Archive for May, 2006

Heat flow through walls of copper tube


Today I calculate how much heat will flow through the walls of a standard 2 inch copper tube, when the temperature difference is 1 kelvin. It turns out to be quite a large number.

Thermal conductivities at 300K:

Properties of 2 inch nominal copper tubing from

  • Outside diameter 2.125 inch
  • Inside diameter 1.959 inch
  • Wall thickness 0.083 inch
  • The average diameter is (2.125inch+1.95inch)/2 = 2.0375 inch.
  • The circumferance is (pi 2.0375 inch) = 6.400995 inch.
  • The surface area of a 10 foot tube is (pi 2.0375 inch)(10 foot) = 0.49555991 meter^2
  • The heat flow in kilowatts per kelvin is (401 W/m K)(pi 2.0375 inch)(10 foot)/(0.083 inch) = 94.260282 kW/K

Thus, a 2 inch by 10 foot copper tube with 0.083 inch wall thickness will pass 94 kilojoules of heat per second with a temperature drop of only one kelvin.

The volume of this tube is (pi (1.95 inch)^2) (10 foot) = 23.490999 liters.

Calculations as usual are done using the Gnu units program.

$ units -V
units version 1.80 with readline, units database in /usr/share/units.dat
$ grep Version /usr/share/units.dat
# 16 June 2002 Version 1.34

Vanilla cinnamon dark chocolate


I prepare a mixture of unsweetened chocolate, cinnamon, and vanilla for breakfast each day. It doesn't raise my blood sugar much. It takes some getting used to, but no more than black coffee did. I no longer consume any caffeine other than whatever might be in the chocolate, which apparently is not much.

I do not recommend starting out with this quantity. This much chocolate has effects on the nervous system of the gut as well as the brain. I drink lots of water. Hardly any used to be left in my gut, even though I also used to eat a lot of wheat bran. Now a reasonable amount remains, sometimes a little extra, and I had to drop the wheat bran, hence the warning to start slow and make necessary adjustments. I also observe an antidepressant effect.

Here is how I prepare it:

1. Turn electric stove on high, put empty stainless steel teakettle on burner. Add 1 cup water.

2. In small glass mixing bowl (I use a dessert bowl but would like something just slightly larger because it spills sometimes) put 2 teaspoons of cinnamon, and 2 squares (2 ounces) of Kraft Baker's unsweetened chocolate (that's what my grocery carries, the orange box). Put in microwave full power 2 minutes (1500 watt microwave, I think — it's about the biggest microwave you can run on a 15 amp circuit).
3. Use a whisk to mix the melted chocolate and cinnamon. Turn off the stove and add boiling water. The chocolate is usually heated hotter than boiling water, and a little water will boil away when added to the chocolate. Mixing with the whisk generally ends up with a ball of chocolate inside the whisk and I have to bang it on the bowl to get it out. I plan to try a bigger mixing bowl so I can just dump all the water in and stir rapidly with the whisk without splashing it all out. Then it shouldn't dry out.

4. Add 2 capfulls of artificial vanilla and stir more. Natural vanilla is about 8 times the cost and I can't taste the difference. But I really do notice it when I forget the vanilla. I don't add the vanilla at first because I'm not sure it takes the heat well, but I don't actually know that it makes any difference.

Icerocket tag: . The icerocket tag was so hard to add I wouldn't do it again.

wordpress tag: dark-chocolate. I copied the links below from there.


Technorati Tag

Icerocket Tags. Nothing.  Never heard of them before I saw it on the wordpress tags page.
Flickr Photos  So far, only tags on are for clothing of that color.  Ironic. 

Wink.  Wow.  Wink has a lot of interesting links on this page. 

There were also some interesting things on the wordpress chocolate tag.  WordPress tags get created from the categories in a blog.  All blog items posted to a category also get posted to that global tag.

