Aerodynamic heating — recovery temperature, heat flux: Worked examples

3.1.28 · D3 · Physics › Compressible Flow & Aerodynamics › Aerodynamic heating — recovery temperature, heat flux

Yeh page parent topic ke liye drill floor hai. Parent ne teen tools banaye the:

stagnation ratio $\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2}$ (ek ruka hua gas kitna garam hota hai),
recovery temperature $T_{r} = T_{e} (1 + r \frac{γ - 1}{2} M_{e}^{2})$ (ek uncooled wall actually kitni garam hoti hai — $T_{0}$ se thodi kam, kyunki recovery factor $r < 1$ ),
heat flux $\overset{q}{˙}_{w} = h (T_{r} - T_{w})$ ( $T_{r}$ se thandi wall mein heat kitni tezi se jaati hai).

Yahan hum har case class ko un teen formulas par throw karte hain — flux ki har sign, zero/degenerate inputs, low- aur high-Mach limits, ek real word problem, aur ek exam twist — taaki koi bhi scenario aisa na ho jo tumne pehle work out na dekha ho.

Neeche use hone wale symbols, plain words mein:

Definition Characters ka cast

$M_{e}$ = Mach number at the boundary-layer edge = flow speed ÷ local speed of sound. Dimensionless.
$T_{e}$ = edge static temperature (woh thandi free-stream air jis mein vehicle fly kar raha hai), kelvin (K) mein.
$T_{0}$ = stagnation temperature = temperature agar tum flow ko bina heat loss ke rok do.
$T_{r}$ = recovery temperature = woh temperature jis par ek insulated (adiabatic) wall settle ho jaati hai.
$T_{w}$ = wall temperature = actual surface temperature (tum ise cooling se control kar sakte ho).
$r$ = recovery factor, 0 aur 1 ke beech — stopping-heat ka woh fraction jo wall rakhti hai.
$γ$ = ratio of specific heats (air ke liye $γ = 1.4$ ).
$h$ = heat-transfer coefficient $[W m^{- 2} K^{- 1}]$ — flow kitni achhi tarah heat wall mein push karta hai.
$\overset{q}{˙}_{w}$ = wall heat flux $[W m^{- 2}]$ = heat energy jo wall ke ek square metre se per second cross karti hai.

Scenario matrix

Is topic mein jo bhi poochha ja sakta hai, woh in cells mein se kisi ek mein aata hai. Neeche har example apni cell ke saath tagged hai.

#	Cell (case class)	Kya special hai	Example
A	Flux ka sign `> 0` ( $T_{w} < T_{r}$ )	wall heat absorb karti hai — cooling zaroori	Ex 1
B	Flux ka sign `= 0` ( $T_{w} = T_{r}$ )	adiabatic wall — flux zero ho jaata hai	Ex 2
C	Flux ka sign `< 0` ( $T_{w} > T_{r}$ )	hot wall heat gas ko wapas dump karti hai	Ex 3
D	Zero / degenerate input ( $M_{e} \to 0$ )	koi speed nahi → koi heating nahi	Ex 4
E	Low-Mach limit (chhota $M_{e}$ )	$T_{r} - T_{e}$ $M_{e}^{2}$ ke saath scale karta hai	Ex 4
F	High-Mach limit (hypersonic)	bahut bada $T_{r}$ , laminar vs turbulent split	Ex 5
G	Real-world word problem	$M_{e}, T_{e}, T_{w}$ khud pick karo	Ex 6
H	Exam twist (flux se $h$ ya $T_{w}$ dhundho)	flux law ko invert karo	Ex 7
I	Stanton-number route	$h = ρ_{e} u_{e} c_{p} S t$	Ex 8

Figure — Aerodynamic heating — recovery temperature, heat flux

Worked examples

Example 1 — Cell A: wall $T_{r}$ se thandi, heat andar aati hai

Worked example Mach 3 jet par ek cooled fin

$M_{e} = 3$ , $T_{e} = 223$ K, $γ = 1.4$ , turbulent toh $r = P r^{1/3} = 0.7 1^{1/3} = 0.892$ . Wall ko $T_{w} = 350$ K par rakha gaya, $h = 900 W m^{- 2} K^{- 1}$ . $T_{r}$ aur $\overset{q}{˙}_{w}$ nikalo.

