Chapman-Jouguet detonation; deflagration vs detonation: Exercises

This is the practice deck for the parent topic. Every problem is graded L1 → L5. Read the problem, try it on paper, then open the collapsible Solution to check yourself. All the algebra is spelled out from scratch — nothing skipped.

Before we start, one reminder of the toolkit you will keep reaching for:

Formula The five relations you will use everywhere

$ρ_{1} u_{1} = ρ_{2} u_{2} (mass)$ $p_{1} + ρ_{1} u_{1}^{2} = p_{2} + ρ_{2} u_{2}^{2} (momentum)$ $h_{1} + \frac{1}{2} u_{1}^{2} + q = h_{2} + \frac{1}{2} u_{2}^{2} (energy)$ $p_{2} - p_{1} = - \overset{m}{˙}^{2} (v_{2} - v_{1}) (Rayleigh line, slope = - \overset{m}{˙}^{2})$ $D_{C J} \approx 2 (γ^{2} - 1) q (strong-detonation CJ speed)$ Here $v = 1/ ρ$ is specific volume (how much space 1 kg of gas takes up), $\overset{m}{˙} = ρ_{1} u_{1}$ is the mass flux (kg through 1 m² of front per second), $q$ is heat released per kg of mixture, and $γ$ is the ratio of specific heats. $D_{C J}$ is the detonation front speed in the lab frame — how fast the burning shock physically travels down the tube (in m/s); in the front's own frame the unburnt gas rushes in at exactly this speed, so $D_{C J} = u_{1}$ . See Rankine-Hugoniot jump conditions and Speed of sound in gases if any symbol feels new.

Intuition Where the Rayleigh line comes from (WHY

$p_{2} - p_{1} = - \overset{m}{˙}^{2} (v_{2} - v_{1})$ ) We don't just hand you this line — it falls straight out of mass + momentum, no energy needed.

From mass conservation the same mass flux $\overset{m}{˙} = ρ_{1} u_{1} = ρ_{2} u_{2}$ passes each side. Since $v = 1/ ρ$ , the inflow speed is $u_{1} = \overset{m}{˙} v_{1}$ and the outflow speed is $u_{2} = \overset{m}{˙} v_{2}$ — speed equals mass flux times the room each kg takes up.
Put these into momentum, $p_{1} + ρ_{1} u_{1}^{2} = p_{2} + ρ_{2} u_{2}^{2}$ . Note $ρ u^{2} = \overset{m}{˙} u = \overset{m}{˙}^{2} v$ , so the momentum balance becomes $p_{1} + \overset{m}{˙}^{2} v_{1} = p_{2} + \overset{m}{˙}^{2} v_{2}$ .
Rearrange: $p_{2} - p_{1} = - \overset{m}{˙}^{2} (v_{2} - v_{1})$ .

Why we bother: this is the only relation linking the final pressure to the final volume using nothing but how fast gas is shoved through the front. In the $(v, p)$ plane it is a straight line through the initial state $(v_{1}, p_{1})$ whose slope is $- \overset{m}{˙}^{2}$ — negative, and steeper the faster the front. That single geometric fact is what powers every branch and tangency argument on this page.

Level 1 — Recognition

L1·Q1 — Name the mechanism

A reacting front moves through a tube at 3 m/s. Pressure behind it is very slightly lower than ahead. Is this a deflagration or a detonation? Which physical mechanism carries it forward?

Recall Solution

WHAT to check: speed and pressure change are the two fingerprints.

Speed 3 m/s is far below the speed of sound in a gas (≈340 m/s in air). So the front is subsonic.
Pressure dropped slightly ( $p_{2} < p_{1}$ ).

Both signatures point to a deflagration. Its propagation mechanism is heat conduction + diffusion of radicals — the hot burnt layer warms and ignites the next sliver of unburnt gas. See Laminar flame speed (deflagration).

Answer: deflagration, carried by diffusion of heat/radicals.

L1·Q2 — Read the sign of the Rayleigh slope

The Rayleigh line has slope $- \overset{m}{˙}^{2}$ . A student measures a slope of $+ 8$ for a supposed reacting front. What is wrong?

