FEEP, MEMS thrusters — micro-propulsion: Worked examples

3.3.43 · D3 · Physics › Rocket Propulsion › FEEP, MEMS thrusters — micro-propulsion

Hamare teen tools (sab parent note mein derive kiye gaye hain):

The scenario matrix

Har micro-propulsion problem in cells mein se kisi ek mein aata hai. Neeche diye examples pe cell label laga hua hai.

Cell	Kya vary/degenerate hota hai	Example
A. Baseline	ordinary $q, m, V, I$ → find $v_{e}$ , $F$	Ex 1
B. Low- $V$ limit	small voltage → slow, weak thrust	Ex 2
C. High- $V$ limit	kilovolt → bahut fast ions	Ex 3
D. Zero / degenerate input	$I = 0$ , ya $Δ v = 0$ , ya $V = 0$	Ex 4
E. Heavy vs light ion	Cs ↔ In swap at fixed $I, V$	Ex 5
F. Real-world word problem	mission $Δ v$ , propellant mass, burn time	Ex 6
G. MEMS / low- $R e$ twist	gas expansion + Reynolds-number sanity	Ex 7
H. Exam twist (inverse)	$F$ aur $I_{s p}$ diya, $V$ aur $I$ nikalo	Ex 8

Poore examples mein do ions use karenge:

Indium In $^{+}$ : $m = 1.9 \times 1 0^{- 25} kg$ , $q = 1.6 \times 1 0^{- 19} C$ .
Caesium Cs $^{+}$ : $m = 2.2 \times 1 0^{- 25} kg$ , same $q$ .

Cell A — Baseline

Worked example Ex 1 — Indium ion at 6 kV, 10 µA beam

$m = 1.9 \times 1 0^{- 25}$ , $q = 1.6 \times 1 0^{- 19}$ , $V = 6000$ V, $I = 10 μ A = 1 0^{- 5}$ A. $v_{e}$ , $I_{s p}$ , $F$ nikalo.

Forecast: $v_{e} \sim 1 0^{5}$ m/s? $F$ kuch µN?

$v_{e} = \frac{2 q V}{m} = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 6000 )}{1.9 \times 1 0 ^{- 25}} \approx 1.005 \times 1 0^{5} m/s$ . Yeh step kyun? Energy conservation: saari electric potential energy $q V$ kinetic energy $\frac{1}{2} m v_{e}^{2}$ mein convert hoti hai.
$I_{s p} = v_{e} / g_{0} = 1.005 \times 1 0^{5} /9.81 \approx 1.02 \times 1 0^{4} s$ . Yeh step kyun? $I_{s p}$ basically $v_{e}$ ko "gravity ke seconds" mein measure karta hai — fuel economy ka standard yardstick (Specific Impulse).
$F = I \frac{2 mV}{q} = I \frac{m}{q} v_{e} = (1 0^{- 5}) (1.1875 \times 1 0^{- 6}) (1.005 \times 1 0^{5}) \approx 1.19 \times 1 0^{- 6} N$ . Yeh step kyun? $F = \overset{m}{˙} v_{e}$ aur $\overset{m}{˙} = (m / q) I$ (charge/sec → mass/sec).

Verify: $F$ ke units: $[A] [kg/C] [m/s] = [C/s] [kg/C] [m/s] = kg\cdotm/s^{2} = N$ ✓. Ek microamp se ek micro-newton milna exactly "eyedropper" regime hai.

Cell D — Zero / degenerate inputs

Worked example Ex 4 — Teen "kya ho agar zero ho?" traps

(a) Beam current $I = 0$ . (b) Accelerating voltage $V = 0$ . (c) Required $Δ v = 0$ .

Forecast: kaunse zero thrust dete hain? Kaunse zero propellant dete hain?