Forecast: andaza lagao — kya $T_{r}$ 224 K ke kareeb hoga ya 600 K ke? Aur kya $\overset{q}{˙}_{w}$ positive hoga ya negative?

Step 1 — stopping-heat fraction. $\frac{γ - 1}{2} M_{e}^{2} = 0.2 \times 3^{2} = 0.2 \times 9 = 1.8$ . Yeh step kyun? Yeh dimensionless number batata hai ki agar hum flow ko bina losses ke rok dein toh static temperature kitne guna badhegi. Yeh har heating formula ka engine hai.

Step 2 — recovery factor apply karo. $T_{r} = T_{e} (1 + r \cdot 1.8) = 223 (1 + 0.892 \times 1.8) = 223 (1 + 1.606) = 223 \times 2.606 = 581 K$ . Yeh step kyun? Wall us stopping-heat ka sirf fraction $r$ rakhti hai, isliye $1.8$ ko $r$ se multiply karte hain base $1$ wapas add karne se pehle.

Step 3 — flux. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 900 (581 - 350) = 900 \times 231 = 2.08 \times 1 0^{5} W/ m^{2} \approx 208 kW/ m^{2}$ . Yeh step kyun? Wall (350 K) post $T_{r}$ (581 K) se thandi hai, toh heat andar aati hai — positive flux, cell A.

Verify: $T_{r} = 581$ K, $T_{0} = 223 (1 + 1.8) = 624$ K se thoda neeche baith ta hai, bilkul $r < 1$ ke liye expected jaisa. Units: $[W m^{- 2} K^{- 1}] \times [K] = [W m^{- 2}]$ ✔. Flux positive ⇒ wall mein jaata hai ⇒ cooling justified. ✔

Example 2 — Cell B: adiabatic wall, zero flux

Worked example Fin ko float karne do (koi cooling nahi)

Same flow jaise Ex 1 ( $T_{r} = 581$ K). Ab cooling band karo aur wall ko steady state tak pahunchne do. $T_{w}$ aur $\overset{q}{˙}_{w}$ kya hoga?

Forecast: agar kuch bhi heat draw nahi karta, toh wall kahan settle hogi?

Step 1 — steady state impose karo, koi heat bahar nahi. "Adiabatic wall" matlab surface se zero net heat cross karti hai: $\overset{q}{˙}_{w} = 0$ . Yeh step kyun? Equilibrium mein ek insulated wall na heat gain karti hai, na lose — yahi recovery temperature ki definition hai.

Step 2 — $h (T_{r} - T_{w}) = 0$ solve karo. $h \neq = 0$ ke saath, yeh force karta hai $T_{w} = T_{r} = 581$ K. Yeh step kyun? Driving difference $T_{r} - T_{w}$ ka zero hone ka ek hi tarika hai — wall bilkul post par baith jaaye.

Verify: wapas plug karo: $\overset{q}{˙}_{w} = 900 (581 - 581) = 0$ ✔. Isliye $T_{r}$ ko recovery / adiabatic-wall temperature kaha jaata hai — yeh woh unique wall temperature hai jo zero flux deti hai.

Example 3 — Cell C: wall $T_{r}$ se garam, heat bahar jaati hai

Worked example Ground par ek soak-heated wall, engine idling

Landing ke baad, ek surface $T_{w} = 650$ K par hai (flight se heat-soaked) jabki local flow almost still lekin warm hai: $T_{r} = 581$ K lo, $h = 900 W m^{- 2} K^{- 1}$ . $\overset{q}{˙}_{w}$ aur uski direction nikalo.

Forecast: wall ab recovery post se garam hai — flux ka sign kya hoga?

Step 1 — same law, naye numbers. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 900 (581 - 650) = 900 \times (- 69)$ . Yeh step kyun? Newton's cooling law signed hai: $T_{r} - T_{w}$ ab negative hai kyunki $T_{w} > T_{r}$ .

Step 2 — sign padho. $\overset{q}{˙}_{w} = - 6.21 \times 1 0^{4} W/ m^{2} = - 62.1 kW/ m^{2}$ . Yeh step kyun? Negative flux matlab heat wall se gas mein jaati hai — wall khud ko $T_{r}$ ki taraf neeche cool kar rahi hai.

