Visual walkthrough — Electric pump-fed cycle — modern innovation

3.3.26 · D2 · Physics › Rocket Propulsion › Electric pump-fed cycle — modern innovation

Hum parent Rocket Propulsion se link karte hain, aur raaste mein Bernoulli's principle, Battery specific energy, aur Rocket Lab Rutherford engine se bhi.

Step 1 — "Pressure" asal mein hai kya?

WHAT. Kisi bhi formula se pehle, hum us ek word ko pin down karte hain jis par sab kuch tika hai: pressure.

WHY. Poori derivation kehti hai "pump fluid mein energy add karta hai." Iska use karne ke liye humein pata hona chahiye ki pressure hai kya — energy terms mein. Agar hum yeh skip kar dein, toh $\Delta p$ sirf ek mystery letter ban ke reh jaata hai.

PICTURE. Figure dekho. Area $A$ ka ek piston liquid ko force $F$ se push kar raha hai. Pressure wo force hai jo us area par spread hoti hai:

$p=\underset{\text{push, newtons}}{\underbrace{F}}÷\underset{{\text{area, m}}^2}{\underbrace{A}}$

Ab piston ko thodi si distance $d$ push karo. Kiya gaya work (force × distance) $F\cdot d$ hai. Swept volume $A\cdot d$ hai. Work ko us volume se divide karo:

$\frac{Fd}{Ad}=\frac{F}{A}=p$

Step 2 — Pump pressure ko $\Delta p$ se raise karta hai

WHAT. Ek pump liquid ko low tank pressure $p_{\text{tank}}$ par leta hai aur use high pump-exit pressure $p_{\text{exit}}$ par deliver karta hai. Energy ki cost sirf rise par aati hai:

$\Delta p=\underset{\text{pump ke baad, high}}{\underbrace{p_{\text{exit}}}}-\underset{\text{pump se pehle, low}}{\underbrace{p_{\text{tank}}}}$

WHY. Tank ne $p_{\text{tank}}$ already free mein provide kar diya. Pump sirf extra ke liye pay karta hai — yani difference ke liye. Poora $p_{\text{exit}}$ use karna free tank pressure ko double-count karna hoga. Yeh "difference hi matter karta hai" wala idea wohi hai jo Bernoulli's principle mein bhi hai, jahan sirf pressure differences fluid ko push karti hain.

PICTURE. Red column exit pressure hai, chhota violet column tank pressure hai, aur unke beech ka orange gap $\Delta p$ hai — sirf yahi woh part hai jiske liye motor pay karta hai.

Step 3 — Ek blob se steady stream tak (power)

WHAT. Ek rocket ek blob pump nahi karta — woh propellant ki ek continuous river pump karta hai. Isliye hum power maangte hain: energy per second, blob per energy nahi.

WHY. Thrust continuous hai; motor continuously run karta hai. Power (watts) woh hai jisme actually motor aur battery ki rating hoti hai. Hum "energy per blob" ko "energy per second" mein convert karte hain, har blob ko lagne waale time $dt$ se divide karke.

PICTURE. Pipe dekho: ek second mein, volume $\dot{V}$ (the volume flow rate, m³/s) ka liquid ka ek slug slide karta hai. Har aisa volume $\Delta p\cdot dV$ energy carry karta tha, toh per second:

$P_{\text{ideal}}=\frac{dW}{dt}=\frac{\Delta pdV}{dt}=\Delta p\underset{{\text{m}}^3\text{/s}}{\underbrace{\dot{V}}}$

Step 4 — Density $\rho$ kyun aa jaati hai (woh term jo sabse zyada bhulte hain)

WHAT. Engineers propellant flow ko kilograms per second ($\dot{m}$) mein quote karte hain, cubic metres per second mein nahi, kyunki thrust depend karta hai bahar phenko gaye mass par. Toh hum $\dot{V}$ ko $\dot{m}$ se replace karte hain.

WHY. Humein "kitne m³ per second" aur "kitne kg per second" ke beech ek bridge chahiye. Woh bridge hai density $\rho$ — har cubic metre mein packed kilograms:

$\rho =\frac{\text{mass}}{\text{volume}}\Rightarrow \dot{m}=\rho \dot{V}\Rightarrow \dot{V}=\frac{\dot{m}}{\rho }$

PICTURE. Equal volume ke do cubes: dense liquid (heavy, bahut kg) vs light liquid (kam kg). Same $\dot{V}$, alag $\dot{m}$. Figure ek heavy metre-cube aur ek light wala balance par dikhata hai.

