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

Throughout we use air with $γ = 1.4$ unless a problem says otherwise. Here is the complete toolkit you will reach for — every symbol was built in the parent note, but let us restate them in plain words so nothing is assumed:

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

Look at the figure: every ratio is a curve rising from 1 at $M_{1} = 1$ . Pressure and temperature race off to infinity; density flattens onto its ceiling $\frac{γ + 1}{γ - 1} = 6$ ; the downstream Mach number sags toward its floor. Keep this picture in mind — it tells you instantly whether an answer is plausible.

Level 1 — Recognition

L1.1

State whether each quantity rises, falls, or stays the same across a normal shock, and give the physical reason: (i) static pressure $p$ , (ii) total temperature $T_{0}$ , (iii) total pressure $p_{0}$ , (iv) Mach number $M$ , (v) entropy $s$ .

Recall Solution

(i) $p$ rises — the shock is a wall of compression.
(ii) $T_{0}$ stays the same — adiabatic, no work ⇒ total enthalpy conserved ⇒ $T_{0}$ constant.
(iii) $p_{0}$ falls — the shock is irreversible, entropy rises, so $p_{02} / p_{01} = e^{- Δ s / R} < 1$ .
(iv) $M$ falls (supersonic → subsonic).
(v) $s$ rises — Second Law; this is why only compression shocks exist.

L1.2

For air ( $γ = 1.4$ ) with $M_{1} = 1$ (an infinitely weak shock), compute $p_{2} / p_{1}$ , $ρ_{2} / ρ_{1}$ , $T_{2} / T_{1}$ , and $M_{2}$ . What do you notice?

Recall Solution

Plug $M_{1}^{2} = 1$ :

$p_{2} / p_{1} = 1 + \frac{2.8}{2.4} (1 - 1) = 1$ .
$ρ_{2} / ρ_{1} = \frac{2.4 ( 1 )}{0.4 ( 1 ) + 2} = \frac{2.4}{2.4} = 1$ .
$T_{2} / T_{1} = \frac{[ 2.8 - 0.4 ] [ 0.4 + 2 ]}{2. 4 ^{2} ( 1 )} = \frac{2.4 \times 2.4}{5.76} = 1$ .
$M_{2}^{2} = \frac{1 + 0.2}{1.4 - 0.2} = \frac{1.2}{1.2} = 1 \Rightarrow M_{2} = 1$ . Notice: at $M_{1} = 1$ every ratio is exactly 1 — there is no shock at all. A shock only does something once $M_{1} > 1$ . This is the anchor point of all the curves in the figure.

Level 2 — Application

L2.1

Air enters a normal shock at $M_{1} = 3$ . Find $M_{2}$ , $p_{2} / p_{1}$ , $ρ_{2} / ρ_{1}$ , and $T_{2} / T_{1}$ .

Recall Solution

With $M_{1}^{2} = 9$ , $γ = 1.4$ :

$M_{2}^{2} = \frac{1 + 0.2 ( 9 )}{1.4 ( 9 ) - 0.2} = \frac{2.8}{12.4} = 0.22581 \Rightarrow M_{2} = 0.4752$ .
$p_{2} / p_{1} = 1 + \frac{2.8}{2.4} (9 - 1) = 1 + 1.16667 \times 8 = 10.333$ .
$ρ_{2} / ρ_{1} = \frac{2.4 ( 9 )}{0.4 ( 9 ) + 2} = \frac{21.6}{5.6} = 3.857$ .
$T_{2} / T_{1} = \frac{p _{2} / p _{1}}{ρ _{2} / ρ _{1}} = \frac{10.333}{3.857} = 2.679$ . Sanity: $M_{2} < 1$ ✓, all ratios exceed those at $M_{1} = 2$ ✓ (steeper up the curves).

L2.2

Upstream conditions are $p_{1} = 50 kPa$ , $T_{1} = 250 K$ , $M_{1} = 2.5$ . Find the actual downstream static pressure $p_{2}$ and temperature $T_{2}$ (not just ratios).

Recall Solution

With $M_{1}^{2} = 6.25$ :

$p_{2} / p_{1} = 1 + \frac{2.8}{2.4} (6.25 - 1) = 1 + 1.16667 \times 5.25 = 7.125 \Rightarrow p_{2} = 7.125 \times 50 = 356.25 kPa$ .
$ρ_{2} / ρ_{1} = \frac{2.4 ( 6.25 )}{0.4 ( 6.25 ) + 2} = \frac{15}{4.5} = 3.3333$ .
$T_{2} / T_{1} = 7.125/3.3333 = 2.1375 \Rightarrow T_{2} = 2.1375 \times 250 = 534.4 K$ .

Level 3 — Analysis

L3.1

For air, find the upstream Mach number $M_{1}$ that produces a static pressure ratio $p_{2} / p_{1} = 4.5$ . (Invert Relation 3.)

