Expander cycle — hydrogen-cooled nozzle drives turbine: Question bank

3.3.24 · D5 · Physics › Rocket Propulsion › Expander cycle — hydrogen-cooled nozzle drives turbine

Shuru karne se pehle, ek-sentence ka picture yaad rakho: cold liquid hydrogen ko coolant ki tarah use kiya jaata hai, woh nozzle heat absorb karta hai aur high-pressure gas mein boil ho jaata hai, aur woh gas turbopumps ko spin karti hai before jalaaye jaane se. Turbine ko feed karne ke liye kuch bhi nahi jalaya jaata.

Symbols aur sign conventions (pehle inhe build karo)

Koi bhi trap samajhne se pehle, har symbol ko earn karna hoga. Ise ek baar padho; phir neeche ke reveals inhe freely use karte hain.

Definition Is page ke har symbol ko simple words mein

$\overset{m}{˙}$ — mass flow rate, "har second kitne kilograms fluid ek point se guzarte hain" (units kg/s). Dot ka matlab hai "per second". Ek stopwatch imagine karo jo streaming kilograms count kar raha ho.
$c_{p}$ — specific heat capacity constant pressure par: woh energy (joules mein) jo ek kilogram ko ek kelvin warm karne ke liye chahiye. Bada $c_{p}$ = fluid ek greedy heat sponge hai. Hydrogen ka ~ $14000 J/kg\cdotK$ hai.
$ρ$ (Greek "rho") — density, ek cubic metre mein packed kilograms (kg/m³). Dense = apne size ke liye heavy (oxygen), light = apne size ke liye fluffy (hydrogen).
$Δ p$ (triangle $Δ$ ka matlab hai "change in") — pressure rise jo ek pump add karta hai, yaani exit pressure minus inlet pressure, pascals (Pa) mein.
$γ$ (Greek "gamma") — specific-heat ratio $c_{p} / c_{v}$ , ek pure number (~1.4 diatomic gases jaise H₂ ke liye). Yeh measure karta hai ki gas kitna "springy" hai jab compress ya expand hota hai; yeh tab appear hota hai jab hum gas ko bina heat add kiye expand karte hain.
$η_{p}$ , $η_{t}$ (Greek "eta") — pump efficiency aur turbine efficiency, 0 aur 1 ke beech pure numbers jo bataate hain ki ideal work ka kitna fraction actually achieve hota hai. Real machines kuch waste karte hain, isliye ye $< 1$ hain.
$p_{1}, p_{2}$ — turbine inlet aur outlet par pressures; $T_{1}, T_{2}$ matching temperatures.

Figure — Expander cycle — hydrogen-cooled nozzle drives turbine

Upar ki figure hydrogen loop ko energy arrows ke saath trace karti hai: jacket mein heat in (green, $+ Q$ ), shaft par turbine se work out, aur usi shaft se pumps mein work in. Notice karo ki single shaft turbine aur pumps ko couple karta hai — ek spinning rod, isliye sab same rotational speed par ghoomte hain; turbine ko exactly utna torque produce karna hoga jitna do pumps us speed par demand karte hain.

Formulas kahan se aate hain (taaki traps obvious ho jaayein)

Intuition Isentropic relation

$T_{2} = T_{1} (p_{2} / p_{1})^{(γ - 1) / γ}$ kyun appear hota hai Jab turbine gas ko fast aur bina heat add kiye expand karta hai, hum ise isentropic kehte hain (koi heat exchange nahi, koi waste nahi). Ek ideal gas ke liye do facts hold karte hain: ideal-gas law $p V \propto T$ , aur "koi heat add nahi" ka matlab hai $p V^{γ} = constant$ (springiness rule with exponent $γ$ ). In dono ko combine karo — $V$ eliminate karo — aur tum $T p^{(1 - γ) / γ} = constant$ par pahunch jaate ho, yaani $\frac{T _{2}}{T _{1}} = (\frac{p _{2}}{p _{1}})^{\frac{γ - 1}{γ}} .$ Yeh tool kyun, aur simpler proportion kyun nahi? Kyunki pressure aur temperature same step mein nahi girte — exponent $(γ - 1) / γ$ (about $0.29$ for $γ = 1.4$ ) batata hai ki temperature pressure se kam steeply girta hai. Kyunki $p_{2} < p_{1}$ , ratio $< 1$ hai, isliye $T_{2} < T_{1}$ : gas kaam karte waqt cool ho jaati hai. Dekho Isentropic flow relations.

Square-cube limit ko assert karne se zyada dekhne layak hai — neeche ka plot heat supply ( $\propto L^{2}$ ) aur pump demand ( $\propto L^{3}$ ) ko ek definite size par cross karte dikhata hai, jiske aage koi closed expander close nahi ho sakta.

True ya false — justify karo

Expander cycle turbine drive karne ke liye ek preburner mein thoda sa fuel jalaata hai.

False. Ek closed expander turbine ke liye kuch nahi jalataa; turbine gas woh hydrogen hai jo sirf nozzle-wall heat se heated hoti hai. Ise baad mein main chamber mein jalaya jaata hai. Preburners Staged combustion cycle aur Gas generator cycle se belong karte hain.

Woh heat jo turbine drive karti hai "free" hai iss sense mein ki isme koi extra propellant nahi lagta.

True. Nozzle ko waise bhi cool karna padta hai (dekho Regenerative cooling); fuel ko walls ke through route karke waste heat recycle hoti hai jo warna wall ko melt kar deti. Uss heat ko create karne ke liye koi propellant spend nahi hota.

Hydrogen turbine se nikalte waqt enter karne se zyada pressure par hoti hai.

False. Turbine gas ko expand karke kaam nikalta hai, isliye pressure

p_{1}

p_{2} < p_{1}

tak girta hai. Woh lower-pressure gas phir bhi chamber pressure se upar honi chahiye taaki inject ho sake.

Ek closed expander cycle mein, saari heated hydrogen eventually combustion chamber tak pahunchti hai.

True. "Closed" ka matlab hai kuch dump nahi hota: 100% hydrogen turbine → chamber se guzarti hai. Ek open (bleed) expander mein ek fraction overboard vent ho jaata hai, jisse thodi si Specific impulse ka nuksan hota hai.

Kyunki turbine hydrogen mein energy add karta hai, hydrogen turbine se nikalte waqt enter karne se zyada hot hoti hai.

False. Turbine energy remove karta hai (

W_{t u r b} > 0

gas se nikalta hai), isliye expansion ise cool karta hai:

T_{2} = T_{1} (p_{2} / p_{1})^{(γ - 1) / γ} < T_{1}

. Heating pehle ho chuki thi, cooling jacket mein.

Ek expander cycle ko arbitrarily high thrust tak scale kiya ja sakta hai sirf nozzle bada karke.

False. Heat supply surface area ke saath badhti hai (

\propto L^{2}

) lekin pump demand propellant flow ke saath badhti hai (

\propto L^{3}

). Square-cube law se demand aakhirkar supply ko outrun kar leti hai, closed expanders ko kuch sau kN ke paas cap kar deti hai.

Oxidizer pump generally fuel pump se kilogram per kilogram sasta (less power) hota hai.

True. Pump power volume flow

\overset{m}{˙} / ρ

ke saath scale karta hai. Liquid oxygen ki density

ρ

hydrogen se kaafi zyada hai, isliye same

Δ p

ke liye per kg kam power lagti hai despite zyada mass flow ke.

Kerosene coolant/working fluid ke roop mein hydrogen jitna hi kaam karega agar tum slightly lower efficiency accept karo.

False. Kerosene heat hone par coke ho jaata hai (carbonize aur channels block kar deta hai) aur iska

c_{p}

low hai, isliye yeh turbine chalane ke liye enough clean heat absorb nahi kar sakta. Hydrogen ka huge

c_{p}

(~14 kJ/kg·K) hi cycle ko possible banata hai — dekho Liquid hydrogen properties.

Error dhundo

Q = \overset{m}{˙} c_{p} T_{o u t}

jacket dwara absorb ki gayi heat hai."

