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Brian Ambrose
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Do you both agree (or not) that this completely closed system with a constant supply of optimal conditions (from my point of view) will reach an equilibirum position |
Agreed
and that equilibirum position will tend to be where the exchange of evaporation and condensation will be at a minimum |
Agreed
and that minimum position will tend to be where the air is saturated (whatever that means) rather than not. |
Just trying to be helpful but... no. Your equilibrium or minimum position is not proportional to 'saturation' - you can have equilibrium (minimum position) at any level of 'saturation'. For example, if it's cold enough equilibrium will occur with no humidity at all (probably).
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Mick Harper
Site Admin
In: London
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For example, if it's cold enough equilibrium will occur with no humidity at all (probably). |
You're not really trying to be helpful, Brian. I am specifically asking you not to put in obstructive devices at this stage. With the first box I merely need you to acknowledge that in helpful conditions saturation (or high humidity or lots-of-water-vapour or whatever the hell you want to call) it will be reached.
As you know, you have the second box for entering caveats so you won't be selling the orthodox pass.
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Brian Ambrose
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This is the same as Box One but this time you can introduce any conditions you feel best mimic the real world. |
To mimic the real world, make the source of heating local, variable, and mobile. This results in hot spots and cold spots, convection currents in the water and the box's atmosphere, varying pressures, and various local conditions of non-equilibrium.
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Mick Harper
Site Admin
In: London
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Sigh. I don't want the inputs, I want your belief as to the output once you have put in your chosen inputs.
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Brian Ambrose
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I merely need you to acknowledge that in helpful conditions saturation (or high humidity or lots-of-water-vapour or whatever the hell you want to call) it will be reached. |
Acknowledged.
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Brian Ambrose
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I don't want the inputs, I want your belief as to the output once you have put in your chosen inputs. |
The output is non-equilibrium.
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Mick Harper
Site Admin
In: London
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Yes, Bri, I know that (or at any rate I know that you believe that). But please nevertheless describe the situation. Remember these are your chosen inputs.
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Brian Ambrose
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Mick Harper wrote: | Yes, Bri, I know that (or at any rate I know that you believe that). But please nevertheless describe the situation. Remember these are your chosen inputs. |
I thought I had?
"hot spots and cold spots, convection currents in the water and the box's atmosphere, varying pressures, and various local conditions of non-equilibrium."
Did you want more along the same lines or am I barking up the wrong tree?
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Mick Harper
Site Admin
In: London
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Goddamnit, Brian, why do you keep describing your inputs. I want to know the situation when you open the box. Is there the same amount of liquid water as when you started the experiment? Does the air have more or less water vapour (overall, as an average, as a cline, as a boom-and-bust regime etc etc etc). Is it at a stable maximum, minimum? What, if anything, has changed in the extended time period while your inputs were happening.
Just say what you expect to find when you 'open the box'.
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Brian Ambrose
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Is there the same amount of liquid water as when you started the experiment?
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Yes. Although I heated the box locally I arranged it so that the box overall radiated away about the same amount overall (to best mimic the real world). The net energy input was therefore zero.
Does the air have more or less water vapour (overall, as an average, as a cline, as a boom-and-bust regime etc etc etc). Is it at a stable maximum, minimum? |
Overall (ie averaging the local differences) there is the same amount of water vapour.
What, if anything, has changed in the extended time period while your inputs were happening.
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The system at a molecular level has become far more chaotic but the total water/vapour ratio has remained about the same.
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Mick Harper
Site Admin
In: London
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Phew, at last. Now we only need Chad's verdict and we can proceed, as they say in Shakespeare, to the third box.
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Chad
In: Ramsbottom
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Agreed.
The second box has a rather heterogeneous atmosphere, but the net ratio of water to vapour is the same.
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Mick Harper
Site Admin
In: London
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Goodee! Now with the third box, which I remind you again is 100 metres by 100 metres by 100 metres, you have to feed in water until it reaches almost to the ceiling, in fact to within one metre of the top. You will then apply all those things you applied with Box 2 and await equilibrium (or not as the case may be, but a goodly length of time).
Report on the state of the slither of air when the box is finally opened.
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Brian Ambrose
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Hmmm, tricky.
Since the surface area of the water is the same, but the heat input is distributed over a much larger volume, the surface is at a lower temperature and therefore less vapour is produced. So we could say there'd be less humidity.
On the other hand, since the air volume is so much smaller, even if there is less vapour produced due to a lower surface temperature, we'd expect a higher humidity.
Then again, it's possible that the two factors cancel out and the humidity remains the same, but that would be a boring result. I'm going to guess that the humidity will be higher, overall, in this last scenario because my heat input is coming from above the surface of the water, not from below.
(I have assumed that the air pressure in the remaining gap is the same in all scenarios)
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Chad
In: Ramsbottom
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I'm assuming similar net temperature and pressures in box 3 as existed in box 2 (given similar energy input and losses... and a billon years of stabilization).
The vastly reduced depth of the air layer would inhibit thermal conduction currents in the gaseous part of the system... but because all the air is in close proximity to the water surface I think I would expect a higher humidity, due to sea spray.
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