Grain geometry — BATES, star, wagon wheel; neutral - progressive - regressive burn: Exercises

Throughout, remember the master chain (from the parent): $A_{b} \Rightarrow K = \frac{A _{b}}{A _{t}} \Rightarrow p_{c} \propto K^{\frac{1}{1 - n}} \Rightarrow F = C_{F} p_{c} A_{t}$

Intuition Why this exact chain?

Why form the ratio $K = A_{b} / A_{t}$ ? Chamber pressure is a tug-of-war between gas being made (proportional to the burning area $A_{b}$ ) and gas escaping (proportional to the throat area $A_{t}$ ). It is not $A_{b}$ or $A_{t}$ alone that sets pressure but their ratio — double both and the pressure is unchanged. That single dimensionless number $K$ (often called $K_{n}$ ) is therefore the natural knob (see Chamber pressure & throat area (Kn ratio)). How does $C_{F}$ enter $F$ ? Once you know $p_{c}$ , the nozzle turns that pressure into thrust. The thrust coefficient $C_{F}$ is the bookkeeping factor that says "how much thrust this particular nozzle squeezes out of a given $p_{c} A_{t}$ " — it packages the nozzle's expansion geometry (see Thrust coefficient and nozzle). So $F = C_{F} p_{c} A_{t}$ : pressure times throat area gives a force scale, and $C_{F}$ scales it up for the nozzle.

Here:

$A_{b}$ = burning surface area (m², the area currently on fire).
$A_{t}$ = throat area of the nozzle (m², the smallest hole the gas escapes through).
$w$ = web burned = how far, perpendicular to itself, the flame front has eaten into the solid (metres).
$r$ = linear burn rate (m/s), how fast that front recedes.
$ρ_{p}$ = propellant density (kg/m³).
$n, a$ = the two Saint-Robert constants (see Saint-Robert burn rate law).
$c^{*}$ = characteristic velocity (m/s), a propellant-and-combustion property measuring how efficiently the burnt gas builds chamber pressure; the higher $c^{*}$ , the more pressure per unit mass flow (see Rocket equation & specific impulse).

Level 1 — Recognition

L1.1

State the burn class (neutral / progressive / regressive) and one sentence why: (a) a single-face end-burner, (b) a BATES segment burning on its inner cylinder only, (c) a late-stage rounded star near burnout.

Recall Solution

(a) Neutral. Only one flat circular face burns; its area $π R^{2}$ never changes as the face marches down the length. (b) Progressive. The port perimeter $2 π (R_{i} + w)$ grows as the surface recedes outward, so $A_{b}$ rises. (c) Regressive. Once the star's points are gone the outer burning boundary is a shrinking arc; perimeter falls → $A_{b}$ falls.

L1.2

For a case-bonded grain (ends inhibited, so only the port perimeter burns) of length $L$ and port perimeter $P (w)$ , write $A_{b} (w)$ . What single geometric quantity do you therefore "design" to shape the whole thrust curve?

Recall Solution

$A_{b} (w) = P (w) \cdot L$ Because $L$ is a fixed constant, the entire time-shape of $A_{b}$ (and hence thrust) is carried by ==the port perimeter $P (w)$ ==. Designing the burn = designing how the perimeter changes as the port opens outward.

Level 2 — Application

L2.1

A BATES segment burns on its inner cylinder only (ends inhibited). Given $R_{i} = 0.025$ m, $L = 0.40$ m, $ρ_{p} = 1800$ kg/m³, $r = 6 \times 1 0^{- 3}$ m/s. Compute $A_{b}$ and mass generation rate $\overset{m}{˙} = ρ_{p} A_{b} r$ at (a) $w = 0$ and (b) $w = 0.015$ m. By what percent did $\overset{m}{˙}$ rise?

Recall Solution

Inner surface is a cylinder of radius $R_{i} + w$ and length $L$ : $A_{b} = 2 π (R_{i} + w) L$ . (a) $w = 0$ : $A_{b} = 2 π (0.025) (0.40) = 0.06283 m^{2}$ . $\overset{m}{˙} = 1800 \times 0.06283 \times 0.006 = 0.6786 kg/s$ . (b) $w = 0.015$ : $R_{i} + w = 0.040$ , $A_{b} = 2 π (0.040) (0.40) = 0.10053 m^{2}$ . $\overset{m}{˙} = 1800 \times 0.10053 \times 0.006 = 1.0857 kg/s$ . Rise $= \frac{0.10053 - 0.06283}{0.06283} = 0.60 = 60%$ . Same $r$ , bigger radius → longer circumference → more surface → more gas. Progressive. ✔

L2.2

Use the amplification law $p_{c} \propto A_{b}^{1/ (1 - n)}$ with $n = 0.35$ . If $A_{b}$ rose by the 60% found above, by what factor does $p_{c}$ rise? (Assume $A_{t}$ fixed.)

