Rayleigh Flow Calculator
Heat addition in a constant-area duct — both flows it can choke, together, at any γ.
T₀/T₀* is the heat added, and it is what the two flows share — the same role the duct length plays in Fanno flow. Both are always shown, so there is no branch to choose.
Results
| Quantity | Subsonic heated up to sonic | Supersonic heated down to sonic |
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What this is computing
Frictionless flow down a constant-area duct with heat crossing the wall. Momentum is conserved because there is no friction, so p + ρV² is constant along the duct — and that single constraint, with mass conservation, is what produces every relation below.
T₀/T₀* is the heat added, non-dimensionalised, and it is what the two flows on this page share — the role duct length plays in Fanno flow. It rises to 1 at M = 1 from both sides, so heating always drives the flow towards sonic and cooling always drives it away.
Adding heat can cool the gas
T/T* peaks at M = 1/√γ = 0.845 in air, below sonic, at a value of 1.0286. Between that point and choking you are adding heat and the static temperature is falling: the flow is accelerating fast enough that kinetic energy is taking more than the heat supplies.
The one value that can mean two things
That peak makes T/T* the only quantity here that a column does not always pin down. Supersonically it falls straight from 1 to 0, so a value has one answer; subsonically anything at or below 1 is reached once, on the way up. Only between 1 and 1.0286 is it met twice — climbing to the peak and coming back — and the page says so, naming both Mach numbers, rather than picking one quietly.
Thermal choking, and the supersonic floor
T₀/T₀* = 1 is all the heat the duct will take. Add more and the flow does not cross sonic: the inlet backs up and the mass flow falls. Supersonically there is a floor as well as a ceiling — T₀/T₀* cannot fall below 0.4898 in air, so a supersonic combustor has a minimum heat release before the solution exists at all.