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no excess heat in June 14 Rossi demo, as no invisible dry steam at end of hose, just feeble mist, perhaps liquid water -- many unbiased critical comments on Vortex-L: Rich Murray 2011.06.25

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  • Rich Murray
    no excess heat in June 14 Rossi demo, as no invisible dry steam at end of hose, just feeble mist, perhaps liquid water -- many unbiased critical comments on
    Message 1 of 1 , Jun 25, 2011
      no excess heat in June 14 Rossi demo, as no invisible dry steam at end
      of hose, just feeble mist, perhaps liquid water -- many unbiased
      critical comments on Vortex-L: Rich Murray 2011.06.25
      Saturday, June 25, 2011
      [at end of each long page, click on Older Posts]
      [you may have to Copy and Paste URLs into your browser]

      I've suggested Rossi, searching blindly the last 2-3 years for a way
      to prevent thermal runaway, has stumbled into using an input electric
      power that vaporizes some of the water flow, without initiating any
      nuclear reactions, while managing to believe that the output mist is
      "visible steam", as he clearly stated at the end of the 13:24 minute
      video demo by Steven Krivit June 14, 2011...

      After 11:00 minutes, Rossi lifts up the last 1 m of the 3 m black
      water outlet hose, then drains water from the end of the hose into the
      blue bucket, then shows the end of the hose against a black sweater --
      clearly the white mist slowly coming directly from the end of the hose
      is not steam, which is invisible, but water mist.

      This proves that very little steam must be coming out of the reactor
      at the start of the 3 m hose.

      Both water and water mist are flowing at 7 liters per hour, 1.94 cubic
      centimeter per second [ 7/3600 = .001.94 liters per second ].

      If all the water was turned into steam, that would be 1.94X1700 =
      3300 cc per second of steam, 3.3 liters per second from a hose with an
      inside diameter of about 2 cm.

      13:24 minutes June 14, 2011

      2011 - Andrea Rossi Explains His Energy Catalyzer

      Uploaded by StevenKrivit on Jun 20, 2011
      Link to New Energy Times reports: http://tinyurl.com/4362kl9
      "Steam": 11:30
      [ mist emerging feebly directly from 3 m black hose after 11:00 minutes... ]

      10:45 minutes June 14, 2011

      2011 - Andrea Rossi Crunches the Numbers for His Energy Catalyzer
      Uploaded by StevenKrivit on Jun 23, 2011
      Link to New Energy Times reports: http://tinyurl.com/4362kl9


      from Rich Murray rmforall@...
      to vortex-l@...
      date Mon, Jun 20, 2011 at 7:01 PM
      subject Re: [Vo]:[Video] Andrea Rossi Explains His Energy Catalyzer
      (NET - June 14, 2011)
      Jun 20 (5 days ago)

      I agree the gas flow out the end of the black hose seems to be visible
      right at the end -- whereas steam would be invisible for a short

      Trained as a dishwasher since age 10, 80 miles E of Houston, Texas, I
      am sure that hot water gives off mist in low altitude, warm, humid

      Rossi seems to be saying that "cool" steam is slightly visible as a
      mist, while "hot" steam is invisible!
      All steam is invisible, by definition.

      Rossi seems to me to be natural, relaxed, matter of fact, genuine.

      Isn't it possible for the pump to fill the reactor up totally with
      water, which would then overflow and exit as water just below boiling,
      or water exactly at boiling, mixed with variable amounts of steam?
      Would any bubbling at the outlet of the reactor be audible?
      How noisy is the background?

      Since about 1 m of the hose lies on the floor, before rising about1.5
      m to pass through a hole in the wall, wouldn't that part of the hose
      on the floor fill up completely with water, with a flow of 7 kg/hour?
      How much pressure results from the 1.5 m rise in the hole?
      Also the hose on the floor, if full to 1.5 m, would be equally full on
      both arms of its "U" bend...
      If so, then would that ensure that all steam is condensed while
      passing through a full "U" bend?

      How much output heat is there if very little of the water is boiled
      within the reactor?

