Solid propellants — fuel + oxidizer in polymer matrix: Worked examples

Before we start, here are the only tools we will use, each stated in plain words so no symbol is unearned:

Recall The four relations we lean on (from the parent note)

Thrust (ideal): $F = \overset{m}{˙} v_{e}$ — force equals how much mass leaves per second, times how fast it leaves.
Thrust (full): $F = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ — add a push/pull from the pressure mismatch at the nozzle mouth.
Mass burn rate: $\overset{m}{˙} = ρ_{p} A_{b} r$ — density $\times$ burning surface area $\times$ how fast the flame eats inward.
Burn-rate law: $r = a p_{c}^{n}$ — flame speed rises with chamber pressure; stable only when $n < 1$ .

Here $\overset{m}{˙}$ (read "m-dot") means "kilograms of solid turned to gas per second". The dot is shorthand for "rate of change with time".

The scenario matrix

Every solid-motor numerical problem is one of these cells. The examples below are tagged with the cell they cover so you can see the whole board is filled.

#	Cell class	What makes it tricky	Example
A	Baseline — plug into $\overset{m}{˙} = ρ_{p} A_{b} r$ , $F = \overset{m}{˙} v_{e}$	nothing; warm-up	Ex 1
B	Pressure exponent scaling ( $n < 1$ )	ratio, exponent, $a$ cancels	Ex 2
C	Degenerate: $A_{b} \to 0$ (burnout)	thrust falls to zero — limiting case	Ex 3
D	Runaway: $n \geq 1$	unstable feedback — the forbidden case	Ex 4
E	Geometry: progressive (cylindrical port grows)	$A_{b}$ rises with time	Ex 5
F	Geometry: neutral (star / end-burner)	$A_{b}$ held constant	Ex 5
G	Pressure term: vacuum vs sea level	sign of $(p_{e} - p_{a}) A_{e}$ flips	Ex 6
H	Real-world word problem (mission burn time + total impulse)	multi-step, unit-heavy	Ex 7
I	Exam twist ( $I_{s p}$ from $v_{e}$ ; feeds Tsiolkovsky)	connects three formulas	Ex 8

Cells A–I are all covered by Examples 1–8.

Statement. A motor has $ρ_{p} = 1750 kg/m^{3}$ , burning area $A_{b} = 0.040 m^{2}$ , burn rate $r = 9 mm/s$ , exhaust speed $v_{e} = 2500 m/s$ , and is running matched ( $p_{e} = p_{a}$ ). Find $\overset{m}{˙}$ and thrust $F$ .

Forecast: Guess the thrust to the nearest kN before reading on. (Hint: $\overset{m}{˙}$ will be under 1 kg/s, $v_{e}$ is 2500 → so $F$ a couple of kN.)

Convert the burn rate: $r = 9 mm/s = 0.009 m/s$ . Why this step? Every other quantity is in metres and kilograms; mixing millimetres would make the answer wrong by $1000 \times$ .
Mass burn rate: $\overset{m}{˙} = ρ_{p} A_{b} r = 1750 \times 0.040 \times 0.009 = 0.63 kg/s$ . Why this step? In one second the flame recedes $0.009$ m over the whole $0.040 m^{2}$ face, carving out $0.00036 m^{3}$ of solid; times density gives the kilograms.
Thrust: $F = \overset{m}{˙} v_{e} = 0.63 \times 2500 = 1575 N$ . Why this step? Since $p_{e} = p_{a}$ the pressure term is exactly zero, so ideal thrust is just momentum flux.

Verify: Units: $(kg/s) (m/s) = kg\cdotm/s^{2} = N$ ✓. Magnitude: $\approx 1.6$ kN, a small-motor value — sensible. ✓

Example 2 — Pressure exponent scaling (cell B)

Worked example

Statement. A propellant burns at $r_{1} = 9 mm/s$ when the chamber pressure is $p_{1} = 6 MPa$ . Its pressure exponent is $n = 0.40$ . What is the burn rate at $p_{2} = 9 MPa$ ?

Forecast: The pressure jumped $50%$ . Because $n < 1$ , will the burn rate rise by more or less than $50%$ ?

