Aerodynamic heating during reentry — stagnation point heat flux Chapman equation: Worked examples

3.4.21 · D3 · Physics › Rocket Flight Mechanics › Aerodynamic heating during reentry — stagnation point heat f

Neeche use kiya gaya har symbol parent note mein already define hai. Ek line ka reminder taaki tumhe scroll na karna pade:

Scenario matrix

Is formula se jo bhi question aa sakta hai, woh in case classes mein se ek hoga. Aakhri column mein likha hai kaunsa worked example us cell ko cover karta hai.

#	Case class	Kya test ho raha hai	Covered by
A	Baseline full calculation	Teeno knobs plug karo, number nikalo	Example 1
B	Ek knob upar/neeche karo (ratios)	$V^{3}$ sensitivity — "speed double karo"	Example 2
C	Nose radius scale karo (ratios)	$R_{n}^{- 1/2}$ sensitivity — bluntness	Example 3
D	Zero / degenerate input	$V = 0$ , $ρ = 0$ , $R_{n} \to \infty$ , $R_{n} \to 0$ limits	Example 4
E	Trajectory ke saath limiting behaviour	Girta $V$ × badhta $ρ$ → ek peak (derived)	Example 5
F	Inverse problem	$\overset{q}{˙}$ diya hai, ek knob solve karo	Example 6
G	Real-world word problem	Mars entry, unit juggling	Example 7
H	Exam-style twist	Convective vs radiative crossover	Example 8

Hum inhe order mein karte hain. Cells D aur E mein figures hain kyunki unka behaviour geometric hai (curves, limits).

Example 1 — Cell A: baseline plug-in

Step 1 — Speed ko cube karo. $V^{3} = (7.8 \times 1 0^{3})^{3} = 4.746 \times 1 0^{11} .$ Yeh step kyun? $V$ formula mein $V^{3}$ ke roop mein aata hai, isliye poore expression ko cube karna hai — pehle cube karo, powers of ten tidy rakho.

Step 2 — Dono square roots lo. $R_{n} = 0.3 = 0.5477, ρ_{\infty} = 1 0^{- 3} = 0.03162.$ Yeh step kyun? Dono ek root ke neeche hain; inhe alag karna arithmetic ko honest rakhta hai (parent ki salah).

Step 3 — Assemble karo. $\overset{q}{˙}_{s} = \frac{1.83 \times 1 0 ^{- 4}}{0.5477} \times 0.03162 \times 4.746 \times 1 0^{11} .$ $= 3.341 \times 1 0^{- 4} \times 0.03162 \times 4.746 \times 1 0^{11} \approx 5.01 \times 1 0^{6} W/m^{2} .$ Yeh step kyun? Left-to-right multiply karo; $1 0^{11}$ magnitude carry karta hai.

Step 4 — Convert karo. $5.01 \times 1 0^{6} W/m^{2} \div 1 0^{4} = 501 W/cm^{2} .$

Example 2 — Cell B: speed double karo

Step 1 — Ratio use karo, nayi calculation mat karo. $\frac{q ˙ _{2}}{q ˙ _{1}} = (\frac{V _{2}}{V _{1}})^{3} = 2^{3} = 8.$ Yeh step kyun? Baaki sab unchanged hai, isliye cancel ho jaata hai — $R_{n}$ ya $ρ$ dobara plug karne ki zaroorat nahin. Ratios fast, error-proof tarika hai.

Step 2 — Apply karo. $501 W/cm^{2} \times 8 = 4008 W/cm^{2} .$

Example 3 — Cell C: nose radius 4 guna karo

Step 1 — Phir se ratio. $\frac{q ˙ _{2}}{q ˙ _{1}} = (\frac{R _{n, 1}}{R _{n, 2}})^{1/2} = (\frac{0.3}{1.2})^{1/2} = (\frac{1}{4})^{1/2} = \frac{1}{2} .$ Yeh step kyun? $R_{n}$ denominator mein root ke neeche hai, isliye factor-4 growth factor-2 shrink banti hai — exponent ka sign direction batata hai.

