Transpiration cooling: Exercises — Hinglish deep dive

3.3.31 · D4 · Physics › Rocket Propulsion › Transpiration cooling

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Level 1 — Recognition

Recall Solution

Effectiveness batati hai ki wall ko gas se coolant ki taraf kitna neeche khaincha gaya hai: $ϕ = \frac{T _{g} - T _{w}}{T _{g} - T _{c}} = \frac{3000 - 1500}{3000 - 300} = \frac{1500}{2700} = 0.556.$ Matlab: wall ne total possible temperature gap $(T_{g} - T_{c})$ ka $55.6%$ close kar liya. $ϕ = 0$ matlab koi cooling nahi ( $T_{w} = T_{g}$ ); $ϕ = 1$ matlab perfect cooling ( $T_{w} = T_{c}$ ).

Recall Solution

Gas side: $η h_{0} = 0.5 \times 5000 = 2500 W m^{- 2} K^{- 1}$ .
Coolant side: $G c_{p} = 3 \times 1200 = 3600 W m^{- 2} K^{- 1}$ .

Coolant conductance ( $3600$ ) $>$ gas conductance ( $2500$ ), isliye weighted average $T_{w}$ coolant side ki taraf jhukta hai — wall aaram se thandi rehti hai ( $ϕ = 3600/6100 = 0.59$ ).

Level 2 — Application

Recall Solution

Step 1 — gas conductance. $η h_{0} = 0.35 \times 6000 = 2100$ . Step 2 — coolant conductance. $G c_{p} = 2.5 \times 1000 = 2500$ . Step 3 — weighted average. $T_{w} = \frac{2100 ( 3200 ) + 2500 ( 400 )}{2100 + 2500} = \frac{6 720 000 + 1 000 000}{4600} = \frac{7 720 000}{4600} \approx 1678 K .$ Wall $\approx 1678 K$ par chalti hai, jyadatar nickel superalloys ke melting point se kam — yeh survive ho sakta hai jabki $3200 K$ impossible hota.

Recall Solution

$ϕ = \frac{G c _{p}}{η h _{0} + G c _{p}} = \frac{2500}{2100 + 2500} = \frac{2500}{4600} = 0.543.$ Cross-check L2.1 ke $T_{w}$ se: $\frac{3200 - 1678}{3200 - 400} = \frac{1522}{2800} = 0.544$ . ✔ (Tiny gap $T_{w}$ ki rounding ki wajah se hai.)

Level 3 — Analysis

Recall Solution

Step 1 — target effectiveness. $ϕ = \frac{3000 - 1000}{3000 - 300} = \frac{2000}{2700} = 0.741$ . Step 2 — $ϕ$ ko invert karo. $ϕ (η h_{0} + G c_{p}) = G c_{p}$ se: $G c_{p} = \frac{ϕ}{1 - ϕ} η h_{0} = \frac{0.741}{0.259} \times (0.4 \times 5000) = 2.857 \times 2000 = 5714.$ Step 3 — $G$ solve karo. $G = 5714/1200 \approx 4.76 kg m^{- 2} s^{- 1}$ . Koi bhi $G \geq 4.76 kg m^{- 2} s^{- 1}$ wall ko $1000 K$ par ya neeche rakhega.

Recall Solution

$η h_{0} = 2000$ fixed ke saath:

$G = 2$ : $G c_{p} = 2400$ , $T_{w} = \frac{2000 ( 3000 ) + 2400 ( 300 )}{4400} = \frac{6 720 000}{4400} \approx 1527 K$ .
$G = 4$ : $G c_{p} = 4800$ , $T_{w} = \frac{6 000 000 + 1 440 000}{6800} = \frac{7 440 000}{6800} \approx 1094 K$ .
$G = 8$ : $G c_{p} = 9600$ , $T_{w} = \frac{6 000 000 + 2 880 000}{11600} = \frac{8 880 000}{11600} \approx 766 K$ .

Drops per doubling: $1527 \to 1094$ matlab $433 K$ ; $1094 \to 766$ sirf $328 K$ . Har doubling kam return deti hai, kyunki $T_{w}$ apne floor $T_{c} = 300 K$ ki taraf squeeze ho raha hai.

Level 4 — Synthesis

Recall Solution

(a) Constant $η = 0.4$ . $η h_{0} = 2000$ , $G c_{p} = 3 \times 1200 = 3600$ . $T_{w} = \frac{2000 ( 3000 ) + 3600 ( 300 )}{5600} = \frac{6 000 000 + 1 080 000}{5600} = \frac{7 080 000}{5600} \approx 1264 K .$ (b) Variable $η$ . $η (3) = \frac{1}{1 + 0.3 \times 3} = \frac{1}{1.9} = 0.526$ . Ruko — yeh $0.4$ se zyada hai, matlab is $G$ par weaker cooling. Toh $η h_{0} = 0.526 \times 5000 = 2632$ . $T_{w} = \frac{2632 ( 3000 ) + 3600 ( 300 )}{2632 + 3600} = \frac{7 896 000 + 1 080 000}{6232} = \frac{8 976 000}{6232} \approx 1440 K .$ Lesson: $G = 3$ par yeh particular model flat $η = 0.4$ assumption se kam blowing benefit deta hai, isliye real wall zyada hot hai ( $1440$ vs $1264 K$ ). Moral yeh nahi ki "variable $η$ hamesha better hai" — moral yeh hai ki "tumhe actual local $η$ use karna hai, aur ek flattering constant tumhe kisi bhi direction mein mislead kar sakta hai."

Recall Solution

Common terms: $η = 0.4$ , $G c_{p} = 3600$ .