Write the ratio of the burn-rate law: $\frac{r _{2}}{r _{1}} = \frac{a p _{2}^{n}}{a p _{1}^{n}} = (\frac{p _{2}}{p _{1}})^{n}$ . Why this step? The messy temperature-dependent constant $a$ is identical in both states, so dividing makes it vanish. We never even need its value.
Plug numbers: $\frac{p _{2}}{p _{1}} = \frac{9}{6} = 1.5$ , so $\frac{r _{2}}{r _{1}} = 1. 5^{0.40}$ . Why this step? Raising to the power $n = 0.40$ is what "which fraction of the pressure change reaches the flame" looks like.
Evaluate the power: $ln 1.5 = 0.4055$ , $\times 0.40 = 0.1622$ , $e^{0.1622} = 1.176$ . So $r_{2} = 9 \times 1.176 = 10.6 mm/s$ . Why this step? We use $ln$ then $e^{(\cdot)}$ because there is no "nice" root here — the exponential undoes the logarithm to recover $1. 5^{0.4}$ .

Verify: The pressure rose $50%$ but the burn rate rose only $\approx 18%$ — less than the pressure change, exactly what $n < 1$ guarantees. This self-limiting behaviour is why the motor doesn't run away. ✓

Example 3 — Degenerate case: burnout, $A_{b} \to 0$ (cell C)

Worked example

Statement. An end-burner has a constant face area while it burns, but in the final instant the last sliver of propellant vanishes and $A_{b} \to 0$ . Take the Example-1 motor ( $ρ_{p} = 1750$ , $r = 0.009$ m/s, $v_{e} = 2500$ ) but let $A_{b} = 0.010 m^{2}$ , then $A_{b} = 0.001 m^{2}$ , then $A_{b} = 0$ . What thrust does each give?

Forecast: As $A_{b}$ shrinks toward zero, does thrust taper smoothly or drop off a cliff?

Thrust as a function of area: $F (A_{b}) = ρ_{p} A_{b} r v_{e} = (1750) (0.009) (2500) A_{b} = 39375 A_{b}$ . Why this step? Chaining the two relations shows thrust is directly proportional to burning area — a straight line through the origin.
Evaluate: at $A_{b} = 0.010 \Rightarrow F = 393.75 N$ ; at $A_{b} = 0.001 \Rightarrow F = 39.375 N$ ; at $A_{b} = 0 \Rightarrow F = 0 N$ . Why this step? Each is just $39375$ times the area; the limit $A_{b} \to 0$ gives thrust $\to 0$ continuously.
Interpret the limit: the moment the burning surface disappears, so does all thrust — the motor cannot coast on "stored force". Thrust exists only while something is burning. Why this step? This is the degenerate boundary of the whole model: no $A_{b}$ , no $\overset{m}{˙}$ , no $F$ .

Verify: $F$ is linear in $A_{b}$ , so halving/tenthing the area exactly tenths the thrust ( $393.75 \to 39.375$ ) ✓, and $F (0) = 0$ with no discontinuity ✓. Units check on the slope $ρ_{p} r v_{e}$ : $(kg/m^{3}) (m/s) (m/s) = kg / (m\cdots^{2})$ ; multiplying by $A_{b}$ in $m^{2}$ gives $(kg / (m\cdots^{2})) (m^{2}) = kg\cdotm/s^{2} = N$ ✓.

Example 4 — The forbidden case: runaway when $n \geq 1$ (cell D)

Worked example

Statement. Two propellants sit at $p_{1} = 5 MPa$ , $r_{1} = 8 mm/s$ . A random flutter pushes pressure up $20%$ to $p_{2} = 6 MPa$ . Compare the burn-rate response for a safe propellant ( $n = 0.4$ ) and a forbidden one ( $n = 1.3$ ). Which one feeds back to blow up?

Forecast: More burn rate makes more gas makes more pressure. When does that loop settle, and when does it explode?

