Data from the Penon report to be discussed

On page 11 of the report by Fabio Penon, available here

http://www.mediafire.com/download/l4vfaky8v9bx90d/105322688-Penon4-1.pdf

one finds this graph. The temperature is that measured for one of Rossi's hot-cats.

The following presentation is of a simulation written to reproduce this figure, and what can be inferred from it.

Input power

On p. 13 one finds the measured power vs time. This data has been transcribed into the simulation, and is plotted on the right.

Note on input power measurements

On p.11 of the Penon report, we find the remarkable paragraph:

“E-Cat power supply was effected through a control box panel provided with a kWh meterwhich did not allow separate evaluation of the voltage and current supplied to the module.For this reason, a voltmeter and a clamp ammeter were installed downstream from the controlbox, so as to monitor power data independently from the panel meter. Due to the fact thatpanel meter data were found to be quite discordant from those measured by the voltmeter andthe ammeter, it was decided to ignore the former and use only the voltmeter and ammeter datarecorded manually in the course of the test.”

I may not have read all the discussions on the web about this report, but this really jumped out at me. Has this not been picked up on by others??? In what direction was the discordance? Did the kWh metermeasure higher or lower input powers??? Why the discrepancy???

Even though it is not mentioned in the report linked on slide 1 of this presentation, we know from elsewhere(<http://pesn.com/2012/10/15/9602206_Penon_High-Temperature_E-Cat_Test-Results_Posted/121009_MIAMI_TEST_FINAL.doc> for example) that Penon used a PCE830 to measure input power, which cannot measure DC current or voltage, and hence cannot measure DC power. Is this the reason for the discrepancy noticed by Penon???

Nevertheless, I will continue the analysis using the table Penon quotes on p.13.

Simulation geometry

To model the reactor we need to assign three material constants for each layer:Density kg/m3, thermal conductivity W/(m.k) and thermal capacity J/(kg.K)

The mass of each layer is calculated from m = 2.π.r.L.dr.ρ

where r is the radius of the layer, L is the cylinder length, dr is the layer thickness and ρ the density.

In this model, the outer diameter is chosen to be 8.5 cm, the innermost one to be 1.9 cmand the layer thicknesses are 0.3 cm. Cylinder length L is 33 cm. Cylinder ends are neglected.

Three layers are of special interest, the outer layer which radiates and has its temperature measured, the resistance heater layer (orange) where input power is applied, and the inner reaction layer (red).

drr

Layer properties

There are different materials in the cylinder, while it is not known what all the layers are,some typical material values are:

Units: conductivity density heat capacity W/m/K kg/m3 J/kg/K

Steel: K=20, ρ = 7700, C = 490Air: K = 0.03, ρ = 1, C = 1010Corundum: K =30, ρ =4000, C = 420Brick: K = 1, ρ = 2000, C = 900Silicon Nitride: K = 30, ρ = 3200, C = 700

We know the outer and inner tubes are steel, but it is not clear what the resistor heater is made from.

By comparing the masses measured by Penon to the volume of material, the following vector of densities were used, each element of the vector corresponding to a different layer, outer layer on the left and inner layer on the right:

Density=[5000 1000 1000 5000 1000 1000 1000 1000 1000 1000 5000 7700]; % kg/m3

This gives: the outer tube a mass of 1.32 kg (Penon=1.27 kg)the resistor block a mass of 2.37 kg (Penon = 2.29 kg)the inner tube a mass of 0.83 kg (Penon = 0.71 kg)

Layer properties

By comparing the inner temperature to the outer temperature, values of thermal conductivity were assigned as:

K0=[20 10 10 10 10 10 10 10 10 10 10 20]; % W/m/K

And values of specific heat assigned as

C=[490 420 420 490 420 420 420 420 420 420 490 490]; % J/kg/K

While the actual values could obviously be somewhat different to these, the values used in the simulation seem not far from what they should be. It was found that small changes to these values did not make much difference to the final result.

Layer 4 was used as the resistive heater layer, and was given slightly different values than its immediate neighbours.

The last two layers represent the inner steel cylinder.

The values of K are then scaled by surface area and thickness of each layer:

Kn = K

0.A

n/dr; % W/K

Internal heat transfer equations

Pheat

m1

K1

C1

T1

m2

K2

C2

T2

Heat is transferred between neighbouring layers according to their material properties.The power transferred by heat conduction is given as:

Pheat

= ((K1 + K

2)/2) * (T

1 – T

2); % W

This power is lost from the hot layer and gained by the cold layer

∆E1 = -P

heat.dt; % J

∆E2 = +P

heat.dt; % J

This calculation is performed between all adjacent layers.

External heat transfer equations

The outer layer is allowed to both radiate and lose heat by air convection.

