Normal shock waves — Rankine-Hugoniot relations (all 5) — derivations: Worked examples

This page is the drill hall for the normal-shock relations. The parent note built the five formulas. Here we throw every kind of input at them — weak shocks, strong shocks, the exact boundary, the degenerate "no shock" case, a real-world word problem, and an exam-style trap — so that when you meet one in the wild you have already seen its twin.

Before we start, keep the five workhorses in view. All of them take one input, the upstream Mach number $M_{1}$ , for a calorically perfect gas with heat-capacity ratio $γ$ (for air $γ = 1.4$ ).

Throughout the examples I cite these by their tags: R1 = downstream Mach, R2 = pressure ratio, R3 = density ratio, R4 = temperature ratio, R5 = stagnation-pressure loss.

Reminders you must not skip:

$M$ ("Mach number") is flow speed divided by the local speed of sound: $M = u / a$ , $a = γ R T$ . See Speed of sound and Mach number.
A normal shock only exists when the incoming flow is supersonic, $M_{1} > 1$ . This is enforced by the Second Law.
Subscript $0$ means stagnation (the value the flow would reach if brought to rest reversibly). See Stagnation properties.

The scenario matrix

Every problem this topic can throw falls into one of these cells. Each worked example below is tagged with the cell it lands in.

Cell	Input class	What is special	Example
C1	Degenerate: $M_{1} = 1$	Infinitely weak shock — all ratios $= 1$	Ex 1
C2	Forbidden: $M_{1} < 1$	No shock allowed — 2nd Law veto	Ex 2
C3	Weak shock: $M_{1}$ just above 1	Small jumps, near-isentropic	Ex 3
C4	Moderate shock: $M_{1} = 2$ – $3$	The "standard" case	Ex 4
C5	Strong-shock limit: $M_{1} \to \infty$	Density saturates, $T$ blows up	Ex 5
C6	Total-pressure / entropy loss	$p_{0}$ drops though $T_{0}$ constant	Ex 6
C7	Real-world word problem	Dimensional data, find actual $p_{2}, T_{2}$	Ex 7
C8	Exam twist: given a downstream quantity, back-solve $M_{1}$	Inverse problem	Ex 8
C9	Different gas ( $γ \neq = 1.4$ )	Check the $γ$ -dependence	Ex 9

The picture below plots the four static ratios against $M_{1}$ so you can see which region every example lives in.

Figure — Normal shock waves — Rankine-Hugoniot relations (all 5) — derivations

Figure s01 — reading it without the image. The horizontal axis is the upstream Mach number $M_{1}$ from $1$ to $6$ ; the vertical axis is the value of a ratio across the shock. Four curves all start at the point $(1, 1)$ — the degenerate shock of Ex 1. Moving right: the blue $p_{2} / p_{1}$ curve rises without limit; the red $T_{2} / T_{1}$ curve rises even faster (like $M_{1}^{2}$ ); the green $ρ_{2} / ρ_{1}$ curve rises but flattens toward a horizontal dashed cap at $6$ ; the orange $M_{2}$ curve falls from $1$ toward a floor near $0.38$ , staying below the dotted sonic line $y = 1$ . Marked dots at $M_{1} = 1.2, 2, 3$ show where Ex 3, Ex 4 and Ex 7 sit. The single message: pressure and temperature run away, density saturates, the flow always ends up subsonic.

Ex 1 — Cell C1: the degenerate shock $M_{1} = 1$

Worked example Air,

$γ = 1.4$ , $M_{1} = 1$ exactly. Find all ratios. Forecast: guess before reading. What should a shock at exactly sonic speed do to the pressure? Nothing, a lot, or something in between?

1. Pressure ratio (R2): $\frac{p _{2}}{p _{1}} = 1 + \frac{2 ( 1.4 )}{2.4} (1^{2} - 1) = 1 + 1.1667 \times 0 = 1.$ Why this step? The factor $(M_{1}^{2} - 1)$ vanishes at $M_{1} = 1$ , killing the whole increment.

