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

Before we start, one shared vocabulary reminder so no symbol is ever used before it is defined:

Definition The cast of characters

$M_{1}$ = Mach number just upstream (flow speed ÷ local sound speed before the shock); $M_{1} > 1$ always.
$M_{2}$ = Mach number just downstream; $M_{2} < 1$ always.
$u$ = the flow speed of the gas (metres per second) — how fast the gas physically streams through the shock. $u_{1}$ is upstream, $u_{2}$ downstream; the gas slows down, so $u_{2} < u_{1}$ .
Static properties $P, T, ρ$ = what a thermometer/barometer riding with the gas reads. $P$ = pressure, $T$ = temperature, $ρ$ (rho) = density (mass per volume).
Stagnation properties $P_{0}, T_{0}$ = what you'd measure if you brought the gas smoothly and reversibly to rest. $T_{0}$ tracks total energy; $P_{0}$ tracks how much of that energy is still usable.
$γ$ (gamma) = the ratio of specific heats, $γ = c_{p} / c_{v}$ . It's a fixed property of the gas that says how "springy" it is; for ordinary air $γ = 1.4$ .
$c_{p}$ = the specific heat at constant pressure — how many joules it takes to raise one kilogram of the gas by one degree while the pressure is held fixed. ( $c_{v}$ is the same at constant volume; their ratio is $γ$ .)
$R$ = the specific gas constant — the per-kilogram constant in the ideal-gas law $P = ρR T$ . It also appears in the entropy formula below.
$Δ s$ = entropy change across the shock; the "irreversibility meter". For a shock $Δ s > 0$ .

True or false — justify

True or false: A normal shock is adiabatic, therefore it is isentropic.

False. Adiabatic means no heat crosses the boundary, but entropy can still be generated internally by the violent, irreversible squeezing; here

Δ s > 0

, so it is adiabatic yet not isentropic.

True or false: Stagnation temperature

T_{0}

is the same on both sides of the shock.

True. The flow is adiabatic and does no shaft work, so total energy (hence

T_{0}

) is conserved — the energy

c_{p} T_{0}

carried in equals the energy

c_{p} T_{0}

carried out.

True or false: Stagnation pressure

P_{0}

is also conserved because the shock is adiabatic.

False.

P_{0}

measures usable energy, which depends on entropy via

P_{02} / P_{01} = e^{- Δ s / R}

(Eq. 11); since

Δ s > 0

, stagnation pressure always drops.

True or false: Because energy is conserved, no "loss" happens across a shock.

False. Total energy is conserved, but availability (work-extractable energy) is lost — the entropy rise degrades high-grade kinetic energy into low-grade thermal energy.

True or false: For any

M_{1} > 1

the downstream flow is subsonic.

True. Equation (7) gives

M_{2} < 1

for every

M_{1} > 1

; the second law forbids the reverse (subsonic → supersonic across a shock).

True or false: A weaker shock (

M_{1}

closer to 1) loses less stagnation pressure.

True. As

M_{1} \to 1^{+}

Δ s \to 0

and

P_{02} / P_{01} \to 1

(Eq. 11); the shock becomes a lossless sound wave in the limit.

True or false: Density can be increased without bound by making the shock arbitrarily strong.

False. As

M_{1} \to \infty

, Eq. (9) gives

ρ_{2} / ρ_{1} \to \frac{γ + 1}{γ - 1} = 6

for air; mass conservation plus a finite velocity drop caps compression, and the extra energy loads into

T

and

P

instead.

True or false: Across a shock the static temperature and the static pressure both rise.

True. The gas is compressed and decelerated, so

P

T

, and

ρ

all jump up while

u

and

M

drop.

True or false: The velocity ratio

u_{1} / u_{2}

equals the density ratio

ρ_{2} / ρ_{1}

True. Continuity

ρ_{1} u_{1} = ρ_{2} u_{2}

rearranges directly to

u_{1} / u_{2} = ρ_{2} / ρ_{1}

(the middle equality in Eq. 9); whatever fraction the gas densifies, its speed slows by.

True or false: Three shocks of

M_{1} = 1.5

each lose the same total stagnation pressure as one shock of

M_{1} = 1.5

False. Each shock generates its own entropy, so three shocks lose more

P_{0}

than one — but far less than a single strong shock that swallows the whole Mach drop, which is why staged oblique shocks are used.

Spot the error

Find the flaw: "The flow is adiabatic, so I'll read

P_{2} / P_{1}

straight off the isentropic relations."

Isentropic relations assume

Δ s = 0

, but a shock has

Δ s > 0

; use the normal-shock relations (8)–(11) through the shock — isentropic relations apply only up to it and after it.

Find the flaw: "

M_{1} = 0.8

, so I'll compute the shock properties from equation (7)."

