Visual walkthrough — Expander cycle — hydrogen-cooled nozzle drives turbine

3.3.24 · D2 · Physics › Rocket Propulsion › Expander cycle — hydrogen-cooled nozzle drives turbine

Ye 3.3.24 Expander cycle — hydrogen-cooled nozzle drives turbine (Hinglish) ka child hai. Agar koi word yahan unexplained lage, toh woh ek bug hai — batao aur main fix kar deta hoon.

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Step 1 — "Mass flow" kya hota hai aur kyun har cheez wahin se shuru hoti hai?

KYA. Kisi bhi physics se pehle, hume ek number chahiye: fuel engine mein kitni tezi se move kar raha hai. Hum ise mass flow rate kehte hain, likha jaata hai $\dot{m}$ (padho "m-dot"). Upar wala chota dot physicist ka shorthand hai "har second kitna". Toh $\dot{m}$ matlab hai har second kitne kilograms of stuff ek point se guzar rahi hai, measure hoti hai $\text{kg/s}$ mein.

YEH quantity kyun, aur koi aur cheez — maano total mass — kyun nahi? Ek rocket ko koi fark nahi padhta ki tank mein ek instant mein kitna fuel hai; usse stream se matlab hai — ek river of fuel jo continuously beh rahi hai. Har downstream cheez (pick up ki gayi heat, kiya gaya kaam) ek rate hai (per second), toh hume mass ki bhi ek rate se shuru karna hoga.

PICTURE. Neeche wala pipe dekho. Har second, hydrogen ka ek slug jiska mass $\dot{m}$ hai, dashed cyan line cross karta hai. Yahi woh river hai jise hum engine mein aakhir tak follow karenge.

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Step 2 — Cooling jacket heat churaata hai: $Q=\dot{m}{c}_p\Delta T$

KYA. Thanda hydrogen temperature $T_{in}$ par nozzle ke around lipti channels mein enter karta hai aur hotter temperature $T_{out}$ par bahar nikalta hai. Woh har second jitni heat absorb karta hai woh hai $Q=\underset{\begin{array}{c}\text{kg per}\\ \text{second}\end{array}}{\underbrace{\dot{m}}}\cdot \underset{\begin{array}{c}\text{joules to heat}\\ \text{1 kg by 1 K}\end{array}}{\underbrace{c_p}}\cdot \underset{\begin{array}{c}\text{kelvin the fuel}\\ \text{warmed up by}\end{array}}{\underbrace{(T_{out}-T_{in})}}$

YEH formula kyun? Symbol $c_p$ specific heat capacity hai: yeh defined hai as woh energy jo ek kilogram ko ek kelvin garam karne ke liye chahiye (constant pressure par — yahi woh choti $p$ ka matlab hai). Toh joules = (kg) × (joules per kg per K) × (K). $\dot{m}$ se "per second" multiply karo aur $Q$ joules per second mein aata hai, jo hai watts ($\text{W}$) — ek power. Hum $c_p$ use karte hain kisi aur property ki jagah kyunki humara sawaal literally hai "is gas ko warm karne ke liye kitni heat?", aur $c_p$ exactly yahi answer karne ke liye bana hai.

PICTURE. Channel ke saath temperature bar follow karo: inlet par cold blue, outlet par hot amber. Andar aane wali heat ka area (flame se red arrows) woh $Q$ hai jo hum ne bank kiya.

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Step 3 — Heat enthalpy banti hai, turbine ki khoraak

KYA. Woh saari heat ne gas ki enthalpy badha di — enthalpy $h$ ko flowing gas ki "usable energy content" samjho. Ek ideal gas ke liye ek beautifully simple link hai: $h=c_{p}T$ Har symbol: $h$ energy per kilogram hai ($\text{J/kg}$); $c_p$ hum abhi mile; $T$ absolute temperature in kelvin hai.

HUM "heat" se "enthalpy" par kyun switch karte hain? Turbine ko seedha heat nahi milti — use ek hot, high-pressure gas stream milti hai. Us stream ki woh property jo turbine shaft work mein convert kar sakta hai woh hai enthalpy. Toh hum woh heat jo humne churayi (Step 2) ko us currency mein translate karte hain jise turbine actually spend karta hai: enthalpy. Equation $h=c_{p}T$ exchange rate hai, aur yeh ideal gas ke liye exact hai.

