Worked examples — Staged combustion cycle — full flow, fuel-rich, oxidizer-rich preburners

3.3.22 · D3 · Physics › Rocket Propulsion › Staged combustion cycle — full flow, fuel-rich, oxidizer-ric

Yeh page ek drill sheet hai. Parent note ne do governing equations banaye the:

• Pump power $P_{\text{pump}}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$
• Turbine power $P_{\text{turb}}={\dot{m}}_{pb}{c}_p{T}_{in}{\eta }_t\left[1-{\pi }_t^{-(\gamma -1)/\gamma }\right]$

Yahan hum har tarah ke input ko cover karenge jo yeh equations receive kar sakti hain — normal, extreme, zero, aur degenerate — taaki koi bhi exam ya reality tumhe surprise na kar sake.

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The scenario matrix

Is topic ke har possible question ka answer in cells mein se kisi ek mein hai. Neeche har example us cell ke saath tagged hai jise woh cover karta hai.

Cell	Kya alag hai	Example
A — Baseline pump	ordinary numbers, ek pump	Ex 1
B — Two pumps summed	$\sum$ se fuel + ox balance karo	Ex 2
C — Limiting ${\pi }_t\to 1$	pressure ratio 1 ke paas (degenerate turbine)	Ex 3
D — Limiting ${\pi }_t\to \infty$	bahut bada expansion, bracket $\to 1$	Ex 3
E — Zero input	$\Delta p=0$ ya $\dot{m}=0$	Ex 4
F — Sign / direction	"pressure drop" vs "pressure rise" trap	Ex 4
G — Temperature control	$T_{in}$ choose karo taaki turbine survive kare	Ex 5
H — Real-world word problem	Raptor-class FFSC, do turbines	Ex 6
I — Exam twist	power budget se ulta $p_c$ nikalo	Ex 7
J — Cross-cycle contrast	same numbers, gas-generator vs staged waste	Ex 8
K — Degenerate efficiency	${\eta }_p\to 0$ (pump power blow up ho jaata hai)	Ex 9

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Example 1 — Baseline single pump (Cell A)

Step 1 — Pressure ko SI mein convert karo. $250$ bar $=250\times 10^5$ Pa $=2.5\times 10^7$ Pa. Yeh step kyun? Formula pascals (J/m³) chahta hai. "bar" ek shortcut hai; $1\text{ bar}=10^5$ Pa. Yahan units mix karna sabse #1 galat answer hai.

Step 2 — $P_{\text{pump}}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$ mein plug karo. $P=\frac{400\times 2.5\times 10^7}{1140\times 0.75}=\frac{1.0\times 10^{10}}{855}\approx 1.17\times 10^7\text{W}$ Yeh step kyun? Pressure $\times$ volume-flow $=$ power. Volume flow $\dot{m}/\rho$ hai; denominator mein efficiency ka matlab hai ki real shaft ko ideal fluid work se zyada deliver karna padta hai.

Step 3 — Human units mein padho. $1.17\times 10^7$ W $\approx 11.7$ MW $\approx 15,700$ hp.

Verify: Units: $\frac{(\text{kg/s})(\text{Pa})}{({\text{kg/m}}^3)}=\frac{(\text{kg/s})({\text{J/m}}^3)}{({\text{kg/m}}^3)}=\text{J/s}=\text{W}$. ✓ Dimensionless $\eta$ units nahi badalta. Magnitude (ek pump ke liye tens of MW) parent ke Example 1 se match karti hai. ✓

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Example 2 — Do pumps, real power balance (Cell B)

Step 1 — Fuel pump power. $\Delta p=3\times 10^7$ Pa. $P_f=\frac{100\times 3\times 10^7}{70\times 0.7}=\frac{3\times 10^9}{49}\approx 6.12\times 10^7\text{W}\approx 61.2\text{MW}$ Yeh step kyun? LH₂ bahut halka hota hai ($\rho =70$). Low density ka matlab hai per kg bahut zyada volume flow, aur volume flow power drive karta hai. Fuel pump monster hai.

