Prandtl-Meyer expansion waves — isentropic, supersonic turning: Question bank

Before you start, two symbols we lean on everywhere below, stated in plain words so nothing is assumed:

$M$ = Mach number = flow speed $V$ divided by the local speed of sound $a$ ; $M > 1$ means "supersonic" (faster than its own sound).
$ν (M)$ = Prandtl–Meyer function, the angle (in degrees or radians) a stream has already turned away to accelerate from sonic ( $M = 1$ ) up to $M$ . Think of it as a running "turn score."
$μ = arcsin (1/ M)$ = Mach angle, the tilt of the weakest wave relative to the flow.

True or false — justify

Turning a supersonic flow around a convex (away-bending) corner raises its Mach number.

True. The wall giving way lets the gas spread out; it converts internal energy into speed, so

M

rises while

p, T, ρ

fall.

An expansion fan raises the entropy of the flow.

False. Each Mach wave is infinitesimally weak and its entropy jump scales like (strength)

^{3} \to 0

, so the total

Δ s = 0

— the fan is isentropic, unlike an oblique shock.

Because static pressure drops across an expansion fan, the stagnation pressure

p_{0}

also drops.

False. Isentropic + adiabatic means

p_{0}

and

T_{0}

are conserved; only the static values and their ratios to stagnation change.

The Mach angle

μ

gets larger as the flow moves downstream through the fan.

False.

M

increases, and

μ = arcsin (1/ M)

shrinks because

1/ M

shrinks — the waves get shallower, so the fan opens in the flow direction.

For an expansion turn you write

ν (M_{2}) = ν (M_{1}) - θ

False. Expansion adds the turn:

ν (M_{2}) = ν (M_{1}) + θ

. Subtracting would be the (hypothetical isentropic) compression direction.

A Prandtl–Meyer fan can turn a flow by any angle you like if you just make the corner sharp enough.

False. The turn is capped at

ν_{m a x} - ν (M_{1})

; even from sonic the total is

ν_{m a x} \approx 130.4 5^{\circ}

for

γ = 1.4

, reached only as

M \to \infty

(vacuum).

The individual Mach waves in the fan are all parallel to each other.

False. Each wave sits at

μ = arcsin (1/ M)

for the local

M

; since

M

changes across the fan, every wave has a different angle — that spread is exactly why it is a fan.

Reversing an expansion fan (running the same geometry backwards) gives you a valid isentropic compression.

Half-true. Mathematically

ν_{2} = ν_{1} - θ

describes an isentropic compression fan, but in reality a concave corner makes characteristics converge into a shock, so the smooth reverse is only an idealised limit.

The velocity component along a Mach wave is unchanged as the flow crosses it.

True. Only the component normal to the wave changes (like a very weak oblique shock); this fact is the seed of the whole

d θ = M^{2} - 1 d V / V

derivation.

A subsonic flow (

M < 1

) can also form a Prandtl–Meyer expansion fan around a corner.

False. Mach waves and the factor

M^{2} - 1

require

M > 1

; a subsonic flow adjusts smoothly and continuously with no wave structure.

Spot the error

"The fan cools the gas, so the total (stagnation) temperature

T_{0}

must fall."

The static temperature

T

falls as energy becomes kinetic, but

T_{0}

(which counts internal plus kinetic energy) is conserved in an adiabatic flow. Confusing static with stagnation is the error.

"Lower pressure means less push on the gas, so the flow slows down through the fan."

Backwards. The favourable pressure gradient (high behind, low ahead) accelerates the gas; lower static pressure is the result of speeding up, not a brake.

"Since a fan and a shock both turn supersonic flow, we can treat a strong expansion as a rarefaction shock."

There is no such thing as a stable rarefaction (expansion) shock — the second law forbids it (

Δ s < 0

). Expansions must spread into a smooth fan.

"The first Mach wave is shallow and the last is steep, since the flow is slowing."

Both facts are inverted. The flow speeds up, so the first wave (at lower

M_{1}

) is steep and the last (higher

M_{2}

) is shallow; the fan opens downstream.

"To find

M_{2}

, subtract the isentropic pressure ratio from

ν (M_{1})

Nonsense units — you cannot add a pressure ratio to an angle. You add the deflection angle

θ

ν (M_{1})

, invert

ν

to get

M_{2}

, and then use isentropic relations for pressure.

"A centred expansion fan has finite thickness because it contains infinitely many waves."

A fan centred at a sharp corner is geometrically zero-thickness at the vertex — infinitely many waves all emanate from one point; they only spread apart away from the corner.

