It is true that talking about depression is a bit abusive.
There would be a vacuum if the reactor were blocked at its entrance.
By measuring the pressure in the reactor it should be negative (relative pressure patmo = 0bar)
As we do an air intake at the entrance of the reactor the depression in the reactor is reduced.
The greater the air intake, the lower the depression.
By manufacturing a GV, which feeds the reactor and by installing a Tee with an air intake between the two, we do not cancel the vacuum in the reactor while avoiding to draw water from the GV.
This also applies to the bubbler and its air intake.
As far as I am concerned, I cannot separate the high temperature in the reactor and the suction - "depression" - strong.
From what I have observed, the behavior of the reactor without vacuum is very different.
I would say that the reactor does not hang.
Although I haven't really understood what "hang" means yet. On several occasions I managed to "hang" the reactor in terms of temperature but not necessarily in terms of modifying the behavior of the car.
For me, hanging the reactor is only valid if the two phenomena are linked.
An ideal venturi
hi Pitmix
A reactor hangs when it starts to heat with a large internal fluid passage.
If it was a simple exchanger, the more air and water fluid is passed, this should lose temperature at the outlet,
If less fluid is passed, it should be hotter at the outlet of the reactor.
It is the reverse that we see.
a very hot reactor if you reduce either the vacuum or the quantity of air + water at the inlet of the reactor, its outlet is colder.
(be careful to measure the temperature of the fluid leaving the reactor and not on the pipe)
If you completely block the air filter and leave the engine at idle or slightly accelerated in order to suck only through the reactor, you will notice that the reactor outlet rises in temperature, despite a rather warm exhaust.
I do this test after having heated the engine quite steep, in order to take advantage of the thermal inertia, Zac does something similar to stretch the reactor .. (I would say even more draconian full gas)
If someone has an explanation for this: why when we put more fluid in the reactor it comes out hotter?
The limit would be to get everything downstream from the engine into the reactor and pump the entire exhaust temperature.
If the humid air entering the reactor is saturated (too much water) the reactor outlet does not heat up and the reactor is drowned.
it is better to have an excess of air, in case of doubt.
Once we have found the (good) air / water ratio
we no longer touch these settings, the way to control the reactor is to control the fluid that passes through the reactor depending on the power required from the engine, which the venturi does very well when we have found its good dimensions,
Andre
A reactor hangs when it starts to heat with a large internal fluid passage.
If it was a simple exchanger, the more air and water fluid is passed, this should lose temperature at the outlet,
If less fluid is passed, it should be hotter at the outlet of the reactor.
It is the reverse that we see.
a very hot reactor if you reduce either the vacuum or the quantity of air + water at the inlet of the reactor, its outlet is colder.
(be careful to measure the temperature of the fluid leaving the reactor and not on the pipe)
If you completely block the air filter and leave the engine at idle or slightly accelerated in order to suck only through the reactor, you will notice that the reactor outlet rises in temperature, despite a rather warm exhaust.
I do this test after having heated the engine quite steep, in order to take advantage of the thermal inertia, Zac does something similar to stretch the reactor .. (I would say even more draconian full gas)
If someone has an explanation for this: why when we put more fluid in the reactor it comes out hotter?
The limit would be to get everything downstream from the engine into the reactor and pump the entire exhaust temperature.
If the humid air entering the reactor is saturated (too much water) the reactor outlet does not heat up and the reactor is drowned.
it is better to have an excess of air, in case of doubt.
Once we have found the (good) air / water ratio
we no longer touch these settings, the way to control the reactor is to control the fluid that passes through the reactor depending on the power required from the engine, which the venturi does very well when we have found its good dimensions,
Andre
0 x
Good ok !!!
I hung the reactor well from the start of my experiments.
There is no doubt, we are talking about the same thing.
It is true that one thing is really amazing is the efficiency of heat exchange.
The reactor outlet temperature is almost identical to that of the exhaust.
I have never made a comparison between the two but I am sure there is not much difference.
On the R5 with a very high pressure (4mCLe) I managed to obtain 300 ° C at the reactor outlet.
I hung the reactor well from the start of my experiments.
There is no doubt, we are talking about the same thing.
It is true that one thing is really amazing is the efficiency of heat exchange.
The reactor outlet temperature is almost identical to that of the exhaust.
I have never made a comparison between the two but I am sure there is not much difference.
On the R5 with a very high pressure (4mCLe) I managed to obtain 300 ° C at the reactor outlet.
0 x
André wrote,
If someone has an explanation for this: why when we put more fluid in the reactor it comes out hotter?
the explanation is the transfer of entropy; which explains why the rod can be hotter than the gases, and that it cools on the other side by absorbing the heat of the exhaust
http://fr.wikipedia.org/wiki/Effet_Thomson
When the rod is hot by the GE, and cooled at the entry by the fluid, there is a temperature difference on the rod.
There is therefore a heat exchange on the rod by conduction with the GE, AND a temperature gradient because of the fluid.
This generates a PDD in the stem.
If the fluid is then sufficiently conductive (ionized), it short-circuits this ddp, and a current arises in the rod and closes in the air gap on the fluid.
