Author Topic: Holcomb Energy Systems:Breakthrough technology to the world (Read 451852 times)

bistander · « **Reply #2220 on:** April 30, 2023, 12:42:24 AM »

Quote from: Jimboot on April 30, 2023, 12:18:55 AM

I could have got thinner laminates laser cut but not for an experiment at $300. Ended up attaching the individual laminates to an angle grinder and spinning them against the bench grinder. Serpentine initially it will be a 3 phase rotor 4 strands of 0.4mm in each coil 13m in length. The photo is the 1st phase.

Thanks for reply.
Doesn't that put a 4-pole rotor in a 2-pole stator, or am I missing something?
bi

bistander · « **Reply #2221 on:** April 30, 2023, 02:55:30 PM »

Quote from: bistander on April 30, 2023, 12:42:24 AM

Thanks for reply.
Doesn't that put a 4-pole rotor in a 2-pole stator, or am I missing something?
bi

Thinking further, that serpentine wind results in 4 pole pairs, or 8 poles. Be interesting to see how it behaves.
bi

bistander · « **Reply #2222 on:** May 09, 2023, 06:30:21 AM »

Hello Jimboot,
Any progress? Do any test?
bi

Jimboot · « **Reply #2223 on:** May 09, 2023, 03:09:30 PM »

Thanks for asking Bi. At this stage I have a bad transformer. I'm using the Zafer BLDC 3 phase driver as shared in Ciftas thread. My 3 serpentine coils are 25 turns and I do get a sine wave off a stator phase. The rotor coils are just underr 1 ohm and at least one is damaged. I was hoping to just turn them into bifilar serpentines but it looks like I have conductivity between strands. Any way 4.5 v around 1 amp in measures around 20v on the stator but there is barely enough current to dimly light an led. I'd like to try some coils at around 3-5ohms next. I'm time poor but it's interesting to to learn and now I have a "rotor" I can run some tests with.

bistander · « **Reply #2224 on:** May 09, 2023, 03:41:51 PM »

Thanks for info Jimboot. Is your reason wanting 3-5 ohms on the coils so that you can raise voltage? To get higher coil resistance, can't you simply just use one of the four strands?

As I'm still wondering how the mix of pole count plays, did you record any frequencies, input/output?

Cool work,
bi

bistander · « **Reply #2225 on:** June 03, 2023, 10:48:25 PM »

Recent post on Holcomb's website:
I'll post a screenshot shot as I am unable to copy text (from here https://m.facebook.com/story.php?story_fbid=###)
They show two bar graphs, with and without HES bypassed. They state averages of 21.7 and 10.7 kW, respectfully. Those stated average numbers would indicate a 50+% saving.

To me it looks like bad math, so I printed the respective charts and drew in their average value as a pencil line. Attached below is photo of my pencil average lines. Maybe Holcomb needs to check facts and figures before publishing.
bi

{edit}
So while reviewing this again, I noticed that the charts are for "Demand Usage". I assumed this was the average power used, but NO. "Demand Usage" means something else in the electric utility industry. See:
https://atrius.com/the-difference-between-electricity-demand-and-electricity-consumption/

So this may not be a case of "bad math" as I originally thought. But how many readers are mislead thinking the 10.7/21.7 ratio was an actual energy savings? Think the same thing applies to the chart which impressed so many people a year or so ago was also demand? I suspect it was.

onepower · « **Reply #2226 on:** June 04, 2023, 01:01:26 AM »

bistander

Quote

So this may not be a case of "bad math" as I originally thought. But how many readers are mislead thinking the 10.7/21.7 ratio was an actual energy savings? Think the same thing applies to the chart which impressed so many people a year or so ago was also demand? I suspect it was.

Many people confuse "power" demand in kW and "energy" consumption in kWh.

There are a few things which pop out. The graph is showing the kW demand measured over 24hrs so it's intuitive that each bar is an indication of the power at any given time. A rough guess is 100 power measurements over 24hr or every 15 minutes. Since the power demand is reduced by 50% on average the energy consumption is also reduced by 50%. The energy in kWh is the average power in kW times the hours of operation. So we add up all the 100 measurements in kW, divide by 100 for the average kW's then multiply that number by the hours of operation or 24 to get the energy in kWh.
21.7kW * 24hr = 520.8 kWh before
10.7kW * 24hr = 256.8 kWh after

As well, note how after the device was installed it flat lined the demand peak. Did the load change during the two test days or or is the power gain non-linear?. That is, providing more/less gain above/below a given power threshold.

