A similar question is:

What is the highest operational ceiling for an air-breathing jet engine?

The answers point to scramjets. Cited reports focus on combustion kinetics.

The Wikipedia Scramjet page:

a block of gas entering the combustion chamber must mix with fuel and have sufficient time for initiation and reaction, all the while traveling supersonically through the combustion chamber, before the burned gas is expanded through the thrust nozzle. This places stringent requirements on the pressure and temperature of the flow, and requires that the fuel injection and mixing be extremely efficient.

For a nuke, no combustion is needed. How does this change the altitude limit?

The document cited by mins 3 concludes that speed and altitude are limited by reactor wall temperature for a nuclear-powered, direct air, shield-less, ram-jet. Twall of 1800 Rankine enables flight at "70,000 feet and a flight Mach number of 3.0, a uranium investment of 81 pounds and a missile gross weight of 64,000 pounds." It goes on to say "lift-drag ratios as low as 3.5 could be tolerated. ... lift-to-drag ratio was estimated at about 5.0."

The nuclear-powered ramjet combines the conventional diffuser and exhaust nozzle with a nuclear reactor. The diffuser decelerates the free-stream air before it enters the reactor passages. Reactor inlet air Mach number is about 0.28.

The question then becomes, for a nuclear scramjet, where inlet air is not braked to subsonic speeds, what improvement in ceiling is attainable?

The conventional fuel-burning scramjet would be limited by the oxygen available at altitude. The nuke needs only atmospheric mass. There is plenty of air at 81,000 feet to provide the needed thrust, according to my amateur calcuations:

W = 64,000 lbf, L/D = 4, let L = W so W/D = 4

drag D = W/4 = 64,000/4 = 16,000 lbf

We need thrust T = D

T = mdot*(u - v) = (mass flow rate)(exhaust velocity - intake velocity)

For a nuclear thermal rocket Isp = 900 s = u/g,

so u = (900 s)(9.8 m/s2) = 8,820 m/s

let v = Mach 15 = 15(343 m/s) = 5,145 m/s

mdot = (atm density)(intake area)(velocity) = rhoAv

at 25,000 m = 81,000 ft, density rho = 0.04 kg/m3

let intake diameter = 3 m, so A about 7 m2

mdot = (0.04 kg/m3)(7 m2)(5,145 m/s) = 1,441 kg/s

T = mdot(u - v) = (1,441 kg/s)(8,820 - 5,145) m/s = 5.3 MN

T= (5,300,000 N)(0.2248 lbf/N) = 1,190,000 lbf

  • 1
    $\begingroup$ Determining the ceiling requires knowing not just the type of engine employed, but also knowing about the weight and aerodynamic characteristics of the aircraft. Without such information, there is no way to give any answer. VTC as needing far more detail about the hypothetical aircraft in order to be answerable. $\endgroup$
    – Ralph J
    Aug 7, 2021 at 0:46
  • 1
    $\begingroup$ @Ralph J: The related question and its answers do not include the aero or weight of the aircraft. $\endgroup$
    – DrBunny
    Aug 7, 2021 at 1:47
  • 1
    $\begingroup$ You may be interested in this document for missiles. The problem doesn't seem to be altitude, as for a payload of about 5 tons, it was possible to design a missile flying at 80,000 ft at Mach 3. The less air, the more the required air acceleration, and the hotter temperature for the heater. $\endgroup$
    – mins
    Aug 7, 2021 at 11:13
  • $\begingroup$ The related question is also asking about existing aircraft - what is the highest ceiling of any existing aircraft. Since nuclear-powered aircraft have never flown, there is no empirical answer to their ceiling. Any answer here would either be speculation, or else it would reflect whatever assumptions are made regarding weight, thrust, and aerodynamics. If one assumes, or "fantasizes" might be a better description, a high enough thrust and a low enough weight and good enough aerodynamics, you can get amazing output... but it's all about data in, data out; or garbage in, garbage out. $\endgroup$
    – Ralph J
    Aug 7, 2021 at 12:26
  • 3
    $\begingroup$ I voted to reopen. The question is well formulated and focused. A theoretical maximum limit could be calculated based on pure engine performance, no other data needed (equivalent to saying that a piston engine cannot operate above X altidude because too low air density or whatever) $\endgroup$
    – Federico
    Aug 9, 2021 at 8:28


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