Gas generator cycle — performance penalty vs simplicity: Worked examples

3.3.23 · D3 · Physics › Rocket Propulsion › Gas generator cycle — performance penalty vs simplicity

Yeh page Gas Generator Cycle topic ka exhaustive drill sheet hai. Kuch bhi solve karne se pehle, hum ek matrix banate hain jisme topic ke har possible situation ka type hota hai — phir hum har ek cell ko ek fully worked example se cover karte hain.

Har symbol jo tumhe chahiye woh is page par hi define hai, to kisi step ko follow karne ke liye kahin aur jaane ki zaroorat nahi.

Definition

Is page par use hone wale har symbol ki simple explanation.

$f$ — turbine flow fraction: total propellant ka woh hissa (ek chhoti si number jaise $0.03$ ) jo gas generator mein jalaya jaata hai aur dump kiya jaata hai, $f = \overset{m}{˙}_{g g} / \overset{m}{˙}_{t o t}$ .
$v_{e, c}$ — bade nozzle se nikalne wale main exhaust ki speed (fast).
$v_{e, g g}$ — dump kiye gaye gas-generator exhaust ki speed (slow).
$g_{0} = 9.81 m/s^{2}$ — woh reference gravity jo ek speed ko specific impulse ke seconds mein convert karti hai.
$Δ p$ — pressure rise (pascals mein) jo pump ko propellant mein add karni padti hai taaki woh chamber mein push ho sake.
$ρ$ — propellant ki density (kg/m³); yeh mass flow $\overset{m}{˙}$ ko volume flow $\overset{m}{˙} / ρ$ mein convert karti hai.
$η_{p}$ , $η_{t}$ — pump aur turbine ki efficiencies (fractions $\leq 1$ ): ideal power ka kitna hissa actually deliver hota hai.
$c_{p}$ — gas ki specific heat at constant pressure (J/(kg·K)): ek kilogram constant pressure par ek degree temperature change par kitne joules store karta hai.
$T_{g g}$ — gas-generator temperature (K): turbine chalane wali gas kitni hot hai.
$γ$ — ratio of specific heats (ek dimensionless gas property, ≈1.2–1.4): yeh set karta hai ki gas expand hone par kitni cool hoti hai. Yeh sirf neeche explain kiye gaye bracket ke through appear karta hai.
$p_{in}$ , $p_{o u t}$ — turbine ki inlet aur outlet pressures: gas turbine mein $p_{in}$ par enter karti hai aur lower $p_{o u t}$ par bahar jaati hai.

Intuition

Woh bracket $[1 - (p_{o u t} / p_{in})^{(γ - 1) / γ}]$ kya kar raha hai? Jab gas turbine ke across high pressure $p_{in}$ se lower $p_{o u t}$ tak expand hoti hai, toh woh cool hoti hai, aur woh lost heat shaft work ban jaati hai. Ek smooth (isentropic) expansion ki physics kehti hai ki temperature ratio pressure ratio ko $(γ - 1) / γ$ power tak raise karne par follow karta hai — yahi se $γ$ enter karta hai. To gas ki heat ka jo fraction work mein convert hota hai woh exactly $[1 - (p_{o u t} / p_{in})^{(γ - 1) / γ}]$ hai: agar $p_{o u t} = p_{in}$ toh yeh $0$ hai (koi expansion nahi, koi work nahi) aur jaise $p_{o u t}$ , $p_{in}$ se bahut neeche girta hai, yeh $1$ ki taraf grow karta hai. Hamare problems mein is bracket ko ek number diya jaata hai (jaise $0.5$ = "aadhi heat work banti hai") taaki tum directly plug in kar sako.

Scenario matrix

Gas-generator land ka har problem inhi case classes mein se ek hota hai. Isse ek checklist ki tarah socho: agar tum har row se ek example solve kar sako, toh exam mein koi bhi cheez tumhe surprise nahi kar sakti.

Cell	Case class	Kya special hai	Example
A	Ideal-limit $v_{e, g g} = 0$	dump ki gayi gas bilkul waste → penalty exactly $f$ hai	Ex 1
B	Non-zero dump $v_{e, g g} > 0$	dump ki gayi gas thoda sa kick deti hai → penalty $< f$	Ex 2
C	Degenerate $f = 0$	closed-cycle limit, koi dump nahi	Ex 3
D	Degenerate $f = 1$	sab kuch dump, koi main chamber nahi	Ex 3
E	Power-balance sizing	pump/turbine physics se $f$ nikalo	Ex 4
F	Sign/limit of $Δ p$	high $p_{c}$ → bada $f$ ; trend, koi ek number nahi	Ex 5
G	Temperature cap	target $f$ hit karne ke liye GG kitna hot hona chahiye?	Ex 6
H	Real-world word problem	Saturn-V-style thrust bookkeeping	Ex 7
I	Exam twist	$I_{s p, eff}$ diya hai, hidden $f$ nikalo	Ex 8

Neeche diya figure poori matrix ke peeche ki mental picture hai: vehicle se do streams nikalte hain, ek fast (orange main stream, fraction $1 - f$ , speed $v_{e, c}$ ke saath) aur ek slow (plum dump stream, fraction $f$ , speed $v_{e, g g}$ ke saath). Ise dhyan mein rakho — har example jisme "do streams" ka zikr ho (khaaskar Ex 2, 3 aur 7) inhi do arrows ki bookkeeping kar raha hai.

Figure — Gas generator cycle — performance penalty vs simplicity

Worked examples

Worked example

Example 1 — Cell A: ideal-limit penalty ( $v_{e, g g} = 0$ ). Ek engine ka ideal $I_{s p} = 340$ s hai aur woh turbine ke through apne flow ka $f = 5%$ dump karta hai. Dump ki gayi gas ko zero useful exhaust speed contribute karne wali maano. Effective $I_{s p}$ nikalo.

Forecast: compute karne se pehle guess karo — answer 340, 323, ya 170 ke karib hoga?

Step 1. $I_{s p, eff} \approx (1 - f) I_{s p, ideal}$ use karo. Yeh step kyun? $v_{e, g g} = 0$ ke saath weighted average ka poora doosra term vanish ho jaata hai, to sirf surviving fraction $(1 - f)$ of flow momentum banata hai.

Step 2. $I_{s p, eff} = (1 - 0.05) (340) = 0.95 \times 340 = 323 s$ . Kyun? Weight-flow ka 5% denominator mein baitha hai (tumne usse carry kiya) lekin numerator mein ~kuch nahi daalta.

Verify: penalty $= 340 - 323 = 17 s$ , jo $f \times I_{s p, ideal} = 0.05 \times 340 = 17$ s ke barabar hai. Consistent hai, aur answer forecast ki tarah 323 ke paas hai — woh wild 170 nahi (woh $f = 50%$ hota).

Worked example

Example 2 — Cell B: dump ki gayi gas USELESS nahi hai ( $v_{e, g g} > 0$ ). Yeh figure ke do arrows ko numeric banata hai. Main exhaust (orange) $v_{e, c} = 3000$ m/s, dumped-gas (plum) $v_{e, g g} = 700$ m/s, aur $f = 0.05$ . $I_{s p, eff}$ nikalo aur "dump-everything" estimate se compare karo. $g_{0} = 9.81$ use karo.

Forecast: kya real penalty Example 1 ke $5%$ se zyada hogi ya kam?

Step 1. Effective exhaust speed $v_{eff} = (1 - f) v_{e, c} + f v_{e, g g} = 0.95 (3000) + 0.05 (700)$ . Yeh step kyun? Thrust har stream ke momentum flux ka sum hai; total flow se divide karne par flow-weighted exhaust speed milti hai.

Step 2. $v_{eff} = 2850 + 35 = 2885 m/s$ . Yeh step kyun? Hum bas weighting arithmetic kar rahe hain — orange stream $0.95 \times 3000 = 2850$ contribute karta hai aur plum stream sirf $0.05 \times 700 = 35$ add karta hai, jo ek nazar mein dikhata hai ki slow dump arrow kitna kam matter karta hai.

Step 3. $I_{s p, eff} = 2885/9.81 = 294.1 s$ , jabki ideal $= 3000/9.81 = 305.8 s$ . Compare kyun? Gap $305.8 - 294.1 = 11.7$ s asli penalty hai.

Verify: naive "dump everything" estimate $0.05 \times 305.8 = 15.3$ s ka loss deta hai. Hamara real loss $11.7 < 15.3$ s — dump ki gayi gas ke chhote kick ne uska kuch hissa recover kar liya, bilkul forecast ki tarah (kam penalty). Units: m/s ÷ (m/s²) = s ✓.

Worked example

Example 3 — Cells C & D: do degenerate limits ( $f = 0$ aur $f = 1$ ). Usi figure ke do arrows use karte hue, $v_{e, c} = 3000$ m/s aur $v_{e, g g} = 700$ m/s, do extremes $f = 0$ (koi plum arrow nahi) aur $f = 1$ (koi orange arrow nahi) par $I_{s p, eff}$ evaluate karo.

Forecast: kaun sa limit ideal closed-cycle-jaisi number reproduce karta hai, aur kaun sa pure gas-generator plume par collapse karta hai?

Step 1 — Case C, $f = 0$ . $v_{eff} = (1 - 0) 3000 + 0 \cdot 700 = 3000$ m/s → $I_{s p} = 3000/9.81 = 305.8$ s. Yeh step kyun? $f = 0$ matlab koi propellant dump nahi hota — plum arrow gayab ho jaata hai aur poora flow main nozzle se expand hota hai. Yeh "no dump loss" idealization hai.

Step 2 — Case D, $f = 1$ . $v_{eff} = 0 \cdot 3000 + 1 \cdot 700 = 700$ m/s → $I_{s p} = 700/9.81 = 71.4$ s. Yeh step kyun? $f = 1$ matlab sab kuch turbine se guzarta hai aur dump ho jaata hai — orange arrow gayab ho jaata hai, koi main chamber nahi bachta, isliye tumhe woh feeble dump-plume $I_{s p}$ milta hai.

Verify: formula $f$ mein linear hai, isliye yeh inhi endpoints ke beech straight interpolate karta hai. $f = 0.05$ par humne 294.1 s paya (Ex 2), aur indeed $305.8 + 0.05 (71.4 - 305.8) = 305.8 - 11.7 = 294.1$ s ✓. Dono degenerate limits physically sensible hain.

Worked example

Example 4 — Cell E: power balance se turbine flow size karo. $\overset{m}{˙}_{t o t} = 300$ kg/s, $Δ p = 12$ MPa, $ρ = 1000$ kg/m³, pump efficiency $η_{p} = 0.7$ , turbine efficiency $η_{t} = 0.6$ , $c_{p} = 2000$ J/(kg·K), $T_{g g} = 1050$ K, aur pressure-ratio bracket $[1 - (p_{o u t} / p_{in})^{(γ - 1) / γ}] = 0.5$ (yaad rakho: "GG gas ki aadhi heat shaft work banti hai"). $\overset{m}{˙}_{g g}$ aur $f$ nikalo.

Forecast: kya $f$ "few percent" band mein aayega jo parent note ne predict kiya tha?

Step 1. Pump power $P_{p u m p} = \frac{m ˙ _{t o t} Δ p}{ρ η _{p}} = \frac{300 \times 12 \times 1 0 ^{6}}{1000 \times 0.7}$ . Yeh step kyun? Power = (pressure rise)×(volume flow), aur volume flow $\overset{m}{˙} / ρ$ hai; efficiency se divide karo real shaft demand ke liye.

Step 2. $P_{p u m p} = \frac{3.6 \times 1 0 ^{9}}{700} = 5.143 \times 1 0^{6}$ W. Yeh step kyun? Sirf numbers evaluate kar rahe hain — yeh woh hydraulic power hai jo turbine ko supply karni chahiye.

Step 3. Turbine specific work $w = η_{t} c_{p} T_{g g} \times 0.5 = 0.6 \times 2000 \times 1050 \times 0.5 = 6.3 \times 1 0^{5}$ J/kg. Yeh step kyun? GG gas ka har kilogram $c_{p} T_{g g}$ heat hold karta hai; bracket ( $0.5$ ) woh share hai jo expansion work mein turn hota hai, aur $η_{t}$ real-turbine losses ke liye trim karta hai.

