Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂ - P₀₁: Worked examples

This page is the drill ground for the parent note. There we derived the shock relations; here we use them until every possible flavour of problem feels familiar. We keep $γ = 1.4$ (air) unless a problem says otherwise.

The scenario matrix

Before we solve anything, let's list the kinds of situation the topic can hand you. Every worked example below is tagged with the cell it lands in.

Cell	Case class	Defining feature	Example
A	Standard moderate shock	$1 < M_{1} ≲ 3$ , plug-and-chug	Ex 1
B	Weak-shock limit	$M_{1} \to 1^{+}$ , everything $\to 1$	Ex 2
C	Strong / hypersonic limit	$M_{1} \to \infty$ , density saturates	Ex 3
D	Degenerate: $M_{1} = 1$ exactly	The "shock" is a sound wave, no jump	Ex 4
E	Forbidden direction	Given $M_{1} < 1$ — does a shock exist?	Ex 5
F	Inverse / back-solve	Given a ratio, find $M_{1}$	Ex 6
G	Non-air gas	$γ \neq = 1.4$ (helium, combustion gas)	Ex 7
H	Real-world word problem	Dimensional data, find actual $P, T, u$	Ex 8
I	Exam twist	Stagnation-pressure loss + inlet design	Ex 9

The tools we need are only the boxed formulas from the parent. A few symbols need naming before we start. $u$ = the flow speed (how fast the gas moves, in m/s — distinct from the Mach number $M$ , which is that speed divided by the local speed of sound). $Δ s$ = the entropy change across the shock (a measure of irreversibility; positive means "messier, less recoverable"). $c_{p}$ = the specific heat at constant pressure (the energy in joules needed to warm 1 kg of the gas by 1 K while it is free to expand; for air $c_{p} \approx 1005 J/kg\cdotK$ ). $R$ = the specific gas constant (for air $R = 287 J/kg\cdotK$ ); the two are linked by $c_{p} = \frac{γ}{γ - 1} R$ . For quick reference:

Intuition Where the

$P_{02} / P_{01}$ formula comes from (the "why") Start from the entropy rise of an ideal gas, $Δ s = c_{p} ln \frac{T _{2}}{T _{1}} - R ln \frac{P _{2}}{P _{1}}$ — where (as just defined) $c_{p}$ is the constant-pressure specific heat and $R$ the specific gas constant. The parent note shows that stagnation pressure and entropy are tied by $\frac{P _{02}}{P _{01}} = exp (- \frac{Δ s}{R})$ — more entropy means less stagnation pressure, always. Now substitute the shock's own $T_{2} / T_{1}$ and $P_{2} / P_{1}$ into $Δ s$ and simplify: the exponential collapses into the two-bracket product above. So the first bracket carries the temperature side of the entropy and the second bracket the pressure side; the exponents $\frac{γ}{γ - 1}$ and $\frac{1}{γ - 1}$ are just $c_{p} / R$ and $1/ (γ - 1)$ in disguise. You never need to re-derive it — but now you know each piece is an entropy term, not a magic constant.

Here $M$ = Mach number (flow speed ÷ speed of sound), subscript $1$ = upstream (before the shock, always supersonic), subscript $2$ = downstream (after, always subsonic). $P$ = static pressure, $T$ = static temperature, $ρ$ = density, $P_{0}$ = stagnation pressure. See Speed of Sound and Mach Number and Stagnation Properties T0 and P0 if any of these feel shaky.

The figure below is our map of the whole topic (this is the "alt text": a labelled plot of all five downstream/upstream property ratios versus the single input $M_{1}$ ). The horizontal axis is the upstream Mach number $M_{1}$ (from $1$ to $6$ ). The vertical axis is the downstream-to-upstream ratio of a property. Each coloured curve is one property: magenta $P_{2} / P_{1}$ and orange $T_{2} / T_{1}$ climb steeply (pressure and temperature soar with shock strength); violet $ρ_{2} / ρ_{1}$ climbs then flattens against the dotted density ceiling at $6$ ; the dashed navy curve $M_{2}$ dips below the horizontal line at $1$ — proof the flow is always subsonic after the shock; and the teal $P_{02} / P_{01}$ curve sinks toward zero as the shock gets stronger (the entropy loss growing). Read it left-to-right as "turning up the shock strength." Every worked example below is just a single vertical slice through this one picture.

