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

3.3.25 · D4 · Physics › Rocket Propulsion › Pressure-fed cycle — simplest, used in upper stages

Intuition Is page ko kaise use karein

Har problem ko pehle khud try karo, phir uska Solution fold kholo. Problems recognition (kya tum parts ka naam jaante ho?) se shuru hokar mastery tak jaate hain (kya tum scratch se poora pressure budget aur mass tradeoff design kar sakte ho?). Yahan har symbol parent note the parent topic mein build kiya gaya hai; agar koi term unfamiliar lage, pehle woh note dobara padho.

Symbols jo hum baar baar use karte hain, plain words mein:

$p_{c}$ — combustion chamber ke andar ka pressure (jahan burning hoti hai). Units: pascals (Pa) ya bar; $1 bar = 1 0^{5} Pa$ .
$p_{tank}$ — woh pressure jo propellant tank ko hold karna padta hai.
$Δ p_{inj}$ — injector holes se liquid squeeze karne mein jo pressure lost hoti hai.
$Δ p_{feed}$ — tank se engine tak ke pipes (feed lines) mein friction se jo pressure loss hoti hai.
$Δ p_{cool}$ — chamber ke around liptee cooling jacket mein liquid flow karne par jo pressure loss hoti hai (unke narrow channels mein friction).
$ρ$ — liquid ki density (kg/m³). $σ$ — woh stress jo ek wall yield hone se pehle le sakti hai (Pa). $t$ — wall thickness (m). $r$ — tank radius (m). $ρ_{m}$ — tank wall material ki density (kg/m³). $V$ — tank ke andar propellant ka volume (m³).
$a$ — liquid column par kaam karne wala effective acceleration (m/s²); ek coasting-then-firing stage par yeh stage ka apna acceleration hota hai, jo ordinary gravity $g$ ki jagah leta hai.
$M$ — kisi gas ka molar mass (kg/mol); $R = 8.314$ J/(mol·K) — universal gas constant; $T$ — absolute temperature (K); $n$ — moles ki sankhya.
$m_{p}$ — propellant mass (kg); $m_{s}$ — structure (mostly tank) mass (kg); $v_{e}$ — exhaust velocity (m/s); $Δ v$ — ek stage jo velocity change deliver kar sakta hai (m/s).

Key tools jinka hum sahara lete hain: Ideal gas law, hoop stress, Bernoulli's equation, aur Tsiolkovsky rocket equation.

Tank-mass formula — pictures se samjho

Mass exercises se pehle, aao $m_{tank} = \frac{3 ρ _{m} p V}{2 σ}$ formula ko sirf quote karne ki bajaye khud earn karte hain — aur yeh pin down karte hain ki har number kahan se aata hai.

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

Formula

$m_{tank} \propto p V / σ$ aur uska factor $3/2$ derive karna Step 1 — hoop stress se wall thickness. Pressure $p$ hold karne wali thin spherical shell ke liye, hoop stress deta hai (figure aur Hoop stress and thin-walled pressure vessels dekho): $σ = \frac{p r}{2 t} \Rightarrow t = \frac{p r}{2 σ}$ Step 2 — wall mass = surface area × thickness × material density. Ek sphere ka surface area $A = 4 π r^{2}$ hai (figure mein blue shell), isliye $m_{tank} = ρ_{m} A t = ρ_{m} (4 π r^{2}) (\frac{p r}{2 σ}) = \frac{2 π ρ _{m} p r ^{3}}{σ}$ Step 3 — $r^{3}$ ko enclosed volume $V$ se replace karo. Propellant volume $V = \frac{4}{3} π r^{3}$ hai (figure mein amber ball), isliye $π r^{3} = \frac{3}{4} V$ . Substitute karo: $m_{tank} = \frac{2 ρ _{m} p}{σ} \cdot π r^{3} = \frac{2 ρ _{m} p}{σ} \cdot \frac{3}{4} V$ $m_{tank} = \frac{3 ρ _{m} p V}{2 σ}$ "3" aur " $/2$ " kahan se aate hain: $2$ surface area $4 π r^{2}$ times thickness $\propto 1/2$ se, aur $3/4$ se $π r^{3}$ ko volume mein convert karne se — multiply karo $2 \times \frac{3}{4} = \frac{3}{2}$ . Toh honest factor $\frac{3}{2}$ hai, na ki $3$ . Mass abhi bhi $p$ mein exactly linear hai: yahi physics maayane rakhti hai.

