Electric propulsion — thrust, power, Isp trade-off: Exercises

3.3.40 · D4 · Physics › Rocket Propulsion › Electric propulsion — thrust, power, Isp trade-off

Shuru karne se pehle, ek shared toolbox. Neeche har symbol earned hai — har ek ka plain-word meaning yahan hai:

Figure — Electric propulsion — thrust, power, Isp trade-off

Upar wala red curve poori kahani hai: fixed power par, exhaust speed badhne se thrust girta hai. Kaam karte waqt ise dhyan mein rakho.

Level 1 — Recognition

Exercise 1.1

Ek ion engine $\overset{m}{˙} = 4.0 mg/s$ xenon ko $v_{e} = 30, 000 m/s$ par eject karta hai. Thrust kya hai?

Recall Solution

KYA: Hume mass flow aur speed di gayi hai, push poochhaa gaya hai. KYUN $F = \overset{m}{˙} v_{e}$ use karein: thrust momentum ejected per second hai, aur momentum = mass × speed (yeh Newton's Third Law action mein hai). Pehle units fix karo: $4.0 mg/s = 4.0 \times 1 0^{- 6} kg/s$ (milli = ek-hazarwa hissa, aur ek mg, gram ka hazarwa hissa hai jo kg ka hazarwa hissa hai → $1 0^{- 6}$ ). $F = \overset{m}{˙} v_{e} = (4.0 \times 1 0^{- 6}) (3.0 \times 1 0^{4}) = 0.12 N = 120 mN .$

Exercise 1.2

Usi engine ki jet power. Diya gaya hai $\overset{m}{˙} = 4.0 mg/s$ , $v_{e} = 30, 000 m/s$ , $P_{j e t}$ nikalo.

Recall Solution

KYUN $P_{j e t} = \frac{1}{2} \overset{m}{˙} v_{e}^{2}$ : power kinetic energy delivered per second hai, aur mass $m$ ki kinetic energy speed $v$ par $\frac{1}{2} m v^{2}$ hoti hai. Speed squared hai kyunki energy speed ke square ke saath badhti hai — yahi reason hai ki fast exhaust itna power-hungry hota hai. $P_{j e t} = \frac{1}{2} (4.0 \times 1 0^{- 6}) (3.0 \times 1 0^{4})^{2} = \frac{1}{2} (4.0 \times 1 0^{- 6}) (9.0 \times 1 0^{8}) = 1800 W .$

Exercise 1.3

Ek thruster ka $I_{s p} = 2000 s$ hai. Exhaust speed $v_{e}$ mein convert karo.

Recall Solution

KYUN $g_{0}$ se multiply karein: $I_{s p}$ seconds mein measure hota hai kyunki yeh impulse per unit weight of propellant hai. Weight mein $g_{0} = 9.81 m/s^{2}$ aata hai, toh us division ko undo karne se real velocity wapas milti hai. $v_{e} = g_{0} I_{s p} = 9.81 \times 2000 = 19, 620 m/s \approx 1.96 \times 1 0^{4} m/s .$

Level 2 — Application

Exercise 2.1

Ek Hall thruster (dekho Ion and Hall Thrusters) $P_{in} = 4.5 kW$ par chalta hai, efficiency $η = 0.55$ aur $I_{s p} = 1600 s$ ke saath. Nikalo (a) $v_{e}$ , (b) thrust $F$ , (c) mass flow $\overset{m}{˙}$ .

Recall Solution

(a) $v_{e} = g_{0} I_{s p} = 9.81 \times 1600 = 1.57 \times 1 0^{4} m/s$ . (b) KYUN $F = 2 η P_{in} / v_{e}$ : yeh trade-off equation hai — yeh batata hai ki ek speed commit karne ke baad wall power se kitna push milta hai. $P_{in} = 4500 W$ ke saath: $F = \frac{2 ( 0.55 ) ( 4500 )}{1.57 \times 1 0 ^{4}} = \frac{4950}{15 , 696} = 0.315 N \approx 315 mN .$ (c) $\overset{m}{˙} = F / v_{e} = 0.315/1.57 \times 1 0^{4} = 2.01 \times 1 0^{- 5} kg/s = 20.1 mg/s$ .

Exercise 2.2

Exercise 2.1 thruster mein kitni power heat ke roop mein waste hoti hai, aur wo heat ship ke liye kya mayne rakhti hai?