Step 2 — Apply karo. $501 \times \frac{1}{2} = 250.5 W/cm^{2} .$

Example 4 — Cell D: zero aur degenerate inputs

Figure kaise padhein. Horizontal axis speed $V$ hai (arbitrary units); vertical axis stagnation flux $\overset{q}{˙}_{s}$ hai same arbitrary units mein, $R_{n}$ aur $ρ_{\infty}$ fixed hain. Red curve cube $\overset{q}{˙}_{s} \propto V^{3}$ hai. Ek kala dot origin mark karta hai, $V = 0$ wala limit jo hum Step 1 mein test karte hain — notice karo ki curve wahan se bilkul flat (slope $3 V^{2}$ wahan zero hai) shuru hoti hai aur phir steeply upar jaati hai. Yeh figure Steps 1 aur "cube ki tarah badhta hai" fact ki picture hai; $ρ$ , $R_{n}$ limits algebra hain jo hum neeche karte hain.

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

Step 1 — $V = 0$ (vehicle at rest). $\overset{q}{˙}_{s} = \frac{C}{R _{n}} ρ_{\infty} \cdot 0^{3} = 0.$ Yeh step kyun? Koi motion nahin → koi ordered kinetic energy $\frac{1}{2} V^{2}$ nahi dump hogi → koi convective heating nahin. Figure par yeh red curve ka kala dot ke paas origin mein neeche aa jana hai. ✔

Step 2 — $ρ_{\infty} = 0$ (vacuum / atmosphere ke upar). $\overset{q}{˙}_{s} = \frac{C}{R _{n}} \cdot 0 \cdot V^{3} = 0.$ Yeh step kyun? Koi gas nahin matlab shock karne ke liye koi molecules nahin aur heat wall tak le jaane ke liye koi medium nahin — convective heating ko literally ek fluid chahiye. Yahi reason hai ki orbit mein satellite nahi jalta despite $V \approx 7.8 km/s$ : wahan $ρ_{\infty}$ essentially $0$ hai. ✔

Step 3 — $R_{n} \to \infty$ (perfectly flat nose). $\overset{q}{˙}_{s}$ ko sirf $R_{n}$ ka function maan lo ( $V, ρ$ fix rakho) aur formal limit lo: $lim_{R_{n} \to \infty} \frac{C ρ _{\infty} V ^{3}}{R _{n}} = C ρ_{\infty} V^{3} \cdot lim_{R_{n} \to \infty} \frac{1}{R _{n}} = C ρ_{\infty} V^{3} \cdot 0 = 0.$ Yeh step kyun? $1/ R_{n} \to 0$ . Physically: infinite radius matlab velocity gradient $β \propto 1/ R_{n} \to 0$ , boundary layer bina kisi bound ke barta hai, aur heat iske aar-paar diffuse nahi ho sakti. Blunter hamesha cooler — extreme par, flat sabse thanda. ✔

Step 4 — $R_{n} \to 0$ (perfectly sharp nose). Same one-variable function, doosra end: $lim_{R_{n} \to 0^{+}} \frac{C ρ _{\infty} V ^{3}}{R _{n}} = C ρ_{\infty} V^{3} \cdot lim_{R_{n} \to 0^{+}} \frac{1}{R _{n}} = C ρ_{\infty} V^{3} \cdot (+ \infty) = + \infty.$ Yeh step kyun? Jaise denominator $R_{n}$ zero ki taraf shrink hota hai, fraction unbounded badta jaata hai. Formula predict karta hai ki idealized sharp nose turant pighal jaayega — "reentry nose kabhi sharp mat banao" ka maths. ✔

Example 5 — Cell E: ek real descent ke saath peak (derived)

Calculus touch karne se pehle, hum har symbol aur uski units fix karein, kyunki is example mein chaar alag quantities hain jo sab milti-julti lagti hain. Poora bookkeeping yahan hai.

