Exercises — Nozzle thermodynamics — isentropic expansion from chamber to exit

$P_0,T_0$ ::: chamber (stagnation) pressure and temperature — the gas brought fully to rest, all energy thermal. $M$ ::: Mach number $=V/a$, the flow speed measured in units of the local sound speed $a=\sqrt{\gamma RT}$. $\gamma$ ::: heat capacity ratio $c_p/c_v$; for rocket exhaust we use $\gamma =1.2$. $c_p$ ::: specific heat at constant pressure — the energy needed to raise 1 kg of gas by 1 K while it is free to expand. For an ideal gas it is fixed by $\gamma$ and $R$: $c_p=\frac{\gamma R}{\gamma -1}$. $R$ ::: specific gas constant, $320\text{J/(kg⋅K)}$ for our exhaust; equals $R_u/\mathcal{M}$ with $R_u=8314\text{J/(kmol⋅K)}$ universal and $\mathcal{M}$ the molar mass. $g_0$ ::: standard gravity, $9.81{\text{m/s}}^2$ — the fixed constant that converts an exhaust velocity into seconds of specific impulse. $A^{\ast }$ ::: throat area, the narrowest cross-section, where $M=1$ exactly. starred quantities ($T^{\ast },P^{\ast }$) ::: values at the throat, i.e. evaluated at $M=1$.

At the throat, $M=1$ by definition, so the critical ratio applies directly: $T^{\ast }=T_0\cdot \frac{2}{\gamma +1}=3500\cdot \frac{2}{2.2}=3500\times 0.9091=\boxed{3182\text{K}}$ What we did: substituted $M=1$ into the temperature relation. Why: the throat is defined as the sonic point, so we don't need to guess $M$ — it is fixed.

$P^{\ast }=P_0{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma }{\gamma -1}}=10\times (0.9091{)}^6$ The exponent: $\frac{\gamma }{\gamma -1}=\frac{1.2}{0.2}=6$. $P^{\ast }=10\times 0.5645=\boxed{5.645\text{MPa}}$ Why the exponent is 6: pressure falls faster than temperature because it depends on both the density drop and the temperature drop of the expanding gas.

Exponent $=\frac{1.4}{0.4}=3.5$; base $=\frac{2}{2.4}=0.8333$. $\frac{P^{\ast }}{P_0}=(0.8333{)}^{3.5}=\boxed{0.5283}$ What this shows: the famous "0.528" rule for air — a converging nozzle chokes once the back pressure drops below ~53% of the chamber pressure.

First the bracket $1+\frac{\gamma -1}{2}{M}_e^2=1+0.1\times 9=1.9$. $T_e=\frac{3500}{1.9}=\boxed{1842\text{K}}$ $P_e=\frac{10}{1.9^6}=\frac{10}{47.05}=0.2125\text{MPa}\Rightarrow \boxed{212.5\text{kPa}}$ Now the local sound speed at the exit: $a_e=\sqrt{\gamma R{T}_e}=\sqrt{1.2\times 320\times 1842}=\sqrt{707,328}=841.6\text{m/s}$. $V_e=M_e{a}_e=3\times 841.6=\boxed{2525\text{m/s}}$ Why $T_e$ first: velocity needs the local sound speed, which needs the local temperature — so temperature is always the first domino.

Why this formula at all? The area-Mach relation is not a new law — it is what you get by writing mass conservation $\rho VA={\rho }^{\ast }{V}^{\ast }{A}^{\ast }$ (same gas passes every station) and then replacing every $\rho$, $V$ and $T$ by its isentropic value in terms of $M$. Momentum conservation enters through the isentropic $P/{\rho }^{\gamma }=\text{const}$ that those substitutions rely on. So the formula is just "constant mass flow, expressed in Mach." $\frac{A_e}{A^{\ast }}=\frac{1}{3}{\left[\frac{2}{2.2}\times 1.9\right]}^{\frac{\gamma +1}{2(\gamma -1)}}$ Exponent $=\frac{2.2}{0.4}=5.5$; bracket $=0.9091\times 1.9=1.7273$. $\frac{A_e}{A^{\ast }}=\frac{1}{3}(1.7273{)}^{5.5}=\frac{1}{3}\times 20.21=\boxed{6.74}$ What it means: the diverging cone must open to nearly 7× the throat area to let the gas reach Mach 3.

