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Energy storage without electricity - the rate of heat loss from a vacuum container

  1. Jul 20, 2010 #1
    This isn't actually a homework question, but I didn't want to clutter the other forums because I couldn't do my own calculations.

    This is to try and find viable ways of storing energy away from a modern society, and any help would be greatly appreciated.

    For (some) ease, say this container has the same properties as a standard thermos flask, other than size.

    3 cubic metres of table salt are heated to 800 degrees centigrade, and stored in this container. (we can assume the container is completely filled)

    If we take the temperature outside to be 20 C, and it's stored simply in the air, how long would it take for the salt to reach 100 C?

    I understand people are loathe to simply answer others' questions for them, but I really am not sure how to go about approaching this.

    Whilst I have some degree of knowledge of advanced physics ideas, the level of equations I've properly covered don't go much more advanced than the most basic of kinetic energy's set (I'm 16, heh).

    I would appreciate any help I can get, as I'd really like to figure this out.
     
  2. jcsd
  3. Jul 21, 2010 #2
    Re: Energy storage without electricity - the rate of heat loss from a vacuum containe

    Interested to see where you're going with this...

    The question is, how do you want to model the problem?
    I can think of three significant methods of heat transfer. Do you know them?
     
  4. Jul 21, 2010 #3
    Re: Energy storage without electricity - the rate of heat loss from a vacuum containe

    Well, obviously there's convection, conduction and radiation. I'd assume we don't need to worry too much about the rate of conduction through the first layer of metal, although it's possible the conduction and convection through the salt itself will have some noticable impact on the result.

    The flask is designed to prevent radiation as much as possible, but the infrared radiation is obviously how it's going to escape.

    I'm aware of the graph shape I'd be looking at if I plotted this, but I don't know the equations (even after a quick google on a holistic approach to my problem) that would help me solve this.

    I hope that's what you mean about method of heat transfer, as any more than that and I'll need a hand.

    Thanks for the reply.
     
  5. Jul 21, 2010 #4
    Re: Energy storage without electricity - the rate of heat loss from a vacuum containe

    Ok, so in a thermos flask the main source of heat loss is through the lid - lets assume we don't have one.
    Now what initially happens is that the inner shell and material (and air) inside come to thermal equilibrium. Since the temperature of your material is much higher than that of the presumed initial (room) temperature of the shell, and the shell has a very low thermal capacity, we can to a good approximation say they settle at the material's temperature.

    Radiation is invariably the main source of heat transfer then. The inner shell emits as a black body, albeit one with a very low emissivity. To actually lose heat from the flask, it also needs to pass to the outer shell. Reflections between parallel surfaces messes with the emissivity, but can be calculated.

    You can achieve your calculation by energy (heat) considerations. http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law" [Broken]should point you in the right direction. However this is a specific problem, and will require a value for (or estimate of) the emissivities involved. These may be possible to find on the Internet...
    If you go down the conduction route, similarly you will need some material specific constants.
     
    Last edited by a moderator: May 4, 2017
  6. Jul 21, 2010 #5
    Re: Energy storage without electricity - the rate of heat loss from a vacuum containe

    Thanks, that helped a lot.

    Now my rate of radiation appears to be 19,846 W-1m2.

    Do I need to now work out the amount the second sheet of metal will absorb, and then consequentially emit into the envrionment?
     
  7. Jul 21, 2010 #6
    Re: Energy storage without electricity - the rate of heat loss from a vacuum containe

    You will need to use http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation" [Broken].

    Also, are you sure your rate of radiation is a constant?
     
    Last edited by a moderator: May 4, 2017
  8. Jul 21, 2010 #7
    Re: Energy storage without electricity - the rate of heat loss from a vacuum containe

    Oh, well, what I worked out was the energy flux density at 800C (~1000K) so obviously it'll change as energy is given off.

    The equation I need to use between the two walls is f008e52272e7abf62ebc8d8227da3592.png

    Which works out as 0.21r, though I'm not entirely sure what I need to do with this.

    I don't think I need to consider interference, and the absorptivity equals the emissivity, which, for each individual wall, 0.35.

    EDIT: Well, for the first one at such a high temperature, that could be correct, but actually for the second wall it seems most likely to start at about .02

    Will re-calculate shortly.

    EDIT2: This now gives me 0.02. What would I do with this?
     
    Last edited: Jul 21, 2010
  9. Jul 22, 2010 #8
    Re: Energy storage without electricity - the rate of heat loss from a vacuum containe

    As I said, I haven't come across any of these equations before, or any like them, so I'm not entirely sure what to do with them - so apologies if it seems like I'm being thick.
     
  10. Jul 22, 2010 #9
    Re: Energy storage without electricity - the rate of heat loss from a vacuum containe

    It doesn't seem like the arithmetic is your problem. You should know that physics isn't about plugging numbers into equations - simple models like this are always wrong anyway, the equations are just there to tell you "how it goes."

    Imagine it, what would happen? Why/where would you have to use that equation? Forget the numbers for a minute.
    I'll start you off. The inner shell heats up. It radiates energy outwards...
     
  11. Jul 22, 2010 #10
    Re: Energy storage without electricity - the rate of heat loss from a vacuum containe

    Sorry, you're right, I just meant that it's me coming to terms with some of the more "grown up" physics, which is great fun, actually, now that it isn't being spoonfed to me.

    So a lot of it is reflected by the second shell, which has an initial emissivity (due to much lower temperature) of .02.

    A lot of the radiation is therefore reflected back. I assume the calulated .02 (emissivity between two walls) can therefore be used to work out the absorptivity - using Kirchoff's law.

    Initially, I suppose the second shell would be at thermal equilibrium, meaning that its absoptivity equals the emissivity (.02).

    If that's all correct, then I can work out some rough rates of radiation at different temperatures, and then figure out how quickly the thing will cool down. It only need to be a rough approximation, of course.
     
  12. Jul 23, 2010 #11
    Re: Energy storage without electricity - the rate of heat loss from a vacuum containe

    Remember emissivity is not an intrinsic material "value" as such, it is a fraction, or even better, a probability.
    We want to consider rates. The rate at which the second shell absorbs (and therefore emits) radiation is lowered by reflecting the radiation back and forth. It builds up in between the walls - it can actually be very well thought of as a type of gas.

    Deriving the formula for the drop in emissivity is not very straight forward, and the value will of course change with temperature (actually, the emissivities themselves are also temperature dependent), it is also given for the steady state, which changing temperature is not.
    For a very rough approximation you could use it as a constant. You would then be modeling an infinite amount of reflections at each instance in time. But still, to do this you "pretend" the two walls are a single wall, of emissivity whatever it turns out to be.

    Just a heads up, you will probably be miles out in your estimate if you do not integrate something somewhere. "I can work out some rates of radiation at different temperatures" rings alarm bells.
     
    Last edited: Jul 23, 2010
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