PICTURE. Ek thermometer upar jaana = enthalpy tank ka bharna. Hotter gas = fuller tank = turbine aur zyada extract kar sakta hai.

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Step 4 — Turbine pressure drop karke enthalpy kharach karta hai

KYA. Hot gas pressure $p_1$ par turbine mein flow karta hai aur lower pressure $p_2$ par bahar nikalta hai. Pressure drop hone par woh thanda bhi hota hai, aur jo enthalpy woh khoata hai woh shaft work ban jaata hai. Ideal (loss-free) work rate hai $W_{turb}^{ideal}=\dot{m}{c}_p{T}_1\left[1-{\left(\frac{p_2}{p_1}\right)}^{\frac{\gamma -1}{\gamma }}\right]$

Term by term:

• $\dot{m}{c}_p{T}_1$ — turbine inlet par pahunchne wali enthalpy rate (Steps 1–3, jahan $T_1=T_{out}$).
• $\frac{p_2}{p_1}$ — pressure ratio; 1 se kam kyunki gas lower pressure par expand hoti hai.
• $\gamma$ (gamma) — gas ka heat-capacity ratio; hydrogen-jaise diatomic gas ke liye $\gamma \approx 1.4$. Yeh measure karta hai gas kitni "springy" hai.
• Exponent $\frac{\gamma -1}{\gamma }$ aur bracket neeche wale isentropic rule se aata hai.

YEH exponent kyun — $(p_2/p_1{)}^{(\gamma -1)/\gamma }$ kahan se aata hai? Ek ideal, loss-free expansion isentropic hoti hai (constant entropy — ek fancy word for "koi wasted disorder nahi"). Isentropic flow temperature ko pressure se aise jodhti hai: $\frac{T_2}{T_1}={\left(\frac{p_2}{p_1}\right)}^{\frac{\gamma -1}{\gamma }}$ (dekho Isentropic flow relations). Work enthalpy drop $\dot{m}{c}_p(T_1-T_2)$ hai. $T_2=T_1(p_2/p_1{)}^{(\gamma -1)/\gamma }$ substitute karo aur $T_1$ factor out karo → exactly wahi bracket milta hai. Hum isentropic relation isliye use karte hain kyunki yeh best possible case hai; yeh turbine output ki ceiling batata hai.

PICTURE. Gas balloon ek paddle wheel ko push karta hua jaise woh high (amber) se low (cyan) pressure tak expand hota hai; temperature bar ka shrinkna woh enthalpy hai jo shaft ko di ja rahi hai.

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Step 5 — Real turbines kuch waste karte hain: ${\eta }_t$ se multiply karo

KYA. Koi bhi real machine loss-free nahi hoti. Hum turbine efficiency ${\eta }_t$ (eta-t) lagaate hain, jo 0 aur 1 ke beech ki ek number hai: $W_{turb}=\dot{m}{c}_p{T}_1\left[1-{\left(\frac{p_2}{p_1}\right)}^{\frac{\gamma -1}{\gamma }}\right]{\eta }_t$

MULTIPLY kyun karte hain? ${\eta }_t$ defined hai as (real work) ÷ (ideal work). Toh real work = ideal work × ${\eta }_t$. Agar ${\eta }_t=0.75$ hai, turbine theoretical maximum ka 75% capture karta hai; baaki 25% swirl, friction aur leftover heat ke roop mein nikal jaata hai.

PICTURE. Do bars side by side: tall ideal bar aur shorter real bar, gap labelled "losses".

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Step 6 — Pumps kya maangti hain: $W_{pump}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$

KYA. Woh shaft jo turbine spin karta hai pumps chalata hai. Ek pump fluid ki pressure $\Delta p$ se badhata hai. Iske liye jis power ki zarurat hai woh hai $W_{pump}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$

Term by term:

• $\Delta p$ — woh pressure rise jo pump supply karna chahiye ($\text{Pa}$).
• $\rho$ (rho) — fluid ki density, kilograms per cubic metre. Dense fluids mein kam volume move karna padta hai.
• $\frac{\dot{m}}{\rho }$ — volume flow (${\text{m}}^3/\text{s}$): mass per second ÷ mass per cubic metre = cubic metres per second.
• ${\eta }_p$ — pump efficiency; hum divide karte hain kyunki ek real pump ideal se zyada input maangta hai.

