This is a nit, but the first word in the link is misspelled. Should be Who's Who. a contraction for Who is Who.
If a superconductor has zero resistance, can it be infinitesimally thin, and still transmit a useful amount of power?
Sorry, no. Superconductors have a limiting magnetic field that limits how high you can make the current density. Current carrying capacity increases as you chill the material below its nominal superconducting temperature. There are superconductors that work at LN2 temperature (77K) but do not have useful current carrying capacity at that temperature. This is why MRI machines still use expensive LHe.
Dan21 : Such comments, bare of any knowledge, are very displaced. Oxygen, usually the most active/energetic combustion partner does not deliver any extra energy. People usually calculate the energy content based on a full oxidation of the base material.
It's easy to explain with a thought experiment. Consider the design of a rocket upper stage, operating above the atmosphere. The energy needed to produce thrust is obtained by the combustion of fuel and oxidizer. The hot gas is expanded through a converging/diverging nozzle to maximize the exhaust velocity of the resultant hot gas. Without oxidizer, the fuel is just a cold inert liquid (solid fuel rocket motors are rarely used in upper stages). Maximum velocity is a function of mass ratio; that is the masses of the loaded stage and the empty stage. The mass of oxidizer is frequently over half the total loaded vehicle mass. (Even though most rockets run on the rich side of stoichiometric, the O/F mass ratio is still above 1 and usually between 4 and 7 for most rocket fuels)
As for: "People usually calculate the energy content based on a full oxidation of the base material.", that's the difference between the higher heating value and the lower heating value given for a fuel. The higher heating value assumes the water generated condenses to liquid and the lower heating value assumes the water remains a gas.
I'll repeat my comment: When comparing the energy released by nuclear reactions to chemical reactions, the mass of both fuel and oxidizer needs to be included. The fact that most terrestrial chemical reactions get free oxygen from the ambient air does not mean the oxygen used in the reaction is massless.
This was marvelous:
E. Beiting, “Investigation of the nickel-hydrogen anomalous heat effect,” Aerospace Report No. ATR-2017-01760, The Aerospace Corporation, El Segundo CA, USA, May 15, 2017.
Quoting from the paper: "
Time-integrating the excess power to obtain an excess energy and normalizing to the 20 gram mass of the ZrO2 NiPd sample yields a specific energy of 173 MJ/kg. Assuming that the active material is the 5.44g of Ni+Pd yields a specific energy of 635 MJ/kg. For comparison, the highest specific energy of a hydrocarbon fuel (methane) is 55.5 MJ/kg. The highest chemical specific energy listed [see Energy Density in Wikipedia] is 142 MJ/kg for hydrogen compressed to 700 bar. Based on these results, it is unlikely that the source of heat energy was chemical in origin."
The ratio above chemical processes is much larger than this, as the heat of combustion of the fuels does not include the mass of the oxidizing part of the reaction. Fuel cannot give up its energy without combustion, which includes the mass of the oxidizer.
One of the hard things in this was protection of the RF power amplifier. When the plasma snaps on, the impedance looking into the coil changes as a step, going from high impedance to low impedance. As you search the tuner to re-match into the lower impedance, I have had it burn out a power transistor in the output of the amplifier. Because of this, I put power resistors in the output of the amplifier to limit the range of possible impedance swing seen by the amplifier output transistors. While adding these resistors reduced the maximum achievable output power and reduced the RF conversion efficiency, it has been successful in preventing damage to the amplifier during tuning of the load.
Are these wirewound resistors? If so, each will add another lump of inductance into the equation. That may or may not be important depending on whether they are counted in the 'heat input' part of the experiment.
Containing plasma is no joke but not all of the fusion designs have first wall material in close proximity to plasma.
I was under the impression that (at least the D-T folks) were planning to use a lithium blanket for a first wall. That way they get the tritium needed for the main reaction, and stop the neutrons without making much radioactive waste.
To me, it's a question of where the budgeted government money goes. On one hand, we have ITER and NIF, both with huge amounts of government funding, and both always 20 years in the future. When I was graduating from college in 1974, the career advice was "The science of fusion plants is done, it's all engineering now" That may be true, but insurmountable engineering is just as bad as insurmountable science.
On the other hand, we have many underfunded small fusion projects, both "hot" fusion, and "cold" fusion. They are all struggling and we only get to sit on this forum and cheer them on.
