Prev
Unit 2 · Lesson 2-2
Normal and Oblique Shocks
2h 35m1300 XP
Next

Normal and Oblique Shocks

The normal shock relations, Prandtl's relation, entropy and total-pressure loss, oblique shocks and the θ–β–M relation

Rijeka, 1886
At the request of Ernst Mach, who had a theory of the sound of supersonic bullets but no way to prove it, the physicist Peter Salcher and his assistant Sándor Riegler set up an experiment in Rijeka: a bullet fired past a photographic plate, lit at the right instant by an electric spark. They recorded 80 shots — the first photographs of a projectile in flight (Salcher's experiment). Salcher and Mach published the results in 1887. Mach's photograph of a supersonic bullet shows what nobody had seen before: a sharp, curved wave standing in front of it, a bow shock (Ernst Mach). A shock is a jump in pressure, temperature and density across a layer thinner than a human hair. Every supersonic inlet has to live with one, and every supersonic engine pays for it in the same currency: stagnation pressure.
Why do shocks form at all? Pressure disturbances travel at the speed of sound. In subsonic flow they run ahead and warn the oncoming gas that an obstacle is there, so the flow adjusts smoothly. In supersonic flow they cannot get upstream: the gas arrives with no warning and must adjust all at once, inside a layer a few molecular free paths thick.
Across that layer mass, momentum and energy are still conserved. That is enough to fix the state behind the shock completely, given the state in front of it.
M22=1+γ−12M12γM12−γ−12,p2p1=1+2γγ+1(M12−1),ρ2ρ1=(γ+1)M122+(γ−1)M12M_2^2 = \frac{1 + \frac{\gamma - 1}{2}M_1^2}{\gamma M_1^2 - \frac{\gamma - 1}{2}}, \qquad \frac{p_2}{p_1} = 1 + \frac{2\gamma}{\gamma + 1}\left(M_1^2 - 1\right), \qquad \frac{\rho_2}{\rho_1} = \frac{(\gamma + 1)M_1^2}{2 + (\gamma - 1)M_1^2}

The normal shock relations: downstream Mach number, pressure ratio and density ratio in terms of the upstream Mach number M1. T2/T1 = (p2/p1)/(ρ2/ρ1).

