A De Laval nozzle has a compression section, where the propellant is compressed (and thereby accelerated) as it moves towards a narrow section (the throat). After the throat, the nozzle widens out again. Here's a reference image.

According to what I've been reading, the propellant is moving at subsonic speeds before the throat, and is accelerated to supersonic speeds as it passes the throat. Furthermore, the propellant continues to accelerate as the nozzle expands again. What I don't understand is: what force continues to accelerate the propellant after it passes through the throat? I've been trying to imagine a single molecule of propellant and thinking about the forces from the particles around it, but can't come up with a force "pushing" from upstream (because it's moving at supersonic speeds), or "pulling" from downstream (because at the minimum there would be 0 opposing force in a vacuum, but certainly not a negative attractive force from downstream). Any ideas?


3 Answers 3


This question is related to gas-dynamics/ thermodynamics of compressible flows. The flow in the Laval nozzle is isentropic. This means that there is no irreversibility in the flow. If we consider this, the expansion in the nozzle causes a drop in the temperature and thus, resulting in a rise in the velocity of the fluid. But this happens only for a particular outlet pressure - inlet pressure ratio, 0.528. if the pressure ratio is more than this, the flow in the diffuser section (after the throat) decelerates. This is because of the Area-velocity relation (which is obtained from compressible continuity equation, and Laplace equation for velocity of sound), in terms of the Mach number. Its implications are:

  1. In subsonic flow regime: Convergent duct causes acceleration; divergent, deceleration.
  2. In supersonic regime: Convergent duct causes DECELERATION; AND DIVERGENT, ACCELERATION.

But this also is not applicable for all cases, as already mentioned, the pressure ratio must be below 0.528.

The Area-Mach number relation (which is obtained from compressible continuity equation, and the stagnation density relation to the mach number, again applicable for isentropic flows) gives the area required at the throat to "choke" the flow (i.e. to accelerate it to sonic speed at the throat) at a given pressure ratio.

The area-mach number relation gives a solution which has TWO VALUES for every area-ratio (area-ratio: ratio of the area of section under consideration to the area of the throat when flow is sonic at the throat. NOTE: flow at throat becomes sonic at throat only when Pe/P < 0.528). one value for subsonic; and the other for supersonic flows. So when the flow is supersonic, the flow accelerates; and decelerates otherwise.

So the answer to your question: Supersonic flow is accelerated by the expansion of the flow and reduction in temperature. In a nutshell, acceleration-deceleration with variation in area depends on 2 things: 1. Mach number, and 2. Pressure ratio.

If you need a more detailed explanation, I refer you to Gas Dynamics by E. Rathakrishnan for a more detailed explanation and the derivations of the area-mach number relations and so.

  • $\begingroup$ I don't think the pressure ratio can be lower than 0.528, as that represents Mach = 1 at the throat, which is the maximum velocity (and so minimum static pressure) possible. $\endgroup$
    – Aidenhjj
    Commented Apr 20, 2018 at 8:20

I guess you have to think in terms of relative velocity. Let's put ourselves in the smaller section of the nozzle.Here M=1, so the flow is travelling at sound speed. The strong depression force the air to expand, and this expansion gives to the flow additional velocity (relatively to the flow). This is not contradictory, because the 'information of additional velocity' isn't travelling faster than sound, we have just applied Galilean relativity!


Basic Bernoulli principle for a compressible flow can be stated as : $$ {\frac {v^{2}}{2}} + H={\text{const}} $$ where $H$ is specific enthalpy. Expressing enthalpy in terms of gas temperature, gives : $$ {\frac {v^{2}}{2}} + c_p \,T={\text{const}}$$ where $c_p$ is specific heat capacity. Now as gas goes down through the valve, it expands, while cooling down in temperature. This can be best seen by the temperature distribution of the rocket nozzle wall (red line) : enter image description here

As per equation, as gas cools down and it's enthalpy is decreasing - exhaust gas speed must increase. This explanation may not be the best or accurate, but should give the basic idea.


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