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You are here: > News > April 24, 2014; completed April 27; 9 am MDT

Heat-Loss Accounting Improves the Efficiency Number in Sterling's First H-Cat Experiment

David Haack has done the calculations on several heat loss items in the test H-Cat Sterling ran last month, including the non-perfect insulation (57,000 cal), steel tub (1,300 cal), and half brick (1,250 cal). These push up the efficiency by quite a bit, further pointing to anomalous heat.

Unaccounted Heat Loss Calories

- Water Basin (inadequate insulation): 57,117
- Steel Tub: 1,299.9 
- Half Brick: 1,146.8
- Water Lost to Steam (unknown): 0 to 46,922
Total Calories: 59,564 to 106,485
(Rounding to significant digits came in final step of tabulating the efficiency percentage.)

By David A. Haack
for Pure Energy Systems News

This is a brief summary of my contribution to Sterling's first H-Cat open-source experiment to quantify the output/input efficiency of that system. Those of you members in the H-Cat community who have been following this saga will probably already be familiar with this story . For the readers who are new to this, I will briefly state that this H-Cat open-source project came about through the generous gift to us from Justin Church of JDC Products. He discovered that passing HHO (Brown's gas) through a regular catalytic converter produced a large amount of heat being formed in the converter. He graciously turned this discovery over to the public, so we at PESN have been involved with this as an open-source project for furthering its development.

In the comment section at the bottom of the page where Sterling posted the report summary of his first experiment, I submitted 11 comments (some quite lengthy) doing calculations that would account for the various heat losses on the output side of the system. These heat losses when accounted for by adding in their calorie values would boost the numbers for efficiency of the H-Cat experimental results. Along with these calculations, I went to length developing my reasoning and methods involved in the various calculations. I did not attempt to redo any of the experimental work Sterling had already done, and I used the numbers and information in his experiment summary to do the math.

My work was not intended to prove or disprove overunity, but rather to exhibit a method that could hopefully prove helpful in the design of the next experiment. I studied 5 areas of heat loss in this experiment and did calculations on the ones I could and explained why one couldn't be calculated. One segment remains open ended, which I will explain in that part.

The 5 portions of the output side where heat losses were apparent are as follows:

Heat loss of water basin: 

This is the steel tub, water, and insulation, where the catalytic converter was submerged during the experiment. By using Sterling's cool-down rate data and taking the difference in temperatures of the water at the beginning and end of the experiment, the calories of heat-energy output can be calculated. Even though this steel tub was well insulated, there was still heat loss involved. This heat loss amounted to 57,116 calories. See my comment # 9 for how this value was derived.

Heat loss of the steel tub: 

This is the steel bucket (tub) used as the water basin where the catalytic converter was placed in order to determine the heat output through temperature change in the water. This steel tub went through the same temperature change, with associated heat loss as the water in it. But steel has a different specific heat capacity than water, so it has to be dealt with separately. The heat loss associated with the steel tub was 1,299.9 calories. See Comment #9 for math.

Heat loss of the half brick: 

This was the brick that was used to hold the converter down under the surface of the water, so heat wouldn't be lost to the air. The same explanation can be assigned to the brick as I did for the steel tub. But again, brick has its own unique heat capacity, so we need to again handle it separately. The heat loss associated with this 1/2 brick come out to 1,146.8 calories. Again, refer to Comment #9 for details.

Heat loss involved with the warmer reservoir: 

[This is a matter that addresses the efficiency of the HHO generator, which Steve D. says is 75%.]

The water reservoir was essentially a tank filled with water, and also serving as a bubbler which HHO gas from the generator was passed through. This served as one of the flash arrestors. It maintained a temperature approximately 6 ΊF. degrees higher than the HHO generator. My guess is that this was due to the heat of compression of the HHO gas. Hooking bubblers and flash arrestors in series would create resistance, thus compressing the gas. Regardless of why it was warmer, there is an associated heat loss involved. That loss, if it could be tabulated, would subtract from the input energy. Unfortunately, I was unable to calculate a heat loss due to missing data and loss of heat to the air rather than a water bath.

Heat loss due to water not accounted for in converter: 

This water loss would be explained as water loss showing a lower water level in the reservoir mentioned before, but not all of it showing up in the catalytic converter at the end of the experiment. This difference in water amounts can best be explained as water that escaped the system primarily as un-reacted HHO, but also as either water vapor or steam through the tube at the bottom of the converter that ends just above the water surface in the tub. 

My personal reasoning was that most of the water would likely need to be steam in order to have enough energy to escape. Sterling, to his credit, says that most of the water likely escaped as un-reacted HHO, and that this is an assumption that can't be pinned down in this particular experiment, so we need to provide the full range of consideration. It could have all vented as HHO, or as water vapor, or it could have all vented as steam, or some combination thereof. There is no way to calculate it, since there was no provision in the experiment to measure how much of each there was. While it is technically unlikely, though theoretically feasible, that the water could go through the system and exit as water vapor without a change in temperature, I will assign 0 calories heat loss to that scenario. 

If it all would have escaped as steam [Sterling says this is very unlikely, thinking that most escaped as un-reacted HHO], then each ml of water would have gone through the latent heat phase change which requires a whopping 540 calories of heat to convert one ml of water at boiling point to steam. In other words, each ml of water that was converted to steam would suck 540 calories of heat out of the converter. That would be a remarkable amount of heat loss. But first, each ml of water at the temperature of the reservoir (42.8 °C) must absorb 57.2 calories in order to first get to the 100 °C. at which latent heat phase change takes place. The calories of heat loss, if the water loss is all attributed to steam, would therefore be 46,922 calories. So that the heat loss could vary all the way from as little as zero all the way to 46,922 calories for the 78.57 ml of water loss. This should also make it obvious that there will need to be a way devised to determine the quantities of each. Or, at least a better way to account for this particular heat loss.

Total calories due to heat loss: 

If we add up the heat losses listed above, we get a sum that ranges all the way from 59,563.5 to 106,485 calories. Notice that I carry my numbers at this stage of the calculation to more places past the decimal than the precision of measurement would justify. That is to avoid error due to too much rounding. I do my rounding in the final result. I also didn't get into the + or - a certain percentage, either way.

