lau:
sublimation
An article from Wikipedia, the free encyclopedia.
In physics, sublimation is the passage of a body from the solid state to the gaseous state, or vice versa, without going through the liquid state. This transformation is therefore done without going through a melting step (from solid to liquid), nor an evaporation step (from liquid to gas).
For example, sulfur, used to fight against certain fungi and insects, sublimates, and it is the vapors that are active.
vala!
To all those who have experienced the water doping
Good evening
it's all relative
you have to give about 450 calories to 1 gr of water to evaporate it
but the reverse is just as true
so if we have "dry" steam coming out of the bubbler, okay, it will become wet again because of the possible drop in temperature of the pipe going to the reactor,
but in reality, there is only part of the "dry" vapor which condenses (on itself and on the walls of the pipe) (and which therefore becomes visible)
but we must not forget that what has just become water again (micro-droplets) has just returned these (450 calories per gr of water) in its path inside this tube
by the absurd if we say: holds the pipe is not isolated so the dry steam becomes all water (micro-drops),
virtually, with this donation of calories (450 cal / gr) in less than 2 we would have 1000 ° C
I think the demonstation is clear enough to understand that, in fact, it naturally balances itself, according to the few calories lost in the pipe
nevertheless, at best it is isolated, at least it transforms from dry steam to wet steam
on the other hand it is very interesting to speak from this point of view of the water carburetor of PITMIX:
to have dry steam at the outlet of this fuel you need:
enough calories available before switching to this fuel
it can be calculated:
you will see that it balances itself
and the best part is that it equilibrates, let's say, around 50 ° C, in fact I don't know about the pantone reactor, but if it equilibrates at -1 ° C, it would freeze up and clog the system
ex if water 100 ° C (in water)
450 cal / gr are required (if gr has its cal count, at 100 ° C, it increases its volume by 1860 times)
where to find these 450 cal
it is the air which passes in the carburetor which is of chore
it takes about 1 calorie (to check) to raise 1 gr of air of 1 degree
if we have air beforehand at 100 ° C, that makes 100 cal / gr compared to zero degrees
with 4.5 gr of air to evaporate 1 gr of water, we would have virtually our account, but this air should be at 0 ° C after having stripped of its calories
if we take 9 gr of air at 100 ° C, it becomes at 50 ° C, after the "skin"
in fact, with respect to zero ° C, we have:
for 1 gr of steamed water at 100 ° C: 100 + 450 = 550 calories
for 9 gr of air at 100 ° C: 900 calories
if we suppose the water in vapor, there remain 1000 - 450 = 550 calories for 10 gr mass molecules,
we would therefore have: 10 gr at 55 ° C of humid air at the outlet of the fuel, (containing 1 gr of water in dry vapor)
(which is possible from the point of view saturating mixture ratio at 1 atm) (at 55 ° C: approx 115 gr water maximum per kg of clean air) (1 / 8,7)
we are 1/10
but since we obviously look a little wet in advance, 1/10 may not be enough
in addition we do not have available before the fuel water and air at 100 ° C.
if we have water and air at 75 ° C, finally at 55 ° C downstream of the fuel: we need 450 / (75-55) = 22,5gr of air pure compared to 1 gr of water (if the air is humid in advance, recalculate in view)
if a gasoline fuel is set for a ratio of 1/15 and there is the same ratio for water, (this is the viscosity which decides) we see that it is not enough to have dry steam at the outlet of the fuel
in fact in this case it would balance with a lowering of 30 ° C
there would be 45 ° C at the fuel outlet
the saturation mixing ratio of 1/15 being at approx 45 ° C, it is limit limit, but very little wet vapor
if the water is at 20 ° C and the air at 20 ° This additionally at 90% RH:
not possible without rod wetting,
but by doing a warmed anti-chamber this can make up for this lack of calories
the problem is to supply just enough heat to this anti-chamber, so as not to wet the rod, and not too much to make the greatest temperature difference between the start and the end of the rod to promote the magnetism
the carburetor must imperatively charge regularly
it is certainly more controllable to control what come fuel
by adjusting the carburetor with a 1/30 ratio we can allow water + air to enter at a slightly lower temperature than 75 ° C
(60 ° C?)
there would be only to accelerate where to slow down this fuel to deliver more or less vapor to the reactor, without worrying about wetting of rod, and to feed the reactor as it should be, according to the temperature which can be maintained there , depending on the engine load
Easier said than done
but hey, if my "logic" can help you
bolt
bob_isat wrote:please
If we follow the reasoning saturated vapor-etc, we must avoid dropping below the temperature of the bubbler, over the entire length of the circuit leaving the reactor, otherwise the water vapor condenses and that's less that goes into the engine ...
