Chapman-Jouguet detonation; deflagration vs detonation: Visual walkthrough

Step 1 — Stand still on the burning wall

WHAT. A reaction front is a thin sheet where cold fuel turns into hot exhaust. Instead of watching it rush toward us, we jump onto the front and ride it. Now the front is frozen in place and the gas is the thing moving: it flows in from the cold side and out the hot side.

WHY. In the lab the front moves and the numbers change every instant — hard to write equations. Ride the front and everything becomes steady: what flows in per second equals what flows out per second. Steady = bookkeeping we can actually balance.

PICTURE. Look at the figure. The vertical band is the front. Left (blue) is state 1 — the unburnt gas. Right (red) is state 2 — the burnt gas. The two speeds are:

$u_{1}$ = how fast cold gas rushes into the front (arrow pointing right, into the band).
$u_{2}$ = how fast hot gas leaves out the back.

Figure — Chapman-Jouguet detonation; deflagration vs detonation

Definition The state variables and their meanings

$ρ$ (rho) = density, mass packed into one cubic metre — "how crowded the molecules are."
$p$ = pressure, the push the gas exerts on the walls.
$u$ = flow speed through the front, measured in the front's own frame.
$T$ = temperature.
$c_{p}$ = specific heat at constant pressure: the heat (in joules) needed to warm one kilogram of the gas by one degree while it is free to expand. It is a fixed property of the gas — bigger for gases that "soak up" heat easily.
$h = c_{p} T$ = specific enthalpy, the heat content of one kilogram. It is just temperature $T$ rescaled by $c_{p}$ so it comes out in joules per kilogram — the same units as motion energy, so the two can be added.

Subscript 1 = cold reactants, 2 = hot products. Every symbol from here on is one of these, or built from these.

Step 2 — Three things that cannot disappear

WHAT. Across the front, three quantities are conserved: mass, momentum, and energy. We write one equation for each. The energy equation uses the heat content $h$ we just defined in Step 1.

WHY. A front does not create or destroy matter, it does not get free pushes, and it does not invent energy — it only converts chemical energy into heat and motion. Those three "cannot cheat" rules are the entire physics. Everything else is algebra.

PICTURE. The figure stacks the three balances as a cold column (left, state 1) equalling a hot column (right, state 2). The energy row has an extra brick $q$ dropped in — the chemical fuel.

This is the same as Rankine-Hugoniot jump conditions but with the fuel brick $q$ added.

Step 3 — Rename density as "room per kilogram"

WHAT. Replace $ρ$ everywhere with $v = 1/ ρ$ , the specific volume — the room one kilogram takes up.

WHY. We want to draw the story on a graph of pressure vs volume, because that is the picture engineers already know (it is the same $(v, p)$ plane as an engine cycle, see Constant-volume vs constant-pressure combustion cycles). Density crammed into one axis is awkward; "space per kg" plots cleanly.

PICTURE. The figure shows one kilogram as a stretchy box. Cold gas = a big box $v_{1}$ at low pressure; compressed gas = a small box $v_{2}$ at high pressure. Same mass, less room.

From $\overset{m}{˙} = ρ u = u / v$ we get the two speeds cleanly: $u_{1} = \overset{m}{˙} v_{1}, u_{2} = \overset{m}{˙} v_{2} .$ Each velocity is just "flux × room." We will substitute these next.

Step 4 — The Rayleigh line: momentum draws a straight line

WHAT. Feed $u_{1} = \overset{m}{˙} v_{1}$ and $u_{2} = \overset{m}{˙} v_{2}$ into the momentum law. Out pops a straight line in the $(v, p)$ plane.

WHY. We are hunting for the final state $(v_{2}, p_{2})$ . Momentum alone can't give the answer, but it constrains it to a line — half the puzzle. A line is the simplest possible constraint, so we cash it in first.

PICTURE. In the figure, the dot at $(v_{1}, p_{1})$ is the starting cold gas. Every allowed final state lies on the straight line through it. The line tilts downward — that sign is the whole plot's punchline.

The momentum law $p_{1} + \overset{m}{˙}^{2} v_{1} = p_{2} + \overset{m}{˙}^{2} v_{2}$ rearranges to:

Step 5 — The Hugoniot curve: energy draws a curve

WHAT. Now use the energy law (the one carrying $q$ ) and eliminate the velocities. Result: a curved relation between $p_{2}$ and $v_{2}$ — the Hugoniot curve.

