Transpiration cooling: Worked examples — Hinglish deep dive

3.3.31 · D3 · Physics › Rocket Propulsion › Transpiration cooling

Yeh page Transpiration Cooling ki "koi gap nahi" companion hai. Parent note ne tumhe master result diya tha. Yahan hum deliberately har tarah ka input dhundhte hain jo formula face kar sakta hai — normal numbers, zeros, infinities, aur wo sneaky edge cases — aur har ek ko ek trustworthy number tak grind karte hain.

Recall Do formulas jinpe hum poore time lean karte hain

$T_{w} = \frac{η h _{0} T _{g} + G c _{p} T _{c}}{η h _{0} + G c _{p}}, ϕ = \frac{G c _{p}}{η h _{0} + G c _{p}} = \frac{1}{1 + \frac{η h _{0}}{G c _{p}}} .$ $ϕ$ ko padhо as "wall kitna cool hua hai, 0-se-1 scale par, gas se coolant ki taraf."

Numbers touch karne se pehle, ek baar aur plain words mein ingredients ka naam lete hain taaki koi symbol unexplained na rahe:

Definition Paanch knobs (words mein)

$T_{g}$ — hot gas ka temperature jo wall ko hammer kar raha hai (kelvin, K).
$T_{c}$ — wo temperature jis par coolant enter karta hai, thanda (K).
$T_{w}$ — wall ka steady temperature — yahi woh cheez hai jo hum usually solve karte hain (K).
$h_{0}$ — bare heat-transfer coefficient: kitne watts wall ke har square metre se cross karte hain har 1 K gas-to-wall temperature gap ke liye, bina kisi coolant blowing ke. Units $W m^{- 2} K^{- 1}$ . (Dekho Convective Heat Transfer.)
$η$ — blowing reduction factor: 0 aur 1 ke beech ek pure number jo batata hai ki coolant ke sweat out hone ke baad bhi us heat ka kitna fraction andar aata hai. $η = 1$ = koi help nahi, $η \to 0$ = perfect blanket.
$G$ — coolant mass flux: kilograms of coolant jo wall ke har square metre se har second push hota hai ( $kg m^{- 2} s^{- 1}$ ).
$c_{p}$ — coolant ki specific heat: joules jo uske 1 kg ko 1 K warm karne ke liye chahiye.

Dono products $η h_{0}$ aur $G c_{p}$ ke units $W m^{- 2} K^{- 1}$ hain — ye conductances hain. Isi liye ye ek fraction mein add aur compare ho sakte hain.

Scenario matrix

Is topic mein jo bhi situation aa sakti hai wo neeche ke kisi cell mein hai. Jo examples follow karte hain unhe cell ke hisaab se label kiya gaya hai taaki tum dekh sako ki coverage complete hai.

Cell	Kya special hai	Covered by
A. Normal	saare knobs finite aur typical	Ex 1
B. Design inverse	target $T_{w}$ paane ke liye coolant $G$ solve karo	Ex 2
C. Zero coolant ( $G = 0$ )	degenerate: bilkul bhi blowing nahi	Ex 3
D. Perfect / infinite coolant ( $G \to \infty$ )	limiting behaviour, the floor	Ex 3
E. No blowing benefit ( $η = 1$ )	coolant sirf heat sink hai, koi film nahi	Ex 4
F. Non-linearity / diminishing returns	$G$ double karna ≠ $Δ T$ half karna	Ex 5
G. Real-world word problem	throat area → total coolant, thrust cost	Ex 6
H. Exam twist	$η$ depends on $G$ (variable blowing)	Ex 7
I. Consistency / sanity	$ϕ$ se $T_{w}$ recompute karo, units cross-check karo	Ex 8

Hum pehle poora map visualise karte hain, taaki tum dekh sako ki cooling curve par har case kahan rehta hai.

Cell A — everyday case

Worked example Example 1 — plain wall temperature (Cell A)

$T_{g} = 3000 K$ , $T_{c} = 300 K$ , $c_{p} = 1200 J k g^{- 1} K^{- 1}$ , $h_{0} = 5000 W m^{- 2} K^{- 1}$ , $η = 0.4$ , $G = 2 kg m^{- 2} s^{- 1}$ .

Forecast: dono conductances close honge (2000 vs 2400), toh $T_{w}$ $T_{g}$ aur $T_{c}$ ke beech land karni chahiye, thodi cold side ke kareeb — guess ~1500 K.