      My guess is that the Rossi team actually don't have a clue about what
      is happening between the device outlet and the far end of the hose.


      from Rich Murray rmforall@...
      to vortex-l@...
      date Wed, Jun 22, 2011 at 9:52 PM
      subject Re: [Vo]:E-Cat proven to be a hoax?
      Jun 22 (3 days ago)

      Well, maybe Rossi has spent 2 or 3 years with a setup that really
      generates gross excess heat energy from LENR, but is explosively
      unstable -- as the temperature is raised to the level that initiates
      LENR, the resulting gross nuclear energy release, naturally,
      immediately rises so steeply as to overwhelm such control parameters
      as H2 pressure, H2O flow, heat input from electric heater -- finally,
      he finds a setup that generates 6 to 12 times more energy than input
      heat, BUT --

      1. he started assuming complete boiling of the water flow into dry
      steam, whereas actually only a small fraction of the water is ever
      boiled in his stable runs, so that,

      2. the claimed output heat is exaggerated by 6 to 12 times input
      electric heater power,

      3. and, highly motivated to finally have a complete success as an
      inventor who contributes hugely to humanity and gains praise and
      wealth and opportunity to continue inventing on a grand scale, he very
      humanly falls into unconscious habitual resistance about actually
      double checking the reality of completely dry steam output flow,

      4. so that close associates fall into this unconscious blindness,
      evolving a resiliant group think dynamic that presents a series of
      confusing demos that finally draw enough scrutiny for the possibility
      of the error to be discussed by many,

      5. whereupon Rossi, a good, honest and forthright man, will quickly do
      a simple check, verify the error, and share the discovery immediately
      and openly,

      6. and, since he lacks the expertise and resources to engineer how to
      stabilize the reaction (even if he understands it correctly...), he
      also immediately discloses every detail of the setup, so that the
      world as a whole can properly explore this crucial breakthrough for
      the benefit of all.

      from Joshua Cude joshua.cude@...
      reply-to vortex-l@...
      to vortex-l@...
      date Fri, Jun 24, 2011 at 10:27 AM
      subject Re: [Vo]:Rossi calorimetry, volume vs mass, etc.
      Jun 24 (1 day ago)

      On Fri, Jun 24, 2011 at 10:33 AM, Daniel Rocha <danieldiniz@...> wrote:

      *Just to be sure of my position. I am completely convinced that the
      data that has been provided is coherent with a power generation of

      But the presented data is also consistent with power equal to the
      input electrical power of 800W.

      That's Rossi's con. If he restricts the data to temperatures, and
      input flow rate, and brings the flowing water to a boil, the same data
      can represent output power over a 7-fold range.
      He of course claims the high end of that.
      And until Krivit, he was not directly challenged.
      Even Krivit's challenge (so far) is pretty mild.

      Jeff Driscoll hcarbon2@...
      reply-to vortex-l@...
      to vortex-l@...
      date Sat, Jun 25, 2011 at 7:11 PM
      subject Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
      7:11 PM (1 hour ago)

      You might be thinking of another scenario -- but if I'm guessing what
      you are saying then the best anyone could do is about 1.86 to 1 ratio.
      But this assumes that any liquid hot water needed to cool water vapor
      in a heat exchanger is included in the calculation (otherwise the
      ratio would be worse, less than 1.86 to 1). I did this calculation,
      shown below, weeks ago.

      Basically in this fraudulent set up, the Ecat would do the following:

      1. Create 1 kg of 99.9 C water from 10 C water which requires (99.9 -
      10) x 4.18 kJ/kg/C = 376 kJ

      2. Using same water from step 1, make 1 kg of water *vapor* requiring
      2257 kJ. Total input to Ecat required at this point is 376 + 2257 =
      2633 kJ

      3. Condense water vapor into micro droplets (i.e. fog) deep *inside*
      the Ecat using a heat exchanger and use this heat to heat 6.00 kg of
      cold liquid water from 10 C to 99.9 C. This is because 2257 kJ /376
      kJ/kg = 6.00 kg (note that the units are correct). Also, note that at
      this point the total input energy is still 2633 kJ.

      The actual/real end result is 6.00 kg of 99.9 C water and 1 kg of
      micro liquid water *droplets* (fog or steam with 0% quality).

      A gullible observer would think that the Ecat just produced 6 kg of
      hot water and 1 kg of water *vapor* when it really made 6 kg of hot
      water and 1 kg of hot *liquid* water droplets.

