Aerodynamic heating — recovery temperature, heat flux: Exercises

Level 1 — Recognition

These test whether you can pick the right formula and plug in. No traps in the algebra yet.

Problem 1.1

Air flows at Mach $M = 2$ with static temperature $T = 250 K$ . Find the stagnation temperature $T_{0}$ .

Recall Solution 1.1

WHAT we do: apply the stagnation ratio directly — no boundary layer, so no recovery factor needed. WHY: $T_{0}$ is defined by bringing the gas to rest adiabatically, which is exactly what the ratio encodes. $\frac{γ - 1}{2} M^{2} = 0.2 \times 2^{2} = 0.2 \times 4 = 0.8$ $T_{0} = T (1 + 0.8) = 250 \times 1.8 = 450 K$

Problem 1.2

For air, $P r = 0.71$ . Write down the recovery factor $r$ for a laminar boundary layer and for a turbulent one (3 significant figures).

Recall Solution 1.2

WHAT we do: apply the two definitions of $r$ . WHY the two differ: turbulent mixing recovers more of the dissipated heat near the wall, so $r$ is larger. $r_{lam} = P r = 0.71 = 0.843$ $r_{turb} = P r^{1/3} = 0.7 1^{1/3} = 0.892$ Both are less than 1, so the wall never quite reaches $T_{0}$ .

Problem 1.3

A wall has recovery temperature $T_{r} = 600 K$ and is held at $T_{w} = 400 K$ by cooling. The heat-transfer coefficient is $h = 500 W m^{- 2} K^{- 1}$ . Find the heat flux $\overset{q}{˙}_{w}$ .

Recall Solution 1.3

WHAT we do: Newton's law of cooling with the correct driving temperature. WHY $T_{r} - T_{w}$ and not $T_{0} - T_{w}$ : $T_{r}$ is the temperature an uncooled wall settles at — the true thermal potential the gas pushes toward. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 500 \times (600 - 400) = 500 \times 200 = 1.0 \times 1 0^{5} W/ m^{2} = 100 kW/ m^{2}$ Positive sign → heat flows into the wall (it is colder than $T_{r}$ ).

Level 2 — Application

Now you chain two formulas together and keep the units clean.

Problem 2.1

A cruise missile flies at $M_{e} = 3$ through air at $T_{e} = 223 K$ . The boundary layer is turbulent ( $r = 0.892$ ). Find the recovery temperature $T_{r}$ , and compare it to $T_{0}$ .

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 — compare with $T_{0}$ . $T_{0} = 223 (1 + 1.8) = 223 \times 2.8 = 624 K$ So $T_{r} = 581 K$ sits just below $T_{0} = 624 K$ — the wall recovers about 89% of the temperature rise, exactly as $r = 0.892$ predicts.

Problem 2.2

Same flow as 2.1 ( $T_{r} = 581 K$ ). The skin is cooled to $T_{w} = 350 K$ and $h = 900 W m^{- 2} K^{- 1}$ . Find $\overset{q}{˙}_{w}$ in $kW/ m^{2}$ .

Recall Solution 2.2

$\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 900 (581 - 350) = 900 \times 231 = 2.079 \times 1 0^{5} W/ m^{2} \approx 208 kW/ m^{2}$ This enormous flux (2000× a domestic radiator per square metre) is why fast vehicles need active cooling or a heat shield.

Problem 2.3

A blunt re-entry body has $T_{r} = 2000 K$ , wall held at $T_{w} = 1200 K$ , and the heating is quoted as $\overset{q}{˙}_{w} = 400 kW/ m^{2}$ . Back out the heat-transfer coefficient $h$ .

Recall Solution 2.3

WHAT we do: invert Newton's law of cooling for $h$ . $h = \frac{q ˙ _{w}}{T _{r} - T _{w}} = \frac{400 000 W/ m ^{2}}{2000 - 1200 K} = \frac{400 000}{800} = 500 W m^{- 2} K^{- 1}$ Units check: $\frac{W/ m ^{2}}{K} = W m^{- 2} K^{- 1}$ ✔.

