Lesson 2 of 14

Nozzle Theory

How a bell of metal turns 3500 K gas at rest into a 3 km/s jet: the ideal rocket, why supersonic flow needs a widening duct, choking, shocks, over- and under-expansion, and the two numbers that separate the chamber from the nozzle — c* and C_F.

8 min3 live simulations
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The question

Every garden hose teaches us that squeezing a flow makes it faster. So why does every rocket nozzle, after squeezing the gas through a narrow throat, open up into a huge bell?

By the end of this lesson you should be able to explain the bell's shape from one curve — and predict what the flow does when the air around it changes.

The ideal rocket

Sutton begins nozzle theory with a deliberately simple model, the ideal rocket: the gas is a perfect gas of fixed composition; the flow is steady, one-dimensional and adiabatic; there is no friction and no heat loss to the walls. It sounds crude, but real engines land within 1–6% of it. That makes it the right place to build intuition — and the right yardstick for the losses real engines suffer.

In the chamber the gas is hot, dense, and almost at rest: its energy is all enthalpy. The nozzle's job is to turn that enthalpy into directed kinetic energy. Energy conservation along the flow gives the exhaust velocity directly:

ve=2γγ−1 R TcM[1−(pepc)(γ−1)/γ]\cvel{v_e} = \sqrt{\frac{2\gamma}{\gamma - 1}\,\frac{R\,\cflame{T_c}}{\mathfrak{M}}\left[1 - \left(\frac{\cpress{p_e}}{\cpress{p_c}}\right)^{(\gamma-1)/\gamma}\right]}

Read it slowly, because it is the most important equation in chemical rocketry. The exhaust velocity rises with the chamber temperature Tc\cflame{T_c}, falls with the molecular mass M\mathfrak M of the gas, and rises as the nozzle expands the gas to a lower exit pressure pe\cpress{p_e}. Lesson 4 is about the first two. This lesson is about the third.

Why wider goes faster

Squeeze a subsonic flow and it speeds up: the garden-hose intuition. But a gas is compressible. As it speeds up it expands and cools, and its density drops. Which effect wins?

DeriveWhy does a nozzle get wider to go faster?
SUBSONIC: narrow to speed upSUPERSONIC: widen to speed up0123400.511.522.5Mach numberrelative to sonic / chamberdensity ρ/ρ₀: 0.92speed v/a*: 0.42mass flux ρv (per m²): 0.62
throatM = 0.40 · A/A* = 1.62
0.40
Challenges0/2
  • Find the Mach number where the mass flux per unit area is greatest.
  • Find the Mach number whose duct is ten times the throat area.
Push gas faster and two things happen: its speed rises and, because it expands and cools, its density falls. Below Mach 1 the speed wins, so the same mass fits through less area; above Mach 1 the density loss wins, so the gas needs more room. The mass flux ρv — and hence 1/A — peaks at exactly Mach 1.

Below Mach 1, speed rises faster than density falls, so each square metre carries more mass: the duct must narrow. Above Mach 1 it is the other way round, and the duct must widen to keep accelerating the gas. In between, at exactly Mach 1, the mass flux per unit area reaches its maximum. That is the ThroatThe narrowest section of a nozzle. In a working rocket the flow there is exactly sonic (Mach 1), which fixes the mass flow.. Written as a differential relation (Sutton eq. 3–20 in spirit):

dAA=(M2−1)dvv\frac{dA}{A} = \left(M^2 - 1\right)\frac{dv}{v}

For M<1M < 1, area and velocity change in opposite directions; for M>1M > 1, in the same direction. A converging–diverging nozzle — de Laval's nozzle, invented for steam turbines in the 1880s — is simply the shape this equation demands.

Estimate first
How fast does the gas pass through the throat of a kerosene rocket engine, where it is at exactly Mach 1?
101001,00010,000100,000
Drag along the line to place your guess — each tick is ten times the last.

