Worked examples — Expander cycle — hydrogen-cooled nozzle drives turbine

3.3.24 · D3 · Physics › Rocket Propulsion › Expander cycle — hydrogen-cooled nozzle drives turbine

Yeh page expander cycle ka drill hall hai. Parent note ne chaar equations banaayi theen; yahan hum unpar har tarah ka input daalte hain — normal numbers, zeros, degenerate cases, limiting values, ek word problem, aur ek exam twist — taaki koi bhi scenario tumhe akela na pakde.

Shuru karne se pehle, teen tools ka ek reminder (saare parent se):

Recall Teen equations jo hum baar baar use karte rahenge

Cooling jacket mein absorbed heat ::: $Q=\dot{m}{c}_p(T_{out}-T_{in})$ Ideal turbine work ::: $W_{turb}=\dot{m}{c}_p{T}_1\left[1-{\left(\frac{p_2}{p_1}\right)}^{\frac{\gamma -1}{\gamma }}\right]{\eta }_t$ Pump power ::: $W_{pump}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$ Closing condition ::: $W_{turb}\ge W_{pump,fuel}+W_{pump,ox}$

Yahan $\dot{m}$ mass flow rate hai (kilograms per second), $c_p$ specific heat hai (ek kilogram ko ek kelvin warm karne ke liye joules), $T$ temperature kelvin mein hai, $p$ pressure hai, $\rho$ (Greek "rho") density hai (kilograms per cubic metre), $\gamma$ (Greek "gamma") gas ka heat-capacity ratio hai, aur $\eta$ (Greek "eta") ek efficiency hai 0 aur 1 ke beech jo ek real machine ko uske ideal se discount karti hai. Yahan yeh do flavours mein aati hai: ${\eta }_t$ turbine efficiency hai (ideal enthalpy drop ka kitna fraction turbine actually shaft work mein convert karta hai), aur ${\eta }_p$ pump efficiency hai (ideal pumping power divided by real, badi power jo ek pump consume karta hai). Yeh saare parent mein define kiye gaye the — agar koi shaky lagta hai, wahan jaake re-read karo aur wapas aao.

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Scenario matrix

Is topic ke har problem ko is grid ke ek cell ki tarah socho. Neeche ke hamare examples har cell ko tick off karte hain.

Cell	Kya tricky banata hai ise	Example
A. Plain forward calc	Seedha numbers plug in karo	Ex 1 (heat), Ex 2 (turbine)
B. The closing inequality	Do pumps vs ek turbine — kya yeh sustain karta hai?	Ex 3
C. Zero / degenerate input	$\Delta T=0$, ya $p_2=p_1$ (koi pressure drop nahi) → kya hota hai?	Ex 4
D. Limiting value	$p_2/p_1\to 0$ (full expansion) — turbine work ki ceiling	Ex 5
E. Scaling / square–cube	Engine ko $k$ se scale karo — supply vs demand race	Ex 6
F. Real-world word problem	Cycle close karne ke liye wall temperature rise design karo	Ex 7
G. Exam twist	Open (bleed) cycle: sirf ek fraction flow turbine drive karta hai	Ex 8

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Example 1 — Cell A: Heat picked up (plain forward)

Step 1 — Temperature rise. $\Delta T=T_{out}-T_{in}=400-50=350\text{K}$ Yeh step kyun? $c_p$ per kelvin of rise define hota hai, isliye hume rise chahiye, endpoints nahi.

Step 2 — Multiply through. $Q=\dot{m}{c}_p\Delta T=3\times 14000\times 350=1.47\times 10^7\text{W}=14.7\text{MW}$ Yeh step kyun? Yeh literally specific heat ki definition hai ek flow tak scale up karke: joules-per-kg times kg-per-second deta hai joules-per-second, yaani watts.

Verify: Units: $\frac{\text{kg}}{\text{s}}\cdot \frac{\text{J}}{\text{kg⋅K}}\cdot \text{K}=\frac{\text{J}}{\text{s}}=\text{W}$ ✅. Aur 14.7 MW 10 aur 100 MW ke guesses ke beech hai — ek real engine ke liye sensible jo otherwise apna nozzle melt kar deta.

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Example 2 — Cell A: Ideal turbine work (plain forward)

Step 1 — Pressure-ratio exponent. $\frac{\gamma -1}{\gamma }=\frac{0.4}{1.4}=0.2857$ Yeh step kyun? Yeh exponent isentropic relation $T_2/T_1=(p_2/p_1{)}^{(\gamma -1)/\gamma }$ se aata hai — yeh pressure ratio ko temperature ratio mein convert karta hai. Hum isentropic (constant entropy) use karte hain kyunki ek ideal, loss-free turbine na entropy gain karta hai na lose karta hai.