$\dot{V}=\dot{m}/\rho$ ko Step 3 mein substitute karo:

Step 5 — Real pumps aur motors thoda lose karte hain (efficiencies)

WHAT. Do "leak factors" insert karo: pump efficiency ${\eta }_p$ aur motor efficiency ${\eta }_m$, dono 0 aur 1 ke beech ke numbers.

WHY. Real pumps fluid ko heat karte hain aur bearings mein rub hoti hai; real motors aur controllers heat waste karte hain. Fluid ko $P_{\text{ideal}}$ deliver karne ke liye tumhe zyada supply karna padta hai. 1 se kam number se divide karna demand ko bada karta hai — bilkul wahi jo ek loss ko karna chahiye.

PICTURE. Ek Sankey-style flow: battery power wide enter karti hai, ek slice motor par lose hoti hai (${\eta }_m$), doosri slice pump par (${\eta }_p$), aur narrow survivor fluid mein useful $P_{\text{ideal}}$ hai.

Step 6 — Power energy banti hai, energy kilograms banti hai

WHAT. Poore burn time $t_b$ tak motor chalao. Energy = power × time. Phir energy ko battery mass mein convert karo specific energy $e_b$ (joules stored per kg) use karke.

WHY. Battery ki rating specific energy $e_b$ (J/kg) mein hoti hai. Mass = total joules ÷ joules-per-kg. Yahi woh step hai jahan time enter karta hai — aur time hi hai jo lambe burns ko doom karta hai.

PICTURE. Ek staircase: watts → (× $t_b$) → joules → (÷ $e_b$) → kilograms. Har tread labelled hai.

$E=P_{\text{elec}}{t}_b,m_{\text{batt}}=\frac{E}{e_b}$

Step 7 — Edge cases (reader ko kabhi stranded mat chodo)

WHAT. Extreme knobs check karo taaki koi bhi scenario tumhe surprise na kare.

WHY. Jo formula tum trust karte ho woh apne limits par sane behave karna chahiye, sirf "nice" numbers par nahi.

PICTURE. Char mini-panels, har ek ek knob ko extreme par le jaata hai aur dikhata hai ki $m_{\text{batt}}$ kahan jaata hai.

• $\Delta p\to 0$ (tank already chamber pressure par hai — ek Pressure-fed cycle): $P_{\text{ideal}}\to 0$, $m_{\text{batt}}\to 0$. Koi pumping work nahi chahiye. Formula agree karta hai: pumps tabhi energy cost karte hain jab woh pressure raise karte hain.
• $t_b\to \infty$ (bada booster, lamba burn): $m_{\text{batt}}\to \infty$. Batteries vehicle ko swamp kar deti hain — exactly isliye turbopump boosters ke liye jeet jaata hai.
• $e_b$ tiny (kamzor 2000s battery): $m_{\text{batt}}$ blow up ho jaata hai. Yahi historical reason hai ki electric pumps ~250 Wh/kg lithium cells aane tak real nahi bane.
• ${\eta }_p,{\eta }_m\to 1$ (ideal machines): demand $P_{\text{ideal}}$ tak collapse ho jaati hai; $m_{\text{batt}}$ apna floor hit karta hai. Tum kabhi is floor ko beat nahi kar sakte, sirf approach kar sakte ho.

Ek-picture summary

Liquid ka ek blob → pressure hai energy/volume ($p$) → pump har blob mein $\Delta p$ add karta hai → ise steadily flow karo ($\dot{V}$) → density se rewrite karo ($\dot{V}=\dot{m}/\rho$) → losses account karo (${\eta }_p,{\eta }_m$) → $t_b$ tak chalao → $e_b$ se weigh karo. Do boxes nikalte hain:

$P_{\text{elec}}=\frac{\dot{m}\Delta p}{\rho {\eta }_p{\eta }_m},m_{\text{batt}}=\frac{\dot{m}\Delta p{t}_b}{\rho {\eta }_p{\eta }_m{e}_b}$

Quick reveals: