The 1 - 1.5 hr duration Rossi demonstration experiment typically
involves a preliminary heating phase that produces some steam and
water vapor, with elevated heating power supplied, about 1 kW,
followed by 30-40 minutes of steam production with 0.40 kW supplied.
Supporting background references are appended.
The Rossi experiment could be faked by chemically generating about 2
kW of energy for 30 minutes, or about 1 kWh, or 3.6 MJ. (See
justification appended.) This energy release can be accomplished by
(1) the generation of energy via the adsorbtion of water vapor by
zeolite or other exothermic water adsorbtion reaction, and (2) the
storage of thermal energy during the preliminary heating phase.
The energy supplied in a 30 min pre-heating phase could amount to up
to 0.5 kWh. This is not of major significance, other than the pre-
heating of water and zeolite (in separate compartments, but with
vapor access to the zeolite) in the device to near 100 °C is required
to supply water vapor to the zeolite upon (input power) demand. Now
to examine the chemical energy storage available.
Chemical energy could be produced by exposing zeolite in the device
to water vapor produced by boiling the water using (1) initially heat
from an electric heating element, and then (2) heat from the zeolite
water vapor adsorbtion itself.
One kg zeolite can generate 50 kJ in one minute, or 0.83 kJ/s = 830 W.
The total heat stored per gram of dry zeolite is 1.3 kJ. Total
zeolite required to generate 3.6 MJ is 2.3 kg.
The 2.3 kg can output 1.9 kW, right on target.
The density of zeolite can be about 1.6 gm/cm^3, so 2.3 kg requires
1.4 liters, well within the apparently available volume enclosed in
foil in the experiment.
Note that, if the device is configured properly, with the zeolite in
a container separate from the water, but configured so that heat from
the vapor adsorbtion by the zeolite is transferred to a stored pre-
heated (in first phase) water bath, the device can even exhibit "heat
after death", i.e. power production with no external input. Water
vapor boiled by the heat from zeolite water vapor adsorption then is
passed through the zeolite container wall to boil more water, and
perpetuates the heat generation by passing the resulting vapor
through the zeolite container, until the zeolite is saturated.
The main problem with this scenario is that it is very time limited.
A device which runs for a day or so would eliminate the ability to
reproduce Rossi's results by chemical means.
Supporting background information follows.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
According to: Gopal et al, "THE RATES OF SOLAR ENERGY STORAGE AND
RETRIEVAL IN A ZEOLITE-WATER SYSTEM":
http://engineering.ucsb.edu/~yuen/references/ref-2.pdf
Power released at 0.07 kJ / (min. gm.) = 70 J/(min gm) = 1.2 J/(gm s)
= 1.2 W/gm = 1.2 kW/kg
One kg zeolite can generate 50 kJ in one minute, or 0.83 kJ/s = 830 W.
The total heat stored per gram of dry zeolite is 1.3 kJ.
The density of zeolite can be about 1.6 gm/cm^3.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
From: http://newenergytimes.com/v2/news/2011/36/3623rf-celani.shtml
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
"The experiment started at about 15:30 and ended at about 16:45.
"The measurement of energy emission was based on a modified flow
calorimeter method (peristaltic pump, small size, about 10-20W of
power). They warmed up the water to 102°C, pressurized vapor
condition. I estimate that the experiment consumed 12-14 liters of
water."
"The amount of the reactant wasn't clear, but it could be a few
grams. According to Rossi, it is a complex mixture of nickel and one
or two secret additives, which are the key for the energy emission.
All the material is in the state of nano-particles or colloid."
"In the pressurized (about 2 atm) chamber, the volume is 1-2 liters;
also inside are the cooling pipe and the reactants. Hydrogen gas was
added continuously, at a low flow rate."
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Report on heat production during preliminary tests
on the Rossi “Ni-H” reactor.
Dr. Giuseppe Levi
http://22passi.blogspot.com/2011/01/report-ufficiale-esperimento-
della.html
"In [Test2] the power measured was 12686 +/- 211 W for about 40 min
with a water flux 146.4g +/- 0.1 per 30 +/- 0.5 s. The mean input
power during the test was 1022 W. This means that 11664 * 40 * 60 =
27993600 J were produced."
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
http://www.lenr-canr.org/acrobat/RothwellJbrieftechn.pdf
METHOD
The reservoir water temperature is measured at 13°C, ambient air at
23°C.
The heater is set to about 1000 W to heat up the Rossi device.
Hydrogen is admitted to the Rossi device.
The displacement pump is turned on, injecting water into the Rossi
device at 292 ml/min.
The water comes out as warm water at first, then as a mixture of
steam and water, and finally after about 30 minutes, as dry steam.
This is confirmed with the relative humidity meter.
As the device heats up, heater power is reduced to around 400 W.
RESULTS
The test run on January 14 lasted for 1 hour. After the first 30
minutes the outlet flow became dry steam. The outlet temperature
reached 101°C. The enthalpy during the last 30 minutes can be
computed very simply, based on the heat capacity of water (4.2 kJ/
kgK) and heat of vaporization of water (2260 kJ/kg):
Mass of water 8.8 kg
Temperature change 87°C
Energy to bring water to 100°C: 87°C*4.2*8.8 kg = 3,216 kJ
Energy to vaporize 8.8 kg of water: 2260*8.8 = 19,888 kJ
Total: 23,107 kJ
Duration 30 minutes = 1800 seconds
Power 12,837 W, minus auxiliary power ~12 kW
There were two potential ways in which input power might have been
measured incorrectly: heater power, and the hydrogen, which might
have burned if air had been present in the cell.
