Solid rocket Isp derivation from grain properties: Worked examples

For the underlying formulas see Nozzle Isentropic Expansion, Saint-Robert Burn Rate Law, Characteristic Velocity c-star, and Tsiolkovsky Rocket Equation.

The scenario matrix

Every problem in this topic is one (or a blend) of the cells below. Each worked example is tagged with its cell letter.

Cell	Scenario class	What makes it tricky	Example
A	Baseline ideal $I_{s p}$ ( $p_{e} = p_{a}$ )	plug the master formula, no pressure term	Ex 1
B	Grain scaling (change $A_{b}$ )	check $I_{s p}$ is invariant	Ex 2
C	Mismatched nozzle ( $p_{e} \neq = p_{a}$ )	must keep $(p_{e} - p_{a}) A_{e}$ term	Ex 3
D	Limiting: $p_{e} / p_{c} \to 0$ (vacuum, infinite expansion)	bracket $\to 1$ , max possible $v_{e}$	Ex 4
E	Degenerate: $p_{e} / p_{c} \to 1$ (no expansion)	bracket $\to 0$ , thrust $\to 0$	Ex 4
F	Molar-mass sensitivity ( $M$ halved)	scaling $v_{e} \propto 1/ M$	Ex 5
G	Equilibrium $p_{c}$ from grain + throat, and the $n \to 1$ danger	exponent $1/ (1 - n)$ blows up	Ex 6
H	Real-world word problem (burn time from web)	link geometry to time, not just rate	Ex 7
I	Exam twist (altitude: over- vs. under-expanded)	sign of $(p_{e} - p_{a})$ flips	Ex 8

Example 1 — Cell A: baseline ideal $I_{s p}$

Worked example

A motor burns a propellant with $γ = 1.2$ , chamber temperature $T_{c} = 3000 K$ , exhaust molar mass $M = 0.025 kg/mol$ , pressure ratio $p_{e} / p_{c} = 0.01$ , and it is perfectly expanded ( $p_{e} = p_{a}$ ). Universal gas constant $R = 8.314 J/(mol\cdotK)$ . Find $I_{s p}$ .

Forecast: Solid motors typically hit $200$ – $260 s$ . Guess a number in seconds before reading on.

Step 1 — prefactor $\frac{2 γ}{γ - 1}$ . $\frac{2 ( 1.2 )}{1.2 - 1} = \frac{2.4}{0.2} = 12$ . Why this step? This factor comes from turning chamber enthalpy into kinetic energy (Nozzle Isentropic Expansion); it is the "how much energy per degree" lever.

Step 2 — specific gas energy $\frac{R T _{c}}{M}$ . $\frac{8.314 \times 3000}{0.025} = 9.977 \times 1 0^{5} J/kg$ . Why this step? $\frac{R}{M}$ is the gas constant per kilogram; times $T_{c}$ it is the thermal energy each kilogram of hot gas carries.

Step 3 — expansion bracket. $\frac{γ - 1}{γ} = \frac{0.2}{1.2} = 0.1667$ ; $(0.01)^{0.1667} = 0.4642$ ; bracket $= 1 - 0.4642 = 0.5358$ . Why this step? The bracket is the fraction of thermal energy actually converted to motion by expanding from $p_{c}$ down to $p_{e}$ . Small $p_{e} / p_{c}$ → bigger fraction.

Step 4 — exhaust velocity. $v_{e} = 12 \times 9.977 \times 1 0^{5} \times 0.5358 = 6.416 \times 1 0^{6} = 2533 m/s$ . Why this step? We multiply the three levers (energy conversion × thermal energy × extracted fraction) and take the square root because energy $\propto v^{2}$ .

Step 5 — divide by $g_{0} = 9.81$ . $I_{s p} = 2533/9.81 = 258 s$ . Why this step? $g_{0}$ is a bookkeeping constant that converts m/s into the "seconds" definition of $I_{s p}$ .

Verify: Units: $J/kg = m^{2} / s^{2} = m/s$ ✓. Value $258 s$ sits in the expected $200$ – $260$ band ✓. No grain quantity ( $ρ_{p}, A_{b}, r$ ) appeared — consistent with the parent's key insight ✓.

