Pressure-fed cycle — simplest, used in upper stages: Worked examples

3.3.25 · D3 · Physics › Rocket Propulsion › Pressure-fed cycle — simplest, used in upper stages

Yahan sab kuch parent ke teen ideas par depend karta hai, jo dobara state kiye ja rahe hain taaki koi bhi symbol earn kiye bina use na ho:

The scenario matrix

Is topic ke har possible question ka jawab inhi cells mein se kisi mein hai. Neeche har example uss cell ke saath tagged hai jo woh fill karta hai.

Cell	Case class	Kya special hai	Example
A	Baseline budget	Saare losses positive, normal	Ex 1
B	Sign flip — gravity term	Tank chamber ke neeche ( $- ρ g h$ vs $+ ρ g h$ )	Ex 2
C	Zero / degenerate input	$Δ p_{inj} \to 0$ : kya toot jaata hai?	Ex 3
D	Limiting value	$p_{tank} \to p_{c}$ : flow rate $\to 0$	Ex 4
E	Mass-penalty scaling	double $p_{c}$ , mass ka kya hota hai?	Ex 5
F	Real-world word problem	poora apogee-kick stage sizing	Ex 6
G	Exam twist — units trap	mixed bar / Pa / MPa	Ex 7
H	Blowdown (no regulator)	pressure girta hai jaise tank khalta hai	Ex 8

Cell A — Baseline pressure budget

Worked example Example 1 — roz ka case

Chamber pressure $p_{c} = 12$ bar. Injector deliberately $30%$ of $p_{c}$ khaata hai. Pipe friction $0.8$ bar khaati hai, cooling jacket $0.5$ bar. Tank chamber ke barabar height par hai (gravity ignore karo). $p_{tank}$ nikalo.

Forecast: padhne se pehle guess karo — kya answer $12$ , $16$ , ya $20$ bar ke kareeb hoga?

$Δ p_{inj} = 0.30 \times 12 = 3.6$ bar. Yeh step kyun? Injector drop chamber pressure ka fraction diya gaya hai, isliye hum $p_{c}$ ko scale karte hain.
Har downstream loss ko $p_{c}$ mein add karo: $p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed} + Δ p_{cool} = 12 + 3.6 + 0.8 + 0.5.$ Yeh step kyun? Flow sirf pressure mein "downhill" chalta hai. Har downstream loss tank ko pehle se pay karni padti hai.
$p_{tank} = 16.9$ bar. Yeh step kyun? Step 2 ka sum complete karte hain — yahi woh total pressure hai jo tank ko supply karni chahiye, yahi hamaara answer hai.

Verify: ratio $16.9/12 = 1.41$ , $1.3$ – $1.5$ rule of thumb ke bilkul andar. Units: saare terms bar mein hain, toh sum bhi bar mein hai. ✔

Cell B — Gravity term mein sign flip

Ab woh term wapas laate hain jise baseline ne ignore kiya tha. Launch pad par, liquid column ka weight hota hai; woh weight feed ko help ya hurt kar sakta hai depending on whether tank chamber ke upar hai ya neeche.

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

Worked example Example 2 — gravity pehle help karti hai, phir hurt

Example 1 ka total loss lo ( $p_{tank} = 16.9$ bar bina gravity ke). Ab propellant dense storable hai ( $ρ = 1200 kg/m^{3}$ ) $3$ m column ke saath, aur rocket pad par hai ( $g = 9.81 m/s^{2}$ ). $p_{tank}$ nikalo (a) tank chamber ke upar, (b) tank chamber ke neeche.

Forecast: ek answer $16.9$ bar se neeche hoga, ek upar. Kaun sa kaun sa hai?

Gravity head $ρ g h = 1200 \times 9.81 \times 3 = 35, 316 Pa = 0.353$ bar. Yeh step kyun? Column ka pressure wahi units (bar) mein chahiye jo baaki sab mein hai. $1 bar = 1 0^{5} Pa$ .
Tank upar — head subtract karo (gravity help karta hai): $p_{tank} = 16.9 - 0.353 = 16.55 bar .$ Yeh step kyun? Master equation $- ρ g h$ carry karta hai jab column ka weight flow assist kare, toh base $16.9$ bar se head hata dete hain.
Tank neeche — head add karo (gravity hurt karta hai): $p_{tank} = 16.9 + 0.353 = 17.25 bar .$ Yeh step kyun? Ab sign $+ ρ g h$ hai: tank ko column bhi lift karna hai, toh head base $16.9$ bar ke upar add ho jaata hai.

