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

Throughout, unless a problem says otherwise, the gas is air with $γ = 1.4$ (the ratio of specific heats — how "springy" the gas is). State 1 = upstream (before the shock, supersonic); State 2 = downstream (after, subsonic). One symbol you will meet immediately: $R$ is the specific gas constant — for air $R = 287 J kg^{- 1} K^{- 1}$ — the number that links pressure, density, and temperature through the ideal-gas law $P = ρR T$ . The five workhorse formulas from the parent, restated so this page stands alone:

One more relation is quoted by name in the L2 trap below, so we restate it here too — the isentropic temperature ratio, which follows directly from energy conservation ( $T_{0}$ conserved) and holds up to and after a shock but never through it: $\frac{T_2}{T_1}=\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}\tag{5}$

Here $u$ is the flow speed (metres per second), $ρ$ the density (mass per cubic metre), $P$ the static pressure, $T$ the absolute temperature (in kelvin, K — never celsius, because these formulas use ratios that only work on an absolute scale). $P_{0}$ and $T_{0}$ are the stagnation (total) pressure and temperature — the values the gas would reach if brought smoothly to rest.

Read the map before you climb. The figure below is the mental picture behind every exercise on this page: a single control volume with the shock as a thin amber wall. On the left the gas is fast and supersonic (State 1); crossing the wall it emerges slower and subsonic (State 2), with $P$ , $ρ$ , $T$ all jumped up. Every problem that follows asks you to quantify one arrow or one label in this picture — so keep glancing back at it.

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

Level 1 — Recognition

Goal: pick the right formula and plug in. No chaining.

Exercise 1.1. A normal shock sits in an air flow at $M_{1} = 3.0$ . Find the downstream Mach number $M_{2}$ .

Recall Solution 1.1

WHAT tool and WHY: $M_{2}$ depends on $M_{1}$ alone — that is exactly what equation (7) delivers. No other relation is needed. Compute the two pieces with $γ = 1.4$ , so $\frac{γ - 1}{2} = 0.2$ :

Numerator: $1 + 0.2 (3^{2}) = 1 + 0.2 (9) = 1 + 1.8 = 2.8$ .
Denominator: $γ M_{1}^{2} - \frac{γ - 1}{2} = 1.4 (9) - 0.2 = 12.6 - 0.2 = 12.4$ . $M_{2}^{2} = \frac{2.8}{12.4} = 0.2258 \Rightarrow M_{2} = 0.2258 = 0.4752.$ Sanity check: $M_{2} < 1$ , so the flow is now subsonic — exactly what a shock must do.

Exercise 1.2. For the same shock ( $M_{1} = 3.0$ ), find the static pressure ratio $P_{2} / P_{1}$ .

Recall Solution 1.2

Tool: equation (8) — it turns strength ( $M_{1}^{2} - 1$ ) directly into a pressure jump. $\frac{P _{2}}{P _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1) = 1 + \frac{2.8}{2.4} (9 - 1) = 1 + \frac{2.8}{2.4} (8) .$ $\frac{2.8}{2.4} = 1.1667$ , so $\frac{P _{2}}{P _{1}} = 1 + 1.1667 \times 8 = 1 + 9.333 = 10.33.$ Pressure jumps to over ten times its upstream value.

Exercise 1.3. State, in one sentence each, what happens to $T_{0}$ (stagnation temperature) and $P_{0}$ (stagnation pressure) across any normal shock, and why.

Recall Solution 1.3

$T_{0}$ is unchanged. The shock is adiabatic (no heat crosses it) and does no shaft work, so the total energy carried per kilogram is conserved: $T_{02} = T_{01}$ .
$P_{0}$ drops ( $P_{02} < P_{01}$ ). The shock is irreversible — entropy $s$ rises — and stagnation pressure is tied to entropy by $P_{02} / P_{01} = e^{- Δ s / R}$ . A positive $Δ s$ forces the ratio below 1. Mnemonic from the parent: "PRT up, M & P₀ down, T₀ stays."

Level 2 — Application

Goal: get several properties from one $M_{1}$ , and convert to real numbers.

Exercise 2.1. Air enters a normal shock at $M_{1} = 2.5$ , with upstream static conditions $P_{1} = 40 kPa$ and $T_{1} = 250 K$ . Find $P_{2}$ , $T_{2}$ , and $ρ_{2} / ρ_{1}$ .