2. Density ratio (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 1 )}{0.4 ( 1 ) + 2} = \frac{2.4}{2.4} = 1.$ Why this step? Numerator $(γ + 1) M_{1}^{2} = 2.4$ and denominator $(γ - 1) M_{1}^{2} + 2 = 2.4$ coincide when $M_{1}^{2} = 1$ .

3. Downstream Mach (R1): $M_{2}^{2} = \frac{1 + 0.2 ( 1 )}{1.4 ( 1 ) - 0.2} = \frac{1.2}{1.2} = 1 \Rightarrow M_{2} = 1.$ Why this step? At the boundary the two algebraic roots ( $M_{2} = M_{1}$ and the shock root) merge — here is the algebra proving it: the shock relation R1 can be written $M_{2}^{2} = f (M_{1}^{2})$ , and the other physical root of the original quadratic is always $M_{2} = M_{1}$ . Set $M_{1} = 1$ in R1: it returns $M_{2}^{2} = 1$ , i.e. $M_{2} = M_{1}$ . The two roots have become the same number. That coincidence is exactly what "the roots merge" means.

4. Temperature (R4): $T_{2} / T_{1} = 1/1 = 1$ .

Verify: all four ratios equal 1 — the shock has zero strength. This is the left edge of the figure where all curves pass through $(1, 1)$ . A "shock" at $M_{1} = 1$ is a Mach wave carrying no jump. ✓

Ex 2 — Cell C2: the forbidden expansion shock $M_{1} < 1$

Worked example Suppose someone claims a shock takes air from

$M_{1} = 0.5$ up to some $M_{2} > 1$ . Test it. Forecast: the algebra will happily return a number. Will the Second Law allow it?

1. Push $M_{1} = 0.5$ through R1: $M_{2}^{2} = \frac{1 + 0.2 ( 0.25 )}{1.4 ( 0.25 ) - 0.2} = \frac{1.05}{0.15} = 7.0 \Rightarrow M_{2} = 2.65.$ Why this step? Nothing in the algebra stops us plugging a subsonic value in — the equation is just a rational function.

2. So the pure math says $0.5 \to 2.65$ : subsonic to supersonic. Now compute the pressure ratio (R2): $\frac{p _{2}}{p _{1}} = 1 + 1.1667 (0.25 - 1) = 1 - 0.875 = 0.125.$ Why this step? We need $p_{2} / p_{1}$ to test entropy; here it is less than 1 — pressure would fall.

3. Density (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 0.25 )}{0.4 ( 0.25 ) + 2} = \frac{0.6}{2.1} = 0.2857.$ Temperature (R4): $T_{2} / T_{1} = 0.125/0.2857 = 0.4375.$

4. Entropy change (R5): $Δ s / R = \frac{γ}{γ - 1} ln \frac{T _{2}}{T _{1}} - ln \frac{p _{2}}{p _{1}} = 3.5 ln (0.4375) - ln (0.125) .$ $= 3.5 (- 0.8267) - (- 2.0794) = - 2.8935 + 2.0794 = - 0.8141.$ Why this step? This is the entropy test. An allowed adiabatic process must have $Δ s \geq 0$ .

Verify: $Δ s / R = - 0.814 < 0$ . Negative entropy → forbidden. The expansion shock cannot exist. The only physical direction is supersonic→subsonic. This cell has no valid answer, and that is the answer. ✓

Ex 3 — Cell C3: the weak shock $M_{1} = 1.2$

Worked example Air,

$M_{1} = 1.2$ . Find $M_{2}, p_{2} / p_{1}, ρ_{2} / ρ_{1}, T_{2} / T_{1}$ , and comment on how "gentle" it is. Forecast: just above Mach 1. Expect small jumps — but how small compared to $M_{1} = 2$ ?

1. $M_{1}^{2} = 1.44$ . Pressure (R2): $\frac{p _{2}}{p _{1}} = 1 + 1.1667 (1.44 - 1) = 1 + 1.1667 (0.44) = 1.513.$ Why this step? The increment scales with $M_{1}^{2} - 1 = 0.44$ , small.