Safe propellant: $\frac{r _{2}}{r _{1}} = 1. 2^{0.4}$ . $ln 1.2 = 0.1823$ , $\times 0.4 = 0.0729$ , $e^{0.0729} = 1.0757$ . So burn rate rises $\approx 7.6%$ for a $20%$ pressure bump. Why this step? The response ( $7.6%$ ) is smaller than the disturbance ( $20%$ ): the extra gas can't sustain the pressure spike, so it relaxes back. Stable.
Forbidden propellant: $\frac{r _{2}}{r _{1}} = 1. 2^{1.3}$ . $ln 1.2 = 0.1823$ , $\times 1.3 = 0.2370$ , $e^{0.2370} = 1.267$ . Burn rate rises $\approx 26.7%$ . Why this step? Now the response ( $26.7%$ ) is larger than the disturbance ( $20%$ ): more gas → higher pressure → even more gas. The loop amplifies itself.
Conclude: the boundary is exactly $n = 1$ . For $n < 1$ disturbances shrink; for $n \geq 1$ they grow without bound → detonation. This is why the parent note insists $n < 1$ . Why this step? It pins the design rule to a limiting value, not a vague "keep it low".

Verify: At the knife-edge $n = 1$ : $1. 2^{1} = 1.20$ , i.e. burn rate rises exactly as much as pressure — neither damping nor amplifying (marginal). Safe side $1.076 < 1.20 < 1.267$ forbidden side ✓, straddling the $n = 1$ marginal value as expected.

Example 5 — Grain geometry: progressive vs neutral (cells E & F)

Figure 1 — Thrust ratio vs distance burned outward, for a tube port (progressive) and a star port (neutral).

Figure — Solid propellants — fuel + oxidizer in polymer matrix

Worked example

Statement. A cylindrical-port grain of length $L = 0.5 m$ starts with port radius $R_{0} = 20 mm$ . The burning surface is the inner wall, $A_{b} = 2 π R L$ . It burns $10 mm$ outward. (a) By what factor does $A_{b}$ rise (progressive)? (b) A star grain instead holds $A_{b}$ constant (neutral) — if it starts at the same $A_{b}$ and thrust, what is its thrust ratio at the same moment? Use $ρ_{p} = 1750$ , $r = 0.009$ m/s, $v_{e} = 2500$ .

Forecast: For the tube, thrust follows the port radius. Guess the percentage thrust rise after the radius grows from 20 to 30 mm.

Convert lengths to metres and find the initial area: $R_{0} = 20 mm = 0.020 m$ , so $A_{b, 0} = 2 π R_{0} L = 2 π (0.020) (0.5) = 0.0628 m^{2}$ . Why this step? The burning surface of an internal cylinder is its lateral (side) area, $2 π R L$ — a rectangle rolled into a tube of circumference $2 π R$ and height $L$ . We work in metres so the area comes out in m² to match $ρ_{p}$ in kg/m³.
Final radius after burning 10 mm: $R_{1} = 20 + 10 = 30 mm = 0.030 m$ . Final area $A_{b, 1} = 2 π R_{1} L = 2 π (0.030) (0.5) = 0.0942 m^{2}$ . Why this step? The flame front eats radially outward, so the port radius grows by the 10 mm burned (converted here to 0.010 m) and the burning circumference grows with it.
Progressive ratio: $\frac{A _{b, 1}}{A _{b, 0}} = \frac{R _{1}}{R _{0}} = \frac{30}{20} = 1.5$ . Since thrust $\propto A_{b}$ , thrust rises $50%$ — an undesirable rising curve. Why this step? For a tube, $A_{b}$ is proportional to $R$ , so the area ratio is just the radius ratio; $L$ , $2 π$ cancel (and the mm↔m conversion cancels too, since it applies to both radii). This is the magenta curve reaching 1.50.
Neutral (star): by design $A_{b}$ stays constant as extra star points recede to offset the widening. Thrust ratio $= 1.00$ — a flat curve. Why this step? A star port adds fresh burning surface exactly where a plain tube would lose it, keeping $A_{b}$ and thus thrust constant — the violet dashed line held at 1.00.

Verify: Initial thrust $F_{0} = ρ_{p} A_{b, 0} r v_{e} = 1750 (0.0628) (0.009) (2500) = 2472 N$ . Progressive final $F_{1} = 1.5 F_{0} = 3708 N$ . Ratio $3708/2472 = 1.5$ ✓. Neutral final $= F_{0} = 2472 N$ , ratio $1.00$ ✓.

