Slad said
QuoteI thought there was a back-up array, consisting of the old 50x 20kW e-cats also in the shipping container, along with the 4x 250kW e-cats. One could imagine these could take over on off days, if they were indeed actually present in the container.
No room! I'm pretty sure this was only a 20 foot container.
Hey up Alan.. You might be right that there would be no room in a 20 footer, but looking at these pictures, I'm not so sure that's the case, or that a 20ft container was definitely used.
The graphics on the ecat website show the two setups. 20kW units on the edges, and 250kW units positioned centrally.
The graphics above show a 20ft container, with a bit of spare space inside, then what looks like a 12ft 10ft container.
This photo below seems to show a container long enough to fit both... You can see a bank of something at the side behind the central 250kW units in this photo.
EDIT: on closer inspection, the photo seems to match the top graphic. Not a spare bank of ecats.
Fact is that last weekend Jed was hammering Thomas Cluck with gas stove FUD and when heat management questioned
I don't recall this at all. Do you have a link to the thread?
Endothermic process like sponge processing
I didn't know that. That's interesting... That would reduce the required sizing of a roughly 500mm flue to a smaller diameter.
Regarding backup e-cats, another photo for reference: (from Rossi's "1MW plant" page)
In a post on his blog, Rossi said "1- it took place inside a plant about 20 meters long, 3 meters high, 3 meters wide, obviously closed" - this actually sounds like (loosely) 2X30foot containers. One was said to contain the control system, so that leaves another one spare for reactors.
@Alan
I think you are on the right track: A shipping container is 2.4m (8ft) high...
Which makes the one at the back of this photo at least 30ft long. (It's longer than 20ft, and the next size up is 30ft).
So enough room for a main and a backup inside.
Quote from Argon: “Fact is that last weekend Jed was hammering Thomas Cluck with gas stove FUD and when heat management questioned”
I don't recall this at all. Do you have a link to the thread?
Actually it was in vortex thread i linked before. Here is link again 'https://www.mail-archive.com/vortex-[email protected]/msg109800.html'
I need to say, I feel sorry about Jed since he has not been same Jed I used to read before. I hope he can return back in full to LENR community after all this mud dissolves.
Perhaps you had in mind Peter Gluck? There's Thomas Clarke (a regular on this forum) and Peter Gluck (a regular on Vortex, and not often on this forum). Peter Gluck and Jed Rothwell have been in a dispute about how Peter has been representing Jed's position on Peter's blog.
Eric, yes you are right, he was hammering Peter Gluck. I was little busy writing original post and later didn't notice slip, but link above leads to middle of long thread I was referring.
Slad,
thanks for the confirmation, that my possible scenario is not totally unrealistic.
I admit, that the really relevant facts are not available before the ERV-report is published.
My point (and something to check after the ERV report has been published) is:
If it is true, as implied by the statement in Mats blog, that the return temperature was not monitored continuously and corresponding data were not recorded (and taken into account), then every kind of manipulation (behind the wall in the customer area) is thinkable.
If no information is available, what was behind the wall in the "customer" area and the return temperature was estimated or only measured at special occasions, then for example a valve with different radiator lines might have existed (equal line resistance however different heat consumption) to manipulate the test outcome.
Remember this? I thought this quote sounded weird for several reasons. (Plywood is often used in high-tech fabrication 
)
Slad - Yep - the insulation and plywood for each reactor box are included in the simulation. You should see what happens when the plywood catches fire.
Don't fret - it's a pro model - these folks were previously submarine power plant and turbulent combustion professionals.
Lets have a closer look at this...
Plywood is an insulator of heat, so it combusts when you apply a heat flux of 7.5 kW/m2. Basically, the plywood can't conduct away this quantity of heat, so it's temperature rises above it's ignition point.
How much heat flux can a 250kW ecat supply?
