Worked examples — Regenerative cooling — heat flux, coolant flow, pressure drop

3.3.28 · D3 · Physics › Rocket Propulsion › Regenerative cooling — heat flux, coolant flow, pressure dro

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Scenario matrix

Is topic ke har problem ko in cells mein se kisi ek mein daala ja sakta hai. Neeche ke examples cell se labelled hain taaki tum dekh sako ki poora map cover hai.

Cell	Kya vary / break hota hai	Example
A. Balanced series	teeno resistances comparable	Ex 1
B. Ek resistance dominate karta hai	gas-side $R_g\gg$ baaki → wahi akela $q$ set karta hai	Ex 2
C. Degenerate: thick wall	$t_w/k_w$ blow up hota hai → conduction flux choke karta hai	Ex 3
D. Degenerate: near-zero wall	$t_w\to 0$ → wall term vanish, do convections bachi	Ex 3
E. Coolant $\Delta T_c$ capped	boiling/coking ceiling forces ${\dot{m}}_c$ up	Ex 4
F. Limiting: channel narrow hota hai	$d_h\downarrow$ → $h_c\uparrow$ (achha) lekin $\Delta p\uparrow$ (bura)	Ex 5
G. Sign / direction check	kya heat enter ho rahi hai ya nikal rahi hai? arrow sahi rakho	Ex 6
H. Real-world word problem	"kya engine ke paas itna propellant bhi hai?"	Ex 7
I. Exam twist: do formulas couple karo	flow sets $h_c$ jo sets $q$ jo sets flow	Ex 8

Hum har cell hit karenge.

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Example 1 — Cell A: balanced series (the baseline)

1. Har layer ko specific resistance mein convert karo. $R_g=\frac{1}{h_g}=5.0\times 10^{-5},R_w=\frac{t_w}{k_w}=\frac{0.001}{350}=2.857\times 10^{-6},R_c=\frac{1}{h_c}=2.0\times 10^{-5}$ Ye step kyun? Heat gas-film, metal, coolant-film series mein cross karta hai, toh jaise series resistors simply add hote hain. Resistances mein kaam karne se "teen physics laws" ek addition mein aa jaati hain.
2. Inhe add karo. $R=7.286\times 10^{-5}$ m²K/W. Kyun? Series resistance sum hoti hai; reciprocal overall coefficient $U$ hai.
3. Driving temperature difference se divide karo. $q=\frac{T_{aw}-T_{co}}{R}=\frac{3000}{7.286\times 10^{-5}}=4.12\times 10^7{\text{W/m}}^2\approx 41{\text{MW/m}}^2$ Kyun? Wahi $q$ har layer se flow karta hai, toh total drop $÷$ total resistance.
4. Hot-wall temperature back out karo sirf gas-side drop use karke: $T_{wg}=T_{aw}-q{R}_g=3300-(4.12\times 10^7)(5\times 10^{-5})\approx 1242\text{K}$ Ye step kyun? Metal sirf apni surface temperature "feel" karta hai. Woh surface gas se bilkul gas-side drop $q{R}_g$ ke neeche baithi hai.

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Example 2 — Cell B: ek resistance dominate karta hai

1. $R_g$ recompute karo. $R_g=1/2000=5.0\times 10^{-4}$. Baaki unchanged: $R_w=2.86\times 10^{-6}$, $R_c=2.0\times 10^{-5}$. Kyun? Sirf $h_g$ badla, toh sirf $R_g$ move karta hai.
2. Magnitudes compare karo. $R_g/R_c=5.0\times 10^{-4}/2.0\times 10^{-5}=25\times$ bada, aur $R_g/R_w=5.0\times 10^{-4}/2.86\times 10^{-6}=175\times$ bada. Ye step kyun? Jab series mein ek resistor baaki se bahut bada hota hai, toh wahi akela current set karta hai — ye Newton's Law of Cooling gas film ka bottleneck hona hai.
3. Sirf $R_g$ se approximate flux nikalo. $q_{\text{approx}}=\frac{3000}{5.0\times 10^{-4}}=6.0\times 10^6{\text{W/m}}^2=6{\text{MW/m}}^2$ Ye step kyun? Agar $R_g$ baaki se bahut bada hai, toh $R\approx R_g$, isliye $q\approx \Delta T/R_g$. Do tiny resistances drop karne se answer barely move karna chahiye — ye line exactly test karti hai ki woh shortcut kitna achha hai.
4. Exact flux teeno ke saath. $R=5.229\times 10^{-4}$, $q=3000/R=5.737\times 10^6$ W/m². Kyun compare karein? Dekhne ke liye ki "chhote wale ignore karo" kitna achha hai.

