Nuclear Explosions

[sevenperforce]

4 hours ago, p1t1o said:

They are "shaped charges" but they are still nukes. You get a spherical explosion in terms of radiation, but it will send a pulse of denser debris in one direction, possibly two. Its not like being off the the side is "sorta ok", you still get vaporised.

Im not even sure what happens to high explosive when you vaporise it faster than a detonation wave can travel through it. I guess it still explodes.

I'm no inorganic materials chemist, but my gut is that vaporizing high explosive simply converts it into a gaseous explosive similar to a fuel-air bomb, but without all that unnecessary nitrogen.

The "shaped charge" aspect of a nuclear propellant shot is, IIRC, primarily the result of the x-ray flux impinging on the dense tungsten propellant. Most of the energy of the nuke is lost.

[p1t1o]

1 hour ago, sevenperforce said:

I'm no inorganic materials chemist, but my gut is that vaporizing high explosive simply converts it into a gaseous explosive similar to a fuel-air bomb, but without all that unnecessary nitrogen.

The "shaped charge" aspect of a nuclear propellant shot is, IIRC, primarily the result of the x-ray flux impinging on the dense tungsten propellant. Most of the energy of the nuke is lost.

That would be my gut instinct too for the fomer. For the latter, well it depends on what you mean by "lost" - if you mean for propulsion, then yes. However according to the various studies done for the project, some ridiculous percentage, like 60-80% IIRC of the energy can be put into the propellant - Im kind of skeptical about that, but they have the maths, so...

[kerbiloid]

Near the center of explosion the temperature is much, much greater than the potential of ionization for any element. So, you get ions and electrons.

In the outer layers you should calc that delta-Entropy vs delta-Entalpy equation (i.e. delta-Energy of chemical reaction) to realize if this compound survives under such temperature.

[Gargamel]

I would really have to believe though, there is a non zero chance of secondary fission explosions.  Very unlikely, for all the reasons mentioned, but also non zero for the other reasons mentioned. 

I really think we need empirical evidence for this.  Lets get a bunch of nukes (we have extras laying around), put them in a pile, and set one off.   See if more than one goes off.  Run this test, say a few hundred times, and we'll have a good data set.   Would be pretty useful if do ever design an Orion drive. 

2 minutes ago, kerbiloid said:

Near the center of explosion the temperature is much, much greater than the potential of ionization for any element. So, you get ions and electrons.

In the outer layers you should calc that delta-Entropy vs delta-Entalpy equation (i.e. delta-Energy of chemical reaction) to realize if this compound survives under such temperature.

I think excess neutrons would more of a concern in the vein of this discussion, rather than ionization.  Close enough proximity might actually cook off another nuke. 

[kerbiloid]

9 minutes ago, gargamel said:

I think excess neutrons would more of a concern in the vein of this discussion, rather than ionization.

Not ionization caused by ionizing radiation, but ionization caused by atom collisions. For almost any element it's <10000-20000 K.

[p1t1o]

16 minutes ago, kerbiloid said:

Near the center of explosion the temperature is much, much greater than the potential of ionization for any element. So, you get ions and electrons.

In the outer layers you should calc that delta-Entropy vs delta-Entalpy equation (i.e. delta-Energy of chemical reaction) to realize if this compound survives under such temperature.

If you are referring to chemical explosives adjacent to a detonation, I'd agree that it is fully ionised very rapidly - before any chemical reaction have a chance to take place. But the chemical energy in the bonds must still be released onto the environment, it will be insignificant but it must still be there I suppose.

 

14 minutes ago, gargamel said:

I would really have to believe though, there is a non zero chance of secondary fission explosions.  Very unlikely, for all the reasons mentioned, but also non zero for the other reasons mentioned. 

I really think we need empirical evidence for this.  Lets get a bunch of nukes (we have extras laying around), put them in a pile, and set one off.   See if more than one goes off.  Run this test, say a few hundred times, and we'll have a good data set.   Would be pretty useful if do ever design an Orion drive. 

I think excess neutrons would more of a concern in the vein of this discussion, rather than ionization.  Close enough proximity might actually cook off another nuke. 

There are almost certainly going to be neutron interactions with that much fissile material around, but it will just "fizzle" and wont add more than a few percent to the yield.

Chernobyl was about as close an approximation as we are likely to get, to a nuke going off in a pile of nukes, and that was just a large "fizzle".

 

Edited March 28, 2017 by p1t1o

[Gargamel]

I still say we need empirical evidence!  

 

(Ok yes, I've been Xkcd binging tonight).

