Exercises — Solid rocket Isp derivation from grain properties

• Left grey box (geometry): $A_b$, $A_t$, and the burn law $r=a{p}_c^n$ feed into $p_c$. Follow the blue arrows.
• Blue arrows carry the throughput path: $p_c\to r\to \dot{m}\to F$. Everything the grain touches.
• Green box (chemistry): $\gamma ,T_c,M,p_e/p_c$ feed $v_e$ along the green arrow.
• $F$ (orange) is where blue and green meet: $F=\dot{m}{v}_e$ (matched).
• Red arrow to $I_{sp}$ comes only from the green $v_e$ — no blue arrow reaches it. That is the whole "grain cancels" story in one picture.

Walk the formulas in order (and trace them on the s01 map).

• $\dot{m}={\rho }_p{A}_{b}r$: $A_b$ appears linearly, so doubling $A_b$ doubles $\dot{m}$.
• $F=\dot{m}{v}_e$ (matched nozzle): $\dot{m}$ doubled, $v_e$ unchanged (chemistry fixed) → $F$ doubles.
• $I_{sp}=F/(\dot{m}{g}_0)$: numerator and denominator both doubled → unchanged. Answer: $\dot{m}$ and $F$ double; $I_{sp}$ is untouched. Geometry is a power knob, not an efficiency knob — exactly the "no blue arrow reaches $I_{sp}$" of the map.

Straight substitution into $\dot{m}={\rho }_p{A}_{b}r$: $\dot{m}=1750\times 0.40\times 0.006=4.2\text{kg/s}.$ Units check: $\frac{\text{kg}}{{\text{m}}^3}\cdot {\text{m}}^2\cdot \frac{\text{m}}{\text{s}}=\frac{\text{kg}}{\text{s}}$. ✓ Answer: $4.2\text{kg/s}$.

Saint-Robert: $r=a{p}_c^n$. $p_c^{0.3}=(5\times 10^6{)}^{0.3}.$ Take logs: $\ln (5\times 10^6)=15.42$; times $0.3=4.627$; $e^{4.627}=102.2$. $r=4\times 10^{-5}\times 102.2=4.09\times 10^{-3}\text{m/s}\approx 4.1\text{mm/s}.$ Answer: $r\approx 0.0041\text{m/s}$.

Step 1 — prefactor $\frac{2\gamma }{\gamma -1}=\frac{2.5}{0.25}=10$. Why: this is the enthalpy-to-KE conversion factor. Step 2 — gas energy scale $\frac{R{T}_c}{M}=\frac{8.314\times 3200}{0.022}=1.209\times 10^6\text{J/kg}$. Step 3 — the expansion bracket exponent $\frac{\gamma -1}{\gamma }=\frac{0.25}{1.25}=0.2$; $(0.008{)}^{0.2}$: $\ln 0.008=-4.828$, times $0.2=-0.9657$, $e^{-0.9657}=0.3808$. Bracket $=1-0.3808=0.6192$. Step 4 — exhaust velocity $v_e=\sqrt{10\times 1.209\times 10^6\times 0.6192}=\sqrt{7.487\times 10^6}=2736\text{m/s}$. Step 5 — to seconds $I_{sp}=2736/9.81=278.9\mathbf{s}$. Answer: $I_{sp}\approx 279\text{s}$. Not a single grain number entered — exactly as the parent's key insight predicts.

Step 1 — mass flow $\dot{m}=1800\times 0.75\times 0.007=9.45\text{kg/s}$. Step 2 — effective exhaust velocity $c=g_0{I}_{sp}=9.81\times 279=2737\text{m/s}$ (equals $v_e$ for a matched nozzle, as expected). Step 3 — thrust $F=\dot{m}c=9.45\times 2737=2.586\times 10^4\text{N}\approx 25.9\text{kN}$. Answer: $\dot{m}=9.45\text{kg/s}$, $F\approx 25.9\text{kN}$.