Recall Solution

The area ratio is $\frac{A _{b 2}}{A _{b 1}} = 1.60$ . Then $\frac{p _{c 2}}{p _{c 1}} = (\frac{A _{b 2}}{A _{b 1}})^{\frac{1}{1 - n}} = 1.6 0^{\frac{1}{0.65}} = 1.6 0^{1.5385} \approx 2.06 .$ A 60% area rise becomes a ~106% pressure rise — that is the $\frac{1}{1 - n}$ amplification at work (see Chamber pressure & throat area (Kn ratio)).

Level 3 — Analysis

L3.1

A BATES segment with free (uninhibited) ends has both the inner cylinder growing and the two annular end-faces shrinking: $A_{b} (w) = port 2 π (R_{i} + w) (L - 2 w) + two ends 2 π [R_{o}^{2} - (R_{i} + w)^{2}] .$ Find $\frac{d A _{b}}{d w}$ at $w = 0$ and state the condition on $L$ that makes the segment exactly neutral at the start ( $d A_{b} / d w = 0$ ). Use $R_{i} = 0.02$ m, $R_{o} = 0.06$ m — solve for the neutral $L$ .

Figure — Grain geometry — BATES, star, wagon wheel; neutral - progressive - regressive burn

Recall Solution

Differentiate term by term (see the red = growing port, violet = shrinking ends in the figure). Port term $P = 2 π (R_{i} + w) (L - 2 w)$ : $\frac{d P}{d w} = 2 π [(L - 2 w) - 2 (R_{i} + w)] .$ Ends term $E = 2 π [R_{o}^{2} - (R_{i} + w)^{2}]$ : $\frac{d E}{d w} = 2 π [- 2 (R_{i} + w)] = - 4 π (R_{i} + w) .$ Sum at $w = 0$ : $\frac{d A _{b}}{d w}_{0} = 2 π [L - 2 R_{i}] - 4 π R_{i} = 2 π L - 4 π R_{i} - 4 π R_{i} = 2 π (L - 4 R_{i}) .$ Setting this to zero: $L = 4 R_{i}$ . With $R_{i} = 0.02$ : $L = 0.08$ m. Interpretation: if the barrel is short ( $L < 4 R_{i}$ ) the shrinking ends win → regressive; if long ( $L > 4 R_{i}$ ) the growing port wins → progressive; exactly $L = 4 R_{i}$ balances them at the start → neutral onset. ( $R_{o}$ dropped out of the derivative — it only fixes total web, not the starting slope.)

L3.2

Explain in one short paragraph why a star grain tends toward neutral even though its port clearly opens up as it burns.

Recall Solution

A star has two kinds of features: sharp points of propellant sticking inward, and valleys (slots) between them. As the front recedes: the points get eaten away, which would drop the perimeter; simultaneously the valleys widen and deepen, which raises the perimeter. A well-cut star is proportioned so these two opposite trends roughly cancel over the main burn, giving $P (w) \approx$ const, hence $A_{b} \approx$ const → neutral. Only near burnout, when the points are gone, does the perimeter start to shrink (the regressive sliver tail).

Level 4 — Synthesis

L4.1

A designer wants a flat thrust curve and must choose between (i) a free-ended BATES with $L = 4 R_{i}$ , and (ii) a case-bonded neutral star. The motor is prone to structural buzzing at the head-end. Which choice better resists combustion instability, and why? (One paragraph.)

Recall Solution

The case-bonded star is generally safer. In a free-standing BATES the grain sits loosely in the case, so its ends and the annular gaps present open surfaces that can drive and sustain acoustic oscillations; the grain can also vibrate as a free body (see Case-bonded vs free-standing grains). A case-bonded star is glued to the wall, damping structural modes, and its port shape (many slots) breaks up clean acoustic standing waves that feed Combustion instability. Both give ≈neutral thrust, but the bonded star pairs that flat curve with better damping. Trade-off: case-bonded grains suffer thermal-stress cracking at the bond line, which must be checked separately.