Total efficiency of the H-Cat: 

I will now give the total efficiency range of possibility from the numbers we were able to determine. We can add these heat loss totals to the calories of heat that Sterling had already calculated due to the warming of the water bath. Sterling had listed this output as 253,000 calories for the Set C data. I should point out that all of the above calculations were either taken from or adjusted to work with this data set, as Sterling had determined the other two sets to be non-reliable data. If we factor in the range of calories due to heat loss as stated above and convert to Joules, we arrive at a range of efficiency from 95% to 109 %. This is the stage of calculation where I finally do the rounding. The range is due to the uncertainty of whether the water loss was un-reacted HHO, water vapor or steam. We also have a heat loss totally unaccounted for with the warmer water-bubbler reservoir.

[Sterling wishes to remind you that anything above 64% efficiency indicates anomalous heat. (75% max electrolysis efficiency times 85% max catalysis efficiency)]

Input side of the system: 

My work was primarily focused on heat losses on the output side. One could consider the heating of the electrical supply unit's output side as a factor, but we don't really have the kind of data to do any meaningful calculations. There was, however, some controversy associated with the amperage readings. I had offered a couple of different methods for Sterling to 'recover' that value. The first method that I offered, Sterling replied back that it wouldn't work and gave his reasons. I accepted those reasons, and offered a second method. I have sent that method to him. {As a side-note to this, Sterling has indicated that he plans to do a follow-up test on the original Fluke meter (that he borrowed from Josh) showing the current reading while charging a battery, compared to a known-accurate DC clamp-on meter his dad has. The data he provides for purposes of recalculating the input from the revised amperage, I will go with.} 

The amp data he provided was said by skeptics to be invalid, with the inference being that it made the whole experiment invalid, and nothing short of a redo would be acceptable. I think Sterling has accepted the reality of needing to redo the experiment, so determining what that actual amperage was is no longer germane. {When I submitted this summary for Sterling's approval, he sent me an email clarifying what I had been speculating about here. It reads, in part, "I do not yet concede that my amp readings should be thrown out. They are close enough to the expected value based on the data from Steve, and they move up and down as expected, that I'm not willing to throw them out. My primary curiosity is just how accurate the meter was. My primary reason in wanting to do a repeat is due to the scientific principle of repeating an experiment."} And now we have it. No more speculation. Straight from Sterling.

[See H-Cat amp meter vindication]

Since I brought it up about my indirect method of 'recovering' the actual value, on more than one occasion, even though it is not technically part of Sterling's experiment, it should be submitted as information for adding to the open-source information base. It can be found as Comment #12 in the comment section. It was first posted Mar 31st & finished Apr 1. And finally, I know that both this summary and all my 12 comments submitted to the blog page just mentioned could contain errors. I will gladly entertain challenges to my reasoning or method. I would either offer a rebuttal or a concession. 

Hope this adds to improving the design of future experimentation. Good luck!!! David A. Haack

# # #

Copies of Comments

posted at /2014/03/31/9602464_Sterlings-first_H-Cat_calorimetric-test/ 

The following are copy and pasted here for your convenience and for archive purposes.

Comment 1

April ~ 1, 2014


With respect to the questions raised by Asterix concerning the validity of your amp readings (and thus the entire experiment), my suggestion was going to be essentially the same as given by Obvious. It is theoretically possible to "recover' data that one isn't sure about by recreating the experiment to the point where all conditions that would affect this particular input. Set all your power settings to where they were when you ran the experiment. You would need to re-check with the instrument that did the original reading to make sure it is giving the same reading as last time at that point, and then take a reading with a meter you know you can trust to give you the correct value.

While this method you would certainly not satisfy the critics that demand precise scientific protocol during all aspects of an experiment, but it would tell you closely enough whether the results were good enough to justify the effort of repeating the experiment. And the beauty is that you don't have to go through all the other aspects of the experiment to get those numbers.

The only problem is whether you want to go through the hassle to get the meter back from the guy to redo that re-calibration.

Comment 2

April 2, 2014


I did the math using your cool-down figure of 0.5 deg.C/hour and come up with an improvement in efficiency of 77% to 93.94% for your Set C data. Your had stated that the cool-down rate was ONLY 0.5 deg C/hour. That is a significant amount when comparing it to your temperature change of 5.5 deg. When you take that 1/2 degree over 149 minutes, it comes to 1.424 deg. When you add that back in to your 5.5 deg., it shoots your efficiency up to above 93%. You never explained why you pared your efficiency estimate down from 93% to 77%. I would be inclined to leave the heat loss factored in as it is part of your output.

Comment 3

April 5, 2014


I just sent a super-long comment on how your H-Cat likely has far more efficiency (close to 500%) than have stated. I ran into difficulty when I tried to send it. It didn't have my name down where it says sign in as Dave Haack. I finally had to sign in as guest. Then it seemed to recognize me. I suspect it had to do with running a disk cleaner today.

It is important that you know who sent that comment as I offered to send you an email about a super-simple method for 'recovering' the unknown amperage in your Set C data. It would be for your own personal closure on the experiment.

Dave Haack

Comment 4

April 5, 2014


I don't know if you got a chance or not to read my last comment. In it, I was able to show an improvement for your H-Cat efficiency of 77% (per your Set C data) to 94% by factoring in the heat loss of the water basin. That was a straight forward calculation of the actual experiment, so nothing new here.

I would now like to show some really astounding efficiency estimates as a result of your claim that only 21% of the water from the HHO reservoir ended up in the H-Cat, suggesting that much of the that water ended up as water vapor or steam that vented off. This suggests that you have a mismatch as to your HHO generator versus your H-Cat. In other words, the generator was producing gas far in excess of what the H-Cat could utilize it. being that we were working with a closed system.

There are several hypothetical remedies to solve this problem:
1) Turn down the HHO generator to the level that it puts out only 20% (I'm rounding from 21.6 for convenience of making my point. But, we (I) don't know if this would result in input power dropping to 1/5 in like proportion, or even whether the generator can work on such low output.

2) Increase the H-Cat to five times its size (surface area of substrate matrix). But, we don't know if the H-Cat would put out 5 times as much heat by expanding its surface-area capacity by 5 times.

3) So, our final choice is to leave the generator and H-CAT alone and feed 5 H-Cats from one generator.
This may be problematic from the standpoint of implementation, but it doesn't violate any of the existing parameters. This would generate 5 times the heat into the water basin. Since we are interested in the heat generated by only one of the H-Cats, it would be 1/5 the total ( which is the amount we have already determined). So, 1/5 of the power used would go to creating gas for that one H-Cat.

