Ion engine — ionization, acceleration grid, neutralizer: Exercises

3.3.41 · D4 · Physics › Rocket Propulsion › Ion engine — ionization, acceleration grid, neutralizer

Constants jo poore note mein use honge (ek baar likho, baar baar use karo):

Level 1 — Recognition

L1·Q1

Ion engine ke teen sequential stages batao, order mein, aur har ek ke liye responsible physical part ka naam bolo.

Recall Solution

Stage 1 — Ionization → discharge chamber / hollow cathode karta hai (electrons fire karta hai jo xenon atoms se electrons knock off karte hain). Stage 2 — Acceleration → grids karte hain (screen grid high $+ V$ par, accelerator grid strong $- V$ par). Stage 3 — Neutralization → neutralizer cathode karta hai (exiting beam mein electrons spray karta hai). Parent note se mnemonic: Ionize, Accelerate, Neutralize → "I Am Nimble".

L1·Q2

Ek electric field xenon atom ko push karta hai. True ya false? Ek line mein explain karo.

Recall Solution

False. Electric field sirf charged objects ko push karta hai. Neutral Xe atom koi electric force feel nahi karta — yahi poori wajah hai ki Stage 1 (ionization) pehle zaroor honi chahiye. Jab yeh Xe⁺ ban jaata hai (charge $+ q$ ), tab field usse grab kar sakti hai.

Level 2 — Application

L2·Q1

Ek xenon ion (charge $q$ , mass $m$ ) rest se start hota hai aur grid voltage $V = 1500 V$ se guzarta hai. Uski exhaust speed $v_{e}$ nikalo.

Recall Solution

KYA: saari electrical energy $q V$ kinetic energy $\frac{1}{2} m v_{e}^{2}$ ban jaati hai (work–energy theorem, dekho Work-Energy Theorem). YEH FORMULA KYUN: ideal case mein koi burning nahi, koi heat loss nahi, isliye energy in = kinetic energy out. $v_{e} = \frac{2 q V}{m} = \frac{2 ( 1.602 \times 1 0 ^{- 19} ) ( 1500 )}{2.18 \times 1 0 ^{- 25}} .$ Numerator: $2 \times 1.602 \times 1 0^{- 19} \times 1500 = 4.806 \times 1 0^{- 16}$ . Mass se divide karo: $4.806 \times 1 0^{- 16} /2.18 \times 1 0^{- 25} = 2.204 \times 1 0^{9}$ . $v_{e} = 2.204 \times 1 0^{9} \approx 4.70 \times 1 0^{4} m/s = 46.9 km/s .$

L2·Q2

Wahi engine beam current $I = 2.0 A$ par $V = 1500 V$ pe run karta hai. Thrust $F$ nikalo.

Recall Solution

KYA: thrust = per second nikalne wala momentum. Parent note se $F = I 2 mV / q$ use karo. YEH FORM KYUN: beam current batata hai kitne ions per second nikalte hain ( $N = I / q$ ), aur har ek momentum $m v_{e}$ carry karta hai. $F = I \frac{2 mV}{q} = 2.0 \frac{2 ( 2.18 \times 1 0 ^{- 25} ) ( 1500 )}{1.602 \times 1 0 ^{- 19}} .$ Andar: $2 \times 2.18 \times 1 0^{- 25} \times 1500 = 6.54 \times 1 0^{- 22}$ ; $1.602 \times 1 0^{- 19}$ se divide karo $= 4.082 \times 1 0^{- 3}$ ; square root $= 0.06389$ . $F = 2.0 \times 0.06389 \approx 0.128 N .$ Roughly do paperclips ke weight jaisa.

L2·Q3

Ek xenon atom ko ionize karne ke liye kitni energy (joules mein) chahiye?

Recall Solution

$E_{i} = 12.13 eV = 12.13 \times 1.602 \times 1 0^{- 19} J \approx 1.94 \times 1 0^{- 18} J .$ (Dekho Plasma Physics — Ionization Energy — is specific value ke liye.)

Level 3 — Analysis

L3·Q1

Engine A $V_{A} = 1200 V$ par run karta hai; engine B (identical grids, same xenon) $V_{B} = 2400 V$ par run karta hai. Exhaust speed kitne factor se change hoti hai? Energy cost per ion kitne factor se change hota hai?

Recall Solution

Exhaust speed $v_{e} \propto V$ scale karti hai. $V$ double karne se $v_{e}$ multiply hoti hai $2 \approx 1.41$ se. Energy per ion $q V$ hai, jo linearly scale karta hai: $V$ double karne se energy double ho jaati hai jo har ion pe lagti hai. Analysis: tum double energy pay karte ho sirf 1.41× speed paane ke liye. Speed bahut jaldi mehnga ho jaata hai — square-root ek diminishing return hai. Yahi wajah hai ki $V$ badhana mainly specific impulse khareedta hai, free performance nahi.