WHY. The Rayleigh line gave a line; we need one more independent condition to pin a single point. Energy supplies the curve. Where line meets curve = the actual final state. (Line + curve = one answer, the classic "two equations, one intersection" move.)

PICTURE. The figure adds a curved arc to the previous plot. Notice it does not pass through the cold point $(v_{1}, p_{1})$ — the fuel $q$ lifts the whole curve up and out, away from the starting dot. That gap is the chemical energy, drawn.

Before we write the curve, we need one new gas property:

Substituting $h = \frac{γ}{γ - 1} p v$ for an ideal gas into the energy law gives:

Step 6 — Slide the line: two hits, one graze, or a miss

WHAT. Fix the cold state. Now dial $\overset{m}{˙}$ (the front speed) and watch the Rayleigh line pivot about $(v_{1}, p_{1})$ . As it steepens it meets the detonation branch of the Hugoniot in two points, then one, then none.

WHY. Mathematically many speeds "work." Physics must choose one. To find the special one, we sweep and see which case is the boundary between "solutions exist" and "no solution."

PICTURE. Three coloured Rayleigh lines fan out from the cold point:

Yellow (shallow) → misses the detonation branch entirely: too slow, no detonation.
Blue (medium) → cuts it twice: a "strong" (upper) and "weak" (lower) solution.
Green (steepest that still touches) → just grazes — one tangent point. This is the borderline.

Step 7 — Why the graze means "moving out at the speed of sound"

WHAT. Show that at the tangent point the burnt gas leaves the front at exactly its local speed of sound: $u_{2} = a_{2}$ , i.e. Mach number $M_{2} = 1$ . The flow is choked.

WHY. We must prove that the geometric event "Rayleigh line grazes the Hugoniot" is the same as the physical event " $M_{2} = 1$ " — otherwise the boxed CJ condition is just an assertion.

PICTURE. The figure zooms into the graze. A little sound-wave ripple is drawn behind the front trying to catch up. When $u_{2} = a_{2}$ , the gas is running out as fast as sound can travel backward through it, so the ripple can never reach the front. The front is sealed off from anything behind it — self-sustaining.

The proof that tangency $⟺ M_{2} = 1$ , one line at a time.

(i) Slope of the Rayleigh line. From Step 4 the line is $p = p_{1} - \overset{m}{˙}^{2} (v - v_{1})$ , so its slope is simply $(\frac{d p}{d v})_{Ray} = - \overset{m}{˙}^{2} .$

(ii) Slope of the Hugoniot curve. The Hugoniot ties $p_{2}$ and $v_{2}$ together, so as we move a tiny step $d v_{2}$ along it, $p_{2}$ shifts by $d p_{2}$ . Differentiate the adiabat $\frac{γ}{γ - 1} (p_{2} v_{2} - p_{1} v_{1}) = q + \frac{1}{2} (p_{2} - p_{1}) (v_{1} + v_{2})$ , holding the fixed cold state $(p_{1}, v_{1})$ and $q$ constant. Using the product rule on each side: $\frac{γ}{γ - 1} (p_{2} d v_{2} + v_{2} d p_{2}) = \frac{1}{2} (d p_{2} (v_{1} + v_{2}) + (p_{2} - p_{1}) d v_{2}) .$ Collect the $d p_{2}$ terms on the left and the $d v_{2}$ terms on the right: $d p_{2} [\frac{γ}{γ - 1} v_{2} - \frac{1}{2} (v_{1} + v_{2})] = d v_{2} [\frac{1}{2} (p_{2} - p_{1}) - \frac{γ}{γ - 1} p_{2}],$ so the curve's slope is the ratio of those brackets: $(\frac{d p}{d v})_{Hug} = \frac{\frac{1}{2} ( p _{2} - p _{1} ) - \frac{γ}{γ - 1} p _{2}}{\frac{γ}{γ - 1} v _{2} - \frac{1}{2} ( v _{1} + v _{2} )} .$

(iii) Impose tangency. Tangency means the two slopes are equal, $(\frac{d p}{d v})_{Ray} = (\frac{d p}{d v})_{Hug}$ . At the tangent point the state also lies on both the line and the curve, so we may substitute $p_{2} - p_{1} = - \overset{m}{˙}^{2} (v_{2} - v_{1})$ (the Rayleigh relation) into the numerator. Doing so and simplifying — the algebra is a few lines of cancelling — collapses the whole equality down to the clean statement $\overset{m}{˙}^{2} v_{2}^{2} = γ p_{2} v_{2} .$ The right-hand side is exactly $a_{2}^{2} = γ p_{2} v_{2}$ (the speed-of-sound definition above), so $\overset{m}{˙}^{2} v_{2}^{2} = a_{2}^{2} .$

(iv) Read off the result. But $\overset{m}{˙} v_{2} = u_{2}$ from Step 3, so $\overset{m}{˙}^{2} v_{2}^{2} = u_{2}^{2}$ . Therefore $u_{2}^{2} = a_{2}^{2}$ , i.e. $u_{2} = a_{2}$ .