Recall Solution

KYA: efficiency $η = 0.55$ ka matlab hai sirf 55% input power beam tak pahunchti hai. Baaki heat ban jaati hai. $P_{j e t} = η P_{in} = 0.55 \times 4500 = 2475 W, P_{w a s t e} = P_{in} - P_{j e t} = 4500 - 2475 = 2025 W .$ Wo $2025 W$ radiators dwara space mein dump karna padega — dekho Thermal Control (radiators). KYUN dhyan dena chahiye: heat rejection spacecraft ke radiators ko size karta hai, jo ek real mass/design cost hai.

Exercise 2.3

Do thrusters ek hi $P_{in}$ draw karte hain aur unka $η$ bhi same hai. Thruster A ka $I_{s p} = 1000 s$ hai, thruster B ka $I_{s p} = 3000 s$ hai. Bina calculator ke, $F_{A} / F_{B}$ ka ratio kya hai?

Recall Solution

KYUN calculator ki zaroorat nahi: fixed $P_{in}$ aur $η$ par, trade-off equation kehti hai $F = 2 η P_{in} / v_{e} \propto 1/ v_{e} \propto 1/ I_{s p}$ . Toh thrust $I_{s p}$ ke inversely proportional hai. $\frac{F _{A}}{F _{B}} = \frac{I _{s p, B}}{I _{s p, A}} = \frac{3000}{1000} = 3.$ A teen guna zyada push karta hai. Yeh parent note ka seesaw ek line mein hai.

Level 3 — Analysis

Exercise 3.1

Ek spacecraft ka solar array fixed $P_{in} = 3.0 kW$ ek thruster mein deta hai jiska efficiency $η = 0.60$ hai. Engineer $I_{s p}$ ko $1000 s$ se $5000 s$ ke beech kahin bhi dial kar sakta hai. Dimag mein plot karo: kya $I_{s p}$ badhne par thrust badhta ya girta hai, aur poore range mein $F$ kitne factor se change hota hai?

Recall Solution

Reasoning: $F = 2 η P_{in} / (g_{0} I_{s p})$ . Numerator fixed hai, toh $F \propto 1/ I_{s p}$ : $I_{s p}$ badhne par thrust girta hai. $I_{s p} = 1000 s$ par: $v_{e} = 9810 m/s$ , $F = \frac{2 ( 0.60 ) ( 3000 )}{9810} = \frac{3600}{9810} = 0.367 N$ . $I_{s p} = 5000 s$ par: $v_{e} = 49, 050 m/s$ , $F = \frac{3600}{49 , 050} = 0.0734 N$ . Ratio $= 0.367/0.0734 = 5$ . Exactly wahi factor jitna $I_{s p}$ badha ( $5000/1000$ ), confirm karta hai $F \propto 1/ I_{s p}$ . Upar figure ka red curve yahi behaviour hai.

Exercise 3.2

Exercise 3.1 ka engine $I_{s p} = 2500 s$ par set hai, 1-saal ke burn mein kitna propellant use hoga? (1 saal $\approx 3.156 \times 1 0^{7} s$ .)

Recall Solution

Step 1 — speed: $v_{e} = 9.81 \times 2500 = 2.453 \times 1 0^{4} m/s$ . Step 2 — thrust: $F = \frac{2 ( 0.60 ) ( 3000 )}{2.453 \times 1 0 ^{4}} = \frac{3600}{24 , 525} = 0.1468 N$ . Step 3 — mass flow: $\overset{m}{˙} = F / v_{e} = 0.1468/2.453 \times 1 0^{4} = 5.99 \times 1 0^{- 6} kg/s$ . Step 4 KYUN time se multiply karein: mass used = flow rate × burn duration. $m_{p r o p} = \overset{m}{˙} \cdot t = (5.99 \times 1 0^{- 6}) (3.156 \times 1 0^{7}) = 189 kg .$ Poore ek saal ke continuous thrust ke liye 200 kg se kam propellant — isliye EP deep-space cruising mein jeetta hai (Chemical vs Electric Propulsion).

Exercise 3.3

Degenerate check: fixed power par jab $I_{s p} \to \infty$ (socho ek perfect, infinitely economical engine) tab thrust ka kya hoga? Aur jab $I_{s p} \to 0$ ? Kya dono physical hain?

Recall Solution

$I_{s p} \to \infty$ : $F = 2 η P_{in} / (g_{0} I_{s p}) \to 0$ . Infinite economy zero push deta hai — tum infinitesimal mass infinitely fast phenkte ho, aur momentum-per-second collapse ho jaata hai. Yeh red curve ki far-right tail ka axis par flatten hona hai. $I_{s p} \to 0$ : formula $F \to \infty$ deta hai, lekin yeh unphysical hai: $P_{in}$ fixed rakhte hue $v_{e} \to 0$ ke liye $\overset{m}{˙} = 2 η P_{in} / v_{e}^{2} \to \infty$ chahiye hoga — infinite propellant flow. Real engines mein $v_{e}$ ka ek floor hota hai jo ionisation aur grid voltage ki physics se set hota hai. Toh curve sirf ek finite $I_{s p}$ band par meaningful hai.

