Worked examples — Hypersonic flow — Mach 5+, high temperature effects

Before we start, a quick tool inventory (each was defined in the parent — we only use them):

Recall The five formulas we will lean on

• Stagnation temperature: $\frac{T_0}{T}=1+\frac{\gamma -1}{2}{M}^2$
• Mach number: $M=\frac{V}{a}$, with speed of sound $a=\sqrt{\gamma RT}$
• Limiting density ratio (normal shock): $\frac{{\rho }_2}{{\rho }_1}\to \frac{\gamma +1}{\gamma -1}$
• Degrees of freedom: $\gamma =1+\frac{2}{f}$
• Newtonian pressure coefficient (unified): $C_p=2{\sin }^2\theta$ for $\theta >0$ (surface facing the flow), and $C_p=0$ for $\theta \le 0$ (surface in shadow). We treat these as one rule with two branches, not a formula plus a patch.

Throughout, air has $R=287{\text{J kg}}^{-1}{\text{K}}^{-1}$ and (when calorically perfect) $\gamma =1.4$.

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The scenario matrix

Every hypersonic problem in this chapter is one of these cells. The examples below each carry a tag [Cell #] so you can see the whole space get covered.

Cell	Case class	What makes it tricky	Example
1	Moderate $M$ (just past 5)	plug-and-chug, but sanity-check the temperature	Ex 1
2	Large finite $M$ (re-entry)	perfect-gas answer is an overestimate	Ex 2
3	Limiting input $M\to \infty$	formula must give a finite, meaningful cap	Ex 3
4	Degenerate angle $\theta =0$	grazing flow, $C_p$ should vanish smoothly	Ex 4
5	Negative / shadow angle $\theta <0$	leeward face — must return $C_p=0$, not a negative	Ex 5
6	Full angle sweep (windward face)	check $\theta =90^{\circ }$ gives the stagnation cap	Ex 6
7	Real-gas twist (small $\gamma$)	same $M$, different $\gamma$ → very different physics	Ex 7
8	Density-ratio limit with real gas	why the shock layer collapses	Ex 8
9	Word problem (re-entry heat budget)	pick the right tool, keep units honest	Ex 9
10	Exam twist (Mach-independence)	the trap: "does doubling Mach double the force?"	Ex 10

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Ex 1 — Moderate Mach, just past the threshold [Cell 1]

Forecast: guess before computing — will $T_0$ be around 500 K, 1000 K, or 1500 K?

1. Write the ratio. $\frac{T_0}{T}=1+\frac{\gamma -1}{2}{M}^2$. Why this step? This is the steady-adiabatic energy equation solved for stagnation temperature — the only tool that turns "how fast" into "how hot."
2. Insert numbers. $\frac{\gamma -1}{2}=\frac{0.4}{2}=0.2$, and $M^2=25$, so $\frac{T_0}{T}=1+0.2\times 25=6$. Why this step? Everything on the right is now a pure number; the ratio is $6$.
3. Multiply back. $T_0=6\times 250=1500\text{K}$. Why this step? We wanted a temperature, so we un-divide by the free-stream $T$.

Verify: at exactly $M=5$ the ratio is $1+0.2(25)=6$, a clean small integer — good. $1500\text{K}$ sits right at the doorstep of ${\text{O}}_2$ dissociation, which is exactly why $M\approx 5$ is called the hypersonic threshold. The forecast answer is the third option.

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Ex 2 — Large Mach: where the perfect-gas number lies [Cell 2]

Forecast: will the perfect-gas $T_0$ be nearer 10 000 K, 20 000 K, or 40 000 K?

1. Perfect-gas prediction. $\frac{T_0}{T}=1+0.2\times (20{)}^2=1+0.2\times 400=81$. Why this step? Same formula as Ex 1 — we deliberately apply it out of its comfort zone to expose the error.
2. Multiply back. $T_0=81\times 250=20250\text{K}$. Why this step? That is hotter than the Sun's surface — a red flag that the model is broken.
3. Physical correction. Above $\sim 2500\text{K}$ the air dissociates: energy leaves the thermal pool to break O${}_2$ and N${}_2$ bonds (see Real Gas Thermodynamics & Dissociation). Why this step? Total enthalpy is still conserved, but it is now stored in chemistry, not in temperature — so the real $T$ might be only $\sim 6000\text{–}8000\text{K}$.

