Aerodynamic heating — recovery temperature, heat flux: Exercises

3.1.28 · D4 · Physics › Compressible Flow & Aerodynamics › Aerodynamic heating — recovery temperature, heat flux

Level 1 — Recognition

Ye test karta hai ki kya tum sahi formula pick karke plug in kar sakte ho. Abhi algebra mein koi trap nahi hai.

Problem 1.1

Air $M = 2$ Mach par flow kar rahi hai, static temperature $T = 250 K$ hai. Stagnation temperature $T_{0}$ find karo.

Recall Solution 1.1

KYA karte hain: stagnation ratio seedha apply karo — koi boundary layer nahi, isliye recovery factor ki zaroorat nahi. KYUN: $T_{0}$ define hoti hai gas ko adiabatically rest par laane se, aur yahi ratio exactly encode karta hai. $\frac{γ - 1}{2} M^{2} = 0.2 \times 2^{2} = 0.2 \times 4 = 0.8$ $T_{0} = T (1 + 0.8) = 250 \times 1.8 = 450 K$

Problem 1.2

Air ke liye, $P r = 0.71$ hai. Laminar boundary layer aur turbulent boundary layer ke liye recovery factor $r$ likhो (3 significant figures).

Recall Solution 1.2

KYA karte hain: $r$ ki dono definitions apply karo. KYUN dono alag hain: turbulent mixing wall ke paas zyada dissipated heat recover karta hai, isliye $r$ bada hota hai. $r_{lam} = P r = 0.71 = 0.843$ $r_{turb} = P r^{1/3} = 0.7 1^{1/3} = 0.892$ Dono 1 se kam hain, isliye wall kabhi $T_{0}$ tak bilkul nahi pahunchti.

Problem 1.3

Ek wall ki recovery temperature $T_{r} = 600 K$ hai aur cooling se usse $T_{w} = 400 K$ par rakha gaya hai. Heat-transfer coefficient $h = 500 W m^{- 2} K^{- 1}$ hai. Heat flux $\overset{q}{˙}_{w}$ find karo.

Recall Solution 1.3

KYA karte hain: sahi driving temperature ke saath Newton's law of cooling. KYUN $T_{r} - T_{w}$ aur $T_{0} - T_{w}$ nahi: $T_{r}$ woh temperature hai jahan ek uncooled wall settle hoti hai — gas jis asli thermal potential ki taraf push karti hai. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 500 \times (600 - 400) = 500 \times 200 = 1.0 \times 1 0^{5} W/ m^{2} = 100 kW/ m^{2}$ Positive sign → heat wall ke andar ja rahi hai (yeh $T_{r}$ se thandi hai).

Level 2 — Application

Ab tum do formulas ko chain karte ho aur units clean rakhte ho.

Problem 2.1

Ek cruise missile $M_{e} = 3$ par $T_{e} = 223 K$ wali air mein fly kar raha hai. Boundary layer turbulent hai ( $r = 0.892$ ). Recovery temperature $T_{r}$ find karo, aur $T_{0}$ se compare karo.

Recall Solution 2.1

Step 1 — dynamic factor. $\frac{γ - 1}{2} M_{e}^{2} = 0.2 \times 9 = 1.8$ . Step 2 — recovery temperature. $T_{r} = T_{e} (1 + r \times 1.8) = 223 (1 + 0.892 \times 1.8) = 223 (1 + 1.606) = 223 \times 2.606 = 581 K$ Step 3 — $T_{0}$ se compare karo. $T_{0} = 223 (1 + 1.8) = 223 \times 2.8 = 624 K$ Toh $T_{r} = 581 K$ $T_{0} = 624 K$ se thoda neeche baithti hai — wall temperature rise ka lagbhag 89% recover karti hai, exactly jaisa $r = 0.892$ predict karta hai.

Problem 2.2

2.1 wala hi flow ( $T_{r} = 581 K$ ). Skin ko $T_{w} = 350 K$ tak cool kiya gaya hai aur $h = 900 W m^{- 2} K^{- 1}$ hai। $\overset{q}{˙}_{w}$ $kW/ m^{2}$ mein find karo.

Recall Solution 2.2

$\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 900 (581 - 350) = 900 \times 231 = 2.079 \times 1 0^{5} W/ m^{2} \approx 208 kW/ m^{2}$ Yeh enormous flux (ek domestic radiator se 2000× zyada per square metre) hi wajah hai ki fast vehicles ko active cooling ya heat shield chahiye hoti hai.

Problem 2.3

Ek blunt re-entry body mein $T_{r} = 2000 K$ , wall $T_{w} = 1200 K$ par rakhi gayi hai, aur heating $\overset{q}{˙}_{w} = 400 kW/ m^{2}$ quote ki gayi hai. Heat-transfer coefficient $h$ back out karo.

Recall Solution 2.3

KYA karte hain: $h$ ke liye Newton's law of cooling ko invert karo. $h = \frac{q ˙ _{w}}{T _{r} - T _{w}} = \frac{400 000 W/ m ^{2}}{2000 - 1200 K} = \frac{400 000}{800} = 500 W m^{- 2} K^{- 1}$ Units check: $\frac{W/ m ^{2}}{K} = W m^{- 2} K^{- 1}$ ✔.