(a) $I = 0$ : $F = I 2 mV / q = 0$ . Koi ion per second nahi → koi momentum flow nahi → koi thrust nahi, lekin $v_{e}$ (ek ion ki speed) tab bhi $1 0^{5}$ m/s rahegi agar $V$ on hai. Yeh step kyun? Thrust ke liye cheez ka nikalna zaroori hai; switched-off beam mein accelerating field hota hai, toh $v_{e} \neq = 0$ par $\overset{m}{˙} = 0$ . "Har ion fast hai" aur "thrust exist karta hai" ko kabhi confuse mat karo.
(b) $V = 0$ : $v_{e} = 2 q (0) / m = 0$ , isliye $F = \overset{m}{˙} \cdot 0 = 0$ . Chahe ions leak bhi hoon, woh koi directed momentum nahi uthate. Yeh step kyun? Koi potential drop nahi → koi energy nahi mili → ions ~0 speed se trickle karte hain. Yeh genuine degenerate case hai, math error nahi.
(c) $Δ v = 0$ : $m_{p} = m_{0} (1 - e^{0}) = m_{0} (1 - 1) = 0$ . Yeh step kyun? Koi velocity change nahi chahiye → kuch bhi mat jalao. Rocket equation sahi se zero fuel return karta hai.

Verify: $1 - e^{0} = 0$ ✓; $2 q \cdot 0/ m = 0$ ✓; $I \cdot (\dots) = 0$ jab $I = 0$ ✓. Teeno degeneracies sensibly behave karte hain.

Cell E — Heavy vs light ion (Cs vs In)

Figure — FEEP, MEMS thrusters — micro-propulsion

Worked example Ex 5 — Same

$I$ , same $V$ : kaun zyada thrust deta hai, kaun zyada $I_{s p}$ ? $V = 6$ kV, $I = 10 μ A$ fix karo. In ( $m = 1.9 \times 1 0^{- 25}$ ) ko Cs ( $m = 2.2 \times 1 0^{- 25}$ ) se compare karo.

Forecast: heavier caesium — zyada thrust ya kam? Zyada $I_{s p}$ ya kam?

$v_{e} \propto q / m = 1/ m$ , toh heavier Cs slower hoga: $v_{e, Cs} = v_{e, In} 1.9/2.2 \approx 1.005 \times 1 0^{5} \times 0.929 = 9.34 \times 1 0^{4}$ m/s. Yeh step kyun? Charge ke per zyada mass → har charge zyada inertia drag karta hai → speed kam → $I_{s p}$ kam. Figure mein shorter teal arrow dekho.
$F = I 2 mV / q \propto m$ , toh heavier Cs per amp zyada thrust deta hai: $F_{Cs} = F_{In} 2.2/1.9 = 1.19 \times 1 0^{- 6} \times 1.076 \approx 1.28 \times 1 0^{- 6}$ N. Yeh step kyun? Fixed current pe ion count fixed hai; heavier ions mein har ek zyada momentum carry karta hai → zyada push. Figure mein longer plum arrow.

Verify: product check $v_{e} \cdot F$ $m$ se independent hona chahiye (kyunki $m \cdot 1/ m = 1$ ): In deta hai $1.005 \times 1 0^{5} \times 1.19 \times 1 0^{- 6} = 0.1196$ ; Cs deta hai $9.34 \times 1 0^{4} \times 1.28 \times 1 0^{- 6} = 0.1196$ ✓. Lesson: heavy = worse nahi — yeh thrust-vs- $I_{s p}$ trade hai, parent ke Ion Thrusters logic jaisa.

Cell F — Real-world word problem

Worked example Ex 6 — CubeSat ke liye full mission budget

Ek 3 kg CubeSat (CubeSats & Attitude Control) ko apni life mein $Δ v = 5$ m/s ka drag make-up karna hai, Ex 1 wale indium FEEP se ( $v_{e} = 1.005 \times 1 0^{5}$ m/s, $F = 1.19 μ N$ ). (a) propellant mass, (b) total impulse, (c) burn time agar continuously chalaya jaaye — yeh sab nikalo.