Step 4. $\overset{m}{˙}_{g g} = P_{p u m p} / w = 5.143 \times 1 0^{6} /6.3 \times 1 0^{5} = 8.16 kg/s$ . Yeh step kyun? Needed power ÷ work-per-kilogram = required GG gas ke kilograms-per-second.

Step 5. $f = \overset{m}{˙}_{g g} / \overset{m}{˙}_{t o t} = 8.16/300 = 0.0272 = 2.72%$ .

Verify: $f \approx 2.7%$ 2–5% band mein clearly fit hai — few-percent penalty, jaise forecast kiya tha. Dekho Turbopump Fundamentals ki kyun pump term $Δ p / ρ$ is bookkeeping mein dominate karta hai.

Worked example

Example 5 — Cell F: $Δ p$ trend (kyun high chamber pressure hurt karta hai). Example 4 ka engine lo lekin chamber pressure demand ko double karo to $Δ p = 24$ MPa, baaki sab same. $f$ ka kya hoga?

Forecast: $f$ barhega, ghattega, ya same rahega?

Step 1. $\overset{m}{˙}_{g g} \propto Δ p$ (yeh fraction ke top par akela baitha hai, baaki sab constant). Yeh step kyun? $\overset{m}{˙}_{g g} = \frac{m ˙ _{t o t} Δ p}{ρ η _{p} η _{t} c _{p} T _{g g} [ \dots ]}$ mein, $Δ p$ double karne se sirf numerator double hota hai.

Step 2. New $f = 2 \times 0.0272 = 0.0544 = 5.44%$ . Yeh step kyun? Linear scaling: jis pressure ke against pump karna hai usse double karo → turbine flow jo jalaana aur dump karna hai usse double karo.

Verify: scratch se recompute karo — $P_{p u m p} = \frac{300 \times 24 \times 1 0 ^{6}}{700} = 1.0286 \times 1 0^{7}$ W, $w$ unchanged at $6.3 \times 1 0^{5}$ , to $\overset{m}{˙}_{g g} = 16.32$ kg/s aur $f = 16.32/300 = 0.0544$ ✓. Isliye gas generator cycles bahut high $p_{c}$ par apna edge kho dete hain — penalty pressure ke saath barhti hai. High- $p_{c}$ engines Staged Combustion Cycle ki taraf lean karte hain.

Worked example

Example 6 — Cell G: temperature cap (target $f$ hit karne ke liye kitna hot?). Tum Example 4 ke engine par $f$ ko $0.02$ tak rakhna chahte ho (original $Δ p = 12$ MPa). Baaki sab same rakhte hue, kaun sa $T_{g g}$ required hai — aur kya woh safe hai?

Forecast: 1050 K se hotter hoga ya cooler?

Step 1. $\overset{m}{˙}_{g g} = P_{p u m p} / (η_{t} c_{p} T_{g g} \times 0.5)$ aur $\overset{m}{˙}_{g g} = f \overset{m}{˙}_{t o t}$ se $T_{g g}$ ke liye solve karo: $T_{g g} = \frac{P _{p u m p}}{f m ˙ _{t o t} η _{t} c _{p} \times 0.5} .$ Yeh step kyun? $\overset{m}{˙}_{g g}$ , $T_{g g}$ ke inversely proportional hai, isliye chhota $f$ demand karna (chhota $\overset{m}{˙}_{g g}$ ) bada $T_{g g}$ force karta hai.

Step 2. $T_{g g} = \frac{5.143 \times 1 0 ^{6}}{0.02 \times 300 \times 0.6 \times 2000 \times 0.5} = \frac{5.143 \times 1 0 ^{6}}{3600} = 1428.6 K$ . Yeh step kyun? Known $P_{p u m p}$ aur target $f$ ko rearranged formula mein straight substitution.

Step 3. Safety check: $1429$ K practical $900$ – $1200$ K blade limit se zyada hai. Yeh step kyun? Parent note ki doosri galti — zyada hot chalao aur turbine melt ho jaayega.

