Is any work done by expanding gas lost in the TrombePump Heat Engine?

Update:  probably most of the inefficiency in a bubble pump occurs when raised water falls back down when dynamics is being relied on to do the pumping rather than statics.  If the bubble pump is just supposed to raise the water a tiny amount, it should be quite efficient, even if the input side is not sealed, according to Archimerged's latest thinking…  Need to do some experiments… 

Reading about Dr. Masao Kondo's Geyser Pump in U.S. Patent 6,162,020, I see he reports that ordinary bubble pumps waste a lot of the work done by the isothermal expansion because some of the work fails to raise water. A gas molecule striking the bottom surface of a bubble pushes water downward instead of upward.

So I am checking whether the TrombePump Heat Engine design (which uses an ordinary bubble pump with small bubbles) suffers from the same effect. Of course, not all of the work goes to raising water, but because of the closed loop design, it appears that no work is lost. The purpose of the heat engine is not raising water, and raised water is not used as an intermediate energy storage. All of the work done by isothermal expansion of hot gas goes into compressing gas, either directly (if the force is transmitted downward, failing to raise water but instead compressing gas), or later (if the force is transmitted upward raising hot water which is cooled in the countercurrent heat exchanger (CCHEX) and finally descends the trombe and compresses cold gas).

There is an important difference between the usual application of a bubble pump (where Kondo's measurements were made) and the TrombePump design. In Kondo's setup, the compressed gas admitted to the water column is completely independent of the gas over the surface of the water being pumped, which is actually open to the surroundings. But in the TrombePump, the compressed gas is the gas over the water being pumped. Thus, when a gas molecule collides with the bottom of an air bubble and decelerates the water, the force is transmitted back to the gas above the input water and the energy ends up in another molecule of gas which will later be used to raise water. In Kondo's bubble pump setup, that energy ends up dissipated to the surroundings.

It is interesting to note that in his geyser pump setup, there is a volume of compressed gas in contact with the liquid being pumped and we might speculate that some of the efficiency improvement of the geyser pump over the bubble pump could be due to transmission of energy back into the compressed gas.

Now you might complain that the force transmitted downward compresses hot gas, not cold gas. This is true. The work done by falling cold water compresses cold gas, while the work done by force transmitted downward through hot water to the lower surface of the hot water compresses hot gas. But remember the hot gas is in physical contact with the gas in the CCHEX which is in physical contact with cold gas. So I better check whether or not it matters that the force was originally applied to hot gas, not cold gas.

My immediate answer is that the work which goes back into hot gas does not earn the profit which work spent raising water does, but eventually the energy will be used to raise water and then it will earn the profit. Since that energy is part of the energy absorbed by expanding hot gas, I suspect that because the gas in the bubbles pushes on all surfaces equally, sending some of the work downward to the hot compressed air above the water being pumped, it might be necessary to have twice as much gas to carry a given amount of energy as would be required in a piston engine. The continuous process machine is more complicated to analyze than a batch piston so perhaps only experiment can give a really reliable answer to the question of how much power a given machine will absorb from the hot heat source.

One question we must consider is whether or not this constitutes a short circuit of energy going from hot source to cold sink without having a chance to be converted to useful work. The answer to this is clearly no. Heat in the cold to hot side of the CCHEX does not flow into the cold sink. It all ends up in the hot gas and water leaving the CCHEX. Thus I am reasonably confident that the observed inefficiency of a bubble pump which pumps from an open container to another open container does not apply to this design.

I wrote up the following before drawing that conclusion, and am not about to delete it… Lazy readers should skip to the last paragraph.

There is isobaric (constant pressure) expansion going on in the cold to hot high-pressure gas tubes of the CCHEX, and isobaric compression going on in the hot to cold low-pressure gas tubes. I claim the amount of work done in these two steps cancels exactly.

The work done in any isobaric step is the area under a horizontal line on the PV chart, always easy to evaluate. It is the constant pressure times the change in volume. Using PV = nRT, it follows that P deltaV = nR deltaT. But n, R, and deltaT are constant regardless of pressure. So P deltaV is constant for both the low pressure process and the high pressure process, except the sign is reversed, making them add to zero.

The cold gas absorbs heat which increases its temperature. Since the volume does not instantly change, the pressure increases. But the gas is able to do work. This increases the volume and reduces the pressure and temperature. Since the volume has increased, the temperature does not drop all the way back. The pressure does return to the local ambient. This process continues and is regarded as constant pressure. Heat is flowing in, somewhat less work is being done, and the internal energy of the gas increases to make up the difference. When kelvins are multiplied by 1/2 Boltzmann's constant, the result is joules — the internal energy of the average degree of freedom of the gas. So, isobaric expansion increases the internal energy of the gas as it warms, but not as much as isochoric (constant volume) expansion would.

Finally, running the spell checker on this post, I thought I noticed that trompe is in the wordpress dictionary, but trombe is not. Actually, I noticed tromp as a suggestion for trombe. But because of this, I searched some more, and discovered that a "Trombe Wall" is a kind of solar collector named after the man who popularized it. So the question of the correct name for this pump / heat engine is open. But it is harder to say "trompe pump" than "trombe pump". And I liked the French definition of trombe as a turbulent column of water or air. (That definition I found in a google book search for books where both trompe and trombe appear on the same page). But google translates trombe as "waterspout" and "colonne d’eau qui, poussée par le vent, tourbillonne en renversant tout sur son passage" as "water column which, pushed by the wind, whirls while reversing all on its passage."


One Response to “Is any work done by expanding gas lost in the TrombePump Heat Engine?”

  1. ambien Says:


    […]Is any work done by expanding gas lost in the TrombePump Heat Engine? « Archimedes Submerged[…]…

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