Verify: the ratio $81$ is $1+0.2M^2$ evaluated at $M=20$; check $0.2\times 400=80$, plus $1$ gives $81$ — arithmetic holds. The lesson: the number is mathematically correct but physically an overestimate, exactly the [!mistake] warned about in the parent note.

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Ex 3 — The limiting input $M\to \infty$ [Cell 3]

Forecast: does ${\rho }_2/{\rho }_1$ grow without bound, or settle to a finite ceiling?

1. State the limit. From Rankine–Hugoniot Relations, as $M\to \infty$, $\frac{{\rho }_2}{{\rho }_1}\to \frac{\gamma +1}{\gamma -1}$. Why this step? Unlike temperature (which grows with $M^2$), density is capped — mass conservation plus the momentum jump forbid infinite compression.
2. Evaluate at $\gamma =1.4$. $\frac{1.4+1}{1.4-1}=\frac{2.4}{0.4}=6$. Why this step? This is the famous factor-of-6 ceiling for cold air.
3. Evaluate at $\gamma =1.1$ (real-gas softened). $\frac{1.1+1}{1.1-1}=\frac{2.1}{0.1}=21$. Why this step? Real-gas effects shrink $\gamma$, so the same infinite Mach squeezes the gas ~3.5× harder → an even thinner shock layer.

Verify: plug $\gamma =1.4$: numerator $2.4$, denominator $0.4$, ratio $6.0$ ✓. Sanity check the trend: smaller $\gamma -1$ in the denominator ⇒ bigger ratio ⇒ $21>6$ ✓. The answer is finite (the first forecast is wrong): density never diverges.

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Ex 4 — Degenerate angle $\theta =0$ (grazing flow) [Cell 4]

Forecast: zero, or some small positive residual?

Figure — Ex 4 (grazing flow): the lavender arrows show the stream running parallel to the solid body; a coral note marks that no arrow points into the wall, so no normal momentum is delivered. This is the geometry that makes $\theta =0$ the boundary between the two branches of the unified rule.

1. Read the geometry. In the figure the incoming particles (lavender arrows) move along the surface; none of their momentum points into the wall. Why this step? Newtonian pressure comes from normal momentum destroyed — if the normal component is zero, no momentum is delivered.
2. Apply the unified rule at the boundary. At $\theta =0$ both branches agree: the windward branch gives $2{\sin }^2{0}^{\circ }=0$, and the shadow branch gives $0$. So $C_p=0$. Why this step? $\theta =0$ is exactly the seam between "facing the flow" and "in shadow"; a good rule must be continuous there, and this one is.

Verify: units-free coefficient, and $2{\sin }^2(0)=0$ ✓. The limit is smooth: for a tiny angle $\theta =1^{\circ }$, $C_p=2{\sin }^2{1}^{\circ }\approx 0.00061$ — nearly zero, confirming no jump.

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Ex 5 — Shadow / negative angle $\theta <0$ (leeward face) [Cell 5]

Forecast: does the formula give a negative $C_p$, or do we override it?

Figure — Ex 5 (shadow face): the wedge presents a mint windward face (struck by the lavender stream) and a coral leeward face pointing away from the flow. The coral annotation marks the $\theta \le 0$ branch of the unified rule, where the model predicts zero impacts and hence $C_p=0$.

1. Spot the shadow. In the figure, the coral face faces away from the stream: no particles can reach it (the thin shock layer wraps the front only). Why this step? Newtonian theory is a particle-impact model — a surface receiving zero impacts feels zero excess pressure, by construction.
2. Select the correct branch — no override needed. Because $\theta =-20^{\circ }\le 0$, the unified rule itself hands us $C_p=0$. We never evaluate $2{\sin }^2\theta$ here; that branch simply does not apply. Why this step? The single signed rule already contains the shadow case, so there is nothing ad hoc to patch — we just read off the branch selected by the sign of $\theta$.
3. Why blind squaring would mislead. If someone forgot the branching and plugged into $2{\sin }^2\theta$, they would get $2{\sin }^2(-20^{\circ })=2{\sin }^2(20^{\circ })\approx 0.234$, because ${\sin }^2$ is even and discards the sign. The unified rule prevents this by reading the sign first. Why this step? It shows precisely which mistake the branching guards against.

