Posts by LDM

Quote from IH Fanboy

So it sounds like the head of pressure to the pump inlets was due to gravity.

Since both the internal tank and the outside tank are much lower then the higher rows of Prominents, just gravity alone can not feed the Prominents.

If we had just gravity, those higher Prominents would have had a negative inlet pressure

The Prominents can maybe overcome this negative pressure by the sucking action as they increase their internal volume during the inlet stroke but I find this an unlikely scenario.

Quote from Bruce__H

IH Fanboy

People keep bringing up this idea that the internal reservoir for the E-Cat plant is sealed and has a head of pressure on it. But in this case I don't understand how the external reservoir sitting outside the E-Cat plant, feeds the internal reservoir by gravity as described by Penon and Barry West.

In case the internal tanks have a pressure on it, then that pressure can feed the Prominents with a positive inlet pressure while at the same time due to the hight difference between the external tank and the external tank can feed the internal one. The water pressure is possibly provided by the additional pump on the JM side which AR describes in his notes about the Smith report.

Quote from Alan Fletcher

I thought about asking that .... but decided to limit my questions to cut-and-dried answers.

If I had contact with a pump DESIGNER then I'd ask that question.

If the tech answering your question does not know the answer he will probably contact another person within the organization who knows the answer.

That might be a pump designer.

Quote from Alan Fletcher

Question : Can you please clarify exactly what the manual means by "back pressure" ... is it JUST the pressure at the discharge end (eg a 5 foot head of water = 0.15 bar), or does it depend in some way on the pressure at the suction port?

Can you also ask them under which exact circumstances the pump can achieve several times the stated feed rate.

Quote from Alan Smith

That is correct. The total lift height of the water column has to be paid for- no free rides.

Has to be paid for in electric energy to operate the pump.

However I disagree about adding the negative column seize at the inlet with the column height at the outlet (But correct me if I am wrong)

Why ? Because when the pump is pulling the water in, the outlet valve is closed and the water column at the inlet will not see the column height at the outlet.

And when pushing the water out of the pump, the inlet valve will be closed and the water column at the outlet will not see the column at the inlet.

Quote from Alan Fletcher

But one of the things that confuses me is the picture (2017 Catalog) of a suggested set-up, which is basically what I have

Alan,

Since you are the owner of a Prominent pump you should have the right to contact the technical support and discuss the "confusions" you have with them

Technical Support Tier 1

Please call 412-787-2484 or email [email protected] and you will be directed to the appropriate technical representative.

Quote from Bruce__H

LDM

Hi LDM. Your idea about driving the pump to high rates using external inputs is terrific! But I don't think that this is how the pumps were used during the Doral test.

Post #240 on this thread shows the pumps we are discussing sitting in front of the Big Frankies used in the Doral setup. Focus on the face of any one of these pumps. You may have to blow up the picture somehow. Near the bottom right of the pump face, right beside the power cord, you can see a cluster of 3 light-coloured discs arranged in a triangle. These discs are pieces of (I think) rubber covering up electrical ports that are not currently in use. The port at the bottom left of the cluster is the one that would be used for driving the pumps in the manner you describe.

So the inputs that would drive these pumps in "Contact" mode were never hooked up. That means that the stroke rate of each pump must have been programmed into it using the default selection menu, and I think that the top stroke rate you can achieve this way is 180 per minute.

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I am totally agreeing with you that on the picture you are referring to the external ports for controlling the stroke rate are not connected.

Good point that you looked at the connectors on the picture, should have done that myself.

However I am not totally convinced since on the picture you refer to you can see that besides the black power cables in the cable ducts before the Prominents, there are also grey cables which seems to have no purpose. I wonder if those cables can be connected to the Prominents.

I am afraid that we can't determine that and Rossi won't tell us.

Quote from Alan Fletcher

"When metering at atmospheric pressure the pump can achieve several times the stated feed rate" ...

0.2 bar is pretty close to atmospheric.

The question is under which circumstances it can be achieved. They are not providing additional details, but maybe the following applies.

As I understand the maximum standard stroke rate programmed into the Prominent is 200.
In my opinion this means that for the specified back-pressure of 2 bar, there is enough drive force in the solenoid to maintain the maximum specified stroke rate of 200.
If the back pressure is much lower and less force is needed, higher stroke rates then 200 can possibly be achieved. How can you achieve that higher stroke rate ?
Possibly by controlling the stroke rate yourself by an external signal pulse signal.
The Prominent manual states :

10.5.3 "Contact" operating mode settings

Alongside the setting menus, which are described in more detail in
Chapter 10.6 ‘Programmable function settings ("Settings" menu )’
on page 47, in ‘Contact’ operating mode in the ‘Settings’ menu, the
‘Contact’ menu is also available.

‘Contact’ operating mode allows you to trigger individual strokes or a
stroke series.

You can trigger the strokes via a pulse sent via the "External control" terminal.

The purpose of this operating mode is to convert the incoming pulses with
a reduction (fractions) or small step-up into strokes.

Further on the documentation states :

The number of strokes per pulse depends on the factor which you input.
By use of the factor you can multiply incoming pulses by a factor between
1.01 and 99.99 or reduce them by a factor of 0.01 to 0.99.