Recall Solution

The slope of the Rayleigh line must be $\leq 0$ , because it equals $- \overset{m}{˙}^{2}$ and a real squared mass flux $\overset{m}{˙}^{2} \geq 0$ makes $- \overset{m}{˙}^{2} \leq 0$ . A positive slope $+ 8$ would require $\overset{m}{˙}^{2} = - 8$ , i.e. an imaginary mass flux — physically impossible. Answer: no real front can have a positive Rayleigh slope; the measurement/setup is in error.

Level 2 — Application

L2·Q1 — CJ speed from heat release

Estimate $D_{C J}$ for a mixture with $q = 4.5 \times 1 0^{6} J/kg$ and $γ = 1.25$ .

Recall Solution

WHY this formula: we are given only $q$ and $γ$ , so we use the strong-detonation CJ estimate $D_{C J} = 2 (γ^{2} - 1) q$ . $γ^{2} - 1 = 1.2 5^{2} - 1 = 1.5625 - 1 = 0.5625$ $D_{C J} = 2 (0.5625) (4.5 \times 1 0^{6}) = 5.0625 \times 1 0^{6}$ $\approx 2250 m/s$ Sanity check: supersonic, order km/s — a believable detonation speed. ✓

L2·Q2 — Rayleigh slope → mass flux

A detonation runs from $(v_{1}, p_{1}) = (1.0, 1.0)$ to $(v_{2}, p_{2}) = (0.5, 20.0)$ in consistent units. Find the slope, and hence $\overset{m}{˙}^{2}$ and $\overset{m}{˙}$ .

Recall Solution

WHAT we do: the Rayleigh slope is $\frac{p _{2} - p _{1}}{v _{2} - v _{1}}$ . $slope = \frac{20.0 - 1.0}{0.5 - 1.0} = \frac{19.0}{- 0.5} = - 38.0$ Since slope $= - \overset{m}{˙}^{2}$ : $\overset{m}{˙}^{2} = 38.0, \overset{m}{˙} = 38.0 \approx 6.164$ WHY it's a detonation: $p_{2} > p_{1}$ and $v_{2} < v_{1}$ — compression, the upper-left branch. The large $∣ slope ∣$ means a steep line, i.e. a fast front. ✓

L2·Q3 — Classify from ratios

A front reports $p_{2} / p_{1} = 0.90$ and $v_{2} / v_{1} = 1.6$ . Which branch, and what is the propulsion relevance?

Recall Solution

Pressure dropped ( $p_{2} < p_{1}$ ) and specific volume rose ( $v_{2} > v_{1}$ : the gas expanded). Both are deflagration signatures ⇒ lower-right Hugoniot branch, a deflagration. Relevance: these are ordinary flames — ramjets and burners run on them. Detonation branches (upper-left) are what Rotating Detonation Engine (RDE) and Pulse Detonation Engine (PDE) exploit. ✓

Use the annotated $(v, p)$ map below to keep the two branches straight — it is the same diagram you will lean on in every level from here on.

Figure — Chapman-Jouguet detonation; deflagration vs detonation

Level 3 — Analysis

L3·Q1 — Recover $q$ from the Hugoniot adiabat

For an ideal gas, $h = \frac{γ}{γ - 1} p v$ , and the detonation adiabat reads $\frac{γ}{γ - 1} (p_{2} v_{2} - p_{1} v_{1}) = q + \frac{1}{2} (p_{2} - p_{1}) (v_{1} + v_{2}) .$ Given $γ = 1.4$ , $(v_{1}, p_{1}) = (1.0, 1.0)$ , $(v_{2}, p_{2}) = (0.6, 15.0)$ , find $q$ in these units.

Recall Solution

WHY solve for $q$ : the adiabat ties the two endpoints together through the chemical heat; given both states we can back out $q$ . Left side: $\frac{1.4}{0.4} (15.0 \cdot 0.6 - 1.0 \cdot 1.0) = 3.5 (9.0 - 1.0) = 3.5 \cdot 8.0 = 28.0$ Right side heat term: $\frac{1}{2} (15.0 - 1.0) (1.0 + 0.6) = \frac{1}{2} (14.0) (1.6) = 11.2$ So $28.0 = q + 11.2 \Rightarrow q = 16.8 (same units) .$ Interpretation: the endpoints sit on the heat-shifted adiabat that carries $q = 16.8$ . Positive $q$ ⇒ genuine heat release, consistent with a detonation. ✓

L3·Q2 — Why tangency = sonic

Explain why the CJ point, defined geometrically as the Rayleigh line tangent to the Hugoniot, is the same as $u_{2} = a_{2}$ ( $M_{2} = 1$ ), and why nature selects it.

