Brillouin Energy Corporation (BEC) updates.

• Let me give you another example. Coal or biomass powered power plants burn carbon based fuels to produce steam at a certain energy level and temperature. If I run this steam through one of our reactors and my COP is only 2 (or even 1.5 but I use two for simplicity (much less than 5!) and I double the energy in this steam so that now I can burn half the fuel that I was using before, you think I cannot make money from such a device? As I said, complete nonsense.

Daniel. You are missing the point. Let us assume that you have a LENR device with COP2. It requires some electrical input (otherwise the COP would be infinite of course) If you use the LENR energy to make steam and then to drive a steam turbine coupled to a generator what would be the result?

Sadly, the efficiency of the very best steam turbine gensets is around 50%. So you can only make enough energy to power your LENR machine, since 50% of COP 2 is COP1.

• A simple baseboard heater with COP 1.2 efficiency (normally they are merely 100% efficient), for example, would ultimately replace almost all baseboard heaters in the world. They might even become mandated over standard COP 1 heaters in some places.

A agree: With a powder reactor of a our London type these would also be much cheaper - in theory if all teh support need can be squeezed into a tiny box.

But this will never work for a SUN-CELL. May be for an Elefant heater....

• Quote from A simple thought experiment can illustrate my case. An adiabatic calorimeter with Design A has normal insulation and it requires about 500W to reach an equilibrium temperature of 500C. A second calorimeter with far superior insulation and smaller surface area can reach 500C with only 50W input. Which one will give you the higher COP? If you agree that this is possible then it follows that you must agree that the COP figure is meaningless in this context. There is a factor 10 difference between the two!

In this example, temperature is being compared, not Heat. The factor of 10 is irrelevant in temperature by itself. It is all about what is at a particular temperature.

• Daniel. You are missing the point. Let us assume that you have a LENR device with COP2. It requires some electrical input (otherwise the COP would be infinite of course) If you use the LENR energy to make steam and then to drive a steam turbine coupled to a generator what would be the result?

Sadly, the efficiency of the very best steam turbine gensets is around 50%. So you can only make enough energy to power your LENR machine, since 50% of COP 2 is COP1.

Then only use the LENR system where it makes sense, not somewhere foolish.

• You fail in calculation. To maintain your reaction you need input 2 you can only serve what you add = 1 ... But burning carbon of any kind cannot need more heat as it already burns at top heat.

Your efficiency never changes. What you hope to see is n+2 -1 > n. But for what n ? And which process?

Do you know that this only works for heat? best Carnot cycle today is 60.x% - x < 1 so your 2 is 1.2 for electricity at generation then transformation happens etc. so 1.2 --> 1.1x... just a no issue.

No you fail to understand basic logic and engineering. I have a black box that outputs 2x heat vs. input. (COP2) I burn X amount of carbon to produce Y amount of heat. If that steam then travels through this black box that produces 2x heat and then I run that steam through a turbine, then I can use 0.5X fuel to get the same output from the turbine. Everyone again is assuming that I require electricity input. I don't. It requires heat. For physics experiments I use electricity because its easy to measure the power accurately. In a real commercial application I assure you I can sell COP 1.2 or 2 devices (we have that now), I simply need to know the W Exh per cm2 and I can design any system you like. The conversion of heat to x to 2x means I can burn half my fuel and power the same steam turbine. The conversion efficiency of that steam to electricity is irrelevant in this case. Please open your mind and follow this simple logic.

This is why Mizuno and I are adamant to write a paper on this issue!

• Daniel. You are missing the point. Let us assume that you have a LENR device with COP2. It requires some electrical input (otherwise the COP would be infinite of course) If you use the LENR energy to make steam and then to drive a steam turbine coupled to a generator what would be the result?

Sadly, the efficiency of the very best steam turbine gensets is around 50%. So you can only make enough energy to power your LENR machine, since 50% of COP 2 is COP1.

Alan, why would it require electrical input??? It absolutely does NOT require electrical input. I merely place my black box between the boiler and the steam turbine. Any increase in steam energy will result in a concomitant reduction in the fuel required to reach the same output. You are total mis-framing the problem with false assumptions.

• I can sell COP 1.2 or 2 devices

Yes just for heat...Not for current. But it wont change much 80% carbon or 50% carbon is still 90% to much carbon. So just a waste of time.

• Please tell me what is wrong with this diagram and why I can't burn half the fuel when I amplify heat energy with COP of 2

Images

• Yes just for heat...Not for current. But it wont change much 80% carbon or 50% carbon is still 90% to much carbon. So just a waste of time.

Again utter nonsense coming from you. You first claimed (wrongly) that I you need COP 5 to make money. I told you I can make money with 2. Now you move the goalpost and say 50% carbon reduction is not enough. Perhaps you are correct but changing the argument does not mean that my original point was wrong. Anyway, reducing carbon by 50% and being able to capture all that value with high profit (negative cost) and you think that's not good enough? I can only shake my head.

• Brillouin Energy Corporation should be having public demos of their HHT system by now! If they really have the 'claimed' >1 COP then what is holding them back? Everything they have done is patented to the hilt! Again suggests a certain lack of confidence in their HHT system. Oh well, back to the drawing board!

• Yes just for heat...Not for current. But it wont change much 80% carbon or 50% carbon is still 90% to much carbon. So just a waste of time.

