If you could encase a grenade within layers and layers of reinforced concrete or within massive iron sphere and detonate it (with a timer), what would happen? Would the energy overcome and destroy the surrounding material and if not would a pocket of pressure remain for a length of time?
Methinks the terms that need more definition here are “layers and layers,” “reinforced,” and “massive.”
I doubt, for instance, that hundreds of thousands of layers of steel-rebar-reinforced concrete inside an iron sphere the mass of Earth’s moon would suffer much from having an hand grenade detonated at its core.
But I am not a munitions expert.
I guess the question is would the energy dissipate, like air escaping slowly from a balloon or would it be ‘trapped’ until such time as someone released it?
Thanks for the response.
I think it would severely damage whatever concrete it could. Concrete is hard, but it’s relatively brittle in my experience: it doesn’t resist sudden trauma very well.
In short, if you had sufficient concrete and iron to mask any evidence of the detonation from the outside, you’d have an iron sphere wrapped around an intact concrete sphere wrapped around a (probably roughly spherical) zone of demolished concrete with several loose hand grenade and timer parts.
Again, IANAME. 
Explosions are rapid expansions/releases of gas. The ideal gas law is:
PV=nRT but the V is constant. Only P (the explosion) and T are changing. So rework the equation to:
T = PV/nR
So you introduce a lot of gas --> the pressure goes up --> since the pressure goes up the temperature goes up.
Now the combined system will cool down to a new equilibrium point after a while and the internal pressure would be P = nRT/V
Many explosives operate by rapidly producing a large volume of gas. If your explosive is of this type, and the gases remain gases at room temperature, and cannot escape through pores or fissures, and are not absorbed by the surrounding material, the detonation will produce a high pressure cavity within your steel or concrete. As with a normally filled gas cylinder, the pressure could remain very high for years.
Does this mean that if the encasing concrete was sufficient enough, we wouldn’t experience the effects of the explosion until the concrete was cracked open from outside?
That would depend on how well-sealed the containment vessel is. You’d lose some energy, as the explosion-products cooled off, but even once it reached room temperature, you’d still be left with gas at fairly high pressure.
You would still experience the shockwave from the initial explosion as well as the sound. Both would be transmitted through the steel/concrete structure.
Not asked, but such containers are produced and used during explosive and propellant development. One example is a bubble chamber. A small quantity of the proposed explosive is detonated in the center of a pressure sphere filled with liquid. The size and expansion rate of the bubble generated by expanding gases is measured and determines the “bubble energy” of the explosive. Other chambers are used to measure pressure rises and maximum levels from propellant deflagration.
Other containers are built to contain the explosive fragments and vent the explosive. Examples are grenade pitch-in barricades and Golan 5 and 10 containers. These are for emergencies (pin drops out), storing EOD explosives, storing dog scent kits, and storing grenades and shells found on soldiers brought into hospitals for treatment.
Would it explode?
For example, you have one ounce of C4 inside a steel ball of sufficient thickness to ensure it will not rupture.
Explosives like C4 work by rapid burning turning the explosive into an expanding ball of gas. So you can say that the explosive is the most dense form of gas normally found outside of stars and suchlike. Somewhat like mass is just a very dense form of energy.
When the igniter goes off and there is no change in volume can the explosive become a gas? Would it just sit there waiting for something to happen to the containment vessel?
It’s a big leap to assume that there would be no change in volume. But for the sake of this question I’m assuming it is so.
A key point is that, depending on how exactly the cooled gas is released, the energy release from the gas pressure will be a lot less destructive than the original shock wave (even after accounting for the significant energy loss from cooling and container deformation).
Data from subterranean atomic testing may be instructive:
Project Dribble (PDF)
Does not specifically address any pressure encountered when the cavity was drilled.
n is changing also, as all the nitrates and oxygen are coming out. A lot of explosives are what’s known as CHON, for carbon, hydrogen, oxygen, and nitrogen. These types of organic explosives decompose into nitrogen, water, carbon monoxide, in that order. There is a secondary reaction, which happens almost simultaneously, that converts the carbon monoxide to carbon dioxide if the oxygen balance is high enough, and is known as afterburn. If there’s any carbon left over, it will form solid carbon. Because the reaction happens so fast, all the oxygen for the reaction must come from within the energetic material, there is not enough time for it to use the oxygen in the air.
We can’t use the ideal gas law, either, because the pressures are too high, so we use what’s known as the Nobel-Able equation of state, which replaces the volume term with (V-alpha w), where alpha is the covolume of gas and w is the weight. Like you said, though, ideal gas will get us pretty close to our final pressure, but we’re more interested in the shock pressure.