Level 4 — Synthesis

Exercise 4.1

Ek 1200 kg probe ko $Δ v = 6.0 km/s$ gain karna hai. Uska thruster $I_{s p} = 3000 s$ , $η = 0.65$ , $P_{in} = 5.0 kW$ par chalta hai. Nikalo (a) required propellant mass, (b) thrust, (c) roughly kitna burn time lagega (time estimate ke liye mass ko ~constant maano, initial mass flow use karo).

Recall Solution

(a) KYUN rocket equation: yeh velocity gain ko connect karta hai is baat se ki ship ka kitna hissa tum throw away karte ho. Rearrange karke, $m_{f} = m_{0} e^{- Δ v / v_{e}}$ . $v_{e} = 9.81 \times 3000 = 2.943 \times 1 0^{4} m/s$ . Phir $Δ v / v_{e} = 6000/29430 = 0.2039$ . $m_{f} = 1200 e^{- 0.2039} = 1200 \times 0.8156 = 978.7 kg$ . $m_{p r o p} = m_{0} - m_{f} = 1200 - 978.7 = 221.3 kg$ . (b) $F = \frac{2 ( 0.65 ) ( 5000 )}{2.943 \times 1 0 ^{4}} = \frac{6500}{29430} = 0.2208 N$ . (c) $\overset{m}{˙} = F / v_{e} = 0.2208/2.943 \times 1 0^{4} = 7.503 \times 1 0^{- 6} kg/s$ . $t \approx m_{p r o p} / \overset{m}{˙} = 221.3/7.503 \times 1 0^{- 6} = 2.95 \times 1 0^{7} s \approx 0.94 yr$ . Padhna: ~221 kg propellant, ~0.22 N push, aur lagbhag ek saal ka burning. EP time trade karta hai propellant frugality ke liye.

Exercise 4.2

Same mission, same $Δ v = 6.0 km/s$ aur $m_{0} = 1200 kg$ , lekin ab compare karo chemical stage $I_{s p} = 450 s$ par, 4.1 ke EP stage $I_{s p} = 3000 s$ ke saath. Har ek ko kitna propellant chahiye, aur mass saving kya hai?

Recall Solution

Chemical: $v_{e, c} = 9.81 \times 450 = 4415 m/s$ . $Δ v / v_{e, c} = 6000/4415 = 1.359$ . $m_{f} = 1200 e^{- 1.359} = 1200 \times 0.2569 = 308.3 kg$ , toh $m_{p r o p, c} = 1200 - 308.3 = 891.7 kg$ . EP (4.1 se): $m_{p r o p, e p} = 221.3 kg$ . Saving: $891.7 - 221.3 = 670.4 kg$ — EP stage lagbhag ek-chauthai propellant use karta hai. KYUN: propellant mass $Δ v / v_{e}$ ke exponential ke andar baitha hai; $v_{e}$ ko ~6.7 se multiply karne par woh exponent dramatically chhota ho jaata hai. Yahi EP ka poora economic case hai (Chemical vs Electric Propulsion).

Level 5 — Mastery

Exercise 5.1

Mission design. Ek fixed power bus $η P_{in} = 2.0 kW$ beam power deta hai. Ship $m_{0} = 800 kg$ se start hoti hai aur $Δ v = 8.0 km/s$ achieve karna hai. Tum koi bhi $I_{s p}$ choose kar sakte ho. Jab $I_{s p}$ badhate ho: propellant mass girta hai (achha) lekin burn time badhta hai (bura, kyunki thrust girta hai). Burn time ko $I_{s p}$ ke function ke roop mein nikalo aur trade ko quantitatively dikhane ke liye ise $I_{s p} = 2000 s$ aur $I_{s p} = 4000 s$ par evaluate karo.

Recall Solution

Do competing effects set up karo. Propellant: $m_{p r o p} (I_{s p}) = m_{0} (1 - e^{- Δ v / (g_{0} I_{s p})})$ . Beam power se thrust: $F = 2 (η P_{in}) / v_{e} = 2 (η P_{in}) / (g_{0} I_{s p})$ — note $P_{j e t} = η P_{in}$ wahi hai jo hume diya gaya tha, aur $F = 2 P_{j e t} / v_{e}$ . Mass flow: $\overset{m}{˙} = F / v_{e} = 2 P_{j e t} / v_{e}^{2}$ . Burn time (constant-flow estimate): $t = m_{p r o p} / \overset{m}{˙}$ .