The radiation equation is

Prad

= ε A σ (T4 – T04); % W

with ε=0.8, and the convection equation follows that of the Levi report:http://arxiv.org/ftp/arxiv/papers/1305/1305.3913.pdf

Pconvect

= C'' (T – T0)n D3n-1 A (T – T

0); % W

with C''=1.32, n=0.25 and D=0.085 m. As both of these powers are lost from the outer layer, ∆E

loss1 = - (P

rad + P

convect).dt; % J

Due to the driving heaters of the reactor, power is applied directly to Layer 4, which gains energy ∆E

drive4 = +P

drive.dt; % J

Finally, the net energy entering (or leaving) each layer is summed for that layer, and the new temperature of the layer calculated from

Tn new

= Tn old

+ ∆En/(m

n.C

n); % K

Result from simulation using Penon's power data

Using a time step of dt=0.2 seconds, and increasing the drive power to Layer 4 accordingto the green power plot (normalised on this graph), one finds the reactor outer temperature in red, and the slightly higher core temperature in magenta.

This is compared to Penon's graph on the next slide.

Comparison of simulation to data #1

By copy-pasting the data from the previous slide on top of the Penon figure, and making sure the axes are comparable, one gets the above figure. The trace in blue is Penon's data, while the data in red is from the simulation just using the power schedule from Penon's PCE830 measurements.

The red data is generally below the blue data by about 100-150 K.

There are several things we can try to do to account for this. One way is to add a DC power offset, and see what happens.

Comparison of simulation to data #2

In these figures, a DC offset has been added to the resistive heaters, 200 W in the upper figure and 1550 W in the lower figure. It is clear that neither fits the data well.

Linear increase in DC power

By increasing the excess resistor power linearly from 200 W at the start to 1550 W at turn off, the data can be reproduced almost exactly. The only discrepancy is after power turn-off, where the simulation drops in temperature faster than the data.

It is also noted that even though the controller is on for the first small flat section of the curve (until the arrow), there seems to be no excess power there. The linear excess power is only applied after that time.

What if the power were generated internally?

While we have found a fit to the data, we do not need to imply that it is due to 'hidden' power which failed to be measured by the PCE830.

Let us take the claim at face value, and assume that excess heat is generated in the core of the reactor. The above figure shows in blue the 'hidden' power from the previous slide vs the temperature of the core. The red line is a parabolic fit to that data.

Let us assume in the next few slides that the red line represents a real Ni/H reaction.

Quadratic heat with temperature of core #1

Applying the quadratic heat production curve directly to the innermost layer of the reactor produces the above curve. This fits rather well, but most notably it also fits the temperature trend after power off.

Actually, the core power curve needs to be adjusted down slightly, because the previous 'hidden' power was applied at Layer 4, some of which does not get to the core. Hence the core power is overestimated and needs to be reduced. See next slide.

Quadratic heat with temperature of core #2

Keeping very close to the quadratic curve shown in slide 13, but scaling down the values slightly, one gets the above fit.

Note again that the initial 5 minutes shows no excess power. Is this the time it takes for the reaction to 'ignite'?

The core 'LENR power vs Temperature' curve is shown next.

Fitted LENR power vs Temperature

This is the data used to produce the fit on the previous slide. While I have no information on what it 'should' or 'shouldn't' be like, at least it seems reasonable: it increases with temperature, it is smooth, and it has an exponential 'feel' to it (though it is quadratic in this plot).

There is one puzzling fact (apart from its existence!): at room temperature the excess power is 200 W. Without this value, the fit to the Penon data is very poor at low temperatures.

200 W at room temperature would be very noticeable!!! Perhaps this only happens after the reaction is 'ignited', or requires some secret waveform, etc, etc, ...

Implications of 200 W excess at room temperature

This plot shows the same core power model, but now let's extend the time axis. One finds that if the LENR power curve that fits so well still applies after the input power is cut, then the reactor should self-sustain with an external temperature of 196 °C. This would be very noticeable!

In the Penon report, they weigh the reactor after operation, and it has obviously cooled to room temperature, so for whatever reason, the excess power must have dropped to zero. So this result is also slightly puzzling.

In this figure, one also notes that the core temperature (shown in magenta) reaches 1207 K, in reasonable agreement with the reported value.

T = 469 K

Discussion

A simulation was written in an attempt to match the data presented in the Penon report on Andrea Rossi's Hot-Cat.

Assuming that the correct values were measured for input power and output temperature, one can model the data in two ways:

A linear increase in unmeasured input power vs time from 200 W to 1550 W (slide 12)A quadratic excess heat with temperature of the inner core, presumably due to a Ni/H reaction (slide 15)

Of these two fits, the excess core heat model on slide 15 looks better, as it matches more closely the decaying temperature after the input power is switched off. However, this should be confirmed with a dummy run – after all, what is the actual thermal time constant of the empty reactor? A measurement of this is necessary to distinguish between the two cases.

Several significant puzzles remain:Why did the Watt-hr meter give very different results from the PCE830?Why does the excess power start only after 5 minutes? (slide 12)If the excess heat is due to LENR, why does the reactor not self-sustain at 200 °C?

Now a trend of possible LENR heat production has been found, the next analysis will be of Levi's report.