Recall Solution

Set $4.5 = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1)$ . Rearrange to solve for $M_{1}^{2}$ — why algebra and not guessing? Relation 3 is linear in $M_{1}^{2}$ , so it inverts cleanly: $M_{1}^{2} - 1 = (4.5 - 1) \cdot \frac{γ + 1}{2 γ} = 3.5 \cdot \frac{2.4}{2.8} = 3.5 \times 0.857143 = 3.0.$ So $M_{1}^{2} = 4 \Rightarrow M_{1} = 2.0$ . (Consistent with worked example (a) in the parent note.)

L3.2

A normal shock at $M_{1} = 2$ in air. Compute the entropy rise $Δ s$ (J kg⁻¹K⁻¹) and the total-pressure ratio $p_{02} / p_{01}$ .

Recall Solution

From example (a): $p_{2} / p_{1} = 4.5$ , $T_{2} / T_{1} = 1.6875$ . Why these enter $Δ s$ ? Entropy change of a perfect gas depends only on the two end-state ratios of $T$ and $p$ : $Δ s = c_{p} ln \frac{T _{2}}{T _{1}} - R ln \frac{p _{2}}{p _{1}} .$

$c_{p} = 1005$ : $1005 ln (1.6875) = 1005 \times 0.523248 = 525.86 J kg^{- 1} K^{- 1}$ .
$R ln (p_{2} / p_{1}) = 287 ln (4.5) = 287 \times 1.504077 = 431.67$ .
$Δ s = 525.86 - 431.67 = 94.19 J kg^{- 1} K^{- 1} > 0$ ✓ (shock is allowed).
$p_{02} / p_{01} = exp (- Δ s / R) = exp (- 94.19/287) = exp (- 0.32819) = 0.7203$ . So the flow loses about $28%$ of its total pressure. Compare with Stagnation properties.

Level 4 — Synthesis

L4.1

An aircraft flies at $M_{1} = 3$ where ambient air is $T_{1} = 220 K$ . A normal shock stands in front of a pitot probe. Find (i) the temperature $T_{2}$ right behind the shock, and (ii) the stagnation temperature $T_{0}$ the probe reads. Are they the same? Explain.

Recall Solution

(i) From L2.1, $T_{2} / T_{1} = 2.679$ , so $T_{2} = 2.679 \times 220 = 589.4 K$ . (ii) Stagnation temperature uses the isentropic $T_{0}$ formula (Isentropic flow relations), and since $T_{0}$ is conserved across the shock we may evaluate it upstream: $T_{0} = T_{1} (1 + \frac{γ - 1}{2} M_{1}^{2}) = 220 (1 + 0.2 \times 9) = 220 \times 2.8 = 616 K .$ Not the same. $T_{2} = 589$ K is the static temperature just behind the shock, where the flow is still moving at $M_{2} = 0.475$ . The probe brings that residual motion to rest, adding the last bit of kinetic energy to reach $T_{0} = 616$ K. Because $T_{0}$ is the same on both sides, we could equally have computed it from state 2 — check: $T_{0} = T_{2} (1 + 0.2 M_{2}^{2}) = 589.4 (1 + 0.2 \times 0.22581) = 616$ K ✓.

L4.2

For the same $M_{1} = 3$ shock in air, the upstream total pressure is $p_{01} = 400 kPa$ . Find the total pressure $p_{02}$ read behind the shock. (This is exactly what a supersonic pitot correction handles.)

Recall Solution

First $Δ s$ from L2.1 ratios ( $p_{2} / p_{1} = 10.333$ , $T_{2} / T_{1} = 2.679$ ):

$c_{p} ln (T_{2} / T_{1}) = 1005 ln (2.679) = 1005 \times 0.985637 = 990.57$ .
$R ln (p_{2} / p_{1}) = 287 ln (10.333) = 287 \times 2.335375 = 670.25$ .
$Δ s = 990.57 - 670.25 = 320.32 J kg^{- 1} K^{- 1}$ .
$p_{02} / p_{01} = exp (- 320.32/287) = exp (- 1.11609) = 0.32765$ .
$p_{02} = 0.32765 \times 400 = 131.1 kPa$ . The stronger the shock, the more total pressure is destroyed — at $M_{1} = 3$ we keep only about a third.

Level 5 — Mastery

L5.1 (Strong-shock limit)

Prove that as $M_{1} \to \infty$ (air), $ρ_{2} / ρ_{1} \to 6$ and $M_{2}^{2} \to \frac{γ - 1}{2 γ}$ . Then evaluate both numerically at $M_{1} = 20$ and comment on how close the limits are.