While this is stating the obvious, it is the kind of gymnastics one sometimes needs to show something is valid. We now know that enough gas could be produced with only 1/5 the power to run the H-Cat experiment. And that is good news in that if you plug in 1/5th the power into your Set C data and using the 94% efficiency I had already shown, it has (assuming the amps of input is correct), we are now, at least hypothetically, at an impressive 469.7% efficiency.

Sterling, I can still show you how to 'recover' that unknown amperage value for your Set C data. It is what you were talking about your dad doing. I have had to do the same type of thing on more than one occasion. Since, it is the kind of thing that would upset those steeped in traditional approaches of experimentation, I would suggest sending the method to you privately in your email account at PESN. It would be so you can have personal closure on this amp issue. If you wand this email, reply in kind, and I will sent it under subject:AMP ADJ. If you don't want it, that is okay also. It is super simple, and does not use all the experiment steps I was forced to insert to make the onlookers happy when I suggested a similar approach in my second to last comment here..

Comment 5

H-Cat members,

RE: 1) April 6, 2014 Comment # 5
2) Appeal to Sterling
3) Explain change in comment format

If I said we have reached a crossroads on this blog page, it would perhaps sent the wrong message. I don't really know what to call it, so I will just explain it. It will be a week tomorrow that this page of Sterling's experiment results was posted. That means that if this were a typical posting, it would be dropping off the bottom of the queue -- soon to be relegated to the archives. This experiment generated a huge following and possibly more comments than any story so far. It would be a shame for it to now drop of into the obscurity like most stories do. I therefore appeal to Sterling as a member of the H-Cat open-source community and CEO of PESN to keep this page in a prominent spot on the Blog Page or blog site. Sterling deserves kudos for running the first experiment to determine the efficiency of the H-Cat. Though it had its problems, it was a great first run. There were critics who claim it was botched and thus inconclusive. While this is fair criticism, I fail to see it as botched. The argument is inferred that an experiment is only as good as it is planned and executed. I was formally educated in that thinking, taught students that concept in high school science, and lived it. Over the years, I came to gradually realize that this world is not always the way it seems. Science has been continually modified and rewritten over the centuries as we have come to unravel its complexities.

Sterling got interested in H-Cat when realizing that 'copious amounts' of heat were generated by passing Brown's gas through a catalytic converter, and estimated to be over unity by anywhere from 2x to 10x. His experiment was designed to determine the efficiency of this catalytic process to see if there was anything taking place in excess of what could be explained by reactions explained by chemistry. Though the experiment, as written up in his report, shows an efficiency of 77%, he is convinced that energy is being tapped from the wheelwork of nature. If Sterling feels that I am misrepresenting this exercise, I want him to set me straight at this point.

I entered this process with basically an open mind. My early training would suggest that this would not be possible. But, what I have come to learn in years since, have taught me that it is better to maintain an attitude of let's wait and see how things play out. After my stint in the education community, any experimenting I did was on my own dime. After my wife had a massive brain tumor removed by the leading neurosurgeon in the nation in one of the leading medical facilities in the country, we were cleaned out financially. I was forced to do my experiments on the cheap. This meant getting certain answers in a round about way. I felt like I had pioneered a new way of doing science. When Sterling wrote that his dad, a renowned physicist, often used such indirect means to obtain missing data, I felt relieved and no longer alone in such endeavors.

I started going through this experiment, section by section, analyzing in my way whether there was anything further that could be learned from what has transpired so far. I won't elaborate on what I've learned so far other than that it appears that the efficiency of this process could run as high as 500%. I am not suggesting that final analysis will bear that out, but it clearly points to the possibility of over unity. My earlier comments here (posted below) explain in detail what gains in efficiency I have uncovered so far, and also the theoretical setup needed to show close to a 500% efficiency.

I feel that I am just getting started in my analysis. In other words, it is only halftime with half of the game left to realize the outcome. Since, most of the comments have now dropped off by most readers having now moved on to other topics in the free-energy field, I am the lone voice crying out in the wilderness. It is sort of like after a joyful party has ended and you are the only one left to clean up. You have only yourself for company. I'm aware of the presence of my moderator, Stuart, and hope that there is at least on other reader out there that doesn't get too bored reading lengthy entries designed to mollify the anal retentives out there. I don't expect this effort to have any immediate resonance with the readers. I would hope that when I finish two things will have been accomplished:
1) Anyone considering repeating this experiment can determine whether it is worth the efforts involved to do it over.
2) Anyone redoing the experiment will know what adjustments will need to be made to get results that are credible and acceptable.

I felt the need to change to this format in order to more clearly identify the various entries I am contributing to this discussion. I only wish I had realized that I would be doing this kind of exercise, so I could have tagged my earlier entries for easier means of referencing back to. While this comment is not involved in analyzing part of the experiment, it is halftime intermission. I hope to continue posting comments here as long as I feel there is more to be learned. And, I hope my comments will be okayed even though they are lengthy. If you are interested in seeing a successful redo of this experiment, check back here from time to time to see if there are any new developments. Stay tuned.

Comment 6

H-Cat members,

April 7,2014 Comment #6

RE: Unaccounted for water loss

This is perhaps the thorniest of all aspects of this experiment. Not only do we need to account for the portion of water not showing up in the catalytic converter when compared to the water used by the HHO generator, but a heat calculation would also need to be made on that missing water to factor into the efficiency numbers, if one wishes to determine whether we are getting more out of the system than would normally be expected. I have already done some preliminary work in a previous comment (it would be #3 if I had started numbering sooner) that starts dealing with the missing-water quandary. I had suggested that the fact that, since not all the water ended up in the bottom of the catalytic converter, might suggest that the HHO generator was putting out far more HHO than the H-Cat could handle. My solution was that if one hooked up 5 H-Cats by means of branched tubing, it should even things out, since it appeared that one H-Cat was using only 1/5 of the available gas. In this scenario, one could hope to see the same 1/5 portion in each of the 5 H-Cats, thus using all the gas available. But, we end up with other problems in the process. The water in the basin would get expectedly 5 times hotter, which might result in unpredictable outcomes. Also, the amounts in each H-Cat might not be the 65 ml expected like when dealing with just one converter. We already ruled out cutting the gas back to 1/5 or increasing the size the the H-Cat to where the inner surface-area was increased to five times larger. So, we are back to square one. Work with the original setup and deal with the missing water in an analytical manner.

