Section 9.1 · Introduction
Supersonic flow over a corner
A normal shock makes a 90° angle with the upstream flow. In general a shock makes an oblique angle with the flow; these are oblique shock waves, and the normal shock is the special case where the wave angle is 90°. Supersonic flows also contain expansion waves, across which the pressure decreases continuously.
The flow at a wall must stay tangent to the wall. At a concave corner the wall turns up through the deflection angle θ, so every streamline turns up, into the main bulk of the flow. At a convex corner the wall turns down, and the streamlines turn away from the flow.
Section 9.1 · Source of the waves
Mach waves: why the waves are oblique
Consider a small source of disturbance, a “beeper,” moving through a stagnant gas and emitting sound pulses. Each pulse spreads as a circle at the speed of sound a. When the beeper moves subsonically (V < a), it always stays inside its family of circles and the waves run ahead of it. When it moves supersonically (V > a), it outruns its own waves. The circles then pile up along a straight envelope BC, the Mach wave, tangent to every circle.
In time t the beeper travels Vt from A to B while the pulse emitted at A spreads to radius at. The right triangle ABC gives the Mach angle μ:
If the disturbances are stronger than sound waves, the envelope becomes an oblique shock at angle β > μ. The physical mechanism is the same. A Mach wave is the limiting case of an oblique shock: an infinitely weak one.
The source moves from right to left. B is its current position; A is where it emitted the oldest pulse shown; C is the tangent point of the Mach wave on that pulse.
A supersonic airplane flies at Mach at an altitude of km. Assume the shock pattern quickly coalesces into a Mach wave that reaches the ground behind the airplane. At the instant a bystander hears the boom, how far ahead of the bystander is the airplane?
Section 9.2
Oblique shock relations
Split the upstream velocity V1 into a component u1 normal to the shock and a component w1 tangential to it; do the same downstream. β is the shock (wave) angle and θ the deflection angle, both measured from the upstream flow direction. Apply the integral conservation laws to a control volume around the shock for steady, inviscid, adiabatic flow with no body forces. The faces parallel to the shock carry all the flux; on the other faces
Continuity, tangential momentum, normal momentum and energy. The tangential velocity does not change across the shock, and the tangential component w never appears in (9.2), (9.7) or (9.12).
Derivation from the integral conservation laws (notes pp. 77–80)
Continuity: on face a,
Momentum is a vector equation,
The normal component of momentum:
Energy,
Total enthalpy, and for a calorically perfect gas total temperature T0, is constant across the shock. Since V² = u² + w² and w1 = w2,
Equations (9.2), (9.7) and (9.12) are exactly the normal-shock equations (8.2), (8.6) and (8.10). The same algebra therefore gives the oblique-shock jumps in terms of the normal Mach number:
Section 9.2 · The vital chart
The θ–β–M relation
Equation (9.18) needs θ, but θ is not a third independent parameter. From the shock geometry,
Trigonometric rearrangement gives θ explicitly as a unique function of M1 and β:
Five things the chart tells you
- Maximum deflection. For each M1 there is a θmax. If the geometry demands θ > θmax, no straight attached shock exists; a curved shock detaches from the corner or the nose. θmax grows with M1, but only to sin−1(1/γ) = 45.58° as M1 → ∞ (γ = 1.4).
- Two solutions. For θ < θmax there are two straight shocks. The smaller β is the weak shock, the larger the strong shock. Nature usually picks the weak one; the strong solution needs a high downstream back pressure.
- θ = 0. Then β = 90° (a normal shock) or β = μ (a Mach wave). Neither deflects the streamlines.
- Raise M1 at fixed θ: β decreases and the shock becomes stronger. Lower M1 far enough and the shock detaches; for θ = 20° this happens below M1 = 1.84.
- Raise θ at fixed M1: β increases and the shock becomes stronger, until θ > θmax and the shock detaches; for M1 = 2 that is θ > 22.97°.
The weak branch gives supersonic M2 except in a thin band just below θmax, between the dotted M2 = 1 line and the dashed θmax line. The strong branch is always subsonic.
Worked examples
Oblique shocks: Examples 9.2–9.5
Each example solves the problem from the notes. The Exact column comes from the relations above; the Notes column is the hand solution read from the θ–β–M chart and Appendices A and B. Change any given value and the exact solution updates. The notes column applies to the original values only.
A supersonic flow with M1 = , p1 = atm and T1 = K is deflected at a compression corner through θ = °. Find M, p, T, p0 and T0 behind the oblique shock.
Caution. The p0,2/p1 column in Appendix B cannot be used for an oblique shock. It is computed from Eq. (8.80) with the actual downstream Mach number of a normal shock, not the normal component Mn,2. Use p0,2 = (p0,2/p0,1)(p0,1/p1)p1, as in the table.
An oblique shock has a wave angle β = °. The upstream Mach number is M1 = . Find the deflection angle, the pressure and temperature ratios, and M2.
The chart reading θ ≈ 6.5° is the main source of the difference in M2: with the exact θ = 6.72°, sin(β − θ) is smaller and M2 = 2.13.
An oblique shock has β = ° and a pressure ratio p2/p1 = . Find the upstream Mach number.
Invert (9.16) directly:
A Mach flow must be slowed to subsonic speed. (1) Pass it through a single normal shock. (2) First pass it through an oblique shock with β = °, then through a normal shock. Compare the final total pressures.
Supersonic inlets. A normal shock ahead of a pitot inlet causes a large total-pressure loss. In an oblique-shock inlet, a central cone or ramp creates an oblique shock, and the flow then passes through a relatively weak normal shock at the lip. For the same flight condition the total-pressure loss is smaller, so the engine produces more thrust. This is why most modern supersonic aircraft use oblique-shock inlets.
