Watts do not matter. Its all about wire temp. Read on…

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jasl90

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Yup, however if you were to look at a temperature curve of a coil, it most likely would show a plateau around the boiling point of VG or PG. This is a point mad scientist had in another thread and it is likely true. Technically if operating correctly our coils should fall within a similar range of temperatures when vaping.

I would if i could but I can't so I won't... Until someone pops in here with access to a $100K worth of FLIR thermal imaging equipment, we're only guessing. Logic dictates that the coil itself needs to be substantially hotter than the boiling point of the liquid in order to vaporize it. Just like boiling a pan of water... The burner has to be well over 220 degrees to get the water to boil. The water may evaporate away from the heat source at 220 degrees but that doesn't bring the burner temp (or the bottom of the pan for that matter) all the way down to 220. It's far cooler than it would be without the water but it still well above 220.
 

dr g

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I would if i could but I can't so I won't... Until someone pops in here with access to a $100K worth of FLIR thermal imaging equipment, we're only guessing. Logic dictates that the coil itself needs to be substantially hotter than the boiling point of the liquid in order to vaporize it. Just like boiling a pan of water... The burner has to be well over 220 degrees to get the water to boil. The water may evaporate away from the heat source at 220 degrees but that doesn't bring the burner temp (or the bottom of the pan for that matter) all the way down to 220. It's far cooler than it would be without the water but it still well above 220.

Right the point is not so much the actual temperature but the fact that it will plateau there. Change the previous post to say "plateau around the temperature it needs to be to vaporize VG or PG."

The point here is that it's a bit of a misnomer to say it's about wire temp because all wire temp likely falls within a similar range during proper vaping.
 
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dr g

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If that's the case, please explain how increasing the voltage on a VV devise is able to produce more vapor and warmer vapor than it does at a lower voltage...

I think you're confusing the temperature of the liquid with the temperature of the coil.

The temperature of the liquid and the coil are thermally linked to a certain degree. The ability of the liquid to transfer heat away from the coil is why putting more power through the coil increases vapor.

A larger mass of vapor holds more heat energy (both in its own mass and probably some self-insulation) which leads to a warmer delivered hit. I also think that the surrounding material absorbs some heat from the coil as well, which helps warm the vapor.
 
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jasl90

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The temperature of the liquid and the coil are thermally linked to a certain degree. The ability of the liquid to transfer heat away from the coil is why putting more power through the coil increases vapor.

A larger mass of vapor holds more heat energy (both in its own mass and probably some self-insulation) which leads to a warmer delivered hit. I also think that the surrounding material absorbs some heat from the coil as well, which helps warm the vapor.

Ok... So if what you're saying is true then the max temp of a coil is the max temp of a coil... Regardless of the power supply. If that is the case, I propose an experiment...

Use your 12 volt car battery as a power supply. IF you do that AND you come back and say that you didn't notice the coil getting hotter than it did with your 18mm lithium battery, I will concede.

Deal?

Edit... Video or it never happened. ;)
 
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dr g

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Ok... So if what you're saying is true then the max temp of a coil is the max temp of a coil... Regardless of the power supply. If that is the case, I propose an experiment...

Use your 12 volt car battery as a power supply. IF you do that AND you come back and say that you didn't notice the coil getting hotter than it did with your 18mm lithium battery, I will concede.

Deal?

Edit... Video or it never happened. ;)

You can obviously overpower a coil, a plateau is just a flat spot.
 

dr g

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when the juice reaches vaporization temp., it is no longer in contact with the coil. In a very crude sense, one could say it is no longer heated by the coil (not to any degree of consequence). One might suggest that effective coil temp. is actually a very wide range.

Don't think this is true, the vaporization of the juice definitely has a regulating effect on the temperature of the coil. Otherwise the temperature would trend toward glowing and we would visually see evidence. To the contrary, we see that when the wick is sufficiently wet, the coil will not glow.
 

pdib

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The heat energy in the coil is transferred directly to formation of vapor. As long as you have the juice and the real estate, you can just keep shuffling an incredibly large amount of energy straight to gaseous state. As much juice as you can put there, is as much energy as you can keep adding to the equation. And, yeah, the coil doesn't glow as long as you're in the effective range of "supply and demand" (enough juice being heated)
 

dr g

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I mean to say that, without some crazy extremes, you really can't "overheat" the juice, cause it flies away.

So, as soon as you acheive adequate temp for your preferred heat up time, it becomes a question of juice supply and a ten second window of necessity to fill.

Ok all true but I don't think that contradicts anything I've said. The juice flying away is the phase change and that appears to regulate the coil temperature to some degree.

The heat energy in the coil is transferred directly to formation of vapor. As long as you have the juice and the real estate, you can just keep shuffling an incredibly large amount of energy straight to gaseous state. As much juice as you can put there, is as much energy as you can keep adding to the equation. And, yeah, the coil doesn't glow as long as you're in the effective range of "supply and demand" (enough juice being heated)

Right and that's the range of coil temperature that all atomizers need to be at regardless of coil size, resistance or current being put through them.
 
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jasl90

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That's what overpowering is, when you exceed the capacity of the juice to replenish in the area around the coil, its temperature will basically run away and you get a burnt hit.

But if your contention is true, that would be a nonissue... The liquid in contact with the coil would regulate the temperature meaning that it wouldn't have to do any more "keeping up" than it would normally.

The only way to make the juice leave the coil faster is with more heat. Since, according to your hypnosis, no additional heat is being generated, that shouldn't be a problem...

The time has come to put up or shut up...
 

dr g

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But if your contention is true, that would be a nonissue... The liquid in contact with the coil would regulate the temperature meaning that it wouldn't have to do any more "keeping up" than it would normally.

The only way to make the juice leave the coil faster is with more heat. Since, according to your hypnosis, no additional heat is being generated, that shouldn't be a problem...

The time has come to put up or shut up...

Heat energy is not always manifest as a change in temperature. I didn't say more heat energy was not generated, I said the temperature would not necessarily change.
 
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