Definition Is example ke paanch symbols (units ke saath)

$h$ — zameen se upar altitude, metres mein ( $m$ ). Yeh "giraawat mein kahan" wala sawaal hai; bada $h$ = upar, $h = 0$ = zameen.
$ρ$ — us altitude par local air density, $kg/m^{3}$ mein. Upar chhoti, neeche badi.
$H$ — atmospheric scale height, metres mein ( $m$ ): vertical distance jitne mein density ek factor $e \approx 2.718$ se gir jaati hai. Yeh atmosphere ki ek fixed property hai (Earth ke liye $H \approx 7000 m$ ). Yeh $h$ aur $ρ$ ko model $ρ (h) = ρ_{0} e^{- h / H}$ ke zariye link karta hai.
$γ$ — flight-path angle: trajectory kitni steeply horizontal se neeche dive karti hai, ek angle (dimensionless, radians mein). Constant assume kiya gaya hai ek straight ballistic plunge ke liye.
$β$ — ballistic coefficient, $kg/m^{2}$ mein (mass per frontal area per drag coefficient). Bada $β$ = bhaari, streamlined, brake karna mushkil. Dekho Ballistic coefficient and deceleration.

Note karo koi alag " $H_{eff}$ " nahin hai — woh phantom gaya; sirf ek scale height $H$ hai.

Step 0 — Speed–density relation (Allen–Eggers) derive karo, aur uska domain. Ek steep ballistic entry ke liye hum assume karte hain: (i) gravity drag ke muqable mein negligible hai, (ii) flight-path angle $γ$ constant rehta hai (ek seedha-line plunge), (iii) lift zero hai. Tab path ke saath Newton ka doosra law sirf drag hai, $m \frac{d V}{d t} = - \frac{1}{2} ρ V^{2} C_{D} A, β \equiv \frac{m}{C _{D} A} (units kg/m^{2}) .$ $β$ kyun introduce karein? Yeh vehicle ka mass $m$ , drag coefficient $C_{D}$ aur frontal area $A$ ek number mein bundle karta hai, taaki equation vehicle par sirf $β$ ke zariye depend kare.

Independent variable $t$ ko altitude $h$ mein change karo seedhi dive ki geometry use karke, $d h = - V sin γ d t$ (girna, isliye $t$ badhne ke saath $h$ ghatta hai). Substitute karne par $t$ hat jaata hai: $\frac{d V}{V} = \frac{ρ d h}{2 β s i n γ} .$ Ab atmosphere model $ρ (h) = ρ_{0} e^{- h / H}$ use karo, jiska differential $d ρ = - \frac{ρ}{H} d h$ hai, yaani $ρ d h = - H d ρ$ . Yahi woh jagah hai jahaan scale height $H$ enter karta hai — yeh ek altitude step $d h$ aur ek density step $d ρ$ ke beech conversion factor hai. Substitute karne par: $\frac{d V}{V} = - \frac{H d ρ}{2 β s i n γ} .$ Entry se (jahaan $ρ \approx 0$ upar, $V = V_{e}$ ) neeche general $ρ$ tak integrate karo: $ln \frac{V}{V _{e}} = - \frac{H ρ}{2 β s i n γ} \Rightarrow V (ρ) = V_{e} exp (- \frac{H ρ}{2 β sin γ}) .$ Validity ka domain: steep, high-speed, drag-dominated, constant- $γ$ descent — classic Allen–Eggers regime. Yeh shallow lifting entries (Shuttle-type) ke liye break down karta hai jahaan $γ$ vary karta hai aur gravity matter karti hai. Note karo $H$ ab exponent ke andar hai, isliye woh answer mein carry through karega.

Step 1 — Flux ko single variable $ρ$ ka function likho. $V (ρ)$ ko $\overset{q}{˙} = A_{0} ρ^{1/2} V^{3}$ mein daalo (likhte hain $A_{0} = C / R_{n}$ , yahan constant hai): $\overset{q}{˙} (ρ) = A_{0} V_{e}^{3} ρ^{1/2} exp (- \frac{3 H ρ}{2 β s i n γ}) .$ Yeh step kyun? Maximise karne ke liye hume $\overset{q}{˙}$ ko ek variable ka function chahiye; $ρ$ woh variable hai aur $V$ uske saath aata hai. Shorthand define karo $k \equiv \frac{3 H}{2 β sin γ}$ (units $m^{3} / kg$ , taaki $k ρ$ dimensionless ho — ek achha units check).