Verify: the physical answer is $C_p=0$ (the correct branch), not $0.234$. Sanity check: pressure on a shadowed face sits at free-stream $p_{\infty }$, i.e. $p-p_{\infty }=0\Rightarrow C_p=0$ ✓.

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Ex 6 — Full windward sweep: the $\theta =90^{\circ }$ cap [Cell 6]

Forecast: guess the largest possible $C_p$ this theory allows.

Figure — Ex 6 ($C_p$ sweep): the lavender curve plots the windward branch $C_p=2{\sin }^2\theta$ from $0$ to $90^{\circ }$. Three highlighted dots (mint, butter, coral) mark our three angles, and the dashed coral ceiling shows the stagnation cap $C_p=2$ that the curve reaches at $\theta =90^{\circ }$.

1. $\theta =15^{\circ }$. $C_p=2{\sin }^{2}15^{\circ }=2(0.2588{)}^2=0.134$. Why this step? Matches the parent note's flat-plate example — a good cross-check that our tool is wired correctly.
2. $\theta =45^{\circ }$. $C_p=2{\sin }^{2}45^{\circ }=2(0.7071{)}^2=2(0.5)=1.0$. Why this step? Half-way in angle, but ${\sin }^2$ is nonlinear, so we land at exactly $1$.
3. $\theta =90^{\circ }$ (surface square to the flow — the stagnation point). $C_p=2{\sin }^{2}90^{\circ }=2(1{)}^2=2$. Why this step? This is the maximum windward $C_p$; the whole normal momentum is destroyed. The curve in the figure peaks here.

Verify: $2{\sin }^{2}90^{\circ }=2$ is the ceiling because ${\sin }^2\le 1$ ✓. Monotone rise $0.134\to 1.0\to 2$ matches the figure's climbing curve ✓. The forecast maximum is $C_p=2$.

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Ex 7 — Real-gas twist: same Mach, softer $\gamma$ [Cell 7]

Forecast: does the smaller $\gamma$ raise or lower the predicted ratio?

1. $\gamma =1.4$ case. $\frac{T_0}{T}=1+\frac{0.4}{2}(25)=1+5=6$. Why this step? Baseline from Ex 1 — we reuse it so only $\gamma$ changes.
2. $\gamma =1.2$ case. $\frac{\gamma -1}{2}=\frac{0.2}{2}=0.1$, so $\frac{T_0}{T}=1+0.1(25)=1+2.5=3.5$. Why this step? A smaller $\gamma$ means more degrees of freedom soaking up energy, so the same kinetic energy raises temperature less — the ratio falls from $6$ to $3.5$.
3. Interpret via $f$. From $\gamma =1+2/f$: $\gamma =1.4\Rightarrow f=5$; $\gamma =1.2\Rightarrow f=2/(0.2)=10$. Why this step? Ten active "storage drawers" vs five — twice the places to hide energy, so much lower temperature. Consistent with Real Gas Thermodynamics & Dissociation.

Verify: $1+0.1\times 25=3.5$ ✓ and $f=2/(\gamma -1)=2/0.2=10$ ✓. Trend check: lower $\gamma$ ⇒ lower $T_0/T$ ✓ (the forecast "lower" is correct).

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Ex 8 — Density-ratio limit with a real-gas $\gamma$ [Cell 8]

Forecast: roughly how many times denser than cold air's limit?

1. Apply the limit. $\frac{{\rho }_2}{{\rho }_1}\to \frac{\gamma +1}{\gamma -1}=\frac{2.2}{0.2}=11$. Why this step? Same $M\to \infty$ normal-shock limit as Ex 3, now with the hot-gas $\gamma$.
2. Compare. Cold air caps at $6$; hot gas reaches $11$, a factor $11/6\approx 1.83$. Why this step? More compression means the same post-shock mass occupies less volume.
3. Geometric consequence. More density ⇒ thinner shock layer ⇒ the bow shock hugs the body even closer, intensifying the wall heating problem. Why this step? Mass conservation ties density directly to layer thickness.

Verify: $\frac{2.2}{0.2}=11$ ✓ and $11/6=1.833\dots$ ✓. The forecast "roughly 2×" is right.