Step 1 — the base ${\rho }_{p}a{c}^{\ast }(A_b/A_t)=1800\times 3\times 10^{-5}\times 1550\times 250$. $1800\times 3\times 10^{-5}=0.054$; $\times 1550=83.7$; $\times 250=2.0925\times 10^4$. Step 2 — exponent $\frac{1}{1-n}=\frac{1}{0.6}=1.6667$. Step 3 — raise to power $\ln (2.0925\times 10^4)=9.949$; $\times 1.6667=16.581$; $e^{16.581}=1.588\times 10^7\text{Pa}$. Answer: $p_c\approx 15.9\text{MPa}$.

Isolate the $M$-dependence. Inside the root, only $\frac{R{T}_c}{M}$ carries $M$; everything else (prefactor, bracket) is fixed. So $v_e\propto \sqrt{\frac{1}{M}}⟹I_{sp}\propto \frac{1}{\sqrt{M}}.$ Halving $M$ multiplies $I_{sp}$ by $\sqrt{2}=1.414$. Answer: $I_{sp}$ rises by a factor $\sqrt{2}\approx 1.41$ (a $+41\%$ gain). This is the reason hydrogen-rich exhaust is prized: light molecules move faster for the same energy.

Full thrust: $F=\dot{m}{v}_e+(p_e-p_a)A_e$. In vacuum $p_a=0$: Step 1 — momentum thrust $\dot{m}{v}_e=9.45\times 2737=25,865\text{N}$. Step 2 — pressure thrust $(p_e-0)A_e=1.0\times 10^5\times 0.20=20,000\text{N}$. Step 3 — total $F=45,865\text{N}$. Step 4 — vacuum $I_{sp}$ $=\frac{F}{\dot{m}{g}_0}=\frac{45,865}{9.45\times 9.81}=494.7\text{s}$. Compare to matched (sea-level) $I_{sp}=\frac{25,865}{9.45\times 9.81}=278.9\text{s}$. Answer: the pressure term adds $\approx 216\text{s}$ in vacuum ($279\to 495\text{s}$). This is why $I_{sp}$ is always quoted with an altitude/vacuum tag.

Reading the axes: the horizontal axis is the burn-rate exponent $n$ (dimensionless, $0$ to just under $1$); the vertical axis is the equilibrium chamber pressure $p_c$ in MPa. The red curve is $p_c=(\text{base}{)}^{1/(1-n)}$ — the base is fixed, only the exponent $\frac{1}{1-n}$ changes as we slide $n$.

• Three coloured dots mark $n=0.3$ (green), $0.7$ (orange), $0.9$ (red). Read their exponents off the labels: $1.43$, $3.33$, $10$. As $n$ climbs, $p_c$ shoots up super-linearly — the dots climb far faster than $n$ does.
• Grey dashed vertical line at $n=1$: the curve races toward it and goes vertical — the exponent $\frac{1}{1-n}\to \infty$. An infinitesimal change in the base then sends $p_c$ to infinity. Physically: a small pressure bump raises $r=a{p}_c^n$, which makes more gas, which raises pressure again — the throat can no longer vent fast enough → runaway burst.
• Blue shaded "stable design zone" on the left: here the curve is nearly flat, the loop settles, $p_c$ is well-behaved. This is where real motors live.
• The $n=0$ edge (green marker on the far left): exponent $\frac{1}{1-0}=1$, so $p_c$ equals the base itself — but more importantly $r=a{p}_c^0=a$, a constant burn rate independent of chamber pressure. The controller loop is broken open: pressure no longer feeds back into burn rate at all. Such a "plateau" propellant is maximally stable (no runaway possible) but gives up the self-regulating flexibility that a mild $n$ provides. Answer: below $n=1$ the pressure loop self-settles; at $n=1$ it diverges (explosion); at $n=0$ the loop is fully decoupled — $r$ is flat and pressure-blind. Safe designs sit in the shaded left zone, $0\le n<1$.

