Electric pump-fed cycle — modern innovation: Question bank

3.3.26 · D5 · Physics › Rocket Propulsion › Electric pump-fed cycle — modern innovation

Intuition Is page ko kaise use karein

Ye woh sneaky conceptual questions hain — woh jo unhe trip karte hain jo formulas mein numbers plug kar sakte hain lekin yeh samjhe nahi hain ki formulas ka matlab kya hai. Har item ek one-line prompt hai; pehle khud socho, phir answer reveal karo. Ye sab parent note the parent topic aur wider chapter Rocket Propulsion par build karta hai.

Neeche use kiye gaye symbols (sab is page ya parent par defined hain): $\overset{m}{˙}$ = mass flow rate (kg/s, har second pump se kitne kg propellant cross karta hai), $Δ p$ = pump jo pressure rise add karta hai (Pa), $ρ$ = liquid density (kg/m³), $η_{p}$ = pump efficiency, $η_{m}$ = motor+controller efficiency, $t_{b}$ = burn time (s), $e_{b}$ = battery specific energy (J/kg), $I_{s p}$ = specific impulse, $P$ = power (W). Do aur jinhe hum rely karenge:

$p_{c}$ = combustion-chamber pressure — chamber ke andar ka pressure jahan propellant jalta hai (typically $100 +$ bar). Ise us "back-pressure" ki tarah socho jise pump ko beat karna padta hai taaki fuel inject kar sake.
$p_{tank}$ = tank pressure — pump se pehle tank ke andar propellant jis pressure par baitha hota hai (typically $1$ – $3$ bar, low rakha jaata hai taaki tank walls patli aur halki rahen).
$p_{inlet}$ = pump-inlet pressure — pump ke suction port par liquid jis pressure par pahunchta hai, apne boiling (vapour) pressure se thoda upar taaki liquid bana rahe.

Intuition Woh ek derivation jis par yahan sab kuch tika hai — pehle yeh padho

Pump ko ek piston ki tarah picture karo jo liquid ke ek slug ko aage dhakelta hai. Ek tiny volume $d V$ (fluid ka ek chhota parcel, m³ mein measure kiya gaya) ko pressure rise $Δ p$ ke against push karne ke liye, aap jo kaam karte ho woh hai $d W = Δ p d V,$ jahan $d W$ woh tiny bit of work / energy (joules mein) hai jo us ek parcel par kharch hua. Isi liye "pressure is energy-per-volume" — ise $Δ p = d W / d V$ mein rearrange karo. Ab, power energy per second hai, toh ek tiny time $d t$ se divide karo: $P_{i d e a l} = \frac{d W}{d t} = Δ p \frac{d V}{d t} = Δ p \dot{V},$ jahan $\dot{V} = d V / d t$ volume flow rate (m³/s) hai. Finally, kyunki mass flow $\overset{m}{˙} = ρ \dot{V}$ hai, hum $\dot{V} = \overset{m}{˙} / ρ$ swap karte hain aur yahan pahunchte hain $P_{i d e a l} = \frac{m ˙ Δ p}{ρ} .$ Toh density unit-fixing trick ki tarah nahi balki isliye aati hai kyunki power volume flow track karta hai, aur $ρ$ woh hai jo measured mass flow ko volume flow mein convert karta hai. Is chain ko yaad rakho — pressure→work, work→power, mass→volume — neeche har jawab ke liye.

True or false — justify

Recall T/F bank

Ek pump ki zaroorat hai kyunki combustion chamber tanks se zyada pressure par hota hai. ::: True — propellant inject karne ke liye aapko chamber pressure $p_{c}$ beat karni hoti hai; pump pressure ko chamber se just pehle raise karta hai taaki tanks ko $100 +$ bar par na rakhna pade. Ek electric pump-fed engine ka $I_{s p}$ same chamber ke gas-generator cycle se zyada hota hai, purely isliye kyunki woh electric hai. ::: "Purely because electric" ke baare mein False — $I_{s p}$ chamber combustion se set hota hai. Electric gas generator ke chhote overboard loss ko avoid karta hai, lekin yeh ek modest gain hai, electricity se magic boost nahi. Burn time double karne se, sab kuch equal rakhte hue, required battery mass roughly double ho jaata hai. ::: True — battery mass $= P_{e l ec} t_{b} / e_{b}$ , aur $t_{b}$ linearly aata hai, toh double time mein double energy aur double mass chahiye. Burn time double karne se required motor power roughly double ho jaata hai. ::: False — power $\overset{m}{˙}, Δ p, ρ$ aur efficiencies par depend karta hai, jinmein se koi bhi $t_{b}$ nahi hai. Time energy change karta hai, instantaneous power nahi. Kyunki electric cycle "turbine par koi fuel waste nahi karta," yeh total mass mein turbopump cycle ko hamesha beat karta hai. ::: False — yeh hot-gas machinery ko battery mass se trade karta hai. Lambe burns mein batteries itni badhti hain ki woh turbine ki cost se zyada ho sakti hain. Dekho Turbopump-fed cycle (gas generator vs staged combustion). Ideal pump power $P_{i d e a l} = \overset{m}{˙} Δ p / ρ$ ki units watts hain. ::: True — $(kg/s) (Pa) / (kg/m^{3}) = Pa \cdot m^{3} / s = J/s = W$ . Density exactly wahi hai jo units fix karti hai. Propellant tanks ko 100 bar ki jagah 2 bar par operate karna hi woh reason hai jiske liye pump exist karta hai. ::: True — low pressure par patle halke tanks poora payoff hai; pump pressure-raising downstream karta hai. Isi liye pressure-fed designs bhi itni heavy hoti hain. Ek zyada efficient pump ( $η_{p}$ jo 1 ke karib ho) battery se electrical power draw kam karta hai. ::: True — $P_{e l ec} = \overset{m}{˙} Δ p / (ρ η_{p} η_{m})$ ; $η_{p}$ badhane se denominator ka loss factor chhota hota hai, toh har second kam battery energy chahiye. Agar pump inlet pressure $p_{in l e t}$ propellant ke vapour pressure se neeche gir jaaye, toh pump theek kaam karta rehta hai. ::: False — liquid inlet par vapour bubbles mein boil ho jaata hai (cavitation); bubbles blades par violently collapse hote hain, performance barbad karte hain aur metal erode karte hain. Tanks ko lightly pressurise kiya jaata hai exactly $p_{in l e t}$ ko vapour pressure se upar rakhne ke liye.

