Aerodynamic heating — recovery temperature, heat flux: Worked examples

This page is the drill floor for the parent topic. The parent built three tools:

the stagnation ratio $\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2}$ (how hot a stopped gas gets),
the recovery temperature $T_{r} = T_{e} (1 + r \frac{γ - 1}{2} M_{e}^{2})$ (how hot an uncooled wall actually gets — always a bit below $T_{0}$ because the recovery factor $r < 1$ ),
the heat flux $\overset{q}{˙}_{w} = h (T_{r} - T_{w})$ (how fast heat pours into a wall colder than $T_{r}$ ).

Here we throw every case class at those three formulas — every sign of the flux, the zero/degenerate inputs, the low- and high-Mach limits, a real word problem, and an exam twist — so you never meet a scenario you haven't already seen worked.

Symbols used below, in plain words:

Definition Cast of characters

$M_{e}$ = Mach number at the boundary-layer edge = flow speed ÷ local speed of sound. Dimensionless.
$T_{e}$ = edge static temperature (the cold free-stream air the vehicle flies through), in kelvin (K).
$T_{0}$ = stagnation temperature = temperature if you stop the flow with no heat loss.
$T_{r}$ = recovery temperature = temperature an insulated (adiabatic) wall settles at.
$T_{w}$ = wall temperature = the actual surface temperature (you may control it by cooling).
$r$ = recovery factor, between 0 and 1 — the fraction of the stopping-heat the wall keeps.
$γ$ = ratio of specific heats (for air $γ = 1.4$ ).
$h$ = heat-transfer coefficient $[W m^{- 2} K^{- 1}]$ — how good the flow is at pushing heat into the wall.
$\overset{q}{˙}_{w}$ = wall heat flux $[W m^{- 2}]$ = heat energy crossing one square metre of wall per second.

The scenario matrix

Everything this topic can ask lands in one of these cells. Each example below is tagged with its cell.

#	Cell (case class)	What's special	Example
A	Sign of flux `> 0` ( $T_{w} < T_{r}$ )	wall absorbs heat — cooling needed	Ex 1
B	Sign of flux `= 0` ( $T_{w} = T_{r}$ )	adiabatic wall — flux vanishes	Ex 2
C	Sign of flux `< 0` ( $T_{w} > T_{r}$ )	hot wall dumps heat back to gas	Ex 3
D	Zero / degenerate input ( $M_{e} \to 0$ )	no speed → no heating	Ex 4
E	Low-Mach limit (small $M_{e}$ )	$T_{r} - T_{e}$ scales as $M_{e}^{2}$	Ex 4
F	High-Mach limit (hypersonic)	huge $T_{r}$ , laminar vs turbulent split	Ex 5
G	Real-world word problem	pick out $M_{e}, T_{e}, T_{w}$ yourself	Ex 6
H	Exam twist (find $h$ or $T_{w}$ from flux)	invert the flux law	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 colder than $T_{r}$ , heat flows in

Worked example A cooled fin on a Mach 3 jet

$M_{e} = 3$ , $T_{e} = 223$ K, $γ = 1.4$ , turbulent so $r = P r^{1/3} = 0.7 1^{1/3} = 0.892$ . Wall held at $T_{w} = 350$ K, $h = 900 W m^{- 2} K^{- 1}$ . Find $T_{r}$ and $\overset{q}{˙}_{w}$ .

Forecast: guess — is $T_{r}$ closer to 224 K or to 600 K? And will $\overset{q}{˙}_{w}$ be positive or negative?

Step 1 — the stopping-heat fraction. $\frac{γ - 1}{2} M_{e}^{2} = 0.2 \times 3^{2} = 0.2 \times 9 = 1.8$ . Why this step? This dimensionless number is how many times over the static temperature would rise if we stopped the flow with no losses. It's the engine of every heating formula.

Step 2 — apply the recovery factor. $T_{r} = T_{e} (1 + r \cdot 1.8) = 223 (1 + 0.892 \times 1.8) = 223 (1 + 1.606) = 223 \times 2.606 = 581 K$ . Why this step? The wall keeps only fraction $r$ of that stopping-heat, so we multiply the $1.8$ by $r$ before adding back the base $1$ .