Recall Solution 2.1

Plan: get the ratios from (8), (9), then $T_{2} / T_{1}$ from (10), then multiply by the given upstream values. With $M_{1}^{2} = 6.25$ , $γ = 1.4$ :

Density: $\frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 ) M _{1}^{2}}{( γ - 1 ) M _{1}^{2} + 2} = \frac{2.4 ( 6.25 )}{0.4 ( 6.25 ) + 2} = \frac{15}{4.5} = 3.333.$
Pressure: $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (6.25 - 1) = 1 + 1.1667 (5.25) = 1 + 6.125 = 7.125.$ $\Rightarrow P_{2} = 7.125 \times 40 = 285 kPa .$
Temperature: $\frac{T _{2}}{T _{1}} = \frac{P _{2} / P _{1}}{ρ _{2} / ρ _{1}} = \frac{7.125}{3.333} = 2.1375.$ $\Rightarrow T_{2} = 2.1375 \times 250 = 534.4 K .$ The gas leaves the shock at over 7× the pressure and more than double the absolute temperature.

Exercise 2.2. For the same shock ( $M_{1} = 2.5$ ), find the stagnation-pressure ratio $P_{02} / P_{01}$ and state the percentage of stagnation pressure lost.

Recall Solution 2.2

Tool: equation (11), the entropy fingerprint. Two bracketed factors, exponents $\frac{γ}{γ - 1} = 3.5$ and $\frac{1}{γ - 1} = 2.5$ .

First bracket: $\frac{\frac{( γ + 1 ) M _{1}^{2}}{2}}{1 + \frac{γ - 1}{2} M _{1}^{2}} = \frac{\frac{2.4 ( 6.25 )}{2}}{1 + 0.2 ( 6.25 )} = \frac{7.5}{2.25} = 3.3333.$ Raise to $3.5$ : $3.333 3^{3.5} = 67.72.$
Second bracket: $\frac{γ + 1}{2 γ M _{1}^{2} - ( γ - 1 )} = \frac{2.4}{2.8 ( 6.25 ) - 0.4} = \frac{2.4}{17.5 - 0.4} = \frac{2.4}{17.1} = 0.14035.$ Raise to $2.5$ : $0.1403 5^{2.5} = 0.007345.$
Product: $P_{02} / P_{01} = 67.72 \times 0.007345 = 0.4973.$ So about $50%$ of the stagnation pressure is destroyed by this shock.

Level 3 — Analysis

Goal: reason backwards, and probe why the numbers behave as they do.

Exercise 3.1. A pressure gauge reads $P_{2} / P_{1} = 4.5$ across a normal shock in air. Find the upstream Mach number $M_{1}$ that produced it, then find $M_{2}$ .

Recall Solution 3.1

Idea: invert equation (8). Because $P_{2} / P_{1}$ increases monotonically with $M_{1}$ , one pressure ratio pins down exactly one $M_{1}$ . $\frac{P _{2}}{P _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1) \Rightarrow M_{1}^{2} - 1 = \frac{γ + 1}{2 γ} (\frac{P _{2}}{P _{1}} - 1) .$ $M_{1}^{2} - 1 = \frac{2.4}{2.8} (4.5 - 1) = 0.85714 \times 3.5 = 3.0 \Rightarrow M_{1}^{2} = 4 \Rightarrow M_{1} = 2.0.$ Then equation (7): numerator $1 + 0.2 (4) = 1.8$ ; denominator $1.4 (4) - 0.2 = 5.4$ ; $M_{2}^{2} = 1.8/5.4 = 0.3333$ , so $M_{2} = 0.5774.$ This is the parent's Example 1, reached from the other direction.

Exercise 3.2. Show numerically that at $M_{1} = 5$ the density ratio (5.00) is closer to its infinite- $M_{1}$ ceiling than the pressure ratio is to anything finite. What is the density ceiling, and what fraction of it is reached at $M_{1} = 5$ ?