Ball bearings in the bubble pump


Thinking of ways to reduce the number of small diameter tubes, I am considering adding ball bearings to the fluid. These would be nearly the diameter of the hyperbolic tubes, which must then have uniform circular cross-section. Bubbles would be injected after every other ball bearing. Making the bearings out of tungsten or depleted uranium would increase the hydrostatic pressure possible in a given height. Tubes could be considerably larger diameter without risk of fluid (i.e., the bearings) leaking between slugs of air or air leaking between slugs of fluid and bearings. The fluid would be some kind of oil, which serves as sealant and lubricant, and if a layer of oil on the tube surface flows slowly or not at all, it doesn't matter because it doesn't weigh much.

Still working on the previous post, calculating gas flow rates, heat flow rates, and volumes. I hope using ball bearings will improve the situation considerably. It does mean we can't claim no solid moving parts, but that was just a slogan.

Sizing a 30kW gross heat flow bubble pump air compressor


This design uses water and air and ignores any difficulties with water vapor and other than perfect slug flow in the tubes. Also, it aims for the given arbitrary target of 30kW gross heat flow into the expansion process. The available work will depend on the temperature difference, and on efficiencies. However, 30kW of heat flow ought to be enough to produce 1kW of work, which is probably enough for one family. Note that the amount of work you can extract from a given amount of compressed air depends on how much heat you put into it during expansion, and the starting temperature. Here, the work represents how much work went into compressing the air, not how much might come out of it.

We assume a pressure ratio of 2: the low pressure reservoir is at half the pressure of the high pressure reservoir. If the machine is 30 feet tall, the pressures are 13 psi and 26 psi. Higher pressures with a 30 foot machine would reduce the pressure ratio.

The required moles per second of air to handle 30kW of heat flow is 30kW = nRT ln(pressure ratio). Using gnu units, this gives

  • 30kW / (R 300K ln(26psi/13psi)) = 17.351613 mol / s
  • (29 g / mol) (30kW / (R 300K ln(26psi/13psi))) = 0.50319679 kg / s.
  • 30kW / (R 300K ln(100psi/87psi)) = 86.363948 mol / s

Substituting this formula for the number of moles (which assumes isothermal expansion) into V = nRT/P, the temperature and gas constant cancels out, and the volume per second at a given pressure is just

  • volume / second = power / (P ln(ratio)).
  • 30kW / (26 psi ln(26psi/13psi)) = 241.4368 liters / second
  • 30kW / (100 psi ln(100psi/87psi)) = 312.44202 liters / second

This is the required air flow volume into the expansion hyperbola at the high-pressure end. The gas expands throughout the hyperbola so the total volume will be larger. This value can be used as volume per unit height, as the expansion is accomodated by bending the tube to be more nearly horizontal. The space required for the slugs of liquid is accomodated in the same way. A reasonable minimum amount of liquid would provide equal volume slugs of gas and liquid at the low-pressure end.

The required volume of the expansion hyperbola depends on the expected rate of heat flow per unit area of hyperbola tubing surface, or on the maximum velocity of the fluid and gas, whichever gives the more stringent limit.

Considering 1 meter / second (2.2 mph) a reasonable velocity for the fluid, and 30 feet (9.14 meters) a reasonable height, we first need the hyperbola length.

Still working on this…

Abandoning mercury, again


I have been writing about using mercury as the working fluid in the bubble pump for two reasons:

  • Mercury does not wet glass, so achieving ideal flow of separate slugs of fluid and gas is feasible.
  • Mercury has density over 12, so the pressure produced by a given hydrostatic head is 12 times higher, and the volume required to contain the resulting compressed gas is 12 times smaller than if the fluid were water.

However, there is not much mercury available compared to the amount of water available. Dealing with high pressure is difficult and dangerous. Getting heat to flow through the high pressure container is difficult. Mercury liquid does not compare to a refrigerant in carrying heat. Using low pressure hyperbolas permits the use of thin walls surrounded by refrigerant to carry heat in or out of the gas. And the size reduction possible due to high pressure may not result in a cost reduction because of the necessary materials.

So, I now plan to use a cheap fluid like water, some tubing the fluid does not wet (in the case of water, something hydrophobic), some patterned method of making many hyperbolas at once in a flat sheet with parallel tubes, and refrigerant to carry heat between the heat source or sink and the tubes, evaporating from the hot surface, rising to the cold surface, condensing, and flowing back down to the hot surface.