Figure — Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂ - P₀₁

Figure 1 — The master map: all five normal-shock ratios as functions of $M_{1}$ ( $γ = 1.4$ ). Pick an $M_{1}$ on the horizontal axis and read every downstream property off the curves.

Example 1 — Cell A · Standard moderate shock

Worked example A shock stands in air at

$M_{1} = 2.5$ , $γ = 1.4$ . Find $M_{2}$ , $P_{2} / P_{1}$ , $ρ_{2} / ρ_{1}$ , $T_{2} / T_{1}$ , and $P_{02} / P_{01}$ .

Forecast: Before computing — guess. $M_{1} = 2.5$ is a fairly strong shock. Will $M_{2}$ be close to $1$ or close to $0.4$ ? Will the pressure roughly quadruple or roughly seven-fold? Write your guesses, then check.

Step 1 — Downstream Mach number. With $M_{1}^{2} = 6.25$ : $M_{2}^{2} = \frac{1 + 0.2 ( 6.25 )}{1.4 ( 6.25 ) - 0.2} = \frac{1 + 1.25}{8.75 - 0.2} = \frac{2.25}{8.55} = 0.2632$ so $M_{2} = 0.513$ . Why this step? Every other ratio is anchored to $M_{1}$ , but $M_{2}$ is the headline physical result — it confirms the flow went subsonic, which is the defining behaviour of a shock.

Step 2 — Pressure ratio. $M_{1}^{2} - 1 = 5.25$ : $\frac{P _{2}}{P _{1}} = 1 + \frac{2 ( 1.4 )}{2.4} (5.25) = 1 + \frac{2.8}{2.4} (5.25) = 1 + 6.125 = 7.125$ Why this step? The Rankine–Hugoniot pressure jump (see Rankine–Hugoniot Relations) is the most-used engineering number: it tells you the force the shock exerts.

Step 3 — Density ratio. $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 6.25 )}{0.4 ( 6.25 ) + 2} = \frac{15}{4.5} = 3.333$ Why this step? Density and flow speed are locked together by mass conservation ( $ρ_{2} / ρ_{1} = u_{1} / u_{2}$ ), so this one number also tells us the flow slowed by a factor of $3.333$ .

Step 4 — Temperature ratio via the ideal-gas link $T_{2} / T_{1} = (P_{2} / P_{1}) / (ρ_{2} / ρ_{1})$ : $\frac{T _{2}}{T _{1}} = \frac{7.125}{3.333} = 2.1375$ Why this step? We avoid re-deriving temperature from scratch — the ideal gas law lets pressure and density hand us temperature for free.

Step 5 — Stagnation-pressure loss. Substitute $M_{1} = 2.5$ into the $P_{02} / P_{01}$ formula. First bracket base: $\frac{( 2.4 ) ( 6.25 ) /2}{1 + 0.2 ( 6.25 )} = \frac{7.5}{2.25} = 3.333$ , raised to $\frac{γ}{γ - 1} = 3.5$ gives $3.33 3^{3.5} = 67.6$ . Second bracket base: $\frac{2.4}{2 ( 1.4 ) ( 6.25 ) - 0.4} = \frac{2.4}{17.1} = 0.1404$ , raised to $\frac{1}{γ - 1} = 2.5$ gives $0.140 4^{2.5} = 0.00738$ . Product: $\frac{P _{02}}{P _{01}} = 67.6 \times 0.00738 = 0.499$ so about half the stagnation pressure is destroyed. Why this step? This is the entropy fingerprint — the irreversibility made numeric (see Entropy and Second Law).

Verify: All static ratios exceed 1 (compression ✓), $M_{2} < 1$ (subsonic ✓), $P_{02} / P_{01} < 1$ (2nd law ✓). Cross-check with the energy-conservation form of $T_{2} / T_{1}$ introduced above: $\frac{1 + 0.2 M _{1}^{2}}{1 + 0.2 M _{2}^{2}} = \frac{2.25}{1 + 0.2 ( 0.2632 )} = \frac{2.25}{1.0526} = 2.137$ ✓ — the ideal-gas route (Step 4) and the energy route agree.

Example 2 — Cell B · The weak-shock limit

Worked example Air flows at

$M_{1} = 1.05$ (barely supersonic). Show that every jump is tiny — including $M_{2}$ and $T_{2} / T_{1}$ — and that the shock is almost isentropic.

