@Bruce_H, please consider dependent and independent variables in your analysis. That is what I mean about confusing math and physics. I don’t think I will post anymore about this. The horse is dead
MIZUNO REPLICATION AND MATERIALS ONLY
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Perhaps you need some afterthought to check instantwrite ,,you at least got Planck right.
perhaps you can come up with a P constant too..
"Planck: "He was, by nature, a conservative mind; he had nothing of the revolutionary and was thoroughly skeptical about speculations. Yet his belief in the compelling force of logical reasoning from facts was so strong that he did not flinch from announcing the most revolutionary idea which ever has shaken physics.""
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Max Planck only shook the basic standard model of physics, not Einstein's Relativity! Let the Mathematics take over for a while and get on with practical issues.
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So you are saying the calorimeter makes excess heat by being better. So much for calorimetry.
I don’t see what you mean. The calorimeter that makes the highest temperature with the least amount of heat input will give the highest COP. You are just reinforcing my point that COP is meaningless. It’s Arbitrary. How about we set up a poll: let’s ask who thinks COP is meaningful and why. Then when we publish our next data, with any luck with a high COP, these same people are going to change their tune
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@Bruce_H, please consider dependent and independent variables in your analysis. That is what I mean about confusing math and physics.
OK. It is true that I don't understand what you are saying.
If this is new thinking arising from your work with these systems, then I suggest that when you write things up you should spend some time clearly explaining this idea because I think that others might also fail to understand.
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I don’t see what you mean. The calorimeter that makes the highest temperature with the least amount of heat input will give the highest COP.
How much more nonsense will be written the next few weeks??
A calorimeter does not make a COP it measures a COP....
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There is some confusion here between thermal energy and temperature. Thermal energy (Joules for example) has a time component, temperature is a scalar quantity. Peak temperature may not even occur at the same time as peak energy output.
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We use electrical power for physics experiments due to the fact that it's easy to measure. Do you follow the power of the battery/starter motor when you consider a 300HP Diesel engine?
No input power is required if your control system removes only the excess heat. Envision a system that reaches operating temperature of 600C and continues heating to 650C. A heat transfer medium leaving the reactor at 650C and returning at 600C would require input power? (hint: the answer is no) Just like we don't leave the starter motor running on a Diesel engine...
This is another area where I think you may be mistaken. Fuels, like diesel fuel, or gas, or wood, can be warmed, and might even begin to produce some "excess" heat in response. But once a threshold is passed they also go into a self-sustaining mode of combustion. Theoretically, it is the same with the sort of temperature-dependent LENR energy release you describe. There should be a low-temperature state of excess heat, but then there will be a threshold beyond which the system goes into a high-temperature state which needs radically less input energy (and in some cases maybe none at all) to sustain it. I have been calling that runaway. So far, all the data I have seen says to me that you do not have that high-temperature runaway state and instead are seeing a state of stable excess heat production. It is not in a state comparable to a Diesel engine.
This is all relevant to the system you ask me to envision. If you picture the 650C state as one of these stable, pre-runaway states that you have been displaying so far then I disagree with you. Think about the heat transfer you mention being really tiny in scale. The medium leaves at 650C and returns at 600C having given up only a tiny amount of heat energy. Do you still think that you could take away all of the input power and still have the system sustain itself at 650C? How does removing a tiny amount of heat make the system independent of outputs?
On the other hand, if you are in one of these high-temperature runaway states, then the system is already much less dependent on (or possibly free of) input energy. In that case I definitely think you could harvest some of the developed heat and the whole thing would be sustainable.
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There’s so many errors in your last post I’m not going to unpack them. I have some actual experiments to run.
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There’s so many errors in your last post I’m not going to unpack them. I have some actual experiments to run.
OK. I thought I should tell you though. Everything I said is just a consequence of having a temperature-activated heat source in a thermal mass.
I do urge you, when developing your next calorimeter, to include inside the calibration reactor a joule heater capable of delivering heat in the same sort of way you think that your active nickel mesh does. That way you can physically test out for yourself how your reactor/calorimeter should react in different circumstances and perhaps see if withdrawing some heat via a heat transfer medium will make the system independent of the normal external heaters you are using.