Gas conductance $η h_{0} = 0.4 \times 5000 = 2000$ . Ye step kyun? Ye coolant ke boundary layer thin karne ke baad effective heat delivery per kelvin hai.
Coolant conductance $G c_{p} = 2 \times 1200 = 2400$ . Ye step kyun? Ye batata hai ki coolant per kelvin, per square metre kitni heat soak kar sakta hai.
Weighted average: $T_{w} = \frac{2000 ( 3000 ) + 2400 ( 300 )}{2000 + 2400} = \frac{6 720 000}{4400} \approx 1527 K$ . Ye step kyun? $T_{w}$ exactly dono temperatures ka conductance-weighted average hai.

Verify: $ϕ = \frac{2400}{4400} = 0.545$ . Phir $T_{g} - ϕ (T_{g} - T_{c}) = 3000 - 0.545 (2700) = 1527 K$ ✔. 300 aur 3000 ke beech, cold ke kareeb — forecast se match karta hai.

Cell B — target ke liye design karna

Worked example Example 2 — target hit karne ke liye kitna coolant chahiye (Cell B)

Ex 1 jaisa gas/coolant, $η = 0.4$ constant rakho. Hum demand karte hain $T_{w} \leq 1200 K$ . Sabse chhota $G$ dhundho.

Forecast: 1200 K, $G = 2$ ke saath mile 1527 K se thanda hai, toh hume zyada coolant push karna hoga — guess karo $G$ thoda 3 se upar.

Target effectiveness: $ϕ = \frac{3000 - 1200}{3000 - 300} = \frac{1800}{2700} = 0.667$ . Ye step kyun? Jab target ek wall temperature ho, $ϕ$ sabse clean handle hai — ye temperature ko ek dimensionless goal mein convert karta hai.
$ϕ$ ko invert karo: $ϕ = \frac{G c _{p}}{η h _{0} + G c _{p}}$ ke algebra se milta hai $G c_{p} = \frac{ϕ}{1 - ϕ} η h_{0}$ . Ye step kyun? Hume $ϕ$ pata hai aur $G$ chahiye, toh coolant conductance isolate karo.
Numbers: $G c_{p} = \frac{0.667}{0.333} \times 2000 = 2 \times 2000 = 4000$ , toh $G = \frac{4000}{1200} \approx 3.33 kg m^{- 2} s^{- 1}$ . Ye step kyun? Required conductance ko $c_{p}$ se divide karo mass flux paane ke liye.

Verify: $G = 3.33$ wapas daalo: $G c_{p} = 4000$ , $T_{w} = \frac{2000 ( 3000 ) + 4000 ( 300 )}{2000 + 4000} = \frac{7 200 000}{6000} = 1200 K$ ✔ exactly target.

Cells C & D — dono degenerate ends

Ye "extremes par kya hota hai" wale cases hain. Yahan students panic karte hain kyunki $0$ ya $\infty$ appear hota hai — lekin formula perfectly well-behaved rehta hai.

Worked example Example 3 — coolant band karo, phir infinity tak le jao (Cells C & D)

Ex 1 ke same numbers.

Forecast: $G = 0$ ke saath wall utni hi hot honi chahiye jitna gas hai (kuch bhi heat remove nahi karta). $G \to \infty$ ke saath wall coolant temperature tak chill down karni chahiye. Toh expect karo $T_{w} \to 3000$ aur $T_{w} \to 300$ .

Zero coolant, $G = 0$ (Cell C): $G c_{p} = 0$ , toh $T_{w} = \frac{2000 ( 3000 ) + 0}{2000 + 0} = 3000 K$ , aur $ϕ = \frac{0}{2000 + 0} = 0$ . Ye step kyun? Bina coolant conductance ke, weighted average ka poora weight gas par hai — wall melt ho jaati hai.
Infinite coolant, $G \to \infty$ (Cell D): upar aur neeche dono ko $G c_{p}$ se divide karo: $T_{w} = \frac{\frac{η h _{0}}{G c _{p}} T _{g} + T _{c}}{\frac{η h _{0}}{G c _{p}} + 1} \to \frac{0 + T _{c}}{0 + 1} = T_{c} = 300 K$ , aur $ϕ \to 1$ . Ye step kyun? Jab coolant conductance dominate karta hai, weighted average cold side par collapse ho jaata hai — yahi floor hai.
Floor real hai: $T_{w}$ kabhi bhi $T_{c} = 300 K$ se neeche nahi ja sakta, chahe kitna bhi coolant daal do. Ye step kyun? Ye ek hard limit set karta hai jo designers respect karte hain — tum apne coolant ke inlet temperature se neeche cool nahi kar sakte.

Verify: $G = 0$ par, $ϕ = 0$ ✔; $G \to \infty$ limit deta hai $ϕ = 1$ , $T_{w} = T_{c} = 300$ ✔. Dono endpoints forecast se match karte hain aur har finite answer ko bracket karte hain.

