Electric pump-fed cycle — modern innovation: Worked examples

3.3.26 · D3 · Physics › Rocket Propulsion › Electric pump-fed cycle — modern innovation

Yeh page drill hall hai. Parent note ne formulas build kiye; yahan hum unpe har tarah ke numbers throw karte hain — chhote aur bade flows, zero cases, limits, ek real-world word problem, aur ek exam-style twist.

Sab kuch parent note ke do boxed results pe tika hai. Hum unhe yahan dobara likh rahe hain taaki koi bhi symbol use hone se pehle dikh sake:

Scenario matrix

Is topic mein jo bhi problem aa sakti hai woh in cells mein se ek mein aati hai. Neeche ke examples labeled hain un cells ke saath jo woh cover karte hain, taaki milke poora grid fill ho jaye.

Cell	Kya stress-test karta hai	Covered by
A — baseline plug-in	ek clean pump, saari values given	Ex 1
B — unit trap	$Δ p$ bar mein diya, $e_{b}$ Wh/kg mein	Ex 2
C — $ρ$ ka effect	zyada dense vs kam dense propellant, same $\overset{m}{˙}$	Ex 3
D — limiting case (long burn)	$t_{b} \to$ bada: batteries dominate	Ex 4
E — degenerate / zero inputs	$\overset{m}{˙} = 0$ , $Δ p = 0$ , $η \to 1$	Ex 5
F — real-world word problem	full-engine, drop-battery strategy	Ex 6
G — inverse solve	battery budget given, max $t_{b}$ nikalo	Ex 7
H — exam twist (cycles compare karo)	electric vs pressure-fed break-even	Ex 8

Forecast line padho aur scroll karne se pehle guess karo. Galat guess karna hi woh jagah hai jahan learning hoti hai.

Ex 1 — Cell A: baseline plug-in

Worked example Ex 1 — ek pump, sab kuch given

$\overset{m}{˙} = 4 kg/s$ , $Δ p = 8 \times 1 0^{6} Pa$ , $ρ = 800 kg/m^{3}$ (RP-1 kerosene), $η_{p} = 0.6$ , $η_{m} = 0.9$ . $P_{elec}$ nikalo.

Forecast: ideal power kuch tens of kW hogi; losses use roughly $1/ (0.6 \times 0.9) \approx 1.85$ se inflate karengi. Computing se pehle ek number guess karo.

Step 1 — ideal power. Yeh step kyun? Pehle loss-free core se shuru karo taaki dekh sako losses kahan bite karti hain. $P_{ideal} = \frac{m ˙ Δ p}{ρ} = \frac{4 \times 8 \times 1 0 ^{6}}{800} = 4.0 \times 1 0^{4} W = 40 kW .$

Step 2 — dono efficiencies se divide karo. Yeh step kyun? Pump kuch heat/friction mein waste karta hai, motor+controller aur kuch waste karte hain; battery ko sab cover karna hai. $P_{elec} = \frac{40 kW}{0.6 \times 0.9} = \frac{40}{0.54} \approx 74.1 kW .$

Verify: $P_{ideal}$ ke units: $\frac{( kg/s ) ( Pa )}{kg/m ^{3}} = \frac{kg}{s} \cdot \frac{N/m ^{2}}{kg/m ^{3}} = \frac{N\cdotm}{s} = W .$ ✓ Dimensions correct hain. Aur $74.1 > 40$ as expected (real mein hamesha ideal se zyada cost hoti hai).

Ex 2 — Cell B: unit trap

Worked example Ex 2 — pressure bar mein, energy Wh/kg mein

$\overset{m}{˙} = 5 kg/s$ , $Δ p = 90 bar$ , $ρ = 1000 kg/m^{3}$ , $η_{p} = 0.65$ , $η_{m} = 0.92$ , burn $t_{b} = 180 s$ , battery rated $e_{b} = 250 Wh/kg$ . $m_{batt}$ nikalo.