Spot the error

Recall Flaw dhundo aur fix karo

"Pump power $P = \overset{m}{˙} Δ p$ hai." ::: Density missing hai: $\overset{m}{˙} Δ p$ = (kg/s)(Pa) watts nahi hai. Aapko volume flow $\dot{V} = \overset{m}{˙} / ρ$ chahiye, jisse $P = \overset{m}{˙} Δ p / ρ$ milta hai — exactly opening box mein pressure→work→power chain. "Electrical power find karne ke liye, ideal power ko efficiencies se multiply karo: $P_{e l ec} = P_{i d e a l} η_{p} η_{m}$ ." ::: Galat direction hai — losses ki wajah se aapko zyada supply karna padta hai, toh aap divide karte ho: $P_{e l ec} = P_{i d e a l} / (η_{p} η_{m})$ . Multiply karna yeh claim karta ki losses aapki madad karte hain. "Battery mass = power ÷ specific energy, toh $m_{ba tt} = P_{e l ec} / e_{b}$ ." ::: Units aur physics dono fail karte hain — $P / e_{b}$ = (W)/(J/kg) = kg/s, ek rate hai, mass nahi. Aapko pehle energy $E = P_{e l ec} t_{b}$ form karni hogi, phir $m_{ba tt} = E / e_{b}$ . "Kyunki tanks 2 bar par hain, pump ko sirf 2 bar add karne ki zaroorat hai." ::: Nahi — pump ko tank pressure $p_{t ank}$ se chamber pressure $p_{c}$ tak puri rise add karni hoti hai, yaani $Δ p \approx p_{c} - p_{t ank} \approx 98$ bar, 2 bar nahi. Low tank pressure ek starting point hai, target nahi. "Electric pumps $I_{s p}$ improve karte hain, toh woh Tsiolkovsky rocket equation mein exhaust velocity badhate hain." ::: Tsiolkovsky ke equation mein exhaust velocity chamber conditions se aati hai, how aapne propellant feed kiya isse koi fark nahi padta. Batteries actually mass ratio worsen karti hain, $Δ v$ hurt karta hai. "Pressure = energy per volume sirf ek analogy hai; aap literally $Δ p$ ko energy density ki tarah use nahi kar sakte." ::: Yeh literal hai — $d W = Δ p d V$ (work $d W$ jo volume $d V$ ka ek parcel push karne mein lagti hai) exact hai, isi liye $P_{i d e a l} = Δ p \dot{V}$ directly follow karta hai. Bernoulli's principle se compare karo, jahan pressure bhi ek energy-per-volume term hai.

