I really want Rossi to be the real deal, but this just doesn't look like enough steam for the power they are talking.
http://www.youtube.com/watch?v=m-8QdVwY98E
http://www.youtube.com/watch?v=m-8QdVwY98E
Merged Cold Fusion Claims
- Thread starter JoelKatz
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No, there's a whiff of visible vapour, but to me it doesn't look as if it's mixed with much invisible steam. I'm no expert, however.
And why the white walls? Nothing can be seen until Krivit offers his t-shirt as background to display the steam.
And why the white walls? Nothing can be seen until Krivit offers his t-shirt as background to display the steam.
Are we reaching a dénouement? Defkalion is to hold a press conference Thursday 23rd June, 2011 @ 14:30 Municipality of Palaio Faliro (it's a southern suburb of Athens):
http://energycatalyzer.blogspot.com/2011/06/defkalion-green-technologies-will-hold.html
According to the advance publicity:
A Greek nickel speculative bubble is one of my pet theories about all this; but we'll know everything in two days because we are promised
Ooh, it's all too exciting!
http://energycatalyzer.blogspot.com/2011/06/defkalion-green-technologies-will-hold.html
According to the advance publicity:
A Greek nickel speculative bubble is one of my pet theories about all this; but we'll know everything in two days because we are promised
Ooh, it's all too exciting!
Dancing David
Penultimate Amazing
better would be to just put the unit in a tub of water and measure the heat output.
Yevgen Barsukov
Student
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You are quite right. Rossi says that they are evaporating 7 kg / hr.
This makes it 1.9 gm per second.
Volume of water vapor at room temperature can be calculated
as 0.804 g/liter, e.g. the pipe should exhaust 1.9 / 0.804 = 2.36 liter / sec
of 100C vapor. This is a tremendous flow through what looks like maximum
5 mm radius pipe!
Generously assuming it is actually r= 5mm, we get exhaust area of 5 mm^2*pi = 79*10^-6 m^2, which gives us ( 2.36 *liter/sec / 79*10^-6m^2 ) the exhaust velocity of 29.87 m/sec.
According to the Beaufort scale here:
http://en.wikipedia.org/wiki/Beaufort_scale
..this is a violent storm-level wind!
Not only would it cause a violent roar, it would also apply to the end of the pipe acceleration of ~2 gm / sec which would cause it to twist and turn like a snake. It would be hard to hold it in your hand.
However, video shows that there is no obvious force exerted to
the pipe, neither does it make any sound.
I think this pretty much settles it. Now we have actually seen it with our own
eyes, kudos to Steven Krivit...
BenBurch
Gatekeeper of The Left
Correct. Specifically a tub so large that over the course of the test the water never gets anywhere near boiling.
One calorie (small c) raises one gram of water (a ml at std temp) one degree Celsius.
If you know the volume, the starting temperature, and the heat rise, and the tank is well insulated, you know with certainty the total number of calories generated. Period.
Allowing the thing get warm enough to evolve steam simply complicated the whole measurement.
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a) Rossi says the exact opposite right in this video.
b) The whole point of the "steam dryness meter" in the earlier demo was to show that the hose was carrying steam, not liquid water.
c) The only basis for power measurement is the *assertion* that the hose is carrying dry steam. You now seem to believe that the hose could carry liquid water as well, meaning that the water volume is not a measurement of the power. In other words, you think Rossi's "calorimetry" can't be trusted. Great!
Want to rethink your statement?
Yevgen
I dispute these claims.
A 2g/s^2 acceleration would cause the hose to twist and turn? Isn't that the equivalent acceleration of 2ml of plain water flowing out of the end of the hose per second?
2.3 Liters per second passing through a 1cm aperture makes an violent roar? So if I took a spherical plastic bag with a radius of 8cm (about the size of medium ziplock bag) and expelled the air inside it in one second through a 1cm hole it would make a violent roar?
A very good experiment would be to calibrate the observation by hooking up a common kitchen kettle to a 4 meter hose. They usually have around 1.5kW heating elements in them. I wonder what the steam from a common kettle would look like after travelling through 4 meters of rubber tubing?
I dispute these claims.
A 2g/s^2 acceleration would cause the hose to twist and turn? Isn't that the equivalent acceleration of 2ml of plain water flowing out of the end of the hose per second?
2.3 Liters per second passing through a 1cm aperture makes an violent roar? So if I took a spherical plastic bag with a radius of 8cm (about the size of medium ziplock bag) and expelled the air inside it in one second through a 1cm hole it would make a violent roar?
A very good experiment would be to calibrate the observation by hooking up a common kitchen kettle to a 4 meter hose. They usually have around 1.5kW heating elements in them. I wonder what the steam from a common kettle would look like after travelling through 4 meters of rubber tubing?
Aepervius
Non credunt, semper verificare
Either you have the same steam as at the start of the rubber, and thus the same speed odf 2.3 liter per second of gas, OR you have condensation in the middle of the rubber, then a lower output of steam, but then liquid coming out of the rubber. Which is it you are telling us is ?
Maybe 2.3 liter per second cannot be called a "violent" roar, but it is certainly not an easy-peasy little steam coming out.
Aepervius
I am saying that rather than make wild meaningless claims a more potent use of time would be to hook up a kettle to a hose and observe the output.
I find the calculations presented so far unconvincing. It wouldn't surprise me at all if no steam at all came out the end of a 4 meter hose hooked up to a common kettle.
