Pressure-fed cycle — simplest, used in upper stages: Visual walkthrough

3.3.25 · D2 · Physics › Rocket Propulsion › Pressure-fed cycle — simplest, used in upper stages

Step 1 — Fluid hamesha high pressure se low pressure ki taraf bahta hai

KYA HAI. Do connected containers ko side by side rakho, ek high pressure par, ek low pressure par. Valve kholo. Fluid hamesha high side se low side ki taraf bhagta hai — kabhi ulti taraf nahi.

KYUN. Pressure difference hi push hai. Agar dono sides barabar push karein, toh pushes cancel ho jaate hain aur kuch nahi hota. Agar ek side zyada push kare, toh net push fluid ko kamzor side ki taraf le jaata hai. Yeh ek fact poori derivation ki neenv hai.

PICTURE. Figure mein, blue arrow fluid ko high-pressure box (left) se low-pressure box (right) ki taraf jaate dikhata hai. Pressures ulte karo toh arrow bhi palat jaata hai.

Figure — Pressure-fed cycle — simplest, used in upper stages

Step 2 — Chamber finish line par ek high-pressure dushman hai

KYA HAI. Combustion chamber ke andar propellants jalte hain, pressure $p_{c}$ par hot gas banti hai ("c" for chamber). Yeh gas har direction mein bahar push karti hai — injector holes se vapas upar tanks ki taraf bhi.

KYUN. Step 1 se, fluid high se low ki taraf bahta hai. Agar chamber pressure $p_{c}$ daakhil hone ki koshish kar rahe liquid ke pressure se zyada ho, toh chamber jeet jaata hai aur propellant ko vapas bahar blow kar deta hai. Chamber koi friendly destination nahi hai — woh aapko pushback de raha hai.

PICTURE. Orange arrows hot gas pressure $p_{c}$ ko bahar push karte dikhate hain. Injector hole par chhota sa upar wala red arrow khatre ki nishani hai: chamber gas feed line mein ulti taraf jaane ki koshish kar raha hai.

Step 3 — Injector ek deliberate wall hai jise tumhe cross karna hoga

KYA HAI. Chamber se bilkul pehle, propellant ko injector se force kiya jaata hai aur pressure $Δ p_{inj}$ se drop ho jaata hai, jo typically $p_{c}$ ka quarter hota hai.

KYUN. Yeh drop galti nahi hai — yeh chahiye hoti hai. Holes ke across bada pressure drop liquid ko fast bahar shoot karta hai aur droplets mein tod deta hai (atomize), jo usse cleanly jalaane deta hai. Lekin iska matlab hai ki liquid ko injector par $p_{c} + Δ p_{inj}$ lekar pahunchna hoga, taaki drop ke baad bhi woh chamber se match kare — aur usse beat kare.

PICTURE. Orange step ki height $Δ p_{inj}$ hai. Liquid step ke upar se shuru hota hai aur uske paas itna pressure hona chahiye ki woh step se neeche gire aur phir bhi $p_{c}$ ke upar land kare.

Step 4 — Pipes aur cooling jacket thoda aur pressure chura lete hain

KYA HAI. Tank aur injector ke beech propellant friction se pressure ke do aur chhote chunks khota hai.

KYUN. Har rough pipe wall aur har bend flow ko resist karta hai — yeh woh real energy hai jo heat mein jaati hai, bilkul Bernoulli's equation ka "losses" term. Inhe bhi pehle se pay karna hoga, warna liquid ke paas injector tak pahunchne par itna pressure nahi bachega.

PICTURE. Staircase bada hota hai: har naya step (green = feed, gray = cooling) ek aur chunk hai jo tank ko apne neeche ki sab cheez ke upar supply karna hoga.

Step 5 — Bernoulli: poore pressure staircase ki bookkeeping

KYA HAI. Bernoulli likho tank surface (top) se chamber (bottom) tak: $p_{tank} + \frac{1}{2} ρ v_{tank}^{2} + ρ g h_{tank} = p_{c} + \frac{1}{2} ρ v_{c}^{2} + ρ g h_{c} + losses (Δ p_{inj} + Δ p_{feed} + Δ p_{cool})$

KYUN har term.

$p_{tank}$ ::: woh pressure jo pressurant gas supply karta hai — woh cheez jo hum find karna chahte hain.
$\frac{1}{2} ρ v_{tank}^{2}$ ::: tank top par motion energy. Tank surface barely hilti hai, isliye $v_{tank} \approx 0$ — yeh term drop ho jaata hai.
$ρ g h$ ::: height (gravity) energy. Coasting upper stages mein rocket almost weightless hota hai, isliye gravity head $\approx 0$ — yeh bhi cancel ho jaate hain.
$\frac{1}{2} ρ v_{c}^{2}$ ::: liquid thodi speed zaroor gain karta hai, lekin injector toll ke muqable mein bahut tiny hai, isliye hum ise losses mein fold kar dete hain.
the losses ::: bilkul Steps 3–4 ke teen steps.