Step 2 — Differentiate karo aur zero set karo. $\overset{q}{˙} = A_{0} V_{e}^{3} ρ^{1/2} e^{- k ρ}$ ke saath, $\frac{d q ˙}{d ρ} = A_{0} V_{e}^{3} ρ^{- 1/2} e^{- k ρ} (\frac{1}{2} - k ρ) .$ Yeh step kyun? $ρ^{1/2}$ aur $e^{- k ρ}$ par product rule. Exponential aur $ρ^{- 1/2}$ $ρ > 0$ ke liye kabhi zero nahi hote, isliye peak wahan hai jahan bracket zero ho.

Step 3 — $ρ_{peak}$ solve karo, $H$ ko explicit rakho. $\frac{1}{2} - k ρ_{peak} = 0 \Rightarrow ρ_{peak} = \frac{1}{2 k} = \frac{1}{2} \cdot \frac{2 β s i n γ}{3 H} = \frac{β sin γ}{3 H} .$ Yeh step kyun? Bracket $ρ$ mein linear hai, isliye ek hi root hai — ek interior peak, bilkul forecast jaisa. Scale height $H$ denominator mein aata hai kyunki yeh $k$ ke through aaya tha; koi alag " $H_{eff}$ " nahin hai — hamesha se yahi ek $H$ tha. 3 exponent $3$ ka ghost hai $V$ par. Units check: $[β] / [H] = (kg/m^{2}) / m = kg/m^{3}$ ✔ — ek density, jaisa hona chahiye.

Step 4 — Peak density ko altitude $h_{peak}$ mein convert karo (actual sawaal). Sawaal mein giraawat mein kahan, yaani ek altitude, density nahin maanga gaya tha. Atmosphere model $ρ = ρ_{0} e^{- h / H}$ invert karo: $e^{- h / H} = \frac{ρ}{ρ _{0}} \Rightarrow h = H ln \frac{ρ _{0}}{ρ} .$ $ρ = ρ_{peak} = β sin γ / (3 H)$ plug karo: $h_{peak} = H ln (\frac{3 H ρ _{0}}{β sin γ}) .$ Yeh step kyun? Yahi woh piece hai jo pehle ka draft skip kar gaya tha. Yeh abstract $ρ_{peak}$ ko ek real altitude mein badalta hai jise engineer trajectory par point kar sake. Kyunki $β sin γ / (3 H)$ ek chhoti density hai, logarithm bada aur positive hai, isliye $h_{peak}$ upar hai — forecast confirm hota hai.

Step 5 — Sanity number. Earth-ish values lo $H = 7000 m$ , $ρ_{0} = 1.2 kg/m^{3}$ , $β = 100 kg/m^{2}$ , $γ = 3 0^{\circ}$ to $sin γ = 0.5$ . Tab $ρ_{peak} = \frac{100 \times 0.5}{3 \times 7000} = 2.38 \times 1 0^{- 3} kg/m^{3}$ aur $h_{peak} = 7000 ln (\frac{1.2}{2.38 \times 1 0 ^{- 3}}) = 7000 \times ln (504) = 7000 \times 6.22 \approx 4.35 \times 1 0^{4} m \approx 44 km .$ Yeh step kyun? Ek concrete altitude (~44 km, stratosphere mein kaafi upar) "intermediate, upar" ko tangible banati hai — woh ~10 km se kaafi upar jahan ballistic capsule peak g-load feel karta hai.

Figure kaise padhein. Horizontal axis: descent depth (0 = atmosphere ka top, 1 = gehraai mein). Vertical axis: har quantity apne maximum se normalize ki gayi taaki teeno fit ho sakein. Black dashed = badhta factor $ρ^{1/2}$ ; black dotted = girta factor $V^{3}$ ; red curve unka product $\overset{q}{˙}_{s}$ hai, aur red dot woh interior maximum mark karta hai jo humne abhi algebraically solve kiya.

