Hypersonic flow — Mach 5+, high temperature effects: Exercises

Symbols we reuse (all defined in the parent):

$M = V / a$ — the Mach number, how many times faster than sound the flow moves.
$a = γ R T$ — the speed of sound in a gas at temperature $T$ .
$γ$ — the ratio of specific heats, $γ = c_{p} / c_{v} = 1 + 2/ f$ , where $f$ counts the active "energy storage drawers" (degrees of freedom).
$T_{0}$ — the stagnation temperature, what the gas would reach if brought to rest.
$C_{p} = \frac{p - p _{\infty}}{\frac{1}{2} ρ _{\infty} V _{\infty}^{2}}$ — the pressure coefficient, pressure rise measured in units of the stream's "dynamic push."

Level 1 — Recognition

L1.1 Is it hypersonic?

An aircraft flies at $V = 1700 m/s$ through air at $T = 220 K$ with $γ = 1.4$ and $R = 287 J/(kg\cdotK)$ . Compute $M$ . Is the flow hypersonic?

Recall Solution

WHAT: find the speed of sound, then divide. WHY: "hypersonic" is defined by $M ≳ 5$ , so we need $M = V / a$ .

$a = γ R T = 1.4 \times 287 \times 220 = 88, 396 \approx 297.3 m/s$ $M = \frac{V}{a} = \frac{1700}{297.3} \approx 5.72$

Since $M \approx 5.72 \geq 5$ , the flow is hypersonic. Real-gas heating begins to matter.

L1.2 Stagnation temperature, cold-gas estimate

For the same flight ( $M = 5.72$ , $T = 220 K$ , $γ = 1.4$ ), use the calorically-perfect formula $\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2}$ to estimate $T_{0}$ .

Recall Solution

WHAT: plug numbers into the boxed stagnation-temperature ratio from the parent note. WHY: it converts kinetic energy of the stream into a temperature the wall would feel if the gas stopped adiabatically.

$\frac{T _{0}}{T} = 1 + \frac{1.4 - 1}{2} (5.72)^{2} = 1 + 0.2 \times 32.72 = 1 + 6.544 = 7.544$ $T_{0} = 7.544 \times 220 \approx 1660 K$

That is already above the ~800 K vibration threshold and near O $_{2}$ dissociation — the perfect-gas number is only an upper estimate.

Level 2 — Application

L2.1 Newtonian pressure on an inclined face

A flat surface is inclined at $θ = 2 0^{\circ}$ to a hypersonic stream. Find its pressure coefficient using Newtonian impact theory, $C_{p} = 2 sin^{2} θ$ .

Figure — Hypersonic flow — Mach 5+, high temperature effects

Recall Solution

WHAT: substitute $θ = 2 0^{\circ}$ into the Newtonian formula. WHY this tool: the parent note showed that in a thin shock layer, particles simply lose their momentum normal to the wall; only the inclination $θ$ to the flow matters, so $C_{p} = 2 sin^{2} θ$ needs neither $M$ nor $γ$ . WHAT IT LOOKS LIKE: in the figure, only the velocity component perpendicular to the plate (the red arrow) is destroyed on impact; the parallel part slides along.

$C_{p} = 2 sin^{2} (2 0^{\circ}) = 2 (0.3420)^{2} = 2 \times 0.11698 = 0.2340$

L2.2 Density ratio ceiling

Across a normal shock as $M \to \infty$ , the density ratio approaches $\frac{ρ _{2}}{ρ _{1}} = \frac{γ + 1}{γ - 1}$ . Evaluate this ceiling for cold air ( $γ = 1.4$ ) and for dissociated air ( $γ = 1.15$ ).

Recall Solution

WHAT: plug each $γ$ into the limiting ratio. WHY: the parent note explains this comes from Rankine–Hugoniot Relations with $u_{2} / u_{1} \to (γ - 1) / (γ + 1)$ and mass conservation $ρ u = const$ .

Cold air: $\frac{ρ _{2}}{ρ _{1}} = \frac{1.4 + 1}{1.4 - 1} = \frac{2.4}{0.4} = 6.0$ Dissociated air: $\frac{ρ _{2}}{ρ _{1}} = \frac{1.15 + 1}{1.15 - 1} = \frac{2.15}{0.15} \approx 14.33$

Real-gas $γ$ falls, the ratio more than doubles, and the shock layer squeezes even thinner against the body. See Normal and Oblique Shock Waves.

