Expander cycle — hydrogen-cooled nozzle drives turbine: Exercises

3.3.24 · D4 · Physics › Rocket Propulsion › Expander cycle — hydrogen-cooled nozzle drives turbine

Shuru karne se pehle, har symbol ko pin down karte hain taaki kuch bhi bina explanation ke use na ho.

Definition Symbols, simple shabdon mein

$\overset{m}{˙}$ — mass flow rate: har second kitne kilograms fluid ek point se guzarta hai, $kg/s$ mein. Ek hose imagine karo: $\overset{m}{˙}$ woh "kg per second" hai jo bahar aata hai.
$c_{p}$ — specific heat at constant pressure: energy (joules) jo gas ke $1 kg$ ko $1 K$ garam karne ke liye chahiye. Bada $c_{p}$ = gas heat ke liye ek lalchi sponge hai.
$T$ — temperature kelvin ( $K$ ) mein. Kelvin bas Celsius ko shift kiya hua hai taaki $0 K$ absolute cold ho; hum isse isliye use karte hain kyunki gas laws ko absolute zero chahiye.
$p$ — pressure pascals ( $Pa$ ) mein, ya $MPa = 1 0^{6} Pa$ . Unit area per push.
$ρ$ (rho) — density: ek cubic metre mein packed kilograms, $kg/m^{3}$ .
$γ$ (gamma) — gas ka heat-capacity ratio, ek pure number (H₂ ke liye hum $1.4$ lete hain). Yeh control karta hai ki jab pressure girta hai to temperature kitna girta hai.
$η$ (eta) — ek efficiency, 0 aur 1 ke beech ka number. Turbine ke liye $η_{t}$ , pump ke liye $η_{p}$ . Real machines kuch energy waste karti hain, isliye hum (turbine ke liye) $η$ se multiply karte hain ya (pump ke liye) $η$ se divide karte hain.
$Q$ — heat rate, joules per second = watts ( $W$ ). $1 MW = 1 0^{6} W$ .
$W$ — power (bhi watts mein). $W_{t u r b}$ woh hai jo turbine banata hai; $W_{p u m p}$ woh hai jo pump khaata hai.

Level 1 — Recognition

Kya tum cycle aur symbols padh sakte ho?

L1.1

Ek sentence mein, closed expander cycle mein turbine ko spin karne wali energy kahan se aati hai?

Recall Solution

Garam nozzle/chamber walls se hydrogen coolant mein conduct hone wali heat se. Turbine ke liye kuch extra burn nahi hota. Hydrogen sirf baad mein, main chamber mein jalaya jaata hai.

L1.2

Har symbol ko uske unit se match karo: $\overset{m}{˙}$ , $c_{p}$ , $Q$ , $ρ$ .

Recall Solution

$\overset{m}{˙} \to kg/s$
$c_{p} \to J/(kg\cdotK)$ (yahan $14000 J/(kg\cdotK)$ )
$Q \to W$ (watts)
$ρ \to kg/m^{3}$

L1.3

Sach ya jhooth: closed expander cycle mein, kuch hydrogen turbine ke baad overboard dump kar di jaati hai.

Recall Solution

Jhooth. Yeh open (bleed) expander hai. Closed cycle mein saari garam hydrogen turbine se guzarti hai aur phir chamber mein jaati hai — kuch dump nahi hota. Woh cycle dekhne ke liye Gas generator cycle dekho jo sach mein gas dump karta hai.

Level 2 — Application

Ek equation mein cleanly plug karo.

L2.1

Hydrogen $\overset{m}{˙} = 4 kg/s$ par flow kar rahi hai, jacket mein $T_{in} = 45 K$ par enter karti hai, $T_{o u t} = 420 K$ par nikalti hai, $c_{p} = 14000 J/(kg\cdotK)$ ke saath. Absorbed heat rate $Q$ nikalo.

Recall Solution

$Q = \overset{m}{˙} c_{p} (T_{o u t} - T_{in}) = 4 \times 14000 \times (420 - 45)$ $Q = 4 \times 14000 \times 375 = 2.1 \times 1 0^{7} W = 21 MW$ Humne kya kiya: "kg per second" ko "joules per kg per kelvin" se aur "kelvin risen" se multiply kiya → joules per second = watts. Kyun: yeh literally $c_{p}$ ki definition hai forward direction mein run ki gayi. Woh $21 MW$ heat hai jo flame se churayi gayi hai jo wall ko cool bhi karti hai aur turbine ko fund bhi karti hai.

L2.2

Ek pump $\overset{m}{˙} = 4 kg/s$ hydrogen ( $ρ = 71 kg/m^{3}$ ) ko $Δ p = 14 MPa$ se $η_{p} = 0.7$ ke saath raise karta hai. Pump power nikalo.

Recall Solution

$W_{p u m p} = \frac{m ˙ Δ p}{ρ η _{p}} = \frac{4 \times 14 \times 1 0 ^{6}}{71 \times 0.7}$ $W_{p u m p} = \frac{5.6 \times 1 0 ^{7}}{49.7} \approx 1.127 \times 1 0^{6} W \approx 1.13 MW$ Neeche $ρ$ kyun: pump volume par kaam karta hai, mass par nahi. Volume flow $\overset{m}{˙} / ρ$ hai; kam density ka matlab bada volume hai, isliye hydrogen ki tiny $ρ$ ise pump karna mehnga banati hai. Dekho Liquid hydrogen properties.