Example 6 — Cell F: inverse problem (ek knob solve karo)

Step 1 — Formula ko $R_{n}$ ke liye rearrange karo. $\overset{q}{˙}_{s} = \frac{C ρ _{\infty} V ^{3}}{R _{n}}$ se shuru karke, root isolate karo: $R_{n} = \frac{C ρ _{\infty} V ^{3}}{q ˙ _{s}} \Rightarrow R_{n} = (\frac{C ρ _{\infty} V ^{3}}{q ˙ _{s}})^{2} .$ Yeh step kyun? Unknown denominator mein root ke neeche hai; algebraically solve karo taaki ek hi baar plug kar sakein.

Step 2 — Numerator group compute karo. $C ρ_{\infty} V^{3} = 1.83 \times 1 0^{- 4} \times 0.03162 \times 4.746 \times 1 0^{11} = 2.746 \times 1 0^{6} .$ Yeh step kyun? Yeh "flux jo nose feel karta agar $R_{n} = 1$ hota" — ek handy intermediate.

Step 3 — Divide aur square karo. $R_{n} = \frac{2.746 \times 1 0 ^{6}}{3.0 \times 1 0 ^{6}} = 0.9153, R_{n} = 0.915 3^{2} \approx 0.838 m .$

Example 7 — Cell G: real-world word problem (Mars)

Step 1 — Speed ko SI mein convert karo. $V = 6.0 \times 1 0^{3} m/s$ , to $V^{3} = 2.16 \times 1 0^{11}$ . Yeh step kyun? Formula pure SI hai; ek "km" bach gaya to answer $1 0^{9}$ se inflate ho jaayega. Yeh parent ki top warning hai.

Step 2 — Roots. $R_{n} = 1.5 = 1.2247$ , $ρ_{\infty} = 2 \times 1 0^{- 4} = 0.01414$ .

Step 3 — Assemble karo. $\overset{q}{˙}_{s} = \frac{1.83 \times 1 0 ^{- 4}}{1.2247} \times 0.01414 \times 2.16 \times 1 0^{11} = 4.56 \times 1 0^{5} W/m^{2} = 45.6 W/cm^{2} .$

Example 8 — Cell H: exam twist (convective vs radiative crossover)

Worked example Radiative heating convective ko kab overtake karta hai?

Convective heating $\overset{q}{˙}_{c} = C R_{n}^{- 1/2} ρ^{1/2} V^{3}$ ke roop mein scale karti hai ( $R_{n}^{- 1/2}$ ke saath badhti hai). Radiative heating (shocked gas ki glow — parent ke footnote aur Radiative heating at hypersonic speeds dekho) ek empirical power law follow karta hai $\overset{q}{˙}_{r} = K R_{n} ρ^{a} V^{b}$ , jahan:

$K$ ek empirical constant hai (gas-radiation data se fitted, units choose ki gayi hain taaki $\overset{q}{˙}_{r}$ $W/m^{2}$ mein aaye),
$a \approx 1.2$ density exponent hai, aur
$b \approx 8$ (ho sakta hai $8$ – $15$ ) kaafi steep velocity exponent hai jo radiation ko sirf high speed par matter karata hai.

Ek feature jo hume chahiye woh hai $R_{n}$ exponent ka sign: convective mein $R_{n}^{- 1/2}$ hai (radius ke saath girta hai), radiative mein $R_{n}^{+ 1}$ hai (radius ke saath badhta hai). Fixed $ρ, V$ ke liye, kis nose radius $R_{*}$ par dono equal hain, aur bahut blunt noses ke liye kaunsa dominate karta hai?

Forecast: Convective $R_{n}$ ke saath girta hai, radiative $R_{n}$ ke saath badhta hai — opposite slopes, to ek $R_{*}$ par zaroor cross karenge. Bahut blunt noses (bada $R_{n}$ ) ke liye radiative jeetega. Guess: ek crossover radius exist karta hai; uske baar radiative dominate karta hai.