Verify: dono mein difference exactly $2 \times 0.353 = 0.71$ bar hai — head ka double, jaise pure sign flip se expect karte hain. Space mein ( $g \to 0$ ) dono $16.9$ bar par wapas aa jaate hain. ✔

Cell C — Degenerate input: agar injector drop khatam ho jaaye?

Worked example Example 3 —

$Δ p_{inj} \to 0$ Maano ek naive designer injector restriction hata deta hai toh $Δ p_{inj} = 0$ , baaki same rakhta hai $p_{c} = 12$ bar, feed $0.8$ , cooling $0.5$ bar. $p_{tank}$ compute karo aur explain karo ki engine ab broken kyun hai jabki maths ek valid number de raha hai.

Forecast: number better lagta hai (lower tank pressure). Kya yeh acha hai?

$p_{tank} = 12 + 0 + 0.8 + 0.5 = 13.3$ bar. Yeh step kyun? Wahi master equation, injector term zero set karke.
Physically interpret karo: bina injector pressure drop ke, liquid ek slow, coherent jet ke roop mein enter karta hai. Yeh step kyun? Injector drop wahi cheez hai jo liquid ko accelerate karke fine droplets mein tod deta hai (atomization). Drop nahi → atomization nahi → poor, unstable combustion.

Verify: arithmetic theek hai ( $13.3$ bar), lekin degenerate case ek trap hai: ek chhota tank pressure jo ek aisa engine produce karta hai jo cleanly burn nahi karega. Yeh limit batata hai ki injector drop ek feature hai, koi minimize karne wala loss nahi. ✔

Cell D — Limiting value: $p_{tank} \to p_{c}$

Yahi woh edge hai jiske baare mein parent par pehla [!mistake] chetaavni deta tha. Aao ise Bernoulli's equation use karke quantitative banate hain injector ke liye.

Worked example Example 4 — dead-engine limit ki taraf jaana

Storable propellant $ρ = 1200 kg/m^{3}$ . Jet speed $v$ nikalo jab injector drop hai (a) $3$ bar, (b) $0.3$ bar, (c) $0$ bar. Trend dekho jaise $p_{tank} \to p_{c}$ .

Forecast: jab drop $10 \times$ chhota hota hai, kya speed bhi $10 \times$ chhoti ho jaati hai?

(a) $v = 2 (3 \times 1 0^{5}) /1200 = 500 = 22.4 m/s$ . Yeh step kyun? $3$ bar ko $3 \times 1 0^{5}$ Pa mein convert karo, Bernoulli mein plug karo.
(b) $v = 2 (0.3 \times 1 0^{5}) /1200 = 50 = 7.07 m/s$ . Yeh step kyun? Same formula, $10 \times$ chhota drop — hum test karte hain ki drop chhota hone par speed kaise respond karti hai.
(c) $Δ p_{inj} = 0 \Rightarrow v = 0$ : bilkul flow nahi. Yeh step kyun? Hum limit $Δ p_{inj} \to 0$ lete hain taaki dead-engine endpoint dekh sakein: zero drop force karta hai $v = 0$ .

Verify: $Δ p$ mein $10 \times$ drop se speed sirf $22.4/7.07 = 3.16 \approx 10$ times slower hua — square-root law, confirmed. Limit par $p_{tank} = p_{c}$ drop zero ho jaata hai aur $v \to 0$ : engine dead hai, parent ki warning se match karta hai. ✔

Cell E — Mass-penalty scaling

Worked example Example 5 — chamber pressure double karo, tank mass ka kya hota hai?

Aluminium sphere: $ρ_{m} = 2700 kg/m^{3}$ , yield $σ = 300 MPa$ , propellant volume $V = 1.5 m^{3}$ . Tank $p_{tank} = 1.4 p_{c}$ par run karta hai. $p_{c} = 15$ bar aur $p_{c} = 30$ bar par tank wall mass compare karo.