Level 3 — Analysis

Yahan tum cases compare karte ho, differences lete ho, aur signs interpret karte ho.

Problem 3.1 — hypersonic speed par laminar vs turbulent

Ek vehicle $M_{e} = 6$ , $T_{e} = 250 K$ par fly kar raha hai. Laminar ( $r = 0.843$ ) aur turbulent ( $r = 0.892$ ) boundary layers ke liye $T_{r}$ compute karo, aur temperature difference batao.

Figure — Aerodynamic heating — recovery temperature, heat flux

Recall Solution 3.1

Step 1 — dynamic factor. $\frac{γ - 1}{2} M_{e}^{2} = 0.2 \times 36 = 7.2$ . Step 2 — laminar. $T_{r, lam} = 250 (1 + 0.843 \times 7.2) = 250 (1 + 6.0696) = 250 \times 7.0696 = 1767 K$ Step 3 — turbulent. $T_{r, turb} = 250 (1 + 0.892 \times 7.2) = 250 (1 + 6.4224) = 250 \times 7.4224 = 1856 K$ Step 4 — difference. $Δ T_{r} = 1856 - 1767 = 89 K$ . Turbulence zyada kinetic energy recover karta hai aur $h$ badhata hai, isliye turbulent regions zyada garam chalte hain — yahi wajah hai ki laminar→turbulent transition point predict karna thermal-protection design mein dominate karta hai. Figure dekho: dono bars ka base $T_{e}$ same hai, aur turbulent bar ka extra sliver exactly yahi $89 K$ hai.

Problem 3.2 — heat flux ki sign

Upar wale turbulent case ke liye, $T_{r} = 1856 K$ hai. Teen wall temperatures consider karo: (i) $T_{w} = 1200 K$ , (ii) $T_{w} = 1856 K$ , (iii) $T_{w} = 2100 K$ , sab $h = 300 W m^{- 2} K^{- 1}$ ke saath. Har case mein $\overset{q}{˙}_{w}$ find karo aur batao heat kis direction mein flow karti hai.

Recall Solution 3.2

KYUN teen cases: driving difference $T_{r} - T_{w}$ positive, zero, ya negative ho sakta hai — "all cases" check. (i) $T_{w} = 1200 < T_{r}$ : $\overset{q}{˙}_{w} = 300 (1856 - 1200) = 300 \times 656 = 1.968 \times 1 0^{5} W/ m^{2} \approx + 197 kW/ m^{2} (wall ke andar)$ (ii) $T_{w} = 1856 = T_{r}$ (adiabatic wall): $\overset{q}{˙}_{w} = 300 (1856 - 1856) = 0 (koi net flux nahi — yahi T_{r} define karta hai)$ (iii) $T_{w} = 2100 > T_{r}$ : $\overset{q}{˙}_{w} = 300 (1856 - 2100) = 300 \times (- 244) = - 7.32 \times 1 0^{4} W/ m^{2} \approx - 73.2 kW/ m^{2} (wall se bahar, gas mein)$ Toh $T_{r}$ se garam wall actually heat gas mein wapas dump kar deti hai.

Level 4 — Synthesis

Ab puri pipeline assemble karo: flow speed → density/velocity → Stanton number → flux, ya ise invert karo.

Problem 4.1 — Stanton-number heating

Ek wing ke ek point par, $ρ_{e} = 0.40 kg/ m^{3}$ , $u_{e} = 900 m/s$ , $c_{p} = 1005 J k g^{- 1} K^{- 1}$ hai, aur Stanton number $S t = 0.0015$ hai. Recovery temperature $T_{r} = 700 K$ hai aur wall $T_{w} = 300 K$ par hai. (a) effective $h$ , aur (b) heat flux $\overset{q}{˙}_{w}$ find karo.

Recall Solution 4.1

(a) Convective coefficient. $h = ρ_{e} u_{e} c_{p} S t$ . $h = 0.40 \times 900 \times 1005 \times 0.0015 = 361800 \times 0.0015 = 542.7 W m^{- 2} K^{- 1}$ (b) Heat flux. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 542.7 \times (700 - 300) = 542.7 \times 400 = 2.171 \times 1 0^{5} W/ m^{2} \approx 217 kW/ m^{2}$ Equivalently $\overset{q}{˙}_{w} = ρ_{e} u_{e} c_{p} S t (T_{r} - T_{w})$ ek shot mein same number deta hai.

Problem 4.2 — skin friction se Reynolds analogy

Usi panel par skin-friction coefficient $C_{f} = 0.0030$ hai. Reynolds analogy $S t \approx C_{f} /2$ (valid jab $P r \approx 1$ ) use karte hue $S t$ estimate karo, phir 4.1 wale same $ρ_{e}, u_{e}, c_{p}, T_{r}, T_{w}$ se $\overset{q}{˙}_{w}$ re-estimate karo.

