10KW LENR Demonstrator?

[Giorgio]

I also suspect that running too much water through the cooling jacket might lower the temperature in the reactor vessel to below the optimum level. For this reason, it may not be trivial to change the experiment in this fashion.

About 150 pages ago I suggested (and also suggested to Rossi) to use a low viscosity thermal oil that is stable from 70 to 250'C.

[chrismb]

Ah! But I recommended it 160 pages ago!! (No matter how small the sh!t pile, there's always a bug that wants to be top of it!! )

Let's just say 'we suggested it' ('cos, actually, I don't really know where any of these things were discussed now, the thread is sooo long) along with several other pieces of protocol that would be simple but fully informative.

[Giorgio]

This means that you should still have the equivalent of 4,5-4,3 Kw of steam coming out from that tube, and I think is clear now that this is not happening.

Bear in mind, also, my run was not an attempt at any sort of calorimetry. It was merely what 25g/min of gas from a boiling volume looks like. Doesn't matter what thermal losses, etc. there are from the pipe, matter cannot disappear.

Exactly, not only it cannot disappear, it also cannot undergo phase transformation without loosing heat.
This is why I find your video extremely interesting and a direct proof that the steam coming out from Rossi video cannot justify the production of 5KW/h of thermal energy.
Even allowing a 1KW dissipation on the pipe the rest of the 4KW (as steam) must come put of the line, and clearly this is not the case.

I will also re-iterate that there is no visible steam at the immediate exit to my hose. The hose was HOT-HOT and little opportunity for condensate within it other than what you see expelled intermittently!!

I run a small simulation out of fun with my thermal dissipation software.
456 w/h with a dT of 70 'C and a length of 6 meters.

Edited to fix a unit

[Joseph Chikva]

Next, you claim that the extra heat can be explained because of hydriding.
This cannot be for 2 reasons.

In case if jacket was empty before starting only electric input of 750W can provide about 250-300W heat for evaporation in case of claimed flow 7l/h.
So, Rossi has shown nothing. Thanks.

[Joseph Chikva]

I also suspect that running too much water through the cooling jacket might lower the temperature in the reactor vessel to below the optimum level. For this reason, it may not be trivial to change the experiment in this fashion.

About 150 pages ago I suggested (and also suggested to Rossi) to use a low viscosity thermal oil that is stable from 70 to 250'C.

What do you think why Rossi did not provide to Krivit data of big water flow?
When he cried "snake". Not because to show steam and fake 5kW of power? As that fake could not be shown without evaporation.

[Giorgio]

Next, you claim that the extra heat can be explained because of hydriding.
This cannot be for 2 reasons.

In case if jacket was empty before starting only electric input of 750W can provide about 250-300W heat for evaporation in case of claimed flow 7l/h.
So, Rossi has shown nothing. Thanks.

This is exactly what I am stating.

[Giorgio]

I also suspect that running too much water through the cooling jacket might lower the temperature in the reactor vessel to below the optimum level. For this reason, it may not be trivial to change the experiment in this fashion.

About 150 pages ago I suggested (and also suggested to Rossi) to use a low viscosity thermal oil that is stable from 70 to 250'C.

What do you think why Rossi did not provide to Krivit data of big water flow?
When he cried "snake". Not because to show steam and fake 5kW of power? As that fake could not be shown without evaporation.

This is exactly what I am stating.

[chrismb]

I run a small simulation out of fun with my thermal dissipation software.
456 w/h with a dT of 70 'C and a length of 6 meters.

There would be one other totally telling point, but 'fraid I have not done fluid flow for a couple of decades now - can anyone do this calc:

Even Rossi is saying that the speed of the gas would be in the order of several m/s. For a flow rate like that and a length of 6m or so, there would be an associated pressure drop. What is the pressure drop? Because the way I see it, it is quite possibly be enough that the boiling point would be raised a little in his device - but only if there is the flow rate [he is agreeing] of several m/s.

Can anyone do a quick plot of pressure drop versus gas veloocity for steam through a 1/2" ID 6 metre pipe?

[Joseph Chikva]

Input data have nothing to do here. The metal hydride cannot hold more than a fixed amount oh heat.

Even the most powerful D+He3 fusion reaction releases only the fixed amount of heat (kinetic energy).
So, you are wrong again.

What does D+He3 fusion has to do with your claim that the extra heat can be explained by hydriding?
You seem very confused sometimes.

That is not my problem if I seem so for you.
As you said "fixed amount oh heat" I answered that every reaction including the most powerful produces fixed amount of heat.
Thanks for discussion.

[Giorgio]

Even the most powerful D+He3 fusion reaction releases only the fixed amount of heat (kinetic energy).
So, you are wrong again.

What does D+He3 fusion has to do with your claim that the extra heat can be explained by hydriding?
You seem very confused sometimes.

That is not my problem if I seem so for you.
As you said "fixed amount oh heat" I answered that every reaction including the most powerful produces fixed amount of heat.
Thanks for discussion.

You are confused.

[Giorgio]

Can anyone do a quick plot of pressure drop versus gas veloocity for steam through a 1/2" ID 6 metre pipe?

http://www.engineeringtoolbox.com/steam ... _1093.html

[ladajo]

I did not follow the link, but does it account for moisture level? Or do we assume dry steam? It would seem so, after I looked at the link.

Edit: checked out link...

[chrismb]

Can anyone do a quick plot of pressure drop versus gas veloocity for steam through a 1/2" ID 6 metre pipe?

http://www.engineeringtoolbox.com/steam ... _1093.html

OK, perfect. Well, it's a doddle to push this house of cards over, then!

So we have;



In other words, if there was a flow rate of 7 litres/hr of dry steamcoming out of the pipe, then there would be a pressue drop of 11kPa. As we know the pressure at the end of the pipe is ambient, so the pressure in the E-cat must be just over 1.1 bar, and the boiling point at just over 1.1 bar is... around 104 degrees C, I believe.

So either he's not boiling water at his measured 100C in the E-cat, or he's not flowing 7l/hr. He can't be doing both.

[tomclarke]

chris


There is no way BP can be 104. Because only way to get stable 101 is if BP is 101. Therefore flow rate must be ~2l/h?

Where does the 7l/h come from? What assumptions?

Of course, BP 101 stable proves steam is wet (or it would not be stable, as I pointed out earlier).

Tom

[chrismb]

Where does the 7l/h come from? What assumptions?

No assumptions. This is Rossi's claim, and how he arrives at his power output figures.