Aerodynamic heating — recovery temperature, heat flux: Visual walkthrough

3.1.28 · D2 · Physics › Compressible Flow & Aerodynamics › Aerodynamic heating — recovery temperature, heat flux

Step 1 — Ek moving air blob do tarah ki energy carry karta hai

KYA HAI. Socho ek chhota parcel (ek "blob") air ka jo speed $u$ se right ki taraf ud raha hai. Wo energy do forms mein carry karta hai: uski motion ki organised energy (saare molecules ek hi direction mein chal rahe hain), aur uske molecules ke randomly hilaate rehne ki hidden energy (woh hila-duli hi temperature hai).

YAHAN SE KYUN SHURU KARTE HAIN. Ek fast vehicle par heating, pehli tarah ki energy ko doosri tarah mein convert karne se zyada mysterious kuch nahi hai. Agar hum woh do buckets nahi dekh sakte, toh baad ki koi baat samajh nahi aayegi.

PICTURE. Figure mein, lamba seedha arrow organised motion hai (speed $u$ ); blob ke andar fuzzy scribbles random hila-duli hain (temperature $T$ ).

Step 2 — Blob ko wall par rokna: energy ko kahin na kahin jaana hai

KYA HAI. Ab blob ek solid wall se takraakar rest mein aa jaata hai. Uski marching speed $u = 0$ ho jaati hai. Woh organised energy $u^{2} /2$ gayab nahi ho sakti — energy conserved hoti hai — isliye woh hila-duli ke bucket mein beh jaati hai, temperature badhaa deti hai.

KYUN YEH TOOL: streamline ke saath energy conservation. Hum energy conservation choose karte hain (force ya momentum nahi) kyunki hum sirf kitna hot hai, yani ek energy question, ki parwah karte hain. Steady flow ke liye jisme bahar se koi heat nahi dali aur koi shaft/work nahi kiya gaya, thermodynamics ka pehla law ek clean statement mein collapse ho jaata hai: total (hila-duli + march) energy pehle aur baad mein same rahti hai.

PICTURE. Left: fast blob, thanda (chhoti hila-duli). Right: wall par ruka hua blob, garam (badi hila-duli). Marching arrow ek dot mein simatt gaya hai; scribbles badh gaye hain.

Step 3 — Enthalpy ko temperature mein convert karo: stagnation temperature $T_{0}$

KYA HAI. Har enthalpy ko $c_{p} T$ se replace karo (temperature ke terms mein hila-duli energy). Ruke hue blob ki temperature ka apna naam hota hai: stagnation temperature $T_{0}$ .

KYUN. Hum enthalpy dekh nahi sakte, lekin hum thermometer padh sakte hain. $h = c_{p} T$ woh dictionary hai jo "energy" ko "degrees" mein translate karti hai, isliye hum abhi translation karte hain.

PICTURE. Ek thermometer jo $T$ (flowing) se $T_{0}$ (stopped) tak chadh raha hai. Mercury ki extra height exactly marching energy $u^{2} /2$ divided by $c_{p}$ hai.

$h = c_{p} T$ aur $h_{0} = c_{p} T_{0}$ substitute karne par:

$hila-duli c_{p} T + march \frac{u ^{2}}{2} = sab hila-duli c_{p} T_{0} \Rightarrow T_{0} = T (1 + \frac{u ^{2}}{2 c _{p} T})$

$T_{0}$ — temperature agar tum gas ko zero heat loss ke saath rokote (kelvin).
$T$ — actual flowing temperature.
$\frac{u ^{2}}{2 c _{p} T}$ — fractional temperature boost: march energy ko gas ki apni thermal energy $c_{p} T$ ke units mein measure kiya.

Step 4 — Boost ko Mach number $M$ use karke rewrite karo

KYA HAI. Clumsy fraction $\frac{u ^{2}}{2 c _{p} T}$ tidy $\frac{γ - 1}{2} M^{2}$ ban jaata hai.