Example 2 — Cell B: doubling grain surface

Worked example

Same propellant as Ex 1 (so $I_{s p} = 258 s$ , effective exhaust speed $c = g_{0} I_{s p}$ ). The grain has $ρ_{p} = 1800 kg/m^{3}$ , burning area $A_{b} = 0.5 m^{2}$ , burn rate $r = 0.008 m/s$ . (a) Find thrust $F$ . (b) Now double $A_{b}$ to $1.0 m^{2}$ . What happens to $F$ and to $I_{s p}$ ?

Forecast: Doubling the burning wall — does the rocket get more efficient, or just more powerful?

Step 1 — mass flow. $\overset{m}{˙} = ρ_{p} A_{b} r = 1800 \times 0.5 \times 0.008 = 7.2 kg/s$ . Why this step? Volume burned per second is area × recession speed; times density gives kg/s. This is the only place geometry enters.

Step 2 — effective exhaust speed. $c = g_{0} I_{s p} = 9.81 \times 258 = 2531 m/s$ . Why this step? By definition $I_{s p} = c / g_{0}$ , so we invert it to recover $c$ .

Step 3 — thrust. $F = \overset{m}{˙} c = 7.2 \times 2531 = 1.822 \times 1 0^{4} N \approx 18200 N$ . Why this step? Thrust = mass flow × the speed each kilogram leaves at.

Step 4 — double $A_{b}$ . $\overset{m}{˙}^{'} = 1800 \times 1.0 \times 0.008 = 14.4 kg/s$ , so $F^{'} = 14.4 \times 2531 = 3.645 \times 1 0^{4} N \approx 36400 N$ . $I_{s p}$ still $= F^{'} / (\overset{m}{˙}^{'} g_{0}) = 36450/ (14.4 \times 9.81) = 258 s$ — unchanged. Why this step? In $I_{s p} = F / (\overset{m}{˙} g_{0})$ both $F$ and $\overset{m}{˙}$ scale with $A_{b}$ , so it cancels.

Verify: $F$ doubled ( $18200 \to 36400$ ) ✓; $I_{s p}$ identical ✓. This directly refutes Mistake ① "bigger surface → higher $I_{s p}$ ."

Example 3 — Cell C: mismatched (sea-level, over-expanded)

Worked example

Take the Ex 2 motor ( $\overset{m}{˙} = 7.2 kg/s$ , $v_{e} = 2533 m/s$ ) but now the nozzle exit is over-expanded: exit pressure $p_{e} = 0.04 MPa$ while ambient $p_{a} = 0.101 MPa$ (sea level). Exit area $A_{e} = 0.09 m^{2}$ . Find thrust and the actual $I_{s p}$ .

Forecast: With $p_{e} < p_{a}$ , is the pressure term a bonus or a penalty?

Step 1 — momentum thrust. $F_{mom} = \overset{m}{˙} v_{e} = 7.2 \times 2533 = 1.824 \times 1 0^{4} N$ . Why this step? The first thrust term is always the raw momentum of ejected gas.

Step 2 — pressure thrust. $(p_{e} - p_{a}) A_{e} = (40000 - 101000) \times 0.09 = (- 61000) \times 0.09 = - 5490 N$ . Why this step? When exit pressure is below ambient, the atmosphere pushes back on the exit plane — this term is genuinely negative.

Step 3 — total thrust. $F = 18240 - 5490 = 12750 N$ . Why this step? Full thrust equation $F = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ ; we may NOT drop the second term here (Mistake ④).

Step 4 — actual $I_{s p}$ . $I_{s p} = F / (\overset{m}{˙} g_{0}) = 12750/ (7.2 \times 9.81) = 180.5 s$ . Why this step? $I_{s p}$ always uses the real thrust, so the pressure penalty drags it below the ideal $258 s$ .

Verify: Over-expanded ( $p_{e} < p_{a}$ ) gives a penalty ✓; actual $180.5 s < 258 s$ ideal ✓. Sign of $(p_{e} - p_{a})$ is negative as required ✓.

