Gas generator cycle — performance penalty vs simplicity: Exercises

3.3.23 · D4 · Physics › Rocket Propulsion › Gas generator cycle — performance penalty vs simplicity

Shuru karne se pehle, yeh raha ek toolkit jo tum poori page pe reuse karoge. Ise ek baar padh lo taaki baad mein koi symbol surprise na kare.

Formula

Is poori page ke liye chaar equations $I_{s p} = \frac{v _{e}}{g _{0}} (exhaust kitni tez nikalti hai \div Earth gravity g_{0} = 9.81 m/s^{2})$ $v_{eff} = (1 - f) v_{e, c} + f v_{e, g g} (flow-weighted average exhaust speed)$ $P_{pump} = \frac{m ˙ _{t o t} Δ p}{ρ η _{p}} (propellant ko pressurize karne ki power)$ $\overset{m}{˙}_{g g} = \frac{P _{pump}}{η _{t} c _{p} T _{g g} B}, B = 1 - (\frac{p _{o u t}}{p _{in}})^{\frac{γ - 1}{γ}}$

Yahan $\overset{m}{˙}_{t o t}$ hai total propellant mass-flow rate (kg/s) — har kilogram per second jo engine consume karta hai, main chamber aur gas generator dono milake. $\overset{m}{˙}_{g g}$ us flow ka wo hissa hai jo gas generator ko bheja jaata hai, isliye $f = \overset{m}{˙}_{g g} / \overset{m}{˙}_{t o t}$ hai turbine flow fraction. Aage, $v_{e, c}$ main-chamber exhaust speed hai aur $v_{e, g g}$ dump kiye gaye gas ki exhaust speed hai, $ρ$ propellant density hai, $η_{p}, η_{t}$ pump aur turbine efficiencies hain, $c_{p}$ gas ki heat capacity hai, aur $T_{g g}$ gas-generator temperature hai.

Aakhir mein $B$ hai pressure-ratio bracket — 0 aur 1 ke beech ek pure number jo batata hai "gas ki heat ka kitna fraction turbine actually extract kar sakta hai". Iske andar: $p_{in}$ turbine mein enter karne wala gas pressure hai, $p_{o u t}$ use chhodne wala (kam) pressure hai, aur $γ$ (Greek letter "gamma") gas ka heat-capacity ratio hai — exhaust gas ki ek property, typically lagbhag $1.2$ – $1.3$ , jo yeh set karta hai ki expansion ke dauran pressure aur temperature ek doosre ke saath kitni strongly track karte hain. Jab $p_{o u t}$ lagbhag $p_{in}$ ke barabar hota hai toh bracket 0 ke paas hoti hai (pakadne ke liye kam energy); jab $p_{o u t} ≪ p_{in}$ hota hai toh yeh 1 ki taraf badhti hai.

Level 1 — Recognition

L1.1 Ek sentence mein batao: gas generator cycle, closed cycle ki tulna mein specific impulse kyun khota hai?

Recall Solution

Kyunki turbine-drive gas ko low pressure par overboard dump kar diya jaata hai, main chamber mein feed karne ki bajaye. Yeh almost koi pressure ratio expand nahi karta, isliye kam exhaust speed se nikalti hai aur almost koi thrust contribute nahi karta — phir bhi tum ise carry karke jalate rahe, toh yeh flow-weighted average $I_{s p}$ ko neeche kheenchta hai.

L1.2 Ek engine $\overset{m}{˙}_{t o t} = 300$ kg/s total jalata hai, jisme se $\overset{m}{˙}_{g g} = 12$ kg/s gas generator ko jaata hai. Turbine flow fraction $f$ kya hai?

Recall Solution

$f = \frac{m ˙ _{g g}}{m ˙ _{t o t}} = \frac{12}{300} = 0.04 = 4%.$ Yeh formula kyun? $f$ define hota hai as "total flow ka kitna bada slice turbine par kharch hota hai". Yeh ek pure ratio hai, isliye kg/s ke units cancel ho jaate hain.

L1.3 True ya false: gas generator stoichiometric temperature (~3500 K) par run karta hai taaki turbine power maximize ho.