Example 6 — Pressure term: vacuum vs sea level (cell G)

Worked example

Statement. A motor has $\overset{m}{˙} = 4 kg/s$ , $v_{e} = 2600 m/s$ , nozzle exit area $A_{e} = 0.05 m^{2}$ , and exit pressure $p_{e} = 70 kPa$ . Find the thrust (a) at sea level ( $p_{a} = 101.3 kPa$ ) and (b) in vacuum ( $p_{a} = 0$ ). (Recall from the definition above: $A_{e}$ is the exit opening's area, $p_{e}$ the gas pressure there, $p_{a}$ the outside pressure pushing back.)

Forecast: The full formula adds $(p_{e} - p_{a}) A_{e}$ . Which environment gives more thrust, and does the extra term ever subtract?

Momentum thrust (same in both): $\overset{m}{˙} v_{e} = 4 \times 2600 = 10400 N$ . Why this step? Mass flow and exhaust speed are set by the chemistry/nozzle, unaffected by outside air.
Sea level pressure term: $(p_{e} - p_{a}) A_{e} = (70000 - 101300) (0.05) = (- 31300) (0.05) = - 1565 N$ . Why this step? Here $p_{e} < p_{a}$ (over-expanded nozzle): outside air pushes back on the exit plane harder than the jet pushes out, so the term is negative — it steals thrust.
Sea-level total: $F = 10400 - 1565 = 8835 N$ . Why this step? The full thrust formula is momentum thrust plus the pressure term; adding a negative number is the same as subtracting, so the atmospheric back-push lowers the net force below the ideal $10400 N$ .
Vacuum pressure term: $(70000 - 0) (0.05) = + 3500 N$ , so $F = 10400 + 3500 = 13900 N$ . Why this step? With no ambient air pushing back, the exit pressure acts fully outward — the term flips positive, so the same motor produces more thrust in space.

Verify: Vacuum thrust ( $13900$ N) $>$ sea-level thrust ( $8835$ N) ✓ — real rockets do gain thrust as they climb. The gap between them is exactly $p_{a} A_{e} = 101300 \times 0.05 = 5065 N$ , and indeed $8835 + 5065 = 13900$ ✓. Units of every term: $(Pa) (m^{2}) = (N/m^{2}) (m^{2}) = N$ , matching the momentum term $(kg/s) (m/s) = N$ ✓.

Example 7 — Real-world word problem (cell H)

Worked example

Statement. A booster carries $M_{p} = 900 kg$ of propellant. It burns at a constant $\overset{m}{˙} = 6 kg/s$ with $v_{e} = 2400 m/s$ (matched). Find: (a) burn time, (b) constant thrust, (c) total impulse $J = F t_{b}$ , and (d) confirm $J$ also equals $M_{p} v_{e}$ .

Forecast: Two different routes should give the same total impulse. Guess whether $J$ is nearer $1$ million or $2$ million N·s.

Burn time: $t_{b} = \frac{M _{p}}{m ˙} = \frac{900}{6} = 150 s$ . Why this step? Total mass divided by mass-per-second gives seconds — how long the propellant lasts.
Thrust: $F = \overset{m}{˙} v_{e} = 6 \times 2400 = 14400 N$ . Why this step? Constant $\overset{m}{˙}$ and $v_{e}$ mean constant thrust throughout the burn.
Total impulse via force×time: $J = F t_{b} = 14400 \times 150 = 2160000 N\cdots$ . Why this step? Impulse is the accumulated push; with constant force it's simply force times duration.
Cross-check via mass×speed: $J = M_{p} v_{e} = 900 \times 2400 = 2160000 N\cdots$ . Why this step? Total impulse equals all the momentum thrown backward, which is total ejected mass times its speed — a route that never mentions time.

Verify: Both routes give $2.16 \times 1 0^{6} N\cdots$ ✓. Units: $N\cdots = kg\cdotm/s$ (momentum) ✓. This value feeds Specific Impulse and the Tsiolkovsky Rocket Equation for the delivered $Δ v$ .

Example 8 — Exam twist: from $v_{e}$ to specific impulse and $Δ v$ (cell I)

Two new quantities appear here; both are defined in plain words before we use them.