Level 3 — Analysis

Here you compare cases, take differences, and interpret signs.

Problem 3.1 — laminar vs turbulent at hypersonic speed

A vehicle flies at $M_{e} = 6$ , $T_{e} = 250 K$ . Compute $T_{r}$ for laminar ( $r = 0.843$ ) and turbulent ( $r = 0.892$ ) boundary layers, and state the temperature difference.

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 recovers more kinetic energy and raises $h$ , so turbulent regions run hotter — the reason predicting the laminar→turbulent transition point dominates thermal-protection design. Look at the figure: the two bars share the same base $T_{e}$ , and the turbulent bar's extra sliver is exactly this $89 K$ .

Problem 3.2 — sign of the heat flux

For the turbulent case above, $T_{r} = 1856 K$ . Consider three wall temperatures: (i) $T_{w} = 1200 K$ , (ii) $T_{w} = 1856 K$ , (iii) $T_{w} = 2100 K$ , all with $h = 300 W m^{- 2} K^{- 1}$ . Find $\overset{q}{˙}_{w}$ in each case and say which way heat flows.

Recall Solution 3.2

WHY three cases: the driving difference $T_{r} - T_{w}$ can be positive, zero, or negative — the "all cases" check. (i) $T_{w} = 1200 < T_{r}$ : $\overset{q}{˙}_{w} = 300 (1856 - 1200) = 300 \times 656 = 1.968 \times 1 0^{5} W/ m^{2} \approx + 197 kW/ m^{2} (into wall)$ (ii) $T_{w} = 1856 = T_{r}$ (adiabatic wall): $\overset{q}{˙}_{w} = 300 (1856 - 1856) = 0 (no net flux — this defines T_{r})$ (iii) $T_{w} = 2100 > T_{r}$ : $\overset{q}{˙}_{w} = 300 (1856 - 2100) = 300 \times (- 244) = - 7.32 \times 1 0^{4} W/ m^{2} \approx - 73.2 kW/ m^{2} (out of wall, into gas)$ So a wall hotter than $T_{r}$ actually dumps heat back into the gas.

Level 4 — Synthesis

Now assemble the full pipeline: flow speed → density/velocity → Stanton number → flux, or invert it.

Problem 4.1 — Stanton-number heating

At a point on a wing, $ρ_{e} = 0.40 kg/ m^{3}$ , $u_{e} = 900 m/s$ , $c_{p} = 1005 J k g^{- 1} K^{- 1}$ , and the Stanton number is $S t = 0.0015$ . The recovery temperature is $T_{r} = 700 K$ and the wall is at $T_{w} = 300 K$ . Find (a) the effective $h$ , and (b) the heat flux $\overset{q}{˙}_{w}$ .

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 $\overset{q}{˙}_{w} = ρ_{e} u_{e} c_{p} S t (T_{r} - T_{w})$ done in one shot gives the same number.

Problem 4.2 — Reynolds analogy from skin friction

On the same panel the skin-friction coefficient is $C_{f} = 0.0030$ . Using the Reynolds analogy $S t \approx C_{f} /2$ (valid when $P r \approx 1$ ), estimate $S t$ , then re-estimate $\overset{q}{˙}_{w}$ with the same $ρ_{e}, u_{e}, c_{p}, T_{r}, T_{w}$ as 4.1.

Recall Solution 4.2

Step 1 — Reynolds analogy. $S t \approx \frac{C _{f}}{2} = \frac{0.0030}{2} = 0.0015$ . WHY this works: momentum and heat are transported by the same turbulent eddies, so when $P r \approx 1$ their transfer coefficients are proportional — friction predicts heating. Step 2 — flux. Since $S t = 0.0015$ matches Problem 4.1 exactly, $\overset{q}{˙}_{w} = ρ_{e} u_{e} c_{p} S t (T_{r} - T_{w}) = 0.40 \times 900 \times 1005 \times 0.0015 \times 400 \approx 217 kW/ m^{2}$ The power of the analogy: you measured drag (easy) and got heating (hard) for free.