Error: yeh temperature rise honi chahiye,

Q = \overset{m}{˙} c_{p} (T_{o u t} - T_{in})

. Heat is par depend karti hai ki temperature kitni change hui, absolute exit temperature par nahi.

"Turbine work formula mein

(γ - 1) / γ

power par

p_{1} / p_{2}

use karta hai."

Error: ratio

(p_{2} / p_{1})

hai, yaani exit over inlet, jo 1 se kam hai.

p_{1} / p_{2}

use karne se

> 1

term aata aur negative work — expansion ke liye physically galat hai. Dekho Isentropic flow relations.

"Pump power

W = \overset{m}{˙} Δ p ρ / η_{p}

hai."

Error: density denominator mein hai:

W = \overset{m}{˙} Δ p / (ρ η_{p})

. Denser fluid ka matlab hai chhota volume flow

\overset{m}{˙} / ρ

, isliye isse

Δ p

ke across push karne ke liye kam power — toh

ρ

divide karna chahiye, multiply nahi.

"Efficiency

η_{t} = 0.75

ko poore bracket ko multiply karna chahiye aur thoda aur add karna chahiye kyunki real turbines ideal se behtar hote hain."

Error: real turbines ideal se worse hote hain, isliye

η_{t} < 1

ideal work ko down scale karta hai. Tum kabhi isentropic maximum se zyada nahi paate.

"Self-sustaining condition

W_{t u r b} \leq W_{p u m p}

hai."

Error: inequality doosri taraf point karti hai. Turbine ko pump demand kam se kam supply karni chahiye:

W_{t u r b} \geq W_{p u m p, f u e l} + W_{p u m p, o x}

, warna shaft slow ho jaata hai aur engine stall kar jaata hai.

"Kyunki hydrogen jacket mein heat absorb karta hai, isliye yeh tank ki taraf chamber se bahar flow kar raha hoga."

Error: flow hai tank → pump → cooling jacket → turbine → chamber. Hydrogen poore time chamber ki taraf move karta hai; woh bas raaste mein heat up hota hai.

"Turbine aur har pump apni marzi ki speed par spin kar sakte hain."

Error: woh ek shaft share karte hain, isliye turbine aur dono pumps same rotational speed par ghoomte hain. Design ko turbine ke torque-vs-speed curve ko us common speed par pumps ki demand se match karna hoga — dekho Turbopump design.

Why questions

Hydrogen ko kisi bhi doosre fuel se working fluid ke roop mein kyun choose kiya jaata hai?

Iska bahut bada specific heat

c_{p}

chhoti si mass ko har kelvin par bahut saari enthalpy pick up karne deta hai, aur yeh coking ke bina cleanly boilta hai, isliye nozzle heat abundant, usable high-pressure gas mein convert ho jaati hai.

Expander cycle propellant waste kyun nahi karta, unlike gas generator cycle?

Turbine gas woh hydrogen hai jo wall se heated hai, combustion se bani gas nahi. Yeh main chamber mein flow karte hue jaati hai aur wahan jalti hai — kuch overboard dump nahi hota, isliye Specific impulse high rehti hai.

Density

ρ

pump power ke denominator mein kyun appear karti hai?

Kyunki pressure raise karne ka kaam volume par act karta hai, aur volume flow

\overset{m}{˙} / ρ

hai — same mass ka denser fluid kam volume occupy karta hai, isliye isse

Δ p

ke across push karne ke liye per kilogram kam kaam chahiye.

Turbine ko sirf ek modest pressure drop hi kyun use karna chahiye, gas ko fully expand karne ki bajay?

Turbine se nikalti hydrogen ko phir bhi combustion chamber mein inject karna padta hai, jo high pressure par hai. Agar turbine gas ko bahut zyada expand kar de, toh uska exit pressure chamber pressure se neeche aa jaayega aur woh enter nahi kar sakti.

Exponent

(γ - 1) / γ

(1 nahi) turbine mein temperature kaise girti hai yeh kyun govern karta hai?

Kyunki bina heat add kiye expansion

p V^{γ} =

const follow karta hai, simple proportionality nahi; ise ideal-gas law ke saath combine karne se temperature pressure se kam steeply girti hai, jo exactly woh fractional exponent encode karta hai.