The machine pumps air from a large low-pressure reservoir into a large high-pressure reservoir, at perhaps double the pressure, unless there is a need for producing air at a particular output pressure. It has a simple layout. Bubbles and fluid flow up through the expansion hyperbolas into a separation tank. The top port feeds the hot low-pressure gas countercurrent heat exchanger, and the bottom port feeds the hot low-pressure fluid countercurrent heat exchanger. The cold low-pressure gas output feeds the low-pressure gas reservoir and the cold low-pressure fluid output feeds the top of the compression hyperbola. Low pressure gas is periodically injecteded into the descending fluid at the top of an inverted U before the nearly horizontal entrance of the hyperbola. The hydrostatic pressure at that point is lower than the gas pressure in the low-pressure reservoir.

At the bottom of the compression hyperbola, the bubbles and fluid separate in the high-pressure separation tank. The cold high-pressure gas and cold high-pressure liquid enter the high-pressure sides of the two countercurrent heat exchangers. The hot gas output feeds the high-pressure reservoir, while the hot liquid feeds directly into the expansion hyperbola. High pressure gas from the reservoir is periodically injected into the rising fluid in the expansion hyperbola, at a point where the hydrostatic pressure is lower than the reservoir gas pressure.

The rate of fluid flow is controlled by adjusting the rate of gas injection into the expansion and compression hyperbolas. Work is produced when excess high-pressure gas is allowed to expand in a turbine or other mechanism.

Using water vapor to concentrate and to disperse heat


See the comment. I spoke too soon about the water vapor idea: it doesn't help much.

An interesting posting on by Pranab Jyoti Ghosh led to my response. I mirror it here:

Hello Pranab,I have examined your description, and I see a fatal flaw: you have no cold sink. You hope to create one by allowing compressed air to expand, but it just doesn't happen that way. The condenser will get too hot to condense the water vapor.

On the other hand, the idea of using some work to pump water vapor out of a container of water vapor over water, lowering the water temperature so that heat will flow in from the surroundings, is a good idea. Nearly 1000 times more heat flows in than was invested in work. I can easily believe this.

The problem is how to make use of this latent heat. The water vapor coming out of the vacuum pump will be at approximately the vapor pressure of water at the ambient temperature. Thus, it will not condense on a surface at ambient temperature and deliver heat to it.

What you have really done is to concentrate heat from the surroundings into the water vapor at the same temperature as the surroundings. (I'm ignoring your idea for heating up the water vapor. That just isn't going to work.)

Concentrating heat is a good thing. It increases the heat flow which can be extracted from the surroundings because heat flow goes as the temperature difference, and investing some work in pumping water vapor results in a lower temperature and more heat flow.

Now if you have a lot of compressed air stored up from the night before, you can allow the compressed air to expand doing work, and let the water vapor condense on the tubes holding the compressed air, warming it back up so you get isothermal expansion. You save up the work in some form of potential energy like pumped water. The work available per mole of compressed air is RT ln(Pmax/Pmin), with T the daytime temperature.

At night, you reverse the process. Using work extracted from descending water, you compress air. You cool the tubes containing the air by using the vacuum pump to pump water vapor out of a container, cooling the water, and use the cold water to cool the tubes. By this method you extract heat from the tubes faster than merely exposing them to the ambient temperature would do. The result will be an excess of compressed air, which is useful for many things, or else and excess of pumped water, also useful.

The amount of excess energy available depends on the temperature difference from day to night, and on the capacity of the compressed air and pumped water storage. If night is 275K and day is 305K, you need to store 10 times the output energy as raised water during the day, and as compressed air at night. (The energy isn't really in the compressed air, but was dissipated into the surroundings at night in the form of the latent heat of water vapor exhausted from the vacuum pump).

This text will be mirrored at the Renewable Energy Design wikia, and at Archimerged's blog.

Sustainable power from heat engines.


This was originally posted on the Development and Sustaniability Wikia at Sustainibility.

It has been known since Sadi Carnot's work in the 1830's that heat engines produce work from the transfer of heat from a hot reservoir to a cold reservoir, without any requirement for fuel. Small toy Stirling cycle engines exist which will sit on top of a TV set and turn using heat energy from the TV, but the torque is low and not much energy is involved. Recently a 55kW Stirling engine / generator has been developed, but it still needs a large temperature difference between the hot and cold reservoirs.