Recall Solution

False. 3500 K par turbine blades pighal jaate. GG deliberately fuel-rich (ya ox-rich) run karta hai taaki $T_{g g} \approx 900$ – $1200$ K rahe. Tum har kg mein kam power accept karte ho, lekin blades survive karte hain.

Level 2 — Application

L2.1 Ek engine ka ideal $I_{s p} = 340$ s hai aur $f = 3%$ flow negligible useful exhaust speed ke saath dump karta hai. Effective $I_{s p}$ estimate karo.

Recall Solution

"Dump-everything" approximation use karo: $I_{s p, eff} \approx (1 - f) I_{s p, ideal}$ . Kyun? Agar dumped gas zero thrust deta hai, toh yeh sirf denominator mein mass add karta hai, isliye $I_{s p}$ surviving fraction $(1 - f)$ se scale down ho jaati hai. $I_{s p, eff} = (1 - 0.03) (340) = 0.97 \times 340 = 329.8 s .$ Penalty $\approx 10.2$ s.

L2.2 Main exhaust $v_{e, c} = 3000$ m/s, dumped gas $v_{e, g g} = 600$ m/s, $f = 0.05$ . Effective exhaust speed aur effective $I_{s p}$ nikalo ( $g_{0} = 9.81$ ).

Recall Solution

Step 1 — weighted average. Kyun? Momentum add hota hai; total thrust har stream ka mass-flow times speed hota hai, isliye average speed flow-weighted hai. $v_{eff} = (1 - 0.05) (3000) + (0.05) (600) = 2850 + 30 = 2880 m/s .$ Step 2 — $I_{s p}$ mein convert karo. $I_{s p, eff} = \frac{2880}{9.81} = 293.6 s .$

L2.3 Usi engine ke liye, ideal $I_{s p}$ kya hota (saara flow $v_{e, c}$ par), aur penalty kitne seconds mein hai?

Recall Solution

Ideal: $I_{s p, ideal} = 3000/9.81 = 305.8$ s. Penalty: $305.8 - 293.6 = 12.2$ s. Kyun " $5%$ of 305.8 = 15 s" se chhota hai? Kyunki dumped gas abhi bhi 600 m/s par nikalta hai, jo ek chhoti si kick deta hai. Real penalty hamesha full $f$ estimate se thodi kam hoti hai.

Level 3 — Analysis

L3.1 Ek engine mein $v_{e, c} = 3200$ m/s aur $v_{e, g g} = 750$ m/s hai. Mission rules $I_{s p}$ penalty ko 8 s tak cap karti hain. Sabse bada turbine fraction $f$ kya ho sakta hai jo tum use kar sakte ho? ( $g_{0} = 9.81$ .)

Recall Solution

Step 1 — penalty ko seconds mein likhao. Ideal minus effective: $Δ I_{s p} = \frac{v _{e, c}}{g _{0}} - \frac{( 1 - f ) v _{e, c} + f v _{e, g g}}{g _{0}} = \frac{f ( v _{e, c} - v _{e, g g} )}{g _{0}} .$ Kyun $v_{e, c}$ difference se cancel ho jaata hai? Kyunki $(1 - f) v_{e, c}$ term full $v_{e, c}$ se exactly $- f v_{e, c}$ se alag hoti hai, aur $+ f v_{e, g g}$ thoda wapas add karta hai. Gap hai speed lost per dumped kg, $v_{e, c} - v_{e, g g}$ , fraction $f$ se multiply. Step 2 — 8 s ke barabar set karo aur $f$ solve karo. $f = \frac{Δ I _{s p} \cdot g _{0}}{v _{e, c} - v _{e, g g}} = \frac{8 \times 9.81}{3200 - 750} = \frac{78.48}{2450} = 0.0320.$ Toh $f \leq 3.2%$ (lagbhag 3.2%).

L3.2 L3.1 ke penalty formula ko dekho. Agar tum dumped-gas speed $v_{e, g g}$ badha sako (jaise ise ek chhote nozzle se route karke), toh usi 8 s budget ke liye allowed $f$ upar jaayega ya neeche? Figure use karke explain karo.

