 |
| Peter Ekström |
Today Peter Ekström posted his thoughts on the production of steam.
Quote from the publication;
"I think Steven Krivit’s video from his visit to Bologna is very revealing. In reference (1c) one can see a little puff of mist at the end of the tube. If we had invisible steam rushing out of the tube at a speed of 11 m/s (Calculation 1 below), the puff of mist would certainly be blown away quickly!
The E-Cat runs on 5 kW power. So what we can see coming out of the tube is the total production of E-Cat running at 5 kW! This little trickle is hardly useful for heating or producing electricity! Note that losses in the tube are small according to Rossi (1a) and that all water is converted into dry steam (Calculation 2).
Where does the power go? Out of the E-Cat or the tube? Not very likely since the losses are small, 5 kW is a lot of power and it would heat the room perceptibly.
The only remaining explanation is that the E-Cat does not produce excess energy. The input electrical power is 3.5*220=770 W (2b). It takes 608 W to heat 7 kg of water in an hour from 25 to 100 C and boil it (Calculation 2). The remaining 770-608=162 W is used to evaporate a small fraction of the water. Since the temperature of the output water is not actually measured, it is quite possible that the temperature is less than 100 o C, which would leave some additional power to evaporate water.
So, where is the water then? It should be trickling out of the tube. When Rossi removed the end piece of the tube, he very deliberately emptied a fair portion of the tube from water, and subsequently held the end above the level of the rest of the tube (1b). It would take a significant time for the tube to be filled with water and the water would trickle out (Calculation 3)."
The Calculations referred to can be found in the publication.
Here is (again) the two videos mentioned;
The references mentioned in Ekström´s PDF is in this video is at
1a "very small condensation": 10.40 (minutes.seconds)
1b The tube trick: 11.00
1c Steam against T-shirt: 11.30
1d "In this moment we are making 7 kilograms of water per hour": 12.15
The references mentioned in Ekström´s PDF is in this video is at
2a Temperature difference: 3.35 (minutes.seconds)
2b Ampére meter: 09.20
So what to say? Many would answer "Emperor´s New Clothes".
Andrea Rossi comments on his blog.
I think Ekström is right. In this video there is no evidence for excess energy! The voltage in Italy is 230 V (not 220 V) which is the same voltage used in France or Germany. And I think that the flow rate of the water is less than 7 kg per hour. The steam comes out much too slowly!
ReplyDeleteExactly. Here is the calculation using the numbers he provided:
ReplyDeleteElectrical energy used per hour: 3.4A*230V*3600s=2815200J~=2.8MJ
Energy required to heat water from 26°C to 100°C during this time:
7kg*74K*4200J/(kg*K)=2175600~=2.17MJ
Leaves us about 600kJ energy/hour, or 600kJ/3600s = 166W more in from the grid than used for heating.
If all the water would be evaporated it would provide excess energy, since enthalphy of evaporation is 2257kJ/kg for water, so about 15KJ for the 7kg he claimed to have used in one hour.
BUT the conversion to steam would manifest in mechanical power as pressure rises. That burp of steam should be able to drive a steam engine of several kilowatts if it is in fact all steam. I can't imagine that happening, in fact it looks even too small for the 166W excess power.
This video was very disappointing to watch, as I was hoping this would be the real deal up to this point. So I guess we will have to continue with existing less capable energy solutions for the time being :-/
What's with all the professors, scientists and investors? These have to notice the obvious scam. I can not understand it all.
ReplyDeleteSome professors are full of bullshit and do talk before connecting their brain.
ReplyDeleteIf Ekstrom maths was right there could not be any steam in the hose, because, the thermal losses in the hose would dissipate the energy left to evaporate the water well before the end of the hose.
The reality is that without practical knowledge of applied thermodynamics, heating systems and energy power plants working it is to much easy to look at the textbook and forget that reality is always a bit more complex and before talking is better to know or test in practice.
For example, the hose used by Rossi is rated to be heat resistant (and pressure resistant too IIRC) not heat insulating.
The expansion of the steam just out of the hose is not easy to model (because it is a gas that go out with a linear speed but not an high pressure). Take in account the expansion of the steam in air (in all directions, the linear velocity of the steam, the volume of the steam (dependent on the fraction condensed in the hose), the resistance of the air.
@painlord2K
ReplyDeleteThat is true until the hose got heated past boiling then delta T would be minimal. How long was the machine on for? Too many variables. And the flow would obviously depend on the width of the hose. All in all it seems to point to at least partial evaporation which means that it could be over unity by this video evidence. BUT definately no way to verify 6x over unity.
The steam is never completely invisible so it's probably very close to 100 deg C. It then hits ambient air at 11m/s (allegedly) and loses energy while moving into a free environment. The end result (video evidence) seems plausible. I just can't tell one way or the other. It's not a very good test to prove anything unfortunately.
ReplyDelete