Forecast: jis moment bar aur Wh unconverted chhod doge, tum $1 0^{5}$ aur $3600$ ke factors se off ho jaoge. Pehle convert karo.

Step 1 — $Δ p$ ko pascals mein convert karo. Yeh step kyun? Hamara formula SI hai; $1 bar = 1 0^{5} Pa$ . $Δ p = 90 \times 1 0^{5} = 9.0 \times 1 0^{6} Pa .$

Step 2 — $e_{b}$ ko J/kg mein convert karo. Yeh step kyun? $1 Wh = 3600 J$ , kyunki ek watt ek ghante ke liye $1 \times 3600$ joules hai. $e_{b} = 250 \times 3600 = 9.0 \times 1 0^{5} J/kg .$

Step 3 — electrical power. Yeh step kyun? Battery mass ke liye pehle power chahiye. $P_{elec} = \frac{5 \times 9.0 \times 1 0 ^{6}}{1000 \times 0.65 \times 0.92} = \frac{4.5 \times 1 0 ^{4}}{0.598} \approx 7.525 \times 1 0^{4} W .$

Step 4 — battery mass. Yeh step kyun? Energy $= P_{elec} t_{b}$ , phir joules-per-kg se divide karo. $m_{batt} = \frac{7.525 \times 1 0 ^{4} \times 180}{9.0 \times 1 0 ^{5}} \approx 15.0 kg .$

Verify: Energy $= 7.525 \times 1 0^{4} \times 180 \approx 1.35 \times 1 0^{7}$ J $= 1.35 \times 1 0^{7} /3600 \approx 3765$ Wh. 250 Wh/kg pe woh $3765/250 \approx 15.1$ kg hai. ✓ Dono routes agree karte hain.

Ex 3 — Cell C: density ka effect

Worked example Ex 3 — same mass flow, alag liquid

$\overset{m}{˙} = 6 kg/s$ ko $Δ p = 1 0^{7} Pa$ , $η_{p} = 0.6$ , $η_{m} = 0.9$ pe push karo. Liquid oxygen ( $ρ = 1140 kg/m^{3}$ ) vs liquid hydrogen ( $ρ = 71 kg/m^{3}$ ) compare karo. Kaun se pump ko zyada power chahiye?

Forecast: hydrogen fluffy hai (low $ρ$ ). Same mass = bahut zyada volume. Pumping pressure per unit volume raise karta hai, toh guess karo: hydrogen ko vastly zyada power chahiye.

Step 1 — LOX power. Yeh step kyun? Dense case plug karo. $P_{LOX} = \frac{6 \times 1 0 ^{7}}{1140 \times 0.54} \approx 9.75 \times 1 0^{4} W \approx 97.5 kW .$

Step 2 — LH₂ power. Yeh step kyun? Same formula, tiny $ρ$ . $P_{LH2} = \frac{6 \times 1 0 ^{7}}{71 \times 0.54} \approx 1.565 \times 1 0^{6} W \approx 1565 kW .$

Step 3 — ratio. Yeh step kyun? Physics clearly dekho: power $\propto 1/ ρ$ , toh $\frac{P _{LH2}}{P _{LOX}} = \frac{1140}{71} \approx 16.1.$

Verify: Ratio density ratio $1140/71 = 16.06$ ke barabar hai — shared factors ( $\overset{m}{˙}, Δ p, η$ cancel ho jaate hain) se independent hai. ✓ Yahi reason hai kyun hydrogen turbopumps giant power hogs hote hain.

Ex 4 — Cell D: limiting case, long burn

Worked example Ex 4 — burn time badhne par kya hota hai

Fixed $P_{elec} = 75 kW$ , $e_{b} = 9 \times 1 0^{5} J/kg$ . $t_{b} = 30, 150, 600 s$ ke liye $m_{batt}$ tabulate karo.

Forecast: $m_{batt} \propto t_{b}$ , origin se guzarti ek straight line. Time triple/quadruple karne se mass bhi utna hi ho jaata hai.