We know from the data that a total of 315 ml of water was used by the HHO generator during the entire length of the experiment, and that only 65 ml ended up in the H-Cat at the end of the experiment. That would suggest that a mere 20.6% was actually involved in the process of catalyzing HHO back to water. What happens to the remainder of the missing water could have huge implications on the outcome of the experiment as far as efficiency goes. We can be reasonably assured that the water wasn't vented off as vapor prior to entering the H-Cat according to the report. This means that it was most likely vented off as vapor or steam through the tubing coming out the bottom of the converter. This is the point where Sterling says he wishes he has fastened a plastic bag to catch any liquid that might have vented.

I will here go on record as suggesting that it is quite likely that most of the lost water escaped out this tubing as steam. The reason is that if one compares water vapor with steam, steam is by far the more volatile. Water vapor would essentially be from evaporation of the warm water in this setup. Any steam escaping would be from HHO changing back to water in the H-Cat and converting to steam anywhere it comes into any area hot enough to cause a phase change of water to steam. That is why this possibility is so HUGE! If the water escaped as steam, we are talking about another large heat loss in the system that would really jack up those efficiency numbers. In the comment where I factored the heat loss due to the cooling of the water basin, it amounting to a gain of efficiency from 77% to 94%.

Here we are talking about latent energy of water involved in the phase change of water to steam If I remember my physics accurately after all these years, it would be 540 calories required to change i gram (=1 ml, =1 of water at 100 deg. C. to 1 g of steam at 100 deg. C. This latent-energy phase change is so huge that it is a quantum jump in itself without even considering other anomalous sources. If one multiplies 540 calories to each of the 250 ml of missing water, that is a whopping 135000 calories that can somehow be eventually factored into the output as heat loss. The problem for calculating efficiency comes from the fact that this water loss was attributed to the entire length of the experiment which involved times where we didn't have accurate data. If we had the correct value for the amperage in Set C, then we could do some more exacting calculations for efficiency.

This is getting to be a lengthy comment, so I will take up the job of coming up with an estimate that might shed some light. Keep in mind that I have previously stated that there is a super-simple methodology for 'recovering' the correct value for the questionable amperage in the Set C data set. To be continued...

Comment 7

H-Cat members,

April 8, 2014 -- Comment #7

RE: Math for the unaccounted water loss

I had concluded, in comment # 6, that the water loss was probably a combination of water vapor and steam. If you are confused about what water I am referring to, please review that comment. Any steam formed would have a much easier time getting to the top of the tube and venting than water vapor would. The water vapor would have a better chance of cooling down from the surrounding water bath, and condensing back to water before escaping. After condensing, it would likely just run back and be part of the water found in the bottom of the converter.

So, the math would consist of calculating the calories of the steam due to the coefficient of latent-water-phase change. When a ml of water at 100 deg. C. changes to steam at 100 deg. C., it would draw 540 calories of heat away from the converter, and when the steam vents, this amounts to a heat loss of that amount. Thus, those calories have to be added back into the output calorie, since they are part of the output. If this steam had not escaped through venting, the heat would have remained in the system and warmed the water; and you would have gotten the calorie count that way.

We are dealing primarily with the Set C data. But, that was for only a 149 minute time period, and the lost water was happening over the duration of the experiment. So, we need to figure what portion of the missing 250 ml would be lost during the 149 minutes of the Set C phase. Some would argue that you can't determine that amount because there were stops and starts and electrical-connection problems involved with the Set A and Set B part. But since the converter was in a water bath, conditions would not vary all that much from one time period to the next. Also, the water is a result of the catalytic process and would depend more on gas converting to water. So, again the lost water is more a function of water in versus water out than electricity in versus electricity out. Using a direct proportion will not be completely accurate, but it will come close enough at giving us a good estimate. We aren't doing this exercise in hopes of being published; but rather to get a feel for what took place during the first experiment. This hopefully, will help steer the next experiment in a more optimized direction.

If 250 ml of water is lost to venting during the entire experiment, then 78.57 is lost during the 149 minutes of Set C phase. The water tank feeding the HHO generator went down 315 ml; and only 65 ml was recovered from the converter. leaving 250 ml. Of all the ways to set up the proportion, only the temperature change during Set C compared to the total temperature during the entire experiment, is the only parameter that can give a reliable answer to water loss amount during Set C phase. Using time comparisons would be quite skewed due to vast differences in what was happening in the 2 time periods. If you had data for any other parameter, I believe it also might be fraught with uncertainties.

5.5 (temp change during Set C) / 17.5(total temp change for entire experiment) = x (unknown ml for Set C) / 250 (total ml lost during entire experiment). Solving for x gives us 78.57 ml lost during the Set C phase. I will apply the latent heat of phase change for water of 540 calories/ml to the entire 78.57 since most water vapor would likely have ended back in the bottom of the converter and given up its energy to the surrounding water bath (thus being accounted for there).

78.57 ml X 540 calories/ml= 42427.8 calories. This can be added to the calories we already have for Set C. We see 253000 calories listed in Set C for output. If we add the two numbers for calories we get 295427.8 calories. We need to multiply by 4.18 Joules/calorie, we get 1234888.2 Joules for the combined output from warming the water basin and heat loss due to water loss. If we redo the input/output ratio to determine efficiency, we get 1234888.2 Joules output / 1380000 Joules input = .8948465 X 100 = 89.48465 % efficiency. I would add in the Joules I got from heat loss of the water basin, but the math will all need to be rechecked before I combine them.

Next, I plan to take up the topic of the bubbler where the HHO gas from the generator was consistently warmer than the generator. I will attempt to explain why this might be, and whether we can quantize any heat gain/loss from the data we have. Stay tuned...

Comment 8

H-Cat members,

April 9, 2014 -- Comment #8

RE: Bubbler reservoir at higher than HHO generator

In my last comment I indicated that I would address the topic of Sterling's noting that the bubbler reservoir operated at a temperature about 6 F deg. warmer than the HHO generator. I will now offer what I feel may be a plausible explanation for this happening. Whenever you compress a gas into a vessel, the gas heats up. That would be the most likely scenario of any I can think of. At first, one might envision Brownian Motion, where the gas molecules bump into each other more oftener in the smaller environment. Or they could also create heat merely by slipping and sliding over one another more readily in a more confined environment.

