On 12/14/2009 12:45 PM, Stephen A. Lawrence wrote:
On 12/14/2009 11:11 AM, Terry Blanton wrote:
Hmmm, suppose you have an accordion device that you drop into the ocean
with a rock. The device pressurizes air as it sinks and locks into
place, then you drop the ballast and it floats to the top. You use the
pressurized air to do work. Is possible?
>
If so, is it OU?
Well, hrmmmmph ... that seems to be harder to answer than I expected,
after a page of scribbles failed to get me anywhere.
OK, here we go (and if my company goes bust as a result of the time I
wasted on this, oh well):
The rock is a red herring, because work done to sink it and lift it
balances out. Ditto the accordion. All we need to worry about is the gas.
Imagine that we put the gas in a cylinder with a piston, area(piston)=A,
volume of cylinder = A*L where L is the distance between the piston and
the back wall. The mass of the cylinder etc is neglectable also because
it's the same going down and up; as I said, it's just the gas we need to
worry about. The pressure in the cylinder -- and the surrounding water,
of course -- is 'P'.
Now imagine that we *drag* the cylinder down to the bottom, and then
lift it back up. Imagine further that the density of the gas EQUALS the
density of the water AT THE BOTTOM. Then we do work all the way down,
but it's neutrally buoyant all the way up. So, all we need to worry
about is the work to get it to the bottom.
In fact we don't need to worry about taking integrals or finding total
work; all we need to worry about is the work needed to drag the cylinder
*a* *little* *bit* farther down, and the amount of additional energy
which will be packed into it by that incremental drag due to additional
compression of the gas. If that incremental bit of work balances, then
the total work will balance also.
Call depth in the water 'h'. Then work done by the water in compressing
the gas a little more as it's dragged down a little deeper will be:
dW(pressure) = P*A*dL
that is, the pressure in the cylinder, times the area of the piston,
times the amount the piston moves (ignore sign issues, we can always
patch them up later if we feel moved to do so).
A little fiddling reduces this to:
dW(pressure) = -(nRT/P) * rho * dh
where 'rho' is density of the water, and PV = nRT is the ideal gas law.
I'm not showing the intermediate steps; if anyone wants to see them I
can post them later.
Now let's look at the work done to drag it "dh" deeper in the water.
That's the buoyancy, times the distance dragged:
dW(buoyancy) = (rho * V - k*n) * dh
That is to say, the buoyancy force is equal to the density of the water
times the displacement (that's rho*V), *minus* the weight of the gas,
which will be some constant, 'k', times the number of moles of gas, 'n'.
However, as it happens, we're going to drag the gas back to the
surface again, so just like the weight of the rock and the weight of the
cylinder, the weight of the gas is neglectable -- work done on it going
down and back up will cancel. So, the part we're interested in reduces to
dW(buoyancy) = rho * V * dh
but we also have
V = nRT/P
so
dW(buoyancy) = rho * (nRT/P) * dh
and that's equal to the incremental work we found was done in
compressing the gas in the cylinder.
So, it appears that the energy gained by squeezing the gas is exactly
equal to the energy it took to drag it to the bottom of the ocean.
* * *
Oh -- one other item -- I assumed the temperature was constant. From
the POV of thermodynamics, at least, that's legit, as we can treat the
ocean as a single "heat source/sink" which absorbs or provides heat as
needed to keep everything at one temperature.