Volume ko push karna $\Delta p$ per unit volume kyun cost karta hai? Work hai force × distance; pressure hai force per area. Area $A$ ka ek piston lekr $\Delta p$ pressure step ke upar distance $d$ tak push karo: work $=(\Delta p\cdot A)\cdot d=\Delta p\cdot (Ad)=\Delta p\times \text{volume}$. Toh $\Delta p$ se utha hua har cubic metre $\Delta p$ joules cost karta hai. Volume-per-second se multiply karo aur watts milta hai. Hum ${\eta }_p$ se isliye divide karte hain kyunki friction matlab hamesha extra pay karna.

PICTURE. Ek piston fluid ko pressure "hill" ke upar dhakelta hua; swept volume shaded, hill ki height = $\Delta p$.

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Step 7 — Self-sustaining condition (poora point yahi hai)

KYA. Step 5 (supply) ko Step 6 ki do copies (demand) ke saath, ek per pump, assemble karo. Engine tab tak chalta rehta hai jab tak turbine kam se kam utna banata hai jitna dono pumps khaate hain: $\boxed{\underset{\text{supply — nozzle heat se}}{\underbrace{{\dot{m}}_{H_2}{c}_p{T}_1[1-(p_2/p_1{)}^{\frac{\gamma -1}{\gamma }}]{\eta }_t}}\ge \underset{\text{demand — fuel pump + ox pump}}{\underbrace{\frac{{\dot{m}}_f\Delta p_f}{{\rho }_f{\eta }_{p,fu}}+\frac{{\dot{m}}_{ox}\Delta p_{ox}}{{\rho }_{ox}{\eta }_{p,ox}}}}}$

Demand side par term by term:

• ${\dot{m}}_f,\Delta p_f,{\rho }_f,{\eta }_{p,fu}$ — fuel (hydrogen) pump, exactly Step 6 formula with fuel subscripts. Dhyaan do ${\dot{m}}_f={\dot{m}}_{H_2}$ — wahi same hydrogen stream jise hum jacket aur turbine mein follow karte rahe.
• ${\dot{m}}_{ox},\Delta p_{ox},{\rho }_{ox},{\eta }_{p,ox}$ — oxygen pump, same formula with oxidizer subscripts aur uski apni efficiency.

Inequality kyun hai aur equals sign kyun nahi? Agar supply exactly demand ke barabar ho toh engine bas barely tick over kar sakta hai; koi bhi hiccup use stall kar deta hai. Real engines chahte hain ki $W_{turb}$ demand se zyada ho taaki startup aur throttling ke liye margin ho. Agar left side right se neeche aa jaaye, toh shaft slow hoga, pumps weak hoge, flow drop hoga — engine khud hi slow ho jaayega.

PICTURE. Ek balance beam: ek pan mein turbine power, doosre mein summed pump power. Cycle "close" hoti hai jab turbine wala pan level ya neeche baithe.

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Step 8 — Woh degenerate cases jo beam ko survive karne chahiye

Har scenario cover hona chahiye, toh formula ko uske limits par push karte hain.

Case A — koi heat pick up nahi hui ($T_{out}=T_{in}$). Tab $Q=0$ aur gas kabhi hot nahi hoti, toh turbine inlet temperature $T_1$ low rehti hai. Bracket $[1-(p_2/p_1{)}^{(\gamma -1)/\gamma }]$ unchanged hai (woh sirf pressures aur $\gamma$ par depend karta hai), lekin yeh $T_1$ ko multiply karta hai — aur $T_1\to$ small hone par $W_{turb}\to$ small hota hai → cycle start nahi ho sakta. Isliye ek cold engine ko pehli trickle of hot gas lene ke liye ek start cartridge ya tank-head bleed ki zarurat hoti hai.

Case B — zero pressure drop ($p_2=p_1$). Tab $p_2/p_1=1$, bracket $1-1^{...}=0$ ho jaata hai, toh $W_{turb}=0$. Jo turbine pressure drop nahi karta woh koi kaam nahi karta — obvious hai, aur formula bhi agree karta hai. Expanders is drop ko deliberately modest rakhte hain (Example 2 mein sirf 12→8 MPa use hua) kyunki gas ko baad mein chamber mein enter karne ke liye still high-pressure hona chahiye.

