3.3.26 · Physics › Rocket Propulsion
Intuition Ek saans mein badi baat
Ek rocket engine ko apne combustion chamber mein propellant bahut zyada pressure par (100+ bar) daalna padta hai. Traditionally aap apna kuch fuel jalate ho ek turbine chalane ke liye jo pumps drive karta hai. Electric pump-fed cycle us turbine ko hata deta hai aur pumps ko electric motor powered by batteries se chalata hai. Plumbing simple ho jaati hai, lekin ab aapko heavy batteries carry karni padti hain. Ye tabhi practical bana jab lithium battery energy density kaafi high ho gayi (~2010s mein).
Intuition Kyun nahi sirf pressure se fuel bahar dhakela jaaye?
Combustion chamber high pressure p c par hota hai. Usme propellant inject karne ke liye aapka tank pressure p c se zyada hona chahiye. Agar aap poore tank ko 150 bar tak pressurize kar do ("pressure-fed" approach), toh tank ki walls bahut thick aur heavy karni padenge — tank ek bomb-grade pressure vessel ban jaata hai. Ek pump tanks ko 1–3 bar ke paas (patli, halki walls ke saath) rehne deta hai jabki pump chamber se theek pehle pressure raise karta hai. Yahi puri wajah hai ki turbo/electric pumps exist karte hain.
Woh subsystem jo propellant ko low-pressure tanks se high-pressure combustion chamber tak le jaata hai. Teen main strategies hain:
Pressure-fed : tanks khud high-pressure hote hain. Koi pump nahi. Simple, lekin heavy tanks.
Turbopump-fed (gas-generator, staged combustion, expander): ek turbine jo hot gas se drive hoti hai pumps chalati hai. Light tanks, complex hot machinery.
Electric pump-fed : ek electric motor + battery pumps chalata hai. Light tanks, koi hot gas plumbing nahi, lekin battery mass hoti hai.
Intuition Motor ko KITNI power supply karni chahiye?
Pump ek flowing liquid ka pressure badhata hai. Fluid mein add ki gayi energy per unit volume pressure rise Δ p ke barabar hoti hai (ye sirf pressure ki definition hai = energy/volume). Volume flow rate se multiply karo aur power mil jaati hai.
KAISE — derive karo:
Liquid ke chhote volume d V ko pressure Δ p se upar dhakeline ka kaam:
d W = Δ p d V
Time d t se divide karo. Volume flow rate hai V ˙ = d V / d t :
P ideal = Δ p V ˙
Ab V ˙ ko mass flow rate m ˙ aur density ρ se express karo, kyunki m ˙ = ρ V ˙ ⇒ V ˙ = m ˙ / ρ :
Aur motor battery se draw karta hai , motor+controller efficiency η m ke saath:
P elec = η m P shaft = ρ η p η m m ˙ Δ p
Intuition KYUN batteries hi poora game hain
Total electrical energy ki zaroorat = power × burn time. Battery mass = energy ÷ specific energy e b (J/kg). Agar burn lamba hai, toh energy ikattha hoti rehti hai aur batteries mass dominate karti hain — yahi wajah hai ki electric pumps chote, short-burn stages ke liye suitable hain, giant boosters ke liye nahi.
Burn time t b mein energy:
E = P elec t b = ρ η p η m m ˙ Δ p t b
Aap aksar spent batteries ko burn ke beech mein drop kar sakte ho (jaise Rocket Lab ka Rutherford karta hai) toh dead battery mass ko burnout tak carry nahi karna padta.
Worked example Example 1 — ek Rutherford-class pump ke liye power
Diya gaya: m ˙ = 5 kg/s propellant, Δ p = 100 bar = 1 0 7 Pa , ρ = 1000 kg/m 3 (RP-1 jaisa), η p = 0.6 , η m = 0.9 .
Ideal: P ideal = ρ m ˙ Δ p = 1000 5 × 1 0 7 = 5 × 1 0 4 W = 50 kW .
Ye step kyun? Seedha derived P ideal = m ˙ Δ p / ρ ka plug-in.
Electrical: P elec = 0.6 × 0.9 50 kW = 92.6 kW .
Ye step kyun? Real losses ideal figure ko 1/ ( η p η m ) se inflate karte hain.
Worked example Example 2 — 150 s burn ke liye battery mass
Same engine, t b = 150 s , e b = 0.9 × 1 0 6 J/kg .
E = P elec t b = 92.6 × 1 0 3 × 150 = 1.39 × 1 0 7 J .
Ye step kyun? Energy = power × time.
m batt = E / e b = 1.39 × 1 0 7 /0.9 × 1 0 6 ≈ 15.4 kg per pump.
Ye step kyun? Energy ko divide karo is se ki battery ka har kg kitne joules store karta hai.
Insight: burn time double karo → battery mass double ho jaati hai. Yahi wajah hai ki electric pumps chote upper-stage burns se pyaar karte hain.