Step 2 — Ratio term evaluate karo. ${\left(\frac{8}{12}\right)}^{0.2857}=(0.6667{)}^{0.2857}=0.8900$ Toh bracket $\left[1-0.8900\right]=0.1100$. Yeh step kyun? Bracket is pressure drop ke liye gas ki inlet enthalpy ka fraction hai jo gas dene ko taiyaar hai. Chhota pressure drop → chhota fraction.

Step 3 — Power assemble karo. $W_{turb}=\dot{m}{c}_p{T}_1[0.1100]{\eta }_t=3\times 14000\times 400\times 0.1100\times 0.75$ $W_{turb}\approx 1.386\times 10^6\text{W}\approx 1.39\text{MW}$ Yeh step kyun? $c_p{T}_1$ inlet enthalpy per kg hai; times mass flow se enthalpy rate milta hai; times bracket se ideal extractable rate milta hai; times ${\eta }_t$ real losses ke liye discount karta hai.

Verify: Turbine ne sirf $1.39/14.7\approx 9.4\%$ absorbed heat li — forecast se match karta hai ki gentle pressure drop ek chhota sa slice deta hai. Yeh deliberate hai: gas ko baad mein combustion chamber mein enter karne ke liye high-pressure rehna chahiye.

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Example 3 — Cell B: Kya cycle close hoti hai?

Step 1 — Fuel pump power. $W_f=\frac{{\dot{m}}_f\Delta p}{{\rho }_f{\eta }_p}=\frac{3\times 13\times 10^6}{71\times 0.7}=7.85\times 10^5\text{W}$ Yeh step kyun? Pump power volume-flow ($\dot{m}/\rho$) times pressure jump hai, efficiency se divided. Halka, fluffy hydrogen ki low density hoti hai, isliye uska volume flow bada hota hai — pump karna expensive hai.

Step 2 — Oxidizer pump power. $W_{ox}=\frac{18\times 13\times 10^6}{1140\times 0.7}=2.93\times 10^5\text{W}$ Yeh step kyun? Wahi formula. Bhale hi O₂ ka mass 6× zyaada hai, uski density 16× zyaada hai, toh uska volume flow chhota hai → pump karna sasta hai.

Step 3 — Total demand aur comparison. $W_{pump}=7.85\times 10^5+2.93\times 10^5=1.08\times 10^6\text{W}=1.08\text{MW}$ $W_{turb}=1.39\text{MW}\ge 1.08\text{MW}=W_{pump}✅$ Yeh step kyun? Yeh parent ki self-sustaining condition hai. Surplus $=1.39-1.08=0.31$ MW startup aur throttling ke liye margin hai.

Verify: Forecast galat tha — fuel pump (0.79 MW) ox pump (0.29 MW) ko dwarf karta hai, kyunki pump work volume follow karta hai, mass nahi. Yeh expander cycle ka ek key lesson hai: low-density hydrogen pumping ki headache hai. Turbopump design aur Liquid hydrogen properties dekho.

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Example 4 — Cell C: Degenerate inputs (zeros)

Step 1 — Zero temperature rise. $Q=\dot{m}{c}_p(T_{out}-T_{in})=3\times 14000\times (50-50)=0\text{W}$ Yeh step kyun? Koi warming nahi matlab koi heat fluid mein cross nahi hui. Physically: wall hot nahi hai (engine off, ya perfect insulation) — harvest karne ke liye koi free energy nahi.

Step 2 — Zero pressure ratio term. ${\left(\frac{p_2}{p_1}\right)}^{\frac{\gamma -1}{\gamma }}={\left(\frac{12}{12}\right)}^{0.2857}=1^{0.2857}=1$ $W_{turb}=\dot{m}{c}_p{T}_1[1-1]{\eta }_t=0\text{W}$ Yeh step kyun? Turbine sirf pressure drop se work extract karta hai. Equal pressures = gas bina chhuye guzar jaata hai = koi shaft work nahi. Physically: ek stalled ya bypassed turbine.

Verify: Dono zero, jaise forecast kiya. Bracket $[1-(p_2/p_1{)}^{(\gamma -1)/\gamma }]$ $p_2=p_1$ par elegantly vanish ho jaata hai aur ratio shrink hone par badhta hai — formula degenerate case ke against self-guarding hai. Agar flight mein Example 4b hota, pumps ko zero drive milta aur engine collapse ho jaata — cycle does not close at zero drop.