There is no shock — shocks require

M_{1} > 1

. Feeding

M_{1} < 1

into (7) gives a mathematically valid but physically forbidden

M_{2} > 1

that would require entropy to decrease.

Find the flaw: "

T_{0}

is conserved, and

T

jumps, so

M

must rise to keep

T_{0} = T (1 + \frac{γ - 1}{2} M^{2})

balanced."

Backwards: since

T

rises while

T_{0}

is fixed, the bracket must shrink, so

M

must fall. The rising static temperature is exactly why

M_{2} < M_{1}

Find the flaw: "A stronger shock heats the gas more, so

T_{0}

downstream is higher."

T_{0}

is conserved regardless of shock strength; what a stronger shock does is convert more kinetic energy into static

T

(raising

T_{2}

), while

T_{0}

stays pinned to the incoming total energy.

Find the flaw: "Since

P_{02} < P_{01}

, the shock has removed energy from the flow."

No energy is removed (

T_{0}

constant); the shock has removed order/availability. Same total energy, but less of it can be turned back into useful work.

Find the flaw: "In a supersonic inlet I want one big normal shock to slow the flow efficiently."

A single strong normal shock destroys huge stagnation pressure (e.g. 94% lost at

M_{1} = 5

); efficient inlets use several weak oblique shocks to decelerate with far smaller total entropy rise.

Why questions

Why does a normal shock exist at all instead of a smooth deceleration?

Downstream pressure disturbances travel at the speed of sound and cannot outrun a supersonic flow to "warn" it; pressure piles up until it collapses into a paper-thin discontinuity — the shock.

Why is the momentum equation written as

P + ρ u^{2}

rather than just

P

Because the net pressure force on the control volume changes the flow's momentum flux;

ρ u^{2}

is that momentum flux per area (density times speed times speed), so pressure and momentum-flux together are what's balanced across the shock.

Why does density saturate at

\frac{γ + 1}{γ - 1}

but pressure and temperature grow without bound?

Continuity ties density to the velocity drop, which is finite; once velocity is nearly halted,

ρ

can't rise further, so the shock's remaining energy is dumped into

P

and

T

instead.

Why do we get two algebraic roots from the combined equations, and which is the shock?

Squaring to eliminate the roots introduces the trivial root

M_{2} = M_{1}

(the "no shock"/sound-wave solution); the genuine shock is the other root, (7), which satisfies

Δ s > 0

Why is stagnation pressure the "engineer's enemy" while stagnation temperature is not?

P_{0}

sets how much thrust/work the engine can extract; losing it directly costs performance.

T_{0}

is conserved by energy balance, so it isn't "lost" to fight over.

Why can a shock only turn supersonic into subsonic and never the reverse?

The reverse process would carry

Δ s < 0

, violating the second law; entropy generation is one-directional, so the flow can only jump from

M_{1} > 1

down to

M_{2} < 1

Edge cases

Edge case: What happens to every property ratio as

M_{1} \to 1^{+}

They all approach 1 and

Δ s \to 0

— the shock degenerates into an infinitesimally weak, isentropic sound wave with no jump.

Edge case: What is

M_{2}

in the exact limit

M_{1} \to \infty

for air?

From Eq. (7),

M_{2}^{2} \to \frac{γ - 1}{2 γ} = \frac{0.4}{2.8} \approx 0.143

, so

M_{2} \to 0.378

— the downstream Mach number bottoms out, it does not go to zero.

Edge case: What does

P_{02} / P_{01}

approach as

M_{1} \to \infty

, and what's the physical meaning?

It tends toward 0 — an infinitely strong shock destroys essentially all usable stagnation pressure, which is why hypersonic inlets suffer catastrophic

P_{0}

loss.

Edge case: If the "shock" had

M_{1} = 1

exactly, is it still a shock?

No — at

M_{1} = 1

there is no jump (

Δ s = 0

, all ratios equal 1); it is the boundary case that is just a Mach wave, the weakest possible disturbance.

Edge case: A gas with

γ \to 1

(many internal degrees of freedom) — what happens to the maximum density ratio?

\frac{γ + 1}{γ - 1} \to \infty

, so such a gas can be compressed enormously by a strong shock, because energy hides in internal modes instead of pushing density's ceiling down.

Recall One-line summary of every trap here

The two ideas that resolve almost every card: (1) adiabatic ≠ isentropic — energy ( $T_{0}$ ) survives, availability ( $P_{0}$ ) does not; (2) the second law sets the arrow — always $M_{1} > 1 \to M_{2} < 1$ , never the reverse, and density compression is capped while $P$ and $T$ are not.

Two chalkboard figures make the whole page visual: one shows the structure of the shock and which quantities jump up vs. down, the other plots how the ratios grow with $M_{1}$ (and where density saturates).

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