Automatic isothermal compression

See Automatic isothermal expansion. Isothermal compression is a little harder, because the air has to move downward when and only when it cools down to temperature. Or at least, the hydrostatic pressure must increase then. Liquid is moving downward to provide the work for compression, which heats the air. If we let the liquid keep moving downward when the air is too hot, we do more work than necessary to compress the air, and there is no way to retrieve that work. We must make sure we never do it in the first place.

So, how can a single solid shape achieve this with no moving parts? Liquid flows into a closed chamber with working gas at the cold temperature. The gas is compressed and it heats up. The flow stops before the final volume is reached because the gas is too hot. When the gas cools down, some more liquid can fit in, and the gas gets compressed more.

So, an idea dawns. The liquid must finish filling up a basin just as the gas is compressed to the proper pressure. Then more liquid can flow because it overflows down to the next basin. Somehow this has to cause the hydrostatic pressure to increase — it must be harder to withstand the flow into the lower basin than it was into the upper basin. Well, more work is being done by gravity on the liquid as it flows farther down, so there ought to be more force available to compress gas.

This seems to work, but only for one pass. How to return to the starting point? It has to involve the other half of the cycle, the isothermal expansion. That part raises liquid and lets high pressure gas expand. We do want separate pools of liquid because the expanding gas is hot and it is wasteful to repeatedly heat and cool the liquid. This is not quite worked out…

More details. Looks promising: The top basin fills completely with liquid before any spills down to the next lower level, and furthermore, the spillway is a siphon for use later in emptying the top basin of liquid. I suspect the basin is really a tube sloping upward to a short inverted U at the end. All basins start full of cold low-pressure working gas. Cold liquid flows in at the top and fills the large first basin. The total volume of the remaining basins and connections is the first checkpoint volume, such that the pressure-volume product is the same as it was before any liquid flowed in. Since pressure is controlled by the height difference between the bottom liquid surface and the top liquid surface, this should be possible. After the gas cools enough, the top basin is full of liquid and flow starts into the second basin. The hydrostatic pressure is higher because the basin is lower and there is a continuous column of liquid from the top of the upper reservoir to the surface in the second basin. Flow continues until the second basin is nearly full, and stops because the gas has heated up. Heat flows out of the gas, the pressure drops, and liquid rises to the top of the second inverted U and starts flowing over. The second basin is full of liquid and all of the gas is in the lower basins. The pressure volume product is constant. As liquid flows, the pressure increases by the drop to the third basin and the PV product is a little high. V decreases as flow continues.

So at the end, we have all of the gas compressed in the bottom basin at cold temperature, and all of the upper basins full of liquid.

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2 Responses to “Automatic isothermal compression”

  1. 0xff Says:

    Explain the relationship between the fluid and the gas, such that the fluid does not simply rise boyantly to the top of the tubes and form a vapor block.

  2. archimerged Says:

    There must be some way to get this blog to send me notifications of comments, but I haven’t found it. I haven’t found out how to put an image on the top either, and don’t want to spend time on it…

    Anyway, see the later ideas using small diameter tubes and a fluid such as mercury in glass tubes so the liquid and gas stay in discrete slugs. I’m more interested in that because it runs as a continuous process. But in this idea, the liquid is always below the gas. I think the top of the inverted U (“:cap”) has to be kind of flattend out. In fact, I think the cap has to be made up of a lot of small diameter tubes so the liquid can’t flow thorugh while leaving any gas behind.

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