How close can I get to a nuclear explosion with extreme shielding?

Question

Imagine you are in empty space, and a 15 megaton nuclear bomb is very close to you (D=200 meters), but what's also very close to you is a giant (Tungsten?) cylinder (2? meters in diameter, 100? meters long) aimed directly at the bomb. If you hide behind it, the radiation and plasma should fly by you for the most part. Though I would expect part of the cylinder to vaporize, which will cause it to accelerate, possibly violently towards you.

So how close could we get to this bomb and live (at least 1 year) to tell about it? I'm looking to minimize my distance D to the bomb and live.

Edit: I did a brief energy flux calculation, and at 100 meters, each square meter is going to have the energy of half a million sticks of dynamite, so 200 meters may have been ambitious. 6.276e+16 / (4 * 3.14 * 100^2) / 1,000,000 = 500'000 MJ ~ 120 ton tnt

Answer 1

According to this website, tungsten has a latent heat of vaporization of 824 kJ/mol. One mole of tungsten is about 0.184 kg, so we get about 4,480 kJ/kg. There are $6.276\cdot10^{16}$ joules per 15 mt TNT. By increasing our distance by some factor, $x$, we can reduce our exposure to the blast energy by $\frac{1}{x^{2}}$. That's the inverse-square law for you. If one end of the cylinder is 5 meters from the nuke, it only occupies about 1/25th of its sky and receives 1/25th of the radiant energy.
With these two values, we can calculate the mass of tungsten the applied bomb energy will be able to vaporize. The value is likely an overestimate as atoms of tungsten are unlikely to reach their vaporization temperatures and just stop heating up. They'll continue to absorb energy, acting as insulation to the tungsten behind them.

Tungsten has a density of 19,250 kg/m^3. With mass, $m$, and density, $d$, we can calculate the cylinder volume. Rearranging everything, we can calculate the cylinder height, $h$, instead: $$h=\frac{m}{d\pi r^{2}}$$

Cylinder radius, $r$, is given to us as 2 m. Stuffing in our figures for bomb energy and latent heat of vaporization in place of $m$ (where $m=\frac{E_{blast}}{E_{perkg}}$), and introducing an inverse-square factor of $\frac{1}{x^{2}}$ to account for separation distance, and simplifying a little, we get a general equation for the cylinder thickness, $y$ [m], per distance, $x$ [m], from the 15 mt explosion: $$y=\frac{182286}{x^{2}\pi}$$

At a distance of $x=$ 100 meters from the bomb, the 2 m radius cylinder may be as thin as $y=$ 6 meters. Tungsten is pretty good at dealing with ionizing and particle radiation. Doubling that figure should be more than enough.

Answer 2

First, some numbers. Assuming a reasonably clean fusion design at one meter we are looking at 2.67E15 rad of x-rays and 1.08E14 rad of neutrons. The latter is harder to stop so I'll look at it. 7cm of water halves the neutrons, thus 3 meters of water brings the dose down to just over 1 rad. Of course you have some mighty energetic water to deal with at that point and need a substantial mass between you and the water to avoid being cooked and squashed flat.

If that was all you had to deal with I wouldn't be surprised if you could survive 20 meters from the bomb. (Don't be surprised at this--consider nuclear pulse propulsion--think of the cartoon character putting a stick of dynamite under something and sitting on it to go flying. Scale that up to tactical nuke range and add a shock absorber so you can actually survive it. The bombs were to be a lot smaller but the shielding would likewise be much less--and it was intended to be manned.) However, I see three additional headaches that complicate matters:

1. Your bomb is not a point source. The best we can do currently is about 6kt/kg of bomb, so you're looking at 2,500kg of bomb. I do not think this is going to be a major factor.

2. Not all of the energy of the bomb is liberated immediately. There will be some short-lived isotopes in the expanding shell of vaporized bomb--you'll catch some radiation from this.

3. That water that stopped the bomb energy will itself be extremely hot and the thermal flux from it will be dangerous.

These factors mean you're going to need a shield at least halfway around you. I don't know enough of the physics to figure out what it's going to have to stop, though. It doesn't need to be nearly as substantial as the one between you and the bomb itself.

Answer 3

At LEAST 10km Required

There's a fun website out there called nukemap that lets you explore nuclear explosions.

For a 15MT explosion at ground level, it expects a crater over 200m deep, and nearly 500m in radius. Everything inside this zone will be vaporized.

But the "not possible to survive zone" from a 15 MT blast is actually much larger than that. The nuclear fireball radius is over 4km. So even if a large "protective" cylinder sits between you and the point of detonation, the air around you will still be super-heated to temperatures that you cannot possibly survive.

At a radius of just over 5 km, "heavily built concrete buildings are severely damaged or demolished; fatalities approach 100%."

So it's possible that around 5km from the point of detonation you could shield yourself with a long metal tube and survive a 15 MT blast - but I wouldn't count on it.

A CDC report on blast overpressure mentions in passing that a 3.5psi overpressure wave moved an un-anchored 1,500lb object 80 feet. Since the 5 psi overpressure radius for this explosion is at 11km, I expect that our Tungsten cylinder will probably crush the user at 10 km, but that if it were heavy enough or anchored to the ground it is possibly they could survive.

Direct line of sight on a bomb this large will cause third degree burns on all exposed skin up to 35km away! So the chances that reflected energy will still manage to kill you at 10km seem significant, but if you're trying to push the limit I think it might be around this range.