channel: Exercises

3.3.42 · D4 · Physics › Rocket Propulsion › Hall-effect thruster — cross-field discharge, annular channe

Intuition Is page ko kaise use karein

Har problem ko pehle khud solve karo — tab solution kholo. Levels seedhi chadhti hain jaise stairs: L1 mein bas ek formula pehchanna hai, L5 mein kai formulas ek saath juggle karne padte hain. Yahan use kiya gaya har symbol parent note Hall-effect thruster — cross-field discharge, annular channel (index 3.3.42) mein build kiya gaya hai — agar koi cheez shaky lage, pehle woh padh lo.

Poore note mein kaam aane wale constants (inhe paas rakhna):

Elementary charge $q = 1.6 \times 1 0^{- 19} C$ — ek electron ya ek singly-ionized atom ki charge.
Electron mass $m_{e} = 9.11 \times 1 0^{- 31} kg$ .
Xenon-ion mass $m_{i} = 2.18 \times 1 0^{- 25} kg$ (ek xenon atom jisne ek electron kho diya).
Standard gravity $g_{0} = 9.81 m/s^{2}$ — ek bookkeeping number jo exhaust speed ko specific impulse (seconds mein) mein convert karta hai.

Level 1 — Recognition

L1.1 — Magnetized species ko pehchano

Ek single radial field $B = 0.015 T$ channel mein hai. Same perpendicular speed par, kiski gyro-circle chhoti hogi — electron ki ya xenon ion ki — aur lagbhag kitne factor se?

Recall Solution

Kya use karein: $r_{L} = m v_{⊥} / (q B)$ . Equal $v_{⊥}$ , $q$ , $B$ par, sirf mass ka fark hai, isliye $\frac{r _{L, i}}{r _{L, e}} = \frac{m _{i}}{m _{e}} = \frac{2.18 \times 1 0 ^{- 25}}{9.11 \times 1 0 ^{- 31}} \approx 2.39 \times 1 0^{5} .$ Answer: electron ki circle chhoti hai — lagbhag $2.4 \times 1 0^{5}$ times chhoti. Yahi bada gap hai jo ek field ko electrons ko magnetize karne deta hai jabki ions ko kuch khaas farq nahi padta.

L1.2 — Drift direction padhna

$E$ axially (exit ki taraf) point karta hai aur $B$ radially (ring gap ke across) point karta hai. $E \times B$ drift kis taraf point karta hai, aur ring shape kyun matter karti hai?

Recall Solution

Kya use karein: $v_{d} = (E \times B) / B^{2}$ . Axial $\times$ radial deta hai azimuthal direction — ring ke around. Ring kyun: azimuthal drift ek ring par ek loop mein band ho jaati hai jiska koi end nahi hota, isliye charge kabhi pile up nahi hota aur Hall current steadily chalta rehta hai. Ek seedhi tube mein drift ek wall se takra jaati.

Level 2 — Application

L2.1 — Thermal electron ka Larmor radius

$T_{e} = 15 eV$ ek perpendicular speed $v_{⊥} = 1.6 \times 1 0^{6} m/s$ ke barabar hai. $B = 0.018 T$ ke saath, $r_{L, e}$ nikalo aur ise $1.5 cm$ ke channel gap se compare karo.

Recall Solution

$r_{L, e} = \frac{m _{e} v _{⊥}}{q B} = \frac{( 9.11 \times 1 0 ^{- 31} ) ( 1.6 \times 1 0 ^{6} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.018 )} \approx 5.06 \times 1 0^{- 4} m = 0.51 mm .$ Compare: $0.51 mm ≪ 15 mm$ , lagbhag $30 \times$ chhota. Electrons strongly magnetized hain — woh channel cross karne se pehle kai baar orbit karte hain. Dekho Larmor Radius and Cyclotron Motion.

L2.2 — Cyclotron frequency

Usi electron aur $B = 0.018 T$ ke liye, $ω_{c}$ aur orbital period $T = 2 π / ω_{c}$ nikalo.

