At 02:10 pm 30/06/2005 -0400, you wrote:
>Naudin's latest test run, #83, claims COP of well over 20.
>
>See: http://jlnlabs.imars.com/mahg/tests/mahg2c.htm
>
>Scroll down to the bottom, last test.
>
>Preliminary evidence seems to suggest that decreasing the duty cycle may
>increase COP efficiency even more.
>
>I bare my ignorance for all to witness here, but could some kind soul present
>in layman's term an explanation of how Naudin's recent MAHG COP values of 15 -
>20 are arrived at?
>
>I understand basic electricity values - that wattage is calculated from Volts
>times Amperage. ...that wattage is a form of measured power.
>
>However, in Naudin's most current run (run #82) when I look at the chart
>titled "Power Input/Output Vs Time" it's not exactly clear to me how the
>OUTPUT POWER value is derived. Is this output power value derived directly,
>indirectly, or inferred? Said differently, it's clear to me that the INPUT
>power (expressed in wattage value) is derived directly from the battery source
>hooked up to the MAHG device. But how do these test runs arrive at the
>CALCULATED OUTPUT WATTAGE VALUE, such as 92.97 watts (averaged) as described
>in #82?
>
>One aspect of these MAHG runs that may be hindering my comprehension of the
>claimed high COP values is the temperature differential between input and
>output. At first glance my layman's interpretation of the delta (difference)
>leaves me confused and mostly unimpressed. The "Temperature Input/Output and
>Water Flow Vs Time" chart for #83 shows input temperatures averaging at around
>20 C. Output temperatures are averaging, roughly speaking, around 23 C. That
>doesn't appear to be a very large temperature differential to get all that
>excited over. Granted, if water is flowing through the device very quickly,
>obviously that would decrease the amount of time the exposed water would have
>a chance to "heat" up. I realize Naudin gives the water flow amount, but
>unfortunately I don't have sufficient experience in these kinds of
>water-volume calorimetery values to fully comprehend the ramifications.
>
>Another related question: In practical terms, assuming the goal is to devise a
>way to extract copious amounts of presumed "OU" energy in the form of excess
>heat (steam?) that would be used to generate something like a sterling engine
>that ultimately runs a generator, doesn't the amount of output heat that would
>need to be generated from a device based on the MAHG principle HAVE to be a
>whole lot higher than what these preliminary experiments are currently showing?
>
>Mongo wishes to better understand MAHG COP values!
>
>Regards,
>Steven Vincent Johnson
>www.OrionWorks.com
Steven
To keep up with this rapidly developing story its
best to follow the posts on,
[EMAIL PROTECTED]
For example, this recent posts may answer some of you queries.
===========================================================
To explain the voltage, the duty cycle is .05 and
the charts are labeled average voltage and average current.
It also appears that JLN is multiplying avg(v)*avg(i) to
obtain average power. This is not correct for pulsed
power calculation. Peak power must be calculated first
and them multiplied by duty cycle. For even more
accuracy a wideband power meter should be used.
These average the instantaneous power readings.
Huge errors result at low duty cycle when average
V and I are used.
George Holz
===========================================================
I must admit that I have always been leery of VxA since I
learnt about electric motors and power factors. ;^)
As for the temperature differences they are a lot high for
the recent tests than they were for the earlier tests which
seems to rule out the pump objection someone raised.
Frank Grimer