Why questions

Recall Reason it through

Tank ko sirf pump-inlet pressure se zyada rehne ki zaroorat kyun hai, chamber pressure se nahi? ::: Kyunki pump, tank nahi, $p_{c}$ ke past bada rise provide karta hai. Tank ko sirf liquid ko inlet par vapour pressure se upar rakhna hota hai (kuch bar) taaki cavitation na ho; ise kabhi $p_{c}$ tak pahunchne ki zaroorat nahi. Exactly isi liye tank patla aur halka reh sakta hai. Power formula mein density $ρ$ kyun aata hai lekin akela pressure nahi? ::: Power = energy per second = pressure-energy per volume ( $Δ p$ ) × volume flow ( $\dot{V}$ ). Mass flow $\overset{m}{˙}$ measured quantity hai, toh hum volume flow mein convert karte hain $\dot{V} = \overset{m}{˙} / ρ$ ke zariye; denser fluid har kg mein kam volume move karta hai, usi mass ke liye kam pump power chahiye. Electric cycle chhote-burn upper stages ke liye kyun suit karta hai, giant boosters ke liye nahi? ::: Batteries energy $= P \times t_{b}$ ke saath scale karti hain. Boosters huge $\overset{m}{˙}$ par lambe chalte hain, toh battery mass explode karta hai; chhote high-Isp upper-stage burns battery ko chhota rakhte hain. Dekho Rocket Lab Rutherford engine. Yeh cycle practically 2010s ke around hi kyun possible hua? ::: Battery specific energy finally ~200–280 Wh/kg tak pahunchi; isse neeche, batteries kisi bhi useful burn ke liye bahut heavy theen. Physics hamesha se known tha — chemistry ne pakad liya. Spent batteries ko mid-burn drop karne se help kyun milti hai jabki aap unki energy already kharch kar chuke ho? ::: Ek drained battery ka pura mass hota hai. Ise jettison karna burnout se pehle dead weight remove karta hai, mass ratio improve karta hai aur isliye Tsiolkovsky rocket equation mein $Δ v$ improve hota hai. Zyada powerful motor automatically heavier battery kyun nahi hota? ::: Battery mass total energy track karta hai, peak power nahi. Ek brief high-power burst ek lambe gentle burst se kam energy use kar sakta hai — time factor decide karta hai. Gas-generator turbopump ko kuch propellant "waste" karne wala kyun describe kiya jaata hai? ::: Yeh turbine spin karne ke liye fuel-rich mix jalata hai, phir us partly-burned gas ko low efficiency par overboard dump karta hai — ek chhota $I_{s p}$ penalty jo electric cycle sidestep karta hai.

Edge cases

Recall Boundary aur degenerate scenarios

Agar $Δ p \to 0$ ho (tank already chamber pressure ke karib ho) toh required pump power ka kya hoga? ::: $P_{i d e a l} = \overset{m}{˙} Δ p / ρ \to 0$ — koi pressure add nahi karna matlab koi pump work nahi; aap ek pressure-fed design mein degenerate ho gaye ho. $Δ p < 0$ pump ke liye physically kya matlab rakhta hai? ::: Iska matlab hai downstream pressure upstream se lower hai — pump turbine ki tarah kaam kar raha hoga (energy extract karte hue), add nahi kar raha. Real feed system mein yeh design se kabhi nahi hota; near-zero ya negative net head signal karta hai ki pump stalled hai ya plumbing galat hai. Cavitation kya hai aur yeh inlet pressure par hard lower bound kyun hai? ::: Agar inlet pressure liquid ke vapour pressure se neeche gire, toh propellant vapour bubbles mein flash ho jaata hai jo impeller par implode karte hain, flow choke karte hain aur metal pit karte hain. Toh $p_{in l e t}$ ko arbitrarily low nahi kiya ja sakta chahe tank mass bachana ho — vapour pressure floor set karta hai. Jab $t_{b} \to 0$ (ek impulsive kick) toh battery mass ka kya hoga? ::: $m_{ba tt} \to 0$ — essentially instant mein thrust ke liye koi energy nahi chahiye, toh electric cycle bahut chhote burns ke liye sabse favorable hai. Agar pump efficiency $η_{p} \to 1$ aur motor $η_{m} \to 1$ ho jaaye? ::: $P_{e l ec} \to P_{i d e a l} = \overset{m}{˙} Δ p / ρ$ ; electrical draw apne theoretical floor tak pahunch jaata hai. Aap kabhi is se neeche nahi ja sakte — yeh woh actual energy hai jo fluid ko deliver ki gayi hai. Agar propellant gas hota, liquid nahi (low $ρ$ )? ::: Low $ρ$ $\dot{V} = \overset{m}{˙} / ρ$ ko huge bana deta hai, toh pump power skyrocket karta hai — ek wajah ki liquid propellants (high $ρ$ ) gases se pump karne mein kahin sasta hai. Jab $e_{b} \to \infty$ (kalpnik perfect battery) ho, toh electric cycle ko kya limit karta hai? ::: Battery mass $\to 0$ , toh koi bhi feed-related cheez ise limit nahi karti; yeh cycle turbopumps ko completely beat kar leta. Yeh dikhata hai ki batteries — concept nahi — asli constraint hain. Agar burn time itna lamba ho ki $m_{ba tt}$ bachaye gaye propellant mass se zyada ho jaaye, toh aapko kya switch karna chahiye? ::: Ek turbopump cycle — ek crossover burn time ke baad turbine ka fixed hardware hamesha badhti batteries ko beat kar deta hai. Agar aap battery mass double kar do bina kisi aur change ke, toh $I_{s p}$ ka kya hoga? ::: $I_{s p}$ unchanged rehta hai (chamber-set hai), lekin extra dead mass mass ratio lower karta hai, achievable $Δ v$ cut karta hai — zyada battery ≠ better rocket.