Step 2 — Oxidizer pump power. $P_o=\frac{600\times 3\times 10^7}{1140\times 0.7}=\frac{1.8\times 10^{10}}{798}\approx 2.256\times 10^7\text{W}\approx 22.6\text{MW}$ Yeh step kyun? Same $\Delta p$, lekin dense LOX ka matlab hai chhota volume flow — 6× mass hone ke bawajood kam power lagti hai.

Step 3 — Sum karo (balance mein $\sum$). $\sum P_{\text{pump}}=61.2+22.6=83.8\text{MW}$ Yeh step kyun? Power balance hai $P_{\text{turb}}=\sum P_{\text{pump}}$: turbines ko total supply karna hoga.

Verify: Fuel pump power, 6× kam mass ke bawajood, oxidizer se zyada hai — density ratio $1140/70\approx 16$ mass ratio $6$ ko overwhelm kar deta hai. Forecast confirm hua: fuel wins. ✓

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Example 3 — Turbine bracket ki do limits (Cells C & D)

Turbine ki fraction extracted hai $B({\pi }_t)=1-{\pi }_t^{-(\gamma -1)/\gamma }$. Chaliye ise dono extremes pe todke dekhte hain. $\gamma =1.3$ use karo (methane-rich gas), toh exponent $\frac{\gamma -1}{\gamma }=\frac{0.3}{1.3}\approx 0.2308$ hai.

Step 1 — Cell C: ${\pi }_t=1$ (koi expansion nahi). $B=1-1^{-0.2308}=1-1=0$ Yeh step kyun? ${\pi }_t=p_{in}/p_{out}=1$ ka matlab hai inlet aur outlet pressures equal hain — gas expand nahi karti, isliye blades pe zero kaam hota hai. Bina pressure drop wala turbine kisi kaam ka nahi. Yeh degenerate case hai — figure mein floor pe red dot.

Step 2 — Typical staged value: ${\pi }_t=1.5$. $B=1-1.5^{-0.2308}=1-0.9124=0.0876$ Yeh step kyun? Modest 1.5× drop sirf ~9% enthalpy extract karta hai — doosra dot, floor se thoda utha hua. Yeh everyday staged-combustion operating point hai (Cells C aur D ke beech between hai), koi matrix corner nahi: staged turbines low pressure ratio pe chalte hain taaki gas main chamber ke liye apna zyada pressure bachaye, aur yahi reason hai ki itna zyada mass flow ${\dot{m}}_{pb}$ kyun chahiye hota hai.

Step 3 — Cell D: ${\pi }_t=100$ (bahut bada expansion, gas-generator regime). $B=1-100^{-0.2308}=1-0.3466=0.6534$ Yeh step kyun? Bada pressure drop → bada extraction (~65%), figure mein top dot. Lekin $B$ saturate hota hai 1 se neeche: jaise ${\pi }_t\to \infty$, ${\pi }_t^{-0.2308}\to 0$, toh $B\to 1$ lekin kabhi pahuncha nahi. Yahi red curve ka dashed ceiling ki taraf flatten hona hai.

Verify: $B(1)=0$, $B$ increasing hai, aur $B<1$ sabhi finite ${\pi }_t$ ke liye, ceiling $B\to 1$ sirf limit mein. Monotonic aur bounded — koi unphysical value nahi. ✓

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Example 4 — Zero aur sign traps (Cells E & F)

Step 1 — Cell E, $\dot{m}=0$. $P=\frac{0\times \Delta p}{\rho {\eta }_p}=0$ Yeh step kyun? Pump se koi mass nahi guzrta → koi fluid lift nahi ho raha → zero fluid power. (Real pumps friction se idle power draw karte hain, lekin ideal formula sahi se 0 useful output deta hai.)

Step 2 — Cell E, $\Delta p=0$. $P=\frac{\dot{m}\times 0}{\rho {\eta }_p}=0$ Yeh step kyun? Agar pump koi pressure add nahi karta, toh useful kaam zero hai chahe kitna bhi fluid spin kare — numerator mein zero factor hai, isliye poora product 0 ho jaata hai. "Kuch na karne" ke dono tarike (no flow, ya no pressure rise) $P=0$ dete hain, jo confirm karta hai ki formula gracefully degrade hota hai.