What does "not for the current" mean? A heat to heat COP of 2 reduces fossil fuel use 1/COP future improvements which are much closer than anything ITER is doing of 3 means 66% reduction, 4 will give 75% reduction and 5 will give 80% reduction. Mr. Wyttenbach if you won't invest in technology that can do this, then its quite OK. There are several billion other people that will.

• Please tell me what is wrong with this diagram and why I can't burn half the fuel when I amplify heat energy with COP of 2

It works for heat in the best case but for current you have to apply the conversion factor that is 50..60%...

Remember that also heat exchange consumes about 3% of the input energy.

I would never invest any money and never recommend any investor doing this.

As soon as somebody presents a stable COP 5 reaction all your investment is immediately gone. This can be soon, in a year or 5 year worst.

• Seems you cannot follow a simple diagram. The currently used carbon fuels already are being burned to make a fixed amount of heat. The COP 2 heat amplifier simply doubles the amount of heat in that steam so you can burn half the fuel. The fact that someone as educated as you cannot follow this logic simply proves my point that we have to educate potential investors more. Thank you for helping me make my point.

Yes of course if someone makes a better mousetrap with higher COP (still have to compete on CAPEX/OPEX basis) then likely they can take market share and this will promote more investment in this field and improvements will drive higher and higher COPs. All good things for everyone.

That being said, I still claim that I can produce power with a COP even as low as 1.2 even considering thermodynamic losses in the conversion to electricity. That nobody else sees this is quite baffling to me. Maybe I am crazy but within the next couple quarters hopefully I will be proven right or wrong.

The story behind the story is that so far LENR produces such low absolute power compared to something like ITER. SO they get the big money. But I can assure you all here now, that I can design and build a LENR based generator with COP of 1.2 (by the standard definition) that will be able to produce as much power as you want at very low cost. I guess since others can't see how to do this, hopefully we will be the first to the market in this area. A household combined heat and power generator that produces 15 kW of heat for heating water and hydronic heatings etc. while producing 5kW of continuous power for the home from something the size of an air conditioner. I assure you we will have no problem selling this. I also assure you that is easily done.

Happy Fusing.

• In this example, temperature is being compared, not Heat. The factor of 10 is irrelevant in temperature by itself. It is all about what is at a particular temperature.

No again you are making the same mistake. For the sake of simplicity I equated a 1:1 relationship between temperature and watts. The fusion reaction gives excess heat in an exponential relation to temperature. SO the calorimeter that is better insulated and reaches a higher temperature with 1/10th the power input will give a 10X COP. This is not rocket science. Therefore the COP value is irrelevant. QED.

• n contrast ITER goes around raising billions of dollars by telling funders that they are going to produce 10x more output than input.

comparison with ITER is useless..it is a sacred European cow,, like the Inca mummy Kings

the last refuge for a lineage of collision science worshippers.

• Thank you for helping me make my point.

Obviously it did not work to help you. SUN-CELL needs current as input if SUN-CELL input is =1 and output heat = 2 the current COP = 1.2.

If you cant get this than you should never invest into technology!

You have always to calculate in the units of what you sell. So SUN-CELL input is = current lost in sell! Net current gain of cell is 0.2.

To say teh other way from heat = 2 you get 1.2 current and loose 1 in SUN-CELL...

• Please tell me what is wrong with this diagram and why I can't burn half the fuel when I amplify heat energy with COP of 2

Assumes there is no loss of heat or entropy in the processing step. To be at the same pressure and temperature, your process must generate twice the steam to get twice the energy content. So, you need a water input in that case. If your process keeps the volume the same, the pressure and temperature must increase to add more energy content via steam pressure. Take a look at a steam table to see how doable that is for input and output steam temperature and pressure.

It seems to me that your reactor would have some stable upper temperature limit which would limit the rate of heat transfer from your reactor to the steam.

• Alan, why would it require electrical input??? It absolutely does NOT require electrical input. I

That is a not big assumption, since Mizuno has always used electrical inputs in the work we have seen. And electrical heating always brings something extra to the table, besides being very convenient in a laboratory setting it produces EM fields which I know are a significant factor for triggering LENR.

If your device was used as a space or water heater, then fine, reducing electricity consumption by a factor of (say) 2 is very worthwhile, but for anything more than that you will need to demonstrate excess heat with purely thermal input, meaning zero EM. And I know of nobody in this field who has successfully done that, people have tried and failed so far. If you have done so successfully then you have a chance of commercial success.

• No again you are making the same mistake. For the sake of simplicity I equated a 1:1 relationship between temperature and watts. The fusion reaction gives excess heat in an exponential relation to temperature. SO the calorimeter that is better insulated and reaches a higher temperature with 1/10th the power input will give a 10X COP. This is not rocket science. Therefore the COP value is irrelevant. QED.

No, the better calorimeter melts down in this case, in a short period of time.
If the insulation value is so bad that the reaction can be improved 10x by fixing it, 90% the heat now will not escape in time compared to the usual rate of losses if the calorimeters are otherwise equal.
And obviously improved COP is better if you can.

• We would not have any discussion here if Mills had more understanding of physics and would have gone on with his self sustaining Silver electrode reaction...

Sad to see how many nonsensical decisions he made in the past 2 years...This happens typically in companies where one person claims to know everything...

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