If you ran it at 100% power, and closed the output steam pipe. All 250kW would have to exit through the walls of the reactor. The module measures about 3.5 x 1.4 x 0.5m...Giving a surface area of 14.7m2
250kW / 14.7m2 gives a maximum possible heat flux of 17kW/m2 (which is higher than 7.5 kW/m2), so enough to cause the plywood to combust.
BUT... If you ran it at full power, with the exhaust pipe closed, it would explode! Or at least blow a pressure release valve open.
So, a (slightly) more realistic explanation is that the reactor is running at 100% power of 250kW, and enough waste heat is leaking out of the reactor walls to combust plywood, (14.7m2 x 7.5kW/m2 = ) 110kW.
Which only leaves 140kW for making steam.
Or to put it another way: The reactor would be running at 56% efficiency. Which is worse than even the worst gas fired boiler that Jed R could conjure up.
The only way Dewey could make his pretend models demonstrate combustion of plywood, would be to either ignore any insulation surrounding the reactors, or to and assume all heat being sent to the customer side is returned unused, whilst heating the returned (now superheated*) steam. Neither is a sensible option. More explosions are imminent...
* T(steam)>120C, 100% dry
Rossi's blog claims that the 250kW modules only leak a total of 10kW of heat, that is 2.5kW per reactor. (99% efficiency*)
Knowing the surface area 14.7m2, we can say this gives a heat flux through the reactor wall of 0.17kW/m2. Which is clearly less than the 7.5kW/m2 required to combust firewood.
Furthermore, through the application of Fouriers law, as discussed here, we can say that limiting the waste heat to the level Rossi describes would only require 10mm of PIR foam to be covering each 250kW reactor.
Or, if you are worried about your insulation catching fire / degrading due to heat, you could use 20mm of mineral wool instead.
So Dewey... I ask you again... What insulation does your pretend model use, and how thick is it???
* How can Rossi get an efficiency of 99%, using this small amount of insulation, compared to a normal gas fired boiler?
Because (a) his temperatures are lower, so easier to insulate; (b) he is not wasting heat by exhausting hot combustion products straight up and out of a flue; (c) he can totally surround his reactor with insulation, rather than leaving gaps necessary to ventilate combustion.
I thought 50 mm of rock wool was bog standard boiler insulation.
If it is true, as implied by the statement in Mats blog, that the return temperature was not monitored continuously and corresponding data were not recorded (and taken into account), then every kind of manipulation (behind the wall in the customer area) is thinkable.
Tom P, it might surprise you that I'm not really that into defending Rossi. I prefer to keep an open mind and watch events... I just enjoy correcting (and sometimes insulting) people who like to bollock on about their own unique interpretation of thermodynamics/heat transfer.
But if you read the quote on Lewans Blog, Rossi says:
QuoteThe water heated by the MW plant was circulating in a closed loop, and since the return temperature was varying, due to different load in the process of the customer, Rossi insisted that the energy corresponding to heating the inflowing cooled water (at about 60˚C) to boiling temperature would not be taken into account for calculating the thermal power produced by the MW plant. The ERV accepted. (This was conservative, decreasing the calculated thermal power. The main part of the calculated thermal power, however, derives from the water being evaporated when boiling)..
Which implies to me that the temperature was being measured (it averages 60C), it's just not being taken into account for the purposes of evaluating the COP.
Note that as we discussed before any manipulation based on there being no load on the customer side, would either require an (easily detectable) flooded steam pipe, or supersonic return steam.
However, we do need to see the report to be sure: If the ERV (*in agreement with IH*) did a really, really bad job of selecting which instruments to use in the steam pipe. For instance, it would be possible fool a pitot-static tube, although this would require the flow rate of water to be 600m^3/day, not 36m^3/day... And would be easily detectable if cross-checked with other instrument readings, such as a vortex meter.
Paul:
"
I took the leaked "facts" as a starting point:
- 101°C (wet) steam flow
- 36m^3 per day
- 20 kw (at least) permanent electrical input
"
"And my assumption is 99°C water return."
If the above "leaks" and assumptions where true, then the IH deserve to loose their 11 MUSD, or even more if they loose in Court.
