To Go Supersonic, Widen the Pipe — The De Laval Nozzle and Choking

Pinch the end of a garden hose and the water shoots out faster. Narrow it, speed it up — our intuition stops there. Yet a rocket nozzle looks exactly the opposite. Behind the throat it flares open like a trumpet. Once gas passes the speed of sound, it only accelerates further if the pipe gets wider. This post traces why that reversal happens through the area-Mach relation, then lets you sweep the back pressure and watch the seven stages that play out inside the nozzle.

Narrow to Speed Up, Then Flip Past Mach 1#

In compressible flow (a flow whose density changes), the link between area and velocity flips depending on the Mach number. The area-velocity relation for 1-D isentropic flow says it in one line.

\frac{dA}{A} = (M^2 - 1)\,\frac{dV}{V}

Here $A$ is the cross-sectional area, $V$ the velocity, and $M$ the Mach number (velocity over speed of sound). The sign is everything.

Below Mach 1 ( $M<1$ ), $M^2-1<0$ . To raise the velocity ( $dV>0$ ) you must shrink the area ( $dA<0$ ). That is the hose-pinching intuition.

Above Mach 1 ( $M>1$ ), $M^2-1>0$ . To go even faster you must enlarge the area. The sign flips.

So to push gas past the speed of sound you first narrow the duct to reach Mach 1, then widen it from there. This narrow-then-wide shape is the converging-diverging, or de Laval, nozzle. The narrowest point is the throat.

One Area, Two Mach Numbers#

The area-Mach relation ties cross-sectional area to Mach number in isentropic flow.

\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)}}

Here $A^*$ is the critical area where $M=1$ (the throat area), and $\gamma$ is the ratio of specific heats (constant-pressure over constant-volume; about 1.4 for air).

The right-hand side is a U-shaped curve with a minimum of 1 at $M=1$ . So any area ratio above 1 has two Mach solutions: one subsonic, one supersonic.

Drag the area ratio in the plot below.

area ratio A/A*: 3.0subsonic: M=0.197, p/p0=0.973  |  supersonic: M=2.637, p/p0=0.047

Raise the area ratio and the two roots pull apart. The subsonic root drifts toward zero while the supersonic one keeps climbing. For the same nozzle shape, which branch the flow follows is decided by the pressure at the exit — the back pressure.

Choking — When the Throat Can Take No More#

Lower the back pressure gradually and the mass flow through the nozzle grows. But not forever. The moment the throat velocity reaches the speed of sound ( $M=1$ ), it hits a wall.

The reason lies in how fast information travels. Pressure disturbances propagate upstream at the speed of sound. When the throat is sonic, a downstream pressure change can no longer climb back upstream. The flow upstream of the throat simply cannot "know" that the back pressure dropped further. The mass flow locks at its maximum. This state is called choking.

The maximum mass flow of a choked nozzle is set by the throat area $A^*$ and the stagnation conditions alone.

\dot{m} = A^* \, p_0 \sqrt{\frac{\gamma}{R T_0}} \left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}

Here $p_0$ is the stagnation pressure, $T_0$ the stagnation temperature, and $R$ the gas constant. No matter how much further the back pressure falls, this value will not change. That is why a rocket engine's thrust is designed through the throat size.

The Seven Stages as Back Pressure Falls#

The real drama starts after choking. Lower the back pressure $p_b$ from the stagnation pressure $p_0$ and the internal flow shifts stage by stage.

Try it in the simulation below. Drag the back-pressure slider from 1.0 down to 0.05 and watch the color inside the nozzle (Mach number) and the pressure curve underneath.

back pressure pb/p0: 0.85specific heat ratio gamma: 1.40

In order, here is what happens.

(a) Subsonic venturi. With the back pressure high enough, the whole flow stays subsonic. It peaks in speed at the throat, then slows again in the diverging section. Pressure bottoms out at the throat and nearly recovers at the exit.

(b) First critical point. Drop the back pressure further and the throat reaches exactly $M=1$ . Choking begins here.

(c)–(d) Shock inside the nozzle. Lower it more and the flow accelerates supersonically in the diverging section until it hits a normal shock — a discontinuity that snaps the flow abruptly back to subsonic. Behind the shock the flow is subsonic, so it decelerates and matches the back pressure. The lower the back pressure, the further downstream the shock is pushed.

(e) Overexpanded. Once the shock leaves the exit, the inside of the nozzle is fully supersonic. But the exit pressure is below the back pressure, so oblique shocks outside the nozzle reconcile the difference. This is the overexpanded regime.

(f) Design point. A single point where the exit pressure exactly equals the back pressure. No shocks inside or outside. It is the most efficient operating condition.

(g) Underexpanded. When the back pressure drops below the design value, the exiting gas keeps expanding through expansion waves outside the nozzle. This is the underexpanded regime.

As a rocket climbs and the ambient pressure (back pressure) falls, it passes through (e) → (f) → (g) in order. That is why a nozzle that is overexpanded at sea level passes through its design point and becomes underexpanded at high altitude.

Solving the Nozzle in Code#

This code finds both roots of the area-Mach relation by bisection and computes the design pressure from the exit area ratio.

import numpy as np
 
def area_ratio(mach, gamma=1.4):
    """isentropic area ratio A/A* as a function of Mach"""
    t = 1.0 + 0.5 * (gamma - 1.0) * mach**2
    expo = (gamma + 1.0) / (2.0 * (gamma - 1.0))
    return (1.0 / mach) * (2.0 / (gamma + 1.0) * t) ** expo
 
def invert_area_ratio(target, gamma=1.4, branch="sub"):
    """Mach root for a given A/A* (subsonic or supersonic branch)"""
    lo, hi = (1e-4, 1.0) if branch == "sub" else (1.0, 8.0)
    for _ in range(80):
        mid = 0.5 * (lo + hi)
        f = area_ratio(mid, gamma)
        # subsonic branch decreases, supersonic branch increases
        ascending = (branch == "sup")
        if (f < target) == ascending:
            lo = mid
        else:
            hi = mid
    return 0.5 * (lo + hi)
 
def pressure_ratio(mach, gamma=1.4):
    """static/stagnation pressure ratio p/p0"""
    return (1.0 + 0.5 * (gamma - 1.0) * mach**2) ** (-gamma / (gamma - 1.0))
 
# design condition of a nozzle with exit area ratio 4.0
ae_over_astar = 4.0
m_exit = invert_area_ratio(ae_over_astar, branch="sup")
p_design = pressure_ratio(m_exit)
 
print(f"design exit Mach   = {m_exit:.3f}")
print(f"design pe/p0       = {p_design:.4f}")
print(f"area-ratio check   = {area_ratio(m_exit):.3f}")

The output:

design exit Mach   = 2.940
design pe/p0       = 0.0298
area-ratio check   = 4.000

A nozzle with an area ratio of 4.0 accelerates the flow to Mach 2.94, where the exit pressure is about 3% of the stagnation pressure. In other words, this nozzle is at its design point when the back pressure sits near 3% of $p_0$ . Since the recovered area ratio matches the input 4.0, the two functions are confirmed to be exact inverses.

What to Remember#

Below the speed of sound, narrowing speeds the flow up; above it, you must widen to go faster. Hence the converging-diverging shape.
When the throat reaches $M=1$ , the nozzle chokes. Dropping the back pressure further will not raise the mass flow.
The same nozzle moves through venturi, internal shock, overexpanded, design, and underexpanded regimes depending on back pressure. The single point where exit pressure equals back pressure is the design point.