Example 4 — Cells D & E: the two limits of the expansion bracket

Figure — Solid rocket Isp derivation from grain properties

Worked example

Using Ex 1's chemistry ( $\frac{2 γ}{γ - 1} \frac{R T _{c}}{M} = 12 \times 9.977 \times 1 0^{5} = 1.197 \times 1 0^{7}$ ), evaluate the extreme cases of the bracket $B = 1 - (p_{e} / p_{c})^{(γ - 1) / γ}$ . (D) Vacuum limit $p_{e} / p_{c} \to 0$ . (E) No-expansion limit $p_{e} / p_{c} \to 1$ .

Forecast: One limit gives the biggest possible exhaust speed; the other gives zero thrust. Which is which?

Reading the figure: the horizontal axis is the pressure ratio $p_{e} / p_{c}$ running from $0$ (left) to $1$ (right); the vertical axis is $I_{s p}$ in seconds. The yellow curve is $I_{s p} = 1.197 \times 1 0^{7} B / g_{0}$ . Follow it left-to-right: it starts high (vacuum), passes through the pink dot (Ex 1, $p_{e} / p_{c} = 0.01$ , $258 s$ ), and falls to the origin at the right edge.

Step 1 — Cell D bracket. As $p_{e} / p_{c} \to 0$ , $(0)^{0.1667} \to 0$ , so $B \to 1$ . Why this step? Expanding to zero pressure extracts all available thermal energy — the physical ceiling.

Step 2 — Cell D max speed. $v_{e, m a x} = 1.197 \times 1 0^{7} \times 1 = 3460 m/s$ ; $I_{s p, m a x} = 3460/9.81 = 353 s$ . Why this step? This is the theoretical best this chemistry could ever deliver — it is the blue dot at the top-left where the yellow curve flattens against the left axis.

Step 3 — Cell E bracket. As $p_{e} / p_{c} \to 1$ , $(1)^{0.1667} = 1$ , so $B \to 0$ . Why this step? If the gas exits at chamber pressure it never accelerated — no energy was converted.

Step 4 — Cell E speed. $v_{e} = 1.197 \times 1 0^{7} \times 0 = 0 m/s$ , so $I_{s p} = 0 s$ . Why this step? No pressure drop → no push. This is the degenerate case — the off-white dot sitting exactly at the origin (right edge) of the figure.

Verify: $B$ ranges monotonically from $0$ (no expansion) to $1$ (vacuum) ✓; $I_{s p}$ from $0$ up to $353 s$ ✓; the Ex 1 case ( $p_{e} / p_{c} = 0.01$ , $258 s$ ) lies between the two limits ✓.

Example 5 — Cell F: halving molar mass

Worked example

Keep everything from Ex 1 fixed but switch to a lighter-molecule exhaust: $M$ halved from $0.025$ to $0.0125 kg/mol$ . Predict the new $I_{s p}$ without redoing the whole calculation, then confirm.

Forecast: The Forecast-then-Verify block in the parent claims a factor $2$ . True?

Step 1 — isolate the $M$ dependence. In $v_{e} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M} B$ , only $M$ changes; everything else is a constant $k$ . So $v_{e} = k / M$ , i.e. $v_{e} \propto M^{- 1/2}$ . Why this step? We factor out the untouched terms so the scaling law stands alone.

Step 2 — apply the scaling. Halving $M$ multiplies $1/ M$ by $2$ , so $v_{e}$ (and hence $I_{s p}$ ) multiplies by $2 = 1.414$ . Why this step? Because $v_{e}$ depends on the square root of $1/ M$ , a factor-2 change becomes a factor- $2$ change.

Step 3 — new value. $I_{s p, new} = 258 \times 1.414 = 365 s$ . Why this step? Scale the known baseline instead of recomputing — faster and less error-prone.

Verify: Direct recompute: $\frac{R T _{c}}{M} = \frac{8.314 \times 3000}{0.0125} = 1.995 \times 1 0^{6}$ ; $v_{e} = 12 \times 1.995 \times 1 0^{6} \times 0.5358 = 3582 m/s$ ; $I_{s p} = 3582/9.81 = 365 s$ ✓. Matches the scaling prediction ✓. (This is why $H_{2}$ / $H_{2} O$ -rich exhausts are prized.)