Recall Solution

$ω_{c} = \frac{q B}{m _{e}} = \frac{( 1.6 \times 1 0 ^{- 19} ) ( 0.018 )}{9.11 \times 1 0 ^{- 31}} \approx 3.16 \times 1 0^{9} rad/s .$ $T = \frac{2 π}{ω _{c}} = \frac{2 π}{3.16 \times 1 0 ^{9}} \approx 1.99 \times 1 0^{- 9} s \approx 2 ns .$ Matlab: har loop ~2 nanoseconds leta hai. Ek microsecond mein ek electron ~500 baar circle karta hai — woh collisions ke beech bahut acchi tarah trap rehta hai.

L2.3 — Xenon ki exhaust speed

Ions $Δ V = 250 V$ se guzarte hain. Exhaust speed $v_{e}$ nikalo.

Recall Solution

Energy conservation kyun: ion charge $q$ hai jo potential drop $Δ V$ cross karta hai, kinetic energy $q Δ V = \frac{1}{2} m_{i} v_{e}^{2}$ gain karta hai. $v_{e} = \frac{2 q Δ V}{m _{i}} = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 250 )}{2.18 \times 1 0 ^{- 25}} = 3.67 \times 1 0^{8} \approx 1.92 \times 1 0^{4} m/s .$

Level 3 — Analysis

L3.1 — Hall drift versus axial crawl

$Δ V = 250 V$ channel length $L = 0.025 m$ across aur $B = 0.018 T$ ke saath, azimuthal drift speed $v_{d} = E / B$ nikalo. Phir ise electron ki axial progress se compare karo agar woh ek gyroradius ( $r_{L, e} = 0.51 mm$ L2.1 se) per collision hop karta hai collision rate $ν = 1 \times 1 0^{6} s^{- 1}$ par.

Recall Solution

Step 1 — axial field. $E = Δ V / L = 250/0.025 = 1.0 \times 1 0^{4} V/m$ . Step 2 — drift. $v_{d} = \frac{E}{B} = \frac{1.0 \times 1 0 ^{4}}{0.018} \approx 5.56 \times 1 0^{5} m/s (azimuthal) .$ Step 3 — axial crawl. Har collision guiding centre ko ek $r_{L, e}$ $B$ ke across nudge karta hai (anode ki taraf), isliye axial speed roughly hai $v_{axial} \approx r_{L, e} ν = (5.06 \times 1 0^{- 4}) (1 \times 1 0^{6}) \approx 506 m/s .$ Interpretation: electron ring ke around $\sim 5.6 \times 1 0^{5}$ m/s se ghoomta hai lekin anode ki taraf sirf ~500 m/s se creep karta hai — ratio ~1100 ka. Hall (azimuthal) current leakage current se bahut zyada bada hai, bilkul jaise intended tha. Yeh slow leak wahi anomalous cross-field transport hai jo parent note mein mention kiya gaya hai.

L3.2 — Kya ek fast (accelerated) ion abhi bhi unmagnetized hai?

Ek from-rest xenon ion lein jo $Δ V = 250 V$ se accelerate hua (isliye $v_{i} = v_{e}$ L2.3 se), mostly axially move kar raha hai lekin ek small perpendicular component $v_{⊥} = 0.1 v_{i}$ ke saath. $B = 0.018 T$ ke saath, $r_{L, i}$ nikalo aur $1.5 cm$ gap se compare karo.

Recall Solution

$v_{⊥} = 0.1 \times 1.92 \times 1 0^{4} = 1.92 \times 1 0^{3} m/s$ . $r_{L, i} = \frac{m _{i} v _{⊥}}{q B} = \frac{( 2.18 \times 1 0 ^{- 25} ) ( 1.92 \times 1 0 ^{3} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.018 )} \approx 0.145 m = 14.5 cm .$ Compare: $14.5 cm ≫ 1.5 cm$ — poore channel ka lagbhag 10×. Ion ka "circle" itna bada hai ki channel ke andar uska path essentially straight hai. Ions acceleration ke baad bhi unmagnetized rehte hain. ✔

