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Heat Conduction

  1. Apr 21, 2008 #1
    Suppose that I have an aluminum cube with side lengths 10 cm. Suppose that I uniformly and continuously apply a temperature of 60 degrees celcius to one of its sides. The medium surroudning it is air with a temperature of 27 degrees celcius. After t seconds, what is the temperature of the side opposite to the side to which a temperature of 60 degrees is applied. Notice that the sides are not insulated and so heat can be lost. I would really appreciate it if someone would solve this for me.

    Note: This is not homework, and I am not at the age to receive such homework. This is simply something that I am interested about.
     
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  3. Apr 21, 2008 #2

    mgb_phys

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    It depends! You are balancing three things.
    1, The rate of heat input - just saying you supply heat at 60deg is not enough. Suppose you connect a very thin wire at 60deg to one face it will heat much more slowly than connecting a huge electrical heater even if that was also at 60deg.
    2, Then there is the rate of heat loss. There are 3 main types:
    Radiation will depend on the reflectivity of the surface and the temperature of the surrounding walls - if the block was pointing at outer space (cold black) there would be more heat loss than if it was inside a reflective sphere.
    Convection will depend on the rate of air flow and moisture content of the air.
    Conduction doesn't matter if we assume the block is not touching anything (I know this is impossible but we are allowed to cheat in thought experiments!)

    3, The rate of heat transfer in the block. This is the easiest to work out since it only depends on the properties of aluminium and the temperature difference. Paradoxically since the rate of heat transfer slows as the temperature difference is reduced the two sides of the block will never reach the same temperature ( in theory) - and your coffee will never actually get cold!

    I hope I hit the right technical level - please post back if you want anything explained.
     
  4. Apr 21, 2008 #3

    Hootenanny

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    To solve this problem, one would have to solve the heat equation, which is a second order PDE with some potentially ugly boundary conditions. The mathematics required to obtain such a solution would be equivalent to that of an intermediate level undergraduate mathematics major. What do you hope to gain from us providing a solutions for you, that in all likelihood you won't be able to comprehend? It is not my intention to be rude or patronising, I am simply curious as to why you would want such a solution?
     
  5. Apr 21, 2008 #4
    I am conducting an experiment relating to thermoelectric modules. Also, I don't mind the math for I have enough knowledge to understand heat equations. As a result, would you please enlighten me with the heat equations? Thanks in advance.
     
  6. Apr 21, 2008 #5

    Hootenanny

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    The heat equation is as follows,

    [tex]\frac{\partial u}{\partial t} = k\nabla^2 u[/tex]

    Where [itex]u=u(x,y,z,t)[/itex] is the temperature at a given point and time in the cube. I as said previously (and mgb_physics has alluded to), you could potentially end up with some nasty boundary conditions. You could simplify the boundary conditions considerably by assuming that the face of the cube is uniformally heated and the boundaries of the other faces remain at a constant temperature, but perhaps this would be an over-simplification.

    Since your query is more practical that theoretical, perhaps it would be better addressed by the engineers in their forum, I could get the mentor's attention to move it there if you like?
     
    Last edited: Apr 21, 2008
  7. Apr 21, 2008 #6
    Please move this to a suitable place as you wish; thank you. I have worked with the heat equations for a while; however, I have not reached anything desirable. I was wondering about using Fourier's Law, however, I have trouble taking into account that heat is being lost. Any ideas?
     
  8. Apr 21, 2008 #7

    Hootenanny

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    This is what I was talking about when I said that the boundary conditions could get a little ugly. Your boundary conditions will be temperature dependent, my best guess would be to solve the heat equation subject to the heat lost determined by Fourier's law. However, this would only be a guess, I haven't attempted anything so complicated myself.
     
  9. Apr 21, 2008 #8
    Could you please demostrate it mathematically?
     
  10. Apr 22, 2008 #9

    Hootenanny

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    Aftert thinking about the problem a little more this morning it is clear that no matter how many assumptions you make, your boundary conditions are either going to be really ugly or completely unrealistic. In any case, the PDE which you obtain is unlikely to have any analytical solutions, you're probably going to have to use numerical analysis to determine the temperature. I've managed to find a few decent resources that deal with the numerical analysis of heat transfers,

    http://icb.olin.edu/fall_03/mc/reading/conduction.pdf
    http://www.buildingphysics.com/manuals/avh_TB.pdf
     
  11. Apr 22, 2008 #10
    Thank you for the documents. In the first document, there is a problem, correct? The simplification from (1.5) to (1.6) is incorrect.
     
  12. Apr 22, 2008 #11

    Hootenanny

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    Indeed there is, I hadn't spotted that! Equation (1.6) should read,

    [tex]\frac{dT_i}{dt} = \alpha\frac{T_{i+1}+T_{i-1}}{L_i^2}}[/tex]

    And the final equation (1.7) should read,

    [tex]\frac{dT_i}{dt} = N^2\frac{T_{i+1}+T_{i-1}}{\tau_\text{cond}}[/tex]

    I apologise for referencing incorrect material. The second document is far more comprehensive and will probably tell you what you need to know.
     
  13. Apr 23, 2008 #12
    Assuming that the heat lost is neglegible and the aluminum block has reached steady state, would the following be the correct derivation of the formula (1D heat equation)?

    [tex]\frac{\partial^2 T}{\partial z^2}=0\iff T(z)=az+b .[/tex]

    By Fourier's Law, [tex]-\kappa\frac{\partial T}{\partial z}=h(T-S),[/tex]
    where S is the room temperature, and [tex]\kappa[/tex] and h are constants. Letting [tex]T(0)=T_0[/tex], then the temperature at any point may be obtained. Is the Fourier's Law correctly written? Thanks in advance.
     
  14. Apr 24, 2008 #13

    Hootenanny

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    Yes, your steady state solution is correct, as is your expression of Fourier's Law, provided that the aluminum block is undergoing forced convection (i.e. a constant flow of air at room temperature is passing over it's surface). As an aside Fourier's Law in this form is commonly referred to as Newton's Law of Cooling.
     
    Last edited: Apr 24, 2008
  15. Apr 24, 2008 #14

    GT1

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    If there is no forced air flow on your cube, then it is a unsteady state free convection problem.
    the free convection heat transfer coefficient is dependent on the temperature of the body, and since the temperature of the body is changing as it gets hotter , the coefficient is also changing all the time so you have to solve PDEs numerically for the exact solution.
    for most of the free convection problems the value of the convection coefficient is between about 5-10 W/(M^2*Deg C)- it doesn't change much so if you don't need high precision you may assume it as constant.
    Finally, the most easy, accurate and fast way to solve this problem is with a software, such as COMSOL. you just sketch the model, all the PDFs are already in the software (with the material properties) and you solve it numerically, it doesn't take more then 10-15min if you have some basic background on heat transfer.
     
  16. Apr 24, 2008 #15

    GT1

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    it's 10-15 min. if you know how to work with this software, this software is easy to use,it shouldn't take you more then few hours to learn all the basic options and more.
     
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