Square-cube law expander cycles ko specifically kyun limit karta hai, gas generators ko nahi?

Expanders heat gather karne ke liye wall area (

L^{2}

) par depend karte hain, jo

L^{3}

propellant demand se peeche reh jaata hai. Ek gas generator apna turbine power propellant jalake banata hai, isliye uski supply bhi volume ke saath scale karti hai aur geometric ceiling nahi hit karta.

Same

Q

ko "nozzle cool karna" aur "turbine power karna" dono kyun kaha jaata hai?

Dono ek hi energy hai (

Q > 0

wall se nikalta hai aur fluid mein jaata hai): yeh ek saath wall ko melt hone se rokta hai aur gas enthalpy raise karta hai jise turbine baad mein extract karta hai. Ek heat flow do kaam karta hai.

Ek open (bleed) expander closed wale se zyada turbine power kyun allow karta hai?

Turbine hydrogen ko overboard dump karna yeh constraint hata deta hai ki woh chamber-injectable pressure par survive kare, isliye zyada bada flow aur bada pressure drop use kiya ja sakta hai — vented propellant ke thrust contribution ki cost par. (Schematic mein bleed-port branch dekho.)

Edge cases

Agar hydrogen cooling jacket mein aur nikalte waqt same temperature par ho (

T_{o u t} = T_{in}

), toh kitna turbine power available hai?

Q = \overset{m}{˙} c_{p} \cdot 0 = 0

, isliye koi enthalpy add nahi hoti aur turbine koi kaam produce nahi kar sakta — cycle start nahi ho sakta. Kuch temperature rise zaroori hai.

Engine startup ke waqt kya hota hai, before nozzle hot ho?

Bahut kam wall heat hoti hai, isliye turbine full power nahi bana sakta — expander cycles ko ek startup transient chahiye hota hai (jaise tank pressure, stored energy tap karna, ya power balance mein built-in margin) walls ke hot hone tak flow bootstrap karne ke liye.

Agar pump efficiency

η_{p}

zero ki taraf girne lage, toh power balance kya kehta hai?

Pump demand

\overset{m}{˙} Δ p / (ρ η_{p})

infinity ki taraf blow up ho jaata hai, isliye fixed turbine power usse cover nahi kar sakta aur self-sustaining inequality fail ho jaati hai — engine close nahi ho sakta.

Ek fixed turbine ke liye, agar tum bahut zyada high chamber pressure demand karo (pumps ke across bada

Δ p

), toh kya limit karta hai?

Pump power

Δ p

ke saath linearly badhti hai jabki turbine power available wall heat se cap hai. Kuch

Δ p

ke baad demand supply se exceed kar leti hai, jo ek reason hai ki expanders moderate chamber pressures prefer karte hain.

Vanishingly small pressure ratio

p_{2} / p_{1} \to 1

ki limit mein, turbine work kya hai?

Bracket

1 - (p_{2} / p_{1})^{(γ - 1) / γ} \to 0

, isliye

W_{t u r b} \to 0

. Koi pressure drop nahi matlab koi expansion nahi aur koi extractable work nahi — turbine ko kuch karne ke liye genuine pressure ratio dena hoga.

Recall Quick self-test

Ek closed expander mein turbine gas heated hoti hai ::: nozzle/chamber walls ke through conducted heat se (regenerative cooling), combustion se nahi. Pump power volume flow ke saath scale karti hai, isliye denser propellant ::: pump karna sasta hai per kilogram (oxygen same $Δ p$ ke liye hydrogen se kam power leta hai). Expanders size-limited hain kyunki heat supply area ki tarah badhti hai jabki demand ::: volume/flow ki tarah — square-cube law. Turbine aur dono pumps ek shaft share karte hain, isliye sab spin karte hain ::: same rotational speed par, aur turbine torque combined pump demand se match karna chahiye.

Dekho bhi: Turbopump design · Regenerative cooling · Isentropic flow relations · Square-cube law · Specific impulse · Liquid hydrogen properties