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I don’t see what you mean. The calorimeter that makes the highest temperature with the least amount of heat input will give the highest COP. You are just reinforcing my point that COP is meaningless. It’s Arbitrary. How about we set up a poll: let’s ask who thinks COP is meaningful and why. Then when we publish our next data, with any luck with a high COP, these same people are going to change their tune
I already explained that improving the mass air flow calorimeter efficiency 10% would increase the air temperature about 4 C at near the maximum power it is designed for.
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It wouldn’t take much of a wizard to work out how much less input would make the same delta T output in a mass air flow calorimeter because the delta is so linear. The delta T is pure watts per degree Celsius. The fan speed might also be increased in order to maintain the same precision (the W/degree goes up with improved efficiency, but the thermocouple has a limited absolute precision.
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The calorimeter that makes the highest temperature with the least amount of heat input will give the highest COP.
Aha! Maybe this is why I haven't been understanding you.
I have been defining COP as total output divided by input. You appear to be calculating it as just the part of the output divided by input.
So ...take a nicely insulated calorimeter and let's say that almost all of the heat output stays inside thereby causing a rise in temperature that you can measure. You have translated almost all of the heat output into temperature change and can therefore pretty much exactly calculate COP based just on this.
Now, use the same calorimeter but take out the insulation. A much larger fraction of the overall heat output leaks across the walls than before. Now, the temperature doesn't go up so much. Calculating heat content from this temperature change only gives you part of the actual heat output from the reactor. But what you need to calculate COP, as I have been defining it, is to estimate the total output from the part you have measured. This is done by studying the properties of the calorimeter and inserting a factor into your calculations that allows for how much heat is escaping. This is what Mizuno and Rothwell do in their air flow calorimetry.
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This is done by studying the properties of the calorimeter and putting a factor into your calculations that that allows for how much heat is escaping. This is what Mizuno and Rothwell do in their air flow calorimetry.
Normally known as calibration.
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Normally known as calibration
by those who do calorimetry, in contrast to Brucian ruminators which chew on COP all day long
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Normally known as calibration.
Daniel_G uses "calibration" for other procedures as well. For instance here
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Aha! Maybe this is why I haven't been understanding you.
I have been defining COP as total output divided by input. You appear to be calculating it as just the part of the output divided by input.
So ...take a nicely insulated calorimeter and let's say that almost all of the heat output stays inside thereby causing a rise in temperature that you can measure. You have translated almost all of the heat output into temperature change and can therefore pretty much exactly calculate COP based just on this.
Now, use the same calorimeter but take out the insulation. A much larger fraction of the overall heat output leaks across the walls than before. Now, the temperature doesn't go up so much. Calculating heat content from this temperature change only gives you part of the actual heat output from the reactor. But what you need to calculate COP, as I have been defining it, is to estimate the total output from the part you have measured. This is done by studying the properties of the calorimeter and inserting a factor into your calculations that allows for how much heat is escaping. This is what Mizuno and Rothwell do in their air flow calorimetry.
“you haven’t been understanding me” that’s the truest statement you made so far. Haven’t you trolled this thread enough? Seriously, now you are just simply trolling. My COP is part of the output? Nah. You really just make this crap up as you go along. Here read this and get back to me…https://uowa.edu.iq/filestorage/file_1551541671.pdf
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This is done by studying the properties of the calorimeter and inserting a factor into your calculations that allows for how much heat is escaping. This is what Mizuno and Rothwell do in their air flow calorimetry.
Ideally, that factor should be as small as possible, as it was with McKubre's flow calorimeter. It was larger with Mizuno's calorimeter for various reasons beyond his control, mainly because the reactor is so large and hot. Whether this factor is large or small, it has to be measured and accounted for by calibrating.
Fleischmann felt that McKubre made his calorimeter so accurate, precise and thermally stable that he sacrificed some cold fusion reaction performance. He might have seen a larger reaction if he had let the sample get hotter. This would have degraded calorimeter performance. The calorimeter itself is an integral part of the experiment. It affects the reaction.
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The calorimeter that makes the highest temperature with the least amount of heat input will give the highest COP.
According to the definition I am using, and I think others are using too, a calorimeter equipped with a control reactor (one with no exothermic or exothermic internal processes) should always have COP = 1. Does this fit with the definition you are using?
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Ideally, that factor should be as small as possible,.
Why? I would have thought that as long as one has a good handle on the proportion of the overall heat being captured and used drive a temperature rise somewhere, it would be sufficient to estimate total heat generation.