L3·Q2

Figure dekho. Yeh $v_{e}$ ko $V$ ke against plot karta hai aur Q1 ka point mark karta hai. Curve ki shape use karke explain karo ki bahut high voltages "less and less extra speed per volt" kyun dete hain.

Figure — Ion engine — ionization, acceleration grid, neutralizer

Recall Solution

Curve $v_{e} = 2 q / m V$ hai — ek sideways square-root (apni side par ek parabola). Jaise $V$ badhta hai iska steepness flat hota jaata hai: $V = 0$ ke paas ek small voltage boost bahut speed add karta hai, lekin right side mein door curve almost flat hai, isliye equal voltage steps ever-smaller speed slices add karte hain. Mathematically slope hai $\frac{d v _{e}}{d V} = \frac{q}{2 mV}$ , jo $1/ V$ ki tarah shrink karta hai. Woh shrinking slope hi diminishing return hai.

L3·Q3

Neutralizer exactly beam ion rate par electrons emit karta hai. Electron mass $m_{e} = 9.11 \times 1 0^{- 31} kg$ hai; xenon ion $2.18 \times 1 0^{- 25} kg$ hai. Agar dono ko same speed par bahar phenka jaaye, toh total momentum ka kitna fraction electrons carry karenge? Yeh unke thrust ko ignore karna kyun justify karta hai?

Recall Solution

Equal speed $v$ par, momentum ratio = mass ratio (per particle, aur equal number hain): $\frac{p _{e}}{p _{i o n}} = \frac{m _{e}}{m} = \frac{9.11 \times 1 0 ^{- 31}}{2.18 \times 1 0 ^{- 25}} \approx 4.18 \times 1 0^{- 6} .$ Toh electrons carry karte hain lagbhag 4 millionths momentum — bilkul negligible. Aur actually electrons downstream at low speed add kiye jaate hain, isliye unka contribution aur bhi chhota hai. Thrust momentum hai, charge nahi, isliye neutralizer charge balance karta hai bina push churaye.

Level 4 — Synthesis

L4·Q1

Ek ion engine beam current $I = 1.76 A$ par $V = 1200 V$ pe hai (xenon). Nikalo: (a) mass flow rate $\overset{m}{˙}$ , (b) exhaust speed $v_{e}$ , aur (c) thrust do tareekon se — $\overset{m}{˙} v_{e}$ se aur $I 2 mV / q$ se — aur check karo ki dono agree karte hain.

Recall Solution

(a) Mass flow. Ions per second ki number $N = I / q$ . Har ek ki mass $m$ hai, isliye $\overset{m}{˙} = \frac{I}{q} m = \frac{1.76}{1.602 \times 1 0 ^{- 19}} \times 2.18 \times 1 0^{- 25} .$ $I / q = 1.099 \times 1 0^{19} ions/s$ ; times $m$ : $\overset{m}{˙} = 2.395 \times 1 0^{- 6} kg/s$ ( $\approx 2.4 mg/s$ ). (b) Exhaust speed. $v_{e} = \frac{2 ( 1.602 \times 1 0 ^{- 19} ) ( 1200 )}{2.18 \times 1 0 ^{- 25}} = 1.764 \times 1 0^{9} \approx 4.20 \times 1 0^{4} m/s .$ (c) Thrust, tarika 1: $F = \overset{m}{˙} v_{e} = 2.395 \times 1 0^{- 6} \times 4.20 \times 1 0^{4} \approx 0.1006 N$ . Tarika 2: $F = I 2 mV / q = 1.76 \times 0.05716 \approx 0.1006 N$ . ✅ Match karte hain — dono formulas same physics ko alag tarike se likha gaya hai.

L4·Q2

Q1 ke engine ke liye, specific impulse $I_{s p} = v_{e} / g_{0}$ compute karo aur chemical rocket se compare karo jahan $v_{e}^{c h e m} = 4.5 km/s$ hai. (Dekho Specific Impulse and Exhaust Velocity aur Chemical vs Electric Propulsion.)

Recall Solution

$I_{s p}$ ka matlab: specific impulse (seconds mein) $= v_{e} / g_{0}$ — kitne seconds 1 kg propellant apna khud ka weight hover kar sakta hai; ek fuel-economy score hai. $I_{s p} = \frac{v _{e}}{g _{0}} = \frac{4.20 \times 1 0 ^{4}}{9.81} \approx 4281 s .$ Chemical: $I_{s p}^{c h e m} = 4500/9.81 \approx 459 s$ . Ratio: $4281/459 \approx 9.3$ . Ion engine lagbhag 9× zyada fuel-efficient hai — wohi conclusion jo parent note ne raw speeds se nikala tha.