Example 6 — Cell G: equilibrium chamber pressure and the $n \to 1$ cliff

Worked example

A grain has $a = 5 \times 1 0^{- 5}$ (SI), $ρ_{p} = 1800 kg/m^{3}$ , characteristic velocity $c^{*} = 1500 m/s$ , area ratio $A_{b} / A_{t} = 200$ . Find the equilibrium chamber pressure $p_{c}$ for (a) $n = 0.35$ and (b) contrast with $n = 0.9$ . Then explain why $n \geq 1$ is forbidden.

Forecast: Same grain, higher exponent — does $p_{c}$ rise a little or explode?

Reading the figure: the horizontal axis is the pressure exponent $n$ (from $0$ up toward $1$ ); the vertical axis is the solution exponent $\frac{1}{1 - n}$ . The yellow curve climbs gently at first, then rockets upward as $n \to 1$ ; the pink dashed line marks the forbidden wall at $n = 1$ where the curve shoots to infinity. The blue dot at $n = 0.35$ marks our stable case.

Step 1 — the base quantity. base $= ρ_{p} a c^{*} (A_{b} / A_{t}) = 1800 \times 5 \times 1 0^{- 5} \times 1500 \times 200 = 2.7 \times 1 0^{4}$ . Why this step? Setting mass generated ( $ρ_{p} A_{b} r$ ) equal to mass vented ( $p_{c} A_{t} / c^{*}$ ) with $r = a p_{c}^{n}$ leaves this base term raised to a power (Characteristic Velocity c-star).

Step 2 — exponent for $n = 0.35$ . $\frac{1}{1 - n} = \frac{1}{0.65} = 1.538$ — this is the height of the blue dot in the figure. Why this step? Solving $p_{c} = (base) p_{c}^{n}$ for $p_{c}$ gives $p_{c} = (base)^{1/ (1 - n)}$ .

Step 3 — evaluate. $p_{c} = (2.7 \times 1 0^{4})^{1.538} = 6.7 \times 1 0^{6} Pa = 6.7 MPa$ . Why this step? This is the pressure where gas made = gas vented — a stable balance point.

Step 4 — the danger of large $n$ . For $n = 0.9$ : $\frac{1}{1 - 0.9} = 10$ , so $p_{c} = (2.7 \times 1 0^{4})^{10} \approx 2.06 \times 1 0^{44} Pa$ — physically absurd. Why this step? The exponent $1/ (1 - n)$ diverges as $n \to 1$ (the vertical pink wall in the figure). For $n \geq 1$ a pressure bump raises burn rate faster than the throat can vent → runaway. Stable motors need $n < 1$ .

Verify: $(2.7 \times 1 0^{4})^{1.538} \approx 6.7 MPa$ ✓ (a plausible solid-motor chamber pressure). The $n = 0.9$ pressure is nonphysically huge, confirming the instability argument (Mistake ③) ✓.

Example 7 — Cell H: real-world word problem (burn time)

Worked example

A cylindrical end-burning grain (burns only on one flat face) has web thickness (length to burn through) $w = 0.60 m$ and burns at constant $r = 0.008 m/s$ . The face area is $A_{b} = 0.5 m^{2}$ , $ρ_{p} = 1800 kg/m^{3}$ , and the motor delivers $I_{s p} = 258 s$ . Find (a) burn time, (b) total propellant mass, (c) total impulse.

Forecast: With a constant thrust of $\sim 18 kN$ (from Ex 2), roughly how many seconds does it fire, and does total impulse depend on $I_{s p}$ or on total mass?

Step 1 — burn time. The face recedes at $r$ , so it takes $t_{b} = w / r = 0.60/0.008 = 75 s$ . Why this step? Burn time is web thickness divided by recession speed — pure kinematics.

Step 2 — total propellant mass. Volume $= A_{b} \times w = 0.5 \times 0.60 = 0.30 m^{3}$ ; $m_{p} = ρ_{p} \times 0.30 = 1800 \times 0.30 = 540 kg$ . Why this step? All the propellant is the face area swept through the whole web.

