Transpiration cooling: Foundations — Hinglish deep dive

3.3.31 · D1 · Physics › Rocket Propulsion › Transpiration cooling

Is page mein assume kiya gaya hai ki aapne pehle kuch nahi dekha. Hum parent Transpiration cooling note ke har ek letter, ratio, aur picture ko is order mein build karte hain jahan har block apne pehle wale par tika ho.

0. Koi bhi symbol aane se pehle ka scene

Ek rocket wall ka flat slice imagine karo. Ek taraf: monstrously hot gas tezi se guzar raha hai. Doosri taraf: cool fluid jo hum pump karte hain. Wall mein microscopic pores hain, isliye coolant seep karke hot flow mein mix ho jaata hai.

Figure 1 — Scene. Amber arrows (left) hot gas hain $T_{g} \approx 3000 K$ par porous wall ke paas se guzarte hue. Cyan arrows (right) coolant hain $T_{c} \approx 300 K$ par pores se push hote hue. Left face se nikle chote cyan arrows "sweat" blanket hain. Wall khud kisi middle temperature $T_{w}$ par hai (white label). Figure ko left-to-right padhein: hot in, coolant out, wall beech mein.

Baaki sab is picture mein kisi quantity ka naam hai. Chalo har ek ko earn karte hain.

1. Temperature — "kitna hot hai" wala number

Parent page teen alag temperatures use karta hai. Inhe seedha rakhna — yahi #1 mistake hai:

Symbol	Simple words	Picture mein
$T_{g}$	hot gas ki temperature	left par aag, $\sim 3000 K$
$T_{c}$	coolant ki temperature jab enter karta hai	right par cold fluid, $\sim 300 K$
$T_{w}$	wall ki apni temperature	beech mein metal

2. Heat flux $q$ — "kitni heat, kitni fast, kitni wall se guzarke"

Heat ek tarah ki energy hai, joules ( $J$ ) mein measure hoti hai. Lekin hum rarely total joules ki parwah karte hain; hum rate aur area ki parwah karte hain.

Hume flux kyun chahiye, na ki sirf total heat? Kyunki nozzle ka throat chhota hota hai lekin per unit area bahut zyada hammered hota hai — flux hi metal ko pighalti hai, total nahi. (Yeh wahi quantity hai jo Nozzle Throat Heat Flux mein study hoti hai.)

Parent page do versions use karta hai:

$q_{0}$ — flux bina kisi coolant blowing ke (bare, worst case).
$q$ — actual flux coolant blowing ke saath (chhota).

3. Convection aur heat-transfer coefficient $h$

Heat gas se wall mein kaise pahunchti hai? Gas ek baar touch karke apni jiggling nahi deti — gas flow karti rehti hai, surface par lagataar fresh hot particles deliver karti rehti hai. Flowing fluid ki yeh heat delivery convection kehlati hai (dekho Convective Heat Transfer).

Yeh proportionality — Newton's law of cooling — do flavours mein aata hai. Bare wall (no coolant blowing) apna coefficient $h_{0}$ use karta hai: $q_{0} = h_{0} (T_{g} - T_{w})$ aur blown wall (coolant sweating) same temperature gap ke liye ek chhota coefficient $h$ use karta hai: $q = h (T_{g} - T_{w})$

Wall ke paas flow ke sab messy fluid mechanics — speed, turbulence, slow layer ki thickness — sab is ek number mein pack ho jaate hain. Woh layer Boundary Layer Theory ka subject hai.

Figure 2 — Blowing $h$ kyun shrink karta hai. Horizontal axis wall se $y$ door tak ka distance hai; vertical axis gas temperature hai. Amber curve (no blowing) wall ke bilkul paas steeply rise karti hai — steep temperature gradient matlab heat fast pour hoti hai, yani bada $h_{0}$ . Cyan curve (with blowing) gently rise karti hai, kyunki coolant blanket ne hot gas ko door push kar diya hai; shallow wall-gradient matlab chhota $h$ . Wall par do arrows dekho: steeper amber wala har second zyada heat deliver karta hai. Same $T_{g} - T_{w}$ , chhota coefficient — yahi poori baat hai.

4. Boundary layer — woh thin slow film jo insulating karti hai

Yeh heat ke liye kyun matter karta hai? Heat ko is sluggish layer ko cross karna hota hai metal tak pahunchne ke liye, aur ek thick, slow layer ek achhi insulator hoti hai. Transpiration cooling ki trick yeh hai:

5. Mass flow $\overset{m}{˙}_{c}$ aur mass flux $G$ — "kitna coolant"

Ab coolant side. Hume count karna hai ki hum kitna coolant push karte hain.