Recall Solution

The geometric heart — why tangency forces $M_{2} = 1$ (the derivative-matching argument):

At the CJ point the Rayleigh line does not cross the Hugoniot, it just kisses it. Kissing means the two curves share the same slope there: the Rayleigh slope equals the Hugoniot slope at that single touching point.
Now the two slopes each carry a physical meaning. The Rayleigh slope is $- \overset{m}{˙}^{2}$ (from the momentum bookkeeping). It can be shown that the slope of the Hugoniot at any state, once you use mass + energy, equals $- \overset{m}{˙}^{2}$ only when $u_{2} = a_{2}$ — the sound speed is precisely the special value of the outflow at which the "chemistry-adiabat slope" and the "momentum-line slope" coincide.
So tangency (slopes equal) $\Leftrightarrow$ $u_{2} = a_{2}$ $\Leftrightarrow$ $M_{2} = 1$ . That is the whole content of the CJ condition: the two ways of computing how $p$ changes with $v$ agree exactly at the sound speed. Look at figure s01 — the coral CJ line touches the lavender curve at one point and their tangents align there.

Why nature picks it:

Tangency is also the smallest $\overset{m}{˙}$ (steepest still-touching line) on the detonation branch — the minimum self-sustaining speed.
If $u_{2} < a_{2}$ (strong detonation), expansion (rarefaction) waves from the burnt gas behind catch up to the front and weaken it, dragging it back toward $u_{2} = a_{2}$ . At exactly $u_{2} = a_{2}$ the flow is choked: rearward disturbances travel at the sound speed and can no longer overtake the front — it becomes autonomous and steady. See Shock waves and Mach number.

Answer: tangency ⇔ slopes match ⇔ $M_{2} = 1$ ⇔ choked ⇔ the unique stable, minimum-speed detonation.

Look at the figure: the steep coral line just kissing the lavender curve is the CJ line; steeper lines miss the detonation branch, shallower (green dashed) lines cut it twice.

L3·Q3 — The two intersections: strong vs weak detonation

When a Rayleigh line cuts (rather than kisses) the upper-left Hugoniot branch, it does so at two points. Name the upper and lower one, say how each compares to the CJ point in speed and in $M_{2}$ , and explain why only the CJ tangency is freely self-sustaining.

Recall Solution

WHAT the two points are (trace a line steeper than CJ cutting the detonation branch twice):

The upper intersection (higher $p_{2}$ , smaller $v_{2}$ ) is the strong / overdriven detonation. Here the outflow is subsonic, $u_{2} < a_{2}$ , i.e. $M_{2} < 1$ .
The lower intersection (lower $p_{2}$ , larger $v_{2}$ ) is the weak / underdriven detonation, with supersonic outflow $u_{2} > a_{2}$ , i.e. $M_{2} > 1$ .
The CJ point sits exactly between them at the tangency, $u_{2} = a_{2}$ , $M_{2} = 1$ — and it is the slowest front on the branch that still has any solution.

Why only CJ is freely self-sustaining:

Strong ( $M_{2} < 1$ ): the outflow is subsonic, so rarefaction (expansion) waves from the expanding burnt gas behind overtake the front and bleed energy from it, weakening it until it decelerates down to $M_{2} = 1$ . It can only be held strong by an external piston or a driving shock continuously pushing — remove the drive and it decays to CJ.
Weak ( $M_{2} > 1$ ): the outflow is supersonic. For an ordinary (non-pathological) reaction this point is unreachable — the reacting flow starting subsonic behind the lead shock cannot smoothly accelerate past sound speed without a special "eigenvalue" structure, so nature simply does not land here for a self-sustaining front.
CJ ( $M_{2} = 1$ , choked): disturbances behind travel at exactly the sound speed relative to the front and can never catch up. The front is sealed off from rearward interference — autonomous, steady, minimum-speed. That is the unique free detonation. See figure s01: the coral tangent (CJ) versus the green dashed line that cuts twice (strong = upper cut, weak = lower cut).

Answer: upper cut = strong/overdriven ( $M_{2} < 1$ , needs a piston), lower cut = weak/underdriven ( $M_{2} > 1$ , unreachable freely), CJ tangency ( $M_{2} = 1$ , choked) = the one self-sustaining detonation.