Section 9.4
Shock reflection and intersection
A shock that hits a solid wall must reflect, because the flow behind it, deflected by θ, would otherwise pass through the wall. The reflected shock turns the flow back through θ so that it runs parallel to the upper wall. The reflected shock is weaker than the incident shock, because M2 < M1, and it makes a smaller angle Φ with the wall. If M2 is too low for a straight attached shock to turn the flow by θ (θ > θmax(M2)), regular reflection is impossible. A normal shock, the Mach stem, forms at the wall instead. This is Mach reflection; its wave pattern requires numerical solution.
When a right-running and a left-running shock intersect, each continues as a refracted shock. The two streams behind them must have the same pressure and the same direction, but they have crossed different shocks, so their entropy, density and speed differ. They are separated by a slip line: p4 = p4′ and the flow directions match, while ρ4 ≠ ρ4′ and V4 ≠ V4′. When two left-running shocks intersect, they merge into a single stronger shock, and a weak reflected wave and a slip line adjust the flow so that both streams leave in the same direction.
Section 9.6
Prandtl–Meyer expansion waves
The expansion fan at a convex corner is a continuous region that can be visualized as an infinite number of Mach waves, each at the local Mach angle μ to the local flow direction. The fan is bounded by the forward Mach line at μ1 to the upstream flow and the rearward Mach line at μ2 to the downstream flow. Since ds = 0 across each Mach wave, the expansion is isentropic: p0 and T0 are constant through it, in direct contrast to the entropy rise across a shock.
The problem: given the upstream flow (region 1) and the deflection θ, find the downstream flow (region 2).
Derivation of the Prandtl–Meyer function (notes pp. 94–97)
Consider one Mach wave that turns the flow by an infinitesimal dθ. Momentum across the wave keeps the tangential velocity:
The binomial expansion
This differential equation describes the flow inside the expansion wave exactly.
Express dV/V in terms of M. With V = aM,
Integrate from M1 to M2:
ν is the Prandtl–Meyer function; ν(1) = 0. Appendix C tabulates ν and μ against M.
- For the given M1, find ν(M1).
- Compute ν(M2) = ν(M1) + θ.
- Invert ν to find M2.
- The wave is isentropic, so p0 and T0 are constant:
Streamlines curve smoothly and continuously through the fan; the turn is spread between the forward and rearward Mach lines. Lower chart: θ is the vertical distance between ν(M1) and ν(M2).
A flow with M1 = , p1 = atm and T1 = K expands around a sharp corner through θ = °. Find M2, p2, T2, p0,2, T0,2, and the angles of the forward and rearward Mach lines relative to the upstream flow.
The rearward Mach line sits at μ2 to the downstream flow, which is itself turned θ down, so it makes μ2 − θ with the upstream direction. The notes list T0 in atm; the unit is K.
Examples 9.10 and 9.11
Isentropic compression versus a sharp corner
Turn a supersonic flow into itself gradually, along a smooth concave surface, and each small turn sends out a weak compression Mach wave. The waves are isentropic, so the same Prandtl–Meyer relation applies with the sign reversed:
Downstream, the compression waves converge and coalesce into a shock away from the wall. Near the wall the compression stays isentropic. Turn the same flow through the same angle at a sharp corner and it crosses a single oblique shock instead.
Section 9.7
Shock-expansion theory: supersonic airfoils
Every surface of a sharp-edged airfoil is flat, so the pressure on each face is uniform. March the flow from the leading edge: on each face, decide whether the flow is turned into itself (oblique shock) or away from itself (Prandtl–Meyer expansion), find the face pressure, and integrate the pressures into forces. Region 1 is the freestream, with M1 and p1, and
Flat plate at angle of attack
The upper surface turns the flow away from itself by α: an expansion fan at the leading edge, so p2 < p1. The lower surface turns the flow into itself by α: an oblique shock, so p3 > p1. Per unit span, the pressure difference acts normal to the plate:
Hence cd/cl = tan α. At the trailing edge the two streams meet again: the upper flow turns back through a shock, the lower flow through an expansion, and a slip line separates them.
Wedge with a blunt base
Half-angle δ, chord c, face length ℓ = c/cos δ, base height t = 2c tan δ. The upper face (region 2) turns the flow by δ − α and the lower face (region 3) by δ + α. The base sits in separated flow; take pb = p1, as in Example 9.6. In body axes, with N′ normal to the chord and A′ along it:
At α = 0, p2 = p3 and the drag of Example 9.6 follows:
For a given geometry this inviscid wave-drag coefficient depends only on the Mach number.
Symmetric diamond wedge
Half-angle ε, side length ℓ, thickness t: sin ε = (t/2)/ℓ, and t/c = tan ε with c = 2√(ℓ² − t²/4). Regions 2 and 3 are the front and rear faces on top; 2′ and 3′ are the front and rear faces underneath. For |α| < ε the front faces see shocks; the shoulders always expand the flow through 2ε.
At α = 0, symmetry gives L′ = 0 and
At angle of attack, in body axes:
Region by region
Trailing-edge waves and the slip line are computed by matching pressure and flow direction behind the trailing edge. They do not change any surface pressure. For angle-of-attack sweeps, asymmetric diamonds, base-pressure studies and linear-theory comparisons, open the Supersonic Airfoil Calculator.
Self-check
Check your understanding
Pick an answer to see whether it is right and why. Use the interactives above to test your reasoning.
Summary
Key equations
Mach wave and oblique shock
p2/p1, ρ2/ρ1, T2/T1, Mn,2 and p0,2/p0,1: normal-shock relations evaluated at Mn,1. T0 is constant; p0 falls.