At 10:06 AM 7/3/2011, Jed Rothwell wrote:
Here is an analysis of Rossi's e-Cat steam test
from Ed Storms. Actually, this is a combination
of two messages he sent me, with a clarification inserted into item 2.
- Jed
Thanks for forwarding this, Jed, and thanks to
Dr. Storms for writing it. I think he has missed something.
Before looking at that, I want to emphasize that
from the public evidence, I cannot determine how
much power the E-Cat is generating. This is not
the same as a claim that it does not generate
power. It might, it might not. Or, because one of
the errors here is an excluded middle, I'll point
out that it might be generating power at a
different level than claimed. It might even be
more! (Rossi may have overestimated input power,
because he included the power dissipated in the
electronics.) We don't have the data we'd need.
A variety of ways the Rossi claims might be
wrong have been suggested. Let's examine each. The following values are used:
Cp (H2O at 65°) = 4.18 J/g-deg ,
enthalpy of vaporization @ 100°C = 2.27 kJ/g.
1. Not all of the water is turned to steam.
Dr. Storms has missed something. It is a
practical certainty, backed by the Essen and
Kullander measurements (if they were accurate,
which they almost certainly are not), that there
is some water not being vaporized. Thus we can be
sure that the statement above is true. The
question is not that, but *how much*?
If applied power is making all of steam, the following would be observed.
Applied power = 745 watt
Actually, that overstates applied power. While
total input power is of interest, it is of
interest only for ruling out fraud, because the
real figure for calorimetry analysis would be
heater power. The applied power includes power
dissipated in the electronics. In a convincing
and accurate demonstration, heater power would be
reported. Rossi almost certainly has the data,
but it would reveal how his heat is being
adjusted to maintain energy generation. Is input
power constant? Is applied power constant? We
have only isolated, static figures, in some
cases. (Seems to me I've seen a plot of input power somewhere?)
Flow rate = 7 liter/hr = 1.94 g/sec
Power to heat water to 100° = 73°*4.18*1.94 = 592 watt
Power to make steam = 745 - 592 = 153 watt
Amount of steam produced = 153/2270 = 0.07g/sec
out of 1.94 g/sec = 3.4 % of water flow.
Dr. Storms is assuming that all input power is
used to heat the water. It would be less because
of power consumed by the electronics. Rossi could
have made a very convincing demonstration with a
blank E-Cat, one with no fuel in the chamber, say
no hydrogen. He was asked to do that, he
declined, saying that it wasn't necessary, since
"everyone knows what is produced." I.e., no
excess energy. But this was a clearly deceptive
argument, or at least wrong. I certainly don't
know how much steam will be produced by an
"empty" E-Cat with the same heating and same flow rate!
In the other direction, the input power figure of
745 watts comes from, as I recall, Rossi's
calculation, which was based on a figure of 220V
for the mains. The actual figure is 230 V., so
the actual input power is about 5% higher. Which
would have a large effect on the marginal power, the power used to make steam.
But Storms' figure is of interest. From this, I
conclude that most of the input water would be
flowing through the E-Cat, that it might be at
100 C., and that there might be some steam.
However, I do not know that the water is actually
at 100 C, it would depend on where the
thermometer is placed, whether it was in the
water or not, and that might depend on internal
design, as well as the possibility of live steam
above colder water would depend on internal
design. Absent a deliberately manipulated
internal design, the water and water vapor would
be in equilibrium at 100 C, unless there were not
enough applied power to boil any of the water.
As another problem, I don't know for a fact what
the input flow is. Below, Dr. Storms points out
that some have thought the input flow was
overstated by a factor of two. This would
drastically affect the calculation above.
Assumptions have been made from what seemed to be
a constant rate of clicking of the pump. That
input flow rate would only be correct if there is
no significant resistance to flow. If the E-Cat
incorporates a flow control valve, there goes
that figure! Does the pump continue to click at
the controlled rate even if flow is blocked? I
don't know. What I do see are assumptions being
made but not explicitly stated as what they are.