is it therefore necessary to put insulation on the pipes? ...
it's all relative
you have to give about 450 calories to 1 gr of water to evaporate it
but the reverse is just as true
so if we have "dry" steam coming out of the bubbler, okay, it will become wet again because of the possible drop in temperature of the pipe going to the reactor,
but in reality, there is only part of the "dry" vapor which condenses (on itself and on the walls of the pipe) (and which therefore becomes visible)
but we must not forget that what has just become water again (micro-droplets) has just returned these (450 calories per gr of water) in its path inside this tube
by the absurd if we say: holds the pipe is not isolated so the dry steam becomes all water (micro-drops),
virtually, with this donation of calories (450 cal / gr) in less than 2 we would have 1000 ° C
I think the demonstation is clear enough to understand that, in fact, it naturally balances itself, according to the few calories lost in the pipe
nevertheless, at best it is isolated, at least it transforms from dry steam to wet steam
on the other hand it is very interesting to speak from this point of view of the water carburetor of PITMIX:
to have dry steam at the outlet of this fuel you need:
enough calories available before switching to this fuel
it can be calculated:
you will see that it balances itself
and the best part is that it equilibrates, let's say, around 50 ° C, in fact I don't know about the pantone reactor, but if it equilibrates at -1 ° C, it would freeze up and clog the system
ex if water 100 ° C (in water)
450 cal / gr are required (if gr has its cal count, at 100 ° C, it increases its volume by 1860 times)
where to find these 450 cal
it is the air which passes in the carburetor which is of chore
it takes about 1 calorie (to check) to raise 1 gr of air of 1 degree
if we have air beforehand at 100 ° C, that makes 100 cal / gr compared to zero degrees
with 4.5 gr of air to evaporate 1 gr of water, we would have virtually our account, but this air should be at 0 ° C after having stripped of its calories
if we take 9 gr of air at 100 ° C, it becomes at 50 ° C, after the "skin"
in fact, with respect to zero ° C, we have:
for 1 gr of steamed water at 100 ° C: 100 + 450 = 550 calories
for 9 gr of air at 100 ° C: 900 calories
if we suppose the water in vapor, there remain 1000 - 450 = 550 calories for 10 gr mass molecules,
we would therefore have: 10 gr at 55 ° C of humid air at the outlet of the fuel, (containing 1 gr of water in dry vapor)
(which is possible from the point of view saturating mixture ratio at 1 atm) (at 55 ° C: approx 115 gr water maximum per kg of clean air) (1 / 8,7)
we are 1/10
but since we obviously look a little wet in advance, 1/10 may not be enough
in addition we do not have available before the fuel water and air at 100 ° C.
if we have water and air at 75 ° C, finally at 55 ° C downstream of the fuel: we need 450 / (75-55) = 22,5gr of air pure compared to 1 gr of water (if the air is humid in advance, recalculate in view)
if a gasoline fuel is set for a ratio of 1/15 and there is the same ratio for water, (this is the viscosity which decides) we see that it is not enough to have dry steam at the outlet of the fuel
in fact in this case it would balance with a lowering of 30 ° C
there would be 45 ° C at the fuel outlet
the saturation mixing ratio of 1/15 being at approx 45 ° C, it is limit limit, but very little wet vapor
if the water is at 20 ° C and the air at 20 ° This additionally at 90% RH:
not possible without rod wetting,
but by doing a warmed anti-chamber this can make up for this lack of calories
the problem is to supply just enough heat to this anti-chamber, so as not to wet the rod, and not too much to make the greatest temperature difference between the start and the end of the rod to promote the magnetism
the carburetor must imperatively charge regularly
it is certainly more controllable to control what come fuel
by adjusting the carburetor with a 1/30 ratio we can allow water + air to enter at a slightly lower temperature than 75 ° C
(60 ° C?)
there would be only to accelerate where to slow down this fuel to deliver more or less vapor to the reactor, without worrying about wetting of rod, and to feed the reactor as it should be, according to the temperature which can be maintained there , depending on the engine load
Easier said than done
but hey, if my "logic" can help you
bolt
0 x
Hello Sam
Snow and ice in spring when the solei is more biting even than it is -5 it evaporates it makes holes like in a sponge only the side in the sun and day by day the planking decreases ...
The water saturation curves I had looked at, I don't know if it's Cristh or JK a guy from forum who posted the link.