This is pure clownery, and you need to be a clown to pay out based on these kinds of "facts"
I thought 50 mm of rock wool was bog standard boiler insulation.
For sure, but you still can't completely cover a gas boiler with it.
Also an electrically heated hot-water cylinder would likely use PIR nowadays.
Rossi could even have used closer to 50mm rockwool, further reducing his heat loss through the reactor walls... My fag-packet calcs assume that all the ancillary components / pumps etc release no heat into the container. 50mm rockwool would allow a good margin for a more realistic appraisal of these other heat loses.
Also, on a scorching hot Miami summer day (35C)... You might want that 50mm there, I just took an average figure for external temperature (25C).
The point is, you don't need all that much insulation to prevent poor Rossi from being boiled alive by his megawatt, broken ceiling fans or not.
Actually, my main point is that Dewey is tripping hard. Burnt plywood. Wow.
Which implies to me that the temperature was being measured (it averages 60C), it's just not being taken into account for the purposes of evaluating the COP.
Now, I'm pretty sure, that your mission here is to defend Rossi, otherwise you would have realized, that a correct COP calculation is absolutely impossible, if, in a closed circulating system, the true, continuous measured return temperature is not being taken into account.
Sorry, but you are utterly wrong. EDIT: I just noticed you said a "correct" COP calc. see 'NB' below
The return temperature need not be taken into account, as long as it is water that is being returned...
You just need to know the mass flow rate of water multiplied by it's enthalpy of vaporisation.
This has the effect of making it into a conservative COP calculation.
If water is not being returned, that means that steam is. Due to the pipe cross-sectional areas seen in photographs, this would force the return steam to travel at a supersonic velocity. About Mach 1.5 in fact. (Assuming 36m^3/day water is used, which is about right to transfer 1MW of heat from a boiler)
That's just not happening. There would be a supersonic shock wave at every junction and fitting!
That means water is being returned, that means you can calculate a conservative COP based enthalpy of vaporisation alone.
NB: I have never claimed a "CORRECT" cop calculation, merely a conservative one.
Display More@ShaneD. The reason Rossi/the ERV don't care about water return temperature, is that it takes a relatively small amount of energy to heat water to 100C, but a lot of energy to turn it into steam.
If the returned water was at 4C, the ERV would be underestimating the COP by 17%.
If the returned water was at 60C, the ERV would be underestimating the COP by 7%.
If the returned water was at 80C, the ERV would be underestimating the COP by 3.5%.
...This assumes 100% dry steam, which in a system like this requires a steam temperature of approximately 120C.
From Mats blog:
QuoteThe water heated by the MW plant was circulating in a closed loop
CLOSED CIRCULATING LOOP!
ALL WATER IS RETURNED!!!
You just need to know the mass flow rate of water multiplied by it's enthalpy of vaporisation.
That would only be valid if dry stream had been produced and I doubt that.
Let's wait for the ERV report and disclosed "steam" temperature. I'm pretty sure that I'll be right and you'll be surprised.
P.S.
Reliable test results should be based on continuous measurement of ALL relevant parameters an not on calculations out of thin air or thick fog.
I'm very well aware it's a closed loop, and by definition all "water" has to be returned. It was unlikely to be anything else, that statement on Mats' blog was just the first confirmation of it. One could have worked it out by looking at the photos.
That would only be valid if dry stream had been produced and I doubt that.
Not true. One could measure the dryness of the steam, and calculate the total enthalpy. By doing the reverse I was able to figure out that the scenario you proposed would result in 1.8% dry steam.
Reliable test results should be based on continuous measurement of ALL relevant parameters an not on calculations out of thin air or thick fog.
I completely agree, but remember there's a big difference between not measuring, and not factoring that measurement to your calculations.
Let's wait for the ERV report and disclosed "steam" temperature. I'm pretty sure that I'll be right and you'll be surprised.
Good plan... And don't get your hopes up!