Worked example

Statement. For the Example-7 booster ( $v_{e} = 2400 m/s$ , $M_{p} = 900 kg$ ), take a total vehicle wet mass $m_{0} = 1200 kg$ (so dry mass $m_{f} = 1200 - 900 = 300 kg$ ). Find (a) specific impulse $I_{s p}$ with $g_{0} = 9.81 m/s^{2}$ , and (b) the ideal velocity gain $Δ v$ .

Forecast: $I_{s p}$ in the couple-hundred-seconds range is typical for solids. Guess $Δ v$ — will it exceed $v_{e}$ itself?

Specific impulse: $I_{s p} = \frac{v _{e}}{g _{0}} = \frac{2400}{9.81} = 244.6 s$ . Why this step? Dividing exhaust speed by $g_{0}$ converts "how fast the gas leaves" into the standard efficiency figure engineers quote in seconds — the only step needed once $v_{e}$ is known.
Mass ratio: $\frac{m _{0}}{m _{f}} = \frac{1200}{300} = 4$ . Why this step? The rocket equation cares only about the ratio of full to empty mass — how big a fraction was throwable — so we compute that pure number first.
Rocket equation: $Δ v = v_{e} ln (m_{0} / m_{f}) = 2400 \times ln 4$ . $ln 4 = 1.3863$ , so $Δ v = 2400 \times 1.3863 = 3327 m/s$ . Why this step? We use the natural log because velocity builds up as the fractional mass falls — each equal fractional mass loss adds an equal velocity chunk, and $ln$ is exactly the "sum of equal fractional steps."

Verify: $I_{s p} \approx 245 s$ sits squarely in the solid-motor range ( $\sim$ 200–290 s) ✓. And $Δ v = 3327 > v_{e} = 2400$ : because the mass ratio exceeds $e^{1} \approx 2.718$ , the ship ends up faster than its own exhaust speed ✓. Units of $Δ v$ : m/s (the $ln$ is dimensionless) ✓.

Recall

Recall Which cell was which?

Baseline plug-in ::: Ex 1 (cell A)
$a$ cancels in a burn-rate ratio ::: Ex 2 (cell B)
Thrust $\to 0$ as $A_{b} \to 0$ ::: Ex 3 (cell C)
Runaway boundary at $n = 1$ ::: Ex 4 (cell D)
Tube grows $A_{b}$ (progressive), star keeps it flat (neutral) ::: Ex 5 (cells E, F)
Pressure term flips sign vacuum vs sea level ::: Ex 6 (cell G)
Total impulse two ways ::: Ex 7 (cell H)
$I_{s p}$ and $Δ v$ from $v_{e}$ ::: Ex 8 (cell I)

Recall The three exit-plane symbols

What is $A_{e}$ ? ::: The nozzle exit opening's area (m²).
What is $p_{e}$ ? ::: The exhaust gas pressure right at the exit plane (Pa).
What is $p_{a}$ ? ::: The ambient (surrounding) pressure pushing back; $\approx 101.3$ kPa at sea level, $0$ in vacuum.
What is $I_{s p}$ and its unit? ::: Specific impulse $= v_{e} / g_{0}$ , measured in seconds.

Connections

Parent topic — the four master relations these examples exercise.
Specific Impulse — Ex 8 turns $v_{e}$ into $I_{s p}$ .
Tsiolkovsky Rocket Equation — Ex 7–8 feed total impulse and $Δ v$ .
De Laval Nozzle — sets $v_{e}$ , $p_{e}$ , $A_{e}$ used in Ex 6.
Newton's Third Law — the momentum principle behind every thrust value.
Liquid Propellants — contrast for the throttle/restart limits shown by fixed geometry.

Worked examples — Solid propellants — fuel + oxidizer in polymer matrix

The scenario matrix

Example 1 — Baseline (cell A)

Example 2 — Pressure exponent scaling (cell B)

Example 3 — Degenerate case: burnout, $A_{b} \to 0$ (cell C)

Example 4 — The forbidden case: runaway when $n \geq 1$ (cell D)

Example 5 — Grain geometry: progressive vs neutral (cells E & F)

Example 6 — Pressure term: vacuum vs sea level (cell G)

Example 7 — Real-world word problem (cell H)

Example 8 — Exam twist: from $v_{e}$ to specific impulse and $Δ v$ (cell I)

Recall

Connections