Now IH have lost E-Cat License and IP, who will manufacture E-Cats in the US?

Dewey

Quote from Dewey Weaver

Axil - you'll be sad to know that the pressure tested reactors failed at 2 Psi. But the great news is that the "ERV" pressure level was reported at 0.0 bar for every single data point reported during the test so not a problem - right?

So you are drip feeding us with information from the ERV report, is this correct?

What I cant understand is why two different units of pressure are being used. Also the standard atmospheric pressure is taken as 1.01325 bar so 0.0 bar must be a vacuum, am I reading this wrong? Similarly atmosphere is generally held to be approximately 14.50377 pounds per square inch, so 2 psi would be close to a vacuum, unless these values treat atmosphere as starting at zero.

I don't think you have seen the report at all and are making it all up!

Best regards
Frank

Quote from Dewey Weaver

the "ERV" pressure level was reported at 0.0 bar for every single data point reported during the test

So, at 0.0 bar of pressure, there must have been a slight vacuum internal the reactor. Let's assume 1 bar == 1 atm (which it essentially does). If there was 1 atmosphere of pressure inside a container (so it's the same pressure inside the container as it is outside it), the gauge will not read 1 atm, but rather 0 atm, as the pressure inside the container would just be the same as the pressure outside. If there was 0 atmosphere of pressure measured inside the container, the container would have negative pressure relative to the atmosphere. This assumes sea level.

Is the 0.0 bar value as certain as the 100.1 C measured temperature value?

Please folks, use the engineering standards of gage or absolute pressure when discussing, to avoid confusion.

that is: barg or bara and psig or psia.

a failure pressure of only 2psig or 0,014 barg would not be possible for a boiler. A boiler would produce larger pressure surges just by the internal dynamics of Bubble formation.

I always wondered how the closed loop 1 MW system could maintain near 0 psig and vaporize about 14 oz/sec of water at 100.1 C given that one unit of water expands to 1700 units of steam.

14 oz = 414g and 414g x 2260J/g = 935640 J/sec which is very close to 1 MW. So the volume of water does seem correct.

This is all based on the reported 36 cubic meters of water flow per day which I believe is from a Rossi interview.

Quote from IH Fanboy

Is the 0.0 bar value as certain as the 100.1 C measured temperature value?

Hmm, I thought it was 103.9°C as of late? Or is Dewey now back to the "classic" 100.1°C?

Of course we could all easily verify Dewey's and Jed's conclusions if IH would finally release that ultimately damning ERV. It can only speak of their honourable consideration for Rossi's feelings that they still hesitate to do so. But that is of course how one would expect white knights to act, and you can pride yourself to be their fanboy.

Quote from Dewey Weaver

I have the pictures and report. Rossi claimed 15 Psi was the spec but then again, everything seems to be off by about that factor or higher (think flowmeter). The systems popped open at 2 Psi. According to Barry, the leaks in Miami gave Rossi fits for weeks. The "ERV" says 0.0 bar for every data point each of the 12 final report annex docs.

So, perhaps he fixed this? Regarding the 0.0 bar I assume that this is relative to atmospheric pressure (e.g. "barg"). I would agree that this is strange, but there is no way to confirm this until we actually see the ERV report.

What difference would 103.9 C versus 100.1 C make? Almost all the energy is used to vaporize the water. Unless the pressure is a lot greater than 0 psig which would increase the temperature needed to vaporize the water.

It seems all that is important is that the water was vaporized to do the energy calculations. It is a given that the temperature was high enough to vaporize the water for this system to work.

If you don't know the specific volume of the produced steam, then you know nothing about the energy.

All I am saying is if you have 14 oz (414 g) of water just below the vaporization temperature and you add 1 MW/sec of energy it will all be vaporized in one sec. This can then be compared with what actually happened. If the 1 MW system worked then this is what was happening, based on the reported 36 cubic meter per day water flow number.

If the system did not vaporize all (any) of the water then the power generated would be less than 1 MW.

If the system did not vaporize any of the water, the power generated would be around 70kW, assuming a return temperature of 60C.

This is why a temperature of 100.1C would be concerning, because we wouldn't have a safe degree of certitude that water was vaporized.

However, chances are this 100.1C was BS. Dewey then switched to 103.9C and Rothwell was pretending he reverse engineered it from the publically available data, when in fact you could come up with any temperature value between 100 and 140C from that data.

Still waiting for Rothwell to access his lab computer and give us real values.

Quote from LENR Calender

Still waiting for Rothwell to access his lab computer and give us real values.

Sorry, I forgot. It is ~103 deg C but for various reasons I think this is too high. Indications are it was hot water and not steam.