P-11B fusion is being pushed by other organizations as well. P-11B is particularly interesting because unlike D-D or D-T hot fusion, there are no high energy neutrons produced in the nuclear reaction. Tri-alpha Energy is one: https://www.ialtenergy.com/tri-alpha-energy.html
We can do better than this. W'd expect a low efficiency for a small fan like this. Let us use the manufacturer's data to determine the efficiency. For the given fan (Figure 3 GSVIT or here for whole datasheet). we have an electrical power of 7.2W nominal at 12V for the given pressure/speed graph. power out is proportional to pressure times spped. Eyeballing this for the maximum efficiency point on the operating curve (which as I've said before we do not get, and GSVT indicate this so justifying their point) we have:
0.4 m3/min, 100Pa. Converting to SI units we get 0.0067 m3/s, 100Pa. We finally need the air density, which is in SI units conveniently 1. therefore the power is pressure x airflow x density = 0.67W.
Err, no. Pumping power is Vdot deltaP where power is (in SI units) Watts, Vdot is volumetric flowrate m^3/sec and P is Pascals (N/m2). It doesn't matter what density the fluid is. It could be water or high altitude air. This is why small car engines get better gas mileage than the same car with a bigger engine; the throttle is open wider and the pumping losses are less. Fortunately for your argument, you used air density as 1 and did not affect the equation. A check on this is a dimensional analysis. Vdot deltaP is m^3 sec-1 X N m-2, which is J/sec or Watts. Adding density to the equation would add an incorrect kg/m^3.
Sorry about the nit picking, and BTW, propellers for light airplanes are about 70% to 80% efficient.
Getting back to the Celani paper, it looks to me that they are measuring heat out by calibrating two measured temperatures and reading those temps during the experimental runs. In other words, first they ran a series of measured power with inert gas, then ran the same experiments with gas that showed higher calibrated temps than previously measured. Am I correct in that?
As Me356 suggests, perhaps it would be better to subtract the power for the pump. This would bump up the COP curve a little bit, but not very much. When he gets the reactor working like he previously described, the power for this pump will be insignificant.
I think you should keep the pump power in the equations. Pump power goes to adding energy to the water; energy that eventually shows up as heat. There's friction on the tubing walls, and internal eddy currents that are friction damped.
Gas "pressure" in condensed matter is an interesting quantity. While the gas is monatomic and interstitial, it is more of an alloy and there is not any real gas pressure. When a vacancy in the lattice is filled with more than one atom of hydrogen, then there is "gas" and gas "pressure". The pressure is related to how tightly the orbitals of the molecular form of the gas have been compressed to fit into the vacancy. Historically this was related to explosions of steam boilers. The cast iron boiler would have large bubble-like vacancies in the metal. Some monatomic H would enter from the steam into the iron lattice and would dissolve into the metal. When the monatomic H found a physical vacancy in the iron, it would stay there and accumulate with other H atoms forming H2 molecules. The pressure inside these vacancies in the iron could become >100k PSI and would eventually cause so much internal stress that the boiler would just explode under the combined stress of the internal vacancies and the pressure from the applied steam.
I think you may be using the term 'vacancy' incorrectly here. A vacancy in a lattice is an atom missing from the lattice that can be occupied by a range of things e.g. a similar size and valance metal ion. H ions normally occupy interstitial sites, not vacancies, which is why the loading factor approaches 1:1. H2 gas does not normally collect around a vacancy to cause hydrogen embrittlement. That is caused by H2 gas collecting in cracks, most usually grain boundaries.
For simple differential thermometry, consider wiring two thermocouples in series, with the polarity reversed. The output of the series string will then be proportional to the temperature difference between the two junctions. It is necessary to use a pair of thermocouples that are closely matched across the entire expected temperature range. as an alternative, the system can be calibrated with the two junctions at the same temperature, so the output vs temperature will give a calibration curve for post-correction of the data.
For details, see http://www.nutechengineers.com/dtmwt.html
There are at least two ways to get well-matched thermocouples. One is to pay the extra money to get matched pairs from the supplier, and the other is to make your own. Use from wire that came off the same roll, and preferably wire that came off immediately adjacent during unrolling.
Making your own thermocouples is easy. For wire bigger than about 24 gauge, bare about 1/4 inch of the wire from the insulation. Give the bare wires about 1.5 to 2 full twists. Get your oxy-acetylene torch to a neutral flame and weld the wires into a bead. For small gauge wires, it takes less than one second, so slowly pass the wire through the flame without stopping. It takes some practice, but practice only uses about 1 cm per try.
For wire smaller than about 20 gauge, I have had excellent success with a Variac (at least 5 amp rating). Run one of the Variac leads to a scratch pad. The scratch pad is a steel or copper or stainless steel plate about 5 cm X 5 cm X .3 cm thick with a hole drilled or tapped for the Variac lead. Bare and twist both ends of the thermocouple wires. Attach the other Variac lead to one end of the thermocouple (alligator clips work for this), and physically scratch the other end of the wires across the scratch pad. Try it several times and you will get a good feeling for the correct Variac setting for each size thermocouple wire. Attach a piece of tape as a pointer for each wire size. I have made good thermocouples with wire as small as 40 gauge with this technique. This might work with a voltage regulated power supply, but I have never tried it.
