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

WHAT we do: read the formula symbol by symbol.

• $V$ appears as $V^3$ → exponent $+3$.
• ${\rho }_{\infty }$ appears under a square root, $\sqrt{{\rho }_{\infty }}={\rho }_{\infty }^{1/2}$ → exponent $+\frac{1}{2}$.
• $R_n$ appears as $1/\sqrt{R_n}=R_n^{-1/2}$ → exponent $-\frac{1}{2}$.

Answer: ${\dot{q}}_s\propto V^3{\rho }_{\infty }^{1/2}{R}_n^{-1/2}$.

${\dot{q}}_s$ is a heat flux: energy per unit time per unit area, so ${\text{W/m}}^2={\text{J⋅s}}^{-1}{\text{⋅m}}^{-2}$. It measures how fast heat pours into each square metre of the nose surface. It is not total energy — it is a rate per area.

The stagnation point — the nose point where the oncoming flow is brought fully to rest ($v\to 0$). All that ordered kinetic energy converts into local enthalpy (temperature/pressure), so this point sees the maximum convective heat flux. See Bow shock and blunt-body theory for why a blunt nose forms a detached shock here.

WHY factor the pieces? So each square root and the cube stay honest.

• $\sqrt{R_n}=\sqrt{1.0}=1.0$.
• $\sqrt{{\rho }_{\infty }}=\sqrt{5.0\times 10^{-4}}=2.236\times 10^{-2}$.
• $V^3=(7500{)}^3=4.219\times 10^{11}$.

${\dot{q}}_s=\frac{1.83\times 10^{-4}}{1.0}\times 2.236\times 10^{-2}\times 4.219\times 10^{11}$ $=1.83\times 10^{-4}\times 2.236\times 10^{-2}\times 4.219\times 10^{11}\approx 1.726\times 10^6{\text{W/m}}^2.$ Convert: $1{\text{W/cm}}^2=10^4{\text{W/m}}^2$, so ${\dot{q}}_s\approx 173{\text{W/cm}}^2$.

WHAT governs this: only the $V^3$ factor moves. WHY it matters: this is the single biggest lever in reentry heating. $\frac{{\dot{q}}_s(2V)}{{\dot{q}}_s(V)}=\frac{(2V{)}^3}{V^3}=2^3=8.$ Doubling speed makes heating 8× worse.

Only $R_n^{-1/2}$ moves. $\frac{{\dot{q}}_s(4R_n)}{{\dot{q}}_s(R_n)}=\frac{(4R_n{)}^{-1/2}}{R_n^{-1/2}}=4^{-1/2}=\frac{1}{2}.$ A 4× bigger nose radius halves the heat flux — this is why capsules are blunt. See Thermal Protection Systems (ablatives, tiles).

WHY a ratio? The shared ${\rho }_{\infty }$ and $C$ cancel, so only the $V^3$ and $R_n^{-1/2}$ pieces survive. $\frac{{\dot{q}}_A}{{\dot{q}}_B}={\left(\frac{V_A}{V_B}\right)}^3{\left(\frac{R_{n,B}}{R_{n,A}}\right)}^{1/2}$

• Speed part: $(8000/6000{)}^3=(4/3{)}^3=64/27=2.370$.
• Radius part: $(2.0/0.5{)}^{1/2}=4^{1/2}=2.0$.

$\frac{{\dot{q}}_A}{{\dot{q}}_B}=2.370\times 2.0=\mathrm{4.741.}$ Vehicle A runs about $4.74\times$ hotter — faster and blunter-disadvantaged.

WHAT the product looks like: ${\dot{q}}_s$ is a product of a rising factor ${\rho }^{1/2}$ and a falling factor $V^3$ as the vehicle drops. See the figure.

• Near the top of the atmosphere: $\rho \to 0$, so ${\rho }^{1/2}\to 0$ → almost no heat even though $V$ is huge.
• Near the bottom: the vehicle has braked hard, $V$ is small, and $V^3$ collapses (cubed!) → heat drops again.
• In between, the product is maximised at an intermediate altitude. Because $V$ enters cubed and $\rho$ only as a half-power, the $V^3$ collapse dominates once braking begins, so the peak sits high up, well before peak deceleration (see Ballistic coefficient and deceleration).

• $\sin 30^{\circ }=0.5$.
• ${\rho }_{\text{peak}}=\frac{200\times 0.5}{3\times 7000}=\frac{100}{21000}=4.76\times 10^{-3}{\text{kg/m}}^2\cdot {\text{m}}^{-1}$.