Number of strokes executed = factor x number of
incoming pulses

Could this be the solution to the question how the higher flow rate for the Doral plant was achieved? It was stated that flow rates of 72 Liters/Hour at low back pressure could be achieved. Was that maybe with external controlled stroke rates higher then the rate of 200 in the specification ?

Quote from Paradigmnoia

How hot would the heater coils in the Lugano device be, if the outside surface was 1410 C?

And, how hot is the Lugano thermocouple in it's fairly central Cap position?

First a remark concerning your statement that the fins on the Lugano device contribute little to the overall cooling.

The area of the central tube of the Ecat is 0.0125 square meter without fins.
The area of the central tube with the fins present is about 0.0263 square meter
If we assume a view factor of the fins of .9 the effective area with fins is 0.9 x 0.0263 = 0.0237 square meter.
The ratio of the areas with and without fins is 0.0125/0.0237 = 0.53
According to the Stephan-Bolzmann law the temperature of the central tube thus will be lowered by the forth root of 0.53 or 0.85.
Thus the temperature of the central part, disregarding convection, would be 15 % lower with the fins present.

I consider this not being little.

Now about the question if the outside surface was 1410 C?

As a consequence of using the correct fin area when calculating radiated power for the Lugano device you have to calculate with an area of 0.0237 square meters instead of the 0.0125 meters.
Now with this area of 0.0237 square meters you can with the data in the Lugano report calculate back what the average emissivity of the central part must have been.
If we look at the data of the first actual run, then the calculated emissivity becomes 0.215 for the reported temperature of 1260 degree C. When using the Nasa emissivity curve (Which is about the same for high temperatures as the curve given in the report) the emissivity should have been .402. The error being 46 % !!!
For me this is an indication that something must have been wrong. Possibly the temperature due to the emissivity error the dear professors are said to have been made or that the body temperatures in table 6 of the report was wrongfully reported in degree K instead of degree C and thus 273.15 has to be extracted from those values.
The curious thing is that both type of corrections leads to temperatures which are close.
If we assume that the values in the second column of table 6 where wrong (But assuming all other temperature where reported correctly) , the real temperature would have been 987 degree C. Now if we for this temperature calculate back the emissivity we arrive at a value of .472 and the value according to the Nasa curve is also .472
This is a strong indication that indeed the real temperature must have been 987 degree C.

Note that these lower temperatures makes the use of Kantal A1 possible as the heating wire used in the Lugano device.

Quote from Alan Fletcher

However, it delivers more at lower back pressure. There is a warning (hidden deep in the documentation where even expert witnesses can't find it) that at VERY (undefined) low back pressures the pump can deliver "several times" more than the specification.

Since the pump has to overcome the difference between inlet and outlet pressure, does it mean that increasing the inlet pressure (for example by the recirculation pump) until the difference between inlet and outlet pressure is very low, will have the same effect ?

This could possibly be tested by supplying the water to the pump from a tank positioned above the inlet of the Prominent

Rossi has also indicated that there was a recirculating pump into a pressurized return tank ( (I think - On JONP?) .

That can be found in Rossi's notes concerning the Smith report.
They were published by Mats Lewan and can be found at :

https://animpossibleinvention.…h-supplemental-report.pdf

See points 3 and 7 on the first page of the notes.

Quote from Alan Fletcher

No .. that's 1-10 bar not 0-10. We need 0-2 bar.

The advertisement says 1-10 bar.

However Prominent list the HDV-S-DL as 0-10 Bar in their documentation, so maybe the advertisement states it wrong.

see : https://www.prominent.com/en/P…s-low-pressure-pumps.html

Alan,

Can you use the DL instead of the DK

I found a HDV-S-DL on ebay (uk).

see : https://picclick.co.uk/Promine…-10-Bar-391839957283.html

Quote from Alan Fletcher

In any case, adding a recirculating pump to the input will accomplish something between nothing ... and forcing water through the suction/dosing valves with a little modulation by the diaphragm.

I am not sure.

As earlier stated, I am NOT a pump expert.
But my (maybe bad) reasoning says that the recirculation pump is pumping water through the Prominent at both sides (They are dual stroke, with dual diaphragms ?)
Now if the diaphragm increases the volume in the Prominent pump, it will suck additional water in the pump. This additional water will be forced out during the release stroke because at that time the pressure of the diaphragm will close the inlet check valve despite the pressure at the inlet side from the recirculation pump (Unless the inlet pressure is very high) . At the same time the other diaphragm is in the inlet position and the recirculation pump is forcing water through the check valves at that side.

So I expect that you will get a basic flow from the recirculation pump with an added flow from the Prominent which can be modulated.

If a recirculating pump was used in combination with the metering pump and the re-circulation pump is on the inlet side of the metering pump then it can provide enough pressure to open the inlet and outlet check valves in the metering pump.
This would result in a flow through the metering pump even is it not in operation.
If the metering pump is operating and if there is enough water between the re circulation pump and the the metering pump, the metering pump can in operation suck additional water from the inlet and pump it out.
Thus the flow out will be a combination of the some of flow provided by the re-circulation pump and that of the metering pump and the total flow will likely be higher then that of the metering pump alone.