Step 3 — total impulse. $J = I_{s p} g_{0} m_{p} = 258 \times 9.81 \times 540 = 1.367 \times 1 0^{6} N\cdots$ . Why this step? From the definition $I_{s p} = J / (g_{0} m_{p})$ , total impulse is $I_{s p} g_{0}$ times mass burned.

Verify (cross-check via thrust×time): $\overset{m}{˙} = ρ_{p} A_{b} r = 7.2 kg/s$ ; $F = \overset{m}{˙} g_{0} I_{s p} = 7.2 \times 9.81 \times 258 = 18220 N$ ; $J = F t_{b} = 18220 \times 75 = 1.367 \times 1 0^{6} N\cdots$ ✓. Also $\overset{m}{˙} \times t_{b} = 7.2 \times 75 = 540 kg = m_{p}$ ✓. Two independent routes agree.

Example 8 — Cell I: exam twist (altitude, sign flip)

Worked example

A motor is designed so $p_{e} = 0.101 MPa$ (matched at sea level), $A_{e} = 0.09 m^{2}$ , $\overset{m}{˙} = 7.2 kg/s$ , $v_{e} = 2533 m/s$ . Compute the actual $I_{s p}$ (a) at sea level $p_{a} = 0.101 MPa$ and (b) at high altitude $p_{a} = 0.010 MPa$ . Which is larger and why?

Forecast: As the rocket climbs and air thins, does $I_{s p}$ go up or down?

Step 1 — sea level, matched. $(p_{e} - p_{a}) = 0$ , so $F = \overset{m}{˙} v_{e} = 7.2 \times 2533 = 18240 N$ ; $I_{s p} = 18 240/ (7.2 \times 9.81) = 258 s$ . Why this step? Matched nozzle kills the pressure term — this is the clean ideal case.

Step 2 — altitude, under-expanded. Now $p_{e} > p_{a}$ , so $(p_{e} - p_{a}) A_{e} = (101000 - 10000) \times 0.09 = 91000 \times 0.09 = + 8190 N$ — a bonus. Why this step? When exit pressure exceeds ambient, the exit gas is still pushing outward against thinner air; the sign is now positive (opposite of Ex 3).

Step 3 — thrust and $I_{s p}$ at altitude. $F = 18240 + 8190 = 26430 N$ ; $I_{s p} = 26 430/ (7.2 \times 9.81) = 374 s$ . Why this step? Higher thrust for the same mass flow → higher $I_{s p}$ .

Verify: Altitude $I_{s p}$ ( $374 s$ ) $>$ sea-level ( $258 s$ ) ✓ — this is why rockets are more efficient in the thin upper atmosphere/space. The pressure-term sign flipped from $-$ (Ex 3, over-expanded) to $+$ (here, under-expanded), exactly the exam trap Mistake ④ warns about ✓.

Recall Which cell am I in? (quick decision tree)

Is the nozzle matched ( $p_{e} = p_{a}$ )? ::: If yes → Cell A, use $I_{s p} = v_{e} / g_{0}$ . If no → keep $(p_{e} - p_{a}) A_{e}$ : negative if $p_{e} < p_{a}$ (over-expanded, Ex 3), positive if $p_{e} > p_{a}$ (under-expanded, Ex 8). They changed the grain ( $A_{b}, ρ_{p}, r$ ) but asked for $I_{s p}$ ? ::: Trap — $I_{s p}$ is unchanged (Cell B). Only thrust and burn time move. They pushed a ratio to an extreme ( $p_{e} / p_{c} \to 0$ or $\to 1$ )? ::: Cell D/E limits: bracket $\to 1$ (max $I_{s p}$ ) or $\to 0$ (zero thrust). They gave $a, n, c^{*}, A_{b} / A_{t}$ ? ::: Cell G — equilibrium pressure $p_{c} = (ρ_{p} a c^{*} A_{b} / A_{t})^{1/ (1 - n)}$ ; watch the $n \to 1$ blow-up.