On Renewable Energy Design ideas for heat engines are being discussed (mostly by one person). See also a discussion on halfbakery (where most readers didn't get the point).

The points which everyone should understand are these:

  • confined gas can be used as an energy sponge, and
  • there is a vast amount of energy available at moderate temperatures. The total power from the sun is 174 billion megawatts.

Energy from the sun is captured by Earth's atmosphere in the daytime, exists as heat for a while, and radiates back out into space at night. This amount of energy is so large that a method which can capture only a tiny fraction of it would produce far more energy than we need. So do not scoff at designs which extract a microscopic amount of work from a huge heat flow. The necessary heat is available in the atmosphere.

A few facts are important:

  • Gasses are made up of molecules. These may be regarded as tiny billiard balls.
  • The temperature of a gas is directly proportional to the average kinetic energy of these tiny billiard balls.
  • The gas is to be confined inside a solid cylinder between two liquid pistons.
  • The cylinder and the liquid are also made up of molecules. The temperature of the cylinder and the liqud also reflects how fast the molecules are vibrating.
  • When a gas molecule strikes the cylinder wall, it is most likely to end up moving with the kinetic energy specified by the temperature of the wall.
  • Since the cylinder wall cannot move, the repeated collisions of gas molecules do not result in any coherent motion of the cylinder wall. They only result in heat exchange until the temperatures are equal, with no work exchange.
  • Since the pistons can move, repeated collisions of gas molecules do result in coherent motion.
  • If the pressure of the gas is lower than the pressure applied to the other side of the piston, then the piston will move to confine the gas in a smaller volume.
    • The gas temperature will increase to reflect work done by the piston on the gas.
    • The energy of the work then flows as heat into the cylinder walls and away into the heat sink.
  • If the pressure of the gas is higher than the pressure applied to the other side of the piston, then the piston will move to allow the gas to expand.
    • The gas temperature will decrease to reflect work done by the gas on the piston.
    • Energy required to replace the energy used for work flows as heat from the cylinder walls into the gas.
    • Heat flows from the hot heat source into the cylinder walls to replace the heat which flowed into the gas.

Thus, we conclude:

  • If the applied pressure on a confined gas is slightly lower than the gas pressure, then
    • the gas will expand doing work, and
    • its temperature will drop.
    • Heat will flow from the container walls into the gas until the temperatures are equal.
    • Heat will also flow from the hot heat source into the container walls.
    • The heat which flows into the gas this way is being converted to useful energy.
  • If the applied pressure on a confined gas is slightly higher than the gas pressure, then
    • the gas will be compressed, absorbing work, and
    • its temperature will increase.
    • Heat will flow from the gas into the container walls until the temperatures are equal.
    • Heat will also flow from the container walls into the cold heat sink.
    • The heat which flows out of the gas came from useful energy doing work on the gas.

Now if the compression and the expansion take place at the same temperature, there is no net gain of useful energy, since all of the energy produced by expansion would be used up compressing the gas again. But if the cold heat sink is colder than the hot heat source, it takes less work to compress the cold gas than will be extracted when the hot gas is allowed to expand. So there is a net output of useful energy.

In more detail:

Given a clever enough arrangement of pipes and valves and liquid and gas, energy can be extracted from a warm heat source and an amount (depending on the temperature difference) of it converted to work, while the rest is rejected into a cold heat sink. The fraction extracted as work is probably less than one percent of the heat flow, but that heat flow exists whether we capture any or not.

What is this clever arrangement?

