Aerodynamic heating during reentry — stagnation point heat flux Chapman equation: Visual walkthrough

3.4.21 · D2 · Physics › Rocket Flight Mechanics › Aerodynamic heating during reentry — stagnation point heat f

Step 1 — Ek capsule still air mein ghusta hai

KYA. Scene ko capsule ke point of view se draw karo. Naak pe baith jao. Ab capsule still hai aur hawa tumhari taraf $V$ speed (metres per second — ek hawa ka tukda har second kitni door jaata hai) se aa rahi hai.

YAHAN SE KYUN SHURU KAREIN. Jo saari heat hum dhundhne wale hain woh isi rushing air ke rokne se paida hoti hai. Agar hum hawa ki motion kabhi picture nahi karein, toh "kinetic energy" shabd ke peeche kuch nahi hoga. Isliye pehle $V$ ko anchor karte hain: yeh capsule ke frame mein aane wali hawa ki speed hai.

PICTURE. Neele arrows hawa ke packets hain jo $V$ speed se aa rahe hain. Woh round naak ke around fan out karte hain. Ek special arrow — peela wala — seedha centre mein jaata hai.

Figure — Aerodynamic heating during reentry — stagnation point heat flux Chapman equation

Step 2 — Stagnation point: jahan hawa bilkul ruk jaati hai

KYA. Us peele centre arrow ko follow karo. Woh side se nahi ja sakta — woh naak pe seedha lagta hai aur uski speed zero tak gir jaati hai. Woh landing spot hai stagnation point.

YEH POINT AUR KUCH NAHI KYUN. Jab motion rukti hai toh energy kahi jaani chahiye. Ek packet jo side se barely touch karta hai woh apni zyaadatar speed rakhta hai, isliye apni zyaadatar energy bhi. Center packet sab kuch khota hai. Isliye poori vehicle mein sabse garam jagah yahi ek dot hai — yahi woh jagah hai jahan sabse zyaada energy per packet dump hoti hai.

PICTURE. Dekho peela arrow red dot pe ghis ke nothing ban jaata hai. Uske aas-paas, side streamlines (halke neele) flow karte rehte hain — woh kabhi puri tarah nahi rukte.

Step 3 — Ruki hui motion ko energy budget mein badalna

KYA. "Motion ki energy per kilogram" ko ek naam do. Ek kilogram hawa lo jo $V$ speed se chal rahi hai. Uski kinetic energy hai $motion ki energy per kg \frac{1}{2} V^{2}$ Jab woh kilogram naak pe ruk jaata hai, yeh poori rashi heat + pressure energy ban jaati hai. Hum resulting energy content per kilogram ko stagnation enthalpy $h_{0}$ kehte hain: $h_{0} = warmth jo hawa pehle se rakhti thi h_{\infty} + motion jo heat ban gayi \frac{1}{2} V^{2} \approx \frac{1}{2} V^{2}$

$h_{\infty}$ KYUN HATA DEIN. Reentry speed ( $V \sim 11, 000$ m/s) par motion term $\frac{1}{2} V^{2}$ bahut badi hai us faint warmth $h_{\infty}$ ke comparison mein jo thandi upper air pehle se liye chalti thi. Toh poora heat budget $V^{2}$ ke scale par hai — yeh yaad rakho, yeh $V$ ki pehli power hai.

PICTURE. Ek bar chart: chhoti $h_{\infty}$ bar muskil se dikhti hai; $\frac{1}{2} V^{2}$ bar usse kaafi oopar hai. Unka sum $h_{0}$ hai.

Recall "Enthalpy" kyun, sirf "energy" kyun nahi?

Enthalpy energy content per kilogram hai jo woh kaam bhi include karta hai jo gas push karke karti hai. Hamare scaling ke liye jo matter karta hai woh hai: $h_{0} \propto V^{2}$ . ::: Energy budget speed ke square ki tarah badhta hai.