PICTURE. Figure tank (top) se chamber (bottom) tak poora staircase line up karta hai, jisme $v_{tank} \approx 0$ aur $ρ g h \approx 0$ gray mein cross out hain.

Step 6 — Terms collect karo: central result

KYA HAI. Step 5 se vanishing terms cross out karo aur jo bacha use gather karo. Gravity aur tank-velocity terms gone hain; chamber-velocity term chhota hai aur absorb ho gaya. Jo bacha woh hai:

$p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed} + Δ p_{cool}$

KYUN. Ise ek bill ki tarah padho: $p_{c}$ par chamber mein liquid deliver karne ke liye, tank ko chamber pressure khud plus har downstream toll pay karna hoga. Kuch free nahi hai; tank akela payer hai.

PICTURE. Equation stacked bars ke roop mein dikhaya gaya hai: total tank bar $p_{c}$ ke barabar hai plus har loss bar upar pila hua. Numbers mein, $Δ p_{inj} \approx 0.25 p_{c}$ aur chhote doosre losses ke saath, yeh roughly $1.3$ – $1.5 p_{c}$ tak stack hota hai.

Step 7 — Degenerate cases (extremes par kya hota hai)

Har honest derivation ko apne edge cases se survive karna chahiye. Yahan teen hain jo matter karte hain.

Case A — Equal pressures ( $p_{tank} = p_{c}$ ). Pressure difference zero hai, flow ruk jaata hai, injector atomize nahi kar sakta: ek dead engine. Isliye boxed equation ke right side par $p_{c}$ akele se strictly zyada hai.

Case B — Tank low ho jaaye ( $p_{tank} < p_{c}$ ). Ab chamber Step 1 ki tug-of-war jeet jaata hai: hot gas injector se ulti taraf feed line mein push karta hai — ek dangerous burn-back. Figure mein red arrow palat jaata hai.

Case C — Thrust ke neeche vs. coasting (woh $ρ g h$ term). Firing ke dauran rocket accelerate karta hai, isliye liquid ka weight help karta hai ise neeche push karne mein ( $p_{tank}$ ke effect mein add hota hai); weightless coast mein yeh help gaayab ho jaati hai. Isliye humne upper stages ke liye $ρ g h$ drop kiya — lekin ek tall, high-acceleration stage ke liye tum ise rakhoge.

PICTURE. Teen mini-panels: (A) equal bars, koi arrow nahi; (B) tank bar chhota, red arrow ulti taraf; (C) coast vs. thrust jisme gravity term appear/vanish hota dikhata hai.

The one-picture summary

Upar ki sab kuch, ek single staircase mein compress karke: chamber floor ( $p_{c}$ ) se shuru karo, har loss step chadho, aur jis height par tumhe khana hai woh $p_{tank}$ hai. Chamber ko poore stack se beat karo, warna kuch nahi bahega.

Recall Poora walkthrough plain words mein (Feynman retelling)

Paani sirf neeche bahta hai — zyada push se kam push ki taraf. Combustion chamber ek bully hai jo neeche $p_{c}$ pressure par pushback de raha hai. Liquid ko andar karne ke liye, tank ko chamber se zyada push karna padta hai. Lekin tank aur chamber ke beech tolls hain: injector jaanboojhkar liquid ko tiny holes se squeeze karta hai (taaki woh fine mist mein spray ho jo achhe se jale) aur isme pressure lagta hai; pipes aur cooling jacket friction se thoda aur rub kar dete hain. Isliye tank ko chamber pressure aur upar stack kiye saare tolls pay karne padte hain. Still tank surface se chamber tak Bernoulli ki energy-bookkeeping likhne par, "tank kitni tezi se chal raha hai" wala term zero ho jaata hai aur (coasting space stage ke liye) gravity term cancel ho jaata hai, aur exactly $p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed} + Δ p_{cool}$ bach jaata hai. Jodo aur tank roughly chamber pressure ka $1.3$ – $1.5$ guna ho jaata hai. Woh extra squeeze hi poori wajah hai ki pressure-fed tanks heavy hote hain — aur isliye yeh trick low-pressure upper stages ke liye bachaayi jaati hai.

Recall Quick self-check

$p_{tank}$ sirf $p_{c}$ ke barabar kyun nahi ho sakta? ::: Equal pressures ka matlab zero pressure difference hai, isliye koi flow nahi aur koi atomization nahi — ek dead engine. Woh kaunsa term tha jo humne drop kiya kyunki tank surface barely hilti hai? ::: Tank motion energy $\frac{1}{2} ρ v_{tank}^{2}$ . Weightless coast mein kaunsa term vanish ho jaata hai? ::: Gravity head $ρ g h$ . $Δ p_{inj}$ kis kaam ke liye hai? ::: Yeh ek deliberate drop hai jo propellant ko ek fine, well-burning spray mein atomize karta hai.