Cell E — sirf heat sink, koi film nahi

Worked example Example 4 — coolant jo heat soak karta hai lekin blanket nahi banata (

$η = 1$ , Cell E) Socho coolant wall ke andar heat absorb karta hai lekin boundary layer thin NAHI karta, toh blowing se koi benefit nahi: $η = 1$ . $T_{g} = 3000$ , $T_{c} = 300$ , $h_{0} = 5000$ , $c_{p} = 1200$ , $G = 2$ rakho.

Forecast: film lose karne ka matlab hai gas full heat deliver karta hai ( $η h_{0} = 5000$ instead of 2000), toh wall Ex 1 ke 1527 K se hotter chalegi — guess ~1900 K.

Gas conductance $η h_{0} = 1 \times 5000 = 5000$ . Ye step kyun? $η = 1$ ke saath full bare coefficient act karta hai — given $h_{0}$ ke liye ye worst case hai.
Coolant conductance $G c_{p} = 2 \times 1200 = 2400$ (unchanged). Ye step kyun? Heat-sink capacity $η$ par depend nahi karti.
Wall temperature $T_{w} = \frac{5000 ( 3000 ) + 2400 ( 300 )}{5000 + 2400} = \frac{15 720 000}{7400} \approx 2124 K$ . Ye step kyun? Same weighted average naye gas conductance ke saath.

Verify: $ϕ = \frac{2400}{7400} = 0.324$ — Ex 1 ke 0.545 se bahut kam effective, aur $2124 > 1527$ ✔. Isi liye film effect matter karta hai: ye $η$ drop karta hai aur tumhe saikdon kelvin milte hain. Compare karo Film Cooling se, jo is blanket mechanism par heavily rely karta hai.

Cell F — diminishing-returns trap

Worked example Example 5 — coolant double karo, kya wall half ho jaati hai? (Cell F)

Ex 1 ke numbers wapas, lekin ab $G$ ko 2 se 4 karo.

Forecast: linear intuition whisper karta hai "double coolant → half temperature," yaani 1527 se ~760 tak. Ye guess likh lo — phir dekho ye kaise fail hota hai.

New coolant conductance $G c_{p} = 4 \times 1200 = 4800$ . Ye step kyun? Sirf $G$ badla, toh sirf yahi product update hoga.
New wall temperature $T_{w} = \frac{2000 ( 3000 ) + 4800 ( 300 )}{2000 + 4800} = \frac{7 440 000}{6800} \approx 1094 K$ . Ye step kyun? Same weighted average re-evaluate karo.
Drop measure karo: $1527 - 1094 = 433 K$ , na ki woh ~767 K jo halving demand karta. Ye step kyun? Average floor $T_{c} = 300$ ke upar pinned hai, toh extra coolant ever-smaller gains deta hai.

Verify: 1527 ka half hota 763.5 K; sach mein jawaab 1094 K usse kaafi upar hai ✔. Figure s01 ka curve concave hai, $T_{c}$ ki taraf flatten ho raha hai — yahi shape poori warning hai.

Cell G — ek full real-world word problem

Worked example Example 6 — nozzle throat ke liye total coolant, aur uska thrust cost (Cell G)

Ek throat mein porous wall area $A = 0.05 m^{2}$ hai. Ex 1 conditions use karo aur $T_{w} = 1200 K$ require karo. Ex 2 se required mass flux $G = 3.33 kg m^{- 2} s^{- 1}$ hai. Total propellant flow $\overset{m}{˙}_{prop} = 20 kg s^{- 1}$ hai. (a) Total coolant flow $\overset{m}{˙}_{c}$ dhundho. (b) Propellant ka kitna fraction cooling mein spend ho raha hai?

Forecast: ek chhota patch times kuch kg per m² per s → well under 1 kg/s, toh shayad 20 kg/s engine ka 1% se kam. Guess ~0.8%.

Total coolant $\overset{m}{˙}_{c} = G A = 3.33 \times 0.05 = 0.1665 kg s^{- 1}$ . Ye step kyun? Mass flux times area = mass flow; ye definition $G = \overset{m}{˙}_{c} / A$ rearranged hai.
Coolant fraction $\frac{m ˙ _{c}}{m ˙ _{prop}} = \frac{0.1665}{20} = 0.0083 = 0.83%$ . Ye step kyun? Coolant woh propellant hai jo tum thrust se divert karte ho, toh uska fraction real cost hai — ye effective Specific Impulse lower karta hai.
Interpret karo: 1% se kam flow throat ko survivable 1200 K par rakhta hai. Ye step kyun? Exactly isi liye transpiration cooling jeet jaati hai: tiny mass penalty ke baad huge temperature drop milta hai.

