Chapman-Jouguet detonation; deflagration vs detonation: Worked examples

Before the matrix, the symbols we will reuse constantly. Each is earned in the parent note; here is the one-line reminder so you never have to scroll back:

Definition Symbols used on this page (all earned here)

Specific volume $v = 1/ ρ$ — the space one kilogram of gas occupies (m³/kg). Big $v$ = spread out, small $v$ = squeezed. So $v_{1} = 1/ ρ_{1}$ links a gas's density $ρ_{1}$ to its specific volume — we swap freely between $ρ_{1} u_{1}$ and $\overset{m}{˙} v_{1}$ below.
Mass flux $\overset{m}{˙} = ρ_{1} u_{1} = ρ_{2} u_{2}$ — mass crossing one square metre of the front per second (kg/m²·s). It sets how fast the front runs. Since $v = 1/ ρ$ , this is the same as $\overset{m}{˙} = u_{1} / v_{1}$ , i.e. $u_{1} = \overset{m}{˙} v_{1}$ .
Inflow speed $u_{1}$ — how fast unburnt gas flows into the front (front's frame). Two equivalent ways to get it: $u_{1} = \overset{m}{˙} / ρ_{1}$ or $u_{1} = \overset{m}{˙} v_{1}$ (identical, because $v_{1} = 1/ ρ_{1}$ ).
Rayleigh slope $\frac{p _{2} - p _{1}}{v _{2} - v _{1}} = - \overset{m}{˙}^{2}$ , always $\leq 0$ . Its magnitude is $\overset{m}{˙}^{2}$ .
Detonation speed $D_{C J}$ — the lab-frame speed of a Chapman–Jouguet detonation front, i.e. the inflow speed $u_{1}$ in the front's frame. " $D$ " for detonation, subscript "CJ" for the Chapman–Jouguet condition that selects it. Units: m/s.
Heat release $q$ — the chemical energy set free per kilogram of mixture as it reacts (J/kg). Bigger $q$ = hotter, more energetic reaction. It is the source term in the energy jump condition (parent §1).
Ratio of specific heats $γ$ (Greek "gamma") — the dimensionless number $c_{p} / c_{v}$ , i.e. how much easier it is to heat a gas at constant pressure than at constant volume. Roughly $1.67$ for a monatomic gas (Ar), $1.4$ for air, and $\sim 1.2$ – $1.3$ for hot polyatomic combustion products. It controls both the sound speed and how much heat turns into front motion.
Gas constant $R = 8.314 J/(mol\cdotK)$ — the universal constant linking pressure, volume and temperature for an ideal gas.
Molar mass $M$ — the mass of one mole of the gas (kg/mol). Together $R / M$ is the specific gas constant (J/(kg·K)) for that particular gas.

The scenario matrix

Every reacting-front problem this topic can pose falls into one of these cells. The Cell column is the label you will see on each example below.

Cell	Case class	Diagnostic signature	Example
C1	Detonation branch (upper-left)	$p_{2} > p_{1}$ and $v_{2} < v_{1}$	Ex 1
C2	Deflagration branch (lower-right)	$p_{2} < p_{1}$ and $v_{2} > v_{1}$	Ex 2
C3	Rayleigh slope → front speed	slope $= - \overset{m}{˙}^{2}$	Ex 3
C4	CJ velocity from chemistry	given $q, γ$ → $D_{C J}$	Ex 4
C5	CJ sonic check ( $M_{2} = 1$ )	is $u_{2} = a_{2}$ ?	Ex 5
C6	Degenerate / zero limit	$q \to 0$ and $\overset{m}{˙} \to 0$	Ex 6
C7	Forbidden region (no solution)	$p_{2} > p_{1}$ and $v_{2} > v_{1}$	Ex 7
C8	Real-world word problem (engine)	detonation tube timing	Ex 8
C9	Exam twist (which is faster? / limit)	comparative reasoning	Ex 9

Ex 1 — Cell C1: pin a front to the detonation branch (with the $(v, p)$ diagram) A front is measured with $p_{1} = 1 atm$ , $v_{1} = 1.0 m^{3} / kg$ , and downstream $p_{2} = 15 atm$ , $v_{2} = 0.60 m^{3} / kg$ . Which branch, and what is the front's mass flux?

Forecast: pressure went up, volume went down — guess detonation. But confirm with the slope, then read it off the diagram.

