# Re: [Vo]:Problems with energy calculation Rossi device

```Absolutely not. It's a flow.
Water get in and get out. It's not a closed container of water.
With 1 watt you can rise, when dynamic transients go away, an half degree!```
```
1 watt = 1 J/s
1 J/s = 1.8 g/s * 4.2 J/(g*K) * deltaT
deltaT = 0.1 kelvin (= 0.1 celsius)

If you don't trust me, read the report. It's a simple calculation!

From: P.J van Noorden
Sent: Saturday, April 09, 2011 5:41 PM
To: vortex-l@eskimo.com
Subject: Re: [Vo]:Problems with energy calculation Rossi device

Ofcourse. degr/sec is a typo.
But do you agree that if the temperature rise of a waterflow of 6.47 W/h is 30
degr/180sec( 0.15 degr/sec) the thermal power needed is only 1.13 W?

Peter
----- Original Message -----
From: Mattia Rizzi
To: vortex-l@eskimo.com
Sent: Saturday, April 09, 2011 5:05 PM
Subject: Re: [Vo]:Problems with energy calculation Rossi device

>I agree with you that when 300W is used the temperature rise will be: 300W /
1.8 x 4.2 J/gr degrC = 40 degr/ sec.

It's not 40 degr/sec.
It's 40 degress.
1.8cc = 1.8 g.
Flow is 1.8 g/s
1.8 g/s * 4.2 J/(g*K) * 40 K = 300 J/s = 300 watt

From: P.J van Noorden
Sent: Saturday, April 09, 2011 4:36 PM
To: vortex-l@eskimo.com
Subject: Re: [Vo]:Problems with energy calculation Rossi device

Hello Mattia,

When using a flow of 6.47 l/hour = 1.8 cc/ sec, and a temperature rise of
30degrC in 180 seconds (from the curve in the article = is 0.15 degr C/sec) the
thermal power would be only: 1.8cc x 4.2 J/gr degC x 0.15 = 1.13 W.

I agree with you that when 300W is used the temperature rise will be: 300W /
1.8 x 4.2 J/gr degrC = 40 degr/ sec.

To calculate the correct temperature increase one must ofcourse take into
account the mass of the complete Rossi device and the water inside the rest of
the machine.
The claim that the Rossi device would produce 4000W would in my opinion lead
to a continous evaporation of the incoming water, which is BTW also the claim
of Rossi.
This is the reason that I stated that in that case the curve would rise very
steeply.

Because I did a lot of experiments in the past in which I tried to measure
the heat release of systems at the bolinigpoint of water, I know that when the
system is boiling you must
use a very good separation between the boiling vessel and the condensor,
otherwise water will flow from one system into the other and the calculation of
the excess heat will be wrong.
I used in the past colour dyes ( and also radioactive material) to be sure if
all water was evaporated.
In the Rossi device the direct coupling of the "chimney" with the black tube
is not a good way to be sure that no liquid water will flow into the black tube.

To remove all doubts Rossi could easily demonstrate the exact amount of
excess heat by increasing the waterflow. A flow of 100ml/sec would lead to a
continous temperature increase of the water by about 10 degr C
if the power is 4 kW. I heard that he did that but I haven`t seen the report.

Peter

----- Original Message -----
From: Mattia Rizzi
To: vortex-l@eskimo.com
Sent: Thursday, April 07, 2011 6:11 PM
Subject: Re: [Vo]:Problems with energy calculation Rossi device

>When 4kW is added to such a waterflow the temperature would rise
instantaneously to 100 degr C.

You are confusing between static and dynamic condition.
It's physically impossible to have a instantaneously rise. You are missing
intertial thermal mass and dynamic conditions.

>The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part).
This would compare to a power of only a few watts.

Check your math. With 300 thermal watt you can rise around 40 degrees
celsius with 6.47l/hour.
But you will see this 40 degree rise only when the system is stationary,
when dynamic is over.

From: P.J van Noorden
Sent: Thursday, April 07, 2011 11:19 AM
To: vortex-l@eskimo.com
Subject: [Vo]:Problems with energy calculation Rossi device

Hello

In figure 6 of the article

it is stated that the waterflow is 6.47 l/ hour=  about 2 ml/sec. This is
about the waterflow of a espressomachine.
When 4kW is added to such a waterflow the temperature would rise
instantaneously to 100 degr C. The curve in figure 6 would have to rise
vertically.
If only 400 Watts is added the waterflow the temp would rise 50 degr C in
temperature continously.

The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part).
This would compare to a power of only a few watts.
Can anybody look at these calculations and figure out what is wrong?

Further it would be interesting to know if water can flow through the
"chimney" of the reactor directly into the black tube. To figure out what is
going on one have to add a substance (dye) to the water and see if the dye can
be seen in the " condensed" water.
If non vaporised water is carried to the end of the black tube this will
have consequences for the calculation of excess heat.

Peter van Noorden```