Recall Solution 4.2

Step 1 — Reynolds analogy. $S t \approx \frac{C _{f}}{2} = \frac{0.0030}{2} = 0.0015$ . KYUN yeh kaam karta hai: momentum aur heat same turbulent eddies se transport hote hain, isliye jab $P r \approx 1$ ho toh unke transfer coefficients proportional hote hain — friction se heating predict hoti hai. Step 2 — flux. Kyunki $S t = 0.0015$ Problem 4.1 se exactly match karta hai, $\overset{q}{˙}_{w} = ρ_{e} u_{e} c_{p} S t (T_{r} - T_{w}) = 0.40 \times 900 \times 1005 \times 0.0015 \times 400 \approx 217 kW/ m^{2}$ Is analogy ki power: tumne drag measure kiya (aasaan) aur heating (mushkil) free mein mil gayi.

Level 5 — Mastery

Ek extended, design-flavoured problem jahan tumhe khud decide karna hoga ki kaunse formulas apply hote hain aur kis order mein.

Problem 5.1 — Re-entry radiative-equilibrium wall temperature

Ek chhote heat-shield tile ko ek aisa flow dikhta hai jismein recovery temperature $T_{r} = 3000 K$ aur convective coefficient $h = 120 W m^{- 2} K^{- 1}$ hai. Tile mein koi active cooling nahi hai — balki yeh apne front face se black body ki tarah re-radiate karta hai ( $ε = 1$ ), Stefan–Boltzmann constant $σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ ke saath. Steady state par convective heat in radiated heat out ke barabar hoti hai: $h (T_{r} - T_{w}) = σ T_{w}^{4} .$ Equilibrium wall temperature $T_{w}$ estimate karo (nearest 10 K tak), aur us state par heat flux bhi.

Recall Solution 5.1

KYA karte hain: convective inflow ko radiative outflow ke barabar set karo aur resulting equation ko $T_{w}$ ke liye solve karo. Yeh ek balance hai, plug-in nahi — figure mein dono curves jahan cross karti hain wahi answer hai. KYUN: koi coolant nahi hai, toh tile shed karne ka ek hi tarika hai — radiation, aur tab tak heat up hoti hai jab tak dono rates match na ho jayein. $120 (3000 - T_{w}) = 5.67 \times 1 0^{- 8} T_{w}^{4}$ Iteration se solve karo (figure mein left line falling hai, right curve rising — yeh ek baar cross karti hain):

Try $T_{w} = 1300$ : LHS $= 120 (1700) = 204000$ ; RHS $= 5.67 \times 1 0^{- 8} \times 130 0^{4} = 1.62 \times 1 0^{5}$ . LHS > RHS → wall abhi bhi heat up ho rahi hai.
Try $T_{w} = 1400$ : LHS $= 120 (1600) = 192000$ ; RHS $= 5.67 \times 1 0^{- 8} \times 140 0^{4} = 2.18 \times 1 0^{5}$ . Ab RHS > LHS → overshoot ho gaya.
Try $T_{w} = 1360$ : LHS $= 120 (1640) = 196800$ ; RHS $= 5.67 \times 1 0^{- 8} \times 136 0^{4} = 1.94 \times 1 0^{5}$ . Lagbhag equal. $T_{w} \approx 1360 K$ Balance par heat flux. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 120 (3000 - 1360) = 120 \times 1640 \approx 1.97 \times 1 0^{5} W/ m^{2} = 197 kW/ m^{2}$ Radiated side check karo: $σ T_{w}^{4} = 5.67 \times 1 0^{- 8} \times 136 0^{4} \approx 1.94 \times 1 0^{5} W/ m^{2}$ — iteration error ke andar match karta hai. ✔ Tile $\approx 1360 K$ "choose" karta hai: itna garam ki flow jo deliver karta hai sab radiate ho jaye, $T_{r} = 3000 K$ se kaafi neeche.

Connections

Stagnation properties & isentropic relations — har problem mein use hone wala $T_{0} / T$ factor
Prandtl number & thermal boundary layer — jahan $r = P r$ aur $P r^{1/3}$ aate hain
Reynolds analogy & Stanton number — Problems 4.1–4.2
Hypersonic re-entry & thermal protection systems — Problem 5.1 ka radiative balance
Boundary layers & viscous dissipation — kyun $T_{r}$ $T_{e}$ aur $T_{0}$ ke beech hoti hai
Normal & oblique shock heating — in numbers ko feed karne wali post-shock $T_{e}$ set karta hai

Recall Self-test checklist

Kya tum memory se kar sakte ho: (1) $M$ se $T_{0}$ nikalna? ::: air ke liye $T_{0} = T (1 + 0.2 M^{2})$ . (2) $T_{0}$ ko $T_{r}$ mein convert karna? ::: dynamic term ko $r$ se multiply karo: $T_{r} = T_{e} (1 + r \cdot 0.2 M_{e}^{2})$ . (3) flux driver pick karna? ::: hamesha $T_{r} - T_{w}$ , kabhi $T_{0} - T_{w}$ nahi. (4) $S t$ se $h$ tak jaana? ::: $h = ρ_{e} u_{e} c_{p} S t$ . (5) batana kyun ek passive tile $T_{r}$ se neeche rehti hai? ::: yeh heat radiate kaar deti hai, isliye adiabatic nahi hai.