Forecast: $p_{c}$ double karo → mass kitne factor se badhega…?

$p_{c} = 15$ bar par: $p_{tank} = 1.4 \times 15 = 21$ bar $= 2.1 \times 1 0^{6}$ Pa. Yeh step kyun? Tank chamber pressure ka $1.4 \times$ hold karta hai (parent rule).
$m = \frac{3 ρ _{m}}{σ} p_{tank} V = \frac{3 ( 2700 )}{3 \times 1 0 ^{8}} (2.1 \times 1 0^{6}) (1.5) = 85.05 kg .$ Yeh step kyun? Hoop-stress-derived mass law; mass pressure ke saath linear hai.
$p_{c} = 30$ bar par: $p_{tank} = 42$ bar $= 4.2 \times 1 0^{6}$ Pa, giving $m = 170.1 kg$ . Yeh step kyun? Same law; sirf pressure double hua.

Verify: $170.1/85.05 = 2.00$ exactly — mass linearly pressure ke saath scale karta hai. Yahi wajah hai ki pressure-fed engines high $p_{c}$ par chalane se inkaar karte hain: tank mass pressure ke saath one-to-one track karta hai. ✔

Cell F — Real-world word problem

Worked example Example 6 — ek apogee-kick stage size karna

Ek satellite ka apogee engine storable propellant jalata hai. Chamber pressure $p_{c} = 8$ bar, injector drop $25%$ , feed+cooling losses $0.6$ bar, vacuum mein (no gravity term). Propellant volume $V = 0.9 m^{3}$ . Pressurant helium hai ( $M = 4 g/mol$ ) $T = 300 K$ par. (a) $p_{tank}$ , (b) helium mass jo drained tank fill karne ke liye end mein chahiye — nikalo.

Forecast: helium mass kuch kilograms hogi — $3$ kg se zyada ya kam?

$Δ p_{inj} = 0.25 \times 8 = 2$ bar; $p_{tank} = 8 + 2 + 0.6 = 10.6$ bar. Yeh step kyun? Master equation; vacuum matlab $ρ g h$ nahi.
$V$ ko $p_{tank}$ par fill karne ke liye helium moles: $n = \frac{p V}{R T} = \frac{( 1.06 \times 1 0 ^{6} ) ( 0.9 )}{8.314 \times 300}$ . Yeh step kyun? Ideal gas law $p V = n R T$ ko $n$ ke liye rearrange kiya; gas jo tank ke andar end up hoti hai woh $V$ occupy karti hai $p_{tank}$ , $T$ par, aur $R$ wahi constant hai jo abhi define kiya.
$n = \frac{954000}{2494.2} = 382.5 mol$ , toh $m = n M = 382.5 \times 0.004 = 1.53 kg$ . Yeh step kyun? Moles ko kilograms mein convert karo molar mass $M$ se (grams per mole → kg per mole matlab $\times 0.001$ ).

Verify: $10.6/8 = 1.33$ , $1.3$ – $1.5$ band mein. Helium mass $1.53$ kg propellant ka ek chhota fraction hai — pressurant "cheap" hai, jaise parent ne claim kiya tha. (Real designs iska kai × zyada carry karte hain kyunki gas storage bottle mein bhi rehti hai.) ✔

Cell G — Exam twist: the units trap

Worked example Example 7 — mixed units, ek galat conversion fail kara deta hai

Seedha diya gaya: chamber pressure $p_{c} = 1.5 MPa$ , injector drop $Δ p_{inj} = 400 kPa$ , feed loss $Δ p_{feed} = 0.3 bar$ . $p_{tank}$ bar mein nikalo.

Forecast: sneaky term woh hai jo alag unit mein hai — compute karne se pehle usse spot karo.

Sab ko Pa mein convert karo: $p_{c} = 1.5 \times 1 0^{6}$ ; $Δ p_{inj} = 4 \times 1 0^{5}$ ; $Δ p_{feed} = 0.3 \times 1 0^{5} = 3 \times 1 0^{4}$ . Yeh step kyun? MPa, kPa aur bar seedha add nahi kar sakte — pehle sab ko ek base unit mein lao.
Sum: $p_{tank} = 1.5 \times 1 0^{6} + 4 \times 1 0^{5} + 3 \times 1 0^{4} = 1.93 \times 1 0^{6} Pa .$ Yeh step kyun? Master equation ek seedha sum hai jab units agree karein.
Bar mein convert karo: $1.93 \times 1 0^{6} /1 0^{5} = 19.3 bar .$ Yeh step kyun? Final answer bar mein maanga gaya; $1 bar = 1 0^{5} Pa$ .