A whole series of tough problems would disappear
if the water was fed by gravity, so that if there
is no boiling, there is no flow. Instead of this,
Rossi uses a more complex and actually less
reliable method, a pump. With gravity feed, and
the source maintained at the right level (which
could vary some, but be restored by adding
measured water), there would be no flow of liquid
water out the hose except for what actually condenses in the hose.
Water does condense in the hose, at least. The
hose would fill, as a result, and this shows the
defect in Dr. Storms' following argument:
The chimney would fill with water through which steam would bubble.
Yes. It will fill with water even if no water is
flowing out the chimney. Then what?
The extra water would flow into the hose and block any steam from leaving.
No. Not at all, in fact. The water is at 100 C.,
at least when exiting the chimney. It would fill
the hose, yes. However, when the hose is emptied,
as Rossi takes care to do when pulling the hose
out of the drain, Steam would exit the hose until
the hose refilled, which would take some time.
Rossi apparently takes care to replace the hose
in the drain before this happens.
As the water cooled in the hose, the small
amount of steam would quickly condense back to
water. Consequently, the hose would fill with
water that would flow out the exit at the same
rate as the water entered the e-Cat.
Yes. That's equilibrium under some conditions.
However, we have no measurements or observation
of what is coming out of the hose, steady-state.
None. That has not been permitted, on an argument
that continuous exposure of the hose end would be
dangerous, which is preposterous, if properly done.
However, excluded middle. The statement was that
not that all the water is turned into steam. Not
that none is. From Dr. Storms' analysis, I'd
strongly suspect that there is probably excess
power, but the input power is not well-known, nor
is the information adequate to rule out liquid water exit.
CONCLUSION: No steam would be visible at the end
of the hose, which is not consistent with observation.
Observation has been inadequate to conclude this.
When steam has been observed at the end of the
hose, it is quite weak, compared to expectation
from 100% boiling. It is not clear that the
"steam" at the end of the hose is actually steam.
If the water is at 100 C., it would certainly
produce misty vapor, as any very hot water does.
No measure of steam quality at the end of the
hose has been done, and, as far as I can tell,
temperature measurements have not been done at the end of the hose.
Roughly, unless there were a high rate of steam
generation, I'd expect that all the steam might
condense in the hose, leading to the same
conclusion: all water would exit the hose, steady
state, at the same rate as inflow. Only if
uncondensed steam reaches the end, *or for
transient periods when the hose is emptied*,
would this not be the case, and in that case, the
temperature at the end would be at boiling.
Measurement not done. Apparently not permitted.
2. The steam contains water droplets, i.e, was not dry.
Power to heat water to 100° = 592 watt
Power to vaporize all water = 1.94 * 2270 = 4404 watt
Total = 4997 watt if all water is vaporized
Excess power = 4249 watt
The only way steam is wet is when water drops
are present. If too many drops are present, they
fall as rain (precipitate). It is simply
impossible to have a large number of drops
present. A 5% figure is chosen as an example
here
(<http://www.engineeringtoolbox.com/wet-steam-quality-d_426.html>http://www.engineeringtoolbox.com/wet-steam-quality-d_426.html)
because this is a plausible amount.
Nevertheless, the conclusion would be the same
even if 20% water drops were present.
Dr. Storms should consider a crucial question:
how is the power regulated so that the two
extremes are avoided, the emptying of the cell
due to boil-off rate exceeding inflow, or water
exiting the hose unvaporized? It's easy to
understand self-regulation if some level of hot
water runoff is allowed, and the input power
(thus the generated power) would be such as to
merely vary the water runoff rate as the output
power varies, within some range. I conclude from
this analysis that, if there is no runoff, some
input variable must be a control, and the obvious
one is input power. But how is that controlled?
Above, Dr. Storms uses a constant figure for
input power. What's that based on other than a
transient measurment, not a continuous one? Is input power varying?
Further, control requires feedback. There is no
indication that the controller has any
information coming back to it other than reactor
temperature. Is there a level sensor in the cooling chamber?
We are forced into a series of speculations. This
is far from a convincing demonstration!
The steam is almost certainly not dry, i.e., zero
percent water droplets. The only question, again, is *how much*?