But, when you think of Brownian Motion being the source of the heat, it really doesn't hold up -- at least in this case. Anyone familiar with an air compressor probably has noticed how the tank gets warm when you are filling it off the compressor. But, once the tank is filled, it no longer continues to get warmer. If it did, you would have your free-energy device right there. So, it is understandable that the reservoir would maintain a warmer temperature due to forcing gas through the water in the bubbler. I guess that might be called the heat of compression. After the experiment was over, you were no longer creating gas to force through the bubbler, so it gradually cooled down.

Now the question becomes one of whether we can hopefully quantify this heat generated and factor it in to our model as a gain or loss in efficiency. Unfortunately, I'm afraid the answer this time would be NO. We just don't have enough measured data to extract any estimates in the indirect manner. I might note that this could be an important objective for the next person running this experiment to design this part of the setup so as to properly insulate from the generator all the way to the converter. to cut down on heat loss, but to have the whole thing immersed in a separate water bath where you can take temperature measurements on a timely basis to get calories produces in the process of the reservoir warming due to heat of compression.

It is important to know whether this would count as input or output and whether it is a gain or loss in efficiency of the H-Cat. These types of considerations can become quite confusing unless you have a firm grasp on what you consider the device you are measuring, so you can draw a line between input and output. In my estimation, I would say that if you are determining the efficiency of the H-Cat, that would entail everything including the HHO Generator, all your flash arrestors, bubbler reservoirs, and catalytic on one side as output. On the other side all you would have on the input side would be the electrical power supply. So, we would be dealing with output. Would this count as a gain in efficiency or a loss? This heat energy is definitely being lost to the ambient surroundings. A very minute amount is probably carried through the apparatus into the converter where it will be tallied with the other sources warming up the water in the basin. So, what do you do with the amount that is warming the air around it. If you could send all this heat energy through the bubbler apparatus and into the H-Cat, it would raise your temperature in the water basin by a certain amount. So, a heat loss is always a gain in efficiency, when you are measuring and experimenting to determine the efficiency of what a certain device will be able to produce.

That is a totally different thing than determining the efficiency of a machine or device on market. When you are calculating the efficiency in that case, you can't factor losses into the efficiency because you are not able to make use of that part of the output to run your machine. It gets confusing sometimes, but you need to differentiate between whether you are measuring efficiency to determine experimentally what a device is capable of producing, or whether you are measuring the efficiency of what a machine will deliver in output to serve yours needs in a particular application. Enough said. No, sense in beating a dead horse to death.

I think I will try to tie up some miscellaneous loose ends and some math checking in my next comment. We need to do that before I can summarize the total output in Joules. Hang in there. The analysis on the output side is nearing completion. Till next time...

Comment 9

H-Cat members,

April 12, 2014 --Comment # 9

RE: Miscellaneous clean-up details

As stated in my previous comment, this session will be devoted to minor considerations and housecleaning.

First, there are the two pieces of experiment apparatus that Sterling already touched on. I would not normally tread on ground that he has already covered, but since I see an opportunity to squeeze out a few more calories of output from them, I will go through the exercise. These two pieces consist of the metal water basin in which the converter was submerged, and the 1/2 brick used to hold down the converter so it wouldn't rise to the surface and loose heat to the air.

The steel tub weighed 4.25 # = 1928 g. The specific heat capacity of steel is 0.10 c/g, for each degree change in heat. Remember that the adjusted change in heat is now 6.742 degrees when adding in the heat loss of the tub, 1.242 deg. to the 5.5 deg. the tub changed over 149 minutes. 1928 g X 0.1 cal/g = 192.8 cal/deg. So, for a change of 6.742, it would be 192.8 cal X 6.742 = 1299.8576 calories required to warm the tub up that amount.

Now, with the brick, it is fortunate that Sterling weighed it. You will see here how important it is in an experiment to measure everything you can think of that could possibly be needed later for some calculation. The brick weighed 1#, 14 oz. That is 1.875 # in decimal. There are 2.20462 # in a kilogram, so the brick weighed 0.8504867 kg. If we divide by 1000, we get 850.48673 g. The specific heat capacity for the common brick is 0.20, so Sterling was correct is saying the brick was of a higher capacity. If it were a hardened brick, the heat capacity would vary slightly from this figure. If we take the weight 850.48673 cal. X 0.20 g/deg., we get 170.09734 cal. To, warm the brick would require 170.09734 cal. X 6.742 deg = 1146.7962 cal. I will skip converting to Joules at this point, since it makes more sense to convert only once, thus introducing less rounding error.

I will now recheck my math on the heat loss of the water basin. The cool down of the basin after the experiment was over was 0.5 C. deg./hr. The Set C portion of the experiment lasted 149 minutes. That would be 2.483 hr. So, in 2.483 hr., at .5 deg/hr, we have a 1.24167 deg. change. We can add this change in temp due to heat loss to the change in heat that the water in the basin changed. That is because, if the system were perfectly insulated, that is what the temp change of the water basin would have been. Calorimetry is based on the concept of insulating to the max so you can capture all the heat put out by the system being measured. Adding our heat loss of 1.24167 deg to the 5.5, we get a value of 6.742 deg as our heat change when heat loss is factored in. Then all we need to do is multiply by the ml of water in the bath, 46000 ml X 6.742 cal/ml gives us 310132 cal in output.

I am now ready to combine all the calories of output into one sum, so we can convert to total Joules of output. This will be the topic for my next comment. Till then...

Comment 10

H-Cat Members,

April 13, 2014 -- Comment # 10

RE: 1) Combining various losses & adding them into the output; 2) Summarizing what to take away for next time

I will now list the various energy losses & add them to the primary heat energy Sterling calculated from the change in temp of the water basin These heat energy losses are all calculated and/or adjusted to cover the elapsed time period of 149 minutes registered in Sterling's Set C data.

Heat loss of the water basin -- Refer to Comment #9 ********* 57,116.82 cal.
Heat loss due to steam venting of water unaccounted for -- explained in Comment # 7 *** 42,427.8 cal.
Heat output attributed to warming the steel tub -- See Comment # 9 **** 1,299.876 cal.
Heat output due to the 1/2 brick -- also in Comment #9 ****** 1,146.7962 cal.
Total output in calories due to the above list values ******101,991.28 cal.