I am NOT a pump expert, so I would like anybody with a working experience on pumps to tell if the above is a valid scenario.

Quote from THHuxleynew

I'd expect the typical flowmeter rate to vary linearly with pressure, which means +26% from its faceplate rate of 32 l/h or 40.4l/h

The measured flowmeter rate (for most of the time) was 1500l/h or 62.4l/h, Nearly 50% higher.

So I'm going to make a cautious prediction of 40.5 l/h at low back-pressure, with an expected tolerance of no more than 10%. This is still way outside what is possible if the Rossi/Penon figures are good, thus showing that some assumption Penon makes must be wrong, or the data is fabricated.

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If I look closely at the diagrams of several Prominent pumps, then they are sometimes made up of three straight lines, with the slope rising from right to left.

This indicates a somewhat non linear effect.

Also for some Prominent low pressure pumps the curves are running to .5 bar. The diagrams of those pumps clearly show that the at lower pressures the rate is increasing faster then for the higher pressures.

What the actual effects are for the pump type used in Doral is difficult to predict. So I look forward for the results of Alan's test.

I hope Alan will also repeat the test with an added recirculation pump as was said to have been used in Doral because it would show the influence of adding such a pump.

Quote from Eric Walker

LDM, I'm not trying to assert that it was 0.2 bar. I just wanted to have the data point on hand for easy reference, and also so that it could be considered as one of the parameter values being tested.

I personally would not even want to attempt to sort out what the true pressure was on the basis of statements and evidence available to us.

I didn't think that you where trying to assert that the pressure was .2 Bar

The reason for bringing up the .1 bar was the same as yours, being a possible set-point parameter to be tested.

Quote from Eric Walker

Just to collect the source of the "0.2 bar" number for future reference, it was mentioned in a question of Mats Lewan that he put to Rossi. (Presumably the number came from Rossi at some point earlier in the interview.)

Eric,

You are right on the interview with Mats Lewan where indeed .2 Bar is stated

However that contradicts his published notes about the Smith report in which he states that the average water column height was 1 meter, (or .1 Bar)

Quote from Paradigmnoia

I can't see this pump doing more than 45 L/h going flat out 0.2 bar, water practically falling through it.

From the remarks of Rossi in his comments the pressure was .1 Bar.

Given .1 bar my guess is about 55 L/h.

However Rossi stated that there was an additional effect of the re-circulation pump

So if somebody is going to test the Prominent, then don't forget to also add a re-circulation pump, otherwise the test will not be an equivalent one.

Quote from Paradigmnoia

The Lugano twisted calibrated resistance heater wires are very large diameter for heater application and have very low effective resistance (and therefore are capable of huge current flows). The heating circuit could be approximated more closely using single phase and higher resistance wire, but a good understanding of the Lugano device heat distribution is required.

If you for these very low ohmic heater wires calculate the needed wire diameters from the resistivity for heater wires which can handle these temperatures (For example Kantal) and even for copper (low ohmic), then you will see that the number of turns will not fit on the area because the needed wire diameter becomes too large
One possible solution to this problem would be that the "average current" mentioned in the report is not the average RMS current but the average of the peak currents. This has the following consequences :

1. Since the peak current is for small pulses, the RMS current is much lower.

2. As a consequence the calculated Joule heating power in the Lugano report is too high (About a factor 10)

But if the "average current" is indeed the the average peak current, then for the dummy run it results in resistance of each of the three heater wires of about 25 Ohm. For Kantal A1 type wire the diameter will then be small enough to properly fit on the tube and to have enough spacing between the windings.

Quote from Paradigmnoia

So the MFMP thermal comparison seems to have some sort of flaw. Their device concentrates the heat in the main tube, but barely heats the Caps, due to the heater coil design. The Lugano device has fully 1/3 of the entire resistance-heater wire length within the Caps and extending into the Rods at least 4 cm. So it takes more power in total to heat the Lugano Main tube in comparison to the MFMP device.

IMO

If the Lugano device has about 1/3 of the heater wire resistance within the caps (Possibly meaning that the windings continue within the caps) and the MFP device not (Having a straight heater wire within the caps) it would mean that the MFMP device was not a good replication of the Lugano device.

It also explains why the temperatures near the caps decreases much faster and to lower temperatures then the Lugano device. I was already wondering about that,

And it also would indeed mean that the Lugano device needed more power to get to the same temperatures near the center of the device then that of the MFMP replica.

Would be nice if somebody has a finite element thermal simulation program and would model both situations, run some tests and compare the results with the contents of the reports.

Quote from Paradigmnoia

Edit: I think i see what MFMP did now. The 190.6 W is not an estimate for the whole device, it is a relative comparison value for comparable zone of temperature measurements only.

The MFMP wrote in their report : "the Optris temperature measurement was the same as subsequent fueled runs, the input heater power should have been190.6 W, not 479 W (an error of 251%)"

Is your conclusion that the MFMP compared the power of a part of the ECAT (190.6 W) to the value of the total power in the Lugano report (479 W) ?

It would explain the large error of 251 %