It's not common knowledge but the kind of process being used generally yields close to a 100% gain per cycle. Which implies this is a one pass device versus a feedback or multi-stage device which can deliver a higher COP. So they could use two devices in series to lower the utility power demand but it's less cost effective. The first device reduces demand by 10kW, the second device by 5kW and a third by 2.5kW and so on.

Limiting the gain to COP=2 implies to me they don't fully understand how this process works or there playing it safe to cause less disruption to the present utility market. I have no idea what there thinking is but if it works it works.

AC

bistander · « **Reply #2227 on:** June 04, 2023, 02:37:46 AM »

Quote from: onepower on June 04, 2023, 01:01:26 AM

bistander
Many people confuse "power" demand in kW and "energy" consumption in kWh.

There are a few things which pop out. The graph is showing the kW demand measured over 24hrs so it's intuitive that each bar is an indication of the power at any given time. A rough guess is 100 power measurements over 24hr or every 15 minutes. Since the power demand is reduced by 50% on average the energy consumption is also reduced by 50%. The energy in kWh is the average power in kW times the hours of operation. So we add up all the 100 measurements in kW, divide by 100 for the average kW's then multiply that number by the hours of operation or 24 to get the energy in kWh.
21.7kW * 24hr = 520.8 kWh before
10.7kW * 24hr = 256.8 kWh after

As well, note how after the device was installed it flat lined the demand peak. Did the load change during the two test days or or is the power gain non-linear?. That is, providing more/less gain above/below a given power threshold.

It's not common knowledge but the kind of process being used generally yields close to a 100% gain per cycle. Which implies this is a one pass device versus a feedback or multi-stage device which can deliver a higher COP. So they could use two devices in series to lower the utility power demand but it's less cost effective. The first device reduces demand by 10kW, the second device by 5kW and a third by 2.5kW and so on.

Limiting the gain to COP=2 implies to me they don't fully understand how this process works or there playing it safe to cause less disruption to the present utility market. I have no idea what there thinking is but if it works it works.

AC

Hello AC,

You say:

Quote

So we add up all the 100 measurements in kW, divide by 100 for the average kW's then multiply that number by the hours of operation or 24 to get the energy in kWh.
21.7kW * 24hr = 520.8 kWh before
10.7kW * 24hr = 256.8 kWh after

That is clearly incorrect. You've been fooled, which was my point. I think most everybody is fooled, perhaps even Holcomb. The 10.7 and 21.7 numbers are not the averages of the blue bars. Look at area above my pencil line vs the area between tops of blue bars and pencil line for each chart. If what you say is true, those areas would be equal.
bi

onepower · « **Reply #2228 on:** June 04, 2023, 08:08:13 AM »

Bistander

Quote

That is clearly incorrect. You've been fooled, which was my point. I think most everybody is fooled, perhaps even Holcomb. The 10.7 and 21.7 numbers are not the averages of the blue bars. Look at area above my pencil line vs the area between tops of blue bars and pencil line for each chart. If what you say is true, those areas would be equal.

Indeed, the 10.7 graph looks closer to 12.5 on average and the 21.7 graph closer to 17 on average.

AC

rakarskiy · « **Reply #2229 on:** June 04, 2023, 08:55:00 AM »

This is just "grasping at straws" to prove that Holcomb's Energy System does not work with the OverUnity effect.

Bad analyst who made a post. There, obviously, one graph (upper) would be enough, where the left part is with the Holcomb generator, and the right part is without it. To get a full 50/50, it is necessary that the consumers (air conditioning system) are always turned on at full capacity, which in reality cannot be, the cooling system has a temperature maintenance mode. And the excitation of the generator must always be stable in order to maintain generation, otherwise the generation will fail. This is her speciality.

If you're itching to check it out, visit the facility where Holcomb's generators work. Ask the owner to run a test: one day with Holcomb generators, another day without. The cost of energy is assumed to exceed the work without the Holcomb generator, take this cost on your pocket. Then argue, some Mui Ne, about lies from Holcomb.

The most accurate way to compare is your own wallet: it will show the difference in savings or loss.

pix · « **Reply #2230 on:** June 04, 2023, 11:32:57 AM »

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 · « **Reply #2231 on:** June 04, 2023, 12:41:07 PM »

One kilowatt hour = 3,600,000 joules.

pix · « **Reply #2232 on:** June 04, 2023, 01:04:14 PM »

Quote from: bistander on June 04, 2023, 12:41:07 PM

One kilowatt hour = 3,600,000 joules.

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

Beginners Mind · « **Reply #2233 on:** June 06, 2023, 01:28:51 PM »

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 · « **Reply #2234 on:** June 06, 2023, 01:57:05 PM »

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:

Quote

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.

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