Well, maybe not from there exactly, but from a deep hydrothermal vent somewhere. I take an empty bathysphere to 6 miles down and fill it up with 300° C water near a black smoker. I dog down the hatch and winch it back up to the surface, still hot and pressurized. At the surface, the air temperature is 0º C, and so is the surface water.
Now that I have the hot pressurized bathysphere on the deck of my ship, in practical terms how can I extract the most work from the water (which I want to keep at the surface, so dropping it back over the side isn’t an option, nice try though). Do I connect the pressure vessel to a piston or turbine and let the water flash to 300º steam and produce power that way? Can the miniscule expansion of the compressed water be harnessed before it starts to boil?
How much useful energy can I actually extract from, say, 1000kg of such pressurized, heated water? Enough to recoup a significant fraction of the cost of raising it in the first place? It seems like there must be some height at which you would break even.
Is there any significant compression of water at that depth? I know that it doesn’t compress at all at any of the depths we normally go to in the ocean; is the deepest part of the sea deep enough to squish, say, 1 cubic foot of water into a smaller space?
I am almost certain that you just get hot water, probably not very cost effective. AIUI there is no serious compression of the water, the pressure comes from the mass of the water column. 1000kg/2200lb of water plus vessel would have a while to cool off on the way up as you would have to dead lift it or come up with some buoyancy scheme. IANAE I am sure one will weigh in and once again show me the errors of my ways.
That’s what I thought too. The reason that a submarine can get crushed at great depths is not because the water is highly compressed, but because of the great weight of water that is pushing against the outside of the sub. Gases compress quite a bit - you can compress a quantity of nitrogen and get a tank of pressurized nitrogen that takes up less space but weighs the same. Liquids don’t compress nearly as much.
How hot would the water even be by the time it got to the surface? You’re pulling that bathysphere a long way up, and for most of that journey that outside water’s going to be quite cold.
There was a plan I read once to use a pipe from the shore down to the depths of a trench - and exploit the heat difference to run a heat engine (Stirling Cycle?). I’m not sure how effective that would be, but due to the noncompressibility of liquid, it would not be that difficult to pump the water up, according to their calculations.
The bulk modulus of water is 2.2 GPa. The pressure at 4,000 meters (close to the average depth of the world ocean) is .0403 GPa. So water at 4,000 meters is .0403/2.2=.018. In other words, a mass of water at 4,000 meters occupies 1.8% less space than it does in a vacuum. I’m pretty sure the bulk modulus of water actually increases at high pressures, so the real difference is probably a little bit less.
However, even though the water in your sealed, pressurized chamber would expand only slightly when exposed to the surface pressure, it would do so with great force.
You don’t need to pull up water at the temperature of a thermal vent; OTEC and similar technologies worked on a circa 25-50 degree difference. It’s the sheer quantity of the differential heat that makes it workable.
Are you sure this isn’t an ME thermodynamics homework problem? Because the question boils down (no pun intended) to something rather simple academic in nature…which only an ME student would ask how to solve.
I’ll outline the strategy only: Basically, what you are asking is simple enough. You’re equating the enthalpy of water (h-l) at Mariana Trench P,T to the percent of water (x)(h-l) + (1-x)(h-v) steam at let-down conditions, Where:
h-l = enthalpy per poundmass of liquid
h-v = enthalpy per poundmass of vapor
x = a decimal value between 0 and 1, exclusive
And, Mariana Trench water conditions can be evaluated knowing density § at depth (H) from a table of water properties. And, Mariana Trench pressure § = pgH. You can assume the mass of your sample is 1 lb-m since everything is on a per-pound basis, anyhow.
Ah yes, but if it’s particularly hot water, you can make particularly hot tea, and, by hooking the logic circuits of a Bambelweeny 57 Sub-Meson Brain to an atomic vector plotter suspended in your particularly strong Brownian Motion producer, and working out a few key improbabilities, you would, if nothing else, win the Galactic Institute’s Prize for Extreme Cleverness. I’d even by you a djinuntoonic (no Pan-Galactics though, still hung over).
But that appears to represent the compression of the air inside the pressure hull, not of the water. Much like the toy water rockets you pump up, but on a mega-scale.
I’m quite sure it isn’t homework; that’s twenty years in the past. I appreciate the outline of the strategy, but I don’t know what you mean by let-down conditions. I take it that only some fraction of the water will become steam at 1 atm pressure because the heat of vaporization is so high.
Plus, even if I knew how many kJ was released as heat by the water going from Marianas P,T to surface P,T based on enthalpy differences, can I really capture all that energy and turn it into mechanical work?
The water is pretty non-compressible, but it is likely to have a fair amount of dissolved gasses in it, especially if near a thermal vent. Those gasses will want to come out of solution when you reduce the pressure, much like when you open a warm bottle of soda pop. So you can expect a lot of foaming.