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Ex 9 — Word problem: a re-entry heat budget [Cell 9]

Here $T_{\infty }$, $a_{\infty }$, $V_{\infty }$ all carry the subscript $\infty$ to mean "in the undisturbed free stream," far ahead of the body. The sound-speed formula uses the same free-stream temperature $T_{\infty }$ throughout — no symbol switching.

Forecast: will the kinetic energy per kilogram be around 1 MJ, 10 MJ, or 30 MJ?

1. Free-stream speed of sound. $a_{\infty }=\sqrt{1.4\times 287\times 250}=\sqrt{100450}\approx 316.9\text{m/s}$. Why this step? Mach number is a ratio; to get a real speed we need the local (here free-stream) sound speed — the yardstick $M$ is measured against.
2. Free-stream speed. $V_{\infty }=M{a}_{\infty }=25\times 316.9\approx 7923\text{m/s}$. Why this step? $M=V_{\infty }/a_{\infty }$ rearranged; $\sim 7.9\text{km/s}$ is a realistic orbital re-entry speed — a good reality anchor.
3. Kinetic energy per kg. $\frac{1}{2}{V}_{\infty }^2=0.5\times (7923{)}^2\approx 3.14\times 10^7\text{J/kg}=31.4\text{MJ/kg}$. Why this step? By the steady energy equation this is exactly the enthalpy the gas must absorb at stagnation — the "heat budget."
4. Where it goes. Only part becomes temperature; the rest breaks bonds (dissociation) and lifts electrons (ionization). The wall heating and the radio blackout both come from this pool. Why this step? Ties the number back to the physics ladder in the parent note.

Verify: $a_{\infty }=\sqrt{1.4\cdot 287\cdot 250}\approx 316.9$ m/s ✓; $V_{\infty }=25a_{\infty }\approx 7923$ m/s ✓; $\frac{1}{2}{V}_{\infty }^2\approx 3.14\times 10^7$ J/kg ✓. Units: $(\text{m/s}{)}^2={\text{m}}^2/{\text{s}}^2=\text{J/kg}$ ✓. The forecast "≈30 MJ" is correct.

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Ex 10 — Exam twist: does doubling Mach double the force? [Cell 10]

Forecast: factor 1 (unchanged), factor 2, or factor 4?

1. Compute $C_p$ at $M=8$. With $\theta =15^{\circ }>0$ we take the windward branch: $C_p=2{\sin }^{2}15^{\circ }=0.134$. Why this step? We evaluate the actual unified rule before reasoning about scaling.
2. Compute $C_p$ at $M=16$. Still $C_p=2{\sin }^{2}15^{\circ }=0.134$ — $M$ does not appear in the rule. Why this step? This is the Mach-independence principle: at hypersonic speed the pressure coefficient depends only on geometry ($\theta$), not on Mach number.
3. Diagnose the trap. The student confused $C_p$ (a coefficient, already normalized by $\frac{1}{2}{\rho }_{\infty }{V}_{\infty }^2$) with the dimensional pressure. The pressure $p-p_{\infty }$ does scale with $V_{\infty }^2\propto M^2$, so it grows $\times 4$ — but the coefficient is unchanged. Why this step? Separating the coefficient from the raw force is the whole point of the exam twist.

Verify: $C_p(M=8)=C_p(M=16)=0.134$, ratio $=1$ ✓. Meanwhile $p-p_{\infty }\propto V_{\infty }^2$, so doubling $M$ (hence $V_{\infty }$) multiplies pressure by $2^2=4$ ✓. The correct answer to the coefficient question is factor 1 — the student was wrong.

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Recall Quick self-test across the matrix

At $M=5$, $\gamma =1.4$, what is $T_0/T$? ::: $6$. As $M\to \infty$ with $\gamma =1.4$, the shock density ratio caps at? ::: $6$. $C_p$ on a shadowed (leeward, $\theta \le 0$) face? ::: $0$ (shadow branch of the unified rule). Maximum windward $C_p$ (surface square to the flow, $\theta =90^{\circ }$)? ::: $2$. Doubling $M$ changes $C_p$ by what factor? ::: $\times 1$ (unchanged — Mach independence). Doubling $M$ changes the actual pressure $p-p_{\infty }$ by? ::: $\times 4$ (it scales as $V_{\infty }^2$).