  • Mercury serves as liquid piston. (The earth's mercury has to be somewhere. Why not collect it into this machine?)
  • Many small diameter glass tubes serve as cylinder for the liquid piston. Together, they are called a manifold.
  • The tubes are kept in a pool of mercury, so that the hydrostatic pressure outside the tubes is equal to the pressure inside. Therefore the tubes can have thin walls.
  • The tubes are bent into the shape of a hyperbola. They are nearly vertical at the bottom and nearly horizontal at the top.
  • The shape of the tubes is dictated by the equation of state of the gas.
  • For a non-ideal gas, the tubes would deviate very slightly from hyperbolic shape.
  • Gas bubbles are confined in the glass tubes between moving slugs of mercury.
  • Glass is used because mercury is repelled from the glass surface so that when gas bubbles separate slugs of mercury in the tube, the slugs remain separate.
  • The gas and liquid move through the tubes as a unit, at exactly the same rate of speed.
  • Two manifolds are used in the machine.
    • A hot expansion manifold has hot mercury and bubbles of hot gas moving upward through hot glass tubes.
    • The hot manifold is in a pool of hot mercury which carries heat from the hot heat source to the manifold.
    • A cold compression manifold has cold mercury and bubbles of cold gas moving downward through cold glass tubes.
    • The cold manifold is in a pool of cold mercury which carries heat from the manifold to the cold heat sink.

How do the bubbles get created?

  • One glass capillary tube is positioned in the mercury flow inside each glass manifold tube.
  • The hydrostatic pressure at that point is equal to the distance upward to the mercury surface, plus the gas pressure of the gas above the mercury.
  • The pressure of the gas entering the capillary tube is slightly higher than the hydrostatic pressure at the end of the capillary.
  • Hence the gas will flow out of the capillary tube into the mercury flow.
  • After a bubble of the proper size has formed in each tube of the manifold, the gas pressure is reduced to slightly below the hydrostatic pressure and a slug of mercury is allowed to form.
  • The process continues indefinitely.
  • Hot high pressure gas is injected through small diameter tubes into the upward flow of mercury inside glass tubes at the bottom of the expansion manifold.
  • Cold low pressure gas is injected into the downward flow of mercury inside the nearly horizontal glass tubes at the top of the compression manifold.

How does the confined gas serve as an energy sponge?

  • Cool gas absorbs heat from warmer tube walls.
  • Hot gas loses heat to cooler tube walls.
  • One set of long curved glass tubes serves as the expansion manifold.
    • The outside of these tubes is in thermal contact with the hot heat source.
    • The tubes are nearly vertical at the bottom, and nearly horizontal at the top.
    • Hot mercury flows upward constantly into this manifold.
    • At the top, mercury flows into a closed reservoir with gas confined above the surface.
    • Hot high pressure gas bubbles are injected into the mercury stream at the bottom of the manifold.
    • Slugs of mercury separate the gas bubbles, and remain separate as the stream moves up the tube.
    • The gas bubbles up a short distance to the mercury surface as the mercury leaves the manifold and enters the reservoir.
    • The hydrostatic pressure at any point in a tube is equal to the vertical distance up to the mercury surface, plus the pressure of the gas above the mercury.
    • The volume of the gas bubbles will change whenever the gas pressure is not equal to the hydrostatic pressure.
    • As the mercury and gas rise through the long tubes, the gas expands and cools.
    • The level of mercury in the reservoir rises slightly because of the gas expansion.
    • As the gas particles collide with the warm tube walls, they regains heat.
    • This process continues for the duration of the trip through the expansion manifold.
    • If the pressure at the top is half the pressure at the bottom, then the volume of a bubble at the top is twice the volume at the bottom.
  • The hot low pressure gas flows into a countercurrent heat exchanger, exchanging heat with cold high pressure gas.
  • The hot mercury flows into another countercurrent heat exchanger, exchanging heat with cold mercury.
  • Another set of long curved glass tubes serves as the cold compression manifold.
    • The outside of the tubes is in thermal contact with the cold heat sink.
    • The tubes are nearly vertical at the bottom, and nearly horizontal at the top.
    • Cold mercury flows from the countercurrent heat exchanger into the compression manifold.
    • Cold low pressure gas is injected into the cold mercury stream just after it enters the individual small nearly horizontal glass tubes of compression manifold, creating a downward moving stream of gas bubbles separated by slugs of cold mercury.
    • At the bottom of the compression manifold, the gas bubbles are collected in a tank which leads to the gas countercurrent heat exchanger.
    • The cold high-pressure gas is warmed at constant pressure using heat from the hot low-pressure gas which is being cooled.
    • After being warmed, the resulting hot high-pressure gas flows into the capillary tubes at the bottom of the hot expansion manifold, creating bubbles as described earlier.
    • The cold mercury flows into the bottom side of the mercury countercurrent heat exchanger, to be warmed up before entering the hot expansion manifold.

