Solid propellants — AP - HTPB - Al composition; burn rate dependence on pressure (Vieille's law): Worked examples

Before anything, recall the single object we keep re-using:

The scenario matrix

Every question about $r = a P^{n}$ falls into one of these cells. Each worked example below is tagged with the cell(s) it covers.

#	Cell class	What makes it tricky	Example
A	Direct forward — given $a, n, P$ , find $r$	non-integer exponent needs logs	Ex 1
B	Degenerate $n = 0$	$P^{0} = 1$ : pressure-independent burn	Ex 2
C	Extract $n$ from two $(P, r)$ points	ratio kills $a$ , logs linearise	Ex 3
D	Extract $a$ after $n$ is known	back-substitute one point	Ex 4
E	Ratio / scaling — new $P$ , find new $r$ (no $a$ needed)	never compute $a$ at all	Ex 5
F	Limiting/danger case $n \to 1$	equilibrium pressure blows up	Ex 6
G	Real-world word problem — full motor: $r \to \overset{m}{˙} \to$ burn time	chain $r, A_{b}, ρ_{p}$ , unit trap	Ex 7
H	Exam twist — temperature shift changes $a$ , or graph read-off	conceptual, not just plug	Ex 8

Example 1 — Cell A: direct forward

Worked example Given

$a, n, P$ , find the burn rate AP/HTPB propellant: $a = 5.0$ (mm/s at $P$ in MPa), $n = 0.35$ . Chamber pressure $P = 7$ MPa. Find $r$ .

Forecast: $7^{0.35}$ is less than $7$ but more than $1$ (any power between 0 and 1 pulls a number toward 1). Guess $r$ is roughly $5 \times 2 = 10$ mm/s.

Write the law: $r = a P^{n} = 5.0 \times 7^{0.35}$ . Why this step? Direct substitution — this is the whole point of Vieille's law.
The exponent $0.35$ is not a whole number, so we can't multiply $7$ by itself. Rewrite the power using the natural log identity $x^{p} = e^{p l n x}$ : $7^{0.35} = e^{0.35 l n 7} = e^{0.35 \times 1.9459} = e^{0.6811} = 1.976$ Why this step? The exponential/log pair is the only tool that evaluates a non-integer power — logs turn "raise to a fractional power" into "multiply, then un-log". (See Combustion Thermodynamics for where $e^{x}$ keeps appearing.)
Multiply: $r = 5.0 \times 1.976 = 9.88$ mm/s. Why this step? Finish the arithmetic.

Verify: Units: $a$ was quoted "mm/s when $P$ in MPa", so $r$ comes out in mm/s. ✓ Sanity: $9.88$ mm/s matches our forecast of "about 10". ✓

Example 2 — Cell B: the degenerate case $n = 0$

Example 3 — Cell C: extract the exponent $n$

Worked example Measure

$n$ from two firings A test motor gives: at $P_{1} = 5$ MPa, $r_{1} = 8.5$ mm/s; at $P_{2} = 10$ MPa, $r_{2} = 10.8$ mm/s. Find $n$ .

Forecast: Pressure doubled ( $5 \to 10$ ) but burn rate rose only from 8.5 to 10.8 (a factor ~1.27). A gentle rise → small $n$ , guess $n \approx 0.3$ .

Write the law at both points and divide: $\frac{r _{2}}{r _{1}} = \frac{a P _{2}^{n}}{a P _{1}^{n}} = (\frac{P _{2}}{P _{1}})^{n}$ Why this step? ==Taking the ratio cancels the unknown $a$ == — a classic trick, because $a$ divides out cleanly, leaving one equation in one unknown $n$ .
Take natural logs of both sides to pull $n$ down from the exponent: $ln \frac{r _{2}}{r _{1}} = n ln \frac{P _{2}}{P _{1}}$ Why this step? The logarithm is the tool that turns a power into a product — $ln (x^{n}) = n ln x$ — which is the only way to solve for an exponent.
Rearrange and plug in: $n = \frac{l n ( r _{2} / r _{1} )}{l n ( P _{2} / P _{1} )} = \frac{l n ( 10.8/8.5 )}{l n ( 10/5 )} = \frac{l n 1.2706}{l n 2} = \frac{0.2394}{0.6931} = 0.3454$ Why this step? Solve for $n$ . This is literally how labs measure $n$ : plot $ln r$ vs $ln P$ , and the slope is $n$ .

Verify: Forecast said ~0.3; we got $0.345$ . ✓ Check: $(P_{2} / P_{1})^{n} = 2^{0.345} = 1.270$ , and indeed $r_{2} / r_{1} = 10.8/8.5 = 1.271$ . ✓

Figure — Solid propellants — AP - HTPB - Al composition; burn rate dependence on pressure (Vieille's law)

The figure above is the meaning of Ex 3: on log-log axes the power law becomes a straight line, and its slope is $n$ . Two points fix the line.

