Aerodynamic heating — recovery temperature, heat flux: Question bank

True or false — justify

The gas at the wall is brought fully to rest, so the wall must sit at the stagnation temperature

T_{0}

False. No-slip stops the bulk motion, but conduction leaks dissipated heat sideways (

P r < 1

), so the adiabatic wall settles at

T_{r} = T_{e} (1 + r \frac{γ - 1}{2} M_{e}^{2})

with

r < 1

— a bit below

T_{0}

At an uncooled (adiabatic) wall the convective heat flux is zero.

True. Adiabatic means no heat is drawn out, so the wall floats up to

T_{r}

; then

T_{r} - T_{w} = 0

makes

\overset{q}{˙}_{w} = h (T_{r} - T_{w}) = 0

. This is precisely how

T_{r}

is defined. See Boundary layers & viscous dissipation.

The recovery factor

r

is always less than 1.

True for air and most gases because

P r < 1

(heat diffuses faster than momentum, so some dissipated heat escapes before being recovered). For gases with

P r > 1

you can get

r > 1

, so "less than 1" is a

P r

-fact, not a law. See Prandtl number & thermal boundary layer.

If you cool the wall harder, the recovery temperature

T_{r}

drops with it.

False.

T_{r}

depends only on the flow —

M_{e}

T_{e}

r

— never on

T_{w}

. Cooling lowers

T_{w}

, which actually raises the flux

h (T_{r} - T_{w})

Turbulent flow always produces a higher recovery temperature than laminar flow.

True in the usual sense:

r_{turb} = P r^{1/3} \approx 0.89 > r_{lam} = P r \approx 0.84

for air, so turbulent

T_{r}

is slightly higher — and turbulence also raises

h

, so the heat flux jumps far more.

The heat-transfer coefficient

h

is a property of the material the wall is made of.

False.

h

characterises the flow's ability to carry heat to/from the surface (velocity, density, boundary-layer state), packaged e.g. via the Stanton number

S t

. The wall material affects how it responds, not

h

. See Reynolds analogy & Stanton number.

For a fixed Mach number, doubling the free-stream static temperature

T_{e}

roughly doubles

T_{r}

True.

T_{r} = T_{e} (1 + r \frac{γ - 1}{2} M_{e}^{2})

is linear in

T_{e}

with the bracket held fixed, so

T_{r}

scales with

T_{e}

. This is why post-shock heating is severe: shocks raise

T_{e}

before these formulas even apply. See Normal & oblique shock heating.

The recovery factor depends on Mach number.

False (to leading order).

r

is set by the Prandtl number and boundary-layer state (laminar/turbulent), not by

M

. The Mach number enters only through the

\frac{γ - 1}{2} M_{e}^{2}

dynamic-temperature term, separately from

r

Spot the error

"The driving temperature difference for heat flux is

T_{0} - T_{w}

because

T_{0}

is the hottest temperature in the flow."

The correct driver is

T_{r} - T_{w}

T_{0}

over-predicts heating, and it can't be right because it would give a nonzero flux at the adiabatic wall (

T_{w} = T_{r} < T_{0}

) — contradicting the definition of an insulated surface.

"Since air is brought to rest at the wall,

T_{r} = T_{e} (1 + \frac{γ - 1}{2} M_{e}^{2})

The recovery factor

r

is missing. The correct form is

T_{r} = T_{e} (1 + r \frac{γ - 1}{2} M_{e}^{2})

; dropping

r

silently assumes

r = 1

(full recovery), which is the

T_{0}

mistake in disguise.

r = \frac{T _{r} - T _{0}}{T _{e} - T _{0}}

The definition is inverted. It should be

r = \frac{T _{r} - T _{e}}{T _{0} - T _{e}}

— the fraction of the dynamic rise (

T_{0} - T_{e}

) that is actually recovered above the edge temperature.

"For laminar flow use

r = P r^{1/3}

and for turbulent use

r = P r

The two are swapped. Laminar is the square root,

r = P r

; turbulent is the cube root,

r = P r^{1/3}

. Mnemonic: "Lam = root, Turb = cube."

"Because

\overset{q}{˙}_{w} = h (T_{r} - T_{w})

, if we make the wall adiabatic the coefficient

h

must go to zero."

No —

h

is a flow property and stays finite. The flux vanishes because the temperature difference

T_{r} - T_{w}

goes to zero when

T_{w}

rises to

T_{r}

, not because

h

changes.

"Recovery temperature and stagnation temperature are the same thing since both come from stopping the flow."

They differ by the recovery factor:

T_{0}

assumes ideal adiabatic deceleration with no sideways heat loss, while

T_{r}

is what a real boundary layer delivers to the wall, giving

T_{r} < T_{0}

when

P r < 1

"Newton's law of cooling

\overset{q}{˙}_{w} = h (T_{r} - T_{w})

tells us the wall always cools down."