Derivation
The normal shock relations
Take a thin control volume around the shock. With no area change, no heat and no work:
ρ1V1=ρ2V2,p1+ρ1V12=p2+ρ2V22,h1+12V12=h2+12V22.\rho_1 V_1 = \rho_2 V_2, \qquad p_1 + \rho_1 V_1^2 = p_2 + \rho_2 V_2^2, \qquad h_1 + \tfrac12 V_1^2 = h_2 + \tfrac12 V_2^2 .
Divide the momentum equation by the mass flux and use p/ρ=a2/γp/\rho = a^2/\gamma:
a12γV1+V1=a22γV2+V2.\frac{a_1^2}{\gamma V_1} + V_1 = \frac{a_2^2}{\gamma V_2} + V_2 .
The energy equation, written with the sonic speed a∗a^* of the stagnation state, gives a2=γ+12a∗2−γ−12V2a^2 = \frac{\gamma + 1}{2}a^{*2} - \frac{\gamma - 1}{2}V^2 on both sides (the same a∗a^*, since TtT_t does not change). Substituting and simplifying leaves
V1V2=a∗2,V_1 V_2 = a^{*2},
Prandtl's relation. With M∗=V/a∗M^{*} = V/a^* and M∗2=(γ+1)M22+(γ−1)M2M^{*2} = \frac{(\gamma + 1)M^2}{2 + (\gamma - 1)M^2}, the condition M1∗M2∗=1M_1^* M_2^* = 1 gives the M22M_2^2 formula. The density ratio is ρ2/ρ1=V1/V2=M1∗2\rho_2/\rho_1 = V_1/V_2 = M_1^{*2}, and the pressure ratio follows from the momentum equation.
💡What a normal shock does
  • The flow behind a normal shock is always subsonic: M1>1⇒M2<1M_1 > 1 \Rightarrow M_2 < 1.
  • Static pressure, temperature and density all jump up.
  • Stagnation temperature is unchanged — the shock is adiabatic.
  • Stagnation pressure falls, because the shock is irreversible. At Mach 1.2 the loss is 0.7 %; at Mach 2, 28 %; at Mach 3, 67 %.
📐A Mach 2 shock, worked through
The deck's example: a normal shock in a Mach 2.0 flow of air at T1=288T_1 = 288 K.
  • Tt=288×(1+0.2×4)=518.4T_t = 288 \times (1 + 0.2 \times 4) = 518.4 K, and the sonic speed of that stagnation state is a∗=2γγ+1RTt=416.6a^* = \sqrt{\frac{2\gamma}{\gamma + 1} R T_t} = 416.6 m/s.
  • Upstream V1=2×1.4×287×288=680.3V_1 = 2 \times \sqrt{1.4 \times 287 \times 288} = 680.3 m/s, so Prandtl's relation gives V2=a∗2/V1=255.1V_2 = a^{*2}/V_1 = 255.1 m/s.
  • M2=0.5774M_2 = 0.5774, p2/p1=4.5p_2/p_1 = 4.5, T2/T1=1.6875T_2/T_1 = 1.6875, so T2=486.0T_2 = 486.0 K and h2=cpT2=488h_2 = c_p T_2 = 488 kJ/kg.
  • Check: M2a2=0.5774×1.4×287×486.0=255.1M_2 a_2 = 0.5774 \times \sqrt{1.4 \times 287 \times 486.0} = 255.1 m/s.
  • The shock destroys 27.9 % of the stagnation pressure.
Lab 1 reproduces every number.
Why the shock only goes one way. Since TtT_t does not change, the entropy rise across the shock depends only on the stagnation pressures:
s2−s1=cpln⁡Tt2Tt1−Rln⁡pt2pt1=−Rln⁡pt2pt1.s_2 - s_1 = c_p \ln\frac{T_{t2}}{T_{t1}} - R \ln\frac{p_{t2}}{p_{t1}} = -R \ln\frac{p_{t2}}{p_{t1}} .
The normal shock relations give pt2/pt1<1p_{t2}/p_{t1} < 1 — entropy rises — only when M1>1M_1 > 1. For M1<1M_1 < 1 the same algebra would describe an expansion shock that lowers the entropy of an adiabatic flow, which the second law forbids. Supersonic flow can be slowed abruptly; it cannot be accelerated abruptly.
Oblique shocks. A wedge in a supersonic stream turns the flow through an angle θ\theta, and the turn happens across a straight shock inclined at a wave angle β\beta to the oncoming flow. The velocity component along the shock passes through unchanged; the component normal to it, M1sin⁡βM_1 \sin\beta, sees an ordinary normal shock. Because the normal component is smaller than M1M_1, an oblique shock is weaker — and loses far less stagnation pressure — than a normal shock in the same stream.
tan⁡θ=2cot⁡β  M12sin⁡2β−1M12(γ+cos⁡2β)+2\tan\theta = 2\cot\beta\;\frac{M_1^2\sin^2\beta - 1}{M_1^2\left(\gamma + \cos 2\beta\right) + 2}

The θ–β–M relation between the flow turning angle θ, the shock wave angle β and the upstream Mach number.