Lekin heat ki tarah, hum per unit area care karte hain, kyunki ek badi wall naturally zyada coolant carry karti hai. Isliye hum area se divide karte hain:

6. Specific heat $c_{p}$ — "ek kilogram kitni heat soak kar sakta hai"

Coolant garam hokar kaam karta hai: yeh cold enter karta hai aur hot exit karta hai, aur raaste mein jo heat absorb karta hai woh wall se churaai hoti hai. Garam hone mein kitni heat lagti hai?

Toh ek kilogram $T_{c}$ se $T_{w}$ tak jaane mein jo heat carry karta hai woh hai $c_{p} (T_{w} - T_{c})$ , aur saara coolant jo heat per second per square metre carry karta hai woh hai: $kg per m^{2} per s G J per kg per K c_{p} K (T_{w} - T_{c}) = W m^{- 2} = ek heat flux. ✓$

Notice karo ki units multiply hokar ek flux bante hain — yeh tumhara guarantee hai ki equation honest hai.

7. Blowing reduction factor $η$ — "blanket kitna help karta hai"

Ek ratio kyun, scratch se naya coefficient kyun nahi? Kyunki yeh ek clean idea isolate karta hai: "blowing ne heat delivery ko kitna thin kiya?" Chhota $η$ = strong shielding. Aur sabse zaroori, $η$ zyada blowing ke saath shrink hota hai ( $G$ up) — zyada coolant, thicker blanket, weaker heat delivery.

8. Do "conductances", balance, aur $T_{w}$

Poori parent derivation ek sentence hai: steady state mein, gas jo heat deliver karta hai woh coolant jo heat carry karta hai ke barabar hai. Left par blown flux $q = η h_{0} (T_{g} - T_{w})$ use karte hue (kyunki $h = η h_{0}$ ) aur right par coolant pickup $G c_{p} (T_{w} - T_{c}$ ):

$gas delivers, per m^{2} η h_{0} (T_{g} - T_{w}) = coolant carries off, per m^{2} G c_{p} (T_{w} - T_{c})$

Ab $T_{w}$ ke liye step by step solve karo (sirf algebra):

$η h_{0} T_{g} - η h_{0} T_{w} = G c_{p} T_{w} - G c_{p} T_{c}$

Har $T_{w}$ ko ek side move karo, baaki sab doosri side:

$η h_{0} T_{g} + G c_{p} T_{c} = G c_{p} T_{w} + η h_{0} T_{w} = (η h_{0} + G c_{p}) T_{w}$

Bracket se divide karo:

$T_{w} = \frac{η h _{0} T _{g} + G c _{p} T _{c}}{η h _{0} + G c _{p}}$

Pull karne wali donon quantities ke units $W m^{- 2} K^{- 1}$ hain — yeh "conductances" hain:

Quantity	Naam	Kis side ka hai
$η h_{0}$	gas-side conductance	gas kitni strongly heat push karta hai
$G c_{p}$	coolant-side conductance	coolant kitni strongly heat carry karta hai

Figure 3 — Tug-of-war. Horizontal white bar ek temperature scale hai $T_{c} = 300 K$ (cyan, left) se $T_{g} = 3000 K$ (amber, right) tak. White marker resulting $T_{w} = 1527 K$ hai parent ke Example 1 se. Cyan arrow coolant conductance $G c_{p}$ ko marker ko left (cooler) ki taraf kheenchte dikhata hai; amber arrow gas conductance $η h_{0}$ ko right (hotter) ki taraf kheenchte dikhata hai. Jahan yeh balance karte hain exactly wahi boxed formula ka weighted average hai.

Edge cases — tug-of-war ko uske limits tak push karna

Boxed formula mein extreme values dalo aur dekho yeh kaise behave karta hai:

Yeh chaar corners har real case ko bracket karte hain: $T_{w}$ hamesha $T_{c}$ aur $T_{g}$ ke beech hi rehti hai, kabhi bahar nahi.

Cooling effectiveness $ϕ$

Aakhir mein, 0 se 1 tak ek clean scorecard:

9. Specific impulse $I_{s p}$ — coolant free kyun nahi hai

Coolant ka jo bhi kilogram tum sweat out karte ho woh ek kilogram hai jise tum thrust ke liye use kar sakte the. Propellant ko thrust mein convert karne ki efficiency Specific Impulse ( $I_{s p}$ ) se measure hoti hai. Isliye parent page blindly $G$ maximize karne ke against warn karta hai: cooling se performance cost hoti hai.

