Question bank — Aerodynamic heating during reentry — stagnation point heat flux Chapman equation

False. ${\dot{q}}_s\propto R_n^{-1/2}$, so a smaller radius (sharper nose) increases flux. Blunt bodies push the bow shock away and thicken the boundary layer, cooling the surface.

False. Because ${\dot{q}}_s\propto V^3$, doubling $V$ multiplies the flux by $2^3=8$, not 2.

False. Density enters as $\sqrt{{\rho }_{\infty }}$ (two competing boundary-layer effects leave only a square root), so doubling it multiplies flux by $\sqrt{2}\approx 1.41$.

False. Peak heating is higher up, near ${\rho }_{\text{peak}}=\beta \sin \gamma /3H$ (here $\beta$ is the ballistic coefficient, $\gamma$ the flight-path angle, $H$ the scale height); the $V^3$ factor collapses once deceleration begins, so the product ${\rho }^{1/2}{V}^3$ peaks at intermediate altitude.

False. It is where the flow is brought fully to rest ($v\to 0$); that arrested kinetic energy is exactly why the local temperature and heat flux are maximal.

True. ${\dot{q}}_s\propto R_n^{-1/2}$, and $\sqrt{4}=2$, so 4× radius gives half the flux.

False. It gives the instantaneous heat flux (W/m²) at one instant/point. The total load requires integrating ${\dot{q}}_s$ over the whole trajectory.

False. Radiative flux scales as ${\dot{q}}_{rad}\propto R_n{\rho }^a{V}^b$ — it grows with $R_n$, the opposite of convective. Super-blunt trades convective relief for radiative penalty.

The error confuses energy content with transfer rate. Enthalpy scales as $V^2$, but the boundary-layer mass flux across the wall adds another factor $\propto V$ (continuity: ${\dot{m}}^{\prime \prime }\approx {\rho }_e{u}_e$), giving the transfer rate an exponent of 3.

${\rho }_{\infty }$ spans ~5 orders of magnitude over a reentry — e.g. $\approx 1.2{\text{kg/m}}^3$ at sea level, $\approx 4\times 10^{-4}$ at 60 km, $\approx 2\times 10^{-5}$ at 90 km. Since flux $\propto \sqrt{{\rho }_{\infty }}$, using the wrong altitude density wrecks the answer far more than a slightly wrong constant $C$; always plug the local free-stream density.

The stagnation point sees the maximum; flux falls off along the body as the flow accelerates and the boundary layer grows. The nose value sizes the worst case, not the average.

The constant encodes the gas properties ($c_p$, composition, viscosity) of Earth air. Mars (CO₂) or other atmospheres need a different $C$.

Sign error in the exponent. It must be $R_n^{-1/2}$; only a negative exponent makes larger radius reduce flux, which is the actual reason for bluntness.

They are two separate peaks. Peak heating is high and early (thin air, high speed); peak deceleration is deeper and later (dense air, slower speed). The shield and the structure face different worst-case moments.

Two factors compound: available enthalpy $\propto V^2$, times the boundary-layer mass flux delivering it $\propto V$ (from continuity, ${\dot{m}}^{\prime \prime }\approx {\rho }_e{u}_e$). Product $=V^3$.

Bluntness produces a detached bow shock standing off the surface, dumping most kinetic energy into the shocked air; it also stretches the flow over a larger radius, giving a gentler velocity gradient $\beta$ and thus a thicker boundary layer that insulates the wall.

$R_n$ sets the stagnation strain rate $\beta \propto V/R_n$ (how fast air accelerates off the nose), which controls boundary-layer thickness $\delta \propto 1/\sqrt{\beta }$. A larger radius → smaller gradient → thicker $\delta$ → lower conductive flux.

Flux depends on $V^3$; the moment braking bites, $V$ (cubed) plummets faster than rising ${\rho }^{1/2}$ can compensate, so the flux peaks earlier/higher than the g-load, which needs the dense low air.

Heat flows only down a potential gradient. The stagnation enthalpy $h_0$ tracks the hot gas temperature $T_0$; the wall enthalpy $h_w$ tracks the cooler $T_w$. If the wall were as hot as the stagnation gas, the difference (and the flux) would vanish. It's the gap that drives conduction across the boundary layer.

The extra drag is the point — a blunt body brakes early, high in thin air, and its low ${\dot{q}}_s$ (from large $R_n$) keeps peak heating survivable. Streamlining would trade a tolerable heat problem for a lethal one.

Radiation comes from the shock-heated gas glowing: emitted power climbs steeply with temperature (and $T_0\propto V^2$), and hotter gas also gets partly ionized, so more speed both heats and lights up the gas far faster than linearly — hence very large $b$, and a mild $a\sim 1$–$1.5$ from the amount of glowing gas set by density.

${\dot{q}}_s\to 0$ like $V^3$. With no relative motion there is no stagnation compression and no convective heating — consistent with the physics.

It tends to 0 as $\sqrt{{\rho }_{\infty }}$. No air means no gas to heat or transfer through, so convective flux vanishes even at high speed (though radiative and solar effects lie outside this equation).

The formula gives ${\dot{q}}_s\to 0$ as $R_n^{-1/2}$. In reality a truly flat face changes the flow structure and the stagnation-point scaling breaks down, so the equation is only trusted for finite, moderately blunt radii.

In the full form $\dot{q}\propto (h_0-h_w)$: if the wall enthalpy exceeded the stagnation enthalpy, the sign would flip (heat leaving the wall). During reentry $h_0\gg h_w$, so it stays strongly positive — but an ablating or radiating hot shield reduces the effective gap.

Not directly. It is a stagnation-point, blunt-body estimate; sharp leading edges have tiny effective $R_n$, enormous local flux, and need different (often local-flat-plate) heating models.

No. Peak heating sits at ${\rho }_{\text{peak}}=\beta \sin \gamma /3H$ (ballistic coefficient $\beta$, flight-path angle $\gamma$, scale height $H$), which is higher and earlier than peak deceleration. The two extrema are separated in both time and altitude.