Recall Solution 3.2

The ceiling: take $M_{1} \to \infty$ in (9). Divide top and bottom by $M_{1}^{2}$ : $\frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 )}{( γ - 1 ) + 2/ M _{1}^{2}} M_{1} \to \infty \frac{γ + 1}{γ - 1} = \frac{2.4}{0.4} = 6.0.$ At $M_{1} = 5$ : $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 25 )}{0.4 ( 25 ) + 2} = \frac{60}{12} = 5.0.$ That is $5.0/6.0 = 83.3%$ of the ceiling. Contrast: $P_{2} / P_{1} = 1 + \frac{2.8}{2.4} (24) = 29$ and it keeps growing without bound as $M_{1} \to \infty$ . So density saturates while pressure runs away. Why: mass conservation ( $ρ_{2} u_{2} = ρ_{1} u_{1}$ ) plus a velocity that can only drop to a finite floor caps $ρ_{2}$ ; the leftover kinetic energy has nowhere to go but into $P$ and $T$ .

Level 4 — Synthesis

Goal: combine shock relations with an outside idea (speed of sound, inlets, entropy).

Exercise 4.1. A supersonic aircraft flies where $T_{1} = 216.7 K$ (tropopause). A normal shock stands in front of its inlet at $M_{1} = 2.0$ . Find the actual flow speeds $u_{1}$ and $u_{2}$ (in m/s) on each side. Use $R = 287 J kg^{- 1} K^{- 1}$ .

Recall Solution 4.1

Bridge tool — speed of sound. $u = M a$ where $a = γ R T$ (see Speed of Sound and Mach Number). We need $a$ on each side, so we need $T_{2}$ .

Upstream: $a_{1} = 1.4 \times 287 \times 216.7 = 87069 = 295.1 m/s$ . $u_{1} = M_{1} a_{1} = 2.0 \times 295.1 = 590.2 m/s .$
Downstream numbers: from Ex. 3.1 at $M_{1} = 2$ , $M_{2} = 0.5774$ , $T_{2} / T_{1} = 1.6875$ (parent Example 1). $T_{2} = 1.6875 \times 216.7 = 365.7 K$ ; $a_{2} = 1.4 \times 287 \times 365.7 = 146930 = 383.3 m/s$ . $u_{2} = M_{2} a_{2} = 0.5774 \times 383.3 = 221.3 m/s .$ Check with mass: $ρ_{2} / ρ_{1} = u_{1} / u_{2} = 590.2/221.3 = 2.667$ — matches equation (9)'s value $2.667$ exactly. Note the flow slowed from $590$ to $221$ m/s, yet $M_{2} < 1$ not only because $u$ fell but also because $a$ rose (hotter gas ⇒ faster sound). Both effects push $M = u / a$ down.

Exercise 4.2. An engineer must decelerate a $M_{1} = 2.0$ intake flow. Option A: one normal shock. Option B: two successive weaker shocks, the first at effective $M_{1} = 1.5$ leaving $M \approx 1.34$ , then a normal shock at $M = 1.34$ . Compare the stagnation-pressure survival $P_{02} / P_{01}$ of the two options and state which is better. (See Supersonic Inlets and Diffusers.)

Recall Solution 4.2

Tool: equation (11), applied once for A and twice (multiplied) for B, because stagnation ratios chain multiplicatively across successive shocks.

Option A, single shock at $M_{1} = 2.0$ : using (11) (parent Example 1e) $P_{02} / P_{01} = 0.7209.$
Option B, shock 1 at $M_{1} = 1.5$ :
- First bracket: $\frac{2.4 ( 2.25 ) /2}{1 + 0.2 ( 2.25 )} = \frac{2.7}{1.45} = 1.8621$ ; raised to $3.5$ gives $8.884$ .
- Second bracket: $\frac{2.4}{2.8 ( 2.25 ) - 0.4} = \frac{2.4}{5.9} = 0.40678$ ; raised to $2.5$ gives $0.10559$ .
- Product: $0.9382.$
Option B, shock 2 at $M = 1.34$ : (same formula, $M^{2} = 1.7956$ )
- First bracket: $\frac{2.4 ( 1.7956 ) /2}{1 + 0.2 ( 1.7956 )} = \frac{2.1547}{1.3591} = 1.5854$ ; raised $3.5$ → $5.022$ .
- Second bracket: $\frac{2.4}{2.8 ( 1.7956 ) - 0.4} = \frac{2.4}{4.6277} = 0.51861$ ; raised $2.5$ → $0.19365$ .
- Product: $0.9725.$
Combined B: $0.9382 \times 0.9725 = 0.9125.$ Verdict: Option B keeps about $91%$ of stagnation pressure vs. only $72.1%$ for the single strong shock. Two weak shocks beat one strong shock — the engineering rationale behind multi-shock supersonic inlets.