Choked: the throat sets the flow

Once the throat reaches Mach 1, something remarkable happens. Pressure signals travel at the speed of sound, and at the throat the gas is moving downstream exactly that fast, so no information from downstream can get upstream past it. The chamber can no longer "feel" what happens outside. The mass flow is fixed by the chamber and the throat alone:

m˙=pc Atc∗\cfuel{\dot m} = \frac{\cpress{p_c}\,A_t}{c^*}

This is Choked flowOnce the throat reaches Mach 1, lowering the downstream pressure no longer increases the mass flow: information cannot travel upstream faster than sound., and it is why rocket engineers talk about chamber pressure and throat area as if they were the engine. The constant of proportionality c∗c^* — the Characteristic velocity c*p_c A_t/ṁ: a figure of merit of the combustion alone (the propellant and chamber), independent of the nozzle. ≈ √(RT_c/M)/Γ. — depends only on the combustion gas, not on the nozzle.

ExperimentThe nozzle lab: turn down the back pressure
back pressure p_b = 0.991 p₀p₀throatM=0.60exit M = 0.12
chamberthroatexit00.250.50.751position along the nozzlepressure p/p₀subsonic limitshock at exitdesignp_b
0.01%0.1%1%10%100%00.51pressure drop (p₀ − p_b)/p₀mass flow / chokedchokedlowering p_b changes nothing upstream
0.991
4.0
A venturi: the gas speeds up to the throat and slows down again, never reaching Mach 1. Lower the back pressure and more gas flows.
Challenges0/3
  • Find the back pressure at which the nozzle first chokes.
  • Park the normal shock right at the nozzle exit.
  • See all six regimes (0/6).
A converging–diverging nozzle between a high-pressure chamber (left) and a room whose pressure you control (right). The colour of the gas is its Mach number — blue subsonic, white sonic, orange supersonic. Lower the back pressure and walk through every regime a rocket nozzle can be in.

The lab walks through every regime a converging–diverging nozzle can be in. Watch the mass-flow plot on the right: it rises as you lower the back pressure, and then, the instant the throat chokes, it stops. Everything that happens afterwards — the normal shock sliding down the nozzle, being pushed out, the plume necking in and then ballooning out — happens downstream of the throat, without the chamber knowing.

Check your intuition
A rocket engine is running with its nozzle perfectly expanded at 10 km altitude. It climbs to 30 km. What happens to the mass flow of propellant through it?

Separating the chamber from the nozzle

The thrust can be split neatly into a product of two figures of merit:

F=CF  pc AtF = C_F\;\cpress{p_c}\,A_t

The Thrust coefficient C_FF/(p_c A_t): how much the nozzle amplifies the thrust that chamber pressure alone would give on the throat area. Typically 1.3–2. CFC_F measures how much the nozzle amplifies the thrust that the chamber pressure alone would give over the throat area. It depends only on γ\gamma, the area ratio and the pressures:

CF=2γ2γ−1(2γ+1)γ+1γ−1[1−(pepc)γ−1γ]+pe−papc AeAtC_F = \sqrt{\frac{2\gamma^2}{\gamma-1}\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{\gamma-1}}\left[1-\left(\frac{\cpress{p_e}}{\cpress{p_c}}\right)^{\frac{\gamma-1}{\gamma}}\right]} + \frac{\cpress{p_e}-\cair{p_a}}{\cpress{p_c}}\,\frac{A_e}{A_t}

It is typically between 1.3 and 2.0. Combine the two equations and the exhaust velocity factorises beautifully:

c=CF×c∗\cvel{c} = C_F \times c^*

c∗c^* is the chamber's grade; CFC_F is the nozzle's. Test engineers use this every day: measure chamber pressure and mass flow and you get c∗c^* — if it is 97% of theory, the injector is mixing well. Measure thrust too and you get CFC_F — if that is low, look at the nozzle.

Going deeper
Where the 0.56 critical pressure ratio comes from

At the throat M=1M = 1, so the isentropic relation p/p0=(1+γ−12M2)−γ/(γ−1)p/p_0 = (1 + \frac{\gamma - 1}{2}M^2)^{-\gamma/(\gamma-1)} gives the critical pressure ratio

p∗pc=(2γ+1)γ/(γ−1).\frac{p^*}{p_c} = \left(\frac{2}{\gamma+1}\right)^{\gamma/(\gamma-1)}.

For rocket gases with γ≈1.2\gamma \approx 1.2 that is 0.56: the pressure has already fallen by almost half at the throat, and the gas has cooled by about 9% (T∗/Tc=2/(γ+1)=0.91T^*/T_c = 2/(\gamma+1) = 0.91). All the rest of the expansion, from 0.56 pcp_c down to perhaps 0.001 pcp_c, happens in the diverging bell — which is why the bell is so much bigger than the converging section.

