pantone mounting Truck Renault B70

[bolt]

Hello

with a bubbler, when it comes out, the percentage of water (in steam) relative to the air used for bubbling (which already contains a little steam in advance, depending on the weather conditions) depends on the T ° where evaporation is done, and is (probably) never complete (unsaturated)
to saturate, it is necessary that the water of the bubbler has good time to balance its saturation vapor pressure with the air in the presence, and this, for a given pressure

but often, at the outlet of the bubbler, the steam pipe is somewhat cooled (depending on whether it is more or less well isolated) and the dry steam produced becomes wet again (can even condense against the wall of the pipe

The vapor containing air at such T ° with a vapor saturation before the phenomenon of "condensation" (liquefaction) is called "saturating mixing ratio" for a given pressure

some benchmarks: under 1 atm:
at 100 ° C this ratio is infinite per kg of pure air and this saturating mixture comprising only water vapor and no air has a mass of 598 g / m3
at 90 ° C: 1395 g / kg of clean air; 424 g / m3 (424 g water in steam per m3, m3 containing this water in vapor + pure air)
((424 g water + 304 g clean air) / m3); 39,47% more water than air
at 87 ° C: approx 1000 g / kg; 380 g water / m3 mixed with 380 g clean air / m3
at 80 ° C: 545 g / kg; 294 g water / m3 mixed with 539,45 g clean air / m3
at 70 ° C: 341 g / kg; 198 g water / m3 mixed with 580,64 g clean air / m3
at 60 ° C: 152 g / kg; 130 g / m3 which gives 855 g of pure air / m3 (1000 / 152 * 130) is 152 / 855 = 17.7% of water with respect to the air by bubbling at 60 ° C at 1 atm (calculation plus developed for 0,9 atm)

all that is a calculation of the weight or volume, put into play
if for example, the bubbler works at 90 ° C and it is assumed that it has time to evaporate until just saturated
and if this kind of vapor arrives at 87 ° C in the reactor: it gives 395 g / kg of "vapor" being probably in liquid form (micro-drops)
m3 (containing 424 g water in steam + 304 g air) becomes? 0,86 m3 (pif) containing 304 g water in steam + 120 g water in micro-drops + 304 g of pure air (at 87 ° C, there is 50 / 50 ratio between vapor and pure air, tjs to 1 atm)

It would therefore be (in this example) only the 120 g of water which, being in micro-drops, could ionize in the reactor, cause "magnetization" and hook the reactor capable then of treating all the water molecules (in micro-drops and dry vapor) and maybe even the accompanying clean air (containing O2 and N2)

It is by looking at this diagram which I do not remember the name of the author.

it started
here et
leaves
(in a little net)

bolt

[phil 14]

good evening everyone
Thank you for your help with my upgrade, I am starting to understand everything you have tried to explain in recent months.
I think I'll start by putting a valve to regulate the steam output at first, then do some testing.
Since the reactor bubbler distance is relatively long (45 cm), it is quite possible that the vapor condenses on the wall of the pipe, so if I understood correctly it is useless to redo an outside air inlet.
a + phil 14