L4·Q3

Q1 ka engine $M_{p} = 80 kg$ xenon carry karta hai. Yeh continuously kitni der tak fire kar sakta hai (days mein)?

Recall Solution

Burn time = propellant / mass flow: $t = \frac{M _{p}}{m ˙} = \frac{80}{2.395 \times 1 0 ^{- 6}} \approx 3.34 \times 1 0^{7} s .$ Days mein convert karo: $3.34 \times 1 0^{7} /86400 \approx 387 days$ — ek saal se bhi zyada thrusting. Tiny 0.1 N force, lekin ek saal tak chalte rehna hi ion engines ko powerful banata hai.

Level 5 — Mastery

L5·Q1

Ek spacecraft ka dry mass $m_{0} = 500 kg$ hai plus $M_{p} = 80 kg$ xenon, aur Q1 engine use karta hai ( $v_{e} = 42 km/s$ , $\overset{m}{˙} = 2.395 \times 1 0^{- 6} kg/s$ ). (a) Tsiolkovsky Rocket Equation se total $Δ v$ nikalo. (b) Roughly, 0.10 N thrust burn ke start mein average acceleration kitna deta hai?

Recall Solution

(a) Tsiolkovsky: $Δ v = v_{e} ln \frac{m _{ini t ia l}}{m _{f ina l}}$ , jahan $m_{ini t ia l} = 580 kg$ , $m_{f ina l} = 500 kg$ . $Δ v = 4.20 \times 1 0^{4} \times ln \frac{580}{500} = 4.20 \times 1 0^{4} \times ln (1.16) .$ $ln (1.16) = 0.14842$ , isliye $Δ v = 4.20 \times 1 0^{4} \times 0.14842 \approx 6234 m/s \approx 6.2 km/s$ . (b) Starting acceleration: $a = F / m_{ini t ia l} = 0.1006/580 \approx 1.73 \times 1 0^{- 4} m/s^{2}$ . Insight: acceleration bahut chhota hai (~ $1.8 \times 1 0^{- 5} g$ ), phir bhi ek saal firing ke baad yeh itna $Δ v$ accumulate karta hai jo ek enormous chemical propellant load maangega. Deep space mein patience beats power.

L5·Q2

Algebraically dikhao ki thrust ko current ke bina $F = \overset{m}{˙} 2 q V / m$ likha ja sakta hai, aur isliye fixed mass flow ke liye thrust $V$ ke saath scale karta hai, jabki fixed beam power $P = I V$ ke liye thrust $\dots$ ? Power dependence derive karo.

Recall Solution

Pehla form: $F = \overset{m}{˙} v_{e} = \overset{m}{˙} 2 q V / m$ . Kyunki $\overset{m}{˙} = (I / q) m$ , substitute karne par $F = I 2 mV / q$ wapas milta hai — consistent. Fixed $\overset{m}{˙}$ ke liye, indeed $F \propto V$ . Power form: beam power $P = I V$ . Energy se, har ion ko $q V$ milta hai; power = (ions/s) $\times q V = (I / q) (q V) = I V$ . Saath hi $\frac{1}{2} \overset{m}{˙} v_{e}^{2} = P$ (beam ki kinetic power) hai, isliye $F = \overset{m}{˙} v_{e} = \frac{2 P}{v _{e}^{2}} \cdot v_{e} = \frac{2 P}{v _{e}} .$ Toh fixed power par, $F = 2 P / v_{e}$ : thrust exhaust speed ke inversely proportional hai. Faster exhaust ⇒ same power ke liye kam thrust. Yeh sabhi electric propulsion ka fundamental thrust–efficiency trade-off hai.

L5·Q3

$F = 2 P / v_{e}$ use karo $P = 2.1 kW$ aur $v_{e} = 42 km/s$ ke saath, $F$ compute karo aur confirm karo ki yeh Q1 ke ~0.10 N se match karta hai.

Recall Solution

$F = \frac{2 P}{v _{e}} = \frac{2 \times 2100}{4.20 \times 1 0 ^{4}} = \frac{4200}{42000} = 0.100 N .$ ✅ Momentum-based answer se match karta hai. Teen independent routes ( $\overset{m}{˙} v_{e}$ , $I 2 mV / q$ , $2 P / v_{e}$ ) teeno 0.10 N par land karte hain — ek strong self-consistency check.