Verify: $0.1665/20 = 0.008325 \approx 0.83%$ ✔, forecast se match; aur $\overset{m}{˙}_{c} = G A$ ke units $(kg m^{- 2} s^{- 1}) (m^{2}) = kg s^{- 1}$ ✔.

Cell H — exam twist: $η$ $G$ ke saath girta hai

Parent ne warn kiya tha ki real blowing mein $η$ $G$ ke saath decrease karta hai (thicker blanket). Yahan woh twist concrete ho jaata hai.

Worked example Example 7 — variable blowing,

$η = 0.4 - 0.05 G$ (Cell H) Ex 1 conditions, lekin ab $η$ fixed nahi: $η (G) = 0.4 - 0.05 G$ (hamare range ke liye valid). $G = 2$ lo.

Forecast: $G = 2$ par, $η$ 0.4 se neeche gir jaata hai, toh gas kam heat deliver karta hai Ex 1 se → wall 1527 K se cooler honi chahiye. Guess ~1350 K.

$η$ evaluate karo: $η = 0.4 - 0.05 (2) = 0.3$ . Ye step kyun? $G = 2$ par blowing ne blanket pehle se thick kar diya hai, gas coupling weak kar diya hai.
Gas conductance $η h_{0} = 0.3 \times 5000 = 1500$ . Ye step kyun? Kam $η$ matlab kam effective delivery — yahi parent note ka "double win" hai.
Coolant conductance $G c_{p} = 2 \times 1200 = 2400$ (pehle jaisa). Ye step kyun? Heat-sink capacity $η$ se unbothered hai.
Wall temperature $T_{w} = \frac{1500 ( 3000 ) + 2400 ( 300 )}{1500 + 2400} = \frac{5 220 000}{3900} \approx 1338 K$ . Ye step kyun? Improved gas conductance ke saath same weighted average.

Verify: $1338 < 1527$ ✔ — variable blowing constant- $η$ estimate ko beat karta hai, exactly jaisa parent ne predict kiya tha; $ϕ = \frac{2400}{3900} = 0.615 > 0.545$ ✔ (zyada effective). Forecast ~1350 close tha.

Cell I — consistency par loop close karna

Worked example Example 8 —

$T_{w}$ do independent tareekon se recover karo (Cell I) Ex 1 ke answer use karke, prove karo ki boxed formula aur $ϕ$ -definition se $T_{w}$ agree karte hain, aur units check karo.

Forecast: ye zaroor match karenge — $ϕ$ same balance se derive hua tha — lekin dono taraf se karna arithmetic slips pakad leta hai.

Box se: $T_{w} = 1527 K$ (Ex 1). Ye step kyun? Hamara reference value.
$ϕ$ se: $ϕ = \frac{G c _{p}}{η h _{0} + G c _{p}} = \frac{2400}{4400} = 0.5 \overline{45}$ , phir $T_{w} = T_{g} - ϕ (T_{g} - T_{c}) = 3000 - 0.5455 (2700) = 1527 K$ . Ye step kyun? $ϕ$ is tarah define hota hai ki $T_{g} - T_{w} = ϕ (T_{g} - T_{c})$ — ek independent route.
Units audit: $η h_{0}$ aur $G c_{p}$ dono $W m^{- 2} K^{- 1}$ hain; fraction mein $\frac{( W m ^{- 2} K ^{- 1} ) ( K )}{W m ^{- 2} K ^{- 1}} = K$ . Ye step kyun? Ek temperature formula ko kelvin return karna chahiye, aur karta hai.

Verify: dono routes 1527 K dete hain ✔; final units K hain ✔. Formula internally consistent hai — upar ke saare cells mein iske upar trust karo.

Recall Coverage self-check

Kya har cell A–I cover hua? ::: A(1) B(2) C&D(3) E(4) F(5) G(6) H(7) I(8) — saare nine cells covered. $T_{w}$ kabhi $T_{c}$ se neeche kyun nahi gir sakta? ::: Kyunki $T_{w}$ ek conductance-weighted average hai $T_{g}$ aur $T_{c}$ ka; do numbers ke weighted average ki minimum value khud chhota number hi ho sakta hai. Woh single dimensionless group kaunsa hai jo sab kuch control karta hai? ::: Conductance ratio $\frac{G c _{p}}{η h _{0}}$ , jo set karta hai $ϕ = \frac{1}{1 + η h _{0} / ( G c _{p} )}$ .

Compare karne laayak related schemes: Regenerative Cooling, Ablative Cooling, aur $h_{0}$ ke peeche boundary-layer machinery Boundary Layer Theory mein.