Forecast: kitne grams fuel? Kitne din ya saal firing?

(a) $m_{p} = m_{0} (1 - e^{- Δ v / v_{e}})$ . Kyunki $Δ v / v_{e} = 5/1 0^{5} = 5 \times 1 0^{- 5} ≪ 1$ , $e^{- x} \approx 1 - x$ use karo: $m_{p} \approx m_{0} \frac{Δ v}{v _{e}} = 3 \times 5 \times 1 0^{- 5} = 1.5 \times 1 0^{- 4}$ kg $= 0.15$ g. Yeh step kyun? Rocket Equation; linear approximation justified hai kyunki $x$ bahut chhota hai.
(b) Total impulse $J = m_{p} v_{e} = 1.5 \times 1 0^{- 4} \times 1.005 \times 1 0^{5} \approx 15.1$ N·s. (Ya $J = m_{0} Δ v = 3 \times 5 = 15$ N·s.) Yeh step kyun? Impulse = momentum delivered = fuel mass × exhaust speed = spacecraft mass × $Δ v$ . Dono routes agree karne chahiye.
(c) $t = J / F = 15/ (1.19 \times 1 0^{- 6}) \approx 1.26 \times 1 0^{7}$ s $\approx 146$ days. Yeh step kyun? Impulse = force × time (constant thrust), toh $t = J / F$ .

Verify: dono impulse routes agree karte hain: 15.1 vs 15.0 N·s (0.7% linear approx se) ✓. $t$ ke units: $N\cdots / N = s$ ✓. Lesson: 0.15 g metal se poora mission ho jaata hai — huge $v_{e}$ ka yahi payoff hai.

Cell G — MEMS / low-Reynolds twist

Worked example Ex 7 — Cold-gas MEMS exhaust speed aur

$R e$ reality check Micro-nozzle se nitrogen: $γ = 1.4$ , $R = 8.314$ J/mol·K, molar mass $M = 0.028$ kg/mol, $T_{0} = 300$ K, pressure ratio $p_{e} / p_{0} = 0.01$ . $R e$ bhi estimate karo: $ρ = 0.05 kg/m^{3}$ , $v = 300$ m/s, channel $L = 100 μ m$ , $μ = 1.8 \times 1 0^{- 5}$ Pa·s.

Forecast: ideal $v_{e}$ 700–800 m/s ke paas? $R e$ bada (inviscid) hai ya chhota (viscous)?

Bracket term: $1 - (p_{e} / p_{0})^{(γ - 1) / γ} = 1 - (0.01)^{0.4/1.4} = 1 - (0.01)^{0.2857} = 1 - 0.2683 = 0.7317$ . Yeh step kyun? Yeh pressure drop ke across release hone wali enthalpy ka fraction hai — expansion ka engine.
$v_{e} = \frac{2 γ}{γ - 1} \frac{R T _{0}}{M} \times 0.7317 = 7.0 \times \frac{8.314 \times 300}{0.028} \times 0.7317$ . Inner: $R T_{0} / M = 8.314 \times 300/0.028 = 8.908 \times 1 0^{4}$ . Toh $v_{e} = 7.0 \times 8.908 \times 1 0^{4} \times 0.7317 = 4.563 \times 1 0^{5} \approx 676$ m/s. Yeh step kyun? Enthalpy → directed kinetic energy nozzle ke through; $\frac{2 γ}{γ - 1} = 7$ jab $γ = 1.4$ .
$R e = \frac{ρ v L}{μ} = \frac{0.05 \times 300 \times 1 0 ^{- 4}}{1.8 \times 1 0 ^{- 5}} = \frac{1.5 \times 1 0 ^{- 3}}{1.8 \times 1 0 ^{- 5}} \approx 83$ . Yeh step kyun? Reynolds Number inertia ko viscosity se compare karta hai; $R e \sim 80$ low hai → thick boundary layer, toh real $v_{e}$ ideal 676 m/s se neeche girega.