Recall Solution

Figure — Gas generator cycle — performance penalty vs simplicity

v_{e, g g}

badhane se denominator mein gap

(v_{e, c} - v_{e, g g})

chhota ho jaata hai, isliye fixed 8 s budget ke liye allowed

f

upar jaata hai — tum zyada flow dump afford kar sakte ho kyunki ab har dumped kg kam hurt karta hai. Geometrically (figure mein red bar), "lost speed per kg" bar chhota hota jaata hai jaise yellow

v_{e, g g}

marker blue

v_{e, c}

marker ki taraf chadhta hai.

Level 4 — Synthesis

L4.1 Ek GG engine mein $\overset{m}{˙}_{t o t} = 250$ kg/s, pressure rise $Δ p = 20$ MPa, propellant density $ρ = 1000$ kg/m³, pump efficiency $η_{p} = 0.75$ , turbine efficiency $η_{t} = 0.65$ , $c_{p} = 2200$ J/(kg·K), $T_{g g} = 1050$ K, aur bracket $B = 0.55$ hai. Required turbine flow $\overset{m}{˙}_{g g}$ aur fraction $f$ nikalo.

Recall Solution

Step 1 — pump power. Kyun? Turbine ka poora kaam yeh hai ki propellant ko pressurize karne wali power supply kare. Power = pressure-rise × volume-flow, pump efficiency se degrade hokar. $P_{pump} = \frac{m ˙ _{t o t} Δ p}{ρ η _{p}} = \frac{250 \times 20 \times 1 0 ^{6}}{1000 \times 0.75} = \frac{5 \times 1 0 ^{9}}{750} = 6.667 \times 1 0^{6} W .$ Step 2 — turbine work per kg. Kyun? GG gas ka har kg temperature mein girte waqt itni energy de sakta hai. $w = η_{t} c_{p} T_{g g} B = 0.65 \times 2200 \times 1050 \times 0.55 = 8.258 \times 1 0^{5} J/kg .$ Step 3 — divide karo. Zaroori total power ÷ har kg ki supply ki gayi work: $\overset{m}{˙}_{g g} = \frac{P _{pump}}{w} = \frac{6.667 \times 1 0 ^{6}}{8.258 \times 1 0 ^{5}} = 8.073 kg/s .$ Step 4 — fraction. $f = \frac{8.073}{250} = 0.0323 = 3.2%.$

L4.2 Usi engine ke liye, agar ideal $I_{s p}$ 335 s hai aur dumped gas ko useless maana jaaye, toh yeh $f$ effective $I_{s p}$ par kitna cost karta hai?

Recall Solution

Step 1 — sahi model chuno. Kyun? Problem kehta hai dumped gas useless hai, toh iska exhaust speed weighted average ke numerator mein kuch contribute nahi karta — sirf iska mass denominator mein bachta hai. Yeh full weighted average ko simple scaling $I_{s p, eff} = (1 - f) I_{s p, ideal}$ mein collapse kar deta hai. Step 2 — substitute karo. Kyun yeh numbers? $f = 0.0323$ seedha L4.1 ke power balance se aaya, aur $I_{s p, ideal} = 335$ s wo value hai jo saara flow deta agar kuch dump na hota. $I_{s p, eff} = (1 - f) I_{s p, ideal} = (1 - 0.0323) (335) = 0.9677 \times 335 = 324.2 s .$ Penalty $\approx 10.8$ s — ek typical GG-cycle tax.

Level 5 — Mastery

L5.1 Tumhe ek fixed mission ke liye $v_{e, c} = 3300$ m/s, $g_{0} = 9.81$ ke saath do designs mein se ek choose karna hai:

Design A (GG cycle): $f = 0.045$ , dumped-gas speed $v_{e, g g} = 700$ m/s.

Design B (closed cycle): koi dump nahi, lekin extra plumbing dry mass add karta hai jo effectively ideal se flat 6 s se usable $I_{s p}$ gira deta hai.

Kaun sa design zyada effective $I_{s p}$ deta hai, aur kitne se?