Prerequisite map

Poori picture, assembled

Parent page jo single equation derive karta hai woh bas yeh kehta hai: gas jo heat deliver karta hai (left) equals coolant jo heat carry karta hai (right), aur $T_{w}$ ke liye solve karna (§8 mein kiya gaya) tug-of-war average deta hai. $gas delivers η h_{0} (T_{g} - T_{w}) = coolant carries off G c_{p} (T_{w} - T_{c})$

Related cooling cousins jinhein ab tum compare kar sakte ho: Film Cooling, Regenerative Cooling, Ablative Cooling.

Quick recall

Recall Teen temperatures kya hain aur unka order kya hai?

$T_{c} \leq T_{w} \leq T_{g}$ : coolant cold, wall beech mein, gas sabse hottest.

Recall Positive

$q$ ka direction kya hai, aur yah yahan positive kyun rehta hai? Positive = heat gas se wall mein flow karti hai; yeh positive rehta hai kyunki $T_{g} > T_{w}$ hamesha hota hai.

Recall Area se divide karke flux

$q$ aur mass flux $G$ kyun nikalte hain? Kyunki melting aur cooling per unit area hote hain — ek chhota hot throat mein modest total heat ke saath bhi huge flux ho sakta hai.

Recall

$η = 0.4$ physically kya matlab hai, aur $η = q / q_{0}$ kab hold karta hai? Blown coefficient bare wale ka $40%$ hai ( $h = 0.4 h_{0}$ ); $η = q / q_{0}$ sirf tab hold karta hai jab $Δ T$ dono cases mein identical ho.

Recall

$T_{w}$ kya karta hai jab $G \to 0$ aur $G \to \infty$ ? $G \to 0$ : $T_{w} \to T_{g}$ (no cooling). $G \to \infty$ : $T_{w} \to T_{c}$ (perfect cooling).

Recall

$ϕ = 1$ ka kya matlab hai aur yah impossible kyun hai? Wall entering coolant jitni cold — perfect cooling, jiske liye infinite coolant chahiye.

Equipment checklist

Right side cover karo aur zor se answer do. Agar kar sako, tum parent derivation ke liye ready ho.

3000 K ko "kisi melting metal ke comparison mein kitna hot hai" mein convert kar sakta hoon

Haan — kelvin absolute zero se straight scale hai; 3000 K kisi bhi alloy ke melting point (~1900 K) se kaafi upar hai, isliye koi bhi material bina help ke survive nahi kar sakta.

Heat flux

q

ke units aur uska sign convention bata sakta hoon

W m^{- 2}

; positive matlab heat gas se wall mein flow kar rahi hai.

Dono convection laws

q_{0} = h_{0} (T_{g} - T_{w})

aur

q = h (T_{g} - T_{w})

likh sakta hoon

Same temperature gap, alag coefficient;

h < h_{0}

blowing ke baad.

Boundary layer aur blowing ise kaise change karta hai explain kar sakta hoon

Wall par gas ki thin slowed layer; coolant ise thicken aur cool karti hai, isliye wall gradient (aur

h

) drop karta hai.

\overset{m}{˙}_{c}

aur

A

se mass flux

G

compute kar sakta hoon

G = \overset{m}{˙}_{c} / A

kg m^{- 2} s^{- 1}

mein.

c_{p}

kya measure karta hai explain kar sakta hoon

Coolant ke ek kilogram ko ek kelvin raise karne ke liye joules — uski heat-sponge capacity.

η

define kar sakta hoon aur bata sakta hoon ki

η = q / q_{0}

kab hold karta hai

η = h / h_{0}

;

q / q_{0}

ke barabar sirf tab hota hai jab

Δ T

blown aur un-blown dono cases mein same ho.

Energy balance se

T_{w}

derive kar sakta hoon

η h_{0} (T_{g} - T_{w}) = G c_{p} (T_{w} - T_{c})

set karo,

T_{w}

collect karo, weighted-average box milta hai.

T_{w}

ke do limiting cases bata sakta hoon

G \to 0

deta hai

T_{g}

;

G \to \infty

deta hai

T_{c}

Zyada coolant free kyun nahi hai bata sakta hoon

Coolant propellant mass hai jo lost ho jaati hai, specific impulse

I_{s p}

ghatata hai.