Level 4 — Synthesis

L4·Q1 — Derive the strong-detonation CJ speed

Starting from $q = \frac{1}{2 ( γ ^{2} - 1 )} D_{C J}^{2}$ (the strong-limit result quoted in the parent note), rearrange to isolate $D_{C J}$ , then evaluate for $q = 3.0 \times 1 0^{6} J/kg$ , $γ = 1.3$ .

Recall Solution

WHAT we do — the algebra: $q = \frac{D _{C J}^{2}}{2 ( γ ^{2} - 1 )} \Rightarrow D_{C J}^{2} = 2 (γ^{2} - 1) q \Rightarrow D_{C J} = 2 (γ^{2} - 1) q .$ Evaluate: $γ^{2} - 1 = 1.69 - 1 = 0.69$ . $D_{C J} = 2 (0.69) (3.0 \times 1 0^{6}) = 4.14 \times 1 0^{6} \approx 2035 m/s .$ This matches the parent note's worked example. ✓

L4·Q2 — Compare two propulsion cycles

Mixture A (RDE, detonation) releases the same $q$ as mixture B (ramjet, deflagration). Using the branch picture, argue why detonation approaches constant-volume heat release and why that raises cycle efficiency.

Recall Solution

Deflagration sits on the lower-right branch: $v_{2} > v_{1}$ (gas expands) while $p_{2} ≲ p_{1}$ . Heat is added while the gas is free to expand at roughly constant pressure — a constant-pressure cycle.
Detonation sits on the upper-left branch: the shock compresses the gas ( $v_{2} < v_{1}$ ) before/as it burns, so heat is dumped into a nearly fixed small volume — closer to constant-volume heat addition.
From Constant-volume vs constant-pressure combustion cycles: for the same $q$ , adding heat at (near) constant volume produces a larger pressure rise and extracts more work per cycle — higher thermodynamic efficiency. That's the payoff behind Rotating Detonation Engine (RDE) and Pulse Detonation Engine (PDE).

Answer: detonation's pre-compression makes heat release quasi-constant-volume, which the ideal cycle analysis shows is more efficient than the constant-pressure deflagration case.

L4·Q3 — Numeric efficiency intuition

An ideal constant-volume heat addition raises pressure from $p_{1} = 1$ to $p_{2}$ with $q = 16.8$ (units of L3·Q1) at $v = 1.0$ fixed, $γ = 1.4$ . Using $q = \frac{v ( p _{2} - p _{1} )}{γ - 1}$ for constant-volume heating, find $p_{2}$ .

Recall Solution

WHY this relation: at constant volume, every joule of heat goes into raising the gas's internal energy (none is spent pushing a moving boundary, because the volume doesn't move). For an ideal gas the internal energy per kg is $\frac{v p}{γ - 1}$ , so a jump in pressure at fixed $v$ costs heat $q = \frac{v ( p _{2} - p _{1} )}{γ - 1}$ . (No need to introduce any specific-heat symbol — this form comes straight from the ideal-gas internal energy.) $16.8 = \frac{1.0 ( p _{2} - 1.0 )}{0.4} \Rightarrow p_{2} - 1.0 = 16.8 \times 0.4 = 6.72 \Rightarrow p_{2} = 7.72.$ Interpretation: the same $q$ at constant volume drives a large pressure rise — the raw material of extra work. ✓

Level 5 — Mastery

L5·Q1 — Full CJ chain from raw jump conditions

A stoichiometric mix has $q = 5.0 \times 1 0^{6} J/kg$ , $γ = 1.3$ , and the ideal-gas relation $a = γ p v$ . (i) Estimate $D_{C J}$ . (ii) Given the pressure ratio $p_{2} / p_{1} \approx 18$ observed for such mixtures, comment on whether the strong-detonation assumption ( $p_{2} ≫ p_{1}$ ) is justified.