Level 3 — Analysis

L3.1 Why real gas lowers temperature

At $M = 12$ , $T = 250 K$ , compare the calorically-perfect stagnation temperature ( $γ = 1.4$ ) with a real-gas estimate that uses an effective $γ_{eff} = 1.20$ (energy has been draining into vibration and dissociation). By what factor does $T_{0} / T$ drop?

Recall Solution

WHAT: compute $T_{0} / T$ twice, once per $γ$ , and take the ratio. WHY: the parent's steel-man warned that vibration and bond-breaking soak up energy, so a lower $γ$ models a smaller temperature rise for the same kinetic energy.

Perfect gas: $\frac{T _{0}}{T}_{1.4} = 1 + \frac{0.4}{2} (12)^{2} = 1 + 0.2 \times 144 = 1 + 28.8 = 29.8$ Effective: $\frac{T _{0}}{T}_{1.20} = 1 + \frac{0.20}{2} (12)^{2} = 1 + 0.10 \times 144 = 1 + 14.4 = 15.4$ Drop factor: $\frac{15.4}{29.8} \approx 0.517$

The naive tables overpredict the temperature rise by roughly a factor of two. This is exactly why thermal-protection design must use Real Gas Thermodynamics & Dissociation, not calorically-perfect Stagnation Properties & Isentropic Relations.

L3.2 Degrees of freedom bookkeeping

Air at moderate temperature has $γ = 1.40$ . When vibration fully activates, $γ$ falls to $1.29$ . Using $γ = 1 + 2/ f$ , find the effective degrees of freedom $f$ in each case and confirm that vibration added the expected amount.

Recall Solution

WHAT: invert $γ = 1 + 2/ f$ to get $f = \frac{2}{γ - 1}$ . WHY: $f$ counts active "storage drawers." Rearranging shows how many are open.

Cold air: $f = \frac{2}{1.40 - 1} = \frac{2}{0.40} = 5$ (3 translational + 2 rotational — a diatomic molecule with no vibration.)

Vibrating air: $f = \frac{2}{1.29 - 1} = \frac{2}{0.29} \approx 6.90$

$f$ rose from $5$ to about $7$ : a vibrational mode contributes two half- $R$ pieces (kinetic + potential of the oscillator), i.e. $Δ f = 2$ , matching the parent note's table entry ( $γ \to 1.29$ , $f \to 7$ ).

Level 4 — Synthesis

L4.1 Lift of a flat plate at angle of attack

A flat plate flies at Mach 8, angle of attack $α = 1 5^{\circ}$ . Using Newtonian theory: (a) find the windward $C_{p}$ ; (b) the leeward $C_{p}$ ; (c) the normal-force coefficient $C_{N} \approx (C_{p, wind} - C_{p, lee}) cos α$ ; (d) the lift coefficient $C_{L} \approx C_{N} cos α$ . Comment on the role of Mach number.

Recall Solution

WHAT: apply $C_{p} = 2 sin^{2} θ$ to each face, then project onto lift/normal directions. WHY the geometry (figure): the windward face is inclined at $θ = α = 1 5^{\circ}$ to the stream; the leeward face is in shadow. Projecting the pressure onto the body-normal and then onto the vertical gives normal force and lift.

(a) Windward: $C_{p, wind} = 2 sin^{2} (1 5^{\circ}) = 2 (0.2588)^{2} = 2 \times 0.066987 = 0.13397$ (b) Leeward (shadow): $C_{p, lee} = 0$ . (c) Normal force: $C_{N} = (0.13397 - 0) cos (1 5^{\circ}) = 0.13397 \times 0.96593 = 0.12941$ (d) Lift: $C_{L} = C_{N} cos (1 5^{\circ}) = 0.12941 \times 0.96593 = 0.12500$

Comment: nowhere did $M = 8$ appear. This is the Mach-independence principle: at high hypersonic Mach numbers the aerodynamic coefficients stop depending on $M$ . Geometry rules.