Step 3 — the flux. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 900 (581 - 350) = 900 \times 231 = 2.08 \times 1 0^{5} W/ m^{2} \approx 208 kW/ m^{2}$ . Why this step? The wall (350 K) is colder than the post $T_{r}$ (581 K), so heat pours in — positive flux, cell A.

Verify: $T_{r} = 581$ K sits just below $T_{0} = 223 (1 + 1.8) = 624$ K, exactly as expected for $r < 1$ . Units: $[W m^{- 2} K^{- 1}] \times [K] = [W m^{- 2}]$ ✔. Flux positive ⇒ into wall ⇒ cooling justified. ✔

Example 2 — Cell B: adiabatic wall, zero flux

Worked example Let the fin float (no cooling)

Same flow as Ex 1 ( $T_{r} = 581$ K). Now stop cooling and let the wall reach steady state. What is $T_{w}$ and $\overset{q}{˙}_{w}$ ?

Forecast: if nothing draws heat away, where does the wall settle?

Step 1 — impose steady state, no heat drawn out. "Adiabatic wall" means zero net heat crosses the surface: $\overset{q}{˙}_{w} = 0$ . Why this step? At equilibrium an insulated wall neither gains nor loses heat — that's the definition of the recovery temperature.

Step 2 — solve $h (T_{r} - T_{w}) = 0$ . With $h \neq = 0$ , this forces $T_{w} = T_{r} = 581$ K. Why this step? The only way the driving difference $T_{r} - T_{w}$ can vanish is for the wall to sit exactly on the post.

Verify: plug back: $\overset{q}{˙}_{w} = 900 (581 - 581) = 0$ ✔. This is why $T_{r}$ is called the recovery / adiabatic-wall temperature — it's the unique wall temperature giving zero flux.

Example 3 — Cell C: wall hotter than $T_{r}$ , heat flows out

Worked example A soak-heated wall on the ground, engine idling

After landing, a surface sits at $T_{w} = 650$ K (heat-soaked from flight) while the local flow is nearly still but warm: take $T_{r} = 581$ K, $h = 900 W m^{- 2} K^{- 1}$ . Find $\overset{q}{˙}_{w}$ and its direction.

Forecast: wall is hotter than the recovery post now — sign of flux?

Step 1 — same law, new numbers. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 900 (581 - 650) = 900 \times (- 69)$ . Why this step? Newton's cooling law is signed: $T_{r} - T_{w}$ is now negative because $T_{w} > T_{r}$ .

Step 2 — read the sign. $\overset{q}{˙}_{w} = - 6.21 \times 1 0^{4} W/ m^{2} = - 62.1 kW/ m^{2}$ . Why this step? Negative flux means heat leaves the wall into the gas — the wall is cooling itself down toward $T_{r}$ .

Verify: magnitude sanity: same $h$ , temperature gap 69 K vs 231 K in Ex 1, so flux should be $231/69 \approx 3.35 \times$ smaller in size: $208/3.35 \approx 62$ kW/m² ✔. Sign negative ⇒ wall loses heat ⇒ it drifts back down to the $T_{r} = 581$ K post. ✔

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

Worked example Standing still, then crawling

Air at $T_{e} = 288$ K, $γ = 1.4$ , laminar $r = 0.71 = 0.843$ . (i) $M_{e} = 0$ . (ii) $M_{e} = 0.3$ . Find $T_{r}$ each time and the rise $T_{r} - T_{e}$ .

Forecast: at zero speed should the wall feel any extra heating at all?

Step 1 — degenerate case $M_{e} = 0$ . $\frac{γ - 1}{2} M_{e}^{2} = 0.2 \times 0 = 0$ , so $T_{r} = 288 (1 + 0) = 288$ K. Why this step? No motion ⇒ no kinetic energy to dissipate ⇒ the wall just sits at the air temperature. The formula must (and does) collapse to $T_{r} = T_{e}$ . This is the sanity anchor for the whole topic.