Chamber: $η h_{0} = 0.4 \times 4000 = 1600$ . $T_{w} = \frac{1600 ( 3000 ) + 3600 ( 300 )}{5200} = \frac{4 800 000 + 1 080 000}{5200} = \frac{5 880 000}{5200} \approx 1131 K .$
Throat: $η h_{0} = 0.4 \times 9000 = 3600$ . $T_{w} = \frac{3600 ( 3000 ) + 3600 ( 300 )}{7200} = \frac{10 800 000 + 1 080 000}{7200} = \frac{11 880 000}{7200} = 1650 K .$

Comment: uniform coolant ke saath throat chamber se $\sim 519 K$ zyada hot chalta hai. Yahi reason hai ki designers throat par zyada coolant inject karte hain — heat flux sabse worst wahan hoti hai jahan geometry sabse tight hoti hai.

Level 5 — Mastery

Recall Solution

(a) Target $ϕ = \frac{3000 - 1200}{3000 - 300} = \frac{1800}{2700} = 0.667$ . $G c_{p} = \frac{ϕ}{1 - ϕ} η h_{0} = \frac{0.667}{0.333} \times (0.4 \times 9000) = 2 \times 3600 = 7200.$ $G = 7200/1200 = 6 kg m^{- 2} s^{- 1}$ . (b) $\overset{m}{˙}_{c} = G A = 6 \times 0.02 = 0.12 kg s^{- 1}$ . (c) Fraction $= \frac{m ˙ _{c}}{m ˙} = \frac{0.12}{50} = 0.0024 = 0.24%$ . Reading: throat ko protect karne ke liye sirf roughly quarter-percent flow lagta hai — yeh ek achha deal hai, lekin yeh dumped coolant bahut kam thrust produce karta hai, aur Specific Impulse ko thoda neeche dhakelta hai. $G$ ko sirf target tak minimize karo, isse zyada nahi.

Recall Solution

Step 1 — target effectiveness. $ϕ = \frac{3000 - 1000}{3000 - 300} = \frac{2000}{2700} = 0.741$ . Step 2 — $G$ mein balance likhna. $η h_{0} = \frac{5000}{1 + 0.3 G}$ aur $G c_{p} = 1200 G$ ke saath: $\frac{ϕ}{1 - ϕ} = \frac{G c _{p}}{η h _{0}} \Rightarrow \frac{0.741}{0.259} = \frac{1200 G ( 1 + 0.3 G )}{5000} .$ Left side $= 2.857$ . Toh $1200 G (1 + 0.3 G) = 2.857 \times 5000 = 14286$ . Step 3 — quadratic. $360 G^{2} + 1200 G - 14286 = 0$ , yaani $G^{2} + 3.333 G - 39.68 = 0$ . $G = \frac{- 3.333 + 3.33 3 ^{2} + 4 \times 39.68}{2} = \frac{- 3.333 + 11.11 + 158.73}{2} = \frac{- 3.333 + 169.84}{2} = \frac{- 3.333 + 13.03}{2} \approx 4.85 kg m^{- 2} s^{- 1} .$ Compare: constant $η = 0.4$ ke saath (L3.1) hume $G \approx 4.76$ chahiye tha. Yahan falling blowing benefit ka matlab hai ki hume thoda zyada coolant chahiye, $G \approx 4.85$ — $η (G)$ model karne se ek chhota lekin honest correction.

Recall Solution

(a) $G \to 0$ (koi coolant nahi): $G c_{p} \to 0$ , toh $ϕ = \frac{0}{η h _{0}} = 0$ aur $T_{w} \to T_{g}$ . Wall utni hi hot jitni gas — bilkul koi protection nahi.
(b) $G \to \infty$ (coolant se bhara hua): $\frac{η h _{0}}{G c _{p}} \to 0$ , toh $ϕ \to 1$ aur $T_{w} \to T_{c}$ . Wall utni hi thandi jitna inlet coolant — perfect lekin propellant-hungry.
(c) $η \to 0$ (blanket itna thick ki almost koi heat nahi guzarti): $ϕ \to \frac{G c _{p}}{G c _{p}} = 1$ , $T_{w} \to T_{c}$ . Same perfect limit, absorbing ki jagah blocking se reached.
(d) $T_{c} = T_{g}$ : driving gap $T_{g} - T_{c} = 0$ , isliye $ϕ$ undefined hai (0/0) lekin $T_{w} = T_{g} = T_{c}$ trivially — simply koi temperature difference hi nahi hai jo fight karo. Cooling meaningless hai.

Yeh chaar corners har real operating point ko bracket karte hain: wall hamesha $T_{c}$ aur $T_{g}$ ke beech rehti hai.

Recall Self-test checklist

Conductances se $T_{w}$ compute karo ::: $T_{w} = \frac{η h _{0} T _{g} + G c _{p} T _{c}}{η h _{0} + G c _{p}}$ Required coolant ke liye invert karo ::: $G c_{p} = \frac{ϕ}{1 - ϕ} η h_{0}$ $G$ double karne se $T_{w}$ aadhi kyun nahi hoti ::: $T_{w}$ neeche se $T_{c}$ se bounded hai; returns diminish hote hain Throat par kya hota hai ::: $h_{0}$ spike karta hai, isliye $T_{w}$ badhti hai jab tak local $G$ nahi badhaya jaata Extra coolant ki cost ::: dumped coolant specific impulse girata hai

Connected notes

← Back to parent · Film Cooling · Regenerative Cooling · Ablative Cooling · Convective Heat Transfer · Boundary Layer Theory · Nozzle Throat Heat Flux · Specific Impulse