[sevenperforce]

12 minutes ago, gargamel said:

I would really have to believe though, there is a non zero chance of secondary fission explosions.  Very unlikely, for all the reasons mentioned, but also non zero for the other reasons mentioned. 

I really think we need empirical evidence for this.  Lets get a bunch of nukes (we have extras laying around), put them in a pile, and set one off.   See if more than one goes off.  Run this test, say a few hundred times, and we'll have a good data set.   Would be pretty useful if do ever design an Orion drive. 

I think excess neutrons would more of a concern in the vein of this discussion, rather than ionization.  Close enough proximity might actually cook off another nuke. 

Modern nukes are implosion devices; implosion devices cannot "cook off". They can predetonate in a fizzle easily enough, and that is what would happen for quite a few of the nukes due to the dizzying neutron flux. But they cannot and do not cook off.

If we were dealing with gun-type nukes like the Little Boy dropped on Hiroshima, then that would be a different matter. But implosion devices can only predetonate.

I suppose there is a vanishingly small chance that an incidental EMP might conceivably produce an electrical flux with precisely the right amount of current to trigger the exploding-bridgewire detonators, but that's sort of a fringe effect.

[sevenperforce]

1 hour ago, p1t1o said:

That would be my gut instinct too for the fomer. For the latter, well it depends on what you mean by "lost" - if you mean for propulsion, then yes. However according to the various studies done for the project, some ridiculous percentage, like 60-80% IIRC of the energy can be put into the propellant - Im kind of skeptical about that, but they have the maths, so...

Did a little digging, and I remember now. They surround the spherical implosion core with a non-fissionable uranium tamper open on one end, which reflects virtually all energy not heading toward the hole, boosting the x-ray yield. As a result, 80% of the total energy production ends up as x-rays headed roughly in one direction. This hits the shaped beryllium tamper, which absorbs the x-rays and transfers their kinetic energy to the dense tungsten.

So yeah, over 60%, which is quite impressive compared to the 10-11% energy efficiency you get from using a bare nuke.

[p1t1o]

49 minutes ago, sevenperforce said:

Did a little digging, and I remember now. They surround the spherical implosion core with a non-fissionable uranium tamper open on one end, which reflects virtually all energy not heading toward the hole, boosting the x-ray yield. As a result, 80% of the total energy production ends up as x-rays headed roughly in one direction. This hits the shaped beryllium tamper, which absorbs the x-rays and transfers their kinetic energy to the dense tungsten.

So yeah, over 60%, which is quite impressive compared to the 10-11% energy efficiency you get from using a bare nuke.

I know the math is supposed to add up, and I know roughly what is going on inside.

I just find it very counter-intuitive that there isnt at least an equal amount of energy (whether EM radiation or kinetic energy of matter) projected in the other direction, I mean newtons 3rd law and everything.

Unless the 60-80% is emitted along one axis rather than one singular direction? Which would make more sense since that is how reaction forces work.

[kerbiloid]

Not in direction, but in wide sector. And this depends on diameter/thickness ratio. Dyson's book describes this.

[sevenperforce]

1 hour ago, p1t1o said:

I know the math is supposed to add up, and I know roughly what is going on inside.

I just find it very counter-intuitive that there isnt at least an equal amount of energy (whether EM radiation or kinetic energy of matter) projected in the other direction, I mean newtons 3rd law and everything.

Unless the 60-80% is emitted along one axis rather than one singular direction? Which would make more sense since that is how reaction forces work.

Equal and opposite forces mean equal and opposite momentum, because momentum must be conserved. Conservation of energy is not vector-dependent.

There is a conservation of momentum between the heavy non-fissionable uranium tamper and the lightweight beryllium accelerator which absorbs the x-ray flux. The only way for momentum to be conserved is for the lightweight beryllium accelerator to be moving much faster than the dense uranium. However, because kinetic energy is proportional to the square of velocity, the beryllium contains much more kinetic energy than the uranium; as a result, much more than 50% of the total energy ends up being used for propulsion.

[p1t1o]

42 minutes ago, sevenperforce said:

Equal and opposite forces mean equal and opposite momentum, because momentum must be conserved. Conservation of energy is not vector-dependent.

There is a conservation of momentum between the heavy non-fissionable uranium tamper and the lightweight beryllium accelerator which absorbs the x-ray flux. The only way for momentum to be conserved is for the lightweight beryllium accelerator to be moving much faster than the dense uranium. However, because kinetic energy is proportional to the square of velocity, the beryllium contains much more kinetic energy than the uranium; as a result, much more than 50% of the total energy ends up being used for propulsion.