Step 1 — Unhe equal set karo (symbolically). Fixed $ρ, V$ ke teeno pieces ko constants mein group karo $a_{c} = C ρ^{1/2} V^{3}$ aur $a_{r} = K ρ^{a} V^{b}$ (to $a, b, K$ single number $a_{r}$ ke andar dab jaate hain). Tab $a_{c} R_{*}^{- 1/2} = a_{r} R_{*}^{+ 1} \Rightarrow \frac{a _{c}}{a _{r}} = R_{*}^{3/2} \Rightarrow R_{*} = (\frac{a _{c}}{a _{r}})^{2/3} .$ Yeh step kyun? $R_{n}$ powers collect karna ( $- 1/2$ minus $+ 1$ exponent $3/2$ deta hai) do curves ko ek clean crossover equation mein badal deta hai. Messy exponents $a, b$ dobara surface nahin karte — sirf $a_{c} / a_{r}$ matter karta hai.

Step 2 — Numeric example. Maano kisi altitude par grouped constants $a_{c} = 2.0 \times 1 0^{6}$ aur $a_{r} = 1.0 \times 1 0^{5}$ (SI, to $\overset{q}{˙}$ $W/m^{2}$ mein aur $R_{n}$ metres mein). Tab $R_{*} = (\frac{2.0 \times 1 0 ^{6}}{1.0 \times 1 0 ^{5}})^{2/3} = 2 0^{2/3} \approx 7.37 m .$ Yeh step kyun? Ek concrete crossover radius trade-off ko tangible banata hai.

Step 3 — Interpret karo. $R_{n} < 7.37 m$ ke liye convective dominate karta hai (ise blunter banana help karta hai). $R_{n} > 7.37 m$ ke liye radiative dominate karta hai (ise blunter banana ab hurt karta hai). Ek optimum hai, "blunter hamesha better" nahin.

Recall

Recall Har cell par khud test karo

Reentry speed double karne par $\overset{q}{˙}_{s}$ kis factor se badlega? ::: $2^{3} = 8$ . Nose radius chaar guna karne par $\overset{q}{˙}_{s}$ ka kya hoga? ::: $4^{- 1/2} = \frac{1}{2}$ se multiply hoga (aadha ho jaayega). Ek orbiting satellite ( $V \approx 7.8$ km/s) kyun nahin jalta? ::: Wahan $ρ_{\infty} \approx 0$ hai, isliye $\overset{q}{˙}_{s} \propto ρ_{\infty} \to 0$ — shock karne ke liye koi gas nahin, heat carry karne ke liye koi medium nahin. Formula perfectly sharp nose ( $R_{n} \to 0$ ) ke liye kya predict karta hai? ::: $\overset{q}{˙}_{s} \to + \infty$ — diverge karta hai, isliye sharp reentry noses forbidden hain. $ρ^{1/2} V^{3}$ ka peak density ke roop mein kahan hai, aur kaise nikalta hai? ::: $ρ^{1/2} e^{- k ρ}$ ko $ρ$ ke respect mein differentiate karo, zero set karo, $ρ_{peak} = 1/ (2 k) = β sin γ /3 H$ milega. Us peak density ko altitude mein kaise convert karte hain? ::: $ρ = ρ_{0} e^{- h / H}$ invert karo to $h_{peak} = H ln (3 H ρ_{0} / β sin γ)$ milega — ek ucha altitude. Peak heating peak deceleration se upar kyun hoti hai? ::: $V^{3}$ (girta hua) $ρ^{1/2}$ (badhte hue) se tez collapse karta hai, isliye product jaldi peak karta hai, upar. Bahut blunt noses ke liye kaun si heating eventually dominate karti hai? ::: Radiative — yeh $R_{n}^{+ 1}$ ke saath badhti hai jabki convective $R_{n}^{- 1/2}$ ke saath girta hai; dono $R_{*} = (a_{c} / a_{r})^{2/3}$ par cross karte hain.