Follow the s01 map left-to-right. Step 1 — chamber pressure (from L2.3, same numbers): $p_c=1.588\times 10^7\text{Pa}=15.9\text{MPa}$. Step 2 — burn rate $r=a{p}_c^n=3\times 10^{-5}\times (1.588\times 10^7{)}^{0.4}$. $\ln (1.588\times 10^7)=16.581$; $\times 0.4=6.632$; $e^{6.632}=759.6$. $r=3\times 10^{-5}\times 759.6=0.02279\text{m/s}$. Step 3 — mass flow $\dot{m}={\rho }_p{A}_{b}r=1800\times 1.2\times 0.02279=49.2\text{kg/s}$. Step 4 — specific impulse (chemistry identical to L2.1): $I_{sp}=278.9\text{s}$, so $c=g_0{I}_{sp}=2736\text{m/s}$. Step 5 — thrust $F=\dot{m}c=49.2\times 2736=1.346\times 10^5\text{N}\approx 135\text{kN}$. Answers: $p_c\approx 15.9\text{MPa}$, $r\approx 0.0228\text{m/s}$, $\dot{m}\approx 49.2\text{kg/s}$, $I_{sp}\approx 279\text{s}$, $F\approx 135\text{kN}$. Notice the causal chain: grain+throat set $p_c$ → $p_c$ sets $r$ → $r$ sets $\dot{m}$ → chemistry sets $I_{sp}$ → the two combine into $F$.

Left side (gas made): ${\rho }_p{A}_{b}r=\dot{m}=49.2\text{kg/s}$ (from L4.1 Step 3). Right side (gas vented): $\frac{p_c{A}_t}{c^{\ast }}=\frac{1.588\times 10^7\times 4.8\times 10^{-3}}{1550}=\frac{7.622\times 10^4}{1550}=49.2\text{kg/s}$. Both sides equal $49.2\text{kg/s}$. ✓ The operating point is self-consistent — the grain manufactures gas at exactly the rate the throat can expel it, which is what "equilibrium pressure" means.

Shrinking $A_t$ raises $p_c$ (from $p_c\propto (A_b/A_t{)}^{1/(1-n)}$), which raises $r$, which raises $\dot{m}$ — so $A_t$ certainly lives in $\dot{m}$ and $F$. Does it survive into $I_{sp}$? $I_{sp}=\frac{v_e}{g_0},v_e=\sqrt{\frac{2\gamma }{\gamma -1}\frac{R{T}_c}{M}\left[1-{\left(\frac{p_e}{p_c}\right)}^{\frac{\gamma -1}{\gamma }}\right]}.$ The only place $A_t$ could enter $v_e$ is through the ratio $p_e/p_c$. Now the key fact from Nozzle Isentropic Expansion: for isentropic flow the exit-to-throat area ratio $A_e/A_t$ fixes the pressure ratio $p_e/p_c$ through the one-to-one isentropic area–Mach–pressure relation $\frac{A_e}{A_t}=\frac{1}{M_e}{\left[\frac{2}{\gamma +1}\left(1+\frac{\gamma -1}{2}{M}_e^2\right)\right]}^{\frac{\gamma +1}{2(\gamma -1)}},\frac{p_e}{p_c}={\left(1+\frac{\gamma -1}{2}{M}_e^2\right)}^{-\frac{\gamma }{\gamma -1}},$ where $M_e$ is the exit Mach number. Read these together: pin the geometric ratio $A_e/A_t$ and $M_e$ is fixed, which pins $p_e/p_c$ — regardless of the absolute value of $p_c$. Both chamber and exit pressure scale up together when $p_c$ rises, so their ratio is frozen by geometry alone. With $p_e/p_c$ held constant, $v_e$ carries no dependence on the absolute $A_t$. Hence $I_{sp}$ is independent of $A_t$. Answer: $A_t$ scales the throughput ($\dot{m}$, $F$) but cancels out of efficiency ($I_{sp}$) — the same 80/20 split as $A_b$. The reason is that only the ratio $p_e/p_c$ (set by the fixed area ratio $A_e/A_t$), not the absolute pressure, reaches $v_e$.