Someone who conducted the experiment and showed experimentally how much energy a 4 meter rubber hose might dissipate along its length would do the first actual work towards debunking the e-cat since the initial announcement.
I am saying that rather than make wild meaningless claims a more potent use of time would be to hook up a kettle to a hose and observe the output.
I find the calculations presented so far unconvincing. It wouldn't surprise me at all if no steam at all came out the end of a 4 meter hose hooked up to a common kettle.
Someone who conducted the experiment and showed experimentally how much energy a 4 meter rubber hose might dissipate along its length would do the first actual work towards debunking the e-cat since the initial announcement.
BenBurch
Gatekeeper of The Left
There is no need to do any experiments to debunk a setup that cannot be replicated, and was stupidly, incompetently wrong on its face...
BenBurch
What? You don't want to definitively disprove to Rossi experiment? What about all the old ladies you "rational skeptics" were crowing about protecting from evil scammers in this matter.
If you aren't motivated by altruism and a love of old ladies what is the explanation for all the snide comments and rude remarks?
Are you unwilling to do the experiment because you don't have a kettle? Is that also why you are such an angry man?
What? You don't want to definitively disprove to Rossi experiment? What about all the old ladies you "rational skeptics" were crowing about protecting from evil scammers in this matter.
If you aren't motivated by altruism and a love of old ladies what is the explanation for all the snide comments and rude remarks?
Are you unwilling to do the experiment because you don't have a kettle? Is that also why you are such an angry man?
Kid Eager
Philosopher
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- 7,296
Hi,
by my quick calculations (which may be wrong but feel free to correct), a flow of 2.3litres per second would be pushing a column of *something* over 2.8 metres long out of the end of the 1cm hose each second. You cannot replicate that flow rate with your garden variety kitchen kettle (and yes, I just had a play with my kettle).
by my quick calculations (which may be wrong but feel free to correct), a flow of 2.3litres per second would be pushing a column of *something* over 2.8 metres long out of the end of the 1cm hose each second. You cannot replicate that flow rate with your garden variety kitchen kettle (and yes, I just had a play with my kettle).
BenBurch
Gatekeeper of The Left
I have offered to do a competent honest job of calorimetry. Free. And trust me, you do not want to pay my hourly rate.
But trying to replicate a part of a magician's trick (for thats all this was) merely by trying to blow steam through a hose is simple stupidity.
As it happens, the behavior of steam is something I know rather a lot about, and can tell you what happens when you use a rubber hose to move live (dry) steam around; not much. You get a little condensation until you get a temperature equilibrium, but then you get live steam for pretty much as long as you would like assuming you have a real source of live steam. Steam is moving though the hose fast enough that it cannot lose much energy to the heated hose, and rubber is a bad conductor of heat.
Kid eager
I think Yevgen's calculation (which assumes the entire 1.9ml/s of water is converted to dry steam) shows 29.8m/s at the opening of the hose. How do you calculate 2.8 meters? I suppose the stream of gas would expand significantly after it exits the tube?
If you apply Yevgen's calculation to the case of an electric kettle (1.5kW) you would vaporize around 0.6 grams of water per second. Generating ~0.75l/s of dry steam. The velocity of steam exiting the hose would thus be around 9.5m/s. Is that the rate of dry steam which you observed in your kettle experiement?
I think Yevgen's calculation (which assumes the entire 1.9ml/s of water is converted to dry steam) shows 29.8m/s at the opening of the hose. How do you calculate 2.8 meters? I suppose the stream of gas would expand significantly after it exits the tube?
If you apply Yevgen's calculation to the case of an electric kettle (1.5kW) you would vaporize around 0.6 grams of water per second. Generating ~0.75l/s of dry steam. The velocity of steam exiting the hose would thus be around 9.5m/s. Is that the rate of dry steam which you observed in your kettle experiement?
BenBurch
Excellent qualitative description. What is the nature of this equilibrium and how much of the steam's energy does it take to maintain the equilibrium? How much water initially condenses? What is the effect of the condensed water entrained in the hose on the steam quality?
Do you honestly think that if the e-cat was genuine that Rossi would take you up on this offer? You constantly repeat this little quip, if you were doing stand-up comedy at this point the audience would be dead silent.
Excellent qualitative description. What is the nature of this equilibrium and how much of the steam's energy does it take to maintain the equilibrium? How much water initially condenses? What is the effect of the condensed water entrained in the hose on the steam quality?
Do you honestly think that if the e-cat was genuine that Rossi would take you up on this offer? You constantly repeat this little quip, if you were doing stand-up comedy at this point the audience would be dead silent.
BenBurch
Gatekeeper of The Left
My opinion;
I honestly believe that if Rossi were even remotely honest, that proper calorimetry would have been done already. The fact that it hasn't been says he is a crook and a swindler.
Now, you want me to calculate for you what every steam plant engineer already knows empirically? I can do that for you!
I need;
1. The type of hose, specific manufacturer, and length.
2. The orientation of the hose.
3. Your signed contract for my regular hourly rate $150/hr.
I honestly believe that if Rossi were even remotely honest, that proper calorimetry would have been done already. The fact that it hasn't been says he is a crook and a swindler.
Now, you want me to calculate for you what every steam plant engineer already knows empirically? I can do that for you!
I need;
1. The type of hose, specific manufacturer, and length.
2. The orientation of the hose.
3. Your signed contract for my regular hourly rate $150/hr.
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