Level 5 — Mastery

Goal: derive a result, reason about limits and the second law.

Exercise 5.1. Using equations (7) and (8), prove that as $M_{1} \to 1^{+}$ every static ratio $\to 1$ (the shock becomes a sound wave). Then evaluate $P_{2} / P_{1}$ , $ρ_{2} / ρ_{1}$ , $T_{2} / T_{1}$ at $M_{1} = 1.001$ to confirm.

Recall Solution 5.1

The proof. In (8), the driver is $(M_{1}^{2} - 1)$ . As $M_{1} \to 1$ , $M_{1}^{2} - 1 \to 0$ , so $\frac{P _{2}}{P _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1) \to 1 + 0 = 1.$ In (9), set $M_{1}^{2} = 1$ : $\frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 )}{( γ - 1 ) + 2} = \frac{γ + 1}{γ + 1} = 1.$ Then by (10), $\frac{T _{2}}{T _{1}} = \frac{P _{2} / P _{1}}{ρ _{2} / ρ _{1}} \to \frac{1}{1} = 1.$ All three ratios collapse to 1 — no jump, no entropy rise, the "shock" is just an infinitesimal sound wave (isentropic, see Isentropic Flow Relations). Numerical confirmation at $M_{1} = 1.001$ ( $M_{1}^{2} = 1.002001$ ):

$P_{2} / P_{1} = 1 + 1.1667 (0.002001) = 1.002335.$
$ρ_{2} / ρ_{1} = \frac{2.4 ( 1.002001 )}{0.4 ( 1.002001 ) + 2} = \frac{2.404802}{2.400800} = 1.001667.$
$T_{2} / T_{1} = 1.002335/1.001667 = 1.000667.$ All within a fraction of a percent of 1 — precisely the "vanishingly weak shock."

Exercise 5.2. For $M_{1} = 2.0$ compute the specific entropy rise $Δ s = c_{p} ln \frac{T _{2}}{T _{1}} - R ln \frac{P _{2}}{P _{1}}$ using $c_{p} = 1004.5 J kg^{- 1} K^{- 1}$ , $R = 287$ . Then verify it reproduces the stagnation-pressure ratio via $P_{02} / P_{01} = e^{- Δ s / R}$ , and confirm the sign obeys the second law.

Recall Solution 5.2

Tool + why: entropy is the bridge between the static ratios (which we have) and the stagnation-pressure loss. This closes the loop with Entropy and Second Law and Rankine–Hugoniot Relations. From parent Example 1: $T_{2} / T_{1} = 1.6875$ , $P_{2} / P_{1} = 4.5$ . $Δ s = 1004.5 ln (1.6875) - 287 ln (4.5) .$

$ln (1.6875) = 0.52343 \Rightarrow 1004.5 \times 0.52343 = 525.79.$
$ln (4.5) = 1.50408 \Rightarrow 287 \times 1.50408 = 431.67.$ $Δ s = 525.79 - 431.67 = 94.12 J kg^{- 1} K^{- 1} .$ Second law: $Δ s = + 94.1 > 0$ — the shock is irreversible, entropy rises, as it must. ✓ Reproduce $P_{0}$ ratio: $\frac{P _{02}}{P _{01}} = e^{- Δ s / R} = e^{- 94.12/287} = e^{- 0.32796} = 0.7204.$ This matches the direct formula (11) value $0.7209$ (tiny rounding difference) — the two roads meet, as they must. The lesson: the positive entropy rise you computed is the stagnation-pressure loss, written a different way. Whenever a question mentions "irreversibility," "loss," or "wasted work," this is the quantity behind it.

Recall One-line self-test (cover the answers)

$M_{1} = 3$ gives $M_{2}$ ::: $\approx 0.475$ $M_{1} = 2.5$ gives $P_{2} / P_{1}$ ::: $\approx 7.125$ Density ceiling as $M_{1} \to \infty$ (air) ::: $\frac{γ + 1}{γ - 1} = 6$ Sign of $Δ s$ across any real shock ::: strictly positive Two weak shocks vs one strong (at overall $M_{1} = 2$ ): better for $P_{0}$ ? ::: two weak (91% vs 72%)