KYUN YEH TOOL: Mach number. Compressible flow mein ek hi cheez matter karti hai — tum us speed ki comparison mein kitna fast jaate ho jis par gas message pass kar sakti hai — sound ki speed $a$ . Woh ratio $M = u / a$ natural yard-stick hai. Hum $M$ mein rewrite karte hain taaki formula kisi bhi altitude, kisi bhi gas par kaam kare, bina $u$ , $c_{p}$ , $R$ alag-alag carry kiye.

PICTURE. Do speed bars side by side: tumhari speed $u$ aur sound speed $a$ ; unka ratio $M$ label kiya gaya hai. Neeche, algebra "gears" dikhate hain ki $c_{p}$ aur $a$ $γ$ aur $R$ se kaise bane hain.

Humein do facts chahiye, har ek picture-able definition:

Ab fraction grind karo:

$\frac{u ^{2}}{2 c _{p} T} = \frac{u ^{2}}{2} \cdot \frac{γ - 1}{γ R T} = \frac{γ - 1}{2} \cdot \frac{u ^{2}}{= a ^{2} γ R T} = \frac{γ - 1}{2} = M^{2} (\frac{u}{a})^{2}$

jo clean master relation deta hai:

$\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2}$

$\frac{γ - 1}{2}$ — ek fixed number ( $= 0.2$ air ke liye) jo is baat se set hota hai ki gas energy kaise store karti hai.
$M^{2}$ — speed ke square ke saath badhta hai: Mach double karo, heating boost chaar guna.

Yahi result Stagnation properties & isentropic relations note bhi nikalta hai.

Step 5 — Reality check: wall $T_{0}$ tak NAHI pahunchti

KYA HAI. Surface se chipki air ki paper-thin layer mein zoom karo — boundary layer. Do cheezein wahan ladhti hain: sliding air layers ke beech friction heat banata hai (viscous dissipation), jabki heat ek saath garam region se sideways conduct bhi hoti rehti hai. Woh cancel nahi hote, isliye real wall $T_{e}$ aur $T_{0}$ ke beech settle hoti hai.

YEH KYUN MATTER KARTA HAI. Agar hum Step 4 par ruk jaate toh hum wall temperature thodi zyada predict karte — design ke liye khatarnaak. Humein neeche correct karna hoga.

PICTURE. Boundary layer sliding cards ke stack ki tarah: top par fast, wall par frozen. Red scribbles = friction heat ban rahi hai; blue arrows = heat conduct ho rahi hai. Woh unequal hain.

Yeh tug-of-war Prandtl number $P r = μ c_{p} / k$ decide karta hai — literally "momentum kitna fast spread hota hai vs heat kitna fast spread hoti hai." Air ke liye $P r < 1$ : heat friction se zyada fast escape karti hai, isliye poora boost recover nahi hota. Poori origin ke liye Boundary layers & viscous dissipation dekho.

Step 6 — Imperfection ko ek number mein bottle karo: recovery factor $r$

KYA HAI. $r$ define karo us fraction ke roop mein jo full stagnation boost ka wall par actually bachta hai.

KYUN. Har baar poori boundary-layer heat balance solve karne ki bajaye, engineers result ko ek dimensionless dial $r$ mein pack karte hain jo $0$ aur $1$ ke beech hota hai.

PICTURE. $T_{e}$ (bottom) se $T_{0}$ (top) tak ek ruler. Wall temperature $T_{r}$ us ruler par fraction $r$ upar baithti hai.

Theory (agle note mein) woh do values deti hai jo tum yaad karte ho: $r_{lam} = P r \approx 0.84, r_{turb} = P r^{1/3} \approx 0.89.$

Step 7 — Recovery temperature $T_{r}$ assemble karo

KYA HAI. "Kitna rise ho sakta tha" gap $T_{0} - T_{e}$ lo, sirf fraction $r$ rakho, aur $T_{e}$ par wapas add karo.

KYUN. Yeh Step 6 par sirf algebra hai, lekin yeh ek abstract factor ko woh number banaata hai jo designer actually chahta hai: woh temperature jis par surface rahegi.