The problem is that these values are given for static gases, it must be quite different when it circulates at 300kmh
in a duct, but the principle remains there, we speak of humid air and saturated air, this distorts the perception of still having drops of water in hot air at more than 150c.
we also forget the time factor between a drop which is in very hot air, we have the example when we inject gas oil in an air at 600c the drops take a certain time to evaporate and again there are some which do not does not evaporate, they are found in the engine oil, there is the effect of walls like in a tube.
Andre
Snow and ice in spring when the solei is more biting even than it is -5 it evaporates it makes holes like in a sponge only the side in the sun and day by day the planking decreases ...
The water saturation curves I had looked at, I don't know if it's Cristh or JK a guy from forum who posted the link.
The problem is that these values are given for static gases, it must be quite different when it circulates at 300kmh
in a duct, but the principle remains there, we speak of humid air and saturated air, this distorts the perception of still having drops of water in hot air at more than 150c.
we also forget the time factor between a drop which is in very hot air, we have the example when we inject gas oil in an air at 600c the drops take a certain time to evaporate and again there are some which do not does not evaporate, they are found in the engine oil, there is the effect of walls like in a tube.
Andre
0 x
Bolt wrote:Good evening
the problem is to supply just enough heat to this anti-chamber, so as not to wet the rod, and not too much to make the greater temperature difference between the beginning and the end of the rod to promote magnetism
bolt
Ah good
there is even a lot of luck for the magnetism to be due to the friction of water vapor in the pipeline, as in the thomasi experiment:
http://bobisat.ifrance.com/data/thomasi2.gif
Or if you think that the thermal gradient has an influence, could you explain to me
0 x
Hello Bolt
Without entry in all detailed calorie and moisture calculations
it's exactly as you describe it is happening with
a water carburetor,
several options are to be considered
1- the air and water components are heated before the carburetor
2- the components are heated in the anteroom after the carburetor
Now limit according to the quantity of air, the temperature and its humidity levels before entering the carburetor. We are limited by saturation to evaporate water, and if we put too much water we pick up with liquid on the nose of the stem, rarely on the end of the stem in normal operation.
However when the rod is wet at the inlet it can only be cold and the reactor does not work well in this situation.
Another observation which can be different if a 100mm rod and a 300mm rod are used.
The hot water is more fluid to pass through the carburetor and it forms smaller drops when sprayed (I even try to add a few drops of dish soap)
this is where we see that a bubbler can only produce humid air and nothing else, whatever the relative humidity that the air swallowed by the bubbler. If the design of the size and temperature conduits is to respect the air which arrives on the rod of the reactor should never be saturated to 100%, normally while entering the reactor the short anteroom slightly elevates the temperature so the saturation is pushed back.
For bolt could you calculate an approximate order of magnitude
cubic meters of air passing through the reactor in one hour
a 12,7mm rod in an internal 14mm tube, a 4mm diameter carburetor inlet, or a 100mm level bubbler, this would give us an idea according to the water consumption as a function of the air swallowed if we exceed saturation or if it is far below, considering the air at 100c reactor outlet.
Normally on all the air swallowed by a diesel, what comes from the reactor is a small percentage in the best of multi-jet cases this should not exceed 10%.
Andre
Without entry in all detailed calorie and moisture calculations
it's exactly as you describe it is happening with
a water carburetor,
several options are to be considered
1- the air and water components are heated before the carburetor
2- the components are heated in the anteroom after the carburetor
Now limit according to the quantity of air, the temperature and its humidity levels before entering the carburetor. We are limited by saturation to evaporate water, and if we put too much water we pick up with liquid on the nose of the stem, rarely on the end of the stem in normal operation.
However when the rod is wet at the inlet it can only be cold and the reactor does not work well in this situation.
Another observation which can be different if a 100mm rod and a 300mm rod are used.
The hot water is more fluid to pass through the carburetor and it forms smaller drops when sprayed (I even try to add a few drops of dish soap)
this is where we see that a bubbler can only produce humid air and nothing else, whatever the relative humidity that the air swallowed by the bubbler. If the design of the size and temperature conduits is to respect the air which arrives on the rod of the reactor should never be saturated to 100%, normally while entering the reactor the short anteroom slightly elevates the temperature so the saturation is pushed back.
For bolt could you calculate an approximate order of magnitude
cubic meters of air passing through the reactor in one hour
a 12,7mm rod in an internal 14mm tube, a 4mm diameter carburetor inlet, or a 100mm level bubbler, this would give us an idea according to the water consumption as a function of the air swallowed if we exceed saturation or if it is far below, considering the air at 100c reactor outlet.
Normally on all the air swallowed by a diesel, what comes from the reactor is a small percentage in the best of multi-jet cases this should not exceed 10%.