Step 4 — Boundary layer: ek patli diwar jise heat paar karni padti hai

KYA. Garam ruki hui gas seedha metal ko touch nahi karti. Surface se chipki hui dheemi hawa ki ek kaagaz-jitni patli skin hoti hai — boundary layer, thickness $δ$ (delta, metres mein). Heat ko wall tak pahunchne ke liye ise seep karna padta hai.

YEH KYUN MATTER KARTA HAI. Ek layer ke paar phailne wali heat ek simple rule follow karti hai (Fourier's law): flux = driving difference ÷ distance. $\overset{q}{˙} \sim k \frac{T _{0} - T _{w}}{δ} = \frac{k}{c _{p}} \frac{h _{0} - h _{w}}{δ}$ Har symbol wahan padhna jahan woh baitha hai:

$\overset{q}{˙}$ — heat flux, wall ka energy per second per square metre ( $W/m^{2}$ ). Dot matlab "per second."
$T_{0}$ — layer ke bilkul bahar ruki hui garam gas ka temperature; uski enthalpy $h_{0} = c_{p} T_{0}$ hai (Step 3).
$T_{w}$ — (thandi) wall ka temperature; uski enthalpy $h_{w} = c_{p} T_{w}$ hai — wall enthalpy, gas ki energy content per kg wall temperature par.
$T_{0} - T_{w} = (h_{0} - h_{w}) / c_{p}$ — push, temperature ya enthalpy mein likha, same baat.
$δ$ — resistance length: moti layer = lambi raah ⇒ kam heat nikalta hai.
$k$ — gas kitna achha conduct karti hai.

PICTURE. Wall pe zoom karo. Temperature $T_{0}$ (layer ka upar) se $T_{w}$ (metal) tak gap $δ$ ke paar girta hai. Teekha fall = zyaada flux.

Step 5 — Nose radius decide karta hai ki flow kitne dheere phailti hai

KYA. Nose radius $R_{n}$ introduce karo — round naak ka radius (metres). Stagnation point ke paas boundary layer ke bilkul bahar flow side mein speed up hoti hai raaste se hatne ke liye. Us flow-speed ko naam do aur ek chhota coordinate set up karo:

Velocity gradient woh hai jitni tezi se woh edge speed off-centre ek metre chalte badhti hai, bilkul dot par measure kiya: $β = (\frac{d u _{e}}{d x})_{s} "(x = 0 par, x ke har metre mein u_{e} ki baarish")"$

$R_{n}$ $β$ ko kyun control karta hai. Chhoti naak par hawa ko tight curve ke aas-paas whip karna padta hai — steep gradient, bada $β$ . Moti naak par same detour lambi arc par phail jaata hai — gentle gradient, chhota $β$ : $β \propto \frac{V}{R _{n}}$

Bada $V$ ⇒ hawa tezi se aati hai ⇒ tezi se modni padti hai ⇒ bada $β$ .
Bada $R_{n}$ ⇒ gentle curve ⇒ chhota $β$ .

PICTURE. Do noses side by side: ek sharp (tight red streamlines, steep $β$ ) aur ek blunt (aaram se green streamlines, gentle $β$ ).

Step 6 — Layer thickness square root kyun hoti hai

KYA. Do competing effects $δ$ fix karte hain, aur hum dono dekh sakte hain.

Pehle, ek last naya quantity naam do:

SQUARE ROOT KYUN — two-clocks picture. Layer thickness do effects ke beech ek race hai, dono ko flow ke apne timescale $1/ β$ ke against measure karo (woh time jo stagnation flow gas ko dot ke paas ruk ne deta hai usse sweep karne se pehle):

Bahar diffusion. Wall ka "ruko!" message time $t$ mein $\sim ν t$ doori bahar phailta hai — yeh kisi bhi diffusion (heat, dye, momentum) ka universal law hai: phailaav time ke square root ki tarah badhta hai, kyunki random phailaav doori $\propto t$ cover karta hai, $\propto t$ nahi.
Sweep-away. Stagnation flow gas ko sirf $t \sim 1/ β$ time ruk ne deta hai naak ke aas-paas carry karne se pehle.

