The Hugoniot & the CJ Point

The question

Start a hydrogen–air detonation with a spark and a long tube, with a blasting cap, or with another detonation, and within a few centimetres it settles to the same speed: about 1970 metres per second. Change the starting pressure tenfold and the speed barely moves. How does a detonation choose its own speed?

Two conservation laws make a line

Ride along with a steady wave. Fresh gas streams in from ahead at the wave speed $\cdet{D}$ , crosses the front, and leaves behind it at some slower speed $u_2$ . Whatever happens inside the front, mass and momentum must be conserved between the two sides:

$\rho_1 \cdet{D} = \rho_2 u_2, \qquad \cshock{p_1} + \rho_1 \cdet{D}^2 = \cshock{p_2} + \rho_2 u_2^2.$

Eliminate $u_2$ and something elegant falls out: the final pressure and specific volume ( $v = 1/\rho$ ) must lie on a straight line through the initial state,

$\cshock{p_2} - \cshock{p_1} = -(\rho_1 \cdet{D})^2 \,(v_2 - v_1).$

This is the . Its slope is set by the wave speed: a faster wave has a steeper line.

The third law makes a curve

Energy conservation gives a second relation, between the end states alone: the . For a shock with no chemistry, the curve passes through the initial state. If the gas also releases its heat of reaction $q$ , the end states have extra energy to account for, and the curve shifts outward, away from the starting point. Every possible end state of a steady wave must lie on both the Rayleigh line and the Hugoniot.

ExploreWhere the line touches the curve

Wave speedMach 6.00

Heat release q/(RT₁)25.0

about 29 for hydrogen–air

CJ Mach number4.50

Crossings2

CJ pressure11.5 × p₁about half the von Neumann spike

Two crossings: a strong and a weak solution. Only a piston pushing from behind could hold the strong one.

Challenges0/3

Find a wave speed with two crossings.
Make the Rayleigh line miss the Hugoniot.
Swing the line until it just touches: the CJ point.

The pressure–volume plane for a gas ahead of a wave (white dot) and behind it. Straight lines are Rayleigh lines (mass and momentum conservation) for different wave speeds; the curves are Hugoniots (energy conservation) — purple for a shock with no reaction, orange for fully burnt gas. Drag on the plot to swing the Rayleigh line.

Play with it and you will find three cases. A fast wave crosses the burnt-gas Hugoniot twice. A slow wave misses it altogether: it cannot release all its heat and stay steady. And there is exactly one wave speed whose line just touches the curve — the , found independently by David Chapman in 1899 and Émile Jouguet in 1905.

Why the tangent point wins

Of all the speeds that conservation allows, why do real detonations pick the tangent one? Because of what happens behind them. In a tube closed at one end, the burnt gas must eventually come to rest, so an expansion wave — a rarefaction — always follows the detonation, trying to catch it and drain its pressure.

At a strong point (the upper crossing), the burnt gas flows away from the front subsonically. The rarefaction can catch up and weaken the front, which slows down until it is no longer strong. A strong detonation needs a piston pushing from behind to survive.
At the CJ point, the burnt gas leaves the front at exactly the local speed of sound. A rarefaction behind it can never overtake it. The wave is shielded from whatever happens downstream: it is self-sustained.
Weak solutions would require the gas to burn without first being shocked to the von Neumann state, which (for ordinary mixtures) the chemistry does not allow.

A detonation calculator

The geometry gives the speed; the same equations then give the pressure, temperature and density behind the wave. For real mixtures, the heat released is reduced by dissociation — at 3000 K and above, water and carbon dioxide partly break apart — and the products' properties change with temperature. The calculator below includes both, in a calibrated two-γ model.

CalculatorWhat a detonation does to a mixture

Equivalence ratio φ1.00 · stoichiometric

Initial pressure1.0 atm

Initial temperature300 K

CJ speed1967 m/sMach 4.81

CJ pressure14.9 atmspike 27 atm

CJ temperature2895 Kspike 1635 K

Gas speed behind840 m/sfollowing the wave

Challenges0/3

Find a mixture that detonates faster than 2800 m/s (now: H₂–air).
Change the initial pressure by a factor of 20 and watch how little the speed moves.
Reach a CJ pressure above 50 atm.

The Chapman–Jouguet state for real fuel–oxidizer mixtures, from a two-γ model calibrated against equilibrium calculations (Shepherd's Shock & Detonation Toolbox): within about 1% on speed and 3.5% on pressure and temperature at stoichiometric conditions. The plot shows the CJ speed of all four fuels against equivalence ratio.

Estimate first

What is the CJ pressure behind a stoichiometric hydrogen–air detonation starting at 1 atm, in atmospheres?

1101001,000

Drag along the line to place your guess — each tick is ten times the last.

Going deeper
Going deeper: the one-γ formulas

If the reactants and products share the same $\gamma$ and gas constant, the CJ Mach number has a closed form:

$M_{CJ} = \sqrt{1 + \mathcal{H}} + \sqrt{\mathcal{H}}, \qquad \mathcal{H} = \frac{(\gamma^2-1)}{2\gamma}\,\frac{q}{R T_1}.$

For a strong detonation ( $\mathcal{H} \gg 1$ ) this gives $\cdet{D_{CJ}} \approx \sqrt{2(\gamma^2 - 1)\,q}$ : the speed depends only on the heat released per unit mass and on $\gamma$ — not on the initial pressure, and only weakly on the initial temperature. The pressure behind the wave is $\cshock{p_{CJ}} \approx \rho_1 \cdet{D_{CJ}}^2/(\gamma+1)$ , which does scale with the initial density. These are the formulas behind the explorer above; the calculator uses separate γ's for reactants and products, and a fitted dissociation model.

Check your intuition

An engineer doubles the initial pressure of a hydrogen–air mixture. What happens to the detonation?

Recap

What to carry forward

Mass and momentum conservation put every end state on a straight Rayleigh line through the initial state; its slope is $-(\rho_1 D)^2$ .
Energy conservation puts it on the Hugoniot, which shifts outward when heat is released.
A self-sustained detonation runs where the line just touches the curve: the Chapman–Jouguet point, where the products leave at the local speed of sound.
$D_{CJ}$ depends on the heat released per unit mass, not on initial pressure; $p_{CJ}$ scales with the initial density.

Next: the CJ theory treats the front as a thin sheet. What actually happens inside it?