"Method of characteristics can't handle expansion fans because the characteristics collide."

In an expansion the characteristics diverge, which is exactly why Method of characteristics works cleanly; collision (convergence) is the compression/shock case.

Why questions

Why does the factor

M^{2} - 1

appear in

d θ = M^{2} - 1 d V / V

It comes from

tan μ = 1/ M^{2} - 1

in the velocity triangle across a Mach wave; it is real only for

M > 1

, encoding that supersonic flow is required for a wave to exist.

Why do we integrate

d θ

from

M = 1

specifically to define

ν

M = 1

is the natural zero — the slowest speed that supports Mach waves — so

ν (1) = 0

gives a clean reference, letting any real turn be handled as a difference

ν (M_{2}) - ν (M_{1})

Why is

ν (M)

monotonically increasing, and why does that matter?

The integrand

\frac{M ^{2} - 1}{M ( 1 + \frac{γ - 1}{2} M ^{2} )}

is positive for all

M > 1

, so

ν

only rises; monotonicity guarantees each

ν

maps to exactly one

M

, making the inversion

ν \to M

unique.

Why can the simple rule

ν_{2} = ν_{1} + θ

ignore the details of the corner geometry?

Because

ν

measures turn-from-sonic and the isentropic state depends only on

M

; the flow "forgets" its path and cares only about the total angle turned, so you just add

θ

Why does the maximum turn angle depend on

γ

ν_{m a x} = \frac{π}{2} (\frac{γ + 1}{γ - 1} - 1)

comes from the

M \to \infty

limit;

γ

(the heat-capacity ratio) sets how much the gas can accelerate before pressure hits vacuum.

Why is expansion tied to a convex corner and shock to a concave one?

Convex bends the wall away, letting characteristics diverge (fan); concave bends into the flow, forcing them to converge and pile up into a shock.

Why does the pressure ratio use an exponent

γ / (γ - 1)

rather than just

γ

That exponent comes from the isentropic relation

p / p_{0} = (T / T_{0})^{γ / (γ - 1)}

; it ties pressure to temperature along a constant-entropy path, which is exactly what a fan is.

Edge cases

What is

ν

and

μ

exactly at

M = 1

ν (1) = 0^{\circ}

(no turning yet) and

μ = arcsin (1) = 9 0^{\circ}

— the "Mach wave" is normal to the flow, i.e. a sound front the flow can barely outrun.

What happens to

μ

M \to \infty

μ = arcsin (1/ M) \to arcsin (0) = 0^{\circ}

; the waves become parallel to the flow, so an infinitely fast stream carries no forward-tilting disturbance.

A flow starts exactly at

M_{1} = 1

and turns

θ = 2 0^{\circ}

— how do you get

M_{2}

Since

ν (1) = 0

, simply

ν (M_{2}) = 2 0^{\circ}

; read

M_{2}

directly from the inverse of

ν

(about

M_{2} \approx 1.78

for

γ = 1.4

What physically limits the flow if you try to turn past

ν_{m a x}

The gas would need

M = \infty

, zero static pressure and zero temperature — a vacuum; the stream simply separates and cannot fill the extra angle, leaving a vacuum region.

Is a fan possible for a flow that is only just supersonic, say

M_{1} = 1.01

Yes, but the available turn is nearly the full

ν_{m a x}

(since

ν (1.01) \approx 0

), and the first Mach wave is almost normal (

μ

near

9 0^{\circ}

) — the fan is very "steep" at its start.

What is the entropy change of a single Mach wave in the limit of zero strength?

Exactly zero to leading order;

Δ s

scales as (wave strength)

^{3}

, so an infinitesimal wave contributes no entropy — this is why the integral of infinitely many still gives

Δ s = 0

If two expansion fans from opposite walls meet, does the flow suddenly shock?

No — two expansions crossing stay isentropic; Method of characteristics resolves the intersection smoothly. It is compressive characteristics meeting that coalesce into shocks.

How does an over-expanded nozzle relate to these fans?

An under-expanded jet exits and turns away through expansion fans to match the lower ambient pressure; an over-expanded one instead compresses through oblique shocks — same corner, opposite pressure mismatch.

Recall One-line summary of the traps

Expand = Away = Add: $M ↑$ , $p, T ↓$ , $μ ↓$ , $p_{0}, T_{0}$ fixed, $Δ s = 0$ . Everything is a difference of $ν$ , the turn cap is $ν_{m a x}$ , and only convex corners fan while concave corners bang.