L4.2

Take the L2.1 progressive BATES ( $R_{i} = 0.025$ , $L = 0.40$ ) but now demand that chamber pressure stay within $\pm 10%$ of its mean while web goes from $w = 0$ to $w = 0.015$ . Using $p_{c} \propto A_{b}^{1/ (1 - n)}$ with $n = 0.35$ , show why this inner-only BATES fails the requirement, and name one geometry that would pass.

Recall Solution

From L2.2 the pressure ratio end-to-start is $1.6 0^{1.5385} \approx 2.06$ — pressure more than doubles. Even measured about the mean, that is roughly $\pm 34%$ swing, far outside $\pm 10%$ . Numerically: $p_{c} (0) \propto 0.0628 3^{1.5385}$ and $p_{c} (w) \propto 0.1005 3^{1.5385}$ ; ratio $2.06$ . A $\pm 10%$ band spans a factor $1.10/0.90 = 1.22$ , and $2.06 ≫ 1.22$ → fails. A neutral star (case-bonded, $P (w) \approx$ const) keeps $A_{b}$ nearly flat, so $p_{c}$ stays within a few percent — that passes. This is exactly why constant-thrust missiles use stars, not plain inner-burning cylinders.

Level 5 — Mastery

L5.1

Design task. You need a two-phase thrust curve: a high boost for the first third of the burn, then a lower sustain. Sketch (in words + the master chain) a grain concept using the geometries from the parent note, and state the burn class in each phase. Then give the qualitative $A_{b} (w)$ shape.

Recall Solution

Concept: a wagon-wheel (deep thin spokes) forward section transitioning into a plain cylindrical (BATES inner) or slightly regressive aft section.

Boost phase: the wagon wheel's huge initial perimeter → huge initial $A_{b}$ → (through the chain $A_{b} \Rightarrow K \Rightarrow p_{c} \propto K^{1/ (1 - n)} \Rightarrow F$ ) a big early thrust. As the thin spokes burn through and slivers form, $A_{b}$ falls sharply — regressive — dropping thrust to the sustain level.
Sustain phase: the remaining cylindrical bore holds $A_{b}$ roughly flat (near-neutral) at a lower value, giving a long steady tail. $A_{b} (w)$ shape: high at $w = 0$ , a steep drop as spokes vanish, then a low plateau — the classic "boost-sustain" staircase (violet curve in the figure). This is the standard way to get two thrust levels from one propellant grain without changing the chemistry.

L5.2

Prove the stability requirement $n < 1$ from the mass balance, and explain physically what goes wrong if $n \geq 1$ .

Recall Solution

Steady mass balance: gas generated = gas exhausted. Gas is made over the burning area at rate $ρ_{p} A_{b} r = ρ_{p} A_{b} a p_{c}^{n}$ , and escapes through the throat at rate $p_{c} A_{t} / c^{*}$ , where $c^{*}$ (the characteristic velocity, defined in the symbol list above) sets how much mass flow a given pressure drives out. Equate them: $ρ_{p} A_{b} a p_{c}^{n} = \frac{p _{c} A _{t}}{c ^{*}} .$ Collect powers of $p_{c}$ : divide both sides by $p_{c}^{n}$ , $ρ_{p} A_{b} a c^{*} = A_{t} p_{c}^{1 - n} \Rightarrow p_{c} = (\frac{ρ _{p} a c ^{*} A _{b}}{A _{t}})^{\frac{1}{1 - n}} .$ For this to give a finite, single positive pressure the exponent must be finite and the map self-correcting: $1 - n > 0$ , i.e. $n < 1$ . Physical meaning: start from equilibrium and nudge $p_{c}$ up slightly. Generation grows like $p_{c}^{n}$ ; exhaust grows like $p_{c}^{1}$ . If $n < 1$ , exhaust outruns generation — the extra pressure is bled off faster than it's made, so the system falls back to equilibrium (self-regulating). If $n \geq 1$ , generation grows at least as fast as exhaust: any upward nudge feeds itself → runaway pressure and burst. (This is the same $\frac{1}{1 - n}$ that appears in Chamber pressure & throat area (Kn ratio).)

Recall Quick self-check clozes

A free-ended BATES is neutral-at-onset when ::: $L = 4 R_{i}$ . A 60% rise in $A_{b}$ with $n = 0.35$ raises $p_{c}$ by a factor ::: $1.6 0^{1/0.65} \approx 2.06$ . Wagon-wheel forward + cylindrical aft gives a ::: boost-then-sustain (regressive→neutral) thrust curve. Stability requires ::: $n < 1$ , so nozzle exhaust ( $\propto p_{c}$ ) outruns generation ( $\propto p_{c}^{n}$ ).