Verify: $T_{g g} = 1428.6$ K back plug karo: $w = 0.6 \times 2000 \times 1428.6 \times 0.5 = 8.571 \times 1 0^{5}$ J/kg, $\overset{m}{˙}_{g g} = 5.143 \times 1 0^{6} /8.571 \times 1 0^{5} = 6.0$ kg/s, $f = 6.0/300 = 0.02$ ✓. Conclusion: tum $f = 0.02$ is tarah safely nahi reach kar sakte — blades melt ho jaayenge, isliye Example 4 ka bada $f = 0.027$ accept karo.

Worked example

Example 7 — Cell H: real-world thrust bookkeeping (Saturn-V style). Ek single gas-generator engine $\overset{m}{˙}_{t o t} = 2600$ kg/s ke saath chalta hai jisme $f = 0.03$ , main exhaust (figure ka orange arrow) $v_{e, c} = 2650$ m/s, dumped exhaust (plum arrow) $v_{e, g g} = 600$ m/s. Pressure terms ignore karte hue, total thrust do tarike se compute karo: (a) stream-by-stream momentum, (b) $v_{eff}$ ke through. Dono match karne chahiye.

Forecast: roughly kitne meganewtons — 3, 7, ya 20 MN ke paas?

Step 1 — main stream flow. $\overset{m}{˙}_{c} = (1 - f) \overset{m}{˙}_{t o t} = 0.97 \times 2600 = 2522$ kg/s; $\overset{m}{˙}_{g g} = 0.03 \times 2600 = 78$ kg/s. Yeh step kyun? Momentum sum karne se pehle total ko do physical streams (figure ke do arrows) mein split karo.

Step 2 — momentum thrust (method a). $F = \overset{m}{˙}_{c} v_{e, c} + \overset{m}{˙}_{g g} v_{e, g g} = 2522 (2650) + 78 (600)$ . Kyun? Mass expel karne se thrust $\sum \overset{m}{˙} v_{e}$ hai (Rocket Thrust Equation, pressure-matched case).

Step 3. $F = 6, 683, 300 + 46, 800 = 6, 730, 100 N \approx 6.73 MN$ . Yeh step kyun? Sirf do momentum fluxes add kar rahe hain — orange arrow almost sab kuch supply karta hai.

Step 4 — method b, $v_{eff}$ ke through. $v_{eff} = 0.97 (2650) + 0.03 (600) = 2570.5 + 18 = 2588.5$ m/s; $F = \overset{m}{˙}_{t o t} v_{eff} = 2600 \times 2588.5 = 6, 730, 100$ N. Kyun? $v_{eff}$ defined hi flow-weighted average ke roop mein isi liye tha taaki $\overset{m}{˙}_{t o t} v_{eff}$ stream sum reproduce kare.

Verify: dono methods $6.73$ MN dete hain — "7 MN" forecast ke paas, ek single bade first-stage engine ke liye sahi family mein. Units: (kg/s)(m/s)=N ✓.

Recall

Jab $v_{e, g g} = 0$ hota hai, toh $I_{s p}$ par penalty kaun si quantity ke barabar hoti hai? ::: Turbine flow fraction $f$ khud: $I_{s p, eff} = (1 - f) I_{s p, ideal}$ . Jab $f \to 1$ , toh $I_{s p, eff}$ kis value ki taraf jaata hai? ::: Feeble dump-plume value $v_{e, g g} / g_{0}$ — koi main chamber nahi bachta. $Δ p$ (chamber pressure demand) double karne se $f$ ka kya hota hai? ::: Double ho jaata hai — $\overset{m}{˙}_{g g} \propto Δ p$ , isliye high- $p_{c}$ engines badi dump penalty pay karte hain. Tum $f$ ghataane ke liye arbitrarily $T_{g g}$ kyun nahi badha sakte? ::: Blade material ~1200 K se upar melt ho jaata hai; fuel-rich chalane se $T_{g g} \approx 900$ – $1200$ K rehta hai. Bracket $[1 - (p_{o u t} / p_{in})^{(γ - 1) / γ}]$ kya represent karta hai? ::: GG gas ki heat ka woh fraction jo expansion ( $p_{in}$ se $p_{o u t}$ tak) turbine shaft work mein convert karta hai.