PICTURE. Teen stacked bars: $T_{e}$ (base), full boost $T_{0} - T_{e}$ ghosted, aur rakha gaya slice $r (T_{0} - T_{e})$ solid — unka top $T_{r}$ hai.

Step 4 ke result se gap ke roop mein shuru karo:

$T_{0} - T_{e} = T_{e} \frac{γ - 1}{2} M_{e}^{2} (subtract T_{e} from T_{0} = T_{e} (1 + \frac{γ - 1}{2} M_{e}^{2}))$

$r$ se multiply karo (sirf recovered slice rakho) aur $T_{e}$ wapas add karo:

$T_{r} = T_{e} + r (T_{0} - T_{e}) = T_{e} + r T_{e} \frac{γ - 1}{2} M_{e}^{2}$

$T_{r} = T_{e} (1 + r \frac{γ - 1}{2} M_{e}^{2})$

$T_{e}$ — local edge temperature (woh base jis par tum build karte ho).
$r$ — Step 6 ka recovery dial.
$\frac{γ - 1}{2} M_{e}^{2}$ — Step 4 ka full stagnation boost, ab edge Mach $M_{e}$ use karke.

Step 8 — Har case, taaki tumhe kabhi koi unshown scenario na mile

KYA HAI. Wall temperature $T_{w}$ upar-neeche slide karo aur dekho heat flux $\overset{q}{˙}_{w} = h (T_{r} - T_{w})$ ka sign kaise badalta hai. Degenerate speeds bhi explore karo.

KYUN. Formula ko sab inputs survive karne chahiye — thandi wall, garam wall, zero speed, bahut zyada speed — warna tum ek real vehicle ko galat design karoge.

PICTURE. $T_{r}$ par ek number line. Wall uske left mein → arrows wall ke andar point karte hain (heating); wall $T_{r}$ par → arrows gayab; wall uske right mein → arrows bahar point karte hain (wall heat kho rahi hai).

Formula Heat-flux sign map (Newton's law of cooling)

$\overset{q}{˙}_{w} = h (T_{r} - T_{w})$

Case	Condition	Kya hota hai
Cooled wall	$T_{w} < T_{r}$	$\overset{q}{˙}_{w} > 0$ : heat wall ke andar flow karta hai → cooling chahiye
Adiabatic wall	$T_{w} = T_{r}$	$\overset{q}{˙}_{w} = 0$ : yahi $T_{r}$ ko define karta hai
Over-hot wall	$T_{w} > T_{r}$	$\overset{q}{˙}_{w} < 0$ : wall gas ko heat lose karta hai

Degenerate / limiting speeds:

$M_{e} = 0$ (still air): boost $= 0$ , isliye $T_{r} = T_{e}$ . March energy nahi → recover karne ko kuch nahi. ✔
$M_{e} \to \infty$ (hypersonic): $T_{r} \approx T_{e} r \frac{γ - 1}{2} M_{e}^{2} \propto M_{e}^{2}$ — temperature speed ke square ke saath explode karta hai; yeh Hypersonic re-entry & thermal protection systems ka regime hai.
$r \to 1$ (imaginary perfect gas, $P r = 1$ ): $T_{r} \to T_{0}$ — recovery complete hai, wall sach mein stagnation tak pahunch jaati. Real air ka $P r < 1$ use iske theek neeche rakhta hai.
Shock ke peeche: $T_{e}$ khud already raise ho chuka hai (dekho Normal & oblique shock heating), isliye tum post-shock $T_{e}$ aur $M_{e}$ usi boxed formula mein feed karte ho.

Ek-picture summary

Sab kuch ek canvas par: kinetic energy $\to$ stagnation $\to$ fraction $r$ rakho $\to$ recovery temperature $\to$ heat flux drive karo. Arrows trace karo aur tumne parent note dobara derive kar liya.