Verify: magnitude sanity: same $h$ , temperature gap 69 K vs Ex 1 mein 231 K, toh flux $231/69 \approx 3.35 \times$ chhota hona chahiye: $208/3.35 \approx 62$ kW/m² ✔. Sign negative ⇒ wall heat lose karti hai ⇒ woh $T_{r} = 581$ K post ki taraf wapas neeche drift karti hai. ✔

Example 4 — Cells D & E: zero speed aur low-Mach limit

Worked example Khada rehna, phir creep karna

Air at $T_{e} = 288$ K, $γ = 1.4$ , laminar $r = 0.71 = 0.843$ . (i) $M_{e} = 0$ . (ii) $M_{e} = 0.3$ . Har baar $T_{r}$ aur rise $T_{r} - T_{e}$ nikalo.

Forecast: zero speed par kya wall ko koi extra heating feel honi chahiye?

Step 1 — degenerate case $M_{e} = 0$ . $\frac{γ - 1}{2} M_{e}^{2} = 0.2 \times 0 = 0$ , toh $T_{r} = 288 (1 + 0) = 288$ K. Yeh step kyun? Koi motion nahi ⇒ dissipate karne ke liye koi kinetic energy nahi ⇒ wall sirf air temperature par baith ti hai. Formula (sahi tarah se) $T_{r} = T_{e}$ tak collapse ho jaata hai. Yeh pure topic ke liye sanity anchor hai.

Step 2 — low speed $M_{e} = 0.3$ . $\frac{γ - 1}{2} M_{e}^{2} = 0.2 \times 0.09 = 0.018$ ; $T_{r} = 288 (1 + 0.843 \times 0.018) = 288 (1.01517) = 292.4$ K. Rise $\approx 4.4$ K. Yeh step kyun? Quadratic growth dikhata hai: heating rise $M_{e}^{2}$ ke saath scale karti hai. $M = 0.3$ par rise bahut tiny hai — isliye low-speed aircraft aerodynamic heating ko ignore karte hain.

Verify: do Mach numbers ke beech rise ratio $(0.3/0)$ … zero par undefined hai, toh instead $M_{e} = 0.3$ vs $M_{e} = 0.6$ compare karo: rise $M^{2}$ ke saath scale karti hai, yaani $\times 4$ . Rise at 0.3 is 4.4 K ⇒ predict $\approx 17.5$ K at 0.6. Direct: $0.2 \times 0.36 \times 0.843 \times 288 = 17.5$ K ✔. Limit $M_{e} \to 0 \Rightarrow T_{r} \to T_{e}$ ✔.

Example 5 — Cell F: hypersonic limit, laminar vs turbulent

Worked example Mach 6 par re-entry heating

$M_{e} = 6$ , $T_{e} = 250$ K, $γ = 1.4$ . Laminar ( $r = 0.71 = 0.843$ ) ko turbulent ( $r = 0.7 1^{1/3} = 0.892$ ) se compare karo. Dekho Hypersonic re-entry & thermal protection systems.

Forecast: konsa flow hotter wall deta hai — laminar ya turbulent — aur kitne se?

Step 1 — stopping-heat fraction. $\frac{γ - 1}{2} M_{e}^{2} = 0.2 \times 36 = 7.2$ . Yeh step kyun? High Mach par yeh " $1 +$ " ko dominate karta hai, toh $T_{r}$ roughly $7 \times T_{e}$ hota hai — thousands-of-kelvin walls ki wajah yahi hai.

Step 2 — laminar. $T_{r} = 250 (1 + 0.843 \times 7.2) = 250 (1 + 6.07) = 250 \times 7.07 = 1768$ K. Step 2b — turbulent. $T_{r} = 250 (1 + 0.892 \times 7.2) = 250 (1 + 6.42) = 250 \times 7.42 = 1856$ K. Yeh step kyun? Turbulent $r$ bada hai, toh us bade stopping-heat ka zyada portion recover hota hai — turbulent wall hotter chalti hai, yahan ~88 K se.

Step 3 — $T_{0}$ se compare karo. $T_{0} = 250 (1 + 7.2) = 2050$ K. Dono $T_{r}$ values iske neeche hain, turbulent zyada kareeb. Yeh step kyun? Confirm karta hai $r < 1$ ⇒ $T_{r} < T_{0}$ hamesha, extreme regime mein bhi.