Read the signs. $p_{2} - p_{1} = 14 > 0$ (compression), $v_{2} - v_{1} = - 0.40 < 0$ (squeezed). Why this step? The parent note's branch rule is purely sign-based: up-and-in = detonation (upper-left of the $(v, p)$ plane).
Compute the Rayleigh slope (display units). $\frac{p _{2} - p _{1}}{v _{2} - v _{1}} = \frac{14}{- 0.40} = - 35$ . Why this step? The slope must be negative for any real front; $- 35$ passes that test. This is in mixed atm-per-(m³/kg) units — good for the sign, not yet physical.
Convert to SI for a physical $\overset{m}{˙}$ . Put pressures in Pa: $p_{1} = 1.013 \times 1 0^{5}$ , $p_{2} = 15 \times 1.013 \times 1 0^{5} = 1.520 \times 1 0^{6}$ Pa. Slope $= \frac{1.520 \times 1 0 ^{6} - 1.013 \times 1 0 ^{5}}{0.60 - 1.0} = \frac{1.418 \times 1 0 ^{6}}{- 0.40} = - 3.546 \times 1 0^{6}$ . Then $\overset{m}{˙} = 3.546 \times 1 0^{6} \approx 1883 kg/m^{2} s$ . Why this step? Only pascals give $\overset{m}{˙}$ in real kg/m²·s (see the units callout above). This is the honest, dimensionally-consistent flux.
Read the geometry. On the diagram, the red Rayleigh line runs from the black initial-state dot $(v_{1}, p_{1})$ steeply up to the left, hitting the red detonation state dot. Why this step? Steep negative slope ⇔ large $\overset{m}{˙}$ ⇔ fast front — the visual signature of detonation.

Verify: signs (up, in) + steep negative slope both say detonation. In the figure, follow the red Rayleigh line: it climbs from the black initial dot to the upper-left detonation dot. ✓

Read the figure below like this: the horizontal axis is specific volume $v$ (m³/kg, rightward = more spread out); the vertical axis is pressure $p$ (atm, upward = higher). The black dot at $(v_{1}, p_{1}) = (1, 1)$ is the initial unburnt state. The black curve is the schematic Hugoniot. The red straight line is the Ex 1 Rayleigh line; where it meets the curve up-and-left is the red detonation dot $(0.60, 15)$ . The black square down-and-right is the Ex 2 deflagration state. The two labelled corner regions mark the detonation (up-left) and deflagration (down-right) branches; the other two corners are forbidden (Ex 7).

Figure — Chapman-Jouguet detonation; deflagration vs detonation

Worked example

Ex 2 — Cell C2: the deflagration branch (pressure DROPS, shown on the diagram) Now $p_{1} = 1.0 atm$ , $v_{1} = 1.0 m^{3} / kg$ , and downstream $p_{2} = 0.95 atm$ , $v_{2} = 4.0 m^{3} / kg$ (an ordinary open flame). Which branch, and where does it sit on the diagram?

Forecast: an open candle flame — pressure barely changes and gas expands hugely. Guess deflagration, lower-right of the initial point.

Read the signs. $p_{2} - p_{1} = - 0.05 < 0$ , $v_{2} - v_{1} = + 3.0 > 0$ . Why this step? Down-and-out = deflagration, the lower-right branch. The gas got hotter, so at nearly the same pressure it expands (ideal gas: $v \propto T$ at fixed $p$ ).
Slope check (display units). $\frac{- 0.05}{3.0} = - 0.0167$ — still negative, still a valid front. Why this step? Even a slow flame obeys $slope = - \overset{m}{˙}^{2} \leq 0$ ; here the tiny magnitude means a tiny $\overset{m}{˙}$ , i.e. a slow subsonic front.
Convert to SI. Pressures in Pa: slope $= \frac{0.95 \times 1.013 \times 1 0 ^{5} - 1.013 \times 1 0 ^{5}}{3.0} = \frac{- 5.065 \times 1 0 ^{3}}{3.0} = - 1688 (Pa per m^{3} / kg)$ . So $\overset{m}{˙} = 1688 \approx 41.1 kg/m^{2} s$ . Why this step? The physical flux (~41) is tiny next to Ex 1's ~1883 — a factor of ~46 — which is exactly why detonation is fast and deflagration is slow. Compare only consistent SI numbers.