Step 3 — Cell F, negative $\Delta p$. $\Delta p=-20$ bar $=-2\times 10^6$ Pa ek negative $P$ deta hai. Yeh step kyun? $\Delta p$ ka sign energy flow ki direction batata hai. Negative result ka matlab hai ki device pump nahi balki turbine/expander hai — fluid energy de raha hai, le nahi raha. Isliye balance mein turbine term pressure drop (${\pi }_t>1$) use karta hai: energy bahar aa rahi hai. Kabhi bhi pressure drop ko pump formula mein plug karke pump power mat kaho.

Verify: $P=0$ jab numerator mein koi bhi factor 0 ho. $P<0\iff \Delta p<0\iff$ energy extracted, yaani turbine behaviour. $P_{\text{turb}}$ ke drop ${\pi }_t>1$ ke saath consistent. ✓

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Example 5 — $T_{in}$ choose karo taaki turbine survive kare (Cell G)

Step 1 — $Q=m{c}_p\Delta T$ se $Q$ nikalo stoichiometric rise ke saath. Yahan $m=1$ kg hai aur stoichiometric rise $\Delta T=3300-100=3200$ K hai, toh $Q=1\times 5000\times 3200=1.6\times 10^7\text{J}.$ Yeh step kyun? Hum $m=1$ kg define karte hain aur same $c_p=5000$ J/kg·K yahan aur baad mein bhi use karte hain, toh koi hidden second heat capacity nahi hai. Heat law $Q=m{c}_p\Delta T$ us fixed energy $Q$ ko recover karne deta hai jo chemistry deliver karti hai, chahe baad mein hum ise kitni bhi mass mein dilute kar dein.

Step 2 — Same $Q$ ko tripled mass pe re-spread karo. $\Delta T_{\text{rich}}=\frac{Q}{m^{\prime }{c}_p}=\frac{1.6\times 10^7}{3\times 5000}=\frac{1.6\times 10^7}{1.5\times 10^4}\approx 1067\text{K}.$ Yeh step kyun? $Q$ chemistry se fixed hai; $c_p$ unchanged hai; sirf mass 1 kg se badh ke $m^{\prime }=3$ kg ho gayi. Same energy ÷ 3× mass = one-third rise. Extra H₂ heat soak kar leta hai bina khud zyada energy add kiye — yahi reason hai ki "rich" cool karta hai.

Step 3 — 100 K inlet wapas add karo. $T_{out}\approx 100+1067=1167\text{K}.$ Yeh step kyun? $\Delta T$ inlet se ek rise hai; gas actually inlet + rise pe baahir jaati hai.

Verify: $m=1$ kg, $c_p=5000$ ke saath: $Q=1.6\times 10^7$ J, 3300 K reproduce karta hai ($100+1.6\times 10^7/5000=3300$ ✓). Mass triple karne par $100+1.6\times 10^7/(3\times 5000)=1167$ K, parent ki turbine-survivable band (~900–1200 K) ke andar, 3300 K se kaafi neeche jo blades melt kar deta. Forecast confirm hua. ✓

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Example 6 — Real-world FFSC engine (Cell H)

Step 1 — Exponent aur bracket. $\frac{\gamma -1}{\gamma }=\frac{0.25}{1.25}=0.2$. $B=1-1.6^{-0.2}=1-0.9099=0.0901$ Yeh step kyun? $B$ woh fraction hai jitna enthalpy turbine actually extract karta hai (Ex 3). Low ${\pi }_t$ ⇒ chhota $B$ ⇒ bahut zyada flow chahiye hoga.

Step 2 — Power per kg of gas. $P/\dot{m}=c_p{T}_{in}{\eta }_{t}B=3200\times 750\times 0.78\times 0.0901\approx 1.687\times 10^5\text{J/kg}$ Yeh step kyun? $c_{p}T$ enthalpy per kg hai; turbine efficiency ${\eta }_t$ aur extracted fraction $B$ se multiply karo to get usable work per kg.