2. Density (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 1.44 )}{0.4 ( 1.44 ) + 2} = \frac{3.456}{2.576} = 1.342.$ Why this step? Direct substitution into R3.

3. Temperature (R4): $\frac{T _{2}}{T _{1}} = \frac{1.513}{1.342} = 1.128.$ Why this step? Ideal-gas identity $T = p / ρ$ ratio.

4. Downstream Mach (R1): $M_{2}^{2} = \frac{1 + 0.2 ( 1.44 )}{1.4 ( 1.44 ) - 0.2} = \frac{1.288}{1.816} = 0.7093 \Rightarrow M_{2} = 0.842.$

Verify: all ratios are close to 1 and $M_{2} = 0.842$ is only slightly subsonic — a weak shock, almost isentropic. Compare to Ex 4 where $M_{1} = 2$ gives $p_{2} / p_{1} = 4.5$ . On the figure this point sits near the left, curves barely lifted off the $(1, 1)$ corner. ✓

Ex 4 — Cell C4: the standard shock $M_{1} = 2$

Ex 5 — Cell C5: the strong-shock limit $M_{1} = 10$ and $M_{1} \to \infty$

Worked example Air,

$M_{1} = 10$ , then the mathematical limit $M_{1} \to \infty$ . Compare density and temperature. Forecast: at Mach 10, does density keep climbing without bound like pressure does?

1. Density at $M_{1} = 10$ (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 100 )}{0.4 ( 100 ) + 2} = \frac{240}{42} = 5.714.$ Why this step? R3 with $M_{1}^{2} = 100$ .

2. Infinite-Mach cap: $M_{1} \to \infty lim \frac{ρ _{2}}{ρ _{1}} = \frac{γ + 1}{γ - 1} = \frac{2.4}{0.4} = 6.$ Why this step? Divide top and bottom of R3 by $M_{1}^{2}$ ; the $M_{1}^{2}$ -independent terms vanish, leaving $\frac{γ + 1}{γ - 1}$ . Density saturates.

3. Temperature at $M_{1} = 10$ (R4): $\frac{T _{2}}{T _{1}} = \frac{[ 2.8 ( 100 ) - 0.4 ] [ 0.4 ( 100 ) + 2 ]}{2. 4 ^{2} ( 100 )} = \frac{279.6 \times 42}{576} = 20.39.$ Why this step? R4 written out as the product of the pressure and density relations, evaluated at $M_{1}^{2} = 100$ .

4. As $M_{1} \to \infty$ , $T_{2} / T_{1} \sim \frac{2 γ ( γ - 1 )}{( γ + 1 ) ^{2}} M_{1}^{2} \to \infty.$ Why this step? Both bracket terms grow as $M_{1}^{2}$ but the denominator only as $M_{1}^{2}$ once — net growth $\propto M_{1}^{2}$ . Temperature is unbounded.

Verify: density $5.71$ is already within 5% of its hard cap of $6$ , while $T$ has ballooned to $20 \times$ . Extra shock energy dumps into heat, not compression — the physics of re-entry heating. ✓

Ex 6 — Cell C6: total-pressure loss and entropy rise ( $M_{1} = 2$ )

Worked example For the

$M_{1} = 2$ air shock (Ex 4), find $Δ s / R$ and $p_{02} / p_{01}$ . Confirm $T_{0}$ is unchanged. Forecast: the flow is adiabatic. Does stagnation pressure survive? Does stagnation temperature?

1. Entropy (R5): $\frac{Δ s}{R} = \frac{γ}{γ - 1} ln \frac{T _{2}}{T _{1}} - ln \frac{p _{2}}{p _{1}} = 3.5 ln (1.687) - ln (4.50) .$ $= 3.5 (0.5232) - 1.5041 = 1.8312 - 1.5041 = 0.3272.$ Why this step? This measures irreversibility. Positive ⇒ physically allowed shock (contrast Ex 2).

2. Stagnation-pressure ratio (R5): $\frac{p _{02}}{p _{01}} = exp (- Δ s / R) = exp (- 0.3272) = 0.7209.$ Why this step? Because $T_{0}$ is conserved (next step), the stagnation pressure loss is governed entirely by entropy: $p_{02} / p_{01} = e^{- Δ s / R}$ .