As $p_e/p_c\to 0$, the bracket $\left[1-(p_e/p_c{)}^{(\gamma -1)/\gamma }\right]\to 1-0=1$ (its ceiling). So $v_{e,\max }=\sqrt{\frac{2\gamma }{\gamma -1}\frac{R{T}_c}{M}}=\sqrt{10\times 1.209\times 10^6\times 1}=\sqrt{1.209\times 10^7}=3477\text{m/s}.$ $I_{sp,\max }=3477/9.81=354.4\text{s}.$ Answer: ceiling $v_e\approx 3477\text{m/s}$, $I_{sp}\approx 354\text{s}$. Physically this is all the chamber enthalpy converted to kinetic energy — you cannot beat it without hotter or lighter gas. A real finite nozzle only recovers the fraction $0.62$ of it (L2.1), giving $279\text{s}$.

• Mass flow: $\dot{m}={\rho }_p{A}_{b}r\to 0$ — the grain stops making gas.
• Chamber pressure: $p_c=({\rho }_{p}a{c}^{\ast }{A}_b/A_t{)}^{1/(1-n)}\to 0$ as $A_b\to 0$ (base $\to 0$, positive power) — the chamber depressurises (tail-off).
• Specific impulse: $I_{sp}=v_e/g_0$. But $v_e$ depends only on $\gamma ,T_c,M,p_e/p_c$. As $p_c\to 0$ the ratio $p_e/p_c\to \infty$, the bracket $\to 1-(\text{large}{)}^{+}<0$ — the formula breaks because expansion is no longer physical (the nozzle can't expand from a dead chamber). In the valid regime, before that, $v_e$ and thus $I_{sp}$ stay near their design value while $F=\dot{m}{v}_e\to 0$. Answer: thrust and $\dot{m}$ fade smoothly to zero at burnout; $I_{sp}$ holds its chemistry-set value until the chamber pressure collapses below the point where the nozzle can still expand. Efficiency dies last, and abruptly, not gradually — the classic thrust "tail-off."

With the bracket held constant, $v_e\propto \sqrt{T_c/M}$.

• Raise $T_c$ by $20\%$: factor $\sqrt{1.20}=1.0954$ → $+9.54\%$.
• Lower $M$ by $20\%$ (i.e. $M\to 0.8M$): factor $\sqrt{1/0.8}=\sqrt{1.25}=1.1180$ → $+11.80\%$. Answer: lowering $M$ by $20\%$ wins ($+11.8\%$ vs $+9.5\%$). The asymmetry: $T_c$ and $M$ enter as $\sqrt{T_c/M}$, but a $20\%$ reduction in a denominator ($\times 1/0.8$) is a bigger multiplier than a $20\%$ increase in a numerator ($\times 1.2$). Light exhaust beats hot exhaust, dollar for dollar — and it's easier on the chamber walls too.

The causal chain, one line ::: geometry $(A_b,A_t)$ + burn law $\to p_c\to r\to \dot{m}\to$ (with chemistry $v_e$) $\to F$; while $I_{sp}=v_e/g_0$ is chemistry-only. Why $A_b$ and $A_t$ cancel from $I_{sp}$ ::: both scale $\dot{m}$ and $F$ together, and $I_{sp}=F/(\dot{m}{g}_0)$ divides them out; only the ratio $p_e/p_c$ (fixed by area ratio $A_e/A_t$) reaches $v_e$. Ceiling on $I_{sp}$ for fixed chemistry ::: $v_{e,\max }=\sqrt{\frac{2\gamma }{\gamma -1}\frac{R{T}_c}{M}}$ at $p_e/p_c\to 0$ (full enthalpy conversion). What happens at $n=0$? ::: burn rate $r=a$ becomes pressure-independent (plateau propellant); the pressure feedback loop is decoupled — maximally stable, no runaway. Better lever: $+20\%$ $T_c$ or $-20\%$ $M$? ::: $-20\%M$ ($\times 1.118$) beats $+20\%T_c$ ($\times 1.095$).