Verify: $T_{r, turb} - T_{r, lam} = 250 \times 7.2 \times (0.892 - 0.843) = 250 \times 7.2 \times 0.049 = 88.2$ K ✔. Dono $T_{0} = 2050$ K se neeche ✔. Aluminium ~933 K par melt hota hai ⇒ dono use melt kar denge ⇒ TPS mandatory. ✔

Example 6 — Cell G: real-world word problem

Worked example SR-71 ka leading edge

"Ek reconnaissance jet wahan cruise karta hai jahan bahar ki hawa $- 56. 5^{\circ} C$ hai Mach 3.2 par. Uska titanium leading edge actively $30 0^{\circ} C$ par rakha jaata hai. Turbulent boundary layer, $r = 0.89$ , $γ = 1.4$ , $h = 1100 W m^{- 2} K^{- 1}$ . Edge ko kitna heat kiya ja raha hai?"

Forecast: pehla kaam — koi formula touch karne se pehle sab kuch kelvin mein convert karo.

Step 1 — extract aur convert karo. $T_{e} = - 56.5 + 273.15 = 216.65 \approx 216.7$ K; $T_{w} = 300 + 273.15 = 573.15 \approx 573.2$ K; $M_{e} = 3.2$ . Yeh step kyun? Stagnation/recovery formulas absolute temperatures ke ratios hain — Celsius nonsense dega. Hamesha kelvin.

Step 2 — stopping-heat fraction. $\frac{γ - 1}{2} M_{e}^{2} = 0.2 \times 3. 2^{2} = 0.2 \times 10.24 = 2.048$ .

Step 3 — recovery temperature. $T_{r} = 216.7 (1 + 0.89 \times 2.048) = 216.7 (1 + 1.8227) = 216.7 \times 2.8227 = 611.7$ K. Yeh step kyun? Yeh woh temperature hai jis par edge uncooled hoti to pahunch jaata — woh post jis se hum flux drive karte hain.

Step 4 — flux. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 1100 (611.7 - 573.2) = 1100 \times 38.5 = 4.24 \times 1 0^{4} W/ m^{2} \approx 42.4 kW/ m^{2}$ . Yeh step kyun? Wall (573.2 K) post (611.7 K) se thoda neeche hai, toh ek modest positive flux — cell A, lekin mild.

Verify: $T_{r} = 611.7$ K, $T_{0} = 216.7 (1 + 2.048) = 660.6$ K se neeche ✔. Gap $T_{r} - T_{w} = 38.5$ K chhota hai ⇒ yeh edge apni natural recovery temperature ke kareeb rakhi ja rahi hai, toh cooling load light hai — SR-71 ke genuinely hot skin chalane se match karta hai. Units ✔.

Example 7 — Cell H: exam twist, flux law ko invert karo

Worked example Required wall temperature ke liye back-solve karo

"Ex 1 ke Mach 3 jet ke liye ( $T_{r} = 581$ K), material sirf $\overset{q}{˙}_{w} = 150 kW/ m^{2}$ ka flux survive kar sakta hai. $h = 900 W m^{- 2} K^{- 1}$ ke saath, wall ko kitna thanda rakhna hoga?"

Forecast: flux ko 208 se 150 kW/m² neeche laane ke liye, kya tum $T_{w}$ zyada karte ho ya kam?

Step 1 — Newton's law rearrange karo. $\overset{q}{˙}_{w} = h (T_{r} - T_{w})$ se, $T_{w}$ ke liye solve karo: $T_{w} = T_{r} - \frac{q ˙ _{w}}{h}$ . Yeh step kyun? Exam flux deta hai aur wall temperature mangta hai — ek formula ko invert karo. $T_{w}$ chhod ke sab kuch known hai.

Step 2 — numbers plug karo. $T_{w} = 581 - \frac{150000}{900} = 581 - 166.7 = 414.3$ K. Yeh step kyun? Allowable temperature gap $150000/900 = 166.7$ K hai; fixed post $T_{r}$ se subtract karo.

Step 3 — direction interpret karo. Flux reduce karne ke liye tumhe ek chhota gap $T_{r} - T_{w}$ chahiye, yaani ek garam wall (yahan 414 K vs Ex 1 mein 350 K). Zyada cooling karne se flux badhega, ghategga nahi. Yeh step kyun? Yeh classic counter-intuitive result hai — parent note mein ek mistake ke roop mein flag kiya gaya hai. Ise yahan numerically verify karo.