Pressure staircase — compute karne se pehle padho

Intuition Staircase figure kaise padhein

Horizontal axis liquid ka safar hai: yeh tank mein shuru hota hai (far left) aur chamber mein khatam hota hai (far right). Vertical axis pressure hai bar mein. Har cyan platform us point par pressure hai; har amber arrow ek drop hai — ek loss jo tumhe tank mein upar se pre-pay karni hoti hai. Sample reading: figure mein tank $17.0$ bar par hai. Feed line cross karne par $0.8$ bar loss hota hai (neeche $16.2$ tak), cooling jacket aur $0.6$ lose karta hai (neeche $15.6$ tak), aur phir bada amber injector drop $3.6$ bar ka hamein chamber platform par $12.0$ bar $= p_{c}$ par giraa deta hai. Dhyaan do ki injector step abhi tak sabse bada hai — woh drop jaanbujhkar hai, kyunki yeh atomization ko power karta hai. Neeche ke har mass aur budget problem mein bass aise staircase se numbers padni hain.

L1 — Recognition

Recall Solution

Isse kya define karta hai: turbopumps ka absent hona. Propellants sirf ek stored inert gas ke dwara drive hote hain jo tank tops par press karta hai. Gas: helium (ya nitrogen). Helium kyun: yeh inert hai (fuel/oxidizer se react nahi karega), halka hai (kam mass penalty), aur propellant mein condense ya dissolve nahi hoga. Is alternative ke liye jo yeh replace karta hai, Turbopump-fed cycles dekho.

Recall Solution

Rule: $p_{tank} \approx (1.3 - 1.5) p_{c}$ . Yeh range kyun? Tank pressure ko injector drop (~ $p_{c}$ ka 25%) plus feed/cooling losses cover karne chahiye. $1.3 \times 8 = 10.4 bar 1.5 \times 8 = 12.0 bar$ Answer: roughly $10.4$ – $12.0$ bar.

L2 — Application

Recall Solution

Step 1 — injector drop. $Δ p_{inj} = 0.30 \times 12 = 3.6$ bar. $p_{c}$ ka fraction kyun? Injector drop ko deliberately chamber pressure ke relative size kiya jaata hai taaki atomization guarantee ho. Step 2 — saare downstream losses add karo. Yahan $Δ p_{feed}$ feed-line friction loss hai aur $Δ p_{cool}$ cooling-jacket friction loss (dono symbol list mein defined hain). $p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed} + Δ p_{cool} = 12 + 3.6 + 0.8 + 0.6 = 17.0 bar$ Step 3 — ratio. $17.0/12 = 1.417$ . Answer: $p_{tank} = 17.0$ bar $\approx 1.42 p_{c}$ ✔ (1.3–1.5 rule ke andar). Yahi exactly woh tank pressure hai jo upar staircase figure mein draw ki gayi hai.

Recall Solution

Tool — hoop stress kyun? Ek pressurized sphere burst hone ki koshish karta hai; wall ek tension carry karti hai jise hoop stress kehte hain. Burst force ko wall strength se balance karne par (Hoop stress and thin-walled pressure vessels dekho) $σ = \frac{p r}{2 t}$ milta hai, jo thin wall $t ≪ r$ ke liye valid hai. Step 1 — $t$ ke liye solve karo. $t = \frac{p r}{2 σ} = \frac{( 1.7 \times 1 0 ^{6} ) ( 0.5 )}{2 ( 3 \times 1 0 ^{8} )}$ Step 2 — compute karo. $t = \frac{8.5 \times 1 0 ^{5}}{6 \times 1 0 ^{8}} = 1.417 \times 1 0^{- 3}$ m. Answer: $t \approx 1.42$ mm. Assumption check karo: $t / r = 1.42 \times 1 0^{- 3} /0.5 = 0.0028 ≪ 1$ , isliye thin-wall valid hai.