Verify: pressure fell, volume rose, flux tiny — textbook laminar deflagration. On the figure this state is the black square lower-right of the initial dot. This is why open flames don't shatter windows. ✓

Worked example

Ex 3 — Cell C3: from slope to physical front speed A detonation in a gas of density $ρ_{1} = 1.2 kg/m^{3}$ has measured mass flux $\overset{m}{˙} = 3400 kg/m^{2} s$ . How fast is the front travelling into the unburnt gas?

Forecast: $\overset{m}{˙} = ρ_{1} u_{1}$ , so bigger flux ÷ density = the inflow speed = the front speed in the lab. Guess a few km/s.

Recall the mass relation. $\overset{m}{˙} = ρ_{1} u_{1} \Rightarrow u_{1} = \overset{m}{˙} / ρ_{1}$ . (Equivalently $u_{1} = \overset{m}{˙} v_{1}$ with $v_{1} = 1/ ρ_{1}$ — same thing, two spellings.) Why this step? In the front's frame the unburnt gas flows in at $u_{1}$ ; that inflow speed equals how fast the front eats into still gas in the lab frame — that is exactly $D_{C J}$ for a CJ detonation.
Plug in. $u_{1} = 3400/1.2 \approx 2833 m/s$ . Why this step? Now we have a physical detonation velocity, not just a flux.

Verify: $\approx 2.83$ km/s is right in the H₂–O₂ CJ band ( $\sim 2840$ m/s quoted in the parent note). Units: $\frac{kg/m ^{2} s}{kg/m ^{3}} = m/s$ . ✓ — see Shock waves and Mach number for why this is supersonic.

Worked example

Ex 4 — Cell C4: CJ velocity straight from the chemistry A fuel releases heat $q = 4.5 \times 1 0^{6} J/kg$ (energy per kilogram, defined in the symbols box), and the products have ratio of specific heats $γ = 1.25$ . Estimate the detonation speed $D_{C J}$ .

Forecast: more heat → faster detonation; expect a couple km/s.

Choose the tool. Use $D_{C J} \approx 2 (γ^{2} - 1) q$ . Why this step? We are given only $q$ and $γ$ — this strong-detonation estimate is the exact formula built for that input (parent §2). No other formula fits the data given.
Evaluate the factor. $γ^{2} - 1 = 1.5625 - 1 = 0.5625$ . Why this step? $(γ^{2} - 1)$ measures how much of the heat becomes ordered kinetic energy of the front; smaller $γ$ (more atoms per molecule) → smaller factor.
Finish. $D_{C J} = 2 (0.5625) (4.5 \times 1 0^{6}) = 5.0625 \times 1 0^{6} \approx 2250 m/s$ . Why this step? The square root turns an energy-per-mass into a velocity (units: $J/kg = m^{2} / s^{2} = m/s$ ).

Verify: supersonic, order km/s, sane for a hydrocarbon detonation. Units check out. ✓

Worked example

Ex 5 — Cell C5: confirm the CJ sonic condition $u_{2} = a_{2}$ At a CJ state the burnt gas has ratio of specific heats $γ = 1.30$ , $p_{2} = 18 atm = 1.824 \times 1 0^{6} Pa$ , $v_{2} = 0.35 m^{3} / kg$ . Its outflow speed is measured $u_{2} = 913 m/s$ . Is this really CJ?

Forecast: CJ means the burnt gas leaves at exactly the local sound speed, $M_{2} = 1$ . So compute $a_{2}$ and compare.

Recall the sound speed. For an ideal gas $a_{2} = γ p_{2} v_{2}$ . Why this step? The ideal-gas law is $p V = n R T$ ; dividing by mass and using molar mass $M$ (both defined in the symbols box) gives the specific form $p v = \frac{R}{M} T$ . Substituting that into $a = γ R T / M$ yields the tidy $a_{2} = γ p_{2} v_{2}$ — it needs only $γ$ , $p_{2}$ , $v_{2}$ , with $R$ and $M$ hidden inside the product $p_{2} v_{2}$ . Sound speed is the yardstick the CJ condition uses.
Compute $a_{2}$ . $γ p_{2} v_{2} = 1.30 \times 1.824 \times 1 0^{6} \times 0.35 = 8.30 \times 1 0^{5}$ , so $a_{2} = 8.30 \times 1 0^{5} \approx 911 m/s$ . Why this step? This is the target the outflow must hit for tangency.
Compare. $u_{2} = 913 \approx a_{2} = 911 \Rightarrow M_{2} = u_{2} / a_{2} \approx 1.00$ . Why this step? $M_{2} = 1$ is the algebraic fingerprint of the Rayleigh line grazing the Hugoniot — the CJ point.