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. Why this step? Shows the quadratic growth: heating rise scales as $M_{e}^{2}$ . At $M = 0.3$ the rise is tiny — this is why low-speed aircraft ignore aerodynamic heating.

Verify: rise ratio between the two Mach numbers should be $(0.3/0)$ … undefined at zero, so instead compare $M_{e} = 0.3$ vs $M_{e} = 0.6$ : rise scales as $M^{2}$ , i.e. $\times 4$ . 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 Re-entry heating at Mach 6

$M_{e} = 6$ , $T_{e} = 250$ K, $γ = 1.4$ . Compare laminar ( $r = 0.71 = 0.843$ ) with turbulent ( $r = 0.7 1^{1/3} = 0.892$ ). See Hypersonic re-entry & thermal protection systems.

Forecast: which flow gives the hotter wall — laminar or turbulent — and by how much?

Step 1 — stopping-heat fraction. $\frac{γ - 1}{2} M_{e}^{2} = 0.2 \times 36 = 7.2$ . Why this step? At high Mach this dominates the " $1 +$ ", so $T_{r}$ is roughly $7 \times T_{e}$ — the source of thousands-of-kelvin walls.

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. Why this step? Turbulent $r$ is larger, so more of the huge stopping-heat is recovered — the turbulent wall runs hotter, here by ~88 K.

Step 3 — compare to $T_{0}$ . $T_{0} = 250 (1 + 7.2) = 2050$ K. Both $T_{r}$ values sit below it, turbulent nearer. Why this step? Confirms $r < 1$ ⇒ $T_{r} < T_{0}$ always, in the extreme regime too.

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 ✔. Both below $T_{0} = 2050$ K ✔. Aluminium melts ~933 K ⇒ both would melt it ⇒ TPS mandatory. ✔

Example 6 — Cell G: real-world word problem

Worked example The SR-71's leading edge

"A reconnaissance jet cruises where the outside air is $- 56. 5^{\circ} C$ at Mach 3.2. Its titanium leading edge is actively kept at $30 0^{\circ} C$ . Turbulent boundary layer, $r = 0.89$ , $γ = 1.4$ , $h = 1100 W m^{- 2} K^{- 1}$ . How hard is the edge being heated?"

Forecast: first job — convert everything to kelvin before touching a formula.

Step 1 — extract and convert. $T_{e} = - 56.5 + 273.15 = 216.65 \approx 216.7$ K; $T_{w} = 300 + 273.15 = 573.15 \approx 573.2$ K; $M_{e} = 3.2$ . Why this step? The stagnation/recovery formulas are ratios of absolute temperatures — Celsius would give nonsense. Always 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. Why this step? This is the temperature the edge would reach uncooled — the post we drive the flux from.

Step 4 — flux. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 1100 (611.7 - 573.2) = 1100 \times 38.5 = 4.24 \times 1 0^{4} W/ m^{2} \approx 42.4 kW/ m^{2}$ . Why this step? Wall (573.2 K) is just below the post (611.7 K), so a modest positive flux — cell A, but a mild one.

Verify: $T_{r} = 611.7$ K is below $T_{0} = 216.7 (1 + 2.048) = 660.6$ K ✔. Gap $T_{r} - T_{w} = 38.5$ K is small ⇒ this edge is being held close to its natural recovery temperature, so cooling load is light — matches the SR-71 running its skin genuinely hot. Units ✔.

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

Worked example Back-solve for the required wall temperature

"For the Mach 3 jet of Ex 1 ( $T_{r} = 581$ K), the material can only survive a flux of $\overset{q}{˙}_{w} = 150 kW/ m^{2}$ . With $h = 900 W m^{- 2} K^{- 1}$ , how cold must you keep the wall?"

Forecast: to lower the flux from 208 to 150 kW/m², do you make $T_{w}$ higher or lower?

Step 1 — rearrange Newton's law. From $\overset{q}{˙}_{w} = h (T_{r} - T_{w})$ , solve for $T_{w}$ : $T_{w} = T_{r} - \frac{q ˙ _{w}}{h}$ . Why this step? The exam gives flux and asks for wall temperature — invert the one formula. Everything is known except $T_{w}$ .