There is more energy than simply the expansion of gas. Ripping apart the generally triple bonded nitrogen and them forming together releases a huge amount of energy in the form of heat and pressure. This all culminates at the Chapman-Jouguet pressure, as given by the Kamlet-Jacobs method, which gives:
P[sub]CJ[/sub]=K rho[sup]2[/sup] N M[sup]1/2[/sup] -delta H[sup]1/2[/sup]
Where K is a constant, rho is the initial density, N is the moles of gas per gram of explosive, M is the grams of gas per mole of gas, and delta H is the change in heats of formation of the original reactant and the product gasses.
As a final note, the pressure through steel or concrete is going to be much stronger than the same amount and type of explosive would have imparted onto air. This is because they are both more dense than air, and they both have higher bulk speeds of sound. Water has the same effect.
As smithsb said, it is not difficult to make a container that would contain the blast.
Explosives react at a much higher pressure when they’re fully contained. Even something as weak as paper can sufficiently increase the pressure, enough so to take it from a deflagration to an explosive event. For example, if you were to pour out the gunpowder in a firework and light it, it would simply burn very quickly. Once you wrap it into a bottle rocket or BlackCat, it will pop.
This is because as the shock front (compressive wave) moves forward, the rarefaction (expansive) wave starts to catch up, eventually overtaking the initial shockwave (because it moves through a higher density material than the initial wave). When a system is under pressure, the pressure from the already reacted material is used to increase the pressure in front of the wave, allowing it to move faster, giving a higher pressure, and staving off the attenuation caused by the rarefaction wave. The higher the pressure, the greater this effect. We have to be careful to take this into consideration when designing our vessel, as one that has a yield strength sufficient to contain your hypothetical of a pound of C4 in free atmosphere would easily burst if it was affecting the internal pressure of the C4 during the blast.
[Nitpick]It’s a detonator, not an ignitor. The difference is that a detonatior is a primary (more sensitive) explosive that imparts pressure into the secondary (more powerful) explosive. Tertiary explosives need both a primary and secondary detonator to set them off. Again, this will be easier if the system is contained.[/Nitpick}
It just takes a lot of mass. And time and money.
Stranger
I guess it just depends on what kind of blast you’re talking about. 
ETA: I dig the time and money quote.
I’m confused.With my very limited understanding of physics and the like, I’d assume the following would be more or less correct. However, judging by your answers (which assert that a relatively stable pocket of high pressure is maintained in the cavity formed by the explosion), I must be quite wrong.
Here is my ‘thought process’ (or lack of it! ):
The heat generated by the explosion cannot be contained. That’s just a basic law of thermodynamics. But, since temperature is, in some sense, equivalent to pressure, it means that the pressure will also go down. In other words, to say there is a preserved pocket of high pressure implies that there is a preserved pocket of heat. Such heat must dissipate. Hence, the pressure cannot be retained in the cavity formed by the explosion.
Where am I wrong? (Please go easy on me!)
Look here:
Volume of products of explosion
A mole of nitroglycerin (227 grams, having a volume of 0.142 liters) will detonate to form 3 moles of CO2, 2.5 moles of H2O, 1.5 moles of N2, and 0.25 mole of O2.
Ignoring the water*, which will liquify as the cavity cools, that’s still 3.25 moles of gas in a 0.142 liter space, 0.044 liters per mole. Given that a mole of gas takes up 22.4 liters at standard atmospheric temperature and pressure, the gas in the cavity will be under about 509 atmospheres of pressure after things cool down.
*Water should not be ignored as even after it condenses, those 2.5 moles will fill 45/142 = 31.6% of the initial nitro cavity with an incompressible fluid. That’ll bump the final pressure even higher than I calculated.
Rereading my question I see I may not have phrased it as clearly as I might have. Don’t know if it changes things, but hey, this is how we learn.
My supposition was that there was no extra space, the C4 took up all the space and the container was strong enough to suffer no damage. No expansion was possible.
I’m pretty sure the increased pressure would just drive the detonation wave front to travel at a faster velocity and at a higher pressure. The explosive “likes” to be under high pressure containment, and will only react more strongly at higher pressures. The gas generated would simply compress the bulk of the remaining material, causing a higher pressure at the unreacted interface. As more and more gas was generated, the pressure would just get higher and higher until the container failed or the reaction was over.
And, Karl, to answer your question, and as Squink succinctly stated, our n term is increasing; we have more gas in the container than we started with.