Step 1 — general form. Yeh step kyun? Time dependence isolate karo: $m_{batt} = \frac{P _{elec}}{e _{b}} t_{b} = \frac{75000}{9 \times 1 0 ^{5}} t_{b} = 0.0833 t_{b}$ (kg, $t_{b}$ seconds mein).

Step 2 — evaluate karo. Yeh step kyun? Table fill karo. $t_{b} = 30 \Rightarrow 2.5 kg; t_{b} = 150 \Rightarrow 12.5 kg; t_{b} = 600 \Rightarrow 50.0 kg .$

Step 3 — limit padho. Yeh step kyun? Jaise $t_{b} \to \infty$ , $m_{batt} \to \infty$ linearly. Koi plateau nahi hai — batteries kabhi "catch up" nahi karti, sirf badhti rehti hain. Yahi mathematical reason hai kyun electric pumps giant, long-burn first stages ke liye ban hain aur short upper-stage burns mein thrive karti hain. Dekho Tsiolkovsky rocket equation: dead battery mass tumhare mass ratio ko hurt karta hai.

Verify: Slope check: $50/600 = 0.0833 = P_{elec} / e_{b}$ . ✓ Perfectly linear, origin-passing.

Ex 5 — Cell E: degenerate & zero inputs

Worked example Ex 5 — boundary cases jinhe formula ko survive karna chahiye

Har extreme pe master equations evaluate karo. $Δ p = 1 0^{7}$ , $ρ = 1000$ , $t_{b} = 100$ , $e_{b} = 9 \times 1 0^{5}$ jahan relevant ho.

Forecast: zeros ko zero power/mass pe collapse karna chahiye; efficiency $\to 1$ ko $P_{elec}$ ko $P_{ideal}$ tak collapse karna chahiye; efficiency $\to 0$ ko blow up karna chahiye. Apna intuition check karo.

Step 1 — no flow, $\overset{m}{˙} = 0$ . Yeh step kyun? Agar kuch move nahi karta, pump koi kaam nahi karta. $P_{elec} = \frac{0 \cdot Δ p}{ρ η _{p} η _{m}} = 0 W .$ Physically: motor idle hai, battery kuch drain nahi karta. ✓

Step 2 — no pressure rise, $Δ p = 0$ . Yeh step kyun? Agar exit pressure inlet ke barabar hai, toh energy per unit volume add nahi hoti. $P_{elec} = \frac{m ˙ \cdot 0}{\dots} = 0 W .$ Free flow ko koi pump work nahi chahiye (friction ignore karke). Yeh Bernoulli's principle baseline hai: pressure rise hi energy term hai.

Step 3 — perfect machine, $η_{p} = η_{m} = 1$ . Yeh step kyun? Sanity check: loss factor vanish ho jaana chahiye. $\overset{m}{˙} = 4$ ke saath: $P_{elec} = \frac{4 \times 1 0 ^{7}}{1000 \times 1 \times 1} = 4 \times 1 0^{4} = 40 kW = P_{ideal}$ . ✓ Electrical demand ideal floor tak gir jaata hai.

Step 4 — dying motor, $η_{m} \to 0$ . Yeh step kyun? Singularity samjho. $P_{elec} = \frac{m ˙ Δ p}{ρ η _{p} η _{m}} \to \infty$ . Ek useless motor infinite battery power demand karega — physically woh pump ko bilkul spin nahi kar sakta. Formula ka blow-up wahi warning hai. ✓

Verify: Dono zero-cases $0$ dete hain; $η = 1$ $P_{ideal}$ recover karta hai; $η \to 0$ diverge karta hai. Chaaron physical expectation se match karte hain.