Andre
0 x
good evening André
I know the water flow rates which pass through the circular conduits not blocked by a rod:
there are charts for this (for common water pipes),
but for air I don't have one, and it's also a lot more complicated,
because there is always a question of pressure drop, and for the air (which is compressible) it implies different densities along the course
and in addition there is a rod
there may be formulas to calculate all this, but i haven't come across it yet
we can make an estimate by measuring the water consumption in a bubbler taking into account the T °, depression, RH% outgoing
for a fuel, the estimation could be done by calculating the air / water masses of the mixing ratio, but for this, we must first see how much this ratio changes to compare with gasoline (1/15)
then you have to know the specific weight of the air used
bolt
I know the water flow rates which pass through the circular conduits not blocked by a rod:
there are charts for this (for common water pipes),
but for air I don't have one, and it's also a lot more complicated,
because there is always a question of pressure drop, and for the air (which is compressible) it implies different densities along the course
and in addition there is a rod
there may be formulas to calculate all this, but i haven't come across it yet
we can make an estimate by measuring the water consumption in a bubbler taking into account the T °, depression, RH% outgoing
for a fuel, the estimation could be done by calculating the air / water masses of the mixing ratio, but for this, we must first see how much this ratio changes to compare with gasoline (1/15)
then you have to know the specific weight of the air used
bolt
0 x
Hello
It would almost be necessary to have a hygrostat which measures the humidity level at the inlet of the reactor and which acts on a flow control valve for a water carburetor, on a bypass for a bubbler or on a computer. injector.
For the calculations I know abacuses which make it possible to calculate the quantity of a fluid in an installation according to the length of the pipes. But know the length of the pipes according to a determined operating regime ...
Do not know
I'll look...
It would almost be necessary to have a hygrostat which measures the humidity level at the inlet of the reactor and which acts on a flow control valve for a water carburetor, on a bypass for a bubbler or on a computer. injector.
For the calculations I know abacuses which make it possible to calculate the quantity of a fluid in an installation according to the length of the pipes. But know the length of the pipes according to a determined operating regime ...
Do not know
I'll look...
0 x
Hello
put a thin hose, to limit the "aerosol" flow of water carburetor is probably the easiest to do
but if we reflect on what is desirable for the "pantone"
we must aim for a good depression in the reactor:
the air being relaxed, for the same amount of gas passed, the molecules (more widely spaced) necessarily go faster, along the stem, to promote "magneto-plasmosis"
and for vaporization in the fuel, at most the pressure drop is done at this location, at best it nebulizes (finer drops)
put a thin hose, to limit the "aerosol" flow of water carburetor is probably the easiest to do
but if we reflect on what is desirable for the "pantone"
we must aim for a good depression in the reactor:
the air being relaxed, for the same amount of gas passed, the molecules (more widely spaced) necessarily go faster, along the stem, to promote "magneto-plasmosis"
and for vaporization in the fuel, at most the pressure drop is done at this location, at best it nebulizes (finer drops)
0 x
A thing
According to Asguard bone tyr you have to suck it up in the reactor and nothing else.
In your opinion, is bubbling used to avoid forcing excessive depression in the bubbler?
The dip tube allows the reactor to suck the vapor that is above the water before sucking the outside air. The small holes are used to pass air which slows down the flow rather than leaving the dip tube open to its initial diameter.
According to Asguard bone tyr you have to suck it up in the reactor and nothing else.
In your opinion, is bubbling used to avoid forcing excessive depression in the bubbler?
The dip tube allows the reactor to suck the vapor that is above the water before sucking the outside air. The small holes are used to pass air which slows down the flow rather than leaving the dip tube open to its initial diameter.
0 x
Hello,
The level of the bubbler determines the pressure in front of the reactor if one neglects the losses in the duct
Small holes this favors small bubbles so + contact with water, + humidity in the air
The same principle for the finesse of a carburetor, the thinner it is
the better it makes contact with air
With a tampon kept moist in a big reducer it works too.
Whatever the method, what we want is (spray) this water.
Pitmix recovers old filters that have in the freon circuit,
They are effective in filtering out large drops! and in addition they are warmer than the aquarium sponge ...
Andre
The level of the bubbler determines the pressure in front of the reactor if one neglects the losses in the duct
Small holes this favors small bubbles so + contact with water, + humidity in the air
The same principle for the finesse of a carburetor, the thinner it is
the better it makes contact with air
With a tampon kept moist in a big reducer it works too.
Whatever the method, what we want is (spray) this water.
Pitmix recovers old filters that have in the freon circuit,
They are effective in filtering out large drops! and in addition they are warmer than the aquarium sponge ...
Andre
0 x
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