Step 3 — Fuel turbine demand ko us se divide karo. ${\dot{m}}_{pb,f}=\frac{60\times 10^6}{1.687\times 10^5}\approx 356\text{kg/s}$ Yeh step kyun? Required power ÷ power-per-kg = kg per second chahiye. FFSC mein yeh 356 kg/s waste nahi hote — yeh fuel gasify karte hain aur chamber mein jaake jalte hain.

Verify: Units $\frac{\text{W}}{\text{J/kg}}=\frac{\text{J/s}}{\text{J/kg}}=\text{kg/s}$. ✓ Kuch sau kg/s gas bilkul wahi reason hai kyun FFSC "full flow" push karta hai — lagbhag saara propellant preburners se guzarta hai. ✓

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Example 7 — Exam twist: chamber pressure ulta nikalo (Cell I)

Step 1 — Pump equation ko $\Delta p$ ke liye rearrange karo. $P=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$ se: $\Delta p=\frac{P\rho {\eta }_p}{\dot{m}}$ Yeh step kyun? Exam tumhe power deta hai aur pressure chahta hai; algebra unknown ko isolate karta hai. Yeh balance ulta run karna hai — available power achievable pressure ko cap karta hai.

Step 2 — Numbers substitute karo. $\Delta p=\frac{30\times 10^6\times 1140\times 0.75}{500}=\frac{2.565\times 10^{10}}{500}=5.13\times 10^7\text{Pa}$ Yeh step kyun? Hum given power budget, LOX density, aur efficiency seedha rearranged formula mein daalta hain. Har quantity already SI mein hai (watts, kg/m³, dimensionless, kg/s), isliye result pascals mein aata hai bina kisi unit juggling ke — upar aur neeche multiply karo, phir divide karo.

Step 3 — Bar mein convert karo. $\Delta p=5.13\times 10^7/10^5=513\text{bar}$ Yeh step kyun? $1\text{ bar}=10^5$ Pa, toh divide karo. Isliye staged engines 250–350 bar tak pahunche — recovered turbine exhaust ke saath, zyada power budget zyada pressure "buy" karta hai.

Verify: Plug back karo: $\frac{500\times 5.13\times 10^7}{1140\times 0.75}=\frac{2.565\times 10^{10}}{855}=3.0\times 10^7=30$ MW. ✓ Round-trip close ho jaata hai.

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Example 8 — Cross-cycle contrast: staged combustion kya bachata hai (Cell J)

Step 1 — Har stream se thrust ($F=\dot{m}v$, Rocket thrust equation se). Main flow agar sab recover ho: $F_{\text{ideal}}=520\times 3500=1.82\times 10^6$ N. Yeh step kyun? Momentum thrust mass flow × exhaust speed hai. "Sab $v_c$ pe" ideal ko reality se compare karo.

Step 2 — Gas-generator actual thrust. Main $468$ kg/s at $3500$ + dumped $52$ kg/s at $1500$: $F_{gg}=468\times 3500+52\times 1500=1.638\times 10^6+7.8\times 10^4=1.716\times 10^6\text{N}$ Yeh step kyun? Dumped stream kuch thrust to banata hai, lekin bahut kam speed pe — yahi waste hai.

Step 3 — Fully-recovered (staged) ideal ke against fractional loss. $\frac{F_{\text{ideal}}-F_{gg}}{F_{\text{ideal}}}=\frac{1.82\times 10^6-1.716\times 10^6}{1.82\times 10^6}=\frac{1.04\times 10^5}{1.82\times 10^6}\approx 0.0571=5.7\%$ Yeh step kyun? Hum shortfall $F_{\text{ideal}}-F_{gg}$ lete hain (woh thrust jo dumped stream deliver karne mein fail hua kyunki woh 3500 ki jagah 1500 m/s pe gaya) aur fraction express karne ke liye ideal se divide karte hain. Staged combustion us 52 kg/s ko chamber mein recover karta hai, toh woh full $3500$ m/s pe exit karta hai — yeh ~5.7% reclaim karta hai, aur upar se higher $p_c$ enable karta hai. Gas-generator cycle aur Specific impulse and chamber pressure se compare karo.

