Re: [Vo]:other tweets
Where did you find that value? 2011/10/6 Peter Gluck peter.gl...@gmail.com See the tweets of the other journalist from Italy. http://twitter.com/#!/raymond_zreick It seems the FatCat has worked at ~ 3.5 kW. Till we will not discover something tricky and if this experiment can be repeated with many generators, it seems this day was a Sweet Thursday for the Rossi believers. Peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
Re: [Vo]:other tweets
3.5kW is lower value than what I estimated, therefore it must be false. . . But anyway, perhaps we can take it as absolute minimum. If then E-Cat produced about 64 kJ excess heat. That would translate into 3kg ethanol need to fake the results. Therefore, test is not conclusive!! However, i still think that Raymond's value for output power is too low and may present e.g. power level at 110°C. So perhaps demonstration was sufficient, if they will examine the fat cat carefully. —Jouni On Oct 6, 2011 9:54 PM, Peter Gluck peter.gl...@gmail.com wrote: See the tweets of the other journalist from Italy. http://twitter.com/#!/raymond_zreick It seems the FatCat has worked at ~ 3.5 kW. Till we will not discover something tricky and if this experiment can be repeated with many generators, it seems this day was a Sweet Thursday for the Rossi believers. Peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
Re: [Vo]:other tweets
between 3 p.m. till 7 p.m. the temperature average delta has been of 5°C (water input/output) for 0,6 cubic meters per hour According to the husband of the cute brunette. :-) T
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Well, that means 600,000/3600*5*1W = 833W. That's the old electric heater hypothesis. Odd coincidence, although there was no input electricity. 2011/10/6 Terry Blanton hohlr...@gmail.com between 3 p.m. till 7 p.m. the temperature average delta has been of 5°C (water input/output) for 0,6 cubic meters per hour According to the husband of the cute brunette. :-) T
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On Thu, 2011-10-06 at 15:11 -0400, Terry Blanton wrote: between 3 p.m. till 7 p.m. the temperature average delta has been of 5°C (water input/output) for 0,6 cubic meters per hour According to the husband of the cute brunette. :-) This is what I get. 0.6 cubic meters / hour = 600 liters / hour = 10 liters / minute = 167 ml /sec, with a 5 deg temp diff. If all these numbers are correct then 5 * 167 ml /s = 835 cal /s = 3.5kw for this demo. That's a lot of heat for a unit running with no power. That's in the range of what he's been getting with these latter units. Craig
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3000 kcal per hour = 3.49 kW True? I am very tired after this day of info-hunting Peter On Thu, Oct 6, 2011 at 10:11 PM, Terry Blanton hohlr...@gmail.com wrote: between 3 p.m. till 7 p.m. the temperature average delta has been of 5°C (water input/output) for 0,6 cubic meters per hour According to the husband of the cute brunette. :-) T -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
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Oh, right. That's calories!
Re: [Vo]:other tweets
On Thu, Oct 6, 2011 at 3:11 PM, Terry Blanton hohlr...@gmail.com wrote: between 3 p.m. till 7 p.m. the temperature average delta has been of 5°C (water input/output) for 0,6 cubic meters per hour Back of the envelope, that's 50.3 Mjoules.
Re: [Vo]:other tweets
You must not forget the losses due the conversion between the heat exchangers. If it was 70%, that means around 5KW for the core. 2011/10/6 Peter Gluck peter.gl...@gmail.com 3000 kcal per hour = 3.49 kW True? I am very tired after this day of info-hunting Peter On Thu, Oct 6, 2011 at 10:11 PM, Terry Blanton hohlr...@gmail.com wrote: between 3 p.m. till 7 p.m. the temperature average delta has been of 5°C (water input/output) for 0,6 cubic meters per hour According to the husband of the cute brunette. :-) T -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
Re: [Vo]:other tweets
On Thu, Oct 6, 2011 at 3:23 PM, Terry Blanton hohlr...@gmail.com wrote: On Thu, Oct 6, 2011 at 3:11 PM, Terry Blanton hohlr...@gmail.com wrote: between 3 p.m. till 7 p.m. the temperature average delta has been of 5°C (water input/output) for 0,6 cubic meters per hour Back of the envelope, that's 50.3 Mjoules. or 13.97 Kwh in 4 hrs = 3.49 kW T
Re: [Vo]:other tweets
Am 06.10.2011 21:25, schrieb Daniel Rocha: You must not forget the losses due the conversion between the heat exchangers. If it was 70%, that means around 5KW for the core. If the heat exchanger is well isolated, it will not loose energy. It will reduce the temperature, because it has a finite thermal resistance 0 . But the water mass flow on the secondary side will be higher. This gives the same energy, if there are no isolation losses. Anyway, they have to accept this. Many industrial machines use heat exchangers and are working. kind regards, Peter