We will now add this value to the calories Sterling obtained from the change in temp of the water basin in Set C. 253,000 cal. Note how large the various losses add up to when compared to the primary heat output value of the water tub. It is approximately 40 % of what was stated for output originally. That is a significant amount. We get a total output value for the H-Cat of 608316.50 cal. I encourage others to check my math as I intend to do yet. Also note that I tend to carry the subtotals out to a significant number of decimal places. I don't round till I have the final total. If one rounds the all the subtotals, you can introduce error simply from too much rounding.

I had intended to also do an output summary, but something has come up. I will have to do that in the next comment. To be continued ...

Comment 11

April 14, 2014

H-Cat members,

April 13, 2014 -- Comment 11

RE: Output wrap-up

Having gotten called away when I did left sort of a mess with my last comment. In the meantime, I did have time to redo my math. The major change is with the caloric value attributed to heat loss due to venting of steam. I had used 17.5 deg.C. as the temp change from the beginning of the experiment to the end. It gets quite confusing when wading through all the discussion of which thermometer read what at this time or that, and how much one thermometer was off from another, you can loose track of where you are at. I finally arrived at 18 deg.C. as being the change in temp for that time period. It is shocking that that 1/2 deg.pares down the # of calories due to steam venting from the 42,427.8 I had previously calculated, to a shockingly lower 41,250. Any other difference that shows up in my total is due to my making sure to carry the math out to a significant number of places at this stage of the calculation. The latest revised number for the 4 lost outputs combined is now 100813.19 calories. When you add that to the 253,000 calories Sterling already had listed in his Set C output data, the new grand total comes to 353,813.19 calories! We can now convert this to 353,813.19 x 4.18 = 1,478,939.1 Joules

This is a really big thing because, if you use the input Joules from the Set C data, 1,478939.1 / 1,380,000 X 100 = 107.1695 % efficiency! If the input Joules are correct, that would be overunity. But, that remains the big unknown in this experiment. There will always be an * (asterisk) associated with this experiment, so long as we can't be sure about the input amperage. I still do firmly believe that the actual value of that unknown amperage can be 'recovered' acceptable to a high level of scientific scrutiny with a method which involves arriving at the number using an indirect approach. Since this is Sterling's experiment, I feel that he should have first shot at providing this 'recovered' but actual number. Having retrieved this actual amperage value would, of course, never satisfy the purists and the skeptics because it wasn't a measurement arrived at the day of the experiment. But, who we really need to satisfy are the members of the H-Cat community who may wish to build their own device.

There is one output that was lost and could not be calculated due to not enough data. That is the bubbler reservoir that maintained approximately 6 F deg. warmer temp than the HHO generator during the entire experiment. One would, perhaps, need to also surround this entire assembly in an insulated water bath and do the same calculations as for the converter assembly. This is not for me to determine as to how to capture this heat loss, but it is another heat loss that could be added into the H-Cat efficiency.

Sterling's having measured the weight of the steel tub and the brick brings out the importance of taking measurements, even if you don't plan to use them. He didn't use them in his calculations, but I was able to. Taking these kind of measurements allows others who study the experiment to evaluate parts that the experimenter didn't delve into in his/her summary. And that, I believe, is it for the output side of my analysis. There are numerous errors, such as tagging numbers with the wrong units of measure. But the math should be consistent, nonetheless. If anyone has issue with any of my analysis, i will be happy to rebut or concede the point of contention. But, I must say it was delicious to have the freedom to complete this task without interruption. I have aimed more for exposing my method, than being obsessively accurate in detail which is commonly already known. In other words, I didn't spend my time on the proof-reading part of submitting my analysis.

See you on the input side, I hope. If, not, tarry on and I wish you all luck in your endeavors in a very exciting aspect of free-energy development. Kudos to all!

Comment 12

H-Cat members,

April 19,2014

RE: Indirect method for 'recovering' actual value for the amperage reading

Now that I have essentially wrapped up my work with analyzing and doing the calculations for the various heat losses on the output side, I would like to briefly turn my attention to the input side. On the input side, we have nothing more than the electrical-power source. So, that would be the only place we would need to look for heat loss. And indeed, according to Sterling, the back, top end of it reached 135-145 F. deg. I'm assuming that would be the area where the output leads of the supply unit were. While this points to definite heat being produced, there is no real way to quantify the amount, and I doubt there would be an easy way of doing it for next time.

That said, the only item still left undone, as far as my part in this would be, is the presenting my method for 'recovering' the actual value for that much disputed amperage reading. This, of course, would be for the Set C data, since Sterling had already determined that this is the only data set with any reliability, due to poor, loose, and separating electrical connections in the other two sets.

So, here are the steps I would propose for 'recovering' the actual value to replace or verify that questionable reading. I will first caution that anyone who would attempt to actually perform the use of this method would need to understand that they would be doing it at their own risk. I do not consider myself qualified to give advice pertaining to electrical or electronic matters. This is a basic type of method that I have used to indirectly arrive at values that I could not get directly. I have not actually performed this particular activity, so I can't say with certainty that it will work. I therefore present this method as an educational exercise worth consideration for this application or any similar type of data recovery.

If you are not thoroughly grounded in working with electrical circuits, containing the various electrical components that may be needed to accomplish the task, I would suggest getting aid or advice from a qualified electrician or equivalent. The components used in the circuit be damaged if you do not understand this sort of thing. Also, any meters used to measure voltage, amperage, or ohms of resistance could also be damaged. And, not the least of which, you could also be injured or worse. I hope this now serves as an adequate disclaimer to explain the dangers of what could happen, if not done in a responsible way.

I had felt that Sterling should have first shot at seeing this methodology, since it was his experiment. I have made it available for him to see, and now offer it to the H-Cat community to be submitted to the knowledge base of the H-Cat open-source project. I am not personally concerned with whether the amperage value in the Set C data is accurate. I did all my calculations using the data and information presented in the experiment. My goal was not to get a valid and reliable end result, but rather to contribute to the method of evaluating this experiment with and eye to improving the work on H-Cat in the future.

HERE ARE THE STEPS FOR INDIRECTLY DETERMINING A MISSING DATA VALUE; NAMELY AMPERAGE (Refer to aforementioned disclaimer before performing these steps):

As you may surely know, the properties of voltage, amperage, and electrical resistance in a circuit are inter-related in such a way that they can be expressed by a formula. If one knows any two values in a circuit, the third can always be determined experimentally or by calculation.