Level 5 — Mastery

One extended, design-flavoured problem where you must decide which formulas apply and in what order.

Problem 5.1 — Re-entry radiative-equilibrium wall temperature

A small heat-shield tile sees a flow with recovery temperature $T_{r} = 3000 K$ and convective coefficient $h = 120 W m^{- 2} K^{- 1}$ . The tile has no active cooling — instead it re-radiates from its front face like a black body ( $ε = 1$ ), with Stefan–Boltzmann constant $σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ . At steady state the convective heat in equals the radiated heat out: $h (T_{r} - T_{w}) = σ T_{w}^{4} .$ Estimate the equilibrium wall temperature $T_{w}$ (to the nearest 10 K), and the heat flux at that state.

Recall Solution 5.1

WHAT we do: set convective inflow equal to radiative outflow and solve the resulting equation for $T_{w}$ . This is a balance, not a plug-in — the two curves in the figure cross at the answer. WHY: with no coolant, the only way the tile sheds heat is radiation, and it heats up until the two rates match. $120 (3000 - T_{w}) = 5.67 \times 1 0^{- 8} T_{w}^{4}$ Solve by iteration (the figure shows the left line falling, the right curve rising — they cross once):

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 still heating.
Try $T_{w} = 1400$ : LHS $= 120 (1600) = 192000$ ; RHS $= 5.67 \times 1 0^{- 8} \times 140 0^{4} = 2.18 \times 1 0^{5}$ . Now RHS > LHS → overshot.
Try $T_{w} = 1360$ : LHS $= 120 (1640) = 196800$ ; RHS $= 5.67 \times 1 0^{- 8} \times 136 0^{4} = 1.94 \times 1 0^{5}$ . Nearly equal. $T_{w} \approx 1360 K$ Heat flux at balance. $\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 120 (3000 - 1360) = 120 \times 1640 \approx 1.97 \times 1 0^{5} W/ m^{2} = 197 kW/ m^{2}$ Check the radiated side: $σ T_{w}^{4} = 5.67 \times 1 0^{- 8} \times 136 0^{4} \approx 1.94 \times 1 0^{5} W/ m^{2}$ — matches to within iteration error. ✔ The tile "chooses" $\approx 1360 K$ : hot enough to radiate away everything the flow delivers, far below $T_{r} = 3000 K$ .

Connections

Stagnation properties & isentropic relations — the $T_{0} / T$ factor used in every problem
Prandtl number & thermal boundary layer — where $r = P r$ and $P r^{1/3}$ come from
Reynolds analogy & Stanton number — Problems 4.1–4.2
Hypersonic re-entry & thermal protection systems — Problem 5.1's radiative balance
Boundary layers & viscous dissipation — why $T_{r}$ lies between $T_{e}$ and $T_{0}$
Normal & oblique shock heating — sets the post-shock $T_{e}$ feeding these numbers

Recall Self-test checklist

Can you, from memory: (1) get $T_{0}$ from $M$ ? ::: $T_{0} = T (1 + 0.2 M^{2})$ for air. (2) turn $T_{0}$ into $T_{r}$ ? ::: multiply the dynamic term by $r$ : $T_{r} = T_{e} (1 + r \cdot 0.2 M_{e}^{2})$ . (3) pick the flux driver? ::: always $T_{r} - T_{w}$ , never $T_{0} - T_{w}$ . (4) go from $S t$ to $h$ ? ::: $h = ρ_{e} u_{e} c_{p} S t$ . (5) say why a passive tile stays below $T_{r}$ ? ::: it radiates heat away, so it is not adiabatic.