How is useful energy actually extracted?

The short answer is that the flow of mercury would speed up indefinitely unless excess energy were extracted. Energy can be extracted by allowing mercury to flow downward without compressing gas but instead turning a turbine, for example.

The machine as described has four tanks of mercury.

  • The two upper tanks are low-pressure hot and cold.
  • The two lower tanks are high-pressure cold and hot.
  • The cold upper tank and the hot lower tank
    • are fed from the output of the mercury heat exchanger,
    • contain only mercury, and
    • feed into the corresponding manifold.
  • The hot upper tank and the cold lower tank
    • are fed from the output of the corresponding manifold,
    • contain both gas and mercury, and
    • feed from the bottom into the appropriate mercury heat exchanger input, and
    • feed from the top into the appropriate gas heat exchanger input.
  • Mercury from the bottom of the hot upper tank flows through the countercurrent heat exchanger to the cold upper tank.
  • Mercury from the cold upper tank flows down through the compression manifold to the cold lower tank.
  • Cold low-pressure gas is injected into the nearly horizontal downward flow and is carried along with it.
  • The descending mercury does work on the cold low-pressure gas, compressing it.
  • The energy transferred to the cold gas raises its temperature causing heat flow to the tube walls, so all of the energy used compressing the gas is delivered as heat to the cold heat sink.
  • Mercury and gas separate in the cold lower tank.
    • Mercury flows through the fluid countercurrent heat exchanger and into the hot lower tank.
    • Gas flows through the gas countercurrent heat exchanger, through the capillary tubes, and into the expansion manifold.
  • Mercury flows from the hot lower tank up into the expansion manifold.
  • Hot high pressure gas from the countercurrent heat exchanger is injected into the expansion manifold at a point where the hydrostatic pressure is lower than the gas pressure.
  • As the gas expands, it does work on the ascending mercury.
  • All of the energy used to raise the ascending mercury higher than the descending cold mercury would push it comes from heat absorbed by the gas from the tube walls, which came from the hot heat source.
  • Mercury and gas separate in the hot upper tank.

Calculation regarding isothermal bubble pump


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.

Patents, copyrights, and slavery


Patents and copyrights have many things in common with slavery.  For example, all three were sanctioned by the original U.S. constitution.  It took the civil war to get slavery removed from the constitution.  What will it take to get rid of patents and copyrights?

But I am getting ahead of myself.  All three institutions involve artificial ownership:  of human beings, of methods of doing something, or of words and sounds and images.  All were controversial when incorporated into the constitution.  All have evil effects on individuals for the benefit of the so-called owners.  And all are unnecessary.

It is for this reason that I do not patent my inventions or apply restrictions to the use of my writing.  I find the above self-evident, but I expect it may be a long time before patents and copyrights cease to exist.  But clear thinking people can refuse to participate in  artificial ownership which places restrictions on the freedom of others.

Injecting bubbles into moving fluid


The isothermal bubble pump is very simple, but not so flexible. There is no control over the spacing or size of bubbles once they are created when the machine is manufactured. So I propose a machine made of two hyperbolic segments of bubble pump with injectors at the inputs and collectors at the exits. The injectors are under automatic control. One injector feeds cold low-pressure gas bubbles into the downward moving fluid stream at the nearly horizontal top of the cold compression hyperbola. The other feeds hot high-pressure gas bubbles into the upward moving fluid stream in the nearly vertical bottom of the hot expansion hyperbola.

Alternating slugs of fluid and gas flow through the two hyperbolas, but elsewhere in the machine, flows of fluid and gas are separate. Heat is exchanged between high and low pressure fluid using a fluid-only countercurrent heat exchanger. Since there is no gas involved, a U tube easily brings the hot low-pressure fluid down to the level of the high-pressure fluid and returns the cold low-pressure fluid to the original level. Incidentally, inside the heat exchanger the fluids have equal hydrostatic pressure; importantly, the pipes are in direct thermal contact.

