Exercises — Electric pump-fed cycle — modern innovation

Turbine (hot gas se ghoomne wala fan) ki jagah ek electric motor aata hai. Motor ko ek battery pack power karta hai. Pumps chalane ke liye koi hot gas tap ya burn nahi hoti — jis plumbing ki turbine ko zaroorat hoti hai (gas generator, hot ducts) woh simply khatam ho jaata hai. Jo cheez replace ho rahi hai uske liye Turbopump-fed cycle (gas generator vs staged combustion) dekho aur real-world example ke liye Rocket Lab Rutherford engine dekho.

$\dot{m}\Delta p$ ke units hain $\left(\text{kg/s}\right)\times \left(\text{Pa}\right)$. Units se kyun shuru karein: agar units watts tak reduce nahi hoti, toh formula galat hai chahe koi bhi number aaye — units ek free correctness check hai. A pascal is ${\text{N/m}}^2=\text{kg}{\text{m}}^{-1}{\text{s}}^{-2}$. Pascal ko kyun rewrite karein: ise kg/s ke saath combine karne ke liye har unit ko base form (kg, m, s) mein likhna padta hai, warna kuch cancel nahi hoga. Multiply karne par: $\frac{\text{kg}}{\text{s}}\cdot \frac{\text{kg}}{\text{m}{\text{s}}^2}=\frac{{\text{kg}}^2}{\text{m}{\text{s}}^3}$ A watt is $\text{kg}{\text{m}}^2{\text{s}}^{-3}$. Yeh match nahi karte: hamare result mein kg ki do powers aur $1/\text{m}$ hai, lekin watt ko ek kg aur ${\text{m}}^2$ chahiye. Yeh specific mismatch kyun matter karta hai: upar ek extra kg hai aur metres ulte hain — exactly jo ek density (${\text{kg/m}}^3$) se divide karne par remove hota hai. Toh $\rho$ se divide karo: $\frac{{\text{kg}}^2}{\text{m}{\text{s}}^3}÷\frac{\text{kg}}{{\text{m}}^3}=\frac{{\text{kg}}^2}{\text{m}{\text{s}}^3}\cdot \frac{{\text{m}}^3}{\text{kg}}=\frac{\text{kg}{\text{m}}^2}{{\text{s}}^3}=\text{W}✓$ $\rho$ se division kyun fix karta hai: $\rho$ se divide karna ek kg cancel karta hai (sirf ek power of mass bacha ke) aur $1/\text{m}$ ko ${\text{m}}^2$ mein badal deta hai (kyunki ${\text{m}}^{-1}\times {\text{m}}^3={\text{m}}^2$) — exactly woh do edits jo watt tak pahunchne ke liye chahiye. Physically, $\dot{m}$ ko $\rho$ se divide karna mass flow ko volume flow mein convert karta hai, aur power = pressure × volume-flow. Toh sahi form hai $P_{\text{ideal}}=\dot{m}\Delta p/\rho$.

Yeh Pressure-fed cycle hai. Tank ki walls ko har jagah $150\text{bar}$ handle karna padega, isliye tank ek mota, bhaari pressure vessel ban jaata hai. Ek pump tank ko $1$–$3\text{bar}$ par rakha rehne deta hai (patli walls) aur pressure sirf chamber ke pehle ek choti si pipe mein raise karta hai.

Convert karo: $120\text{bar}=120\times 10^5=1.2\times 10^7\text{Pa}$. $P_{\text{ideal}}=\frac{\dot{m}\Delta p}{\rho }=\frac{8\times 1.2\times 10^7}{1000}=\frac{9.6\times 10^7}{1000}=9.6\times 10^4\text{W}=96\text{kW}.$ Hamne kya kiya: seedha $P_{\text{ideal}}=\dot{m}\Delta p/\rho$ mein plug in kiya. Kyun: ideal power bas woh energy per second hai jo fluid ko di gayi hai, aur pressure rise exactly energy per unit volume hai.

Yahan $P_{\text{elec}}$ woh electrical power hai jo battery is ek pump ke liye deliver karti hai: $P_{\text{elec}}=\frac{P_{\text{ideal}}}{{\eta }_p{\eta }_m}=\frac{96}{0.65\times 0.9}=\frac{96}{0.585}\approx 164.1\text{kW}.$ Divide kyun karein: 1 se kam har efficiency ka matlab hai kuch supplied energy waste hoti hai (friction fluid ko garam karta hai, resistance wires ko garam karta hai), toh fluid ko 96 kW deliver karne ke liye zyada feed karna padega.

$P_{\text{ox}}$ akele oxidiser pump ki ideal power hai: $P_{\text{ox}}=\frac{18\times 1.2\times 10^7}{1140}=\frac{2.16\times 10^8}{1140}\approx 1.895\times 10^5\text{W}\approx 189.5\text{kW}.$ $P_{\text{total}}$ dono pumps ki ideal powers ka sum hai (fuel L2.1 se = 96 kW): $P_{\text{total}}=P_{\text{ideal,fuel}}+P_{\text{ox}}=96+189.5=285.5\text{kW}.$ Insight: oxidiser flow fuel flow se double se zyada hai (yeh mixture ratio hai), isliye thoda zyada dense hone ke bawajood yeh pumping power mein dominant hai.