Verify: $M_{2} \approx 1.00$ ⇒ this is a Chapman–Jouguet detonation, sonic in the burnt gas. ✓

Worked example

Ex 6 — Cell C6: BOTH degenerate limits ( $q \to 0$ and $\overset{m}{˙} \to 0$ ) Two "nothing-happens" limits are hiding in the machinery. (i) What happens to $D_{C J}$ and the Hugoniot curve as the heat release $q \to 0$ (an inert "reaction")? (ii) What happens as the mass flux $\overset{m}{˙} \to 0$ ?

Forecast: kill the chemistry → the detonation collapses to an ordinary shock or nothing. Kill the flux → the Rayleigh line goes flat and the front stops.

Limit (i): $q \to 0$ , detonation velocity. $D_{C J} = 2 (γ^{2} - 1) q \to 0$ . Why this step? With no released energy there is no self-sustaining driver; the CJ speed above the plain sound speed vanishes.
Limit (i): $q \to 0$ , Hugoniot curve. The detonation adiabat is the Hugoniot shifted up-and-out by $q$ . Set $q = 0$ : it relaxes onto the ordinary shock Hugoniot through $(v_{1}, p_{1})$ , and the only zero-strength front is the trivial $p_{2} = p_{1}, v_{2} = v_{1}$ — no wave. Why this step? It shows the detonation branch is created by the chemistry; kill $q$ and the branch merges back to the mundane shock adiabat.
Limit (ii): $\overset{m}{˙} \to 0$ , Rayleigh slope. Slope $= - \overset{m}{˙}^{2} \to 0^{-}$ : the Rayleigh line becomes horizontal through $(v_{1}, p_{1})$ . Why this step? A flat Rayleigh line means $p_{2} \to p_{1}$ ; no pressure jump can be supported.
Limit (ii): $\overset{m}{˙} \to 0$ , front speed. Recall $u_{1} = \overset{m}{˙} v_{1}$ (same relation as Ex 3's $u_{1} = \overset{m}{˙} / ρ_{1}$ , since $v_{1} = 1/ ρ_{1}$ ). As $\overset{m}{˙} \to 0$ we get $u_{1} \to 0$ : the front is stationary — no gas flows through it, so no reaction front propagates at all. Why this step? $\overset{m}{˙}$ is the flow that carries fresh gas into the front; zero flux = frozen, extinguished front. Both degenerate limits therefore describe the same physical "nothing propagates" endpoint from two directions (no heat vs no flow).

Verify: both limits are continuous and physical — $q \to 0 \Rightarrow D_{C J} \to 0$ and reacting curve → inert shock curve; $\overset{m}{˙} \to 0 \Rightarrow$ slope $\to 0$ , $u_{1} \to 0$ , front freezes. No paradox. ✓

Worked example

Ex 7 — Cell C7: the forbidden region (no real front) Someone claims a front with $p_{2} = 5 atm$ , $v_{2} = 3.0 m^{3} / kg$ from $p_{1} = 1 atm$ , $v_{1} = 1.0 m^{3} / kg$ . Is this possible?

Forecast: pressure rose and volume rose. That is neither compression-in nor expansion-out cleanly — smell a contradiction.

Slope test. $\frac{p _{2} - p _{1}}{v _{2} - v _{1}} = \frac{4}{2} = + 2$ . Why this step? The Rayleigh slope must equal $- \overset{m}{˙}^{2}$ , which is $\leq 0$ for any real mass flux. A positive slope is impossible.
Interpret. A positive slope would require $\overset{m}{˙}^{2} = - 2 < 0$ , i.e. imaginary mass flux. Why this step? Mass flux is a real physical rate; no real front can produce this pair.
Conclusion. The state lies in the forbidden quadrant (upper-right of $(v_{1}, p_{1})$ ): the claim is unphysical. Why this step? This is the fourth quadrant our branch rules exclude — every case is now covered: down-right = deflagration, up-left = detonation, up-right & down-left = forbidden.

