Trinity and Jumbo

Leading up to the Atomic Test at Trinity, the scientists also constructed a steel bottle for the bomb.

You see, they weren’t totally sure the test would be successful, so they needed to create a robust container to contain the possible “fizzle” so the toxic, costly plutonium would not be scattered across the countryside. If the detonation was successful, the bottle would simply vaporize.

When the scientists were sure the test would work, they disregarded Jumbo and left it to be a side-show of sorts.
My question is, is it possible to create a steel bottle thick and robust enough to contain a Trinity/Hiroshima/Nagasaki-level detonation?

Sure, if you had enough money.
Underground detonations are contained - all you need is a bottle that strong.

Insanely great Username/OP combination! According to this thread [thread=713662]"What objects could survive a close-up nuclear explosion?[/thread], probably not. The fireball vaporizes anything close by - how close depends on the size of the nuclear explosion. To have a steel container the size of the bomb survive, you’d have to get a yield well below one kiloton TNT equivalent. To contain a 10-20 kiloton explosion my guess (based on the survival of the Trinity tower, as discussed in the other thread) is you’d probably need a container a several hundred feet in diameter. Even then, if it didn’t have a heat sink the “container” (more like a stadium) would melt.

Jumbo wasn’t designed to contain an atomic detonation. It merely had to contain the results of the exploding TNT that was used to trigger the fission. Containing a TNT explosion is many orders of magnitude easier.

That’s why no steel bottle could ever work for the actual fission bomb, even one as comparatively small as Trinity.

Now I’m sure that some Dopers will come in and nitpick that if you just used all the steel in the world and put the bomb at the center…

And ninjaed. :slight_smile:

I was thinking of a hollow steel sphere several hundred feet in radius (I said diameter above, didn’t I?) . The “shell” would still have to be very think - possibly feet thick. That’s a lot of steel - the darn thing is the size of a battle ship - but it’s not all the steel in the world.

A few feet thick is still a lot of steel. Say that you have a 10 foot thick ball of steel whose outer skin has a diameter of 500 feet. That comes to a rounded 700,000 tons of steel or 3.5 World Trade Centers.

Wouldn’t a large hollow sphere be less strong than a small one? The explosion would compress the air in front of it.

Up to a point. Obviously, if you could build a sphere large enough the compression wave would be meaningless. I think you’re violating the spirit of the question, though. A balloon would work if it were large enough. A containment bottle needs a lot more that a mere 10 feet of steel.

Here’s some points:

If you have the capacity to make a steel (sphere) bottle, you can evacuate the air out of it, making only blast pressure due to the vaporized bomb mass itself.

Due to the inverse square law, the greater, the distance from ground zero, the less thermal energy is imparted to the inner skin of the bottle.

Actually, due to the inverse square law, the same amount of thermal energy is imparted to the skin no matter how big you make it. You’d still need enough steel to absorb that amount of energy. What this means, I think, is that it wouldn’t make much of a difference how voluminous you made your jar: The smaller it was, the thicker you’d need to make the walls, and the two would largely cancel out.

Until the steel of the bottle itself starts to vaporize. Still, this could at least provide a basis for some theoretical energy calculations.

The halving thickness for gamma radiation passing through steel is very close to 1 inch (2.5cm), according to Wikipedia), so a mere foot of steel should absorb nearly all the gamma released–only about 0.02% would get through. Alpha and beta, of course, are absorbed even more readily. So, for the sake of discussion, we can assume that all of the energy released by the bomb is being absorbed by the steel.

Some of that energy would do fun things like make the steel radioactive, but that doesn’t immediately contribute to the breakdown of the bottle integrity. Let’s assume that all the energy is directed to heating the steel. Different steels have different thermal properties, and not all of them are fully characterized. Let’s say it’s a carbon steel with the following characteristics:

Specific heat: 0.49 kJ/(kg K)
Full melting point: 1540C
Heat of fusion: 272 kJ/kg* (Extra energy required to liquefy the material after the melting point is reached.)
Boiling temperature: 2870C*
Heat of vaporization: 6090 kJ/kg* (Extra energy required to vaporize the material after the boiling point is reached.)

*Values are actually for pure iron, as I was unable to find values specifically for steel.

A little weather research suggests that an air temperature of ~19C would not be out of line for the site and time of the Trinity test, though I didn’t find a record of the actual temperature on the day of the test. The sheer thermal mass of the bottle and the temperature extremes involved make a few degrees of initial temp pretty irrelevant, so let’s set T[sub]0[/sub] to 20C.

So, to melt a kilogram of the steel, we’ll need to raise its temperature by 1520C (or K), then add another 272 kJ. Adding the heat of fusion to the basic formula, we have

Q = cmdT + m*Q[sub]f[/sub]
Plugging in specific heat ©, mass (m), temperature delta (dT), and heat of fusion (Q[sub]f[/sub]), we get:
Q = 0.49 kJ/(kg K) * 1 kg * 1520K + 1 kg * 272 kJ/kg = 1016.8 kJ. For reasons of precision and convenience, 1000 kJ to melt each kilogram of steel.

