# Why should the velocity through the nozzle throat be sonic?

I have read (even on this site), that exhaust gas velocity is (normally?) sonic (exactly Mach 1) through the choke point of a De Laval, or convergent-divergent, nozzle:

At the "throat", where the cross-sectional area is at its minimum, the gas velocity locally becomes sonic (Mach number = 1.0), a condition called choked flow.

Why is that optimal/required?
What would be the effect of exhaust gas being supersonic through the throat?
And what about subsonic (eg Mach 0.9)?

• Possibly worth mentioning that Mach 1.0 in the rocket throat will not be 300m/s, it will be the speed of sound for the exhaust chemical mix at slightly absurd temperatures. Oct 8 '19 at 8:02
• Sonic flow will be acheived only if the pressure drop is big enough. For a small pressure the flow will stay sub sonic.
– Uwe
Oct 8 '19 at 9:32
• "Optimal/required" is probably the wrong question way to think about it. The point is that once the flow reaches Mach 1, there is no way for a pressure change to propagate upstream. Therefore, the inlet side of the nozzle can't "know" anything about what happens after the flow reaches Mach 1 at the throat of the nozzle, so nothing can change to allow more mass flow. That is why the flow is said to be "choked." Oct 8 '19 at 16:13
• @Bohemian pressure = (density)(R)(temperature), so that is an equivalent statement Oct 9 '19 at 23:42
• Probably too shallow for an answer, but because this is the level I think of it at and think it's useful for the novice: the purpose of a nozzle is to accelerate flow. Convergence can only accelerate flow to Mach 1. Divergence is required afterward. So the throat has to be at Mach 1 for a C-D nozzle to be doing its job. Oct 11 '19 at 2:11

Uwe's comment on the question is spot on. The characteristics of the flow through the nozzle depend critically on the pressure ratio - the two pressures being the pressure at the entrance to and exit from the nozzle.

Above the critical pressure ratio flow through the nozzle is subsonic and it is not choked at the throat. Below the critical pressure ratio, the flow chokes at the throat, reaching a maximum.

This figure from The Dynamics and Thermodynamics of Compressible Fluid Flow by Shapiro shows the variation in properties based on the pressure ratio. The most interesting feature of Fig. 4.3 is the maximum in the curve of flow per unit area...The pressure ratio p/p0 where the flow per unit area is a maximum is called the critical pressure ratio and has a value for all real gasses and vapors of approximately one-half.

Pressure ratios greater than the critical correspond to subsonic flow and pressure ratios less than the critical correspond to supersonic flow.

(p. 76)

The next graph from p. 87 shows the same parameters plotted against Mach number. You can see the knee in the area ratio curve at the critical pressure ratio well here. This corresponds to the throat of the nozzle. There's a good discussion of the phenomenon here

If you lower the back pressure enough you come to a place where the flow rate suddenly stops increasing all together and it doesn't matter how much lower you make the back pressure (even if you make it a vacuum) you can't get any more mass flow out of the nozzle. We say that the nozzle has become 'choked'. You could delay this behavior by making the nozzle throat bigger ... but eventually the same thing would happen. The nozzle will become choked even if you eliminated the throat altogether and just had a converging nozzle.

The reason for this behavior has to do with the way the flows behave at Mach 1, i.e. when the flow speed reaches the speed of sound. In a steady internal flow (like a nozzle) the Mach number can only reach 1 at a minimum in the cross-sectional area. When the nozzle isn't choked, the flow through it is entirely subsonic and, if you lower the back pressure a little, the flow goes faster and the flow rate increases. As you lower the back pressure further the flow speed at the throat eventually reaches the speed of sound (Mach 1). Any further lowering of the back pressure can't accelerate the flow through the nozzle any more, because that would entail moving the point where M=1 away from the throat where the area is a minimum, and so the flow gets stuck. The flow pattern downstream of the nozzle (in the diverging section and jet) can still change if you lower the back pressure further, but the mass flow rate is now fixed because the flow in the throat (and for that matter in the entire converging section) is now fixed too.

The changes in the flow pattern after the nozzle has become choked are not very important in our thought experiment because they don't change the mass flow rate.

The answer is much simply presented in the book by John D Anderson - Modern Compressible Flow with historical perspectives. Refer Chapter 5.

$$\begin{equation} \rho = \mbox{density} \end{equation}$$ a = speed of sound, Ma = Mach number, u = speed of fluid, A = area

Area velocity relationship can be deduced from the continuity relationship - $$\begin{equation} \frac{d\rho}{\rho} + \frac{du}{u} + \frac{dA}{A} = 0 \end{equation}$$

Watch now ... $$\begin{equation} \frac{dA}{A} = - \{\frac{d\rho}{\rho} + \frac{du}{u}\} \end{equation}$$

From momentum conservation relationships between two points, we can write $$\begin{equation} dp = -\rho udu \\ \implies \frac{dp}{\rho} = \frac{dp}{d\rho}\frac{d\rho}{\rho} = -udu \end{equation}$$

and

$$\begin{equation} \frac{dp}{d\rho} = a^2 \end{equation}$$

Therefore,

$$\begin{equation} \frac{d\rho}{\rho} = -\frac{udu}{a^2} = -Ma^2 \frac{du}{u} \end{equation}$$

From these relations, we can write

$$\begin{equation} \frac{dA}{A} = (Ma^2 - 1)\frac{du}{u} \end{equation}$$

Now check for Ma = 1, you will get

$$\begin{equation} \frac{dA}{A} = 0 \end{equation}$$

This means the area function is either at maxima or minima. The practical solution is that this is minimum. This minimum area is called throat of the nozzle.

• Welcome to Space! Not everyone believes that a mathematical derivation answers a "Why?" question completely, but I am sure many will still find your answer informative!
– uhoh
Oct 28 '20 at 8:36
• It isn't clear where your Ma term comes from after the "Therefore"
– user20636
Oct 28 '20 at 13:14
• u/a^2 = u^2/a^2 x 1/u = Ma^2 x 1/u
– ArKE
Oct 29 '20 at 4:10