Forecast: With $M_{1}$ a whisker above 1, guess: are the ratios near $1.0$ or near $1.5$ ? Will $M_{2}$ be just under 1? And is the stagnation-pressure loss "a few percent" or "a few tenths of a percent"?

Step 1 — Downstream Mach number. $M_{1}^{2} = 1.1025$ : $M_{2}^{2} = \frac{1 + 0.2 ( 1.1025 )}{1.4 ( 1.1025 ) - 0.2} = \frac{1.2205}{1.3435} = 0.9084 \Rightarrow M_{2} = 0.953$ Why this step? $M_{2}$ is only just below 1 — the flow barely crossed the sonic line, confirming this is the gentlest possible shock.

Step 2 — Pressure ratio. $M_{1}^{2} - 1 = 0.1025$ : $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (0.1025) = 1 + 0.1196 = 1.1196$ Why this step? It directly measures "how far from no-shock" we are; a value of $1.12$ is close to 1 — a gentle squeeze.

Step 3 — Density ratio. $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 1.1025 )}{0.4 ( 1.1025 ) + 2} = \frac{2.646}{2.441} = 1.0840$ Why this step? Confirms density also barely moves.

Step 4 — Temperature ratio via the ideal-gas link: $\frac{T _{2}}{T _{1}} = \frac{1.1196}{1.0840} = 1.0328$ Why this step? Temperature rises only about $3%$ — the "tiny jump" claim must hold for every property, not just pressure and density, so we compute $T_{2} / T_{1}$ explicitly to complete the picture.

Step 5 — Stagnation-pressure loss. Substitute $M_{1} = 1.05$ into the two-bracket formula: $\frac{P _{02}}{P _{01}} = 0.99987$ a loss of about 0.013%. Why this step? This is the punchline: entropy rises like the cube $(M_{1}^{2} - 1)^{3}$ near $M_{1} = 1$ , so a weak shock is nearly reversible — it behaves almost like a sound wave in Isentropic Flow Relations.

Verify: As $M_{1} \to 1^{+}$ the parent note says all ratios $\to 1$ . Here $M_{2} = 0.953$ (just below 1 ✓), and $P_{2} / P_{1}$ , $ρ_{2} / ρ_{1}$ , $T_{2} / T_{1}$ all sit just above 1 while $P_{02} / P_{01}$ is just below 1 — every jump is tiny, consistent with the limit. ✓

Example 3 — Cell C · Strong / hypersonic limit

Worked example A re-entry capsule sits behind a shock at

$M_{1} = 12$ . Find the density ratio and compare it to the theoretical ceiling. Find $P_{2} / P_{1}$ and $P_{02} / P_{01}$ .

Forecast: The parent note says density saturates at $\frac{γ + 1}{γ - 1} = 6$ . At $M_{1} = 12$ , guess: is $ρ_{2} / ρ_{1}$ already at $5.9$ , or only $4$ ? And is the pressure jump in the tens or the hundreds?

Step 1 — Density ratio. $M_{1}^{2} = 144$ : $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 144 )}{0.4 ( 144 ) + 2} = \frac{345.6}{59.6} = 5.799$ Why this step? It tests the saturation claim; a value of $5.799$ is within $3.4%$ of the ceiling $6$ — density is essentially maxed out.

Step 2 — Pressure ratio. $M_{1}^{2} - 1 = 143$ : $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (143) = 1 + 166.83 = 167.83$ Why this step? Shows the contrast: pressure keeps climbing without bound while density flattens. Excess energy has nowhere to go but $P$ and $T$ .

Step 3 — Stagnation-pressure loss. Substitute $M_{1} = 12$ into the two-bracket formula: $\frac{P _{02}}{P _{01}} = 0.001645$ — over 99.8% of the stagnation pressure is gone. Why this step? This is precisely why hypersonic inlets never use a single normal shock (see Supersonic Inlets and Diffusers).

Verify: A ratio $ρ_{2} / ρ_{1} < 6$ sits below the ceiling ✓, and it should approach 6 as $M_{1}$ grows; the value $5.799 < 6$ and is larger than the $M_{1} = 5$ value $5.0$ from the parent's Example 2 ✓. And $P_{02} / P_{01} ≪ 1$ ✓.

