Holcomb Energy Systems:Breakthrough technology to the world

[pix]

The most accurate way to measure "power in" vs "power out".
Is to use resistive heating elements and heat up given amount of water in sealed thermo insulating  box at given time.
it is easy, very transparent and I am wondering why such serious looking company can't do it.
On the water side:
Q= Cw x  m x  (T1-T0) [Joule]
Cw=specific heat of water (constant)
m= mass ( constant in sealed vessel)
T0,T1- temperature difference of heated water.


On the electric supply side:
During test  time  t  measure current I , voltage V frequently and calculate average.
P=U x I x t [kWh]


1kWh=1000 Joule


Compare P and Q, that's it.
I hate to see presentations showing burning incadescent lights, it proves nothing  because they can be lit up by high frequency current.
Graphs also are not accurate.
How can you prove that during 24 hrs of HES measurements power consumption wasn't reduced?


Cheers,
Pix

[bistander]

One kilowatt hour = 3,600,000 joules.

[pix]

One kilowatt hour = 3,600,000 joules.

Yes, my typoo;
1kW=1000J/s

[Beginners Mind]

Have people seen the thread in Bob Walker's Linked-In pages in the last 4 day entitled "No longer associated with Holcomb Energy Systems?"  https://www.linkedin.com/feed/update/urn:li:activity:7070168761009332224/?commentUrn=urn%3Ali%3Acomment%3A%28activity%3A7070168761009332224%2C7071172912308707328%29

Bob was the President of Global Business Development at Holcomb Scientific Research and an insider at the firm.  In his Linked-In posts Bob states he was fooled; that the HES actually consumes more power than it produces.

How could HES's President of Global Development have been fooled?  The same way everyone else is: a metering error.  I first mentioned this metering error on this forum months ago after reviewing detailed videos of a demonstration taken at HES labs and subsequently replicating the metering error in our own lab.  My comments were made in hopes of saving time and money for people attempting replications.  That is my hope again in repeating it now.   

Careful review of the video demonstration showed that HES wires their 3-phase power meters incorrectly.  The voltage and current probes are not attached to corresponding phases.  An HES engineer in the video acknowledges this, clearly believing it to be correct, without intention to deceive.  But this is a tremendous error in 3-phase metering, which here results in meters reading more power out than in.

Note that all the third party verifications touted by HES have simply verified HES's meter readings.  HES's claims of OU have not been publicly verified by third parties using their own 3-phase power meters, properly attached.  [A large measured current gain is real but it is measured within an LC circuit and is accompanied by a huge phase shift.  It does not correspond to a power gain.]

And see Bob Walker's Linked-In thread for some comments on power bill claims.

[rakarskiy]

Do not fool a sick head. Of course there is no 50/50. But if you look at the diagram from the electricity metering program for payment, you can immediately see for which kW the consumer will not pay.
The left part of the diagram is with the Holcomb generator, the right part is the consumption without the Holcomb generator.
That's the whole analysis. There is no saving of 50%, but it is there and tangible. They have bad analysts and marketers in their company, apparently since they broadcast such blunders, and perhaps they do it on purpose.

I'm doing the analysis too, here's an example:

I ran the installation again, changed the conditions a little. Analyzed more precisely the reverse impulse. The area (S) was compared with the magnitude of the current and its strength.
Over the entire area of the current value, the excitation and return pulses in one period (T) are equal to each other. And its quality component is very different. Zone "A" - the excitation impulse is equal to the return impulse, they simply superimpose the projections of the current strength on each other. Then the remaining areas were reduced into rectangles, it turned out that the areas "B" and "C" are equal to each other. The average reverse current of the "C" zone is 0.6A, and the excitation zone "B" is 0.21A. Thus: 0.6A - 0.2A = 0.4A - this is the part that we can attribute to the Super Unit. On the whole, the efficiency of this impulse will be low COP-1.08.