Verify: wapas plug karo: $\overset{q}{˙}_{w} = 900 (581 - 414.3) = 900 \times 166.7 = 1.50 \times 1 0^{5} W/ m^{2} = 150$ kW/m² ✔. Aur $414.3 > 350$ K confirm karta hai "garam wall ⇒ kam flux". ✔

Example 8 — Cell I: Stanton-number route

Worked example Skin friction se

$h$ predict karo Reynolds analogy & Stanton number use karte hue: edge par $ρ_{e} = 0.4 kg/ m^{3}$ , $u_{e} = 900 m/s$ , $c_{p} = 1005 J k g^{- 1} K^{- 1}$ , skin-friction coefficient $C_{f} = 0.0020$ . Reynolds analogy deta hai $S t \approx C_{f} /2$ . $h$ nikalo aur phir $\overset{q}{˙}_{w}$ ke liye $T_{r} = 581$ K, $T_{w} = 350$ K.

Forecast: kya yeh $h$ Ex 1 mein use kiye gaye 900 ke same ballpark mein aayega?

Step 1 — friction se Stanton. $S t \approx C_{f} /2 = 0.0020/2 = 0.0010$ . Yeh step kyun? Reynolds analogy heat transfer ko momentum transfer se link karta hai — drag measure karne se tum bina alag thermal experiment ke heating predict kar sakte ho.

Step 2 — $h$ build karo. $h = ρ_{e} u_{e} c_{p} S t = 0.4 \times 900 \times 1005 \times 0.0010 = 361.8 W m^{- 2} K^{- 1}$ . Yeh step kyun? Group $ρ_{e} u_{e} c_{p}$ (units $W m^{- 2} K^{- 1}$ ) maximum possible heat-transfer scale hai; $S t$ us ka woh fraction hai jo actually realise hota hai.

Step 3 — flux. $\overset{q}{˙}_{w} = ρ_{e} u_{e} c_{p} S t (T_{r} - T_{w}) = 361.8 \times (581 - 350) = 361.8 \times 231 = 8.36 \times 1 0^{4} W/ m^{2} \approx 83.6 kW/ m^{2}$ . Yeh step kyun? Confirm karta hai ki dono flux formulas ( $\overset{q}{˙}_{w} = h Δ T$ aur Stanton form) same statement hain, bas $h$ expand karke.

Verify: $ρ u c_{p} S t$ ke units: $kg m^{- 3} \cdot m s^{- 1} \cdot J k g^{- 1} K^{- 1} = J s^{- 1} m^{- 2} K^{- 1} = W m^{- 2} K^{- 1}$ ✔. $h = 361.8$ ek plausible convective coefficient hai. Direct flux check: $h Δ T = 361.8 \times 231 = 83, 576 W/ m^{2}$ ✔.

Recall Matrix ke across quick self-test

Flux sign jab $T_{w} < T_{r}$ ? ::: Positive — heat wall mein jaati hai (cell A). Flux jab $T_{w} = T_{r}$ ? ::: Zero — adiabatic wall (cell B); yahi $T_{r}$ define karta hai. $M_{e} \to 0$ hone par $T_{r}$ ka kya hota hai? ::: $T_{r} \to T_{e}$ ; koi motion nahi, koi heating nahi (cell D). Heat flux kam karne ke liye, wall garm karo ya thanda? ::: Garm karo — chhota gap $T_{r} - T_{w}$ matlab kam flux (cell H). Kaun sa flow hotter recovery temperature deta hai, laminar ya turbulent? ::: Turbulent, kyunki $r = P r^{1/3} > P r$ (cell F).

Connections

Parent: recovery temperature & heat flux — yahan drill ki gayi teen formulas
Stagnation properties & isentropic relations — stopping-heat fraction $\frac{γ - 1}{2} M^{2}$ ka source
Prandtl number & thermal boundary layer — $r = P r$ vs $P r^{1/3}$ set karta hai (Ex 4–5)
Reynolds analogy & Stanton number — Ex 8 ka $S t$ route
Hypersonic re-entry & thermal protection systems — Ex 5 ki Mach 6 wall
Boundary layers & viscous dissipation — kyun $r < 1$
Normal & oblique shock heating — input ke roop mein use hone wala post-shock $T_{e}$ supply karta hai