Example 4 — Cell D: extract the coefficient $a$

Example 5 — Cell E: pure scaling, never touch $a$

Worked example Predict the new burn rate after a pressure change

A motor burns at $r_{1} = 9.0$ mm/s when $P_{1} = 6$ MPa. The exponent is $n = 0.4$ . During operation the chamber pressure rises to $P_{2} = 9$ MPa. What is the new burn rate $r_{2}$ ? (You are NOT told $a$ .)

Forecast: Pressure went up by factor $1.5$ . With $n = 0.4$ the rate rises less than $1.5 \times$ — guess $r_{2} \approx 9 \times 1.18 \approx 10.6$ mm/s.

Use the ratio form so $a$ never appears: $r_{2} = r_{1} (\frac{P _{2}}{P _{1}})^{n} = 9.0 \times (\frac{9}{6})^{0.4}$ Why this step? ==When only a scaling is asked, computing $a$ is wasted work== — the ratio form answers it directly and avoids rounding errors.
Evaluate the bracket: $(1.5)^{0.4} = e^{0.4 l n 1.5} = e^{0.4 \times 0.4055} = e^{0.1622} = 1.1761$ . Why this step? Fractional power → log/exp, our recurring tool.
Multiply: $r_{2} = 9.0 \times 1.1761 = 10.59$ mm/s.

Verify: Forecast ~10.6 ✓. Cross-check direction: $P$ rose, $n > 0$ , so $r$ must rise — and $10.59 > 9.0$ . ✓ If we'd used $n = 0$ it would stay $9.0$ (matches Ex 2 logic). ✓

Example 6 — Cell F: the danger limit $n \to 1$

Worked example Why

$n$ near 1 is a bomb Two propellants have the same $a$ but different exponents: (i) $n = 0.3$ , (ii) $n = 0.95$ . Both are near an equilibrium set by $A_{b} / A_{t} = 200$ (dimensionless group). Compare how the equilibrium pressure exponent $\frac{1}{1 - n}$ amplifies the geometry ratio.

Forecast: $\frac{1}{1 - 0.3}$ is mild; $\frac{1}{1 - 0.95}$ is huge. The $n = 0.95$ motor's pressure will be wildly more sensitive. Guess: one raises 200 to a small power, the other to power 20.

Recall the equilibrium-pressure relation from the parent note: $P_{e q} \propto (\frac{A _{b}}{A _{t}})^{\frac{1}{1 - n}}$ Why this step? This is where $n$ decides stability, from Characteristic Velocity c-star and the mass-balance argument.
Case (i): exponent $= \frac{1}{1 - 0.3} = \frac{1}{0.7} = 1.4286$ . So $P_{e q} \propto 20 0^{1.4286}$ . Why this step? A mild power: a 1% change in $A_{b} / A_{t}$ changes $P_{e q}$ by about $1.43%$ . Tame.
Case (ii): exponent $= \frac{1}{1 - 0.95} = \frac{1}{0.05} = 20$ . So $P_{e q} \propto 20 0^{20}$ . Why this step? A 1% change in the geometry ratio changes $P_{e q}$ by about $20%$ . A tiny perturbation runs away → the physical meaning of instability.
As $n \to 1^{-}$ , the exponent $\frac{1}{1 - n} \to + \infty$ : the limit is a vertical wall — no stable pressure exists. Why this step? Covers the limiting behaviour cell explicitly: the singularity at $n = 1$ is why designers keep $n < 0.5$ .

Verify: Exponents: $1/0.7 = 1.4286$ ✓ and $1/0.05 = 20$ ✓. The ratio of amplification $20/1.4286 = 14$ : the near-1 exponent is 14× more twitchy. ✓

Example 7 — Cell G: full real-world motor (word problem)

Worked example From burn rate to burn time

A cylindrical grain burns on its inner surface. Take burning area $A_{b} = 0.5$ m², density $ρ_{p} = 1800$ kg/m³. The propellant obeys $a = 5.0$ mm/s (MPa), $n = 0.35$ , and the chamber holds steady at $P = 7$ MPa. The web (thickness of propellant to burn through) is $W = 40$ mm. Find (a) burn rate, (b) mass flow rate $\overset{m}{˙}$ , (c) approximate burn time $t_{b}$ .

Forecast: From Ex 1, $r \approx 9.9$ mm/s. Burning through 40 mm at ~10 mm/s → about 4 seconds. Mass flow of order a few kg/s.