The Trinity test released ~84 TJ (terajoules) of energy. This offers us a simple, if intimidating calculation:
84 TJ = 84,000,000,000 kJ.
84,000,000,000 kJ/1000 kJ/kg = 84,000,000 kg of steel. 184,800,000 pounds, or 92,400 tons.

That’s the amount that could be completely melted by the energy release, absent any sort of compression wave or vaporization. So, you need an ablative layer of that much steel on the inside of your bottle, with more outside to contain your new vat of molten metal. Over 90% of the energy will be absorbed by the first 10 cm of steel, so let’s say that’s how thick you want to make your ablative layer. The density of carbon steel is ~7.8 g/cm[sup]3[/sup], or 7800 kg/m[sup]3[/sup], so you need ~11000 m[sup]3[/sup] of steel in your ablative layer.

The volume of a spherical shell is (4/3)π(R[sup]3[/sup]-r[sup]3[/sup]), where R and r are the outer and inner radii, respectively. We also know that (R-r) is 10 cm, or 0.1 m. From that, we see that:

11000 m[sup]3[/sup] = (4/3)π((r+0.1)[sup]3[/sup]-r[sup]3[/sup])

Solving for r gives us an internal radius of 93.5 meters, or ~309 feet.

Now, let’s see how much energy the first halving thickness of the ablative shell absorbs, to see if we have vaporization issues. It’ll absorb ~42 TJ throughout its total volume, which is a shell of volume:
(4/3)π(93.525[sup]3[/sup]-93.5[sup]3[/sup]) = 2747 m[sup]3[/sup]

That’s roughly 21,000,000 kg of steel, so each kilogram would be soaking ~2000kJ of energy. The first 1000 kJ/kg goes to melting it. The remaining 1000 kJ would first go to raise the temperature of the molten layer to the boiling point, an increase of 1330C. That would use

Q = 0.49 kJ/(kg K) * 1 kg * 1330K = 650 kJ

The remaining 350 kJ would go toward heat of vaporization without actually raising the temperature further. Our heat of vaporization is 6090 kJ/kg, so 350 kJ could potentially vaporize up to about 5% of our molten steel from the inner surface of the ablative layer–over 1,000,000 kg of steel vapor bursting from the inner surface. Looks like we have a little concussive force, after all.

So, you have an evacuated 10 cm-thick steel sphere over 600 feet across with a bomb in the center of it. That’s just the part that would be completely destroyed by the bomb; it would melt through and superhot steel vapor would burst out, which, let’s face it, would be embarrassing.

Internal structure can’t be relied on to support it, since it will all melt, so it has to be held up by superstructure, i.e., lots more steel–in fact, just making the shell thick enough to support itself over that kind of span might be enough to complete containment. I’ll leave that part as an exercise for the reader.

Feel free to check my math and/or my reasoning. I am not a physicist, and I am specifically not your physicist. :smiley:

Okay, lets have a little fun with this. What if we instead coated the inner layer with space shuttle-style thermal tiles and/or super hi temp tungsten alloy? (assume the savings in mass…and money…are critical in building the bottle in the first place)

That raises the pressure of an evacuated 600+ diameter void to what exactly? I assume…if the bottle holds and radiated heat from the outer skin cools the bottle, the steel vapor will eventually condense and solidify.

Insulating it doesn’t help: the energy is in there, and it’s not going away. Sooner or later, it’s going to soak through. All you can do is provide enough thermal mass–like, say, the ground–for it to soak into, and enough thickness of material to shield the gamma.

No way am I going to recrunch all that for tungsten, but the melting point of ferrotungsten (your high-temp tungsten alloy) is only about 30% higher. Even if you somehow saved half the total mass, you definitely wouldn’t be saving money. Ferrotungsten is expensive, about $47/kg this month, as opposed to around $0.25/kg for steel. You’d be looking at ~$2 billion, just for the raw metal that’s going to get melted.

Haven’t the (steel) foggiest about the pressure, and ultimately, the vapor pressure inside the bottle isn’t the important part. In the instant the bomb goes off, all that steel is going to vaporize; effectively, every square centimeter of the inner surface of the bottle is suddenly going to have a jet pushing on it. I have no idea how to calculate how much force the vapor jets would apply, but it’s not something I’d want to ignore.

As to condensation: aside from the steel vapor itself, the interior of the bottle is a vacuum, so the only way for the vapor to shed heat is by radiating it away into an environment that’s already mindbogglingly hot–a giant pool of molten metal and runny steel walls. Most of it will probably go into the walls. In order for a kilogram of vapor to condense, it has to shed enough heat to melt about 6 kilograms of room-temperature steel, let alone the already hot walls, so it’s likely that the radiant heat will continue melting the walls for some time. You might want some really big radiator fins on the outside.

If you want to illustrate how insanely hot it would be inside the bottle, the fact that a rain of molten steel is a sign that it’s cooling off should do it.


Like I said, Jumbo was created to contain a fizzle, not a detonation.

FWIW, some underground atomic testing images…with vaporized chambers.

Well, if you put it That way…