Verify: $(0.01)^{0.2857}$ : logs lo, $0.2857 \times ln 0.01 = 0.2857 \times (- 4.605) = - 1.316$ , $e^{- 1.316} = 0.268$ ✓. $R e \approx 83 ≪ 2000$ (laminar, viscous-dominated) ✓. Lesson: ideal formula size ignore karta hai, par $R e$ batata hai ki micro-nozzles efficiency kho dete hain.

Cell H — Exam twist (inverse problem)

Worked example Ex 8 — Performance diya, settings nikalo

Design spec: indium FEEP ko $F = 2.0 μ N$ at $I_{s p} = 8000$ s deliver karna hai. Required $V$ aur $I$ nikalo. ( $m = 1.9 \times 1 0^{- 25}$ , $q = 1.6 \times 1 0^{- 19}$ , $g_{0} = 9.81$ .)

Forecast: kilovolts of $V$ ? tens of µA of $I$ ?

$I_{s p}$ se: $v_{e} = I_{s p} g_{0} = 8000 \times 9.81 = 7.848 \times 1 0^{4}$ m/s. Yeh step kyun? $I_{s p}$ directly $v_{e}$ fix karta hai — yeh voltage ka kaam hai.
$v_{e} = 2 q V / m$ invert karo → $V = \frac{m v _{e}^{2}}{2 q} = \frac{1.9 \times 1 0 ^{- 25} \times ( 7.848 \times 1 0 ^{4} ) ^{2}}{2 \times 1.6 \times 1 0 ^{- 19}} = \frac{1.9 \times 1 0 ^{- 25} \times 6.159 \times 1 0 ^{9}}{3.2 \times 1 0 ^{- 19}} \approx 3.66 \times 1 0^{3}$ V. Yeh step kyun? Energy equation ko ek unknown $V$ ke liye solve karo. Yeh ≈3.7 kV hai — ek realistic FEEP voltage.
$F = \overset{m}{˙} v_{e} = (m / q) I v_{e}$ se → $I = \frac{F q}{m v _{e}} = \frac{2.0 \times 1 0 ^{- 6} \times 1.6 \times 1 0 ^{- 19}}{1.9 \times 1 0 ^{- 25} \times 7.848 \times 1 0 ^{4}} = \frac{3.2 \times 1 0 ^{- 25}}{1.491 \times 1 0 ^{- 20}} \approx 2.15 \times 1 0^{- 5}$ A. Yeh step kyun? $v_{e}$ pata hai, toh thrust current pin down karta hai — dono knobs independent hain (parent ka key point).

Verify: plug back karo: $v_{e} = 2 (1.6 \times 1 0^{- 19}) (3660) /1.9 \times 1 0^{- 25} = 6.16 \times 1 0^{9} = 7.85 \times 1 0^{4}$ ✓; $F = (m / q) I v_{e} = 1.1875 \times 1 0^{- 6} \times 2.15 \times 1 0^{- 5} \times 7.848 \times 1 0^{4} = 2.00 \times 1 0^{- 6}$ N ✓. Dono targets recover ho gaye.

Recall Kaun sa knob kya karta hai?

Voltage set karta hai ::: exhaust speed $v_{e}$ aur isliye $I_{s p}$ . Beam current set karta hai ::: thrust $F$ (ions per second). $V$ double karne se $v_{e}$ multiply hota hai ::: $2$ se, 2 se nahi. Fixed $I, V$ pe heavier ion deta hai ::: zyada thrust, kam $I_{s p}$ (ek trade, koi defect nahi). Micro-nozzle mein $R e ≪ 2000$ ka matlab hai ::: viscous losses dominate karte hain → real $v_{e}$ ideal se neeche.

Worked examples — FEEP, MEMS thrusters — micro-propulsion