3. Stagnation temperature: total enthalpy is conserved across the shock (energy equation with $q = 0$ ), so $T_{02} = T_{01}$ , i.e. $T_{02} / T_{01} = 1$ exactly. Why this step? Adiabatic + no work ⇒ constant total enthalpy ⇒ constant $T_{0}$ . This is why "adiabatic ≠ isentropic."

Verify: $p_{0}$ drops to $72%$ of its upstream value (a 28% loss = drag penalty), yet $T_{0}$ is untouched. See Isentropic flow relations for what "no loss" would look like: there $p_{0}$ would stay constant. The gap is the entropy generated inside the shock. ✓

Ex 7 — Cell C7: real-world word problem

Worked example A supersonic wind tunnel runs air at

$M_{1} = 3$ , static pressure $p_{1} = 20 kPa$ , static temperature $T_{1} = 220 K$ . A normal shock stands in the test section. Find the physical $p_{2}$ (kPa), $T_{2}$ (K), and downstream speed $u_{2}$ (m/s). Take $R = 287 J/kg\cdotK$ . Forecast: upstream the air is cold and thin. After the shock — how hot, how dense, and does it slow below the local sound speed?

1. $M_{1}^{2} = 9$ . Pressure ratio (R2): $\frac{p _{2}}{p _{1}} = 1 + 1.1667 (9 - 1) = 1 + 9.333 = 10.33 \Rightarrow p_{2} = 10.33 \times 20 = 206.7 kPa .$ Why this step? Ratios are dimensionless; multiply by the given $p_{1}$ to get real units.

2. Density ratio (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 9 )}{0.4 ( 9 ) + 2} = \frac{21.6}{5.6} = 3.857.$ Temperature (R4): $\frac{T _{2}}{T _{1}} = \frac{10.33}{3.857} = 2.679 \Rightarrow T_{2} = 2.679 \times 220 = 589.4 K .$ Why this step? $T_{2} = T_{1} \times (T_{2} / T_{1})$ ; the air heats from $220$ K to nearly $590$ K.

3. Downstream Mach (R1): $M_{2}^{2} = \frac{1 + 0.2 ( 9 )}{1.4 ( 9 ) - 0.2} = \frac{2.8}{12.4} = 0.2258 \Rightarrow M_{2} = 0.4752.$ Why this step? We need $M_{2}$ and $a_{2}$ to convert to a real speed.

4. Downstream sound speed: $a_{2} = γ R T_{2} = 1.4 \times 287 \times 589.4 = 236, 800 = 486.6 m/s .$ Then $u_{2} = M_{2} a_{2} = 0.4752 \times 486.6 = 231.3 m/s .$ Why this step? $u = M a$ turns the dimensionless Mach number into m/s. Units: $(J/kg\cdotK) (K) = m^{2} / s^{2} = m/s$ . ✓

Verify: $p_{2} \approx 207$ kPa (10× jump), $T_{2} \approx 589$ K (air more than doubles in absolute temp), and the flow slows to $231$ m/s, which is subsonic since $u_{2} < a_{2} = 487$ m/s ( $M_{2} = 0.475 < 1$ ). All physically sane. ✓

Ex 8 — Cell C8: exam twist (inverse problem)

Worked example Downstream of a normal shock in air the

measured static pressure ratio is $p_{2} / p_{1} = 7.0$ . Work backwards to find $M_{1}$ and hence $M_{2}$ . Forecast: normally $M_{1}$ is the input. Here it is the unknown — can we invert the pressure formula cleanly?

1. Invert R2: $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1) \Rightarrow M_{1}^{2} = 1 + \frac{γ + 1}{2 γ} (\frac{p _{2}}{p _{1}} - 1) .$ Why this step? R2 is linear in $M_{1}^{2}$ , so it inverts algebraically — no quadratic needed.

2. Plug in: $M_{1}^{2} = 1 + \frac{2.4}{2.8} (7.0 - 1) = 1 + 0.8571 (6) = 1 + 5.143 = 6.143 \Rightarrow M_{1} = 2.478.$ Why this step? Numeric substitution with $γ = 1.4$ .

