Chapman-Jouguet detonation; deflagration vs detonation: Question bank

Before you start, here are every symbol these questions lean on — each defined in plain words so no line below uses anything unearned:

Definition The symbols used on this page

Front — the thin moving region where unburnt gas turns into burnt gas; we treat it as a discontinuity (a sudden jump).
Subscript 1 / 2 — state 1 = unburnt reactants entering the front; state 2 = burnt products leaving it.
$p$ (pressure) — the push per unit area of the gas.
$ρ$ (density) — mass packed into each unit of volume.
$v$ (specific volume) — the room per kilogram of gas, $v = 1/ ρ$ . Small $v$ = squeezed/dense; large $v$ = spread out. It is just density flipped over, and it is the horizontal axis of every diagram below.
$u_{1}, u_{2}$ — the speed the gas flows into ( $u_{1}$ ) and out of ( $u_{2}$ ) the front, measured in the frame riding along with the front.
$\overset{m}{˙}$ (mass flux) — the mass crossing one unit of front area per second, $\overset{m}{˙} = ρ_{1} u_{1} = ρ_{2} u_{2}$ . Think "how much gas is being fed through the front each second." A bigger $\overset{m}{˙}$ means a faster, more forceful front.
$h$ (specific enthalpy) — the heat content per kilogram of the gas, $h = c_{p} T$ : the thermal energy it carries plus the "push-work" $p v$ it can do on its surroundings. Bigger $h$ = hotter, more energetic gas. It is the term the released chemical heat $q$ pours into.
$q$ (heat release) — chemical energy set free per kilogram of mixture; fixed by the fuel, not by how the front moves.
$γ$ (ratio of specific heats) — a pure number ( $\approx 1.2 - 1.4$ for hot gases) that tells you how "springy" the gas is under compression; it links pressure, volume and sound speed. See Speed of sound in gases.
$a_{2}$ (downstream sound speed) — how fast a small pressure ripple travels in the burnt gas.
$M_{2}$ (downstream Mach number) — ratio of exit speed to sound speed, $M_{2} = u_{2} / a_{2}$ . $M_{2} < 1$ subsonic, $M_{2} = 1$ sonic ("choked"), $M_{2} > 1$ supersonic.
$D_{C J}$ (Chapman–Jouguet detonation speed) — the single self-sustaining speed at which a detonation front travels through the unburnt gas; numerically equal to the inflow speed $u_{1}$ at the CJ state.

Why tangency $\Leftrightarrow$ sonic exit ( $M_{2} = 1$ )

The one picture behind every question

Almost every item below points at the same $(v, p)$ diagram — specific volume across, pressure up. Study it first; the questions are just "where on this plot?"

Figure — Chapman-Jouguet detonation; deflagration vs detonation

True or false — justify

A deflagration is simply a detonation that hasn't sped up yet.

False — they run on different mechanisms (diffusion vs shock) and sit on different branches of the Hugoniot (see figure), so you cannot smoothly accelerate one into the other.

Both deflagration and detonation release the same chemical heat

q

per unit mass of a given mixture.

True —

q

is fixed by the chemistry, not by the propagation mode; what differs is how the front moves.

Across a detonation front the pressure rises.

True — a detonation is shock-driven compression, so

p_{2} ≫ p_{1}

(factor ~15–40) with the gas squeezed (

v_{2} < v_{1}

) — the upper-left corner of the figure.

Across a deflagration front the pressure rises.

False — the gas expands (

v_{2} > v_{1}

) and accelerates, so

p_{2} ≲ p_{1}

; this is the lower-right corner, and it is why open flames don't shatter things.

The Rayleigh line always has a negative slope.

True — its slope is

- \overset{m}{˙}^{2}

, and

\overset{m}{˙}^{2} \geq 0

, so the slope can only be zero or negative, never positive.

The Rayleigh line is anchored to the initial state, not just defined by its slope.

True — it always passes through

(v_{1}, p_{1})

(intercept

p_{1} + \overset{m}{˙}^{2} v_{1}

); changing

\overset{m}{˙}

merely pivots it about that fixed point, tilting it steeper or shallower.

A steeper (more negative) Rayleigh line corresponds to a faster front.

True — steeper means larger

\overset{m}{˙}^{2} = ρ_{1}^{2} u_{1}^{2}

, hence larger inflow speed

u_{1}

, i.e. a faster front.

At the Chapman–Jouguet point the downstream flow is exactly sonic.

True — CJ is defined by

u_{2} = a_{2}

, so

M_{2} = 1

; the burnt gas leaves at its own local speed of sound. On the plot it is the point of tangency (slope of Rayleigh line = slope of Hugoniot).

The CJ point is where the chemical energy release is maximised.

False — CJ is a tangency/sonic condition (minimum self-sustaining detonation speed);

q

is fixed by chemistry and is not extremised there.

A supersonic reaction front must be a detonation.

True — deflagrations are subsonic by definition (diffusion-limited); supersonic self-sustaining fronts sit on the detonation branch.

The Hugoniot curve without heat release (

q = 0

) is just an ordinary shock adiabat.

True — setting

q = 0

recovers the non-reacting shock Hugoniot; heat

q

shifts the curve up and out.

The strong detonation branch has subsonic exit flow.