Based only on the numbers Rossi gave Lewan, it would be just over 100 deg C, as I said.

Jed, can you elaborate on those "various reasons"? Moreover, how do those various reasons allow you to derive such an exact value as the 100.1°C you have previously stated several times? I have to admit that it seems like a strange coincidence to me that now after Dewey gave up the 100.1 and brought up the 103.9 instead, you suddenly discover that it was "~103" as well.

Jed,

Thanks for giving us that value.

You might have missed my previous post. But, based on the vortex post you've been linking, the output temperature could be 100.1 or it could be 103 or 120 deg C.

It only takes a few kW to heat vapor up 1 degree extra, so there is no reason you would deduce a 100.1C based on the 1MW dimensioning of the plant.

Does that make sense?

Rossi must be totally crazy If he really has stated, what has been quoted by Shane D..

Here is another example (comment by Rossi in his JONP):

Quote

Andrea Rossi

June 9, 2016 at 1:12 PM

Norma Gaylor:

...
...

Let me remember that the ERV is an expert nuclear engineer, with
experience of work in nuclear power plants and in certifications of
industrial plants, mainly in the energy field: for this reason he has
been chosen jointly between Cherokee Fund ( from which eventually has
been born IH ) and Leonardo Corporation.

Warm Regards,

A.R.

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He had signed the License Agreement with Industrial Heat LLC, neither with "Cherokee Fund" nor another entity "from which eventually has been born IH".

The only question is whether he completely has lost touch with reality or it's an example how he deceives his followers.

Quote from JedRothwell

Quote from LENR Calender: “Still waiting for Rothwell to access his lab computer and give us real values.”
Sorry, I forgot. It is ~103 deg C but for various reasons I think this is too high. Indications are it was hot water and not steam.

Based only on the numbers Rossi gave Lewan, it would be just over 100 deg C, as I said.

Jed,

Thanks for sharing that value from your computer.

I need once again to dispel the myth that the tempearature would be just over 100 deg C, based on the numbers Rossi gave Lewan.

You are for some reason ignoring/skipping my other posts where I have done so.

So here we go again.

You base your 100.1C value on the calculations made in the following post (of which I will quote the relevant part):

http://www.mail-archive.com/vo…eskimo.com/msg109919.html

Quote

The circulating water amounted to 36m^3/day, which gives ~0.41l/s, that's
410g/s. 1g of water needs 4.1J/g to increase 1C, so that quantity of water
absorbed ~1700W for every C. Water entered at 60C, so, we have that the
liquid part absorbed ~70kW. The latetent heat of water, assuming, 1 atm,
is ~2.3MJ/kg. So, 410g/s of water can absorb ~930kW by completely
vaporizing.So, we have that an amazing coincidence 1MW (~70kW + ~930) is nearly
*exactly* what is needed to vaporize the flux in the circuit.

Display More

From my own calculations, the results in this post are correct, some values were rounded (no big deal).

However, it is erroneous to deduct a 100.1C value from this.

The specific heat of water, as said above is 4.1 J/g. Specific heat of vapor is 1.890 at 375K.
source: http://www.engineeringtoolbox.com/water-vapor-d_979.html

So, if heating that water up 1 degree takes 1700W, heating the same flow of vapor 1 degree takes around 800W.

Therefore, if heating 60C water to 100.1C does take approximately 1MW:

- heating 60C water to 103C takes 1.002 MW
- heating 60C water to 105C takes 1.004 MW
- heating 60C water to 110C takes 1.008 MW
- heating 60C water to 120C takes 1.016MW
- heating 65C water to 110C takes 0.995 MW (note I upped the input temperature a bit)

So how accurate is your 100.1C extrapolation? The only way you could extrapolate this value is if you had the exact input temperature, as well as the e-cat's power ouput known within a kW!

So will you please stop saying that you obtained this 100.1C value based on Rossi's data? Especially when your own insider data says 103C!

Jed

Are you making this stuff up? We need evidence, not opinion remember.

Best regards
Frank

Quote from LENR Calender

So how accurate is your 100.1C extrapolation? The only way you could extrapolate this value is if you had the exact input temperature, as well as the e-cat's power ouput known within a kW!

That is what Rossi claimed. Exact values. It is 102.8 deg C to be exact.

He also claimed it was exactly 35 tons of cooling water per day, and then he arbitrarily subtracted 10% from that, for no apparent reason. As he told Lewan.

Based on my analysis and his choice and layout of instruments, I don't believe either number. At all. I also don't believe the 60 deg C return water temperature. That is too low.

If Rossi ever gives you the data, you will see for yourself. Perhaps you will agree with him, and perhaps you will agree with I.H. At present, you have no basis to agree or disagree with anyone.