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How much thermal energy, measured in joules per centimeter squared, would be required to:

  • melt (or ignite, if possible) commonly-used commercial airliner paint

  • melt (or ignite, if possible) commonly used aerospace composites

The reason I'm asking:

I'm working on writing a nuclear-jet powered aircraft in a sci-fi setting. It uses an indirect air cycle molten-salt-cooled reactor to heat air to provide thrust without using fuel (other than the occasional reactor overhaul/nuclear fuel change-out). Radiator panels are built into its skin. It's intended to be a subsonic, long-haul cargo vehicle - not some kind of fancy supersonic jet.

I'm writing a scene where it's flying away from a nuclear explosion (at do-not-exceed speed; its radiators are overheating already), but doesn't get far enough away to avoid its radiators exploding/leaking coolant and its paint melting due to the thermal effects. How powerful would those thermal effects need to be? I imagine that radiators filled with molten salt deforming would require a lot more heat per square centimeter than paint melting.

I asked for "common aerospace materials" rather than "cutting edge" because these things are intended to be mass-produced in a relatively futuristic setting.

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  • $\begingroup$ radiators won't be painted or made of aircraft composite. you want as little insulation on radiators as possible. $\endgroup$
    – John
    Commented Jul 24, 2021 at 2:49
  • $\begingroup$ @John I meant the paint as a separate thing getting melted, and used composites as a benchmark. I'm not sure how durable radiator panels would be in comparison. $\endgroup$
    – KEY_ABRADE
    Commented Jul 24, 2021 at 2:51
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    $\begingroup$ better question why does your aircraft have exposed radiators, that is extremely rare, modern aircraft use forced air scoops and internal radiators since it is far more efficient, $\endgroup$
    – John
    Commented Jul 24, 2021 at 2:55
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    $\begingroup$ this paper should be helpful apps.dtic.mil/sti/citations/AD0362112 EFFECTS OF NUCLEAR EXPLOSIONS ON FIGHTER AIRCRAFT COMPONENTS $\endgroup$
    – John
    Commented Jul 24, 2021 at 2:56
  • $\begingroup$ @John I figured that it wouldn't have enough surface area to compensate for cooling the nuclear fission reactor that provides engine heating. IIRC, such a system hasn't actually been tested in flight (although an aircraft with such a system was flown without it activated). Also, thank you for the paper - that'll be immeasurably valuable. $\endgroup$
    – KEY_ABRADE
    Commented Jul 24, 2021 at 3:54

1 Answer 1

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How are airplanes painted? enamel and epoxy.

It's difficult to find any data on the auto-ignition point for enamel, but for epoxy it's 300C.

Carbon composite will burn at 300-500C.

For comparison, wood ignites at 380C. So answers about when wood will spontaneously ignite will also answer the question of when aerospace materials will spontaneously ignite.

Based on this:

The flash burning of the surface of objects, particularly wooden objects, occurred in Hiroshima up to a radius of 9,500 feet from X; at Nagasaki burns were visible up to 11,000 feet from X.

Nukemap listed the "thermal radiation radius" for 3rd degree burns as 1.96km (6266ft) for Hiroshima, and 2.21km (7255 feet) for Nagasaki. 9500ft is 51% greater than 6266ft, and 11000 ft is 51% greater than 7255ft.

So: take the distance Nukemap lists for 3rd degree skin burns, and multiply it by 1.5, and you will get the maximum distance wooden or aerospace materials might receive surface burns.

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  • $\begingroup$ Brilliant answer - thank you. I'm going to point out that NUKEMAP-2 actually has a setting for when dry wood will burst into flames - 35 small calories per cm^2, or about 146.44 joules/cm^2. I'll use that as a conservative estimate for ignition of aerospace composites, while the 3rd-degree-burn radius * 1.5 is the most "out-there" estimate. $\endgroup$
    – KEY_ABRADE
    Commented Jul 24, 2021 at 19:08
  • $\begingroup$ @KEY_ABRADE 9500/6266 = 1.51611, and 11000 / 7255 = 1.51619. These ratios are so exact - especially when compared with the imprecision of 9500ft and 11000ft - that I suspect Nukemap used those figures of 9500ft and 11000ft as calibration. $\endgroup$
    – causative
    Commented Jul 24, 2021 at 19:53

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