Example 4 — Cell D · Degenerate case $M_{1} = 1$

Worked example What does the shock do at exactly

$M_{1} = 1$ ? Compute $M_{2}$ , $P_{2} / P_{1}$ , $ρ_{2} / ρ_{1}$ .

Forecast: $M_{1} = 1$ is the boundary between "shock possible" and "shock forbidden." Guess: does $M_{2}$ come out as $1$ , or as something $< 1$ ? Do the ratios equal $1$ exactly?

Step 1 — $M_{2}$ . $M_{1}^{2} = 1$ : $M_{2}^{2} = \frac{1 + 0.2 ( 1 )}{1.4 ( 1 ) - 0.2} = \frac{1.2}{1.2} = 1 \Rightarrow M_{2} = 1$ Why this step? At the boundary the two roots of the shock equation merge: the "shock branch" and the "no-shock branch" become the same point. There is no jump.

Step 2 — Pressure ratio. Here $M_{1}^{2} - 1 = 0$ : $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (0) = 1$ Why this step? Confirms zero pressure jump — the "shock" is infinitely weak.

Step 3 — Density ratio. $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 1 )}{0.4 ( 1 ) + 2} = \frac{2.4}{2.4} = 1$ Why this step? Density is unchanged too.

Verify: All ratios are exactly $1$ and $M_{2} = M_{1} = 1$ . This is the degenerate "sound wave" case — physically it means no shock at all, just an isentropic sonic point. Consistent with the weak-shock limit of Example 2. ✓

Example 5 — Cell E · The forbidden direction

Worked example A student claims a normal shock decelerates subsonic air from

$M_{1} = 0.6$ to some $M_{2}$ . Use the formula to test whether this is physically allowed.

Forecast: Guess: does the formula spit out a valid $M_{2}$ , a nonsense value, or a warning sign (like a value $> 1$ , which the 2nd law forbids for a compression shock)?

Step 1 — Blindly apply the $M_{2}$ formula. $M_{1}^{2} = 0.36$ : $M_{2}^{2} = \frac{1 + 0.2 ( 0.36 )}{1.4 ( 0.36 ) - 0.2} = \frac{1.072}{0.504 - 0.2} = \frac{1.072}{0.304} = 3.526$ so $M_{2} = 1.878$ . Why this step? The algebra is happy to give an answer — but the answer $M_{2} > 1$ means the flow would go from subsonic to supersonic.

Step 2 — Check the entropy / pressure direction. With $M_{1} = 0.6$ , the pressure "ratio" formula gives $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (0.36 - 1) = 1 + \frac{2.8}{2.4} (- 0.64) = 1 - 0.7467 = 0.2533$ a pressure drop. Why this step? A rarefaction jump like this would require $Δ s < 0$ — the flow spontaneously getting more ordered. The Second Law forbids it (see Entropy and Second Law).

Step 3 — Conclusion. No such shock exists. The math has a mathematically valid root, but it is the unphysical branch. Shocks require $M_{1} > 1$ and always produce $M_{2} < 1$ . Why this step? This is the classic trap: formulas don't know physics; you must apply the 2nd-law filter.

Verify: For any $M_{1} < 1$ , the formula yields $M_{2} > 1$ and $P_{2} / P_{1} < 1$ — a signature of the forbidden expansion shock. Confirmed numerically: $M_{2}^{2} = 3.526 > 1$ and $P_{2} / P_{1} = 0.2533 < 1$ . ✗ physically. ✓ as a demonstration of the trap.

Example 6 — Cell F · Inverse problem (back-solve $M_{1}$ )

Worked example A pressure gauge across a shock reads

$P_{2} / P_{1} = 4.5$ in air. What was the upstream Mach number? (This is Example 1 of the parent, run backwards.)

Forecast: The parent's forward example got $P_{2} / P_{1} = 4.5$ from $M_{1} = 2$ . Guess: will inverting the formula recover exactly $M_{1} = 2$ ?

Step 1 — Rearrange the pressure formula for $M_{1}^{2}$ . Starting from $\frac{P _{2}}{P _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1),$ solve: $M_{1}^{2} = 1 + \frac{γ + 1}{2 γ} (\frac{P _{2}}{P _{1}} - 1)$ Why this step? Pressure ratio is a monotonic function of the Mach number $M_{1}$ , so it can be inverted uniquely — one gauge reading pins down exactly one upstream Mach number.