3. Now go forward for $M_{2}$ (R1): $M_{2}^{2} = \frac{1 + 0.2 ( 6.143 )}{1.4 ( 6.143 ) - 0.2} = \frac{2.229}{8.400} = 0.2653 \Rightarrow M_{2} = 0.5151.$ Why this step? With $M_{1}$ recovered, R1 gives the downstream Mach.

Verify: feed $M_{1} = 2.478$ back into R2: $1 + 1.1667 (6.143 - 1) = 1 + 1.1667 (5.143) = 7.00$ ✓ — matches the given data. And $M_{2} = 0.515 < 1$ , consistent with a real compression shock. ✓

Ex 9 — Cell C9: a different gas ( $γ = 1.667$ , monatomic)

Worked example Argon (monatomic,

$γ = 5/3 \approx 1.667$ ) at $M_{1} = 2$ . Find $ρ_{2} / ρ_{1}$ and its saturation cap, and compare with air. Forecast: heavier $γ$ than air. Does the density compress more or less?

1. $γ - 1 = 0.6667, γ + 1 = 2.6667$ . Density (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.6667 ( 4 )}{0.6667 ( 4 ) + 2} = \frac{10.667}{4.6667} = 2.286.$ Why this step? R3 with argon's $γ$ ; every $γ$ enters explicitly, so we just re-substitute.

2. Saturation cap: $\frac{γ + 1}{γ - 1} = \frac{2.6667}{0.6667} = 4.0.$ Why this step? Same $M_{1} \to \infty$ limit as Ex 5; monatomic gas caps at $4 \times$ , not $6 \times$ like air.

3. Pressure for context (R2): $\frac{p _{2}}{p _{1}} = 1 + \frac{2 ( 1.667 )}{2.6667} (4 - 1) = 1 + 1.25 (3) = 4.75.$ Why this step? R2 with argon's $γ$ ; slightly higher than air's $4.50$ .

4. Temperature (R4) for completeness: $\frac{T _{2}}{T _{1}} = \frac{p _{2} / p _{1}}{ρ _{2} / ρ _{1}} = \frac{4.75}{2.286} = 2.078.$ Why this step? Same ideal-gas identity, so we can compare argon's heating to air's $T_{2} / T_{1} = 1.687$ at the same $M_{1} = 2$ .

Verify: argon compresses less ( $2.286$ vs air's $2.667$ ) and its density cap is lower ( $4$ vs $6$ ), yet it heats more ( $T_{2} / T_{1} = 2.078$ vs air's $1.687$ ). A gas with fewer molecular degrees of freedom (larger $γ$ ) dumps a bigger share of the shock energy into temperature. The formulas correctly track $γ$ . ✓

Recall

Recall Active recall — cover the answers

At $M_{1} = 1$ , what are all four static ratios? ::: All equal 1 — a shock of zero strength (the two roots merge). Why can't a shock take subsonic flow to supersonic? ::: It would give $Δ s < 0$ , forbidden by the Second Law. What is $ρ_{2} / ρ_{1}$ for air at $M_{1} = 2$ ? ::: $2.667$ . What is the density cap as $M_{1} \to \infty$ for air, and for argon? ::: $6$ (air, $γ = 1.4$ ) and $4$ (argon, $γ = 5/3$ ). Across a shock, which stagnation quantity is conserved and which drops? ::: $T_{0}$ conserved, $p_{0}$ drops (here to $0.72$ at $M_{1} = 2$ ). How do you invert to get $M_{1}$ from a measured $p_{2} / p_{1}$ ? ::: $M_{1}^{2} = 1 + \frac{γ + 1}{2 γ} (\frac{p _{2}}{p _{1}} - 1)$ (linear, no quadratic).

See also: Oblique shock waves (the same jumps applied to the velocity component normal to a tilted shock), Rayleigh & Fanno flow (heat and friction driving flow toward $M = 1$ ), and Isentropic flow relations (the loss-free limit these shocks depart from).