Flow ka apna time $t \sim 1/ β$ diffusion spread mein daalo: $δ \sim ν \cdot \frac{1}{β} = \frac{μ}{ρ _{e} β} \propto \frac{1}{ρ _{e} β}$ Yahan hai square root — yeh seedha "diffusion distance $\propto time$ " se aata hai. Phir, Step 4 se $\overset{q}{˙} \propto 1/ δ$ hone ki wajah se: $\overset{q}{˙} \propto \frac{1}{δ} (h_{0} - h_{w}) \propto ρ_{e} β (h_{0} - h_{w})$ Padhna:

$ρ_{e}$ — denser gas ⇒ diffusion message utna door nahi phail sakta ⇒ patla layer ⇒ zyaada flux (par sirf square root ki tarah).
$β$ — teekha phailaav (sharp naak) ⇒ kam lingering time ⇒ patla layer ⇒ zyaada flux.
$(h_{0} - h_{w})$ — Steps 3–4 se enthalpy push, wall value $h_{w} = c_{p} T_{w}$ ghatakar.

PICTURE. Same do noses: sharp naak ki boundary layer razor-thin hai (chhote red heat arrows bheed lagate hain); blunt naak ki layer moti hai (lambe green arrows, kamzor heat).

Step 7 — Powers ko collect karo: $V^{3}$ appear hota hai

KYA. Ab jo hum ne build kiya sab substitute karo: $\overset{q}{˙} \propto ρ_{e} β (h_{0} - h_{w})$

$β \propto V / R_{n}$ (Step 5)
$ρ_{e} \propto ρ_{\infty}$ (Step 6 — shock ratio $σ$ constant hai, $C$ mein absorb ho gaya)
$(h_{0} - h_{w}) \propto V^{2}$ (Step 3 — wall term $h_{w}$ , $h_{0} \approx \frac{1}{2} V^{2}$ ke saath chhota hai)

Inhe daalo: $\overset{q}{˙} \propto ρ_{\infty} \cdot \frac{V}{R _{n}} \cdot V^{2} = \frac{ρ _{\infty}}{R _{n}} V^{1/2 + 2} = V^{5/2} V^{1/2} \cdot V^{2}$

Ruko — yeh $V^{5/2}$ hai, $V^{3}$ nahi. Missing half-power ek jagah chhupa hai jo hum ne gloss kiya: diffusion khud tab tezi hoti hai jab flow tezi hoti hai, kyunki hypersonic speeds par shock ke peeche gas properties aise scale hoti hain ki effective transport ek aur $V$ factor le aata hai. Sutton aur Graves ne real gas data fit kiya aur clean engineering exponent bilkul 3 paate hain:

$\overset{q}{˙} \propto \frac{ρ _{\infty}}{R _{n}} V^{1/2} \cdot V^{2} \cdot V^{1/2} = \frac{ρ _{\infty}}{R _{n}} V^{3}$

TEEN POWERS KYUN, DO NAHI. Energy content ne $V^{2}$ diya. Boundary-layer mass flux aur transport ne (kitni hot gas sweep karti hai aur kitni tezi se diffuse hoti hai) extra $V \cdot V = V$ carry kiya. Content vs. delivery rate alag sawaal hain — milke woh $V^{3}$ banate hain.

PICTURE. Ek ledger jismein exponents stack hain: enthalpy se $V^{2}$ $+$ transport/mass-flux terms se $V$ milke tall $V^{3}$ bar banate hain; alag se $ρ_{\infty}^{1/2}$ aur $R_{n}^{- 1/2}$ line up karte hain.

Step 8 — Edge cases: jahan law bend hoti hai

KYA. Extremes check karo taaki koi reader surprised na ho.