Level 4 — Synthesis

L4.1 — Thrust, specific impulse, aur efficiency budget

Ek xenon Hall thruster $Δ V = 300 V$ , propellant flow $\overset{m}{˙} = 4 mg/s$ , discharge current $I_{d} = 4.5 A$ par chalta hai. (a) Exhaust speed $v_{e}$ . (b) Thrust $F = \overset{m}{˙} v_{e}$ . (c) Specific impulse $I_{s p}$ . (d) Input electrical power $P = I_{d} Δ V$ aur "jet" power $P_{jet} = \frac{1}{2} \overset{m}{˙} v_{e}^{2}$ ; efficiency $η = P_{jet} / P$ report karo.

Recall Solution

(a) $v_{e} = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 300 )}{2.18 \times 1 0 ^{- 25}} = 4.40 \times 1 0^{8} \approx 2.10 \times 1 0^{4} m/s .$ (b) $F = \overset{m}{˙} v_{e} = (4 \times 1 0^{- 6}) (2.10 \times 1 0^{4}) \approx 0.0840 N = 84.0 mN .$ (c) $I_{s p} = \frac{v _{e}}{g _{0}} = \frac{2.10 \times 1 0 ^{4}}{9.81} \approx 2140 s .$ (d) $P = I_{d} Δ V = (4.5) (300) = 1350 W .$ $P_{jet} = \frac{1}{2} \overset{m}{˙} v_{e}^{2} = \frac{1}{2} (4 \times 1 0^{- 6}) (2.10 \times 1 0^{4})^{2} \approx 882 W .$ $η = \frac{P _{jet}}{P} = \frac{882}{1350} \approx 0.653 = 65%.$ Iska matlab: do-tihai electrical power ordered exhaust kinetic energy ke roop mein jaati hai; baaki ionization, wall losses, aur electron leakage mein — yeh ek realistic Hall-thruster number hai.

L4.2 — Magnetic field design karo

Tumhe electrons ko magnetize karna hai ( $r_{L, e} < 1 mm$ ) lekin ions ko free rakhna hai ( $r_{L, i} > 5 cm$ ) same particles ke liye jaise L2.1 ( $v_{⊥, e} = 1.6 \times 1 0^{6}$ m/s) aur L3.2 ( $v_{⊥, i} = 1.92 \times 1 0^{3}$ m/s). Woh $B$ ka range nikalo jo dono conditions satisfy kare.

Recall Solution

Electron condition ( $r_{L, e} < 1 0^{- 3}$ m): $B > \frac{m _{e} v _{⊥, e}}{q ( 1 0 ^{- 3} )} = \frac{( 9.11 \times 1 0 ^{- 31} ) ( 1.6 \times 1 0 ^{6} )}{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{- 3} )} \approx 9.11 \times 1 0^{- 3} T .$ Ion condition ( $r_{L, i} > 0.05$ m): $B < \frac{m _{i} v _{⊥, i}}{q ( 0.05 )} = \frac{( 2.18 \times 1 0 ^{- 25} ) ( 1.92 \times 1 0 ^{3} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.05 )} \approx 5.23 \times 1 0^{- 2} T .$ Answer: koi bhi $B$ jo $9.1 mT < B < 52 mT$ mein ho, kaam karega. Yahi window hai isliye real Hall thrusters $15$ – $25 mT$ ke around run karte hain — band ke andar comfortably.

Level 5 — Mastery

L5.1 — Rocket equation ke saath mission burn

L4.1 ka 84-mN thruster continuously fire karta hai ek spacecraft ko velocity change $Δ v = 2.0 km/s$ dene ke liye. Spacecraft wet mass $m_{0} = 500 kg$ se start karta hai. (a) Tsiolkovsky Rocket Equation use karke $v_{e} = 2.10 \times 1 0^{4} m/s$ ke saath, propellant mass burned $m_{p} = m_{0} (1 - e^{- Δ v / v_{e}})$ nikalo. (b) $\overset{m}{˙} = 4 mg/s$ par, thruster ko kitne time (days mein) fire karna padega?

