Posts by Slad

I think Dylan would be a better comparison

Didn't like that last one ZephirAWT? Perhaps you should develop a sense of humour?

Try this one on for size:

Cognitive Dissonance bites and bites deep,
Wits my, about I, how do should keep?
And who the feck are Airbus? Those european choads,
Probably all researching coffee*, and shivering in the cold.
More ‘unclear reactions, reactionary crap,
A complete mental breakdown, a psychiatric flap,
Synapses snap, took a nap just went black can’t go back.

Mental intrusions, demented delusions, psychic lattice occlusions,
Excess heat in your head.
Chinese connivance, contrivance of the science,
Kuhn out of Creedmore on licence,
A strange feeling of dread.
What next, Hydrinos? Eskimos wearing chinos?
Can’t tell arsehole from elbow, sanity drips from my nose,
The last throes of my life too much strife where’s the knife?

The Tesla coil of reason, blasts scalar waves through your cranial lead,
Soon it will be open season, the sweet relief of hot metal to your head,
‘Thing is we all knew the answers, to all of the problems posed,
I should have seen this coming,
I should have kept my mind closed.

* An oft-repeated Mary Yugo ad-hom against Levi

An oldie but a goodie:

Parkhomov, you crafty Russian,
Mates with Putin, cold war cousins.
I heard your apartment got even more damp,
Since you pinched some more lead, off the roof, for a clamp.
And why is it, that only you,
Can make LENR out of your wife's stew?
C'mon. Stop it. You must think we're mental,
You can't do LENR with a kitchen utensil!
What COP did you get before it began to misbehave?
You should've put that fucker in your microwave.
It's just like baking a cake can't be done in a lab...
All just an obvious publicity grab.

Quote from Dewey Weaver

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.

Hey Dewey, after a bit of research, I managed to track down your submarine engineer:

And your turbulent combustion specialist:

Quote from stephenrenzz

What? He said QuarkX is the size of a pen.....but he didn't say how big the pen was.....

Stephen, some homework for you:

What's the critical heat flux of water?
What's the surface area of a normal pen?
Based upon your previous answers, is Rossi's claim credible?

Quote from stephenrenzz

it is now known that Rossi uses MANY sock puppets on all of the forums. randombit0 is one of Rossi's characters on here. I can give you a full rundown on how to spot Rossi fake users, it is quite easy. If you don't believe me look around on the topic of all Rossi's fake users.

Name them or shut up forever hold your peace. I actually agree with you that Randombit0 might be Rossi. But until 15min ago, I was sure that you were a Dewey sock-puppet... So maybe that shows how reliable these sort of guesses are.

Quote from stephenrenzz

From what I see Dewey has provided WAY more actual fact regarding Rossi/IH situation that I have see you even speculate on.

Quote from stephenrenzz

but there have been many facts that Dewey has brought to light that have in fact been substantiated.

Quote from stephenrenzz

What does interest me about listening to Dewey is that he has put forth information that even the Rossi-camp has had to admit is true.

See what I'm getting at?! But rather than go down that rabbit-hole, lets take things a face value: In a pseudonymous environment, trust and truth is king. When someone tells obvious lies (link), it normally causes everything else they say be viewed as equally suspicious.

Quote from JedRothwell

I do not understand why anyone still believes Rossi. In the interview with Lewan, he came right out and said the whole test was fake. He didn't just admit it, he bragged about it! Right here:
"IH never had access to the customer’s area. At the end of the test, an expert hired by IH, insisted that it was important to know where the water came from and where it was used. The ERV explained that this had no importance."

Quote from JedRothwell

There are thousands of qualified HVAC engineers and equipment engineers in Florida who are licensed to perform such tests.

Quite a big jump in logic there from Rossi's quoted statement, to hold it up as the kind of "evidence" you claim it is.
Secondly, whilst "Oystler" is no doubt too humble to say it, he more than qualified to hold the much vaunted qualification of 'Florida HVAC engineer', in fact he could likely design from first principles, calibrate, and fit the necessary measurement equipment himself.

Quote from JedRothwell

It cannot be used for nothing. Releasing 1 MW of heat in 6,500 sq. feet of enclosed warehouse space would heat the building enough to kill the occupants. There has to be heat removal and ventilation equipment

Come on Jed! This kind of thing is on the Florida HVAC technicians exam!! You need at 500mm diameter flue, with a mechanical draught of 12m/s!!!
Your conjectures hold no weight... Back to your studies for another year I think. But keep trying: You'll get your HVAC licence eventually.

Quote from oystla

Prove? It would only prove that some heat of unknown quantity was used for something. And that would say nothing about the e-cat plant heat claim.

What is needed, and what would "irrefutably prove the issue one way or the other. " is:
- a log of fluid pressure delivered from the e-cat plant
- A log of fluid temperature delivered from the e-cat plant
- A log of measured steam quality of fluid delivered from the e-cat
- A log of steam mass rate delivered from the e-cat plant
- A log of fluid temperature at inlet to e-cat plant
- A log of fluid pressure at inlet of e-cat plant
- A log of fluid massrate at inlet of e-cat plant

With These data we will excactly be able to calculate heat produced by e-cat plant.

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It is not necessary to measure "fluid pressure at inlet of e-cat plant"
And (3) can be derived from (1), (2) and (4)

Quote from ozcar

First post here...

And as valid as (nearly) anyone else's, given the information available.