Recall Solution

Step 1 — Design A effective speed (weighted average). Yahan se kyun shuru karo? Design A flow ka real fraction real speed par dump karta hai, isliye $I_{s p}$ mein convert karne se pehle dono streams ka flow-weighted average chahiye — simple $(1 - f) v_{e, c}$ scaling wo chhoti kick ignore kar deti jo 700 m/s gas abhi bhi deta hai. $v_{eff, A} = (1 - 0.045) (3300) + (0.045) (700) = 3151.5 + 31.5 = 3183.0 m/s .$ Step 2 — Design A ki speed ko $I_{s p}$ mein badlo. $g_{0}$ se kyun divide karo? Yeh definition hai $I_{s p} = v_{e} / g_{0}$ ; yeh m/s mein speed ko "seconds" mein convert karta hai jisme specific impulse measure hoti hai. $I_{s p, A} = \frac{3183.0}{9.81} = 324.5 s .$ Step 3 — Design B. Sirf 6 s kyun ghatao? Design B kuch dump nahi karta, isliye iska exhaust full ideal $I_{s p}$ tak pahunchta hai; akela loss woh flat 6 s hai jo problem iske bhaari plumbing ko assign karta hai, toh hum ise seedha subtract karte hain. $I_{s p, B} = \frac{3300}{9.81} - 6 = 336.4 - 6 = 330.4 s .$ Step 4 — compare karo. Kyun subtract karo? Question poochh raha hai "kitne se", yaani dono effective values ke beech ka gap. $330.4 - 324.5 = 5.9 s .$ Design B lagbhag 5.9 s se jeetta hai. GG dump penalty yahan (~11.9 s) closed-cycle ki complexity tax (6 s) se badi hai, isliye closed cycle better performer hai — un higher complexity ki cost par jo Staged Combustion Cycle note describe karta hai.

L5.2 Upar Design A ke liye, $v_{e, g g}$ ki kaun si dumped-gas speed par dono designs tie karte?

Recall Solution

Step 1 — tie condition state karo. Kyun? "Tie" ka matlab hai dono effective specific impulses barabar hain, isliye hum $I_{s p, A} = I_{s p, B} = 330.4$ s set karte hain aur wo exhaust speed padhte hain jo Design A ko chahiye hogi. Step 2 — us target $I_{s p}$ ko wapas speed mein convert karo. $g_{0}$ se kyun multiply karo? $I_{s p} = v / g_{0}$ rearrange hokar $v = I_{s p} g_{0}$ deta hai; hum required $v_{eff}$ paane ke liye definition invert karte hain. $v_{eff, A}^{tie} = 330.4 \times 9.81 = 3241.2 m/s .$ Step 3 — weighted average se $v_{e, g g}$ solve karo. $v_{e, g g}$ isolate kyun karo? Weighted average mein baki sab kuch fixed hai ( $f = 0.045$ , $v_{e, c} = 3300$ ), toh akela unknown dumped-gas speed hai; hum jaane-maane term ko hata dete hain aur uske coefficient $f$ se divide karte hain. $3241.2 = (0.955) (3300) + 0.045 v_{e, g g} = 3151.5 + 0.045 v_{e, g g},$ $0.045 v_{e, g g} = 3241.2 - 3151.5 = 89.7 \Rightarrow v_{e, g g} = \frac{89.7}{0.045} = 1993 m/s .$ Toh GG cycle ko apne dumped gas ko ~1993 m/s tak recover karna hoga (lagbhag ek real nozzle ke barabar) taaki closed cycle se match kar sake — isliye practice mein closed cycles pure performance par jeette hain. Dekho Specific Impulse and Exhaust Velocity aur Nozzle Expansion and Pressure Ratio ki itni zyada dumped-gas speed recover karna kyun itna mushkil hai.

Recall

One-line self-check: penalty in seconds ::: $Δ I_{s p} = f (v_{e, c} - v_{e, g g}) / g_{0}$ One-line self-check: turbine flow ::: $\overset{m}{˙}_{g g} = P_{pump} / (η_{t} c_{p} T_{g g} B)$ , with $P_{pump} = \overset{m}{˙}_{t o t} Δ p / (ρ η_{p})$

Dekho bhi: Turbopump Fundamentals · Rocket Thrust Equation · Specific Impulse and Exhaust Velocity.