Separately, heat is exchanged between high and low pressure gasses in a gas-only countercurrent heat exchanger which can be placed anywhere in any orientation. Typically, it would comprise a series of insulated low pressure tanks of descending temperature, filled with coils of high-pressure pipes flowing in the opposite direction so cold gas enters the coldest low pressure tank, absorbs some heat, flows to the next warmer low pressure tank, absorbs more heat, etc. The volume of these tanks is unimportant so long as very little heat is lost and there is very little net heat flow in or out either end. The amount of heat gained by the high pressure gas should be essentially equal to the heat lost by the low pressure gas. The more tanks provided, the longer the gas stays in the tanks, and the larger the volume (so the relative surface area is lower) the better this goal is achieved. A single heat exchanger can serve many pairs of bubble pump hyperbolas.

This machine does not require a jump-start because the bubble injectors and collectors are inherently irreversible, and it cannot function as a heat pump unless the temperatures of the heat reservoirs are reversed. The liquid always flows down the compressor and up the expander, heat always flows from heat source to heat sink, and the motor always runs forward, but if the heat source is colder than the heat sink, forcing the motor to turn forward will pump heat from the cold source to the hot sink. Forcing the motor to turn backward will just pump fluid around the circuit with no gas flow.

The bubble collectors are simply vertical tubes leading to gas reservoirs. These tubes have such large diameters that the bubbles detatch from the walls of the tube and float freely (and irreversibly) upward. The gas pressures are always high enough to prevent fluid flow upward, so fluid never fills the gas reservoirs or flows into the gas-only tubes above the reservoirs. Hot fluid leaving the nearly horizontal upper end of the hot expansion hyperbola takes a sharp bend downward to the hot input port of the fluid countercurrent heat exchanger, but the bubbles escape upward into a short vertical pipe leading up to a gas reservoir. Cold fluid and bubbles arriving at the bottom of the cold compression hyperbola flow around a sharp bend from nearly vertical to horizontal. After a short horizontal segment, the bubbles escape up into the high pressure gas reservoir while the fluid flows to the cold input port of the fluid countercurrent heat exchanger.

Hot gas from the hot low-pressure reservoir and cold gas from the cold high-pressure reservoir flow slowly through the gas countercurrent heat exchanger and into the cold low-pressure reservoir and hot high-pressure reservoir. From there, the hot high-pressure gas feeds the bubble injector at the nearly vertial bottom of the hot expansion hyperbola, and the cold low-pressure gas feeds the bubble injector at the nearly horizontal top of the cold compression hyperbola.

A positive displacement pump somewhere in the fluid-only circuit converts fluid flow to rotary motion or rotary motion to fluid flow, always in the forward direction (the pump cannot run backward). A motor-generator converts electricity to or from shaft motion.

Thinking of gas at the level of moving molecules, one can imagine how and why the expansion hyperbola provides motive force to the liquid. At the bottom where the gas bubbles are injected, the hydrostatic pressure is so high that the bubbles are just big enough to touch the walls of the tube, and they are not expanding. But the fluid is moving upward in the nearly vertical hyperbolic tube, and the hydrostatic pressure drops as the amount and weight of the fluid above decreases. So the gas expands, doing work on the slug of liquid behind and ahead. Molecules of gas collide with the liquid, accelerating the liquid and decelerating the gas molecules, so the temperature and pressure drop. Molecules of gas also collide with the hot walls of the tube, gaining energy so the temperature and pressure of the gas rise. The faster heat flows into the gas, the more work it can do on the liquid and the faster the liquid moves. Nearer the top of the hyperbola, the slope of the tube decreases, so that a larger increase in volume is permitted without reducing the pressure so much. The goal is isothermal expansion: the pressure-volume product is constant and all of the energy removed from the hot walls of the tube is used to accelerate the moving fluid while the gas stays at constant temperature.

Thus, if no energy were being extracted from the fluid motion, the fluid would continue to accelerate. This is generally true of heat engines — they speed up as long as heat (microscopic disordered kinetic energy) is supplied faster than energy is removed, storing the excess as macroscopic ordered kinetic energy. Here, the moving fluid plays the role of a flywheel.