Case C — scale up karna (square–cube trap). Har length ko $k$ se multiply karo. Heat nozzle surface area ke upar pick up hoti hai, jo $k^2$ se grow karti hai, toh heat supply $\propto k^2$ hai. Lekin propellant mass flow jo pumps ko move karni hai woh chamber ke throughput ke saath scale hoti hai — volume/throat ke saath, yaani $\propto k^3$. Kyunki pump power $W_{pump}\propto \dot{m}\Delta p$ hai aur $\dot{m}\propto k^3$ (fixed $\Delta p$ par), demand $k^3$ se grow karta hai. Compare karo:

$\frac{\text{demand}}{\text{supply}}\propto \frac{k^3}{k^2}=k.$

Toh jaise-jaise engine bada hota hai, demand supply se aage nikal jaata hai aur beam ka right pan eventually crash kar jaata hai. Yeh Square-cube law hai aur yeh closed expanders ko kuch hundred kN thrust tak cap karta hai — ek geometric limit hai, laziness nahi. (Isliye bade engines gas generator ya staged combustion use karte hain.)

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Ek-picture summary

Yeh poori derivation compress ki gayi hai: hydrogen river (Step 1) jacket mein enter karta hai aur heat peeta hai (Step 2 → enthalpy, Step 3), use turbine mein pressure drop karte hue kharach karta hai (Steps 4–5), shaft woh power do pumps tak le jaata hai (Step 6), aur beam decide karti hai ki kya loop close hoti hai (Step 7), hamesha square–cube law ki chhaya mein (Step 8).

Recall Feynman retelling — apne words mein bolke batao

Thanda hydrogen ek stream mein flow karta hai ($\dot{m}$). Woh roasting nozzle ke around pipes mein bhaagta hai aur heat peeta hai — kitni heat uski enormous appetite $c_p$ aur kitne degrees woh warm hua ($Q=\dot{m}{c}_p\Delta T$) par depend karta hai. Woh heat "usable content" banti hai (enthalpy, $h=c_{p}T$). Hot gas phir ek turbine mein squeeze hoti hai: jaise-jaise pressure $p_1$ se $p_2$ tak girti hai woh thandi hoti hai, aur khoya hua enthalpy ek shaft ghoomata hai ($W_{turb}$). Real turbines ideal ka sirf ek fraction ${\eta }_t$ hi rakhte hain. Woh shaft do pumps chalata hai — ek hydrogen fuel ke liye (${\dot{m}}_f$, efficiency ${\eta }_{p,fu}$), ek oxygen ke liye (${\dot{m}}_{ox}$, efficiency ${\eta }_{p,ox}$); har pump ki cost hai "main jitna pressure add karta hoon times kitna volume main move karta hoon, divided by main kitna leaky hoon" ($\dot{m}\Delta p/\rho {\eta }_p$). Engine forever tab hi chalta hai jab turbine ka output kam se kam dono pumps ka combined bill ho. Bina heat ke start nahi hoga (kyunki $T_1\to 0$), bina pressure drop ke kuch nahi karta, aur — kyunki heat area se aati hai ($k^2$) jabki pump work mass flow se ($k^3$) — ise kabhi bada nahi banaya ja sakta.

Recall Quick self-test

Step 2 mein hum $c_p$ ki jagah koi aur property kyun nahi use kar sakte? ::: Kyunki $c_p$ defined hai as heat per kg per kelvin — yahi exactly woh number hai jo answer karta hai "is gas ko warm karne ke liye kitni heat?" Exponent $(\gamma -1)/\gamma$ kahan se aata hai? ::: Isentropic $T$–$p$ relation $T_2/T_1=(p_2/p_1{)}^{(\gamma -1)/\gamma }$ se, jo best-case loss-free expansion hai. Pumps ke liye ${\eta }_p$ se divide kyun karte hain lekin turbines ke liye ${\eta }_t$ se multiply? ::: Pumps ko ideal se zyada input chahiye (÷), turbines ideal se kam output dete hain (×). Subscripts $f$ aur $ox$ ka kya matlab hai, aur har pump ko apna ${\eta }_p$ kyun milta hai? ::: $f$ = fuel (hydrogen), $ox$ = oxidizer (oxygen); har ek alag machine hai jo alag fluid move karti hai, isliye har ek ki apni efficiency hai ${\eta }_{p,fu}$, ${\eta }_{p,ox}$. Expanders chhote kyun rehte hain? ::: Heat supply area se grow karti hai ($k^2$), pump demand mass flow se ($k^3$); demand/supply $\propto k$ eventually balance todh deta hai.

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