Nozzle Area Ratio and Mach Number

A subsonic flow speeds up when the duct narrows — that is the garden hose, and everyone's intuition is built on it. A SUPERSONIC flow does the opposite: it speeds up when the duct widens. The reason is density. Above Mach 1 the gas thins out faster than the area grows, so the flow has to accelerate to carry the same mass through, and the arithmetic of that trade is this relation. It is why a rocket nozzle is a bell and not a hole, and why the narrowest point — the throat, where the flow passes exactly through Mach 1 — sits in the middle rather than at the end.

At $\gamma = 1.4$ the exponent $(\gamma+1)/(2(\gamma-1))$ collapses to exactly 3 and the relation becomes hand-workable. At Mach 2 the bracket is $(5/6)(1.8) = 1.5$, so $A/A^{*} = 1.5^{3}/2 = 1.6875$. At Mach 3 it is $(5/6)(2.8) = 7/3$, giving $(7/3)^{3}/3 = 343/81 = 4.235$. Rocket exhaust runs nearer $\gamma = 1.2$, where the same Mach numbers demand much larger bells: Mach 5 at $\gamma = 1.2$ needs an area ratio of 116.

There is no closed-form inverse, and there are two answers. Every area ratio above 1 is produced by two different Mach numbers, one subsonic and one supersonic, and both are physically real — they are not spurious roots to be discarded. An area ratio of 1.6875 at $\gamma = 1.4$ occurs at Mach 0.3722 and again at Mach 2, and in a real converging-diverging nozzle BOTH stations exist: the subsonic one upstream of the throat, the supersonic one downstream of it. Which root applies at a given station is decided by which side of the throat you are standing on, and which root the nozzle EXIT runs at is decided by the pressure ratio across the nozzle, not by the geometry alone. The solver here returns the supersonic root by convention, because that is the one an exit is normally asked about; for the subsonic station, work from the pressure ratio instead.

Two conditions have to hold before any of this means anything. The relation is ISENTROPIC — no shock, no friction, no heat loss — so it describes a well-behaved nozzle and stops describing one the moment a normal shock stands inside the diverging section, because the flow behind that shock is subsonic and obeys none of it. And the throat must actually be CHOKED. $A^{*}$ is defined as the area where $M = 1$, and if the pressure ratio across the nozzle is too small the throat never reaches Mach 1, there is no $A^{*}$, and the ratio has nothing to measure against. For $\gamma = 1.4$ choking needs a pressure ratio of about 1.893 across the throat; for $\gamma = 1.2$, about 1.772.