Step 2 — plug numbers. $T_{w} = 581 - \frac{150000}{900} = 581 - 166.7 = 414.3$ K. Why this step? The allowable temperature gap is $150000/900 = 166.7$ K; subtract from the fixed post $T_{r}$ .

Step 3 — interpret the direction. To reduce flux you need a smaller gap $T_{r} - T_{w}$ , i.e. a warmer wall (414 K here vs 350 K in Ex 1). Cooling harder would raise the flux, not lower it. Why this step? This is the classic counter-intuitive result — flagged as a mistake in the parent note. Verify it numerically here.

Verify: plug back: $\overset{q}{˙}_{w} = 900 (581 - 414.3) = 900 \times 166.7 = 1.50 \times 1 0^{5} W/ m^{2} = 150$ kW/m² ✔. And $414.3 > 350$ K confirms "warmer wall ⇒ less flux". ✔

Example 8 — Cell I: the Stanton-number route

Worked example Predict

$h$ from skin friction Using Reynolds analogy & Stanton number: at the edge $ρ_{e} = 0.4 kg/ m^{3}$ , $u_{e} = 900 m/s$ , $c_{p} = 1005 J k g^{- 1} K^{- 1}$ , skin-friction coefficient $C_{f} = 0.0020$ . Reynolds analogy gives $S t \approx C_{f} /2$ . Find $h$ and then $\overset{q}{˙}_{w}$ for $T_{r} = 581$ K, $T_{w} = 350$ K.

Forecast: will this $h$ land in the same ballpark as the 900 used in Ex 1?

Step 1 — Stanton from friction. $S t \approx C_{f} /2 = 0.0020/2 = 0.0010$ . Why this step? Reynolds analogy links heat transfer to momentum transfer — measuring drag lets you predict heating without a separate thermal experiment.

Step 2 — build $h$ . $h = ρ_{e} u_{e} c_{p} S t = 0.4 \times 900 \times 1005 \times 0.0010 = 361.8 W m^{- 2} K^{- 1}$ . Why this step? The group $ρ_{e} u_{e} c_{p}$ (units $W m^{- 2} K^{- 1}$ ) is the maximum possible heat-transfer scale; $S t$ is the fraction of it actually realised.

Step 3 — flux. $\overset{q}{˙}_{w} = ρ_{e} u_{e} c_{p} S t (T_{r} - T_{w}) = 361.8 \times (581 - 350) = 361.8 \times 231 = 8.36 \times 1 0^{4} W/ m^{2} \approx 83.6 kW/ m^{2}$ . Why this step? Confirms the two flux formulas ( $\overset{q}{˙}_{w} = h Δ T$ and the Stanton form) are the same statement, just with $h$ expanded.

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

Recall Quick self-test across the matrix

Flux sign when $T_{w} < T_{r}$ ? ::: Positive — heat flows into the wall (cell A). Flux when $T_{w} = T_{r}$ ? ::: Zero — adiabatic wall (cell B); this defines $T_{r}$ . What happens to $T_{r}$ as $M_{e} \to 0$ ? ::: $T_{r} \to T_{e}$ ; no motion, no heating (cell D). To lower the heat flux, warm or cool the wall? ::: Warm it — a smaller gap $T_{r} - T_{w}$ means less flux (cell H). Which flow gives the hotter recovery temperature, laminar or turbulent? ::: Turbulent, because $r = P r^{1/3} > P r$ (cell F).

Connections

Parent: recovery temperature & heat flux — the three formulas drilled here
Stagnation properties & isentropic relations — source of the stopping-heat fraction $\frac{γ - 1}{2} M^{2}$
Prandtl number & thermal boundary layer — sets $r = P r$ vs $P r^{1/3}$ (Ex 4–5)
Reynolds analogy & Stanton number — the $S t$ route of Ex 8
Hypersonic re-entry & thermal protection systems — the Mach 6 wall of Ex 5
Boundary layers & viscous dissipation — why $r < 1$
Normal & oblique shock heating — supplies the post-shock $T_{e}$ used as input