Recall Feynman: poori story plain words mein bolo

Ek air ka blob jo wall ki taraf rush kar raha hai woh do tarah ki energy carry karta hai — fast jaane ki marching energy, aur woh jiggling energy jo hum warmth ki tarah feel karte hain. Jab wall blob ko rokte hai, marching energy ke paas extra jiggle mein jaane ke siwa aur koi jagah nahi hoti, isliye air aur garam ho jaati hai. Agar woh perfectly bina kisi leak ke ruk jaata, toh woh "stagnation temperature" $T_{0}$ tak pahunchta, aur humne nikala $T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}$ — jitna fast jaao (bada $M$ ), utna garam, speed squared ke saath badhta hua. Lekin surface ke bilkul paas air ki ek thin skin hoti hai jahan friction heat banata hai jabki conduction quietly use carry kar le jaati hai. Kyunki air heat ko momentum drag karne se thoda fast escape karne deti hai ( $P r < 1$ ), wall sirf us boost ka ek fraction $r$ (lagbhag 85–90 %) hi rakhti hai. Woh fraction rakho, local air temperature mein add karo, aur tumhe recovery temperature $T_{r}$ milti hai — woh temperature jis par ek uncooled surface sach mein drift karti hai. Finally, heat sirf hot se cold ki taraf flow karti hai, aur jo "hot" matter karta hai woh $T_{r}$ hai: agar tumhari wall $T_{r}$ se thandi hai toh heat andar pourta hai, agar woh exactly $T_{r}$ par baith jaaye toh kuch flow nahi karta, aur agar tum kisi tarah use aur garam banao toh woh heat wapas air mein dump kar dega. Woh single chain — stagnate, fraction $r$ rakho, $T_{r} - T_{w}$ se drive karo — yahi poora aerodynamic heating hai.

Connections

Stagnation properties & isentropic relations — Steps 1–4 yahan poore detail mein hain
Boundary layers & viscous dissipation — Step 5 ki friction-vs-conduction fight
Prandtl number & thermal boundary layer — kyun $r$ 1 se neeche land karta hai
Reynolds analogy & Stanton number — Step 8 mein $h$ ko skin-friction correlations mein turn karta hai
Hypersonic re-entry & thermal protection systems — Step 8 ka $M_{e} \to \infty$ limit
Normal & oblique shock heating — woh post-shock $T_{e}, M_{e}$ supply karta hai jo tum Step 7 mein feed karte ho
Parent: Recovery temperature & heat flux

Kaun sa conservation law

T_{0}

deta hai, aur kyun woh?

Energy conservation along a streamline (total enthalpy

h + u^{2} /2

constant), kyunki "kitna hot" ek energy question hai jisme koi heat add nahi hui aur koi work nahi kiya gaya.

Temperature boost ko Mach number use karke kyun rewrite karte hain?

Kyunki compressible flow mein sirf wahi matter karta hai ki tum speed of sound ke relative kitna fast ho;

M = u / a

formula ko altitude- aur gas-independent banata hai, deta hai

\frac{γ - 1}{2} M^{2}

T_{r}

strictly

T_{e}

aur

T_{0}

ke beech kyun hai?

Friction boundary layer mein heat banata hai lekin conduction (

P r < 1

) kuch carry kar le jaati hai, isliye wall par stagnation boost ka sirf ek fraction

r < 1

bachta hai.

Recovery-temperature formula mein

r = 1

physically kya matlab hai?

Complete recovery — wall full stagnation temperature

T_{0}

tak pahunch jaayegi; real air ka

P r < 1

hota hai isliye

r < 1

aur

T_{r} < T_{0}

Heat-flux sign jab

T_{w} > T_{r}

ho?

\overset{q}{˙}_{w} = h (T_{r} - T_{w}) < 0

: wall recovery temperature se zyada garam hai, isliye woh gas ko heat lose karta hai.

M_{e} \to 0

par

T_{r}

ki limit?

T_{r} \to T_{e}

: koi bulk motion nahi matlab recover karne ke liye koi kinetic energy nahi, isliye wall bas edge temperature par baithti hai.