(a) Burn rate: identical to Ex 1: $r = 5.0 \times 7^{0.35} = 9.88$ mm/s $= 0.00988$ m/s. Why this step? Vieille's law gives the surface regression speed first — everything else flows from it.
(b) Mass flow: $\overset{m}{˙} = ρ_{p} A_{b} r = 1800 \times 0.5 \times 0.00988$ . Why this step? ==Mass generated per second = density × burning area × regression speed== — this is the bridge from geometry to thrust (feeds Specific Impulse). $\overset{m}{˙} = 1800 \times 0.5 \times 0.00988 = 8.89 kg/s$ Unit trap: $r$ MUST be in m/s (not mm/s) here, else you'd be off by 1000. $kg/m^{3} \cdot m^{2} \cdot m/s = kg/s$ . ✓
(c) Burn time: the surface must travel the web $W = 40$ mm $= 0.040$ m at speed $r$ : $t_{b} = \frac{W}{r} = \frac{0.040}{0.00988} = 4.05 s$ Why this step? Distance ÷ speed = time — the web is the distance the flame front covers.

Verify: Units of $t_{b}$ : $m \div (m/s) = s$ ✓. Forecast said "about 4 s" and "a few kg/s" — got 4.05 s and 8.89 kg/s ✓. Cross-check: total propellant mass $\approx \overset{m}{˙} \cdot t_{b} = 8.89 \times 4.05 = 36.0$ kg; also $ρ_{p} A_{b} W = 1800 \times 0.5 \times 0.040 = 36.0$ kg — the two independent routes agree. ✓✓

Example 8 — Cell H: exam twist (temperature shift)

Worked example A cold-day firing

A motor's coefficient depends on grain temperature. At the reference temperature $a = 5.0$ mm/s (MPa). The temperature sensitivity is $σ_{p} = 0.002$ per K (meaning $a$ increases by $0.2%$ per Kelvin rise). The grain is fired $25$ K colder than reference, with $n = 0.35$ , at $P = 7$ MPa. Does the motor burn faster or slower, and by how much is $r$ ?

Forecast: Colder grain → harder to heat the surface → burn rate should drop. A 25 K drop at 0.2%/K is a 5% drop. Guess $r \approx 9.88 \times 0.95 \approx 9.4$ mm/s.

Adjust $a$ for temperature. A change of $Δ T = - 25$ K gives $a_{co l d} = a (1 + σ_{p} Δ T) = 5.0 (1 + 0.002 \times (- 25)) = 5.0 \times (1 - 0.05) = 4.75$ Why this step? ==Temperature enters through $a$ , not through $n$ == — the exponent is a chemistry/geometry property, but $a$ carries the thermal state (initial temperature $T_{0}$ ). Colder → smaller $a$ .
Apply Vieille's law with the corrected coefficient: $r = a_{co l d} P^{n} = 4.75 \times 7^{0.35} = 4.75 \times 1.976 = 9.39 mm/s$ Why this step? Same $7^{0.35} = 1.976$ as Ex 1 (reused); only $a$ changed.
Interpret: burn rate fell from $9.88$ to $9.39$ mm/s — the motor is slower and lower-thrust in the cold. This is exactly why launch vehicles have temperature limits. Why this step? Connects the number to a real engineering constraint.

Verify: Forecast "drops ~5% to ~9.4" ✓. Ratio: $9.39/9.88 = 0.9504 \approx (1 - 0.05)$ ✓ — the 5% drop propagated straight through, since $P^{n}$ was untouched. ✓

Recall The whole matrix in one breath

Forward: plug in ( $r = a P^{n}$ ). Degenerate $n = 0$ : $P$ drops out. Find $n$ : ratio kills $a$ , logs pull it down. Find $a$ : divide one point by $P^{n}$ . Scaling: use the ratio, never compute $a$ . Danger $n \to 1$ : exponent $\frac{1}{1 - n} \to \infty$ , explode. Word problem: $r \to \overset{m}{˙} = ρ_{p} A_{b} r \to t_{b} = W / r$ . Temperature twist: shift $a$ , keep $n$ .

Active Recall

Which quantity in

r = a P^{n}

carries the effect of grain temperature?

The coefficient

a

(not the exponent

n

Why does taking a ratio

r_{2} / r_{1}

help find

n

Because

a

cancels, leaving

(P_{2} / P_{1})^{n}

, one equation in one unknown.

For a plateau propellant

n = 0

, what happens to

r

if pressure doubles?

Nothing —

P^{0} = 1

, so

r = a

regardless of pressure.

In a full motor, how do you get burn time from burn rate?

t_{b} = W / r

, the web thickness divided by the regression speed.

n \to 1

, what happens to the equilibrium-pressure exponent

1/ (1 - n)

It diverges to

+ \infty

— no stable pressure, the motor runs away.