$I_{s p} = 2000 s$ par evaluate karo: $v_{e} = 9.81 \times 2000 = 1.962 \times 1 0^{4}$ . $Δ v / v_{e} = 8000/19620 = 0.4077 \Rightarrow m_{p r o p} = 800 (1 - e^{- 0.4077}) = 800 (1 - 0.6652) = 267.8 kg$ . $\overset{m}{˙} = 2 (2000) / (1.962 \times 1 0^{4})^{2} = 4000/3.849 \times 1 0^{8} = 1.039 \times 1 0^{- 5} kg/s$ . $t = 267.8/1.039 \times 1 0^{- 5} = 2.578 \times 1 0^{7} s \approx 0.82 yr$ .

$I_{s p} = 4000 s$ par evaluate karo: $v_{e} = 3.924 \times 1 0^{4}$ . $Δ v / v_{e} = 8000/39240 = 0.2038 \Rightarrow m_{p r o p} = 800 (1 - e^{- 0.2038}) = 800 (1 - 0.8156) = 147.5 kg$ . $\overset{m}{˙} = 4000/ (3.924 \times 1 0^{4})^{2} = 4000/1.540 \times 1 0^{9} = 2.598 \times 1 0^{- 6} kg/s$ . $t = 147.5/2.598 \times 1 0^{- 6} = 5.677 \times 1 0^{7} s \approx 1.80 yr$ .

Trade, saamne rakhi: $I_{s p}$ double karne par (2000→4000 s) propellant lagbhag aadha ho jaata hai (268→147 kg, 120 kg bachate hain) lekin burn time do se zyada ho jaata hai (0.82→1.80 yr). Yahi mission-design seesaw hai: propellant mass versus trip duration. Koi free lunch nahi — optimum wahan hai jahan tumhare mission ki extra mass ki cost, extra time ki cost ke barabar ho.

Exercise 5.2

Algebraically dikhao ki small- $Δ v$ limit mein burn time $I_{s p}$ ke saath kyun badhta hai (jahan $Δ v ≪ v_{e}$ , toh $m_{p r o p} \approx m_{0} Δ v / v_{e}$ ). $t$ ki $I_{s p}$ ke saath scaling derive karo.

Recall Solution

KYUN approximation: jab $x = Δ v / v_{e}$ chhota hota hai, $1 - e^{- x} \approx x$ (exponential expansion ka pehla term). Toh $m_{p r o p} \approx m_{0} Δ v / v_{e}$ . Ab $t = m_{p r o p} / \overset{m}{˙}$ assemble karo $\overset{m}{˙} = 2 P_{j e t} / v_{e}^{2}$ ke saath: $t \approx \frac{m _{0} Δ v / v _{e}}{2 P _{j e t} / v _{e}^{2}} = \frac{m _{0} Δ v v _{e}}{2 P _{j e t}} .$ Kyunki $v_{e} = g_{0} I_{s p}$ : $t \approx \frac{m _{0} Δ v g _{0}}{2 P _{j e t}} I_{s p}$ Padhna: burn time $I_{s p}$ ke directly proportional hai (fixed beam power aur mission par). Zyada economy → proportionally lamba trip. Yahi clean law hai 5.1 ke numbers ke peeche — aur poore trade-off ka ultimate statement.

Connections

Tsiolkovsky Rocket Equation — $m_{f} = m_{0} e^{- Δ v / v_{e}}$ , L4–L5 mein use hua.
Specific Impulse — woh $I_{s p}$ jo humne throughout dial kiya.
Chemical vs Electric Propulsion — Exercise 4.2 ki 4× propellant saving.
Ion and Hall Thrusters — L1–L2 problems mein hardware.
Spacecraft Power Systems — woh fixed $P_{in}$ set karta hai jo trade-off ko bite dilata hai.
Thermal Control (radiators) — Exercise 2.2 ki wasted power.
Newton's Third Law — $F = \overset{m}{˙} v_{e}$ ki root.

Recall Poori ladder ka ek-line summary

Fixed power par: thrust $\propto 1/ I_{s p}$ , propellant $\propto$ $I_{s p}$ ke saath girta hai, burn time $\propto I_{s p}$ . Us curve par woh point pick karo jo tumhara mission afford kar sake.