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Newton's law of cooling

  1. Dec 22, 2005 #1
    I'm having trouble thinking of an instance when newton's law of cooling is no longer valid. could it have something to do with the surface of the system or the surroundings not having uniform temperature.

    thanks very much
     
  2. jcsd
  3. Dec 22, 2005 #2

    FredGarvin

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    What do you mean by no longer valid? If you can expand on that just a bit it might make this a bit easierr. The only way there would be no heat transfer is if the surface temperature is equal to the surrounding temperature. Even if there wasn't an even distribution, it would still be valid. I guess the other way it could not be valid is if you had a non-real heat transfer coefficient, but I can not see how that would be physically possible.
     
  4. Dec 22, 2005 #3

    Clausius2

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    A generalized version of the Newton law of cooling may be:

    [tex]q(x,y,z,t)=h(x,y,z,t)(T(x,y,z,t)-T_\infty(x,y,z,t))[/tex]

    The Newton law always hold, IF AND ONLY IF you calculate the convection coefficient h in a suitable way for your problem. In fact, one always is able to calculate h from this equation, by merely working it out. The value of the Newton law is relationed with experiments, because it is easy to calculate the heat flux based on experimental correlations, BUT this equation does NOT add any additional physical law to Navier-Stokes equations.
     
  5. Dec 22, 2005 #4

    siddharth

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    I think Newton's law of cooling is applicable without restriction to a body cooling by convection only.

    So, in case a body of surface area [itex] A [/itex] at an absolute temperature [itex] T [/itex] is also losing heat by radiation, then Newton's law of cooling, ie, [tex] \frac{dT}{dt} = -bA(T-T_0) [/tex] will be valid only for small temperature differences between [itex] T [/itex] and [itex] T_0 [/itex]. ([itex] T_0 [/itex] is the temperature of surroundings which is less).

    This will be because, by stefan's law, the net loss of energy due to radiation is
    [tex] \Delta u = \sigma e A(T^4 -{T_0}^4) [/tex]

    If the temperature difference is small,ie, [tex] T=T_0 + \Delta T [/tex] and [tex] T^4 -{T_0}^4 [/tex] can be approximated as [tex] 4{T_0}^3(T-T_0) [/tex].
    So, the net rate of cooling in this case will also be [tex] \frac{dT}{dt} = -bA(T-T_0) [/tex]
     
    Last edited: Dec 22, 2005
  6. Dec 22, 2005 #5

    Clausius2

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    The What? I disagree completely with you. Radiation and Convection are two different mechanisms no matter how large is [tex]\Delta T[/tex]. The fact that radiation behaves in a similar fashion than convection when [tex]\Delta T[/tex] becomes small does NOT mean both things are the same. The convection coefficient has a STRONG dependence on the external flow, and has NOTHING to do with radiation, which is by the way relationed with the electronic structure of the solid.

    A very wrong advice you gave this time.
     
  7. Dec 22, 2005 #6

    siddharth

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    I agree that convection and radiation are very different.

    What I meant to say to the OP was that the formula for the net rate of cooling may not be applicable in when there is heat loss due to radiation also.

    I think the confusion is because I used the words Newton's law of cooling while working with the case where heat is lost by radiation and convection.

    What I meant was that the rate of fall of temperature due to radiation will also be [tex] bA(T-T_0) [/tex] when the temperature difference is small.

    So the net rate of fall of temperature due to both radiation and convection will be [tex] \frac{dT}{dt} = -(b_1 + b_2)A(T-T_0) [/tex] and this formula is valid only when [tex] \Delta T [/tex] is small

    Do you agree with what I am saying now?
     
    Last edited: Dec 22, 2005
  8. Dec 23, 2005 #7

    Clausius2

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    If you mean that b1 and b2 are two different coefficients without any relation, and that newton cooling law always hold despites there is radiation or not (ok, there is gonna be some portion of q not taken into account due to radiation), then I am fine with your post.
     
  9. Dec 23, 2005 #8

    siddharth

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    Yes, that's what I mean.
    I probably should have made it clearer in my first post though.
     
  10. Dec 23, 2005 #9

    Clausius2

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    Well, sorry, maybe I misunderstood your reply too. But to say the truth I remember one guy here who was convinced that at small [tex]\delta T[/tex] both things were the same. It was long time ago.
     
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