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Example 3 — Cells C & D: thick wall vs vanishing wall

1. Case C, thick wall. $R_w=0.050/350=1.429\times 10^{-4}$. Ab $R=R_g+R_w+R_c=5\times 10^{-5}+1.429\times 10^{-4}+2\times 10^{-5}=2.129\times 10^{-4}$. $q_C=3000/2.129\times 10^{-4}=1.409\times 10^7{\text{W/m}}^2\approx 14{\text{MW/m}}^2$ Ye step kyun? Moti wall conduction (Fourier's Law of Conduction) ko bottleneck bana deti hai; flux Ex 1 ki value ke ek tihaii tak gir jaata hai. Coolant tak kam heat pahunchti hai — lekin wall ka interior dangerously hot ho jaata hai kyunki drop $q{R}_w$ bahut bada hai.
2. Case D, zero wall. $R_w\to 0$, toh $R=R_g+R_c=7.0\times 10^{-5}$. $q_D=3000/7.0\times 10^{-5}=4.286\times 10^7{\text{W/m}}^2\approx 43{\text{MW/m}}^2$ Kyun? Conduction resistance bilkul hata do aur sirf do films bachi rahti hain — maximum flux jo ye film pair pass kar sakti hai.

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Example 4 — Cell E: coolant temperature-rise capped

1. Jacket mein total power. $Q=qA=4.1\times 10^7\times 0.5=2.05\times 10^7$ W. Kyun? Flux power per area hai; total ke liye wetted area se multiply karo.
2. Energy balance, $200$ K case. ${\dot{m}}_c=\frac{Q}{c_p\Delta T_c}=\frac{2.05\times 10^7}{4200\times 200}=24.4\text{kg/s}$ Ye step kyun? Wall mein jaane wala har joule coolant ki rising temperature mein land karna chahiye — energy conservation, no boiling.
3. Stricter cap, $120$ K case. ${\dot{m}}_c^{\prime }=\frac{2.05\times 10^7}{4200\times 120}=40.67\text{kg/s}$ Kyun? $\Delta T_c$ par cap real constraint hai; tighter cap same $Q$ remove karne ke liye bada flow force karta hai.
4. Ratio. ${\dot{m}}_c^{\prime }/{\dot{m}}_c=200/120=1.667$.

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Example 5 — Cell F: channel narrow karna (the cruel trade-off)

1. $d_h=2$ mm par baseline $\Delta p$ (blue bar). $\Delta p_1=f\frac{L}{d_h}\frac{\rho v^2}{2}=0.02\cdot \frac{0.6}{0.002}\cdot \frac{800\cdot 30^2}{2}=2.16\times 10^6\text{Pa}=21.6\text{bar}$ Ye step kyun? Ye Darcy–Weisbach seedha hai — channel length par wall friction se pressure lost.
2. $d_h=1$ mm, $v=60$ m/s par naya $\Delta p$ (pink bar). $\Delta p_2=0.02\cdot \frac{0.6}{0.001}\cdot \frac{800\cdot 60^2}{2}=1.728\times 10^7\text{Pa}=172.8\text{bar}$ Kyun? Do cheezein ek saath kharab hui: $d_h$ half hua ($L/d_h$ mein factor $2$) aur $v$ double hua ($v^2$ mein factor $4$).
3. Growth factor (yellow arrow). $\Delta p_2/\Delta p_1=172.8/21.6=8$. Ye step kyun? $\Delta p\propto v^2/d_h$; $d_h$ half karne se $\times 2$, $v$ double karne se $\times 4$; milke $\times 8$.