L4.2 Stagnation heating and nose radius

Stagnation heat flux scales as $q_{w} \propto \frac{1}{R _{n}}$ , with $R_{n}$ the nose radius. Capsule A has $R_{n} = 1.0 m$ ; a sharp probe B has $R_{n} = 0.02 m$ . What is the ratio $q_{w, B} / q_{w, A}$ ? Interpret.

Recall Solution

WHAT: form the ratio of the scaling law. WHY: the parent note argues blunt bodies survive re-entry because heating falls with larger nose radius. See Boundary Layers & Aerodynamic Heating.

$\frac{q _{w, B}}{q _{w, A}} = \frac{R _{n, A}}{R _{n, B}} = \frac{1.0}{0.02} = 50 \approx 7.07$

The sharp probe suffers about 7 times the stagnation heat flux of the blunt capsule — a vivid quantitative reason capsules are round and the Shuttle had a blunt belly.

Level 5 — Mastery

L5.1 Reverse-engineer an effective $γ$ from a measured density ratio

A hypersonic experiment measures a limiting normal-shock density ratio of $\frac{ρ _{2}}{ρ _{1}} = 12.0$ (much higher than the $γ = 1.4$ ceiling of 6). Deduce the effective $γ_{eff}$ describing the real, partly dissociated gas, and the effective degrees of freedom $f$ .

Recall Solution

WHAT: invert $\frac{ρ _{2}}{ρ _{1}} = \frac{γ + 1}{γ - 1}$ to solve for $γ$ . WHY: the density ratio is a direct, measurable fingerprint of $γ$ ; because real-gas chemistry lowers $γ$ , a high ratio tells us how far chemistry has progressed.

Let $r = 12.0$ . Then $r (γ - 1) = γ + 1 \Rightarrow γ (r - 1) = r + 1 \Rightarrow γ = \frac{r + 1}{r - 1} .$ $γ_{eff} = \frac{12 + 1}{12 - 1} = \frac{13}{11} \approx 1.1818$ Degrees of freedom: $f = \frac{2}{γ _{eff} - 1} = \frac{2}{0.1818} \approx 11.0$

The gas behaves as if it had ~11 active degrees of freedom — far beyond the 5 of cold diatomic air — consistent with vibration fully excited plus dissociation opening chemical storage.

L5.2 Full re-entry temperature chain

A capsule enters at $V = 7000 m/s$ through air at $T = 250 K$ , $γ = 1.4$ , $R = 287 J/(kg\cdotK)$ . (a) Find $M$ . (b) Find the calorically-perfect $T_{0}$ . (c) The real gas dumps energy into dissociation so that the actual peak temperature is only $40%$ of the perfect-gas rise above ambient. Estimate the real peak temperature and name which real-gas effects are active there.

Recall Solution

WHAT & WHY: chain the tools — Mach from speed of sound, stagnation temperature from the boxed ratio, then correct downward for real-gas energy absorption.

(a) Speed of sound and Mach: $a = 1.4 \times 287 \times 250 = 100, 450 \approx 316.9 m/s$ $M = \frac{7000}{316.9} \approx 22.09$

(b) Perfect-gas stagnation temperature: $\frac{T _{0}}{T} = 1 + 0.2 \times (22.09)^{2} = 1 + 0.2 \times 488.0 = 1 + 97.6 = 98.6$ $T_{0} = 98.6 \times 250 \approx 24, 650 K$

(c) Rise above ambient $= T_{0} - T = 24, 650 - 250 = 24, 400 K$ . Real rise $= 0.40 \times 24, 400 = 9760 K$ , so $T_{real} = 250 + 9760 = 10, 010 K .$ At ~10,000 K, air is fully dissociated and ionizing — a plasma forms, causing the re-entry communications blackout. The perfect-gas 24,650 K is physically impossible; energy is instead locked in broken bonds and free electrons. See Real Gas Thermodynamics & Dissociation.

Recall Self-test recap

Which formula is Mach-independent? ::: The Newtonian pressure coefficient $C_{p} = 2 sin^{2} θ$ — geometry only. Why does a high measured density ratio imply a low $γ$ ? ::: Because $ρ_{2} / ρ_{1} = (γ + 1) / (γ - 1)$ increases as $γ$ decreases; real-gas chemistry lowers $γ$ . What conserved quantity stays fixed even as temperature drops below the perfect-gas prediction? ::: Total enthalpy (energy) — it just redistributes into internal/chemical modes.