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Example 5 — Cell D: Limiting value (full expansion)

Step 1 — Ratio term ka limit lo. ${\lim }_{p_2\to 0}{\left(\frac{p_2}{p_1}\right)}^{0.2857}=0\Rightarrow [1-0]=1$ Yeh step kyun? 1 se neeche koi bhi positive base ek positive power tak, jaise base $\to 0$, 0 tend karta hai. Toh bracket apni ceiling 1 tak pahunch jaata hai — turbine poori inlet enthalpy de deta hai.

Step 2 — Ceiling evaluate karo. $W_{max}=\dot{m}{c}_p{T}_1(1){\eta }_t=3\times 14000\times 400\times 0.75=1.26\times 10^7\text{W}=12.6\text{MW}$ Yeh step kyun? Yeh 400 K gas stream ka absolute maximum hai jo woh deliver kar sakti hai: uski full enthalpy rate $\dot{m}{c}_p{T}_1=16.8$ MW, ${\eta }_t=0.75$ se discounted.

Verify: $12.6\text{MW}<14.7\text{MW}$ absorbed heat — achha, tum kabhi bhi jitni heat daali hai usse zyaada work extract nahi kar sakte (yeh energy budget todega). Example 2 ke gentle drop (1.39 MW) se compare karo: full expansion 9× zyaada de sakta tha, lekin tab gas near-vacuum hoti aur chamber feed ke liye useless. Yahi design tension hai — expanders chamber-feed pressure preserve karne ke liye turbine power sacrifice karte hain.

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Example 6 — Cell E: Square–cube scaling

Step 1 — Har side scale karo. Baseline supply aur demand dono $P_0$ hone do. $\text{Supply}(k)=P_0{k}^2,\text{Demand}(k)=P_0{k}^3$ Yeh step kyun? Yeh Square-cube law hai: area size ke square ke roop mein badhta hai, volume (aur uske command wala flow) cube ke roop mein.

Step 2 — $k=2$ plug in karo. $\text{Supply}=P_0\cdot 4,\text{Demand}=P_0\cdot 8$ $\frac{\text{Supply}}{\text{Demand}}=\frac{4}{8}=0.5$ Yeh step kyun? Hum dono rates ko directly compare karte hain. $k=1$ par ratio 1 tha (just closing); $k=2$ par yeh 0.5 tak gir gaya hai.

Verify: Doubled engine apne pumps ki demand ka sirf half power supply kar sakta hai — yeh close karne mein fail karta hai. Forecast galat hai: bada closed expander ke liye worse hota hai. Isi liye RL10 few-hundred-kN class mein rehta hai jabki ek Staged combustion cycle engine bahut bada ho sakta hai.

Neeche ka figure dono curves ko scale factor $k$ ke against plot karta hai. Chalk-blue supply line ($k^2$) aur chalk-pink demand line ($k^3$) follow karo: yeh exactly $k=1$ par cross karte hain (pale-yellow dot jahan engine just close hota hai), aur us point ke baad pink demand curve blue ke upar pull away karti hai — shaded pink region power deficit hai. Yellow dots ki doosri pair $k=2$ mark karti hai, jahan supply 4 tak pahunchi hai lekin demand 8 tak race kar gayi hai.

Figure s01 — Square–cube law: heat supply ($\propto k^2$, blue) versus pump demand ($\propto k^3$, pink). Yeh $k=1$ par milte hain; uske baad demand supply se aage nikal jaati hai (pink shaded deficit), toh scale-up closed expander apne khud ke pumps ko power nahi de sakta.

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Example 7 — Cell F: Real-world design problem

Step 1 — Example 2 ka bracket reuse karo (same pressures aur $\gamma$): ratio term $=0.8900$, toh bracket $=0.1100$. Yeh step kyun? Pressure ratio ke baare mein kuch nahi badla, toh yeh dimensionless factor identical hai — recompute karne ki zaroorat nahi.

Step 2 — Turbine equation ko $T_1$ ke liye solve karo. $W_{turb}=\dot{m}{c}_p{T}_1[0.1100]{\eta }_t\ge 2.0\times 10^6$ $T_1\ge \frac{2.0\times 10^6}{4\times 14000\times 0.1100\times 0.75}=\frac{2.0\times 10^6}{4620}=432.9\text{K}$ Yeh step kyun? Hum forward formula invert kar rahe hain — yeh ek requirement reverse-engineer karna hai. $T_1$ ke alawa har factor fixed hai, toh $T_1$ forced hai.