Recall Solution

Tool — ideal gas law kyun? In conditions par helium nearly ideally behave karta hai, isliye $p V = n R T$ jahan $n$ moles hai (Ideal gas law dekho). Step 1 — moles. $n = \frac{p V}{R T} = \frac{( 1.7 \times 1 0 ^{6} ) ( 1.5 )}{8.314 \times 290} = 1057.7 mol$ Step 2 — mass. $m = n M = 1057.7 \times 0.004 = 4.23$ kg. Answer: $\approx 4.23$ kg helium drained volume mein (real tanks isse kaafi zyada carry karte hain, kyunki gas storage bottle mein bhi rehti hai).

L3 — Analysis

Recall Solution

Tool — $m \propto p$ kyun? Wall thickness $t = p r /2 σ$ (thin-wall, $t ≪ r$ ) pressure ke saath badhti hai, aur mass = (area)× $t$ ×density hai, isliye tank mass jo pressure hold karta hai usmein linear hai. Step 1 — Design A ( $p = 1.8 \times 1 0^{6}$ Pa): $m_{A} = \frac{3 ( 2700 ) ( 1.8 \times 1 0 ^{6} ) ( 2 )}{2 ( 3 \times 1 0 ^{8} )} = 48.6 kg$ Step 2 — Design B ( $p = 1.2 \times 1 0^{7}$ Pa): $m_{B} = \frac{3 ( 2700 ) ( 1.2 \times 1 0 ^{7} ) ( 2 )}{2 ( 3 \times 1 0 ^{8} )} = 324 kg$ Step 3 — ratio. $m_{B} / m_{A} = 120/18 = 6.67$ . Step 4 — Design B ke liye thin-wall check. $V = \frac{4}{3} π r^{3}$ se, $r = (\frac{3 V}{4 π})^{1/3} = (\frac{6}{4 π})^{1/3} = 0.782$ m. Phir $t = \frac{p r}{2 σ} = \frac{( 1.2 \times 1 0 ^{7} ) ( 0.782 )}{6 \times 1 0 ^{8}} = 0.0156$ m, isliye $t / r = 0.0156/0.782 = 0.020 ≪ 1$ ✔ thin-wall 120 bar par bhi hold karta hai. Answer: high-pressure tank $\approx 6.67 \times$ zyada heavy hai ( $324$ vs $48.6$ kg). Yahi core reason hai pressure-fed engines low $p_{c}$ par kyun rehte hain: unke tanks hi pump hain aur pump mass pressure ke saath explode karta hai.

Recall Solution

Tool — Bernoulli kyun? Bernoulli's equation kehta hai ki pressure energy ek streamline ke along kinetic energy mein convert hoti hai. Injector ka pressure drop liquid ko ek fast jet mein speed up karne mein kharch hota hai jo droplets mein break hota hai. Step 1 — $v$ ke liye solve karo. $v = \frac{2 Δ p _{inj}}{ρ} = \frac{2 ( 3 \times 1 0 ^{5} )}{1000}$ Step 2 — compute karo. $v = 600 = 24.49$ m/s. Answer: $v \approx 24.5$ m/s. Yeh fast jet isliye hi hota hai ki hum deliberately injector drop bada rakhte hain — yeh atomization ko fine droplets mein power karta hai jo cleanly burn karte hain.

L4 — Synthesis

Recall Solution

Step 1 — injector. $Δ p_{inj} = 0.25 \times 15 = 3.75$ bar. Step 2 — gravity/acceleration head. $a$ effective acceleration hai (symbol list mein defined), $ρ ah = 1200 \times 30 \times 1.2 = 43, 200$ Pa $= 0.432$ bar. Subtract kyun? Accelerating liquid column bottom par (chamber ki taraf) pressure add karta hai, isliye tank ko kam supply karna padta hai. Step 3 — budget assemble karo. $p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed+cool} - ρ ah$ $p_{tank} = 15 + 3.75 + 1.5 - 0.432 = 19.818 bar$ Answer: $p_{tank} \approx 19.82$ bar $\approx 1.32 p_{c}$ . Helpful acceleration head lagbhag aadha bar shave off kar deta hai.