Ex 6 — Cell F: real-world word problem

Worked example Ex 6 — Rutherford-class stage with drop-batteries

Ek stage 9 pumps chalata hai (LOX + RP-1 split), total propellant flow $\overset{m}{˙}_{tot} = 45 kg/s$ , $Δ p = 100 bar$ , effective $ρ = 1030 kg/m^{3}$ (mixture), $η_{p} = 0.6$ , $η_{m} = 0.9$ , burn $t_{b} = 160 s$ , $e_{b} = 0.95 \times 1 0^{6} J/kg$ . Aadhi batteries burn midpoint pe drop ho jaati hain. Nikalo (a) total electrical power, (b) installed battery mass, (c) battery mass jo burnout tak carry hoti hai. Rocket Lab Rutherford engine se compare karo.

Forecast: power hundreds of kW mein; installed batteries ~tens of kg; mid-burn pe aadhi drained cells drop karne se carried mass noticeably kam hoti hai.

Step 1 — total power. Yeh step kyun? Saare pumps ko ek aggregate flow maano. $P_{elec} = \frac{45 \times 1 0 ^{7}}{1030 \times 0.6 \times 0.9} = \frac{4.5 \times 1 0 ^{9}}{556.2} \approx 8.09 \times 1 0^{5} W \approx 809 kW .$

Step 2 — installed battery mass (full burn). Yeh step kyun? Tumhe physically itna physically carry karna hai ki poore 160 s power mile. $m_{batt} = \frac{P _{elec} t _{b}}{e _{b}} = \frac{8.09 \times 1 0 ^{5} \times 160}{0.95 \times 1 0 ^{6}} \approx 136.2 kg .$

Step 3 — mid-burn drop ke saath carried-to-burnout mass. Yeh step kyun? Model: first-half cells (pack ka aadha) $t = 80$ s tak deplete ho jaate hain aur jettison ho jaate hain; sirf second-half pack ( $\approx 68.1$ kg) burnout pe aboard hoti hai. $m_{burnout} \approx \frac{1}{2} \times 136.2 \approx 68.1 kg .$

Verify: Energy check: $E = 8.09 \times 1 0^{5} \times 160 \approx 1.294 \times 1 0^{8}$ J. Mass $= 1.294 \times 1 0^{8} /0.95 \times 1 0^{6} \approx 136.2$ kg. ✓ Dead cells drop karna exactly woh Rutherford trick hai jo burnout mass ratio ko favourable rakhti hai — dekho Tsiolkovsky rocket equation.

Ex 7 — Cell G: inverse solve

Worked example Ex 7 — tumhare paas battery budget hai, max burn nikalo

Engineering constraint: sirf $m_{batt} = 20 kg$ cells fit hoti hain. $\overset{m}{˙} = 5 kg/s$ , $Δ p = 95 bar$ , $ρ = 1000$ , $η_{p} = 0.62$ , $η_{m} = 0.90$ , $e_{b} = 0.9 \times 1 0^{6} J/kg$ ke saath, sabse lamba burn $t_{b}$ kya hai jo tum power kar sakte ho?

Forecast: mass formula ko $t_{b}$ ke liye rearrange karo; couple of hundred seconds expect karo.

Step 1 — formula ko $t_{b}$ ke liye solve karo. Yeh step kyun? $m_{batt} = \frac{m ˙ Δ p t _{b}}{ρ η _{p} η _{m} e _{b}}$ invert karo: $t_{b} = \frac{m _{batt} ρ η _{p} η _{m} e _{b}}{m ˙ Δ p} .$

Step 2 — numbers plug karo ( $Δ p = 9.5 \times 1 0^{6}$ Pa). Yeh step kyun? Direct evaluation. $t_{b} = \frac{20 \times 1000 \times 0.62 \times 0.90 \times 0.9 \times 1 0 ^{6}}{5 \times 9.5 \times 1 0 ^{6}} = \frac{1.0044 \times 1 0 ^{10}}{4.75 \times 1 0 ^{7}} \approx 211.5 s .$

Verify: $t_{b} = 211.5$ ke saath forward-check: $P_{elec} = \frac{5 \times 9.5 \times 1 0 ^{6}}{1000 \times 0.62 \times 0.9} \approx 8.512 \times 1 0^{4}$ W; $m = \frac{8.512 \times 1 0 ^{4} \times 211.5}{0.9 \times 1 0 ^{6}} \approx 20.0$ kg. ✓ Budget pe round-trip hota hai.