Recall Solution

Density limit. Why divide by $M_{1}^{2}$ ? To expose the dominant terms: $\frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 ) M _{1}^{2}}{( γ - 1 ) M _{1}^{2} + 2} = \frac{γ + 1}{( γ - 1 ) + 2/ M _{1}^{2}} M_{1} \to \infty \frac{γ + 1}{γ - 1} = \frac{2.4}{0.4} = 6.$ Mach limit. Divide top and bottom of $M_{2}^{2}$ by $M_{1}^{2}$ : $M_{2}^{2} = \frac{1/ M _{1}^{2} + \frac{γ - 1}{2}}{γ - \frac{γ - 1}{2} / M _{1}^{2}} M_{1} \to \infty \frac{( γ - 1 ) /2}{γ} = \frac{0.4/2}{1.4} = 0.142857 \Rightarrow M_{2} = 0.3780.$ At $M_{1} = 20$ ( $M_{1}^{2} = 400$ ):

$ρ_{2} / ρ_{1} = \frac{2.4 ( 400 )}{0.4 ( 400 ) + 2} = \frac{960}{162} = 5.926$ — within $1.2%$ of the ceiling 6.
$M_{2}^{2} = \frac{1 + 0.2 ( 400 )}{1.4 ( 400 ) - 0.2} = \frac{81}{559.8} = 0.14469 \Rightarrow M_{2} = 0.3804$ — within $0.6%$ of the limit. Comment: density is essentially saturated while temperature (and pressure) keep climbing — the extra energy of an ever-faster flow goes into heat, not compression. This is why hypersonic re-entry is a thermal problem, not a compression problem.

L5.2 (Reverse-shock check)

Suppose someone claims a shock takes air from $M_{1} = 0.6$ to a downstream $M_{2} > 1$ ("expansion shock"). Test this with the entropy relation and explain what the Second Law says.

Recall Solution

Formally plug $M_{1} = 0.6$ ( $M_{1}^{2} = 0.36$ ) into the ratios:

$p_{2} / p_{1} = 1 + \frac{2.8}{2.4} (0.36 - 1) = 1 + 1.16667 \times (- 0.64) = 0.25333$ (pressure would drop).
$T_{2} / T_{1} = \frac{[ 2.8 ( 0.36 ) - 0.4 ] [ 0.4 ( 0.36 ) + 2 ]}{2. 4 ^{2} ( 0.36 )} = \frac{0.608 \times 2.144}{2.0736} = 0.62864$ .
$Δ s = 1005 ln (0.62864) - 287 ln (0.25333) = 1005 (- 0.46428) - 287 (- 1.37333)$ $= - 466.60 + 394.15 = - 72.45 J kg^{- 1} K^{- 1} < 0$ . The entropy decreases — this violates the Second Law. Such an "expansion shock" is impossible. A subsonic flow cannot be shocked to supersonic; the only physically real branch is compression, $M_{1} > 1 \to M_{2} < 1$ . To accelerate to supersonic you need a smooth isentropic expansion (a nozzle or a Prandtl–Meyer fan), never a shock.

L5.3 (Cross-check via Fanno/Rayleigh intuition)

Confirm that the $M_{1} = 2$ air shock lies on the correct branch by checking that total enthalpy (hence $T_{0}$ ) is identical on both sides, computing $T_{0} / T_{1}$ from state 1 and from state 2 and showing they match.

Recall Solution

From example (a): $M_{2}^{2} = 1/3$ , $T_{2} / T_{1} = 1.6875$ .

Upstream: $T_{0} / T_{1} = 1 + 0.2 (4) = 1.8$ .
Downstream: $T_{0} / T_{1} = (T_{2} / T_{1}) (1 + 0.2 M_{2}^{2}) = 1.6875 (1 + 0.2/3) = 1.6875 \times 1.066667 = 1.8$ . They agree exactly — total temperature is conserved, confirming the energy equation. This is the shared backbone with Rayleigh & Fanno flow: a shock is the point where the Fanno and Rayleigh lines cross, jumping from the supersonic to the subsonic intersection while conserving mass, momentum, and total enthalpy.

Recall

Recall Which relation do I reach for? (cover the answers)

Given $M_{1}$ , how do I get downstream Mach? ::: $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{1}^{2}}{γ M _{1}^{2} - \frac{γ - 1}{2}}$ Given a target $p_{2} / p_{1}$ , how do I find $M_{1}$ ? ::: Invert the linear-in- $M_{1}^{2}$ pressure relation. What is conserved across every shock? ::: mass, momentum, total enthalpy (so $T_{0}$ ). How do I get $p_{02} / p_{01}$ ? ::: compute $Δ s = c_{p} ln \frac{T _{2}}{T _{1}} - R ln \frac{p _{2}}{p _{1}}$ , then $e^{- Δ s / R}$ . How do I reject an impossible shock? ::: check $Δ s \geq 0$ ; if negative, it violates the Second Law.