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Originally Posted by C4 Diesel
It would be compressed, but think of the shape of the water bottle. The height is significantly greater than the diameter, even at 75% full. Therefore the radial temperature profile is much more significant than the vertical one because most of the heat transfer is going to occur through the sides. That's what I meant... Not that it doesn't change the profile, just that the change is relatively insignificant.
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C4 - Ok, now I understand your position more clearly. Without having aspect ratios, it's tough for me to say exactly what the radial vs. axial resistance will be, etc. You are probably correct, unless we're talking about one of those little weirdo bottles that I've seen. Let's assume we're talking standard 20 oz. bottles, in which case I would agree. I also think that the radial distribution will be, more or less, the same in the full bottle versus the 1/2 empty / 1/4 empty bottle, so I'm not sure how much of a factor that makes. I'm taking it as a separable solution (big assumption, but common in practice I believe).
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So much for convection....Wait a minute... I believe we're all forgetting something here... Water is a LIQUID!!! Therefore it flows within itself (because of simple molecular motion) which to a great degree evens out whatever temperature gradient may have occured if the mass was solid.
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This is an important assumption as well...I believe that for significant flow to occur via natural convection, rather large density changes are required. I agree that this affect occurs, no doubt. The fluid is also tightly confined by the bottle, if it's a 20 oz., which also has affects on the flow. Natural convection, as I understand it, is much more pronounced in homogeneous gases than in homogeneous liquids.
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This being the case the rate of the temperature (energy) change is based only on the conductivity, thermal mass and surface area / volume ratio.
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This is the statment that I am not so sure about, at least in the way you mean it. I still believe that the temperature difference between, say 25 deg. C, and 0 deg. C is going to be a larger factor, at least at first. However, I will agree with you that the center of the liquid will be cooling due to conduction affects, and some convection affects, and this will tend to decrease the temperature gradient. However, any type of diffusive/conductive transport will be a function of the conductivity, the thermal mass (which is essentially the specific heat, correct?) and SA/V. I just don't think that the water is isothermal.
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Either way, I would expect the reduction in thermal mass to play a greater role than the increase in SA/vol.
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I'm not so concerned about the reduction in SA/V ratio, as the increase in heat flux on the smaller volume. I will wholeheartedly agree that there will be affects related simply to the smaller volume of the cylinder that is 1/4 full...as I said before, I just don't think that the cylinder is isothermal, and that the rate of cooling / phase change is strongly affected, although not completely dominated, by the rate of the heat flux within the liquid. Note that if the water is considered to be isothermal, then the heat flux within the cylinder is more or less zero, and this would imply that the entire bottle would freeze more or less instantly. As I said before, I've never actually performed the experiment in a controlled fashion, but I believe that the water tends to freeze first at the edges, and the phase boundary moves into the center of the volume. Is this not correct?
Either way, it's been fun discussing (I would say arguing, but that term has certain negative connotations that I would prefer to avoid) the subject.
