I wrote the text at the end (below) a couple of days ago in a comment box on halfbakery, but never posted it. I was getting ridiculous amounts (thousands of kilometers) of tubing needed, assuming low pressure operation at several meters of mercury. I have determined that it will help to operate the pump at maximum possible pressure, and I plan to reduce the amount of material required for tubes by keeping the tubes inside a large pressure vessel and surrounding them with fluid at the same pressure as that inside the tubes. I think the fluid should be mercury and the tubes glass. They can be thin glass if they are surrounded by mercury as well. Heat will be carried to the tubes by flowing mercury. Heat will get into and out of the mercury at the top of the machine where the pressures are much lower. The gas will never come near the top of the machine, because the number of moles of gas needed takes up too large a volume at low pressure.
It would have been nice to have expansion tubes surrounded by low-pressure refrigerant which would be condensing on the tubes to keep them hot, but it appears that is not to be.
With really high pressure, Van der Walls' equation applies,
(P + n^2 a / V^2) (V – n b) = nRT instead of PV = nRT.
For helium, a = 0.03412, b = 0.02370, while for nitrogen, a = 1.390 and b = 0.03913, with a in liters^2 atm / mole^2, b in liters / mole.
So if you use helium working gas, pressures can be about 30 times higher to get about the same non-ideal effects as nitrogen. I haven't worked out exactly how this affects things.
The important point is that to produce a given power p using pressure ratio r = P2/P1, the number of moles of gas per second entering and leaving the expansion hyperbola must be p / (R T ln(r)), with R the molar gas constant and T the temperature in kelvins. This doesn't depend on pressure, just the pressure ratio. For a pressure ratio of 2, heat power of 33 kW to get 1kW out, and temperature 300K, we need 19 moles/second gas flow, and have to handle this on a budget of $5000.
Original text never posted to halfbakery:
Here are some calculations per kW of capacity.
Suppose we have 90K difference between hot and cold temperature reservoir, but we expect to maintain only 30K difference between the compressed gas and the expanding gas. That is, we allow a 30K temperature drop from the outside of the hyperbola to the gas inside. This gives very fast heat flow. Also we are going to have a high surface area for the hyperbolas required for 1kW. So it is reasonable to expect to get rapid flow of heat into and out of the expanding and compressing gas.
Given 10% of heat flow goes to work (30K/300K), and 30% of work goes to electricity (quite conservative, I think), I need 33 kW of heat flow to produce 1kW of electricity. If the pressures are 250 psi and 500 psi, so the machine is about 13 meters tall, then one mole per second of air represents (1 mol / sec) (R) (300K) ln(500 psi / 250 psi) = 1.73 kW, and 19 moles per second represents 33 kW. The air flow is 28 liters/sec at 250 psi and 14 liters/sec at 500 psi.
The bubbles injected at the bottom of the expansion tube expand to twice the length by the time they reach the top of the hyperbola, while the slug of mercury between bubbles stays the same length.
So if we want the low pressure air slug the same length as a mercury slug, we need 28 liters/sec mercury flow to carry 33 kW of heat producing 1kW of electricity. Suppose the throughput of the hyperbola is 1 second for 33 kJ of heat to flow into one second's worth of gas. Then the hyperbola has to hold 28 liters of mercury and somewhat less of air because its volume changes, say 21 liters of air, so round that to 50 liters total volume of the expansion hyperbola and another 50 liters for the compression hyperbola.
Thus, allowing 1 second for transfer of 33kJ of heat into the expanding gas and another 1 second for the transfer of 30 kJ of heat out of the compressing gas, we need 100 liters inside volume of hyperbolic bubble pump tubing to produce 1 kW of electricity. If the cross section is 10 mm^2, that volume fits inside 10 km of tubing. If each hyperbola is 20 m, we need 500 hyperbolas, half compression and half expansion. Now a nuclear steam plant gets a budget of $5000/kW. Matching that, we need to build each hyperbola for ten dollars. What does 20m of 500 psi rated tubing cost?
Further, look at the surface area of these 500 hyperbolas. It may well be that the actual flow rate could be higher than 50 liters/second, or 100 mL / sec per hyperbola.