Verify: Loss fraction $=\frac{52(3500-1500)}{520\times 3500}=\frac{104000}{1820000}=0.0571=5.71\%$. ✓ Units: velocities cancel, mass flows cancel — ek pure dimensionless fraction, jaise fraction honi chahiye. Kuch-percent $I_{sp}$ gain jo puri missions decide karta hai — yahi reason hai FFSC exist karta hai. ✓

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Example 9 — Degenerate efficiency: ${\eta }_p\to 0$ hone par kya hota hai (Cell K)

Step 1 — Accha pump, ${\eta }_p=0.75$. $P=\frac{400\times 2.5\times 10^7}{1140\times 0.75}=\frac{1.0\times 10^{10}}{855}\approx 1.17\times 10^7\text{W}\approx 11.7\text{MW}$ Yeh step kyun? Yeh Ex 1 baseline hai — hamara reference point. ${\eta }_p$ denominator mein hai, isliye ise shrink karne se answer bada hoga.

Step 2 — Bura pump, ${\eta }_p=0.25$ (teen-guna kam efficient). $P=\frac{1.0\times 10^{10}}{1140\times 0.25}=\frac{1.0\times 10^{10}}{285}\approx 3.51\times 10^7\text{W}\approx 35.1\text{MW}$ Yeh step kyun? ${\eta }_p$ ko one-third karne par required power triple ho jaati hai ($P\propto 1/{\eta }_p$). Wasteful pump turbine ko — aur isliye preburner flow ${\dot{m}}_{pb}$ ko — teen guna mehnat karwata hai. Isliye turbopump efficiency ke liye itni mehenat ki jaati hai.

Step 3 — Degenerate limit ${\eta }_p\to 0$. ${\lim }_{{\eta }_p\to 0}P={\lim }_{{\eta }_p\to 0}\frac{\dot{m}\Delta p}{\rho {\eta }_p}=+\infty$ Yeh step kyun? Ek fixed nonzero numerator ko zero ki taraf jaate denominator se divide karne par woh bina bound ke blow up ho jaata hai. Physically: ek pump jo apna saara input heat mein waste karta hai (${\eta }_p=0$) kisi bhi finite power pe real pressure rise deliver nahi karta — toh real $\Delta p$ maangne ke liye infinite shaft power chahiye. Yeh Ex 4 ke zero cases ki mirror image hai lekin doosri taraf se: wahan numerator zero ho gaya tha jisse $P=0$ mila; yahan denominator zero ho raha hai jisse $P\to \infty$ milta hai. Dono milke formula ke pure behaviour ko bracket karte hain.

Verify: $P(0.75)=1.17\times 10^7$ W aur $P(0.25)=3.51\times 10^7$ W: ratio exactly $0.75/0.25=3$ hai, jo $P\propto 1/{\eta }_p$ confirm karta hai. ✓ Jaise ${\eta }_p\to 0^{+}$, $P\to +\infty$ (denominator $\to 0^{+}$, numerator fixed positive). ✓ Units Ex 1 se unchanged (W). ✓

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Recall Self-test

${\pi }_t=1$ pe Bracket $B$ ::: $B=0$ — koi expansion nahi, koi kaam nahi. Jaise ${\pi }_t\to \infty$, $B\to$ ::: 1 (lekin finite ${\pi }_t$ ke liye kabhi reach nahi karta). Jab $\Delta p<0$ ho toh pump power ka sign ::: negative — yeh turbine ki tarah act kar raha hai, energy extract kar raha hai. LH₂ pump power LOX pump se zyada kyun hai despite kam mass? ::: Iska density ~16× kam hai, isliye volume flow (jo power set karta hai) kaafi zyada hoti hai. FFSC mein, kya preburner kg/s waste hote hain? ::: Nahi — yeh propellant gasify karte hain aur chamber mein dobara jalte hain. Jaise pump efficiency ${\eta }_p\to 0$, required power $\to$ ::: $+\infty$ ($P\propto 1/{\eta }_p$).

Yeh bhi dekho: Turbopump design · Combustion stoichiometry and flame temperature · Regenerative cooling · Expander cycle.