I agree that this is largely a red herring. The
strong issue is the larger possibility of
unboiled water exiting the chamber, and for
massive water to leave by droplets this way, at
boiling temperature, would be unlikely.
Power to vaporize 95% of water = 4183 watt
Excess power = 3736 watt
CONCLUSION: Significant excess power is being
made regardless of how dry the steam may be.
3. Energy is stored in the apparatus that is
being released during the demonstration.
What if the behavior observed is due to a
combination of factors? Each factor alone could seem implausible.
Assume e-Cat contained 2 kg of material having
an average heat capacity equal to that of
copper. Copper has a heat capacity of 0.385 J/g*K.
Assume steam is made for 15 min, i.e. the e-Cat
remains above 100° C during this time.
During 15 min, 1750 g of water is converted to
steam = 1.94*15*60*2270 = 3963 kJ
Applied energy = 745 *60*15 = 672 kJ
Amount of energy that has to be stored = 3291 kJ
Energy stored in Cu/degree = 2000*.385 = 770 K/°
Initial temperature of e-Cat = about 4400°
The e-cat would have to weight over 20 kg to
contain enough energy to make steam for only 15
min. after being heated initially to over 500° C.
CONCLUSION: The e-Cat cannot retain enough
energy to account for the observed behavior
during cooling from high temperatures.
I agree with this conclusion as applied to an
assumed energy value, based on no unvaporized
water. This is a house of cards, though. If there
is unvaporized water -- and there is *some*, the
question is how much, the required stored energy
is lowered. How much? Without good calorimetry,
and we don't have good calorimetry due to a pile
of unconfirmed or not even done measurements, we can't tell.
4. The flow rate is wrong by a factor of 2.
Power to heat water to 100° = 296 watt
Power to vaporize all water = 2204 watt
Total = 2500 watt if all water is vaporized
Excess power = 1752 watt
CONCLUSION: Excess power is being generated even
if the flow rate is misrepresented by a factor of 2.
The problem here is even clearer. An extreme
result from one sector of the argument is then
used as the basis for an argument in the other.
The extreme result used is based on an assumption
of total vaporization. The argument for "no
liquid water" was based on an assumption that the
input flow rate was constant. If the flow rate
were halved, then the energy available from the
input power is far larger, and steam production
would be, with no excess heat, far greater.
BASIC CONCLUSION: None of the plausible
assumptions are consistent with the claim for excess energy being wrong.
The conclusion is defective. Rather than looking
at assumptions, how about backing up and looking
at what we know, and what we don't know? There
are "plausible assumptions" incorporated in Dr.
Storms' arguments. Plausible assumptions are
based on normal experience, and where science
often goes awry is where experience is not
normal, where something new is being examined.
Dr. Storms is really comparing two sets of
plausible assumptions, but he's assuming that one
set, the one he used, is correct, and the sets
appear to contradict, in his analysis, so the others are not true.
We do not know that there is no liquid water
flowing through the E-Cat, presumably hot, but
unboiled. Quite simply, we have been excluded
from seeing the evidence, what actually comes out
the hose. The physical arrangement of the E-Cat
would allow observation of live steam at the
chimney steam relief valve, as apparently
observed by Kullander and Essen, *at the same
time* as there is liquid water flow out the hose.
At the low rates involved, the hose can be
emptied by Rossi, as we have observed in the
Krivit video, and then observers can see the
(modest) steam coming out for a short time. (Did
other observers see this emptying? If so, why did
they not report it? All this shows how the
reports of even skilled observers may be
incomplete, with significant actions missed, because of expectations.)
My conclusion from the data: no conclusion can be
made about the level of excess power. It may be
between the extremes of no excess power to excess
power based on 100% vaporization, and we cannot tell where it falls.
Rossi was unexpected to those in the field
because of the amount of excess power claimed,
not because of a claim of *some* small level of
excess power. If that excess power is being
overestimated, it impacts the import of his results.
I do not trust raw calculation unless confirmed
by experiment or independent data, it is far too
easy, even necessary, to incorporate defective assumptions in calculations.