Recall Solution

(i) $γ^{2} - 1 = 0.69$ : $D_{C J} = 2 (0.69) (5.0 \times 1 0^{6}) = 6.9 \times 1 0^{6} \approx 2627 m/s .$ Order 2.6 km/s — right in the measured band for H₂-based detonations. ✓ (ii) The strong-detonation estimate assumes $p_{2} ≫ p_{1}$ . With $p_{2} / p_{1} \approx 18$ , the ratio is comfortably $≫ 1$ , so dropping $p_{1}$ relative to $p_{2}$ introduces only a few-percent error. The assumption is reasonable, and the formula gives a good first estimate. ✓

L5·Q2 — Diagnose an impossible claim

A paper claims a self-sustaining detonation with downstream Mach number $M_{2} = 1.4$ (supersonic outflow, a "weak/underdriven" state) that persists indefinitely with no external piston or confinement and no special reaction structure. Explain why this violates the CJ selection principle, and what actually happens.

Recall Solution

WHY it's suspicious: $M_{2} = 1.4 > 1$ places the outflow on the weak/underdriven side of the detonation branch — the lower of the two intersections a cutting Rayleigh line makes.

For an ordinary reaction, the flow behind the lead shock starts subsonic and releases heat. To reach $M_{2} > 1$ it would have to accelerate smoothly through the sound speed ( $M_{2} = 1$ ) — but the CJ point is exactly a choke point: at $M_{2} = 1$ the flow is sonic and, without a very special "eigenvalue" reaction structure, cannot pass to the supersonic weak branch. So the weak point is unreachable for a normal self-sustaining front.
Even if one imagined arriving there, a supersonic outflow means rearward disturbances travel slower than the flow leaves — this is the overdriven-style regime that only survives with an external drive; freely, the front relaxes.
What actually happens: with no piston, no confinement, no special reaction structure, the free front settles at the CJ state $M_{2} = 1$ (sonic, choked), where rearward rarefactions can never overtake it and it runs autonomously at the minimum self-sustaining speed.

Answer: a freely propagating detonation is pinned at $M_{2} = 1$ (CJ, choked). $M_{2} = 1.4$ is the weak/underdriven branch, unreachable without special structure; the claim of an unsupported, self-sustaining $M_{2} = 1.4$ is inconsistent — the real front relaxes to $M_{2} = 1$ .

L5·Q3 — Reverse-engineer $q$ from a measured detonation speed

A detonation is measured at $D_{C J} = 2840 m/s$ (stoichiometric H₂–O₂), with $γ = 1.2$ for the hot products. Using the strong-detonation estimate, infer $q$ . Is the value physically sensible for H₂–O₂?

Recall Solution

WHAT we do: invert $D_{C J} = 2 (γ^{2} - 1) q$ for $q$ : $q = \frac{D _{C J}^{2}}{2 ( γ ^{2} - 1 )} .$ $γ^{2} - 1 = 1.44 - 1 = 0.44$ ; $D_{C J}^{2} = 284 0^{2} = 8.0656 \times 1 0^{6}$ . $q = \frac{8.0656 \times 1 0 ^{6}}{2 ( 0.44 )} = \frac{8.0656 \times 1 0 ^{6}}{0.88} \approx 9.16 \times 1 0^{6} J/kg .$ Sensible? H₂–O₂ releases on the order of $\sim 1 0^{7} J/kg$ of mixture, so $\approx 9 \times 1 0^{6} J/kg$ is the right order of magnitude. ✓ (The crude strong-limit formula and the assumed $γ$ make this an estimate, not a precise value.)

The plot below shows exactly this inverse logic — read a detonation speed off the vertical axis and trace back to the $q$ that produces it; the slate dot marks this H₂–O₂ point.

Recall One-line self-test

Free detonation outflow Mach number ::: $M_{2} = 1$ (sonic / choked, the CJ tangency) Definition of Mach number $M_{2}$ ::: $M_{2} = u_{2} / a_{2}$ , outflow speed over local sound speed Sign of the Rayleigh slope ::: always $\leq 0$ (equals $- \overset{m}{˙}^{2}$ ) Detonation branch location in $(v, p)$ ::: upper-left ( $p_{2} > p_{1}$ , $v_{2} < v_{1}$ ) Deflagration pressure change ::: $p_{2} < p_{1}$ (slight drop) Strong-detonation CJ speed ::: $D_{C J} = 2 (γ^{2} - 1) q$ Strong vs weak detonation ::: strong = upper cut $M_{2} < 1$ (needs a piston); weak = lower cut $M_{2} > 1$ (unreachable freely); CJ tangency $M_{2} = 1$ is the self-sustaining one