It only cools if

T_{w} > T_{r}

. When

T_{w} < T_{r}

the flux is positive — heat flows into the wall and it heats up; "cooling" is just the name of the law, not the guaranteed direction.

Why questions

Why is

T_{r}

between

T_{e}

and

T_{0}

rather than at one extreme?

Inside the boundary layer viscous dissipation adds heat (pushing toward

T_{0}

) while conduction carries heat away sideways (pushing toward

T_{e}

); the wall settles at the balance point, so

T_{e} < T_{r} < T_{0}

(for

P r < 1

Why do we use

T_{r}

— not

T_{e}

T_{0}

— as the reference for heat flux?

T_{r}

is the temperature an uncooled wall would actually reach, so it is the true thermal "potential" the gas drives the wall toward; using it makes the adiabatic case (

T_{w} = T_{r}

) correctly give zero flux.

Why does

r < 1

follow from

P r < 1

P r = μ c_{p} / k

compares momentum diffusion to heat diffusion;

P r < 1

means heat diffuses faster, so dissipated energy escapes the near-wall region before it can all be recovered as wall heating — recovering less than 100%. See Prandtl number & thermal boundary layer.

Why is transition from laminar to turbulent flow so critical for thermal-protection design?

Turbulence raises both the recovery factor (slightly) and the heat-transfer coefficient

h

(often 3–5×), so the location where transition happens can multiply the local heat load, dictating where the shield must be thickest. See Hypersonic re-entry & thermal protection systems.

Why does the Stanton-number form

\overset{q}{˙}_{w} = ρ_{e} u_{e} c_{p} S t (T_{r} - T_{w})

get used in engineering?

It packages

h

into a dimensionless number that collapses onto correlations (e.g. Reynolds analogy

S t \approx C_{f} /2

), letting you predict heating directly from skin friction. See Reynolds analogy & Stanton number.

Why does the stagnation relation

T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}

come out purely in terms of Mach number?

Energy conservation gives

T_{0} = T (1 + \frac{u ^{2}}{2 c _{p} T})

; substituting

c_{p} = \frac{γ R}{γ - 1}

and

a^{2} = γ R T

turns

u^{2} / a^{2}

into

M^{2}

, the natural compressible-flow variable. See Stagnation properties & isentropic relations.

Why can cooling the wall increase the heat you must remove, seemingly making the problem worse?

Because

T_{r}

is fixed by the flow, lowering

T_{w}

enlarges the difference

T_{r} - T_{w}

, and

\overset{q}{˙}_{w} = h (T_{r} - T_{w})

grows — colder walls draw a bigger heat flux, though they keep the structure safe.

Edge cases

What is

T_{r}

in the limit

M_{e} \to 0

(very slow flow)?

The dynamic term

r \frac{γ - 1}{2} M_{e}^{2} \to 0

, so

T_{r} \to T_{e}

— no appreciable heating, the wall just sits at the local static temperature.

What happens to the heat flux if the wall is deliberately held above

T_{r}

Then

T_{r} - T_{w} < 0

, so

\overset{q}{˙}_{w} < 0

: the flux reverses and the wall loses heat to the gas (e.g. a hot surface radiating/convecting into cooler recovered flow).

If a hypothetical gas had

P r = 1

exactly, what would

r

be?

Both formulas give

r = 1 = 1^{1/3} = 1

, so

T_{r} = T_{0}

— momentum and heat diffuse equally, nothing escapes, and full recovery is achieved.

For a gas with

P r > 1

(some oils/dense gases), can

T_{r}

exceed

T_{0}

Yes — then

r > 1

, and

T_{r} = T_{e} (1 + r \frac{γ - 1}{2} M_{e}^{2})

can rise above

T_{0}

; the "recovery" over-shoots because heat diffuses slower than momentum, trapping dissipated energy near the wall.

At the exact adiabatic operating point

T_{w} = T_{r}

, is the boundary layer still dissipating energy?

Yes — viscous dissipation continues; it is exactly balanced by sideways conduction so that net flux into the wall is zero. Zero flux does not mean zero physics, just a steady balance.

γ \to 1

(a gas with many internal energy modes), what happens to the heating term?

The factor

\frac{γ - 1}{2} \to 0

, so both

T_{0}

and

T_{r}

approach

T_{e}

— such a gas soaks kinetic energy into internal modes with little temperature rise.

Recall One-line self-test before you leave

Cover and answer: Which temperature drives the wall heat flux, and why not the hotter $T_{0}$ ? $T_{r} - T_{w}$ drives it, because $T_{r}$ (not $T_{0}$ ) is what an uncooled wall actually reaches, so it correctly gives zero flux at the adiabatic wall. ::: Correct if you named $T_{r}$ and gave the adiabatic-consistency reason.