Step 2 — Plug in. $\frac{P _{2}}{P _{1}} - 1 = 3.5$ : $M_{1}^{2} = 1 + \frac{2.4}{2.8} (3.5) = 1 + 3 = 4 \Rightarrow M_{1} = 2.0$ Why this step? Confirms the inverse recovers the forward answer.

Verify: Substitute the recovered $M_{1} = 2$ back into the forward formula: $1 + \frac{2.8}{2.4} (4 - 1) = 1 + 3.5 = 4.5$ ✓ — round trip closes exactly.

Example 7 — Cell G · Non-air gas ( $γ \neq = 1.4$ )

Worked example A shock forms in

helium ( $γ = 5/3$ , monatomic) at $M_{1} = 2.0$ . Find $M_{2}$ , $ρ_{2} / ρ_{1}$ , and the maximum possible density ratio. Compare to air.

Forecast: Helium's $γ = 1.667$ is larger than air's $1.4$ . The density ceiling is $\frac{γ + 1}{γ - 1}$ . Guess: is helium's ceiling higher or lower than air's 6?

Step 1 — Density ceiling. $\frac{ρ _{2}}{ρ _{1}}_{m a x} = \frac{γ + 1}{γ - 1} = \frac{2.667}{0.667} = 4.0$ Why this step? Establishes the physical limit — helium can be compressed at most $4 \times$ in one shock, less than air's $6 \times$ . A higher $γ$ means a stiffer gas.

Step 2 — $M_{2}$ at $M_{1} = 2$ . With $\frac{γ - 1}{2} = 0.3333$ and $M_{1}^{2} = 4$ : $M_{2}^{2} = \frac{1 + 0.3333 ( 4 )}{1.6667 ( 4 ) - 0.3333} = \frac{2.3333}{6.3333} = 0.3684 \Rightarrow M_{2} = 0.607$ Why this step? Shows the same upstream Mach number $M_{1}$ gives a different $M_{2}$ once $γ$ changes — never memorize air-only numbers.

Step 3 — Density ratio at $M_{1} = 2$ . $\frac{ρ _{2}}{ρ _{1}} = \frac{2.6667 ( 4 )}{0.6667 ( 4 ) + 2} = \frac{10.667}{4.6667} = 2.286$ Why this step? Compared to air's value of $2.667$ at the same $M_{1}$ , helium compresses less — consistent with its lower ceiling.

Verify: $ρ_{2} / ρ_{1} = 2.286 < 4.0$ ceiling ✓; $M_{2} < 1$ ✓; helium ceiling $4 <$ air ceiling $6$ ✓ (stiffer gas, less squeezable).

Example 8 — Cell H · Real-world word problem

Worked example A pitot probe on a supersonic wind-tunnel model measures upstream conditions

$P_{1} = 40 kPa$ , $T_{1} = 250 K$ , ahead of a normal shock at $M_{1} = 3.0$ in air. Find the actual downstream pressure $P_{2}$ , temperature $T_{2}$ , and the downstream flow speed $u_{2}$ .

Forecast: $M_{1} = 3$ is a strong shock. Guess: does $P_{2}$ land near $200$ kPa or $400$ kPa? Does $T_{2}$ exceed $600$ K? Does $u_{2}$ drop below the speed of sound (which it must, since $M_{2} < 1$ )?

Step 1 — Ratios from $M_{1} = 3$ . With $M_{1}^{2} = 9$ : $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (8) = 1 + 9.333 = 10.333, \frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 9 )}{0.4 ( 9 ) + 2} = \frac{21.6}{5.6} = 3.857$ $\frac{T _{2}}{T _{1}} = \frac{10.333}{3.857} = 2.679$ Why this step? Get dimensionless ratios first — they carry all the physics; dimensions come in last.

Step 2 — Actual static values. Multiply each ratio by the measured upstream value: $P_{2} = 10.333 \times 40 = 413.3 kPa, T_{2} = 2.679 \times 250 = 669.8 K$ Why this step? This is where "just a formula" becomes a real number a sensor would read; a shock at Mach 3 raises static pressure tenfold and roughly triples the temperature.