The nozzle geometry this describes — throat at $M=1$, exit opened to $A_e/A^{\ast }=6.74$ — is drawn below. Notice the throat (teal dashed) is the pinch point and the exit (plum dashed) is where the gas has reached Mach 3.

Mass flow is easiest to evaluate at the throat, where all conditions are fixed. Throat: $T^{\ast }=3182\text{K}$, $P^{\ast }=5.645\text{MPa}$. Density from ideal gas: ${\rho }^{\ast }=\frac{P^{\ast }}{R{T}^{\ast }}=\frac{5.645\times 10^6}{320\times 3182}=5.545{\text{kg/m}}^3$. Velocity $V^{\ast }=a^{\ast }=\sqrt{\gamma R{T}^{\ast }}=\sqrt{1.2\times 320\times 3182}=1105.7\text{m/s}$. $\dot{m}={\rho }^{\ast }{V}^{\ast }{A}^{\ast }=5.545\times 1105.7\times 0.01=\boxed{61.3\text{kg/s}}$ Why the throat: $\dot{m}=\rho VA$ is constant everywhere, but at the throat every factor is known without solving for $M$.

Compare exit pressure to ambient: $P_e=212.5\text{kPa}>P_a=101\text{kPa}$ Since the jet leaves at higher pressure than its surroundings, the gas is under-expanded — it still has pressure energy left to convert, so it keeps expanding after leaving the nozzle. At the lip this produces expansion fans (Prandtl–Meyer waves) that turn the flow outward. Contrast: if $P_e<P_a$ the jet would be over-expanded and would form oblique shocks squeezing it inward. Perfect expansion is $P_e=P_a$. The bar chart below makes the comparison concrete: the teal bar ($P_e$) stands taller than the plum ambient line ($P_a$), which is exactly the signature of under-expansion. See Nozzle Flow Regimes for the full family of cases.

For each, $T=T_0/(1+0.1M^2)$ then $V=M\sqrt{\gamma RT}$. A ($M=2$): bracket $=1.4$, $T=2500\text{K}$, $V=2\sqrt{1.2\times 320\times 2500}=2\times 979.8=1960\text{m/s}$. B ($M=4$): bracket $=2.6$, $T=1346\text{K}$, $V=4\sqrt{1.2\times 320\times 1346}=4\times 718.9=2876\text{m/s}$. B wins: $\frac{2876}{1960}=\boxed{1.47\times }$ faster. The subtlety: doubling Mach did not double velocity, because the sound speed $a$ drops as $T$ falls. Velocity is $M\times a$, and $a$ works against you at high $M$ — the reason ever-larger nozzles give diminishing returns.

The energy equation caps total energy: $c_{p}T+V^2/2=c_p{T}_0$. Velocity is largest when all thermal energy has become kinetic, i.e. $T\to 0$: $V_{\max }=\sqrt{2c_p{T}_0}=\sqrt{2\cdot \frac{\gamma R}{\gamma -1}{T}_0}=\sqrt{2\cdot \frac{1.2\times 320}{0.2}\times 3500}$ $V_{\max }=\sqrt{2\times 1920\times 3500}=\sqrt{1.344\times 10^7}=\boxed{3666\text{m/s}}$ As $T\to 0$, sound speed $a\to 0$, so $M=V/a\to \infty$. The maximum velocity is finite even though Mach diverges — a beautiful reminder that Mach number is not speed, it is speed relative to a shrinking yardstick.

Two unknowns ($M_e,T_e$) but two equations. Combine them: $V_e=M_e\sqrt{\gamma R{T}_e},T_e=\frac{T_0}{1+\frac{\gamma -1}{2}{M}_e^2}.$ Substitute the second into the first and square: $V_e^2=M_e^2\gamma R\frac{T_0}{1+0.1M_e^2}.$ Plug numbers ($V_e^2=8.649\times 10^6$, $\gamma R{T}_0=1.2\times 320\times 3500=1.344\times 10^6$). Let $x=M_e^2$: $8.649\times 10^6(1+0.1x)=1.344\times 10^{6}x\Rightarrow 8.649+0.8649x=1.344x\Rightarrow 8.649=0.4791x.$ $x=18.05\Rightarrow M_e=\sqrt{18.05}=\boxed{4.25}.$ Area ratio: bracket $=1+0.1(18.05)=2.805$; $\frac{A_e}{A^{\ast }}=\frac{1}{4.25}[0.9091\times 2.805{]}^{5.5}=\frac{1}{4.25}(2.550{)}^{5.5}$. $(2.550{)}^{5.5}=173.3$, so $\frac{A_e}{A^{\ast }}=\frac{173.3}{4.25}=\boxed{40.8}$. Lesson: a modest velocity increase demands a huge extra expansion ratio — nozzle bells grow explosively at high Mach.