Level 3 — Analysis

Do ya teen steps combine karo aur result ke baare mein reason karo.

L3.1

Garam hydrogen $T_{1} = 420 K$ par, $\overset{m}{˙} = 4 kg/s$ , $p_{1} = 13 MPa$ se $p_{2} = 9 MPa$ tak expand hoti hai. $γ = 1.4$ , $η_{t} = 0.75$ , $c_{p} = 14000$ lo. Turbine power nikalo.

Recall Solution

Pehle exponent: $\frac{γ - 1}{γ} = \frac{0.4}{1.4} = 0.2857$ . Pressure ratio: $\frac{p _{2}}{p _{1}} = \frac{9}{13} = 0.6923$ . Power tak raise karo: $0.692 3^{0.2857}$ . $ln (0.6923) = - 0.3677$ use karte hue, times $0.2857 = - 0.1051$ , $e^{- 0.1051} = 0.9003$ . $W_{t u r b} = 4 \times 14000 \times 420 \times (1 - 0.9003) \times 0.75$ $= 23520000 \times 0.0997 \times 0.75 \approx 1.759 \times 1 0^{6} W \approx 1.76 MW$ Pressure-ratio term kyun: ek isentropic (ideal, no-loss) expansion ke dauran, temperature aur pressure $T_{2} / T_{1} = (p_{2} / p_{1})^{(γ - 1) / γ}$ se ek doosre se locked hain — dekho Isentropic flow relations. Bracket $1 - (p_{2} / p_{1})^{(γ - 1) / γ}$ bas temperature ka lost fraction hai, jo ( $h = c_{p} T$ kyunki) shaft ko di gayi enthalpy ka fraction hai.

L3.2

L3.1 ka turbine aur do pumps use karte hue — fuel L2.2 se ( $1.13 MW$ ) aur ek oxygen pump jo $\overset{m}{˙}_{o x} = 22 kg/s$ ( $ρ_{o x} = 1140 kg/m^{3}$ ) ko $Δ p = 14 MPa$ par $η_{p} = 0.7$ ke saath move karta hai — kya cycle close hoti hai?

Recall Solution

Ox pump: $W_{o x} = \frac{22 \times 14 \times 1 0 ^{6}}{1140 \times 0.7} = \frac{3.08 \times 1 0 ^{8}}{798} \approx 3.86 \times 1 0^{5} W \approx 0.386 MW$ Total pump demand: $W_{p u m p} = 1.13 + 0.386 = 1.52 MW$ Turbine $1.76 MW$ se compare karo: $1.76 MW > 1.52 MW ✓$ Cycle khud ko sustain karti hai $\approx 0.24 MW$ ke margin ke saath (startup, throttle, control ke liye use hota hai). Dhyaan do: ox pump sasta hai ( $0.39 MW$ ) bahut zyaada mass move karne ke bawajood, kyunki oxygen hydrogen se $\sim 16 \times$ dense hai → tiny volume flow. Saari pump cost halki, fluffy hydrogen mein rehti hai.

Figure — Expander cycle — hydrogen-cooled nozzle drives turbine

Level 4 — Synthesis

Reasoning ki ek chain banao, limiting/degenerate cases include karo.

L4.1

Ek engineer propose karta hai turbine pressure ratio ko $p_{2} / p_{1} = 9/13$ se badhakar $p_{2} / p_{1} = 5/13$ karna (bahut deeper drop) zyaada turbine power nikalne ke liye. Naya $W_{t u r b}$ compute karo (same $\overset{m}{˙} = 4$ , $T_{1} = 420$ , $c_{p} = 14000$ , $γ = 1.4$ , $η_{t} = 0.75$ ). Phir hidden penalty batao.

Recall Solution

Naya ratio $5/13 = 0.3846$ . $ln (0.3846) = - 0.9555$ , times $0.2857 = - 0.2730$ , $e^{- 0.2730} = 0.7611$ . $W_{t u r b} = 4 \times 14000 \times 420 \times (1 - 0.7611) \times 0.75$ $= 23520000 \times 0.2389 \times 0.75 \approx 4.214 \times 1 0^{6} W \approx 4.21 MW$ Yeh bahut zyaada power hai. Hidden penalty: turbine ke baad, hydrogen ko phir bhi combustion chamber mein inject karna hoga, jo high pressure par hota hai. Agar tum gas ko $5 MPa$ tak drop karo lekin chamber ko $\sim 9$ – $10 MPa$ chahiye, to gas andar nahi ja sakti — pumps ko aur zyaada pre-pressurise karna padega, jo gain kha jaayega. Limiting case: jab $p_{2} \to 0$ to bracket $\to 1$ aur $W_{t u r b} \to \overset{m}{˙} c_{p} T_{1} η_{t}$ (ek hard ceiling — tum saari enthalpy se zyaada extract nahi kar sakte), lekin $p_{2} = 0$ ka matlab gas ka zero pressure hai aur woh chamber mein enter hi nahi kar sakti. Closed expander isliye deliberately sirf ek modest drop rakhta hai.

