Orbiter-Forum Thermal radiation and water freezing
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 08-11-2019, 05:35 PM #2 Linguofreak Orbinaut You're missing the heat of fusion/solidification of water. To freeze, water at 0° C must emit the same amount of energy as it would emit in cooling from 80° C to zero (the amount of energy associated with the water steam transition is even greater). And, of course, the time to emit that energy falling from 80 to 0 will be less than the time to emit that energy at 0 the whole time (because you get more blackbody radiation at higher temperatures). You're also assuming that the water is radiating at 20° C the entire time, and the radiative power at 20° C is about 22% greater than at 0°. Furthermore, freezing causes water to expand, which, depending on the air space in the bottle will increase the pressure by some amount (possibly quite extreme if there is little air space), and pressure causes the freezing point of water to drop, which will increase the amount of cooling required by a little bit (the heat of fusion may also be affected by pressure, which may well be a greater effect, I'm not sure). The water will have to do work on the air in the container and the walls of the container to expand, and I'm not exactly sure exactly how this will effect the cooling time (very likely the freezing point drop and any change in heat of fusion with pressure already take this into account).
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 08-11-2019, 05:46 PM #3 RisingFury OBSP developer Does the astronaut have to take off the helmet to drink from the bottle before or after unscrewing the cap?
 08-11-2019, 08:43 PM #4 Furet Orbinaut Quote: Originally Posted by RisingFury  Does the astronaut have to take off the helmet to drink from the bottle before or after unscrewing the cap? Surely not. I would have to calculate the freezing time of the astronaut. ---------- Post added at 20:43 ---------- Previous post was at 20:01 ---------- Quote: Originally Posted by Linguofreak  You're missing the heat of fusion/solidification of water. To freeze, water at 0° C must emit the same amount of energy as it would emit in cooling from 80° C to zero (the amount of energy associated with the water steam transition is even greater) Oops! I didn't know that. Your remark led me to find these articles: https://en.wikipedia.org/wiki/Freezing https://en.wikipedia.org/wiki/Enthalpy_of_fusion Quote: Originally Posted by Linguofreak  You're also assuming that the water is radiating at 20° C the entire time, and the radiative power at 20° C is about 22% greater than at 0°. I didn't realize that the difference is so significant. I'm missing the math yet to calculate the extra time, but I'll try again. Quote: Originally Posted by Linguofreak   Furthermore, freezing causes water to expand, which, depending on the air space in the bottle will increase the pressure by some amount (possibly quite extreme if there is little air space), and pressure causes the freezing point of water to drop, which will increase the amount of cooling required by a little bit (the heat of fusion may also be affected by pressure, which may well be a greater effect, I'm not sure). The water will have to do work on the air in the container and the walls of the container to expand, and I'm not exactly sure exactly how this will effect the cooling time (very likely the freezing point drop and any change in heat of fusion with pressure already take this into account). Yes, I do agree. But I really have no idea on how to compute this too. Let's assume a standard pressure in the container during the whole freezing process. Thanks for your comments, Linguofreak.
 08-11-2019, 08:58 PM #5 Linguofreak Orbinaut Quote: Originally Posted by Furet   I didn't realize that the difference is so significant. I'm missing the math yet to calculate the extra time, but I'll try again. Never underestimate the difference an x^4 term can make. I'm not very confident with calculus myself, so I'm afraid I can't help you with exactly how much longer it would take. EDIT: TBH, though, the enthalpy of fusion will almost certainly be the dominant factor in the extra time. The difference between radiative power at 20 C and 0 C will make a contribution to the extra time over your estimate that's guaranteed to be less than 22%, while the freezing process after reaching 0 C will add 488% (400% because the heat of fusion is equivalent to an 80 degree temperature change, multiplied by 1.22 because of the fact that freezing happens at 0 C rather than 20 C, and the energy is being carried off radiatively). Last edited by Linguofreak; 08-12-2019 at 12:12 AM.
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 08-13-2019, 07:56 AM #6 Linguofreak Orbinaut Gaak! I made a stupid error with the math: I'd previously had my calculator in hexidecimal input / octal output mode for some computer related stuff, and I remembered to switch the output mode to decimal, but not the input mode, so I calculated the difference between radiative power at 659 K and 627 K (0x293 and 0x273) rather than between 293 K and 273 K The difference in radiative power between 20 C and 0 C is more like 33 percent!
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 08-13-2019, 09:52 AM #7 Urwumpe Not funny anymore Also, doesn't blackbody radiation behave differently when the object is not a perfectly black body? AFAIR, there are some weird effects if the true spectrum of the body is not like in the theory.
 08-13-2019, 10:51 AM #8 Linguofreak Orbinaut Quote: Originally Posted by Urwumpe  Also, doesn't blackbody radiation behave differently when the object is not a perfectly black body? AFAIR, there are some weird effects if the true spectrum of the body is not like in the theory. I believe there are some differences, but I'm not sure that water deviates enough from a blackbody at the wavelengths involved at ~273 K.
 08-14-2019, 02:21 AM #9 Thunder Chicken Fine Threads since 2008 You have most of the physics correct, but the math is a bit more involved. The emission E from the bottle to space (assuming 0K surroundings) will be equal to the rate of change of thermal energy of the water. While it is liquid the equation that needs to be solved is: e A k T^4 = m c dT/dt Where e - emissivity A - surface area k - Stefann-Boltzmann constant m - water mass c - liquid water specific heat T - water temperature t - time Emissive power changes with time as T changes with time. Integrating this out I get an elapsed time of 4285 s to cool from 293K to 273K. Once the water hits the freezing point (T = 273K), the temperature of the water remains constant during the phase change, and the emissive power remains constant. e A k T^4 = constant = m dh/t_f where dh is the enthalpy of fusion of water ice. t_f is the elapsed time of the freezing process and is easy to calculate as everything in this equation is constant: t_f = m dh / e A k T^4 With your numbers: t_f = (1 kg)x(333.55 kJ/kg) / (0.95 x 0.05675 mÂ² x 5.67e-08 W/m^2 K^4 x (273K^4)) t_f = 19,645 s or about 5.5 hrs (if my math is correct). This is just the time required to accomplish the phase change to solid ice at 273K and needs to be added to the time for the liquid cooling process, 4285 s. So the total time would be 23,930 s or about 6 hrs and 40 minutes.* TL;DR - radiation heat transfer is *very* inefficient, ice can absorb a *lot* of energy while melting. *Entirely possible there is a math error, but I think it is correct. Last edited by Thunder Chicken; 08-14-2019 at 03:16 AM.
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 08-15-2019, 10:50 AM #10 Furet Orbinaut Quote: Originally Posted by Thunder Chicken  Entirely possible there is a math error, but I think it is correct. The first part of the calculus with the integration is beyond my math skills so I had to ask for help to a friend of mine. He eventually found the same results as yours. Things are clearer for me now. This is very interesting and I've learnt. Thanks very much to you all.
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 08-15-2019, 03:33 PM #11 Thunder Chicken Fine Threads since 2008 Quote: Originally Posted by Furet  The first part of the calculus with the integration is beyond my math skills so I had to ask for help to a friend of mine. He eventually found the same results as yours. It's always good to get a confirmation.

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