It is at this moment that the reactor hangs;
The current in the rod, generates an entropic transfer (due to the displacement of the electrons) against the direction of the thermal flow. This transfer removes calories from the inlet of the rod which cools and then pumps calories to the GE, and heats on the other side, adding its calories to those of the GE and therefore becomes hotter than the GE.
But beware: there is a limit.
It is the fluid which evacuates the calories at the exit of the rod. If this outlet becomes too hot, the fluid can no longer evacuate them. Then the delta T ° decreases and the current no longer flows. The reactor deactivates.
When the current flows, a magnetic field is created around this current. The ions are then deflected in their linear course and their trajectory is curved in spirals around the rod.
That's what hooks the reactor. But all this is done at the same time as soon as the current flows.
The ions are then amplified; ionization essentially taking place in this magnetic field. It remains to find the impact they have on the engine .......
And to start this damn reactor, which I still can't do.
Jeannot
If someone has an explanation for this: why when we put more fluid in the reactor it comes out hotter?
the explanation is the transfer of entropy; which explains why the rod can be hotter than the gases, and that it cools on the other side by absorbing the heat of the exhaust
http://fr.wikipedia.org/wiki/Effet_Thomson
When the rod is hot by the GE, and cooled at the entry by the fluid, there is a temperature difference on the rod.
There is therefore a heat exchange on the rod by conduction with the GE, AND a temperature gradient because of the fluid.
This generates a PDD in the stem.
If the fluid is then sufficiently conductive (ionized), it short-circuits this ddp, and a current arises in the rod and closes in the air gap on the fluid.
It is at this moment that the reactor hangs;
The current in the rod, generates an entropic transfer (due to the displacement of the electrons) against the direction of the thermal flow. This transfer removes calories from the inlet of the rod which cools and then pumps calories to the GE, and heats on the other side, adding its calories to those of the GE and therefore becomes hotter than the GE.
But beware: there is a limit.
It is the fluid which evacuates the calories at the exit of the rod. If this outlet becomes too hot, the fluid can no longer evacuate them. Then the delta T ° decreases and the current no longer flows. The reactor deactivates.
When the current flows, a magnetic field is created around this current. The ions are then deflected in their linear course and their trajectory is curved in spirals around the rod.
That's what hooks the reactor. But all this is done at the same time as soon as the current flows.
The ions are then amplified; ionization essentially taking place in this magnetic field. It remains to find the impact they have on the engine .......
And to start this damn reactor, which I still can't do.
Jeannot
0 x
Hi Jeannot,
If your rod occupies the same volume in the reactor as on your diagram it should not work.
See André's explanations on the role of the chambers provided upstream and downstream of the rod (points 10 and 11)
Awesome this wiki. No need to browse 50 subject pages to find "the info that kills" ...
I see that your theories are getting closer to mine: +1 for the spiral path of the ions due to self-induction.
By cons I do not understand how the rod can be the seat of a ddp under the sole effect of heat. To create a thermocouple you need two separate metals.
I prefer the appearance of eddy current due to the variation of the flow, over the opening / closing of the valves.
If your rod occupies the same volume in the reactor as on your diagram it should not work.
See André's explanations on the role of the chambers provided upstream and downstream of the rod (points 10 and 11)
Awesome this wiki. No need to browse 50 subject pages to find "the info that kills" ...
I see that your theories are getting closer to mine: +1 for the spiral path of the ions due to self-induction.
By cons I do not understand how the rod can be the seat of a ddp under the sole effect of heat. To create a thermocouple you need two separate metals.
I prefer the appearance of eddy current due to the variation of the flow, over the opening / closing of the valves.
0 x
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Christophe
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Crispus wrote:By cons I do not understand how the rod can be the seat of a ddp under the sole effect of heat. To create a thermocouple you need two separate metals.
https://www.econologie.com/ionisation-de ... -3324.html ou https://www.econologie.com/wiki-moteur-p ... ur_d%27eau
Crispus wrote:I prefer the appearance of eddy current due to the variation of the flow, over the opening / closing of the valves.
If not for the wiki, your participation is welcome ...
0 x
Do a image search or an text search - Netiquette of forum
Regarding ionization, I am convinced, it is also this theory of bob_isat that put me on the track of the electromagnetic field.
But Jeannot's explanation seems to mix causes and effects:
jeannot wrote:When the rod is hot by the GE, and cooled at the entry by the fluid, there is a temperature difference on the rod.
There is therefore a heat exchange on the rod by conduction with the GE, AND a temperature gradient because of the fluid.
This generates a PDD in the stem.
I do not think that the only disorderly thermal agitation can generate a ddp in a homogeneous metal, unlike a thermocouple.
For me the rod is probably the seat of a (weak) tension, but more due to the circulation of ions around the rod. they behave like a coil which generates a magnetic field in the rod.
Fluctuations in this field (due inter alia to variations in flow in the inlet - hence my mention of the valves), cause the appearance of eddy currents in the rod (a constant field does not generate current).
We then find ourselves for the rod with the current (Foucault) + temperature gradient combination, conducive to the Thomson effect mentioned by jeannot.