L4.2 (degenerate case)

Agar $p_{2} = p_{1}$ ho (bilkul koi pressure drop nahi) to $W_{t u r b}$ kya hai? Physically interpret karo.

Recall Solution

$p_{2} / p_{1} = 1$ , isliye $(1)^{0.2857} = 1$ , aur bracket $1 - 1 = 0$ : $W_{t u r b} = \overset{m}{˙} c_{p} T_{1} \times 0 \times η_{t} = 0 W$ Physical meaning: turbine work sirf pressure drop se extract karta hai. Koi drop nahi → koi expansion nahi → koi kaam nahi, chahe gas kitni bhi garam ho. Heat akela turbine ke liye useless hai; gas ke girne ke liye pressure difference chahiye. Isliye fuel pump ko pehle pressure build karna hota hai jise turbine baad mein spend kar sake.

Level 5 — Mastery

Full synthesis, scaling law, design judgement.

L5.1 — The square-cube ceiling

Ek baseline closed expander $Q_{ba se} = 21 MW$ jacket heat produce karta hai aur $W_{ba se} = 1.52 MW$ pump power chahta hai (L2/L3 se). Tum engine ki har linear dimension ko $k = 2.5$ se scale karte ho. Heat surface area $\propto k^{2}$ se scale hoti hai; pump demand propellant/volume flow $\propto k^{3}$ se scale hoti hai. Scaled heat aur demand nikalo, aur ratio (available heat):(baseline ke relative pump demand) nikalo.

Recall Solution

Scaled heat supply factor: $k^{2} = 2. 5^{2} = 6.25$ . Scaled pump demand factor: $k^{3} = 2. 5^{3} = 15.625$ . $Q_{sc a l e d} = 6.25 \times 21 = 131.25 MW, W_{sc a l e d} = 15.625 \times 1.52 = 23.75 MW$ Heat-to-demand advantage baseline ke relative: $\frac{k ^{2}}{k ^{3}} = \frac{1}{k} = \frac{1}{2.5} = 0.4$ Toh ek $2.5 \times$ -bada engine sirf $40%$ relative heat headroom rakhta hai. Interpretation: demand, supply se zyaada tezi se badhti hai jab tum scale up karte ho. Aakhirkar $k$ itna bada ho jaata hai ki walls physically itni heat pass nahi kar sakti ki pumps chala sakein — cycle close hona band ho jaata hai. Yeh Square-cube law hai aur yeh closed expanders ko kuch sau kN thrust ke paas cap karta hai (jaise RL10). Yeh geometry hai, engineering laziness nahi.

L5.2 — Design choice under the ceiling

Tumhe ek bada engine ( $\sim 2 MN$ ) chahiye lekin "kuch extra burn nahi" wala idea pasand hai. L5.1 ke result ko dekhte hue, tum kaunsa cycle choose karoge, aur ek trick kya hai jo larger size par expander-jaisi idea ko alive rakh sakti hai?

Recall Solution

$\sim 2 MN$ par closed expander enough wall heat supply nahi kar sakta (square-cube). Tum Staged combustion cycle (highest performance) ya Gas generator cycle (simpler, thoda gas dump karta hai) choose karoge. Expander idea stretch karne ki trick: open (bleed) expander par jao — sirf hydrogen ka ek fraction turbine se route karo aur usse overboard dump karo. Isse tum zyaada turbine flow ko higher drop par run kar sakte ho (zyaada power), kuch Specific impulse ki cost par, closed-cycle power cap ko sidestep karte hue jabki free wall heat ka use karte raho. Dekho Turbopump design ki shaft power pumps ko kaise size karti hai.

Recall Quick self-check (cloze)

Closed expander mein turbine ki energy nozzle walls se conduct hone wali heat se aati hai, kuch extra burn karne se nahi. Pump power ==volume flow ( $\overset{m}{˙} / ρ$ )== par depend karta hai, isliye dense propellant (oxygen) pump karna sasta hota hai. Zero pressure drop wala turbine zero kaam produce karta hai chahe temperature kuch bhi ho. Heat $k^{== 2 ==}$ se scale hoti hai lekin pump demand $k^{== 3 ==}$ se scale hoti hai, jo closed expander thrust ko cap karta hai.

Turbine power pressure ratio par depend karta hai, sirf temperature par nahi — kyun? ::: Kyunki isentropic expansion ke liye $T_{2} / T_{1} = (p_{2} / p_{1})^{(γ - 1) / γ}$ hai, isliye enthalpy drop pressure ratio se set hoti hai. Closed expander 2 MN tak scale kyun nahi ho sakta? ::: Heat supply surface area ( $k^{2}$ ) ke saath badhti hai lekin pump demand flow ( $k^{3}$ ) ke saath badhti hai; ratio $1/ k$ se girta hai jab tak walls pumps ko feed nahi kar sakti.