Also, the temperature (250 K, -23°C) is a bit chilly to be called “room temperature”. Still, this means you don’t need liquid nitrogen for cooling: a good but standard refrigeration system would be enough (and their insane diamond pressure apparatus, of course).
well the pressure aspect wouldn't require constant energy input. And I assume it is a lot more economical to use a consumer refrigerator than liquid nitrogen.
But yah, you can bathe a much more complicated "circuit" in liquid nitrogen a lot easier than compressing it with diamond anvils. Though no commercial superconducting computers have emerged, cmos is just too cost effective currently, and the known examples are still in the 4K range.
"well the pressure aspect wouldn't require constant energy input."
But you have to put energy in to compress it, energy that is lost (since you're compressing it isothermally). Since we're talking about 1E6 atm, any appreciable volume would require insane amounts of E to compress.
Wouldn't that also be pretty much a bomb if the pressure got relieved? I don't know how to do the math but 1E6 atm seems a lot of energy (is it potential energy?) that has to go somewhere.
How much energy it is depends on the volume change as you compress. If your material is completely incompressible, there is no volume change, no work done as the pressure rises, and therefore no energy stored.
In practice, for a material of bulk modulus K by definition you have
-K dV/V = dP.
as the basic differential equation relating pressure and volume (if we assume that the bulk modulus is constant through our pressure range, which is quite an assumption in this case). Solving that, we get:
V = V_0/exp(P/K)
where V_0 is the original volume and V is the volume after the pressure has been applied. Bulk modulus is commonly measured in GPa, so we're going to do that here. 1atm is about 1e5 Pa or 1e-4 GPa.
The work done while compressing is the integral of -PdV, which per that first differential equation we can rewrite as the integral of P * V/K * dP (we want to integrate dP because we know what the limits are: it's going from 1atm, or 1e-4 GPa, to 1e6 atm = 1e2 GPa).
Plugging in our expression for V we get the integral, from 1e-4 to 100, of P/K * V_0/exp(P/K) dP. The integral there is:
K * V_0 * (-P/K - 1) * exp(-P/K)
For simplicity, I'm going to approximate that 1e-4 by 0, so we get:
or so. If we assume we started with a 1 cm^3 volume of steel, that's an energy of:
21e9 N/m^2 * 1e-6 m^3 = 2.1e4 J
or about equivalent to 5 grams of TNT per <https://en.wikipedia.org/wiki/TNT_equivalent>. For comparison, 1 cm^3 of steel is about 7-8 grams of material, so we're right in the "about as much energy as the same weight of TNT" ballpark. Obviously if you start with a larger volume you end up with more stored energy: if we started with 1m^3 we end up equivalent to about 5 tons of TNT.
If your material is more compressible, the numbers are higher, as long as it's not too compressible (in which case it just collapses into a tiny volume quickly and not much more work gets done after that). If we take K = 50 (a typical number for glass from that table), you get:
If your material is less compressible, you store less energy; for diamond with K = 443, you get something like 9.7*V_0.
Again, all this assumes pressure-invariant bulk modulus, which seems moderately unlikely when you are dealing with pressures that are on the order of the size of the bulk modulus or even larger, as here.