Verify: ratio $19.3/15 = 1.29$ — $1.3$ – $1.5$ rule ki low edge par (yahan feed loss tiny hai). Agar tum kPa→Pa step bhool jaate aur $400$ raw add karte, toh nonsense milta; sanity ratio isse pakad leta hai. ✔

Cell H — Blowdown mode (no regulator)

Pehle ke saare examples mein assume kiya gaya tha ki ek regulator $p_{tank}$ steady rakhta hai. Sabse saste thrusters regulator skip karte hain: gas simply expand hoti hai jaise liquid nikalti hai, toh pressure girta hai burn ke dauran. Yahi blowdown hai.

Worked example Example 8 — blowdown burn ke end mein pressure

Ek blowdown tank gas volume $V_{1} = 0.5 m^{3}$ se $p_{1} = 20$ bar par shuru hota hai (regulator-free). Jaise propellant drain hota hai, gas $V_{2} = 1.5 m^{3}$ tak expand hoti hai. Isothermal assume karo. Final tank pressure $p_{2}$ nikalo, aur check karo ki woh abhi bhi $p_{c} = 12$ bar wale chamber ko $2.5$ bar total loss ke saath beat karta hai ya nahi.

Forecast: blowdown ratio $1.5/0.5 = 3$ hai. Kya pressure $3 \times$ gir jaata hai?

Isothermal ideal gas: $p_{1} V_{1} = p_{2} V_{2} \Rightarrow p_{2} = p_{1} \frac{V _{1}}{V _{2}}$ . Yeh step kyun? Fixed gas, fixed temperature → product $p V$ conserved rehta hai.
$p_{2} = 20 \times \frac{0.5}{1.5} = 20/3 = 6.67 bar$ . Yeh step kyun? Blowdown ratio plug karo; pressure aur volume inversely proportional hain.
Required minimum to feed: $p_{need} = p_{c} + losses = 12 + 2.5 = 14.5 bar$ . Yeh step kyun? Master equation burn ke har instant par apply hoti rehti hai.

Verify: $p_{2} = 6.67 bar < 14.5 bar$ — engine tank khali hone se bahut pehle starve ho jaata hai! Ek blowdown design itna high start karna chahiye ki final pressure bhi budget clear kare. Toh blowdown ratio (yahan $3$ ) limited hai: $V_{2}$ aisa choose karo ki $p_{2} \geq 14.5$ bar rahe. Yeh woh extra constraint hai jiske baare mein regulated systems (Ex 1–7) kabhi nahi sochte. ✔

Recall Scenario checklist — kya tum har case ko matrix mein rakh sakte ho?

Kaun sa cell sign flip involve karta hai, aur kya flip hota hai? ::: Cell B — $\pm ρ g h$ gravity term ka sign flip hota hai (tank upar: minus; tank neeche: plus). Kaun sa cell degenerate input hai, aur valid number trap kyun hai? ::: Cell C — $Δ p_{inj} \to 0$ lower tank pressure deta hai lekin atomization nahi, toh engine cleanly burn nahi karega. Kaun sa cell dikhata hai ki limit par flow → 0 ho jaata hai? ::: Cell D — jaise $p_{tank} \to p_{c}$ injector drop → 0 aur jet speed $v \to 0$ . Kaun sa cell real-world word problem hai, aur woh kya size karta hai? ::: Cell F — apogee-kick stage, tank pressure aur helium pressurant mass size karta hai. Kaun sa cell units trap hai? ::: Cell G — mixed MPa / kPa / bar ko sum karne se pehle sab convert karna padega. Blowdown mode (Cell H) mein tank ko start mein over-pressured kyun rakhna padta hai? ::: Kyunki $p V$ constant hai, pressure girta hai jaise liquid drain hoti hai; final pressure abhi bhi $p_{c}$ + losses se zyada hona chahiye.