Wait — check dimensions: $\beta$ here is a ballistic coefficient ${\text{kg/m}}^2$, and dividing by length $H$ (m) gives ${\text{kg/m}}^3$, a density. So ${\rho }_{\text{peak}}\approx 4.76\times 10^{-3}{\text{kg/m}}^3$. Comparing to sea level ($1.2{\text{kg/m}}^3$), this is a very thin, high-altitude density — confirming L3.2: peak heating is high up. See Reentry trajectory dynamics.

WHY multiply factors? Each variable enters as an independent power, so their effects multiply. $\frac{{\dot{q}}_{\text{new}}}{{\dot{q}}_{\text{old}}}=(0.9{)}^3\times (2{)}^{-1/2}$

• $(0.9{)}^3=0.729$.
• $2^{-1/2}=0.7071$.

$\frac{{\dot{q}}_{\text{new}}}{{\dot{q}}_{\text{old}}}=0.729\times 0.7071=\mathrm{0.5155.}$ Net heat flux drops to about 51.6% — nearly halved. Speed reduction (cubed) does most of the work, bluntness finishes the job.

WHAT we do: invert the formula for $R_n$. WHY invert: the unknown is buried inside $1/\sqrt{R_n}$, so we isolate it algebraically. ${\dot{q}}_s=\frac{C\sqrt{{\rho }_{\infty }}{V}^3}{\sqrt{R_n}}\le {\dot{q}}_{\max }\Rightarrow \sqrt{R_n}\ge \frac{C\sqrt{{\rho }_{\infty }}{V}^3}{{\dot{q}}_{\max }}.$ Compute the numerator first:

• $\sqrt{{\rho }_{\infty }}=\sqrt{8.0\times 10^{-4}}=2.828\times 10^{-2}$.
• $V^3=(9000{)}^3=7.29\times 10^{11}$.
• Numerator $=1.83\times 10^{-4}\times 2.828\times 10^{-2}\times 7.29\times 10^{11}=3.773\times 10^6$.

$\sqrt{R_n}\ge \frac{3.773\times 10^6}{5.0\times 10^6}=0.7546\Rightarrow R_n\ge 0.7546^2=0.569\text{m}.$ Minimum nose radius $\approx 0.57\text{m}$.

(a) Factor the pieces:

• $\sqrt{R_n}=\sqrt{4.7}=2.168$.
• $\sqrt{{\rho }_{\infty }}=\sqrt{10^{-4}}=10^{-2}$.
• $V^3=(11000{)}^3=1.331\times 10^{12}$.

${\dot{q}}_s=\frac{1.83\times 10^{-4}}{2.168}\times 10^{-2}\times 1.331\times 10^{12}=8.44\times 10^{-5}\times 10^{-2}\times 1.331\times 10^{12}$ $\approx 1.12\times 10^6{\text{W/m}}^2\approx 112{\text{W/cm}}^2.$ Right order for lunar return. ✔

(b) Convective: ${\dot{q}}_{\text{conv}}\propto R_n^{-1/2}$ → bigger nose ⇒ thicker boundary layer, gentler velocity gradient ⇒ less convective flux. Radiative: shock-layer radiation scales roughly ${\dot{q}}_{\text{rad}}\propto R_n{\rho }^a{V}^b$ (positive power of $R_n$) — a bigger, thicker glowing shock layer radiates more. So bluntness trades convective relief for radiative penalty; real designs balance the two.

Convective ($\propto V^3$): ${\dot{q}}_{\text{conv}}^{\prime }=110\times (1.1{)}^3=110\times 1.331=146.4{\text{W/cm}}^2$. Radiative ($\propto V^{8.5}$): ${\dot{q}}_{\text{rad}}^{\prime }=40\times (1.1{)}^{8.5}$. Now $(1.1{)}^{8.5}=e^{8.5\ln 1.1}=e^{8.5\times 0.09531}=e^{0.8101}=2.248$. So ${\dot{q}}_{\text{rad}}^{\prime }=40\times 2.248=89.9{\text{W/cm}}^2$.

Interpretation: A mere $10\%$ speed bump raises convective heating by $\sim 33\%$ but more than doubles radiative heating, because its exponent ($\sim 8.5$) dwarfs the convective $3$. This is exactly why radiation becomes the dominant threat at the highest reentry speeds — the $V^b$ with big $b$ overtakes $V^3$.

Formula ${\dot{q}}_s=\frac{C}{\sqrt{R_n}}\sqrt{{\rho }_{\infty }}{V}^3$; $V$ cubed (double $V$ ⇒ 8×), ${\rho }_{\infty }$ half-power, $R_n^{-1/2}$ (4× radius ⇒ half flux). Peak heating is high up (thin air), before peak deceleration. Radiation reverses the $R_n$ sign and uses a huge $V^b$.