True — a strong (overdriven) detonation is the upper crossing with

u_{2} < a_{2}

; relaxing to the tangent CJ point brings it up to

u_{2} = a_{2}

Why questions

Why does the flow being choked (

M_{2} = 1

) make the detonation self-sustaining?

Because rearward disturbances travel at most at the local sound speed

a_{2}

; once

u_{2} = a_{2}

they can never overtake the front, so it propagates steadily and autonomously.

Why do we analyse the front in a frame moving with it?

In that frame the front is stationary and the flow is steady, so the mass/momentum/energy bookkeeping (jump conditions) becomes simple algebraic balances rather than time-dependent equations.

Why is the Rayleigh line straight, while the Hugoniot is curved?

The Rayleigh line comes from the momentum jump

p_{1} + \overset{m}{˙}^{2} v_{1} = p_{2} + \overset{m}{˙}^{2} v_{2}

, which is linear in

(v, p)

; the Hugoniot comes from the energy balance (with

h

and

q

), which is nonlinear.

Why does the enthalpy

h

, not the plain temperature, appear in the energy balance?

Because

h = c_{p} T

already bundles the gas's thermal energy with the flow-work

p v

it does crossing the front; conserving

h + \frac{1}{2} u^{2}

(plus

q

) is the correct steady-flow energy bookkeeping.

Why is a larger heat release

q

associated with a faster CJ detonation?

More chemical energy raises the product temperature and sound speed, pushing the Hugoniot out so the tangent Rayleigh line must be steeper (bigger

\overset{m}{˙}

); in the estimate

D_{C J} \propto q

Why do rotating and pulse detonation engines want detonation rather than deflagration?

Detonation heat release is near constant-volume, which is thermodynamically more efficient than the constant-pressure release of ordinary flames — see Constant-volume vs constant-pressure combustion cycles, Rotating Detonation Engine (RDE), Pulse Detonation Engine (PDE).

Why does tangency of the Rayleigh line to the Hugoniot force

M_{2} = 1

At tangency the Rayleigh slope

- u_{2}^{2} / v_{2}^{2}

equals the Hugoniot slope

- a_{2}^{2} / v_{2}^{2}

, so

u_{2}^{2} = a_{2}^{2}

and the exit flow is exactly sonic.

Why does the naive picture "shock ignites gas, which drives shock" not run away to infinite speed?

The CJ (sonic) condition caps it: only the tangent Rayleigh line has a self-sustaining solution, selecting a unique minimum detonation speed rather than an unbounded one — geometrically, steeper lines miss the tangency and need external driving.

Edge cases

What happens to the Hugoniot curve in the limit

q \to 0

It collapses back through

(v_{1}, p_{1})

onto the ordinary (non-reacting) shock adiabat; no combustion, just a pure shock.

What is the deflagration CJ point, and how does it relate to the detonation one?

A Rayleigh line can also graze the lower-right branch of the Hugoniot; that tangency is the deflagration CJ point, also sonic (

M_{2} = 1

), representing the fastest self-sustaining deflagration — but ordinary lab flames are far slower and stay well subsonic.

If a mixture is too weak to release enough heat, can a Rayleigh line still find the detonation branch?

q

is too small the Hugoniot barely lifts, so the tangent Rayleigh line may not reach the detonation branch — no self-sustaining detonation exists and the mixture only deflagrates (a detonability limit).

As the inflow speed

u_{1}

is pushed above

D_{C J}

, what kind of solution appears?

The Rayleigh line becomes steeper than the tangent and cuts the Hugoniot at two points — a strong (subsonic exit) and a weak (supersonic exit) overdriven detonation, sustained only by an external piston/driver, not self-sustaining.

What does

v_{2} = v_{1}

(no volume change) correspond to?

A vertical move in the

(v, p)

plane — pure constant-volume heat addition, the idealised limit detonation engines approximate; a real CJ detonation compresses slightly (

v_{2} < v_{1}

What distinguishes the strong from the weak detonation crossing?

Both lie on the detonation branch for one steep Rayleigh line: the strong (upper) has

u_{2} < a_{2}

, the weak (lower) has

u_{2} > a_{2}

; they coincide at the CJ tangency where

u_{2} = a_{2}

Recall One-line memory hooks

Rayleigh = momentum = straight line, slope $- \overset{m}{˙}^{2}$ , anchored at $(v_{1}, p_{1})$ ::: line Hugoniot = energy + $q$ (uses enthalpy $h$ ) = curve of end states ::: curve CJ = tangent = slopes match = $M_{2} = 1$ = choked = self-sustaining minimum speed ::: point Strong = upper crossing, $u_{2} < a_{2}$ ; Weak = lower crossing, $u_{2} > a_{2}$ ; merge at CJ ::: crossings Detonation = shock, compress, $p ↑ v ↓$ , upper-left ::: branch Deflagration = diffusion, expand, $p ↓ v ↑$ , lower-right ::: branch

Question bank — Chapman-Jouguet detonation; deflagration vs detonation

Why tangency $\Leftrightarrow$ sonic exit ( $M_{2} = 1$ )

The one picture behind every question

True or false — justify

Spot the error

Why questions

Edge cases

Why tangency ⇔ sonic exit (M2​=1)

The one picture behind every question

True or false — justify

Spot the error

Why questions

Edge cases

Why tangency $\Leftrightarrow$ sonic exit ( $M_{2} = 1$ )