Mach Architect#

A textbook plot of $A/A^*$ against $P_b/P_0$ does not leave the regime change in the hand.

Drag the blue throat point up or down and move the back-pressure slider for $P_b/P_0$. Exit Mach and the shock location move together.

Narrow the throat and watch whether the mass flow locks. Drop the back pressure below 0.5283 and watch the shock slide toward the exit.

Controls#

• Nozzle geometry: drag the blue control points on the SVG up and down to adjust cross-section
• Back pressure: the right-side slider sets $P_b/P_0$
• Level select: switch with the LV1–LV4 buttons at the top

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Level Walkthrough#

Level 1: Subsonic Acceleration#

The most basic case. In a converging nozzle, subsonic flow accelerates as the area shrinks. The isentropic relation:

$$\frac{A}{A^*} = \frac{1}{M}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$$

Since $M < 1$, decreasing $A$ increases $M$. The goal is exit Mach 0.45–0.55.

Level 2: Choking#

When the flow chokes, $M = 1$ at the throat. Lowering back pressure further leaves the upstream mass flow unchanged.

Critical pressure ratio ($\gamma = 1.4$):

$$\frac{P^*}{P_0} = \left(\frac{2}{\gamma+1}\right)^{\frac{\gamma}{\gamma-1}} = 0.5283$$

Drop back pressure below this value, or make the throat narrow enough.

Level 3: Supersonic Nozzle Design#

In a C-D nozzle, the flow accelerates supersonically once the area opens up beyond the throat. To reach $M = 2$:

$$\frac{A_e}{A^*} = \frac{1}{2}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}\cdot 4\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}} \approx 1.6875$$

Plug $M=2$ and $\gamma=1.4$ into the area relation and the printed 1.6875 falls out. $(\gamma-1)/2=0.2$ and $M^2=4$, so the parenthesis is $1+0.8=1.8$. Times $2/(\gamma+1)=1/1.2$ that is 1.5. The exponent $(\gamma+1)/(2(\gamma-1))$ is 3. $1.5^3=3.375$. Times $1/M=1/2$ gives 1.6875. A throat of 1 needs an exit of that multiple to hit $M=2$.

The subsonic target 0.45 to 0.55 is the other side of the same relation. At $M=0.45$, $A/A^*\approx1.45$. At $M=0.55$, $\approx1.25$. The band is narrow. A small throat move throws the exit Mach out of the window. The part where the duct must widen to speed up is the same relation as where a Laval nozzle flips.

Back pressure has to come down to the design value Pb_design before supersonic flow reaches the exit without a shock.

Level 4: Shock Positioning#

When the back pressure is above design but below critical, a normal shock forms inside the nozzle.

Pre/post-shock relations:

$$M_2^2 = \frac{M_1^2(\gamma-1)/2 + 1}{\gamma M_1^2 - (\gamma-1)/2}$$ $$\frac{P_2}{P_1} = \frac{2\gamma M_1^2 - (\gamma-1)}{\gamma+1}$$

Put $M_1=2$ into the shock relations and the fractions cancel cleanly. $M_2^2=(4\times0.2+1)/(1.4\times4-0.2)=1.8/5.4=1/3$, so $M_2\approx0.577$. The pressure ratio is $(2\times1.4\times4-0.4)/2.4=10.8/2.4=4.5$. Design for Mach 2 and then leave the back pressure above design, and somewhere in the duct the pressure jumps by 4.5 while Mach falls to about 0.58.

Raise the back pressure and the shock walks toward the throat. Lower it and the shock walks toward the exit. The level-4 target $x = 0.65 \sim 0.80$ means park that jump near the exit.

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The Simulator's Physics Model#

This simulator rests on these assumptions:

1. 1D quasi-1D flow: only area variation; viscosity/heat transfer ignored
2. Isentropic flow (except across shocks): $P_0$, $T_0$ conserved
3. Calorically perfect gas: $\gamma = 1.4$ (air)
4. Normal shocks: Rankine–Hugoniot relations
5. Nozzle geometry: smooth area distribution via Hermite interpolation

The shock position is found by discretely searching for the location that satisfies the exit pressure condition. Upstream of the shock is isentropic supersonic; downstream is isentropic subsonic with stagnation pressure loss applied.

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Leave the design back pressure and the M = 2 run dies in a shock#

Match the area ratio 1.6875 and then park the back pressure above design. The $M=2$ built in level 3 falls to $M_2\approx0.577$ inside the duct. Pressure jumps by 4.5. The moment back pressure sits between 0.5283 and the design value, the throat stays choked and the supersonic run ends at the shock.

Putting that jump on a mesh means giving it a thickness on purpose. That choice is the artificial viscosity that smears a shock on purpose. This simulator only searches for a discrete station. It does not resolve a shock thickness at all. While the back-pressure slider walks $x$ from 0.65 to 0.80, watch whether the pressure ratio leaves 4.5.