Recall Solution

Tsiolkovsky kyun: thrust mass eject karta hai, isliye ship ka khud ka mass accelerate hone ke saath ghatta hai — exponential us shrinking-mass bookkeeping ko capture karta hai. (a) Exponent: $Δ v / v_{e} = 2000/21000 = 0.09524$ . $m_{p} = m_{0} (1 - e^{- 0.09524}) = 500 (1 - 0.9091) = 500 (0.09091) \approx 45.4 kg .$ (b) Burn time $= m_{p} / \overset{m}{˙} = 45.4/ (4 \times 1 0^{- 6}) \approx 1.136 \times 1 0^{7} s$ . $t = \frac{1.136 \times 1 0 ^{7}}{86400 s/day} \approx 131 days .$ Iska matlab: ek tiny 84-mN thrust ek bada $Δ v$ deta hai — lekin mahine lag jaate hain. Yahi electric-propulsion ka bargain hai: low thrust, superb fuel economy (sirf 45 kg xenon 2 km/s ke liye).

L5.2 — Gridded ion engine ke space-charge limit se compare karo

Ek gridded ion engine (Ion Thruster (Gridded)) jisme grid gap $d = 1 mm$ aur voltage $Δ V = 300 V$ hai, Child–Langmuir Law se limited hai: $J = \frac{4}{9} ε_{0} \frac{2 q}{m _{i}} \frac{Δ V ^{3/2}}{d ^{2}}, ε_{0} = 8.85 \times 1 0^{- 12} F/m .$ (a) Maximum ion current density $J$ (A/m²) calculate karo. (b) L4.1 ka Hall thruster $I_{d} = 4.5 A$ push karta hai annular exit area $A = 5 cm^{2} = 5 \times 1 0^{- 4} m^{2}$ se; uski current density $J_{Hall}$ aur ratio $J_{Hall} / J$ nikalo. Comment karo ki Hall device kyun jeet jaata hai.

Recall Solution

(a) Pehle mobility factor: $\frac{2 q}{m _{i}} = \frac{2 ( 1.6 \times 1 0 ^{- 19} )}{2.18 \times 1 0 ^{- 25}} = 1.468 \times 1 0^{6} \approx 1.212 \times 1 0^{3} .$ $Δ V^{3/2} = 30 0^{1.5} = 5196$ . $d^{2} = (1 0^{- 3})^{2} = 1 0^{- 6}$ . $J = \frac{4}{9} (8.85 \times 1 0^{- 12}) (1.212 \times 1 0^{3}) \frac{5196}{1 0 ^{- 6}} \approx 24.8 A/m^{2} .$ (b) $J_{Hall} = \frac{I _{d}}{A} = \frac{4.5}{5 \times 1 0 ^{- 4}} = 9000 A/m^{2} .$ $\frac{J _{Hall}}{J} = \frac{9000}{24.8} \approx 363.$ Kyun jeet jaata hai: gridded engine ions ko ek vacuum gap ke across accelerate karta hai, isliye uska apna ion cloud khud ko repel karta hai — Child–Langmuir ceiling aata hai. Hall thruster ions ko ek quasineutral plasma ke andar accelerate karta hai jahan trapped electrons space charge cancel kar dete hain, isliye koi aisi ceiling nahi hoti. Result hai current density saikdon times zyada, isliye far higher thrust per unit area — yahi puri wajah hai ki yeh design exist karta hai.

Wrap-up recall

Recall Ek-line takeaways

$r_{L} \propto m$ — same field, electron circle xenon ke circle se $2.4 \times 1 0^{5} \times$ chhota. ::: L1.1
$ω_{c} = q B / m$ mein koi velocity nahi — orbit rate speed-independent hai. ::: L2.2
Drift azimuthal hai aur koi kaam nahi karta; axial leak ek slow collisional crawl hai. ::: L3
Electrons ko magnetize karna lekin ions ko free rakhna $B$ ki ek window define karta hai, maximum nahi. ::: L4.2
Quasineutral plasma Child–Langmuir ceiling se bachta hai → ~ $1 0^{2} \times$ current density. ::: L5.2