Pump ko fluid ko tank pressure se utha ke wahan pahunchana hai jahan chamber plus har downstream loss demand karta hai. Required outlet pressure build karo: $p_{\text{outlet}}=p_c+\Delta p_{\text{inj}}+\Delta p_{\text{line}}=100+20+5=125\text{bar}.$ Losses kyun add karein: injector aur lines pump aur chamber ke beech pressure "khate" hain, isliye pump ko zyada upar se start karna padta hai taaki chamber mein $100\text{bar}$ mile. Pump sirf tank ke upar ka rise provide karta hai: $\Delta p=p_{\text{outlet}}-p_{\text{tank}}=125-3=122\text{bar}=1.22\times 10^7\text{Pa}.$ Corrected ideal power: $P_{\text{ideal}}=\frac{8\times 1.22\times 10^7}{1000}=\frac{9.76\times 10^7}{1000}=9.76\times 10^4\text{W}=97.6\text{kW}.$ Takeaway: L2.1 ka "clean" $120\text{bar}$ losses ki ek chain chhupaata tha; real pumps size karne ka matlab hai har downstream drop add karna aur tank head subtract karna.

Pehle total electrical power nikalo. $P_{\text{elec,total}}$ ka matlab sabhi pumps ke liye total electrical power hai — total ideal power $P_{\text{total}}=285.5\text{kW}$ (L2.3 se) lo aur losses se inflate karo: $P_{\text{elec,total}}=\frac{P_{\text{total}}}{{\eta }_p{\eta }_m}=\frac{285.5}{0.65\times 0.9}=\frac{285.5}{0.585}\approx 488.0\text{kW}.$ Energy over the burn ($E$ = total electrical energy): $E=P_{\text{elec,total}}{t}_b=4.880\times 10^5\times 180=8.784\times 10^7\text{J}.$ Battery mass: $m_{\text{batt}}=\frac{E}{e_b}=\frac{8.784\times 10^7}{0.9\times 10^6}\approx 97.6\text{kg}.$ Chain: power ⟶ energy (×time) ⟶ kilograms (÷specific energy) — page ke upar wala "power ⟶ energy ⟶ kilograms" chant.

Battery mass $t_b$ mein linear hai (baaki sab fixed), aur origin se pass hoti hai — toh mass burn time ke directly proportional hai. Line ka slope hai $97.6/180=0.542\text{kg/s}$.

• $60\text{s}$: $0.542\times 60\approx 32.5\text{kg}$ ($180\text{s}$ ka ek-tehai).
• $360\text{s}$: $0.542\times 360\approx 195.2\text{kg}$ ($180\text{s}$ ka double). Geometry kya bataati hai: origin se straight line ka matlab hai "input double karo, output double ho jaata hai." Poora formula dobara compute karne ki zaroorat nahi — bas scale karo.

$180\text{s}$ par electric system total: $25+97.6=122.6\text{kg}$. Turbopump: $40\text{kg}$. Turbopump yahan halka hai. Break-even: electric mass = turbopump mass set karo. $25+0.542t_b=40⟹0.542t_b=15⟹t_b=\frac{15}{0.542}\approx 27.7\text{s}.$ Toh electric sirf mass mein $\approx 28\text{s}$ se chote burns ke liye jeetta hai. Usse upar batteries turbopump ki fixed machinery se bhaari ho jaati hain. (Electric cost/simplicity mein usse bahut aage bhi jeet sakta hai — mass akela score nahi hai.) Linked Battery specific energy exactly woh hai jo batteries improve hone par is break-even ko baad mein shift karta hai.

Zaroori energy: $E=P_{\text{elec,total}}{t}_b=4.88\times 10^5\times 200=9.76\times 10^7\text{J}.$ Minimum $e_b$ ke liye $m_{\text{batt}}=E/e_b$ rearrange karo (maximum allowed mass minimum required $e_b$ deta hai): $e_b\ge \frac{E}{m_{\text{batt}}}=\frac{9.76\times 10^7}{80}=1.22\times 10^6\text{J/kg}.$ Convert to Wh/kg: $\frac{1.22\times 10^6}{3600}\approx 338.9\text{Wh/kg}.$ Synthesis point: aaj ke best cells (~$280\text{Wh/kg}$) kam padti hain, toh ya mass cap ya burn time relax karna padega — exactly isliye yeh cycle Battery specific energy ke badhne ka intezaar karta raha.