The heater power was measured at 400 W.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
My concerns regarding the Rossi steam calorimetry are repeated here.
The following approximate values for water can be used for rough
estimates:
Liquid Density: 1000 kg/m^3 = 1 gm/cm^3
Heat of vaporization: 40.6 kJ/mol = 2260 J/gm
Heat capacity: 4.2 J/(gm K)
Molar mass: 18 gm/mol
Density of steam at 100 C and 760 torr: 0.6 kg/m^3 = 0.0006 gm/cm^3
Now to examine the importance of mass flow vs volume flow
measurements for the steam.
If x is the liquid portion by volume, then x/((x+(1-x)*0.0006)) is
the portion by mass. This gives the following table:
Liquid Liquid Gas
Portion Portion Portion
by Volume by Mass by Mass
--------- ------- -----------
0.000 0.0000 100.00
0.001 0.6252 0.3747
0.002 0.7695 0.2304
0.003 0.8337 0.1662
0.004 0.8700 0.1299
0.005 0.8933 0.1066
0.006 0.9095 0.0904
0.007 0.9215 0.0784
0.008 0.9307 0.0692
0.009 0.9380 0.0619
0.010 0.9439 0.0560
0.011 0.9488 0.0511
0.012 0.9529 0.0470
0.013 0.9564 0.0435
0.014 0.9594 0.0405
We can thus see from this table that if 1 percent by volume of the
steam is entrained water micro-droplets, easily not seen in tubing or
exhaust ports, that only 5.6 percent of the heat of vaporization is
required to produce that mixture.
Rough calculations based on Rossi specifics:
Suppose for the Rossi experiment the mass flow of a system is 292 ml/
min, or 4.9 gm/s, the inlet temperature 13 °C.
The delta T for water heating is 100 °C - 13 °C = 87 °C = 87 K.
If the output gas is 100% gas, we have the heat flow P_liq given by:
P_liq = (4.9 gm/s)*(87 K)*(4.2 J/(gm K))= 1790 J/s = 1.79 kW
and the heat flow H_gas for vaporization given by:
P_gas = (4.9 gm/s)*(2260 J/gm) = 11.1 kW
for a total thermal power P_total of:
P_total = 1.79 kW + 11.1 kW = 12.9 kW
Now, if the steam is 99% gas, we have:
P_liq = 1.79 kW
P_gas = (0.1066)* (11.1 kW) = 1.18 Kw
P_total = 1.79 kW + 1.18 kW = 2.97 kW
or 23% of the originally estimated power out.
The dryness of the steam was measured by measuring using an
instrument that measures the dielectric constant of the steam.
At 100 ° C, the dielectric constant of water is roughly 55.3, and
steam is 1.0
The capacitance of two capacitors in series is:
1/C = 1/C1 + 1/C2
so using identical area capacitors with water being 1 unit of
thickness, and vapor being 99 units of thickness, we have the
relative capacitance C:
1/C = 1/(55.3/.01) + 1/(99/1) = 0.01028
C = 97.3
and so there is a 2.7% difference in capacitance for 1% water liquid
vs dry steam, by volume. The margin for error becomes very small at
lower liquid concentrations. Using capacitance may be a valid
technique, but depends highly on the accuracy of the capacitance
measurement. For this reason, it seems reasonable to do calorimetry
on the steam-liquid out.
Even dry steam emitting from a hose to ambient conditions will appear
to mist or produce water droplets, so all calorimetry results depend
on the measurements of the dryness of the steam.
Regarding the probe used, the following URL:
http://pdf.directindustry.com/pdf/delta-ohm/hygromyters-
hd-21011/25140-119895-_3.html
http://tinyurl.com/45rwsvh
shows the working range of the HP474AC probe to only be 5% to 98%
RH. It has a resolution of 0.1 %, but that could be very misleading
because it only has an accuracy of +-2.5% in its working range.
It is also notable that the HP474AC probe requires calibration:
http://www.crntecnopart.com/en/images/stories/crntp/
enviroment_analysis/pdf/solpathu_eng.pdf
http://tinyurl.com/4z5985v
It is not clear how relevant any of this is because it appears the
probe is designed to, and calibrated to, detect the percent of water
vapor in air, not percent of water microdrops in pure steam. Pure
water vapor should have more capacitance than 100% humid air, and be
beyond the meter's measuring limits. Droplets would add even more
capacitance, but once the meter is already maxed at 100%, the
question arises can extra water droplets make any difference to that
already maxed 100% reading?
The significance of all this lies in just how much input power and
energy must be accounted for, in terms of electrical power input,
possible caloric retention of the device from the initial period
involving at least 1 kW heating, and possible chemical reactions
within the device.
It is notable that if even all the steam were liquid, the heat flow
of P_liq = 1.79 kW, vs the electrical input of 400 W, provides a COP
of 4.5, which is excellent. Even if there is ambient heating of the
input water to a room temperature of 26 °C prior to entry to the
device, the delta T for water heating is 100 °C - 26 °C = 74 °C, vs
the estimated 87°C, giving a thermal output P_liq = 1.54 kW, and
excess heat of 1.14 kW.
So, there is no need to look for secret large power cables that
provide over 10 kW of heat. It is credible that the steam produces
at least a kW, and the water heating produces about a kW, so about 2
kW of excess power needs explanation if the source of power is said
to be conventional, and if the experiment ran for an hour then 2 kWh
of excess energy needs to be explained. It is thus important not
only to do calorimetry on the steam produced, but also to do a
complete energy balance on the experiment.
This is all academic, however, if a 1 MW reactor is produced which
will operate 6 months without refueling.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