Worked example Example 3 — kyun pressure-fed yahan worse hota
Same Δ p = 100 bar ko bina pump ke feed karne ke liye, tank ko 100 bar hold karna padega. Tank wall mass p ⋅ V ke saath scale karta hai. 2 bar par pump-fed tank ko ~50× patli walls chahiye. Electric pump motor+battery add karta hai (~20–30 kg) lekin potentially hundreds of kg tank steel bachata hai — kuch bhi chote propellant loads ke alawa ke liye net win hai.
Common mistake "Battery mass sirf thrust par depend karta hai."
Kyun sahi lagta hai: zyada thrust ko zyada power chahiye, aur power hi poori kahani lagti hai.
Fix: battery mass energy = power × time par depend karta hai, isliye burn duration utni hi important hai . Ek high-thrust short burn ko kam battery chahiye ho sakti hai low-thrust long burn se. Hamesha t b factor saath rakho.
Common mistake "Electric pumps engine ko zyada efficient banate hain (higher
I s p )."
Kyun sahi lagta hai: "electric = modern = better," aur gas-generator cycles fuel bahar dump karte hain jo kuch waste hota hai.
Fix: electric pumps cycle simplicity aur cost improve karte hain, aur gas generator ka chhota I s p loss avoid karte hain, lekin chamber combustion I s p set karta hai. Battery dead weight hai jo mass ratio ko kam karti hai. Ye ek manufacturing/complexity innovation hai, magic I s p boost nahi.
P = m ˙ Δ p use karna (ρ bhool jaana).
Kyun sahi lagta hai: dimensional sloppiness — m ˙ Δ p mass-flow times pressure jaisa dikhta hai.
Fix: units check karo. m ˙ Δ p = (kg/s)(Pa) watts nahi hai . Aapko V ˙ = m ˙ / ρ chahiye volume flow pane ke liye, jo P = m ˙ Δ p / ρ watts mein deta hai. Density essential hai.
Recall Feynman: ek 12-saal ke bachche ko samjhao
Socho tumhe ek aisa balloon mein paani spray karna hai jo pehle se bahut tight phula hua hai. Balloon ke push ko beat karne ke liye tumhare water pistol mein ek strong pump chahiye. Purane rockets woh pump thoda fuel jalakar ek mini-engine mein chalate the. Naye rockets sirf ek battery aur electric motor use karte hain, jaise cordless drill, pump spin karne ke liye. Ye simple aur clean hai, lekin batteries heavy hoti hain, isliye ye tab hi kaam karta hai jab engine short time ke liye chale.
Mnemonic Power formula yaad rakho
"Mad Pete's Rho" → P = ρ m ˙ Δ p : M ass-flow times Δp , sab kuch ρ ho se divide. Aur battery mass = same cheez × t / e b : "power ⟶ energy ⟶ kilograms ."
Rockets pumps kyun use karte hain instead of sirf high-pressure tanks ke?
P ideal = m ˙ Δ p / ρ ko d W = Δ p d V se derive karo.
Woh ek factor kya hai jo electric pumps ko long-burn boosters ke liye unsuitable banata hai?
Electric pump-fed cycle mein turbine ki jagah kya aata hai? Ek electric motor jo batteries se powered hota hai, propellant pumps drive karta hai.
Ideal pump power formula? P ideal = m ˙ Δ p / ρ (mass flow × pressure rise ÷ density).
Ideal pump power ko η p se kyun divide karte hain? Real pumps energy friction/heat mein lose karte hain, isliye shaft power ideal se zyada honi chahiye = P i d e a l / η p .
Battery mass expression? m ba tt = m ˙ Δ p t b / ( ρ η p η m e b ) , yaani energy needed ÷ battery specific energy.
Pumps tanks ko light kaise rehne dete hain iski main wajah? Pump pressure chamber se theek pehle raise karta hai, isliye tanks 1–3 bar ke paas reh sakte hain patli walls ke saath instead of 100+ bar ke.
Electric pumps short-burn stages tak kyun limited hain? Battery mass = power × burn time / specific energy; lambe burns ko impractically heavy batteries chahiye.
Kya electric pump directly I s p badhata hai? Nahi — chamber combustion I s p set karta hai; electric pumps simplicity/cost improve karte hain aur gas-generator overboard loss avoid karte hain.
Electric pump cycles ko 2010s mein kya cheez possible banayi? High-specific-energy lithium batteries (~200–280 Wh/kg).
Real-world example engine? Rocket Lab ka Rutherford (Electron rocket), pehla orbital electric-pump engine.
Pump power mein units check ka trap? m ˙ Δ p (kg·Pa/s) watts nahi hai; mass flow ko volume flow mein convert karne ke liye density ρ se zaroor divide karo.
Inject propellant needs tank p over pc
Pump raises pressure before chamber
Light thin tanks 1 to 3 bar
Turbopump-fed hot gas turbine
Electric pump-fed motor plus battery
Li battery energy density 2010s
P ideal equals m dot dp over rho