One bell, a whole atmosphere

A nozzle is designed for one exit pressure, but a launch vehicle climbs through the whole atmosphere. At sea level a large bell is Over-expandedExit pressure below ambient: the atmosphere squeezes the jet, shocks form at the lip, and the pressure term of thrust turns negative.: the gas leaves below ambient pressure, the atmosphere squeezes it back through oblique shocks at the lip, and the pressure-thrust term is negative. High up the same bell is Under-expandedExit pressure above ambient: the jet keeps expanding outside the nozzle, where it can no longer push on the walls.: the gas leaves above ambient pressure and keeps expanding outside, where it can no longer push on anything.

FlyOne bell for a whole atmosphere
0204060801.21.41.61.82altitude (km)thrust coefficient C_Falways matched (aerospike ideal)your bell, ε = 16ε = 5ε = 40ε = 150
pₑ/pₐ = 0.67over-expanded: plume pinched
16
0.0 km
Thrust coefficient1.635100% of ideal
Isp (c* = 1800 m/s)300 s
Sea level → vacuum+10%thrust gain of this bell
Challenges0/3
  • Find the largest bell that does not separate at sea level.
  • Reach C_F above 1.95 in vacuum.
  • Launch with the aerospike and compare it with your bell.
The thrust coefficient C_F = F/(p_c A_t) of bells with different area ratios as the vehicle climbs (p_c = 100 bar, γ = 1.2). Small bells are best low down, big bells best in vacuum; no fixed bell is best everywhere. The pink envelope is a nozzle that is always perfectly expanded — the promise of the aerospike.

Push the bell too far at sea level and the pressure near the wall drops so low that the boundary layer cannot climb back up against the atmosphere: it Flow separationIn a strongly over-expanded nozzle the boundary layer cannot climb the adverse pressure and peels off the wall, causing side loads.. A common rule, Summerfield's criterion, puts this at exit pressures below roughly a third of ambient. Separation is not just inefficient: it is usually asymmetric and unsteady, and the side loads it produces have broken nozzles and gimbal actuators in testing.

This is why first-stage engines have modest bells (area ratios of 15–40) and upper-stage engines enormous ones (80–280): the RL10B-2 carries a nozzle extension that unfolds only after staging.

Real nozzles

The ideal rocket is an idealization. Sutton lists the ways real nozzles fall short, each costing a percent or two:

  • Divergence loss. The flow leaves a conical nozzle along the wall angle, not straight back. For a cone of half-angle α\alpha the thrust is multiplied by λ=12(1+cos⁡α)\lambda = \tfrac12(1 + \cos\alpha): 0.983 for the classic 15° cone. Bell (contoured) nozzles turn the flow back towards the axis, recovering most of this in a shorter length.
  • Boundary layers on the walls slow the gas and reduce the effective throat area.
  • Chemistry lag. The gas may not recombine fast enough as it cools (Lesson 4).
  • Particles. Aluminium oxide droplets in solid-motor exhaust lag the gas (Lesson 9).

Together these typically cost 2–6% of ideal Isp, and every engine is characterized by a measured correction factor.

Recap
What to carry forward
  • The ideal rocket (perfect gas, 1-D, isentropic) predicts real engines within a few percent.
  • vev_e grows with Tc/M\sqrt{T_c/\mathfrak M} and with the pressure ratio the nozzle expands through.
  • Subsonic flow speeds up in a converging duct, supersonic in a diverging one. Mass flux per area peaks at Mach 1: the throat.
  • The throat chokes: m˙=pcAt/c∗\dot m = p_c A_t/c^*, independent of what happens downstream.
  • F=CF pcAtF = C_F\,p_c A_t and c=CFc∗c = C_F c^*: c∗c^* grades the chamber, CFC_F the nozzle.
  • Nozzles are over-expanded low down (and may separate) and under-expanded high up. Area ratio is a compromise over the flight.

Next: once the gas leaves the nozzle, what does its velocity buy? The rocket equation, gravity losses, and why every rocket to orbit is mostly propellant.

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