On Jun 24, 2011, at 9:14 PM, Joshua Cude wrote:
On Fri, Jun 24, 2011 at 1:53 PM, Horace Heffner
hheff...@mtaonline.net wrote:
It is notable that the power input varies depending on the
controller actions, that if the power input (plus any nuclear
output heat if any) should become less
On Sat, Jun 25, 2011 at 3:58 AM, Horace Heffner hheff...@mtaonline.netwrote:
The power is noted to be 770 W. If you assume no nuclear reaction then
that is all there is. It should only take minutes to reach equilibrium.
True. Some say it's really 800W (230V), but still only minutes, as you
In reply to Daniel Rocha's message of Fri, 24 Jun 2011 12:19:34 -0300:
Hi,
[snip]
It cannot raise water more than 3 milimeters. That sort of rubber, I
made some calculations elsewhere, does not radiate more than 25W per
meter. The steam must be dry to be pumped out.
You mention the hose
In reply to Horace Heffner's message of Fri, 24 Jun 2011 03:09:22 -0800:
Hi,
[snip]
The thermal conductivity of copper is 386 W/(m K), about 2700 times
that of rubber. Several meters of similar sized copper pipe coiled a
barrel of water at 75 C should easily condense 12 kW of steam.
Do we really know how the e-cat is using all the input power?
Harry
On Fri, Jun 24, 2011 at 1:53 PM, Horace Heffner hheff...@mtaonline.netwrote:
It is notable that the power input varies depending on the controller
actions, that if the power input (plus any nuclear output heat if any)
should become less than that required to convert all the input water to
On Jun 23, 2011, at 1:02 PM, Daniel Rocha wrote:
2011/6/23 Horace Heffner hheff...@mtaonline.net:
Liquid LiquidGas
PortionPortion Portion
by Volume by Mass by Mass
- --- ---
0.000 0. 100.00
0.001 0.6252 0.3747
I will just
I wrote: A couple meters of rubber hose can not radiate away 80% of
12 kW of heat suggested to be produced in the original runs.
To be more specific, it can be expected the heat flow through the
rubber tube walls is about 220 W per m of hose.
Using the thermal conductivity for rubber at
-Original Message-
From: Horace Heffner
Thanks for jumping back into the analysis, as tiresome as it has gotten to
be (even for this particular audience).
Almost everyone agrees that it would be very easy for Levi and his crew to
rectify the wet/dry steam controversy - that his
The pressure of the hose is too small, in another thread I wrote this :
Considering a stream of 10m/s, 1.5g/s out of the hose, with, 5cm2 of
area, the pressure inside above 1atm the chamber is
P=F/A=(1.5*10(-3)*10)/5*10(-4)=(1.5*10(-2)*10(4))/5=1.5*20=30N/m2 or
and increase of 3*10(-4) atm.
It
Just to be sure of my position. I am completely convinced that the
data that has been provided is coherent with a power generation of
2.5KW. My doubt is from where the power is drawn. Rossi does have
control over the current, using his computer, so he can surely change
the power while cheating on
The appended response appears to be nonsensical. Perhaps it is due to
a language barrier?
The calculation provided appears to be meaningless. It appears to
*assume* a priori a free flow of steam, i.e. no percolator effects,
no pressure or flow variations. Also, it would be more
On Fri, Jun 24, 2011 at 6:09 AM, Horace Heffner hheff...@mtaonline.netwrote:
I wrote: A couple meters of rubber hose can not radiate away 80% of 12 kW
of heat suggested to be produced in the original runs.
To be more specific, it can be expected the heat flow through the rubber
tube walls
As for the Krivit's test, there is nearly no condensation inside the
hose. That is visible in any of the video. The water output due vapor
doesn't require a very fast flow, so it is certainly free, with no
turbulence. The kinetic energy is just too small due vapor, 0.2W.
On Fri, Jun 24, 2011 at 10:33 AM, Daniel Rocha danieldi...@gmail.comwrote:
Just to be sure of my position. I am completely convinced that the
data that has been provided is coherent with a power generation of
2.5KW.
But the presented data is also consistent with power equal to the input
The output temperature and flow output, even visually, are convincing.
They are visually equivalent to putting off a candles by blowing them,
that is 0.2W - 0.4W. But to make it only by heating water and
vaporizing requires more than 2000KW.
I don't think the con comes from that. If that was so
On Fri, Jun 24, 2011 at 12:34 PM, Daniel Rocha danieldi...@gmail.comwrote:
The output temperature and flow output, even visually, are convincing.
They are visually equivalent to putting off a candles by blowing them,
that is 0.2W - 0.4W. But to make it only by heating water and
vaporizing
But 2KW does give a very feeble buff, unless it is ousted in a very
thin cavity and accelerated by propellers, like in a hand vaporizer.
On Jun 24, 2011, at 10:08 AM, Joshua Cude wrote:
On Fri, Jun 24, 2011 at 12:34 PM, Daniel Rocha
danieldi...@gmail.com wrote:
The output temperature and flow output, even visually, are convincing.
They are visually equivalent to putting off a candles by blowing them,
that is 0.2W - 0.4W.
It won't change much. I used 1.5g/s which gives 3.300W, so, 1.8g/s
fits the bill. So, it is just a slightly stronger blow.
On Thu, Jun 23, 2011 at 3:12 PM, Horace Heffner hheff...@mtaonline.netwrote:
I believe the HP474AC probe actually measures the capacitance of the air,
and converts that to relative humidity.
Not quite. It measures capacitance with a polymer dielectric which absorbs
water from the air in some
2011/6/23 Horace Heffner hheff...@mtaonline.net:
Liquid Liquid Gas
Portion Portion Portion
by Volume by Mass by Mass
- --- ---
0.000 0. 100.00
0.001 0.6252 0.3747
I will just concentrate in the second entry. Are you suggesting
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