Solve $\frac{A}{A^{\ast }}(M)=25$ with the area-Mach formula, $\gamma =1.2$, exponent $5.5$: $25=\frac{1}{M}{\left[0.9091(1+0.1M^2)\right]}^{5.5}.$ Try $M=3.60$: bracket $=0.9091(2.296)=2.087$; $2.087^{5.5}=62.8$; $/3.6=17.4$ — too low. Try $M=3.90$: bracket $=0.9091(2.521)=2.292$; $2.292^{5.5}=90.1$; $/3.9=23.1$ — still low. Try $M=3.98$: bracket $=0.9091(2.584)=2.349$; $2.349^{5.5}=99.5$; $/3.98=25.0$. ✔ $\boxed{M_e\approx 3.98}$ Why iterate: the area-Mach relation is not invertible in elementary functions — you sweep $M$ until the geometry matches. Real design tables (and Specific Impulse Optimization tools) do exactly this.

Specific impulse in seconds is $I_{sp}=V_e/g_0$ when the pressure term vanishes (perfect vacuum expansion, see Thrust Equation Derivation). Here $g_0=9.81{\text{m/s}}^2$ is the fixed standard gravity that turns a velocity into seconds. Maximum $V_e$ is the $T\to 0$ limit from Exercise 3.3, using $c_p=\frac{\gamma R}{\gamma -1}$: $V_{\max }=\sqrt{2c_p{T}_0}=\sqrt{2\cdot \frac{\gamma R}{\gamma -1}{T}_0}=3666\text{m/s}.$ $I_{sp,\max }=\frac{3666}{9.81}=\boxed{374\text{s}}.$ Why this is the ceiling: no nozzle, however long, can extract more than the total thermal enthalpy $c_p{T}_0$. This single number tells you whether a propellant is worth pursuing before you design any hardware. Real losses (Real Nozzle Losses) pull the achieved value to ~95% of this.

From $V_{\max }=\sqrt{2c_p{T}_0}$ with $c_p$ fixed: $V\propto \sqrt{T_0}$. Doubling $T_0$ multiplies $V$ by $\sqrt{2}=\boxed{1.41}$. Now write $R=R_u/\mathcal{M}$ where $R_u=8314\text{J/(kmol⋅K)}$ is universal and $\mathcal{M}$ is molar mass. Then $V_{\max }=\sqrt{\frac{2\gamma }{\gamma -1}\cdot \frac{R_u}{\mathcal{M}}{T}_0}\propto \sqrt{\frac{T_0}{\mathcal{M}}}.$ Velocity scales as $1/\sqrt{\mathcal{M}}$. Halving the molar mass gives the same $\sqrt{2}$ boost as doubling temperature — but halving $\mathcal{M}$ (e.g. burning hydrogen, $\mathcal{M}\approx 2$) is far easier and safer than doubling an already 3500 K flame that would melt the chamber. The design punchline: this is why hydrogen–oxygen engines dominate high-$I_{sp}$ upper stages — light exhaust beats hot exhaust. Verify: with $\mathcal{M}=10$ vs $20$ and same $T_0$, $V$ ratio $=\sqrt{20/10}=1.41$.

Thrust $F=\dot{m}{V}_e+(P_e-P_a)A_e$. As the cone lengthens, $A_e/A^{\ast }\uparrow$, $M_e\uparrow$, so $V_e\uparrow$ — but with diminishing returns since $V_e\to V_{\max }=3666\text{m/s}$ (a hard ceiling). Meanwhile $P_e\to 0$, and once $P_e<P_a$ the pressure term $(P_e-P_a)A_e$ goes negative — the atmosphere pushes back on the ever-larger exit. Numerically, going from $M=3.8$ ($A/A^{\ast }=25$) to $M=4.25$ ($A/A^{\ast }=41$) adds ~$65\%$ more area but only lifts $V_e$ from ~$2900$ to ~$3010\text{m/s}$, a mere $4\%$. So thrust rises toward an asymptote and the pressure term eventually subtracts: there is an optimum length, not infinite growth. $\boxed{\text{Refuted.}}$ The extra structural mass of the giant bell also cuts vehicle acceleration — the true optimisation lives in Specific Impulse Optimization.