Step 8 — Turn tangency into a speed formula

WHAT. Impose $M_{2} = 1$ plus the strong-detonation shortcut $p_{2} ≫ p_{1}$ on the jump laws, and solve for the front speed. We keep the algebra visible so you see which small terms get dropped and why.

WHY. The graph tells us the CJ point exists; now we want the actual number an engineer plugs into an Rotating Detonation Engine (RDE) design. Algebra converts the picture into m/s.

PICTURE. The figure shows the collapse: three inputs ( $q$ , $γ$ , $M_{2} = 1$ ) funnelling into one output arrow labelled $D_{C J}$ .

Formula The strong-detonation algebra, step by step

1. Sonic exit ( $M_{2} = 1$ ). From Step 7, $u_{2} = a_{2} = γ p_{2} v_{2}$ , so $u_{2}^{2} = γ p_{2} v_{2} .$

2. Momentum. The exact law is $p_{1} + \overset{m}{˙}^{2} v_{1} = p_{2} + \overset{m}{˙}^{2} v_{2}$ . Drop $p_{1}$ next to $p_{2}$ (the rule above): $\overset{m}{˙}^{2} (v_{1} - v_{2}) \approx p_{2} .$ With $\overset{m}{˙} = u_{1} / v_{1} = u_{2} / v_{2}$ this rearranges (dividing by the sonic relation $u_{2}^{2} = γ p_{2} v_{2}$ ) to the compression ratio $\frac{v _{2}}{v _{1}} = \frac{γ}{γ + 1},$ which we now derive rather than assume — see the box below.

3. Energy. The exact law is $\frac{γ}{γ - 1} (p_{2} v_{2} - p_{1} v_{1}) = q + \frac{1}{2} (p_{2} - p_{1}) (v_{1} + v_{2})$ . Drop $p_{1} v_{1}$ and $p_{1}$ (the rule again): $\frac{γ}{γ - 1} p_{2} v_{2} \approx q + \frac{1}{2} p_{2} (v_{1} + v_{2}) .$ Substitute $v_{1} = v_{2} (γ + 1) / γ$ from step 2 and simplify: $q = \frac{p _{2} v _{2}}{γ - 1} \cdot \frac{γ + 1}{2} .$

4. Get $D_{C J}$ . From step 2, $u_{2}^{2} = γ p_{2} v_{2}$ and $\overset{m}{˙}^{2} v_{1}^{2} = u_{1}^{2} = D_{C J}^{2}$ ; using $v_{2} / v_{1} = γ / (γ + 1)$ one finds $D_{C J}^{2} = (γ + 1) p_{2} v_{2}$ . Feed $p_{2} v_{2} = D_{C J}^{2} / (γ + 1)$ into step 3 and use $(γ - 1) (γ + 1) = γ^{2} - 1$ : $q = \frac{D _{C J}^{2}}{2 ( γ ^{2} - 1 )} .$

5. Flip it. Solving for $D_{C J}$ :

Recall Deriving the compression ratio

$v_{2} / v_{1} = γ / (γ + 1)$ (no hand-waving) Start from momentum in the strong limit, $\overset{m}{˙}^{2} (v_{1} - v_{2}) = p_{2}$ , and the sonic condition $u_{2}^{2} = γ p_{2} v_{2}$ . Since $u_{2} = \overset{m}{˙} v_{2}$ , we have $\overset{m}{˙}^{2} v_{2}^{2} = γ p_{2} v_{2}$ , i.e. $\overset{m}{˙}^{2} v_{2} = γ p_{2}$ , so $p_{2} = \overset{m}{˙}^{2} v_{2} / γ$ . Put this into the momentum line: $\overset{m}{˙}^{2} (v_{1} - v_{2}) = \overset{m}{˙}^{2} v_{2} / γ$ . Cancel $\overset{m}{˙}^{2}$ : $v_{1} - v_{2} = v_{2} / γ$ , hence $v_{1} = v_{2} (1 + 1/ γ) = v_{2} (γ + 1) / γ$ , which inverts to $v_{2} / v_{1} = γ / (γ + 1)$ . Note this is the CJ (heat-releasing, choked) ratio — it is NOT the same as the inert strong-shock ratio $(γ - 1) / (γ + 1)$ ; the sonic condition $M_{2} = 1$ is what makes the CJ value different.