Step 3 — Downstream flow speed. First the downstream Mach number: $M_{2}^{2} = \frac{1 + 0.2 ( 9 )}{1.4 ( 9 ) - 0.2} = \frac{2.8}{12.4} = 0.2258$ , so $M_{2} = 0.4752$ . The downstream sound speed uses the hotter downstream temperature, $a_{2} = γ R T_{2}$ with $R = 287 J/kg\cdotK$ : $a_{2} = 1.4 (287) (669.8) = 518.6 m/s, u_{2} = M_{2} a_{2} = 0.4752 (518.6) = 246.4 m/s$ Why this step? Flow speed = Mach number × local sound speed, and the local sound speed depends on the downstream (hotter) temperature — a common slip is using $T_{1}$ here.

Verify: $P_{2} = 413$ kPa, $T_{2} = 670$ K, $u_{2} = 246$ m/s. Sanity: the upstream speed is $u_{1} = M_{1} γ R T_{1} = 3 1.4 (287) (250) = 3 (316.9) = 950.6$ m/s. Mass conservation says $u_{1} / u_{2}$ should equal $ρ_{2} / ρ_{1} = 3.857$ : check $950.6/246.4 = 3.858$ ✓ (rounding). Beautiful closure.

Example 9 — Cell I · Exam twist (inlet design)

Worked example A supersonic engine inlet decelerates air from

$M_{1} = 3$ to subsonic. Design A uses one normal shock at $M_{1} = 3$ . Design B uses two shocks in series: a weak normal shock at $M = 1.5$ followed by a normal shock at $M = 2$ . Compare the total stagnation-pressure recovery $P_{02} / P_{01}$ and explain which is better. (Treat each stage's recovery as multiplying.)

Forecast: The parent note hinted "multiple weak shocks are preferred." Guess: will Design B's total recovery beat Design A's by a little or by a lot?

Step 1 — Design A recovery. A single normal shock at $M_{1} = 3$ has (from the two-bracket formula): $\frac{P _{02}}{P _{01}}_{A} = 0.3283$ Why this step? This is the baseline — one violent shock throws away about two-thirds of the stagnation pressure.

Step 2 — Design B, stage recoveries. The first (weak) shock at $M = 1.5$ and the second at $M = 2$ each have their own single-shock recovery: $\frac{P _{02}}{P _{01}}_{M = 1.5} = 0.9298, \frac{P _{02}}{P _{01}}_{M = 2} = 0.7209$ Why this step? Each stage's entropy generation is smaller because near $M = 1$ entropy grows like the cube $(M^{2} - 1)^{3}$ — spreading the deceleration keeps every jump gentle.

Step 3 — Multiply the stage recoveries. $\frac{P _{02}}{P _{01}}_{B} = 0.9298 \times 0.7209 = 0.6703$ Why this step? Stagnation-pressure recovery multiplies through successive stages, so two gentle shocks ( $0.67$ ) crush one harsh shock ( $0.33$ ).

Step 4 — Verdict. Design B recovers about twice the stagnation pressure of Design A. This is exactly why real supersonic inlets use a ramp of Oblique Shock Waves before a final weak normal shock (see Supersonic Inlets and Diffusers). Why this step? Ties the numeric result to the engineering reason it matters — thrust is proportional to recovered stagnation pressure.

Verify: $0.6703 > 0.3283$ (staged is better ✓); each single-stage factor is $< 1$ (2nd law ✓); the harshest single shock loses the most ✓. The staged recovery cannot exceed 1 ( $0.67 < 1$ ✓).

Recall Quick self-test (cover the answers)

Which cell forbids a shock, and why? ::: Cell E — subsonic inflow ( $M_{1} < 1$ ) would need $Δ s < 0$ , banned by the 2nd Law. As $M_{1} \to \infty$ in air, the density ratio approaches what number? ::: $\frac{γ + 1}{γ - 1} = 6$ . For helium ( $γ = 5/3$ ), the density ceiling is? ::: $4.0$ . At $M_{1} = 2$ in air, $P_{2} / P_{1}$ and $M_{2}$ ? ::: $4.5$ and $0.577$ . Why do two weak shocks beat one strong shock in an inlet? ::: Entropy grows like $(M^{2} - 1)^{3}$ , so gentle jumps waste far less stagnation pressure; recoveries multiply.