If we know the volts and the ohms that existed at the time the Set C data was taken, we could determine the amperage. We know what the voltage was because it is recorded in Sterling's report above. If we can set the resistance in the circuit to what it was during Set C, the amperage would then be re-created.

I would recommend a variable power supply (such as a variable transformer or some such) to set or dial in to the voltage value for Set C. The next task would be to set the resistance in the circuit to where it was on the day of the experiment. We would need something like a variable rheostat that could adjusted to that value. To determine where to set it at, we will need the same amp-meter that was used at that time. In Sterling's case, it would be the Fluke meter he borrowed from Josh Peters. You will need to set (toggle) it to the same setting it was on at that time. If a meter see the same amperage in the circuit on two different occasions, it will read the same, irregardless of whether it is reading the correct value or not. In other words, the same input stimulus elicits the same reaction in the meter. So, once you have the meter set exactly the same way as when the experimental reading was taken, you adjust the resistance on the rheostat until the amp-meter sees the same amperage as before. Once you have the rheostat set to the right resistance, you can record how many ohms there are and calculate the amperage by formula or you can take a reading of the amperage with a reliable amp-meter set to take the direct current reading accurately by having it set to the proper settings.

The only way this would not work is if the Fluke meter was not actually detecting the amperage in the circuit, but rather some other junk or noise. I would leave it to the person doing the recovery to determine whether you need rectifiers, resistors, etc. get the right conditions of setup to both recreate the conditions as to voltage, amperage, and resistance present that day, as well as what would be needed for protecting the various components from frying out or other damage.

And that, essentially concludes my comments pertaining to Sterling's first H-Cat experiment. Good luck, and safe experimenting to all the H-Cat community!!!

Comment 13

H-Cat members,

April 20, 2014 -- Comment # 13

RE: A couple follow-up points to Comment #12

Happy Easter!

First, item is that I forgot to do is label Comment #12. It is the comment immediately preceding this one (not the one I wrote to PA32R). I need to clear that point up for those looking for Comment # 12 as directed here from my report summary.

Next, I failed to mention that if you can't get your hands on the original amp-meter (or it would be too difficult to get the original again), any other amp-meter that has the same 'wrong' setting as the original meter, will suffice. It may not be calibrated to exactly the same tolerances as your original, but it should also get you very close to the actual value. This would be like taking a temp with two different thermometers. They may not give you precisely the same reading, but the two readings should be close enough.

And lastly, I should have stated the formula I was referring to in my explanation in Comment # 12. No matter how simple or trivial something may seem, it should be included in this type of presentation. Ohm's law states that the amperage in a circuit varies directly proportional to the voltage in that circuit, and inversely proportional to its constant resistance. But, any value can actually be held constant and the other two will vary according to this law (relationship). So, this law can be written mathematically as amps = volts / ohms. Or, this law can be stated by either of 2 variants; volt = amps X ohms or ohms = volts / amps. No matter which variant you are using, you simply plug in two known values, and solve for the remaining unknown value.


Comment 14

H-Cat members,

April 25, 2014 -- Comment # 14

Re: Revision of numbers for unaccounted water

In Comment # 7, I did the math for the water loss associated with there being less water in the catalytic converter at the end of the experiment, than what amount of water there was in the water reservoir at the beginning of the experiment. This water loss could be due to any of the following forms of loss: HHO gas, plain water, water vapor, or steam. I had argued for most of the water loss as likely being attributed to steam. Sterling argued, and rightly so, that such an argument could not be substantiated -and I agree. He said that for purposes of presenting a credible analysis, we must err on the side of caution -- to which I again agree.

What I would like to now do is consider the math involved in all four cases:
1) Case involving all water escaping as unconverted HHO gas. I could not come up with any number for the specific heat of HHO gas, so I am not able to do the math. I can't imagine there being much heat loss involved with the gas passing through the system and venting unchanged. So, I would say we have no heat loss to speak of in this case.

2) Water being forced out of the reservoir either due to conversion of HHO back to water, or merely forced out by the bubbler pressure. Here we do have some heat loss due to the fact that the water in the reservoir was constantly at 108 - 110 deg. F. So, 109 would be a good number. We, need Centigrade, so 109-32 X 5/9 = 42.77 deg. C. The temperature range of the water basin during Set C was (30 - 35.5) deg. C, so the average would be (30 + 25.5)/2 = 32.75 deg. C. That would make a temperature drop of 42.77 - 32.75 = 10.02 C. deg. From Comment # 7, we learned that 78.57 ml of water ended up short during the 149 minutes of Set C duration. Since water has a specific heat of 1 cal/ml - C. deg., we take 78.57 ml X 10.02 = 787.2714 calories. That would be the calories lost if all the water escaped in the form of water. ( The specific heat of water with KOH in it may vary from the specific heat of 1.0 that I used.)

3) Water leaving the system as water vapor. This would be the scenario of water leaving the reservoir as water, warming up enough in the heat of the catalytic converter to evaporate (or it could possibly be steam that has cooled down). It would not, in my opinion, be fair to apply the temperature drop as in Case 2, since the water could be leaving the system in a warmed state. There is however, one factor that needs to be addressed. The specific heat of water vapor is 0.95 cal/ml-C.deg., rather than 1.0 for water. So, in this case we would have 78.57 ml X .95 = 74.6415 calories of heat loss if all the water lost during Set C was in the form of evaporated water or cooled steam.

4) Case where all the water lost during Set C interval was lost as steam. The calculations for steam are all the same as the math shows for this scenario of venting due to steam, except for one additional consideration. Steam has the same heat capacity as water vapor as they are essentially the same thing except for their temperatures. I had come up with 42,427.8 calories for the heat loss of the 78.57 ml of water after it had gone through the phase-change conversion of 540 cal/ml. But, those calories were assuming a specific heat of 1.0 (for water, rather than 0.95 for water vapor and steam). So, our calorie value for this scenario should be 95% of what I had figured in Comment # 7. I will have to make an adjustment to my upper limit result to reflect this change.

My calculations for this water loss will need to be expressed as a range form no heat loss with all gas escaping to a high if it all leaves as steam. The correction for the list in Comment # 10 should be revised to reflect these values. Line 2 should now read Heat loss range for water loss of system ****** 0.0 cal. (if all gas) to 40306.41 (if all steam). I would like to point out that either of the other cases, as well as any combination of the cases as a mix would all fall within this range. And the total-outputs line should now read "Total outputs for all additional heat losses, (not accounted in Set C), range from ************59,563.48 cal. to 98,891.716 cal."