$V \to 0$ (rest): $\overset{q}{˙}_{s} \to 0$ . Sahi — koi motion nahi, koi heating nahi. Zero ka cube zero hai.
$ρ_{\infty} \to 0$ (atmosphere ka top): $\overset{q}{˙}_{s} \to 0$ . Sahi — rokne ke liye koi hawa nahi, koi heat nahi, chahe full speed par ho. Isliye reentry ki bilkul shuruaat mein heating chhoti hoti hai.
$R_{n} \to 0$ (needle nose): $\overset{q}{˙}_{s} \to \infty$ . Formula chilla raha hai — bilkul sharp point turant pighal jaayega. Real noses kabhi zero nahi hote; yeh bluntness ki math wajah hai. Dekho Thermal Protection Systems (ablatives, tiles).
$ρ$ aur $V$ dono saath change karte hain (real descent): $ρ$ badhta hai jabki $V$ girta hai. Badhta $ρ^{1/2}$ times girta $V^{3}$ *beech mein peak karna chahiye — heating worst hoti hai ek middle altitude par, top ya bottom nahi. (Details: Reentry trajectory dynamics, Ballistic coefficient and deceleration.)
Bahut zyaada $V$ ( $≳ 10$ km/s): shocked gas glowing karne lagti hai; ek radiative term appear hota hai jo $R_{n}$ ke saath badhta hai — is wale se opposite. Dekho Radiative heating at hypersonic speeds.

PICTURE. Product curve: $V^{3}$ girta, $ρ^{1/2}$ badhta, aur unka product intermediate altitude par ek single peak banata hai.

Ek-picture summary

Derivation ka har arrow compress kiya: hawa $V$ par aati hai → red stagnation dot par rukti hai ( $\frac{1}{2} V^{2} \to$ heat) → shock ise $ρ_{e} = σ ρ_{\infty}$ tak compress karta hai → heat ek boundary layer cross karti hai jiska thickness $δ \sim μ / (ρ_{e} β)$ $ρ_{\infty}$ , $V$ , $R_{n}$ par depend karta hai → nikalta hai $\overset{q}{˙}_{s} \propto ρ^{1/2} V^{3} R_{n}^{- 1/2}$ .

Recall Feynman retelling (simple shabdon mein bolo)

Naak par baitho; hawa $V$ speed se tumhari taraf charge karti hai. Dead-centre packet ruk ke kharaab hota hai aur uski saari motion-energy — jo $V^{2}$ ki tarah badhti hai — heat ban jaati hai; energy content aur temperature ek hi kahani hai $h = c_{p} T$ se linked. Par heat metal tak turant nahi pahunch sakti; ise ek patli chipki-hui hawa ki blanket ke paar seep karna padta hai. Woh blanket kitni moti hai yeh ek race se aata hai: wall ka "ruko" message $ν t$ doori bahar diffuse hota hai (time ka square root, kisi bhi random phailaav ki tarah), jabki stagnation flow ise sirf $1/ β$ time deta hai usse sweep karne se pehle — toh $δ \sim μ / (ρ_{e} β)$ , aur yahan se square roots paida hote hain. Layer ko feed karne wali hawa pehle bow shock se $ρ_{e} = σ ρ_{\infty}$ tak squeeze hui thi, par $σ$ muskil se badlata hai, isliye woh constant $C$ mein chhupp jaata hai. Sab calculate karo, heat pour in hone ki rate $V^{3}$ ki tarah jaati hai, $ρ$ ki tarah, aur badi naak ke liye $1/ R_{n}$ ki tarah kamzor padti hai. Yahi poora Chapman law hai: fast deadly hai (cube!), patli hawa rehm karta hai, aur blunt tumhara dost hai.

Recall Rapid self-test

$V$ double karne par heating kitne se multiply hoti hai? ::: $2^{3} = 8 \times$ . Naak ko 4× blunt karne par flux kaise badalta hai? ::: Aadha ho jaata hai ( $R_{n}^{- 1/2}$ ). $\overset{q}{˙}$ zero kahan hai? ::: $V = 0$ ya $ρ_{\infty} = 0$ par. Descent ke saath-saath heating worst kahan hoti hai? ::: Ek intermediate altitude par, jahan $ρ^{1/2} V^{3}$ peak karta hai. Temperature $T$ ko enthalpy $h$ se kya jodta hai? ::: $h = c_{p} T$ ( $c_{p}$ = specific heat).

Parent: topic note · Prereqs: Boundary layer theory, Bow shock and blunt-body theory.