Step 3 — Wall heat requirement mein translate karo. Agar H₂ jacket mein 50 K par enter karta hai, toh wall ko $\Delta T=432.9-50=382.9\text{K}$ add karna hoga, absorbing $Q=\dot{m}{c}_p\Delta T=4\times 14000\times 382.9=2.144\times 10^7\text{W}\approx 21.4\text{MW}.$ Yeh step kyun? Loop cooling jacket tak wapas close karta hai: required turbine temperature dictate karta hai ki walls ko coolant mein kitni heat dump karni hai.

Verify: $432.9\text{K}>400\text{K}$ — zyaada, jaise forecast kiya, kyunki humne modest flow increase akele se zyaada power maangi. Sanity: wall heat ka $21.4$ MW $>$ extracted $2.0$ MW (turbine sirf ~9% leta hai), Example 2 ke ~9.4% extraction fraction se consistent. Feasible.

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Example 8 — Cell G: Exam twist (open / bleed cycle)

Step 1 — Turbine sirf bled flow dekhta hai. ${\dot{m}}_{turb}=f\dot{m}=0.6\times 3=1.8\text{kg/s}$ Yeh step kyun? Open cycle mein turbine ka working mass sirf bled fraction hota hai — baaki turbine bypass karke seedha chamber jaata hai.

Step 2 — ${\dot{m}}_{turb}$ par turbine formula apply karo. Bracket abhi bhi $0.1100$ hai (same pressures aur $\gamma$ jaise Example 2). $W_{turb}={\dot{m}}_{turb}{c}_p{T}_1[0.1100]{\eta }_t=1.8\times 14000\times 400\times 0.1100\times 0.75$ $W_{turb}\approx 8.32\times 10^5\text{W}\approx 0.83\text{MW}$ Yeh step kyun? Turbine power mass flow mein linear hai, toh working flow ko $f=0.6$ se scale karna power ko 0.6 se scale karta hai. Yahi open cycle ka poora arithmetic hai.

Step 3 — Loss kyun accept karte hain (part b)? Bled gas dump hoti hai, chamber ko nahi jaati, toh ise baad mein high-pressure rehne ki zaroorat nahi. Yeh designer ko turbine ke across bahut bada pressure drop use karne ki freedom deta hai — chhota $p_2$ bracket $[1-(p_2/p_1{)}^{(\gamma -1)/\gamma }]$ ko uski ceiling 1 ki taraf push karta hai (recall Example 5), toh bled hydrogen ka har kilogram bahut zyaada work yield karta hai. Isliye ek chhota engine itna turbine power raise kar sakta hai ki apne pumps drive kar sake, bhale hi usi size ka closed expander close karne mein fail ho jaata. Yeh step kyun? Yeh trade resolve karta hai: tum specific impulse ka ek slice sacrifice karte ho (dumped hydrogen kabhi nahi jalta, toh uski energy unused chali jaati hai) aur badle mein zyaada pump-driving muscle milti hai. Gas generator cycle compare karo, jo exactly usi reason se combustion products dump karta hai.

Verify: $0.83=0.6\times 1.39$ ✅ — exactly forecast fraction, turbine power ki $\dot{m}$ mein linearity confirm karta hai. Physical sanity check: dumped 40% hydrogen (1.2 kg/s) unburned exit karta hai, toh chamber ko closed cycle se kam fuel per second milta hai — yahi $I_{sp}$ penalty hai jo trade away ho rahi hai.

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Recall Self-test: kaunsa cell?

Matrix mein kaunsa cell poochta hai "what if $p_2=p_1$?" ::: Cell C — degenerate/zero input (Ex 4). Kaunsa cell dikhata hai ki bade engines close karne mein fail karte hain? ::: Cell E — square–cube scaling (Ex 6). Ex 3 mein, kaunsa pump demand dominate karta hai aur kyun? ::: Fuel (H₂) pump — hydrogen ki low density matlab huge volume flow. Full expansion ($p_2\to 0$) turbine work ko inlet enthalpy ke kitne fraction par cap karta hai? ::: ${\eta }_t$ times poori enthalpy $\dot{m}{c}_p{T}_1$ (bracket $\to 1$). Ex 8 ke open (bleed) cycle mein, yeh bada turbine pressure drop kyun use kar sakta hai? ::: Bled gas overboard dump hoti hai, toh use chamber feed karne ke liye high-pressure rehne ki zaroorat nahi.