That actually helps, thanks, Im still not 100% on how the dense propellant ends up with more energy than the dense radiation case, but I can live with it being a mystery to me. I can see there being some sweet-spot trade-off with masses, molecular weights, radiation fluxes, opacities and transparencies, emission/absorbtion rates, lots of high numbers regarding temperatures, pressures and velocities, the word "plasma" all over the place etc.

I have faith in the science behind it, I just find it noodle-baking enough that I think I'd have to see it work before I can get that hyped about Orion-style propulsion, its always bugged me.

[Nothalogh]

4 hours ago, sevenperforce said:

Did a little digging, and I remember now. They surround the spherical implosion core with a non-fissionable uranium tamper open on one end, which reflects virtually all energy not heading toward the hole, boosting the x-ray yield. As a result, 80% of the total energy production ends up as x-rays headed roughly in one direction. This hits the shaped beryllium tamper, which absorbs the x-rays and transfers their kinetic energy to the dense tungsten.

So yeah, over 60%, which is quite impressive compared to the 10-11% energy efficiency you get from using a bare nuke.

It's glorious, isn't it?

And if you think about the implications of that as a weapon, you immediately know why it was all blackbagged

[sevenperforce]

8 minutes ago, Nothalogh said:

It's glorious, isn't it?

And if you think about the implications of that as a weapon, you immediately know why it was all blackbagged

Eh, the optimal effective range of a shaped-charge megaton nuclear directed energy weapon (i.e., nuclear lance of fiery destruction) is "only" 2000 km.

[Nothalogh]

Just now, sevenperforce said:

Eh, the optimal effective range of a shaped-charge megaton nuclear directed energy weapon (i.e., nuclear lance of fiery destruction) is "only" 2000 km.

Why bust bunkers when you can bust mountains?

[sevenperforce]

11 minutes ago, Nothalogh said:

Why bust bunkers when you can bust mountains?

And, if you're exoatmospheric...

...a one-megaton nuclear lance will blow a hole 70 meters deep and 200 meters wide...

...through solid aluminum...

...at 10,000 km.

Edited March 28, 2017 by sevenperforce

[Capt. Hunt]

5 hours ago, sevenperforce said:

Did a little digging, and I remember now. They surround the spherical implosion core with a non-fissionable uranium tamper open on one end, which reflects virtually all energy not heading toward the hole, boosting the x-ray yield. As a result, 80% of the total energy production ends up as x-rays headed roughly in one direction. This hits the shaped beryllium tamper, which absorbs the x-rays and transfers their kinetic energy to the dense tungsten.

So yeah, over 60%, which is quite impressive compared to the 10-11% energy efficiency you get from using a bare nuke.

You make it sound like  after one goes off, it'd leave the casing intact except for the end the charge was directed towards.  Using your numbers, you still get 20% of the energy radiating in all directions.  That means that even with a "small" casaba howitzer, you still get a couple thousand tons of TNT equivalent blast in all directions.

Edited March 28, 2017 by Capt. Hunt

[Nothalogh]

10 minutes ago, Capt. Hunt said:

thousand tons of TNT equivalent

At the max.

Vehicles needing kiloton units or larger would be some absolutely impressive pieces of work.

Either by their sheer size, or outrageous performance

[Capt. Hunt]

ok, I guess I assumed 20 kt as a nice round number, but you're probably right.  I've never seen the actual numbers for the proposed Orion "pellets"

Edited March 28, 2017 by Capt. Hunt

[Nothalogh]

1 minute ago, Capt. Hunt said:

I've never seen the actual numbers for the proposed Orion "pellets"

.333 kiloton to 1 kiloton, usually

[Capt. Hunt]

we're still talking tens to hundreds of tons of TNT.  Those are still big booms.

Edited March 28, 2017 by Capt. Hunt

[Nothalogh]

7 minutes ago, Capt. Hunt said:

we're still talking tens to hundreds of tons of TNT.  Those are still big booms.

Correct, but we're also talking about a vehicle built like an aircraft carrier, instead of like a soda can

[sevenperforce]

2 hours ago, Capt. Hunt said:

You make it sound like  after one goes off, it'd leave the casing intact except for the end the charge was directed towards.  Using your numbers, you still get 20% of the energy radiating in all directions.  That means that even with a "small" casaba howitzer, you still get a couple thousand tons of TNT equivalent blast in all directions.

Well, yes. A nuclear shaped charge still produces a local spherical fireball. I thought that was assumed.   

[Nothalogh]

You know what we really need, an above ground test filmed with modern camera technology