Derivation
The θ–β–M relation
Split the velocity into a component ww along the shock and components u1,u2u_1, u_2 normal to it. Momentum along the shock, with no pressure gradient in that direction, keeps ww the same on both sides. The geometry then gives
tan⁡β=u1w,tan⁡(β−θ)=u2w⟹tan⁡(β−θ)tan⁡β=u2u1=ρ1ρ2.\tan\beta = \frac{u_1}{w}, \qquad \tan(\beta - \theta) = \frac{u_2}{w} \quad\Longrightarrow\quad \frac{\tan(\beta - \theta)}{\tan\beta} = \frac{u_2}{u_1} = \frac{\rho_1}{\rho_2}.
The density ratio is the normal-shock one evaluated at the normal Mach number M1sin⁡βM_1\sin\beta:
ρ1ρ2=2+(γ−1)M12sin⁡2β(γ+1)M12sin⁡2β.\frac{\rho_1}{\rho_2} = \frac{2 + (\gamma - 1)M_1^2\sin^2\beta}{(\gamma + 1)M_1^2\sin^2\beta}.
Expanding tan⁡(β−θ)\tan(\beta - \theta) and solving for tan⁡θ\tan\theta gives the relation above. Behind the shock, M2=M2n/sin⁡(β−θ)M_2 = M_{2n}/\sin(\beta - \theta), with M2nM_{2n} from the normal-shock formula.
💡Weak, strong and detached
For a given M1M_1 and θ\theta the relation has two solutions. The weak one has the smaller β\beta and leaves the flow (usually) supersonic; it is what a wedge or a ramp produces in practice. The strong one is nearly normal and leaves the flow subsonic; it appears only when the pressure downstream forces it. Each Mach number also has a largest turning angle — 22.97° at Mach 2, 34.07° at Mach 3. Ask a wedge to turn the flow more than that and no attached shock exists: the shock detaches and stands ahead of the body as a curved bow shock, normal at its centre — the shape on Mach's photograph.
📐Slowing a Mach 2 stream for an engine
An engine's compressor needs subsonic air, so a Mach 2 inlet must end with a normal shock. How much stagnation pressure survives (Lab 3)?
  • One normal shock at Mach 2: 0.721 of the stagnation pressure survives.
  • A 10° ramp first: its oblique shock costs 1.5 % and leaves Mach 1.64, and the normal shock that follows is far weaker — 0.866 survives.
  • The best single ramp, 16.4°: 0.903.
  • Two 8° ramps: 0.935.
Slowing down in steps through weaker shocks is the whole art of the supersonic inlet. Lesson 5-1 turns this into real intake design.
I know of nothing more terrible than the poor creatures who have learned too much. Instead of the sound powerful judgement which would probably have grown up if they had learned nothing, their thoughts creep timidly and hypnotically after words, principles and formulae, constantly by the same paths.
– Ernst Mach, address of 16 April 1886, in Popular Scientific Lectures (1898), translated by Thomas J. McCormack (Wikiquote)
🕰Concorde's intakes
Concorde's four Olympus 593 turbojets sat behind variable intake ramps (Concorde), which changed the shock pattern in front of the engines as the aircraft accelerated. At Mach 2 the intake, not the engine, did much of the compression: the slowing of the air from Mach 2 in the intake raised its pressure severalfold before the compressor did any work at all.
💡Where you will use this
Lesson 5-1 designs external- and mixed-compression inlets with these relations. Lesson 5-2 finds shocks inside over-expanded nozzles. Lesson 6-2 meets them again in the blade passages of transonic fans, whose tips run supersonic relative to the air.
📚Further reading
  • John D. Anderson, Modern Compressible Flow: With Historical Perspective (3rd ed., McGraw-Hill, 2003), chapters 3 and 4.
  • Hans W. Liepmann and Anatol Roshko, Elements of Gasdynamics (Wiley, 1957; Dover reprint, 2001), on shock waves.
  • NASA Glenn Research Center, Normal shock waves and Oblique shock waves.
↓But there is a problem
Shocks are not the only thing that eats stagnation pressure. Walls rub the flow, and flames heat it, and both happen in every engine. Concorde carried an afterburner behind each engine. Crews lit it for take-off and to push through Mach 0.95 to 1.7, then shut it off for the rest of the flight. Fuel was one reason. The other is a piece of gas dynamics: you cannot keep adding heat to a moving gas.