Recall Solution

$p_{tank} = 19.82$ bar $= 1.982 \times 1 0^{6}$ Pa aur $m = \frac{3 ρ _{m} p V}{2 σ}$ use karo (upar derive kiya, thin-wall $t ≪ r$ ). (a) Aluminium. $m_{A l} = \frac{3 ( 2700 ) ( 1.982 \times 1 0 ^{6} ) ( 3 )}{2 ( 3 \times 1 0 ^{8} )} = 80.27 kg$ (b) Titanium. $m_{T i} = \frac{3 ( 4500 ) ( 1.982 \times 1 0 ^{6} ) ( 3 )}{2 ( 9 \times 1 0 ^{8} )} = 44.60 kg$ Ratio. $m_{T i} / m_{A l} = \frac{4500/9 \times 1 0 ^{8}}{2700/3 \times 1 0 ^{8}} = \frac{5 \times 1 0 ^{- 6}}{9 \times 1 0 ^{- 6}} = 0.556$ . (c) Thin-wall check. $r = (\frac{3 V}{4 π})^{1/3} = (\frac{9}{4 π})^{1/3} = 0.895$ m; $t = \frac{p r}{2 σ} = \frac{( 1.982 \times 1 0 ^{6} ) ( 0.895 )}{6 \times 1 0 ^{8}} = 2.96 \times 1 0^{- 3}$ m, isliye $t / r = 0.0033 ≪ 1$ ✔. Answer: Al $\approx 80.3$ kg, Ti $\approx 44.6$ kg; titanium $\sim 0.56 \times$ mass hai. Kyun? Jo matter karta hai woh $ρ_{m} / σ$ ratio hai (density per unit strength) — titanium ki higher strength uski higher density ko beat kar deti hai.

L5 — Mastery

Recall Solution

Tool — Tsiolkovsky kyun? $Δ v$ performance ka true measure hai; yeh exhaust velocity $v_{e}$ ko mass ratio $m_{0} / m_{f}$ (wet over dry) ke saath weigh karta hai. Humein dekhna hai ki faster exhaust heavier tank ko beat karta hai ya nahi. Setup. Wet mass $m_{0} = m_{p} + m_{s} + 500$ . Dry mass $m_{f} = m_{s} + 500$ (propellant burn ho gaya).

Case A ( $p_{c} = 15$ bar): $m_{s} = 160$ kg. $m_{0} = 8000 + 160 + 500 = 8660, m_{f} = 160 + 500 = 660$ $Δ v_{A} = 2900 ln \frac{8660}{660} = 2900 \times ln (13.1212) = 2900 \times 2.5743 = 7465 m/s$

Case B ( $p_{c} = 30$ bar): $m_{s} = 320$ kg. $m_{0} = 8000 + 320 + 500 = 8820, m_{f} = 320 + 500 = 820$ $Δ v_{B} = 3100 ln \frac{8820}{820} = 3100 \times ln (10.7561) = 3100 \times 2.3755 = 7364 m/s$

Answer: $Δ v_{A} \approx 7465$ m/s vs $Δ v_{B} \approx 7364$ m/s. Low pressure jeet jaata hai ~101 m/s se: tank-mass penalty modest exhaust-velocity gain se zyada hai. Yahi exactly reason hai ki upper stages pressure-fed aur low- $p_{c}$ rehte hain — analysis, sirf intuition nahi, isse confirm karta hai.

Recall Solution

Step 1 — equate karo. Humein chahiye $3100 ln \frac{8500 + m _{s}^{B}}{500 + m _{s}^{B}} = 7465$ . Step 2 — log isolate karo. $ln \frac{8500 + m _{s}^{B}}{500 + m _{s}^{B}} = 7465/3100 = 2.4081$ , isliye ratio $= e^{2.4081} = 11.113$ . Step 3 — fraction clear karo. $8500 + m_{s}^{B} = 11.113 (500 + m_{s}^{B}) = 5556.5 + 11.113 m_{s}^{B}$ . Step 4 — like terms collect karo. $m_{s}^{B}$ terms ek side aur constants doosri side move karo: $8500 - 5556.5 = 11.113 m_{s}^{B} - m_{s}^{B} \Rightarrow 2943.5 = 10.113 m_{s}^{B}$ Step 5 — solve karo. $m_{s}^{B} = 2943.5/10.113 = 291.1$ kg. Answer: Case B break even karta hai sirf agar uska tank $\approx 291$ kg tak cut ho sake (actually needed $320$ kg se neeche). Kyunki real tank break-even se zyada heavy hai, high pressure genuinely haarta hai — Exercise 10 confirm ho gaya.