Verify: slope $= + 2 > 0$ ⇒ $\overset{m}{˙}^{2} < 0$ ⇒ impossible. Reject the claim. ✓

Worked example

Ex 8 — Cell C8: real-world word problem (a PDE tube) A pulse detonation engine tube is $1.5 m$ long. A detonation runs the full length at $D_{C J} = 1900 m/s$ (the CJ front speed, defined in the symbols box). How long does one detonation pass take, and roughly how many detonation cycles per second could the tube fire if refill+purge takes $4 \times$ the burn time?

Forecast: burn time = length ÷ speed (sub-millisecond); total cycle is $5 \times$ the burn, so a few hundred Hz.

Burn (transit) time. $t_{burn} = L / D_{C J} = 1.5/1900 = 7.89 \times 1 0^{- 4} s$ . Why this step? Distance ÷ speed is the time for the front to traverse the tube — the core of PDE timing.
Full cycle. Refill+purge $= 4 t_{burn}$ , so $t_{cycle} = t_{burn} + 4 t_{burn} = 5 t_{burn} = 3.95 \times 1 0^{- 3} s$ . Why this step? A real engine cannot fire again until fresh mixture is loaded and old products are cleared.
Firing frequency. $f = 1/ t_{cycle} = 1/3.95 \times 1 0^{- 3} \approx 253 Hz$ . Why this step? Frequency = 1 ÷ cycle time; this sets the thrust pulse rate.

Verify: $t_{burn} \approx 0.79$ ms (sub-millisecond, as expected for km/s over ~1 m); $f \approx 253$ Hz sits in the real PDE range (tens–hundreds of Hz). Contrast the continuous operation of an RDE. ✓

Worked example

Ex 9 — Cell C9: exam twist (which self-sustaining speed does nature choose?) For one fixed heat release $q$ the detonation branch of the Hugoniot admits many Rayleigh lines that intersect it. A "strong" detonation solution gives $u_{2} < a_{2}$ ; a "weak" one $u_{2} > a_{2}$ ; the CJ one $u_{2} = a_{2}$ . Rank their front speeds and state which the tube actually runs at.

Forecast: the parent note says CJ is the minimum self-sustaining speed — guess CJ is the slowest stable detonation and the one nature picks.

Link speed to slope. Steeper Rayleigh line ⇔ larger $\overset{m}{˙}$ ⇔ faster front. Strong detonations sit on a steeper line than CJ. Why this step? $∣ slope ∣ = \overset{m}{˙}^{2}$ , so geometry ranks the speeds directly.
Apply stability. If $u_{2} < a_{2}$ (strong), rearward rarefactions overtake and weaken the front until $u_{2} = a_{2}$ . Why this step? A front that can be caught from behind is dragged down toward the sonic CJ point — that is the parent's self-sustaining argument.
Rank & choose. Speeds: strong $>$ CJ $>$ weak, but only CJ is stable & self-sustaining for a freely propagating detonation. The tube runs at $D_{C J}$ . Why this step? weak solutions require special downstream boundary support (unusual); strong ones decay; CJ is the robust attractor.

Verify: consistent with parent §2 — CJ is the tangency = sonic = minimum self-sustaining speed, the one an unsupported detonation settles onto. ✓

Recall Quick self-test

A front shows $p_{2} > p_{1}$ and $v_{2} > v_{1}$ — which cell? ::: C7, the forbidden region (positive Rayleigh slope ⇒ imaginary $\overset{m}{˙}$ ). You are given only $q$ and $γ$ — which formula? ::: $D_{C J} = 2 (γ^{2} - 1) q$ (Cell C4). How do you confirm a state is genuinely CJ? ::: Compute $a_{2} = γ p_{2} v_{2}$ and check $u_{2} = a_{2}$ , i.e. $M_{2} = 1$ (Cell C5). As $q \to 0$ , what does the detonation branch become? ::: The ordinary inert-shock Hugoniot through $(v_{1}, p_{1})$ ; $D_{C J} \to 0$ (Cell C6). As $\overset{m}{˙} \to 0$ , what happens to the front? ::: Rayleigh slope $\to 0$ , $u_{1} \to 0$ ; the front freezes — no wave (Cell C6).

See also the parent: Hinglish version, and the efficiency angle in Constant-volume vs constant-pressure combustion cycles and Adiabatic flame temperature.

Worked examples — Chapman-Jouguet detonation; deflagration vs detonation

The scenario matrix

The worked examples