Ex 8 — Cell H: exam twist — electric vs pressure-fed break-even

Worked example Ex 8 — pump kab worth it nahi rehta?

Ek Pressure-fed cycle tank jo propellant volume $V$ ko chamber pressure $p_{c}$ pe store karta hai, uski wall mass roughly $m_{wall} = k p_{c} V$ hai jahan $k = 6 \times 1 0^{- 4} kg/(Pa\cdotm^{3})$ (thin-shell scaling). Ek electric pump tank ko 2 bar tak drop karne deta hai lekin motor+battery mass $m_{add}$ add karta hai. $Δ p = 100 bar$ , $V = 1.5 m^{3}$ , aur electric add-on $m_{add} = 45 kg$ ke liye, kya electric pump jeet ta hai?

Forecast: high-pressure tank mass $k p_{c} V$ hai jahan $p_{c} = 1 0^{7}$ Pa — woh ek bada number hai. Guess: pump easily jeet ta hai.

Step 1 — pressure-fed wall mass. Yeh step kyun? Tank ko poora $1 0^{7}$ Pa hold karna hai. $m_{wall,PF} = 6 \times 1 0^{- 4} \times 1 0^{7} \times 1.5 = 9000 kg .$

Step 2 — pump-fed wall mass. Yeh step kyun? Tank ab sirf 2 bar $= 2 \times 1 0^{5}$ Pa hold karta hai. $m_{wall,EP} = 6 \times 1 0^{- 4} \times 2 \times 1 0^{5} \times 1.5 = 180 kg .$

Step 3 — total comparison. Yeh step kyun? Electric penalty ko light tank mein add karo. $m_{EP,total} = 180 + 45 = 225 kg vs m_{PF} = 9000 kg .$ Electric $8775$ kg se jeetta hai — ek landslide. Tank-wall term sab par dominate karta hai.

Step 4 — twist: break-even $V$ nikalo. Yeh step kyun? Exams poochhte hain "kab pressure-fed lighter hogi?" Set karo $k p_{c} V + 0 = k (2 \times 1 0^{5}) V + 45$ aur solve karo — lekin note karo $k p_{c} V$ hamesha low-pressure tank se zyada hogi, toh break-even ke liye $m_{add}$ ko saving se exceed karna hoga: $k (p_{c} - 2 \times 1 0^{5}) V < m_{add} \Rightarrow V < \frac{45}{6 \times 1 0 ^{- 4} \times 9.8 \times 1 0 ^{6}} \approx 7.65 \times 1 0^{- 3} m^{3} .$ Sirf bahut chhote propellant volumes (< 7.65 litres) ke liye fixed 45 kg add-on tank savings se zyada hota hai.

Verify: $V = 7.65 \times 1 0^{- 3}$ pe: saving $= 6 \times 1 0^{- 4} \times 9.8 \times 1 0^{6} \times 7.65 \times 1 0^{- 3} \approx 45.0$ kg $= m_{add}$ . ✓ Break-even confirm; is se upar pump hamesha jeet ta hai.

Recall Self-test: yeh kaun sa cell hai?

Ek problem $Δ p$ bar mein, $e_{b}$ Wh/kg mein deti hai, aur battery mass maangti hai. Kaun sa cell, aur pehla move kya hai? Answer ::: Cell B (unit trap). Formula chhone se pehle $Δ p \to$ Pa ( $\times 1 0^{5}$ ) aur $e_{b} \to$ J/kg ( $\times 3600$ ) convert karo.

Parent topic pe wapas jao parent topic · related: Specific impulse (Isp), Turbopump-fed cycle (gas generator vs staged combustion).