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Example 6 — Cell G: heat kis taraf jaati hai? (sign / direction)

1. Student ka expression compute karo. $h_g(T_{wg}-T_{aw})=2\times 10^4(1242-3300)=2\times 10^4(-2058)=-4.12\times 10^7{\text{W/m}}^2$ Ye step kyun? Sign expose karne ke liye. Is convention se negative flux matlab hai ki flow assumed arrow ke opposite hai.
2. Convention fix karo. Wall mein heat ke liye Newton's Law of Cooling flow direction mein hotter-minus-colder use karta hai: $q=h_g(T_{aw}-T_{wg})>0$. Kyun? Heat hamesha hot → cold flow karta hai. Gas ($3300$ K) wall ($1242$ K) se hotter hai, toh heat wall mein enter karti hai; sahi expression positive hai.
3. Magnitude identical hai, sirf arrow flipped hai: $|q|=4.12\times 10^7$ W/m².

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Example 7 — Cell H: real-world word problem

1. Fuel flow available. ${\dot{m}}_{\text{fuel}}=\frac{1}{3}\times 30=10$ kg/s. Ye step kyun? Sirf fuel cooling channels se route hota hai; oxidizer usually nahi hota (woh hot metal ke saath react kar jaata).
2. Demand se compare karo. Chahiye $24.4$ kg/s, hai $10$ kg/s. Deficit $=24.4-10=14.4$ kg/s. Kyun? Agar demand available coolant se zyada hai, toh pure regen cooling heat carry nahi kar sakti.
3. Design consequence. Shortfall supplement se meet karna hoga: Film Cooling (wall ke saath cool gas layer) ya Ablative Cooling (sacrificial liner) heat load $q$ cut karne ke liye taaki required ${\dot{m}}_c$ $\le 10$ kg/s par aa jaye. Ye step kyun? Tum fuel invent nahi kar sakte; balki demand ghataate ho.

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Example 8 — Cell I: exam twist, do formulas coupled

1. $h_c$ scaling. $h_c\propto \text{Nu}\propto {\text{Re}}^{0.8}$. Re double karne par: factor $=2^{0.8}$. $2^{0.8}=1.741$ Ye step kyun? $k_c,d_h,\text{Pr}$ unchanged hain, toh sirf ${\text{Re}}^{0.8}$ term move karta hai; $h_c$ $1.741\times$ badhta hai, double se kam — $0.8$ exponent tumhe punish karta hai.
2. Naya coolant resistance. $R_c\propto 1/h_c$, toh $R_c^{\text{new}}=R_c^{\text{old}}/1.741$. Kyun? Resistance transfer coefficient ka inverse hai.
3. Naya total resistance. Old total $R=1$ (normalized), $R_c^{\text{old}}=0.40$ ke saath, baaki $=0.60$. $R^{\text{new}}=0.60+0.40/1.741=0.60+0.2298=0.8298$ Ye step kyun? Sirf coolant term shrink karta hai; gas aur wall terms untouched hain, toh improvement diluted ho jaata hai.
4. Flux gain. Kyunki $q\propto 1/R$, flux old total to new total ke ratio se badhta hai: $\frac{q^{\text{new}}}{q^{\text{old}}}=\frac{R^{\text{old}}}{R^{\text{new}}}=\frac{1}{0.8298}=1.205$ Ye step kyun? Same driving $\Delta T$, toh flux total resistance se inversely proportional hai; old-by-new divide karne se improvement factor milta hai — $20.5\%$ gain jabki $h_c$ $74\%$ badha.

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Recall Active recall — answers cover karo

Ex 1 baseline flux ::: $\approx 41$ MW/m², $T_{wg}\approx 1242$ K Ex 3 kaunsi wall zyada flux deti hai, thick ya none ::: none (zero-thickness), $\approx 43$ vs $14$ MW/m² Ex 4 flow jab $\Delta T_c$ $200\to 120$ K ho ::: factor $200/120=1.667$ se badhta hai, $40.7$ kg/s tak Ex 5 $\Delta p$ factor jab $d_h$ half ho aur $v$ double ho ::: $\times 8$ Ex 6 heat direction jab $q$ negative aaye ::: convention ulta tha; heat phir bhi gas → wall flow karti hai Ex 8 $h_c$ factor jab velocity double ho ::: $2^{0.8}=1.741$, lekin flux sirf $\sim 20\%$ upar