Pehli aadhi ($0$–$90\text{s}$): poori $97.6\text{kg}$ aboard. Doosri aadhi ($90$–$180\text{s}$): $97.6-48.8=48.8\text{kg}$ aboard. Time-averaged mass carry ki gayi: $\bar{m}=\frac{97.6\times 90+48.8\times 90}{180}=\frac{97.6+48.8}{2}=73.2\text{kg}.$ Kyun help karta hai: rocket equation $\Delta v=I_{sp}{g}_0\ln (m_i/m_f)$ chhote final mass $m_f$ ko reward karta hai. Burnout tak carry kiya gaya dead battery mass $m_f$ badhata hai aur $\Delta v$ khaata hai. Spent cells eject karna $m_f$ ghatata hai, aur batteries ki cost wala kuch $\Delta v$ wapas dilata hai.

Battery mass $\propto \frac{1}{{\eta }_p{\eta }_m}$. Har option se fractional reduction compare karo.

• ${\eta }_p$ improve karna: mass scale hoti hai $\frac{0.65}{0.72}=0.9028$ se → $9.72\%$ reduction.
• ${\eta }_m$ improve karna: mass scale hoti hai $\frac{0.90}{0.95}=0.9474$ se → $5.26\%$ reduction. Pump-efficiency upgrade jeetta hai kyunki yeh zyada relative jump hai ($0.65\to 0.72$ matlab $+10.8\%$ vs $0.90\to 0.95$ matlab $+5.6\%$). Synthesis lesson: woh efficiency term chase karo jo abhi sabse kam hai, kyunki wahan relative gains sabse bade hote hain.

(a) Total electrical power $P_{\text{elec,total}}$. Ideal powers: fuel $96\text{kW}$ (L2.1), oxidiser $189.5\text{kW}$ (L2.3), total ideal $P_{\text{total}}=285.5\text{kW}$. ${\eta }_p{\eta }_m=0.585$ se divide karo: $P_{\text{elec,total}}=\frac{285.5}{0.585}\approx 488.0\text{kW}.$ (b) Battery mass. Energy pehle, $E=P_{\text{elec,total}}{t}_b=488.0\times 10^3\times 180=8.784\times 10^7\text{J}$, phir $m_{\text{batt}}=\frac{E}{e_b}=\frac{8.784\times 10^7}{0.9\times 10^6}\approx 97.6\text{kg}.$ (c) Propellant burn hua. Total flow $={\dot{m}}_f+{\dot{m}}_{ox}=8+18=26\text{kg/s}$, $180\text{s}$ mein: $m_{\text{prop}}=26\times 180=4680\text{kg}.$ (d) Ratio. Battery mass ko propellant mass se divide karo: $\frac{m_{\text{batt}}}{m_{\text{prop}}}=\frac{97.6}{4680}\approx 0.0209\approx 2.1\%.$ Comment: battery propellant mass ka sirf ~2% hai — yeh stage battery-limited nahi hai; batteries ek modest tax hain, toh yahan electric cycle ek sound choice hai. Yeh battery-limited tab hota hai jab burn time ya $\Delta p$ bahut badh jaaye jabki propellant chhota rahe.

Power $\propto 1/\rho$. Kam density matlab har kg zyada volume occupy karta hai, toh har second zyada volume pressurise hota hai — power badhti hai. Ratio: $\frac{P_{\text{new}}}{P_{\text{old}}}=\frac{{\rho }_{\text{old}}}{{\rho }_{\text{new}}}=\frac{1140}{1083}\approx \mathrm{1.0526.}$ Toh power $\approx 5.26\%$ badhti hai. Mastery point: denominator mein density ek static constant nahi hai — ek real design worst-case (lowest) density ke liye budget karta hai taaki motor kabhi stall na kare.

Ideal power: $P_{\text{ideal}}=\frac{350\times 1.1\times 10^7}{1050}=\frac{3.85\times 10^9}{1050}\approx 3.667\times 10^6\text{W}=3667\text{kW}.$ Electrical power: $P_{\text{elec,total}}=\frac{3667}{0.585}\approx 6268\text{kW}.$ Energy: $E=6.268\times 10^6\times 160=1.003\times 10^9\text{J}$. Battery mass: $m_{\text{batt}}=\frac{1.003\times 10^9}{0.9\times 10^6}\approx 1114\text{kg}.$ Verdict: ek tonne se zyada dead battery — comparison ke liye upper-stage case (L5.1) ko $100\text{kg}$ se kam chahiye tha. Huge flow aur thodi lambi burn energy ko enormously pile up karti hai. Exactly isliye electric pumps chote, short-burn upper stages par rahte hain aur turbopumps bade boosters mein raj karte hain. Rutherford (Rocket Lab Rutherford engine) ek chota first-stage engine hai exactly kyunki iski flow modest hai aur yeh spent packs jettison kar sakta hai.

Is page ki har problem ek idea hai alag kapde mein: har second pump fluid ko jo energy deta hai woh hai $\dot{m}\Delta p/\rho$; losses ise inflate karti hain; aur battery kilograms bas woh energy over time divided by ek kilo battery kitna hold karta hai. Yeh teen moves pakad lo aur poora cycle — ek single pump se ek poore booster ke go/no-go tak — nikal aata hai.