Quote from Tom Paulsen

But you are a moderator here, so if you think that the deletion of 5 comment in different threads just for today is ok, then I have to accept this and to ask myself, wether I'd like to participate in a censored forum, which doesn't allow me, to make my own judgment.

Before making such threats, one should question whether their contribution would actually be missed.

Quote from Alan Smith

Maybe we should have a 'Fuckwit of the Week' banner to apply to the posts of regular offenders.

As long as it's put to a public ballot :evilgrin:

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.

Quote from Tom Paulsen

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.

Quote from Tom Paulsen

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.

Quote from Tom Paulsen

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!

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.

Quote from Slad

@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.

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Quote from Paradigmnoia

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.

Quote from Tom Paulsen

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:

Quote

The 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.

Remember this? I thought this quote sounded weird for several reasons. (Plywood is often used in high-tech fabrication )

Quote from Dewey Weaver

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.

@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.

Regarding backup e-cats, another photo for reference: (from Rossi's "1MW plant" page)

Quote from Argon

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.

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.

Quote from Tom Paulsen

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.

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Tom P, What you say is plausible in some senses... With 20kW electrical power applied, after raising the water temp by 2C you have 16518 J/s 'left over' for generating steam. If the mass flow rate of the water is 0.416kg/s (same as saying 36m^3/day), then the output will be 1.8% dry 'steam'... Which is better known as 'hot water with a few small dispersed bubbles in it'. The system would have to be unpressurised, and there could be no load in the 'production area', just a looped pipe returning the water though the wall..

It's very difficult to produce steam this wet when you boil water, but with a setup like the ecat, you could imagine pumping more water into the reaction chamber than was boiling off, essentially flooding the output pipe with hot water.

This is going to have some bad effects of the ERV's sensors positioned inside the output pipe. It would be detected instantly by whatever method is being used to measure dryness. It would also would play havoc with any pitot-static tubes (the 1700x difference in density would give very odd dynamic pressure readings). Also, the pipe velocity would be much reduced from an expected value of about 25m/s.

So to sum up; it's plausible mathematically, but in practise, would be very easy to detect.

Note that Dewey (in his typical FUD obfuscated style) is not saying directly that the output pipes 'steam' temperature was measured to be 101C (100.1C?). He is saying that 'something' (condensed steam run off perhaps?) was measured to be 101C, even when the e-cat was supposedly down for maintenance, and hence the results must be faked.

I 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.

Saving this one for posterity:

Quote from Dewey Weaver

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.

Dewey, you are so full of it, pal. What's the plywood doing in there?

Quote from Dewey Weaver

Regarding the simulation, at a 250kW setting on the 1MW system, the temperature in the control unit is approx. 88C after 3.5 hours inclusive of the 3kW air conditioner unit

Are these supposed models, modelling the e-cat units as having no internal insulation???

EDIT: I'm was explicitly stating that you are good at VC'ing and trolling. The implicit statement is that you aren't very good at anything else! Although I wanted to leave that as being implied, as it's a bit rude, and unlikely to be true.

Quote from Tom Paulsen

When will anybody realize, that in a closed loop "inflowing water" means "ALL WATER".

Tom P. It seems there is some hope for you... I thought you were suggesting that the return pipe contained steam, not water.

@Tom P, interesting idea... Send steam out, get steam returned back, pretend you just boiled some water. The problem is the pipe diameters as seen in photos of the shipping container.

The outlet pipe looks to be about 125mm diameter. In order to send 1MW of steam down such a pipe, the steam velocity would be approx 25m/s. The return pipe looks to be about 25mm in diameter, a 25 times reduction in cross sectional area.

This means if this pipe contains steam, it's velocity would be approx. 525m/s... 1.55x the speed of sound! (This would be a very bad thing from an engineering/safety perspective)

Dewey Weaver... If you don't mind me saying, I think you should stick to what you are good at, namely venture capitalism, and trolling forums - not pontificating on thermodynamics. To reply to a post full of numbers with a sarcastic comment is pretty bad form anywhere, but particularly so on a science forum.

Sure, I didn't include all my workings as (a) it would have taken to long to type, (b) people don't want to read about Nusselt and Prandtl numbers, (c) anyone with the ability to check my workings will likely have their own spreadsheets or simulators if they want to check what I said is correct.

Really, you should be more grateful. I just saved you $1000's of dollars that you could of blown on having someone model this...

The thing is, thermodynamics isn't an art... it's a science. It's based 100 year old theories and 'simple' calculations. Hence (unless the shipping container is crammed into a tiny little room) anyone you pay to do the calcs will get pretty much the same answers as I gave.

Unless you have some proper drawings for what's inside the 250kW ecats (i.e. insulation type(s) and thickness), it will be a useless exercise anyway... My calculations are only intended to show that it would be exceptionally easy to design a system that doesn't "boil your fish bowl" or whatever expression you wish to use... Between 10 to 50mm of bog-standard insulation is all that's necessary.

One little hint: When you are looking for your expert to do this supposed thermal modelling, make sure they have an engineering qualifications; Japanese language translators, who have once visited a ships engine room, are rarely up to the necessary calibre... Including have been running their own LENR library for years.

EDIT: The fourth picture down this page shows 4.4MW's of boiler stuck in a basement, with a HVAC system smaller than found in a small (500 person) nightclub. Funny that.

http://supaflex-agencies.com/s…t-london-bridge-city.html

@Paradigimoia, don't give Thomas any ideas....