0 x
an explanation of the Thompson effect:
https://www.econologie.com/forums/post813.html#813
if it is really "several thousand amps" it may well be that this produces a certain magnetic field
is this current possibly produced and circulating internally of the rod, which must necessarily come and go (in the center and on the periphery) does it not cancel itself (for the magnetic effect)
bolt
https://www.econologie.com/forums/post813.html#813
One end of a copper rod is heated and the other is cooled. If the hot side is heated high enough, it will thermally increase the kinetic energy of the external orbit electrons to a point where their kinetic energy (1/2 mv2) will be greater than the working function and allow them to discharge into the 'space. Due to the conductivity of copper the electrons, instead of absorbing into the air, will shift in huge quantity towards the side cools in straight lines, after the speed of heat propagation. By the reaction, the electrons exciting for the cool side will travel in the opposite direction to the speed of a particle (1/2 mv2) towards the hot side encircling the copper rod by gyroscopic phenomenon and after the Fleming rule. The product is very low voltage (few millivolts) as a result of electrons traveling in a circular motion. But, as in any electrical circuit, the EMF in the copper rod is governed by Ohm's law (E / R = I) and will be in the range of several thousand amps.
if it is really "several thousand amps" it may well be that this produces a certain magnetic field
is this current possibly produced and circulating internally of the rod, which must necessarily come and go (in the center and on the periphery) does it not cancel itself (for the magnetic effect)
bolt
0 x
Andre wrote:A reactor hangs when it starts to heat with a large internal fluid passage.
If it was a simple exchanger, the more air and water fluid is passed, this should lose temperature at the outlet,
If less fluid is passed, it should be hotter at the outlet of the reactor.
It is the reverse that we see.
a very hot reactor if you reduce either the vacuum or the quantity of air + water at the inlet of the reactor, its outlet is colder.
If you completely block the air filter and leave the engine at idle or slightly accelerated in order to suck only through the reactor, you will notice that the reactor outlet rises in temperature, despite a rather warm exhaust.
If someone has an explanation for this: why when we put more fluid in the reactor it comes out hotter?
Hello André
I tried an explanation resulting only from "fluid mechanics" here:
https://www.econologie.com/forums/post36980.html#36980
but it is possible that something else is happening causing the same effects when the reactor "hangs up"
bolt
0 x
Hi Crispus
thank you for André's diagram; I will make changes to take into account the anteroom it recommends.
For ddp, it is not generated by entropy but by dT ° c. Then when the ionization around the rod is correct it generates entropy which cools the rod at the entrance and absorbs the calories of the GE
If we take the Thomson effect formula, we find
dQ / dx = I. dT. r / dx
Considering an average car that runs at 100 whose engine develops 30 kW and absorbs around 100 kW
of which there are 40 in the GE.
Based on your measurements, which say: when the reactor hangs, the temperature decreases by half (dixit André)
which means that the reactor pumps around 20 kw, a good part of which by entropy (say 18 kw)
taking for "r" 8.4 (given in a table) and dT ° = 200 °
we would have for I = 18.000 / 200. 8.4 or about 10A
It doesn't seem impossible to me; these 10 A must also pass through the air gap, in a conductor of equivalent section of 50 mm2 (for rod of 14 and tube of 16). Especially since this conductor brings his own electrical charge.
The reactor then behaves like a squirrel cage wheel, with the rod as an axis. On the outside of the rod, in the air gap like dozens of current bars, // at the rod. Around these bars the magnetic fields form circles which add up and form only one field around the rod, which turns elsewhere in the same direction as that which forms in the rod due to the current in the rod.
What do you think !!!!! (a drawing would be better, but I don't know how to do it ..)
@++
Jeannot
thank you for André's diagram; I will make changes to take into account the anteroom it recommends.
For ddp, it is not generated by entropy but by dT ° c. Then when the ionization around the rod is correct it generates entropy which cools the rod at the entrance and absorbs the calories of the GE
If we take the Thomson effect formula, we find
dQ / dx = I. dT. r / dx
Considering an average car that runs at 100 whose engine develops 30 kW and absorbs around 100 kW
of which there are 40 in the GE.
Based on your measurements, which say: when the reactor hangs, the temperature decreases by half (dixit André)
which means that the reactor pumps around 20 kw, a good part of which by entropy (say 18 kw)
taking for "r" 8.4 (given in a table) and dT ° = 200 °
we would have for I = 18.000 / 200. 8.4 or about 10A
It doesn't seem impossible to me; these 10 A must also pass through the air gap, in a conductor of equivalent section of 50 mm2 (for rod of 14 and tube of 16). Especially since this conductor brings his own electrical charge.
The reactor then behaves like a squirrel cage wheel, with the rod as an axis. On the outside of the rod, in the air gap like dozens of current bars, // at the rod. Around these bars the magnetic fields form circles which add up and form only one field around the rod, which turns elsewhere in the same direction as that which forms in the rod due to the current in the rod.
What do you think !!!!! (a drawing would be better, but I don't know how to do it ..)
@++
Jeannot
0 x
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