Now go pat yourself on the back for saving money compared to buying pre-made thermocouples, but of course you have to have the acetylene torch and/or Variac. I've only made type K thermocouples this way, so I can't speak for the ability to make other types.
Today its cheaper to buy a whole microwave oven than the ignitioncoil, which would require grid transformer anyway. And the ignition coil will get destroyed with few watts of power, because it's cooling is very bad (it's filled with asphalt).
Older automotive coils were asphalt impregnated. Newer ones are epoxy potted. Some new automotive engine designs have a coil plugged directly onto the spark plug, so there are no plug wires.
I can't argue with something that works. Just trying to be helpful if you saw a leak in the fitting.
Yes, but this is 1/16" OD tubing. It came from McMaster Carr. I just sent them a picture and asked them to send me new tubing and 2 new ferrule sets for my compression fittings. I am sure they will respond quickly. In the mean time I have just put some JB Weld over the holes. Hopefully that will work and tomorrow I an re-start the experiment. That is, heaven forbid, I don't find yet another leak!
I can't tell from the text here, but be careful with compression fittings. Swagelok is NOT compatible with generic fittings. It may work for awhile, but not as reliable as staying within one manufacturer. You have to call them, but the folks at McMaster are good at identifying the manufacturer for individual things that they sell.
BobHiggens said: "I don't understand the comment about some being better than others. If it is not LiAl4,
it is not going to decompose when it is supposed to. Somebody must have
gotten some oregano instead of the real stuff. But, you can be sure if
it comes from Alfa that it is the real stuff."
H2 for automotive use is sometimes stored in the car with metal hydrides. Over the range of 0 to 300C and a few atmospheres, hydrides can store and release more H2 than can be stored as a liquid. LH2 density is only .071 kg/liter. Maybe some of that is taking place.
Go Dan, Please complete this idea on how the reaction would be started/stopped and what would take it to happen. U B tink'n sweet
Startup is relatively easy to explain. The interstitial sites (of which there are many, many compared to the dislocation sites) must be filled first, as they are lower energy sites but are not NAE. Explaining the stopping is more difficult. Could be local overheating from LENR destroying the NAE, or the insoluble He blocking the Deuterium, or maybe some other nuclear low probability process poisoning the NAE, such as the creation of Tritium or 3He, or there's enough local energy to knock a Pd atom into the wrong place which would block the resonance.
It's also hard to explain how other NAE are created during LENR activity. But I have a few ideas on that too. Maybe if the required length of the dislocation is 3 or 4 atoms long, there is a possibility of the lattice vibrations during LENR creating the required morphology. I'm personally not comfortable with the lattice being able to absorb 2 MeV per event without breaking up. Maybe it *does* break up but usually heals itself.
One error not to do in considerin Edmunt Storm's theory is to criticize it to the details to refute it totally.
It is not an upfront theory like many theory I've read.
It is just an experimentator, looking ate data, eliminating what is impossible, refuted, not observed, to propose a narrow corridor of possible theories...
Here we have an experimentalist approach, like Sherlock Holmes, Agatha Christie (or CSI), where common sens is stated, and after removing the impossible, only remains the improbable .
I have an alternative idea to the ‘nanocrack’ location for the NAE. Assume a short edge dislocation in the Palladium. For those of you not metallurgists, visualize a 2 cm wide putty knife sticking into the middle of a 10 cm cube of Jello. Now visualize that with both putty knife and Jello as being Palladium. [That’s an edge dislocation that Axil posted on Thursday] The Deuterium nuclei bounce back and forth along the dislocation and rebound where the dislocation stops in both directions. If there are just two nuclei there, they would bounce off each other with a random speed distribution, then bounce off the ‘wall’ at the end of the dislocation. Or maybe their energy and repulsion that holds them there are such that they occasionally pass each other. During the moment of pass, they be pushed together by the lattice to fuse. That's the key: The 2 nuclei pass each other in the correct location that the Pd lattice pushes them together.
This concept embraces Storms’ NAE and Hydroton ideas, as well as McKubre’s comment that the system needs to ‘breathe’ which would pump deuterium nuclei in and out of the volume. Reaction rate is determined by how fast the D nuclei and He atoms diffuse through the metal. Furthermore, if the D preferentially occupies interstitial sites rather than the dislocation, it explains why loading needs to be high. Finally, invoking the requirement that the dislocation has both a minimum and maximum length, explains why LENR is rarely observed.