If we add each the lower and upper heat loss values to the calories already calculated by Sterling in Set C, we get values of 253,000 + 59,563.48 = 312,563.48 calories for our conservative lower limit; and 253,000 + 96,891.716 = 349,891.71 calories for our most optimistic assumption. We must now multiply each of these limits by 4.182 J/cal to convert the work lost as heat to Joules, since the work of electrical input is also expressed in Joules. For the lower limit, 312,563.48 cal X 4.182 J/cal. = 1,307,140.4 Joules , and 349,891.71 X 4.182 = 1,463,247.1 Joules for the upper limit.

We are now ready to compare input expended in Joules versus output realized in Joules to calculate the range of efficiency from conservative to optimistic. For the conservative efficiency value, we have 1,307,140.4 Joules output / 1,380,000 Joules input (from Set C) = 0.9472031 X 100 = 94 % efficiency. For the optimistic upper range value, we have 1,463,247.1 / 1,380,000 = 1.0603239 X 100= 106 percent efficiency.

I am struck by the results which range from 6% under unity to 6 percent overunity. So, the results might indicate that we missed some heat loss some where ( or some other loss), and this is a demonstration of conservation of energy at its best. Or, it might hint at the possibility that more might be going on here than regular, well understood dynamics. For me, the jury is still out, but looks mighty encouraging. I for one can't wait for the repeat of this experiment to confirm some of what we already learned and hopefully learn more next time. I leave this exercise in hopes it will inspire more people to join the H-Cat community and serve due diligence on a very worthy project. For the most part, this has been a rewarding experience for me. Kudos to the team who have brought it this far. There are too many to mention, and I could end up missing a name.

David A. Haack

Comment 15

H-Cat members,

April 25, 2014 -- Comment # 15

Re: Warmed up reservoir

It seemed I would not get to weigh in on the heat loss due to the water reservoir warming up. Sterling had sent an email to Steve for information about this reservoir, and we hadn't heard back. Well, we have now heard back and I now have enough data to do a calculation on this component.

This is the water reservoir that comes directly off the generator, through which the HHO gas from the generator passed. It serves as the tank for the water and KOH electrolyte mix and also as a flash back arrestor. I hope I have this right. If not, someone will correct me on this point. This reservoir maintained a temperature consistently warmer than the generator which also warmed because of its own characteristics, which we I will not get into in this comment. I go into my own explanation for this warming in Comment # 8.

The temperature of the reservoir maintained a consistent temperature range of (108 - 110) deg.F. The average temperature of 109 is the temperature to work with. We next need to find what the cool down rate was. In 6 hours the tank cooled down to 6 degrees F. above ambient. According to the graph plotting water vs. ambient we can pull the ambient temperature during the Set C phase. It shows 15 deg. C. We, eventually will need this in centigrade, but it might be simpler to first change this centigrade reading to Fahrenheit If we convert 15 degrees C. to Fahrenheit, we get 59 degrees. If the tank cooled to 6 deg.F. above ambient, that would be 65 deg.F. Subtracting 65 from 109 gives us a change of 44 F.deg. We need to know how much the change would be in the 149 minute interval of Set C. If we set up a proportion: 149 / 360 = y / 44 (where 360 is 6 hr. in min., x is the unknown change, and 44 was the total drop in 6 hrs). Solving for y, we get 18.21 deg.F. I used y instead of the usual x so as to not confuse it with X which would mean multiply or times. We need this change to centigrade and that number would be 7.66 C. deg.

Now, we can calculate the various heat losses associated. In this case there are only two: the water in the reservoir and the plastic tank. The water in the tank is a bit tricky in that we need to know how much was in the reservoir at the time of the Set C run. Since the HHO generator was going the entire duration of the experiment, we would have a constant drop in water level in the generator. The generator started at 3:39 P with 3 liters of water, and ended at 12 P From the Calculations section of the experiment report, it show that a total of 315 ml of water was used. We have 3,000 ml (3 liters) at the start and 315 ml at the end is 2685 left in the tank at the end of the experiment. The total time elapsed from start to finish would be 981 minutes. We can use another proportion to get the water used during Set C: 315 / 981 = y /149 (where 315 is total H2O used, 981 is total run time, y is unknown water used during Set C and 149 is Set C run time). Solving for y, we get 47.844 ml used by the HHO generator during the Set C interval. If we add the 47.844 ml to the 2685 left at the end, we get 2732.844 ml in the tank at the start of Set C. If we average 2732.844 and 2685 we would have about 2709 ml as the amount of water in the tank during the Set C run. Every ml of water has 1 calorie of heat lost for each change of 1 centigrade degree. With 2709 ml of water dropping the 7.66 C.deg. we already calculated in the previous paragraph, We end up with 20,751 calories heat loss due to the water in the tank during the Set C run.

Now, for the plastic tank. The weight of the tank was 13.7 oz. (This was not provided in the experiment data, but is available through the supplier). First we need to convert the ounces to pounds. Dividing 13.7 by 16, we get 0.85625 pounds. There are 2.20462 # in a kilogram, so this tank weighs about 0.388388 kg., 388.388 grams. Plastic has a specific heat value of 0.4 cal/gm-C., we have about 388 g going through a heat loss of 7.66 C. deg. (2 paragraphs above) at a specific heat value of 0.4. We multiply the 388 g by 7.66 deg. change by the specific heat of 0.4 and get 1190 calories of heat lost by the tank during the Set C run.

If we add the 20,751 from the water in the tank and the 1190 from the heat lost to the plastic tank, we get a total heat loss attributed to the reservoir as 21,941 calories. This is not a huge sum, but well worth calculating if one is doing an experiment. When you see what was needed to get this value, I hope you can appreciate why it was not determined sooner. I think I will leave the readjusting of values as a result of this new value for another comment. I do think there will be another comment needed to deal more with the input side of this experiment. After, we deal with that issue, there is not much gained by re-tabulating now. Until then ...

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Page posted by Sterling D. Allan Jan. 29, 2011
Last updated April 27, 2014




"It is harder to crack a prejudice than an atom." // "I'd rather be an optimist and a fool than a pessimist and right." -- Albert Einstein

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