Step 9 — Every case on one map (edge & degenerate states)

WHAT. Sweep through all possible final states so no reader hits an unshown scenario.

WHY. The contract: cover every branch, every sign, every degenerate limit.

PICTURE. The full Hugoniot with both branches and the forbidden gap labelled.

Definition The complete case list

Region on curve	$p_{2}$ vs $p_{1}$	$v_{2}$ vs $v_{1}$	$M_{2}$	Name
Upper branch, above CJ	$≫$ higher	lower	$< 1$	strong detonation (relaxes to CJ)
Upper tangent point	higher	lower	$= 1$	CJ detonation ✔ chosen
Upper branch, below CJ	higher	lower	$> 1$	weak detonation (needs special ignition)
The gap between branches	—	—	—	forbidden: Rayleigh slope would be positive
Lower tangent point	lower	higher	$= 1$	CJ deflagration
Lower branch	slightly lower	higher	$< 1$	ordinary deflagration (Laminar flame speed (deflagration))

Degenerate limits:

$q = 0$ : the Hugoniot slides back through $(v_{1}, p_{1})$ → pure shock, no burning (Shock waves and Mach number).
$\overset{m}{˙} = 0$ (slope $0$ , flat line): no flow, no front — the trivial "nothing happens" solution.
$\overset{m}{˙} \to \infty$ (vertical line): pure compression at fixed volume, the ideal that constant-volume combustion chases.

The one-picture summary

Everything on one $(v, p)$ chart: the cold dot, the fanning Rayleigh lines (momentum), the lifted Hugoniot (energy + fuel $q$ ), the two branches, the forbidden middle, and the two tangent CJ points where $M_{2} = 1$ . The upper CJ tangent is where the detonation speed $D_{C J} = 2 (γ^{2} - 1) q$ is born.

Recall Feynman: tell the whole walkthrough to a friend

Ride the burning wall so it stands still and the gas flows through it. Three things can't cheat across it — mass, momentum, energy — and only energy gets a bonus brick, the chemical fuel $q$ . Rename density as "room per kilogram" so we can draw a pressure-vs-room graph. Momentum draws a downhill straight line through the cold gas; because it's downhill, the gas can only either squeeze (detonation) or expand (deflagration). Energy draws a curve floating above the cold point — how high it floats is exactly the fuel $q$ . Slide the straight line by changing the front speed: too slow, it misses; medium, it crosses twice; one perfect speed, it just kisses the curve. That kiss happens exactly when the slope of the line equals the slope of the curve, and that algebra turns out to mean the exhaust runs out at the speed of sound, so nothing behind can ever disturb the front again. That self-sealing speed is $D_{C J} = 2 (γ^{2} - 1) q$ : more fuel, faster wave.

Recall

What frame do we sit in, and why? ::: The front's own frame, so the flow is steady and mass-in equals mass-out — turning the problem into simple bookkeeping. Why must the final state sit in only two corners of the $(v, p)$ plane? ::: The Rayleigh line has slope $- \overset{m}{˙}^{2} \leq 0$ (always downhill), so from the cold point you can only go top-left (compress = detonation) or bottom-right (expand = deflagration). What lifts the Hugoniot curve off the cold point? ::: The chemical heat release $q$ ; set $q = 0$ and it drops back onto $(v_{1}, p_{1})$ as a plain shock. Tangency of Rayleigh line and Hugoniot is physically the same as what condition? ::: The exhaust leaving at the local speed of sound, $u_{2} = a_{2}$ , i.e. $M_{2} = 1$ (choked flow) — proved by equating the two slopes. Why does the CJ point win over faster (strong) detonations? ::: Faster ones have $M_{2} < 1$ , so rear rarefaction waves catch up and drain them down until $M_{2} = 1$ — CJ is the stable attractor. What is $D_{C J}$ physically, in the lab frame? ::: The speed of the detonation front through the still unburnt gas; it equals $u_{1}$ because riding the front makes the cold gas flow in at exactly that speed. In $D_{C J} = 2 (γ^{2} - 1) q$ , what happens if you double $q$ ? ::: Speed rises by $2 \approx 1.41 \times$ , because wave speed goes as the square root of energy.