      The gullible observer would think that the energy normally needed to
      create this is 4890 kJ because:

      (6 kg + 1 kg) x (99.9 - 10) x 4.18 kJ/kg + (1 kg) x 2257 kJ/kg = 4890 kJ

      While in *reality* it took the following amount of electrical energy:

      (6 kg + 1 kg) x (99.9 -10) x 4.18 kJ/kg = 2633 kJ

      So, the gullible observer would see 2633 kJ of electrical energy go
      into the Ecat and 4890 kJ of thermal energy leave the Ecat.
      This is a ratio of 4890/2633 = 1.86

      I can't think of any way of increasing this ratio using any other
      similar method.


      from mixent@...
      reply-to vortex-l@...
      to vortex-l@...
      date Sat, Jun 25, 2011 at 5:58 PM
      subject Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
      5:58 PM (2 hours ago)

      In reply to Joshua Cude's message of Fri, 24 Jun 2011 16:20:48 -0500:

      >I was talking about running it above boiling, but way below the level needed
      >to boil it all. Different thing. And it's easy. The power can range within a
      >factor of 7. In this case, anywhere between 600W and about 5 kW.

      BTW (the latent heat of steam) / (the heat energy required to bring water to the
      boil) is a factor of about 6.7 (depending on starting temperature of water), and
      curiously close to the COP Rossi claims to be aiming for.

      In short, if virtually none of the water were converted to steam, and he was
      assuming that it all was, then it would neatly explain the conversion factor he
      is claiming.

      Regards, Robin van Spaandonk


      from Stephen A. Lawrence salaw@...
      to vortex-l@...
      date Fri, Jun 24, 2011 at 3:36 PM
      subject Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat

      Stephen A. Lawrence

      YOW -- WHAT YOU JUST SAID !!!!

      On 11-06-24 04:20 PM, Jed Rothwell wrote:

      *So the only way for Rossi to make it produce a little steam and a lot
      of hot water would be for him to adjust the anomalous heat output
      It would be a miracle if Rossi has such good control over the
      anomalous heat that he can push the temperature up to 99°C and have
      mostly liquid water go through plus a little steam.
      If he can do that, he has truly mastered cold fusion!

      Jed, man, think about that -- don't just jerk your knee at me in an
      automatic defense of Rossi, really think about it.

      Rossi has a factor of SEVEN in output level in the range he has to hit
      in order to produce SOME steam and SOME hot water, and you have just
      said it would be hard for him to control the anomalous heat well
      enough to do that.

      But Rossi's claiming to have produced exactly enough heat to EXACTLY
      vaporize all the input water, and NOT HEAT THE STEAM beyond boiling
      --- that target is orders of magnitude smaller than the target he'd
      need to hit to produce some steam and some hot water!
      If he overshoots his "dry steam" power level by even a little, the
      steam temperature will go up by a lot;
      the specific heat of steam is very small compared to the heat of
      vaporization of water.
      But the temperature never rises more than about a degree over boiling!

      Jed, the point you just made is the point that's been bugging me all
      along -- it would take a miracle of fine control to generate EXACTLY
      enough anomalous heat to EXACTLY vaporize all the input water, without
      superheating the steam, and without leaving wet steam or having the
      device spit water!

      There's no evidence of that degree of control, no evidence of a
      feedback loop which could be providing it, no reason except wishful
      thinking to believe such control exists ...
      so the conclusion is that he's actually got the power level set
      somewhere within the "factor of 7" window, and he's producing very wet
      steam or a mix of steam and liquid water;
      he does *NOT* have it "right on the edge",
      producing dry steam just over the boiling point.
      It's absurd to think he could exercise the level of precise control
      needed to produce "exactly dry steam".

      (And that about uses up my Friday night send-some-useless-email time...)

      [ Here's a credulous slant... ]

      15 mins 51 secs, 65.83 MB, Flash Video 480x360, 25.0 fps, 44100
      Hz, 567.05 kbits/sec

      Brian Josephson
      June 24th, 2011 at 7:02 AM

      Our ‘video FAQ’ on the Rossi reactor is now available on our media server at
      http://sms.cam.ac.uk/media/1150242 ,
      as well as at the original youtube location
      (where the video has already had more than two thousand views in less
      than 3 days).
      The version at sms.cam.ac.uk is recommended as it has been improved
      somewhat (videos cannot be updated on youtube, unfortunately), and
      also includes a transcript (embedding code is also available, included
      at the end of the transcript).

      Rich Murray, MA
      Boston University Graduate School 1967 psychology,
      BS MIT 1964, history and physics,
      1943 Otowi Road, Santa Fe, New Mexico 87505
      505-819-7388 rmforall@...


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