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Something About My Deduction On the Fourier's Law of Heat Conduction

  1. Jun 13, 2012 #1
    My hypothesis is to use two different stable heat sources with different tempreature T1 and T2 (T1>T2) transmits the heat . Then I let the distance between this two heat sourse filled with idea gas or ideal metal in a tube. So if the distance is L, the heat capacity is Cv (Constant). So the temperature at r. (r is the distance to heat source T2 0<r<L). Then when the system balances, we have the temperature T(r)=T1+(T2-T1)r/L. That is the fourier's law form. I will tell you my way to deduce this formula in details if you are interested in this research.

    Do you guys think I can analyze the heat in this way: Seperating the heat in the tube into many chunks of heat units ΔV. The volume of each units is the same. Then as the area S is constant. So the length of each unit will be the same Δl.

    Here is the hypothesis; if
    I. The heat units are all continuous
    II. The volume of each units cannot be compressed

    Then I will have each unit has the same velocity v0

    And other hypothesis, if
    in a very short time t, a heat ΔQ(t) transmits into the tube, each part in the tube will obsorb the same amount of heat (ΔL/L)ΔQ(t).

    Do you agree with all these hypothesis? If so, I can deduce the Fourier's Law of Heat Conduction. If not, please tell me your thoughts. Thank you!
  2. jcsd
  3. Jun 13, 2012 #2

    Simon Bridge

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    Heat will flow from the hot place to the cold one. This makes the cold one hotter - in order for the cooler part to stay at T2, it would have to gradually crank down the rate it pumps heat until it removes it as fast as heat arrives.
    Gas or metal - it makes a difference. For instance, the gas will flow by convection between the hot and cold parts of the tube.
    At equilibrium - I guess...
    If you are careful sure.
    What do you mean by velocity? You mean the heat flow through each length interval dl will be the same rate?

    You mean the tube warms up - but the hot end of the tube will be a higher temperature than the cold part (conditions due to your setup). Is this consistent with all parts having the same stored heat?

    But what you are talking about sounds like the normal descriptions of heat conduction. If you can get it to do work on the way, then you have a heat engine.
  4. Jun 13, 2012 #3
  5. Jun 13, 2012 #4
    Any ideas?
  6. Jun 14, 2012 #5
    @Simon Bridge

    I need your help
  7. Jun 14, 2012 #6

    Simon Bridge

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    So this is the classical picture of a heat engine ... you have a source and a sink, a hot and cold heat reservoir, always at the same temperature and something to conduct the heat between them.

    The entire tube starts at thermal equilibrium with the cold reservoir.

    Note - if the tube absorbs heat as described then it will eventually be the same temperature as the hot reservoir. Usually the metal rod will be hottest by the hot reservoir and coldest at the cold reservoir ... and have a constant temperature gradient down it's length.

    That would be at steady state - so heat is leaving the tube as fast as it is added.

    In the transient state, you get the normal rules for heat conduction.

    I'm unsure about your reasoning.
    What is it that you believe you have found out?
  8. Jun 14, 2012 #7
    @Simon Bridge

    Definately, I am thinking that the hotter end is hotter because it obsorb more heat units than the cooler end. Suppose in each same time Δt, each heat unit the heat sourse transmits is ΔQ (ΔQ is a constant). So it will cost the first heat unit time L/v to go from the hotter end to the cooler end. Then during this time, the hotter end obsorbs (L/v)(ΔQ/Δt), the cooler end obsorb 0. So the temperature different will be ΔT=Q/Cv= (ΔQ/vΔtCv)L Because (ΔQ/vΔtCv) is a constant, so it will be ΔT=kL. It can be easily known that If we take different point in the tube and their distance is ΔL, we will have ΔT=kΔL when the heat sourse transmits the heat proportionally to time. If it is not propotional to the time ΔQ(t), then I will have different conclusion. Do you think it is correct? I Hope you can reply me soon! Thank you!
  9. Jun 15, 2012 #8

    Simon Bridge

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    You have to decide if your reservoirs are constant temperature OR provide a constant heat flow. This is much how you'll be used to handling electric circuits where the power source can either be a voltage or a current source but not both. Just like we usually have constant-voltage supplies, we usually have constant temperature sources (in idealized systems of course).

    If you bring a hot reservoir in contact with a rod, you can compute the heat flow depending on the temperature of the rod and it's thermal conducting properties.

    If ΔQ is the same for every Δt then you can say the heat current, ΔQ/Δt, is a constant. But this situation will only apply a long time after the system is first hooked up.

    Since you have a simple geometry it is easy to show that[itex]\frac{\Delta Q}{\Delta t} = -kA\frac{\Delta T}{\Delta L}[/itex] which makes your last relation more like:

    [tex]\Delta T = -\frac{1}{kA}\frac{\Delta Q}{\Delta t}\Delta L[/tex]
    ... if the current is a constant then this collapses to what you got: but all it tells you is that the temperature gradient is a constant.

    Your conclusion seems to agree with mine but your method appears a tad confused.
    So what is it that you are trying to work out?
  10. Jun 15, 2012 #9
    @Simon Bridge

    You are right. However, if you agree with the hypothesis that each interval will obsorb the same heat at different time. Then, let's say the heat current is ΔQ(t). At Δt, it arrive to point A(Δl) and it obsorbs Δq(Δt).

    At this time, A totally obsorbs Δq(Δt).

    Then at 2Δt, another heat unit arrives. Point A obsorbs Δq(2Δt) and the neighbor point B (2Δl) obsorbs Δq(Δt) based on the hypothesis.

    At this time A totally obsorbs Δq(Δt)+Δq(2Δt), and B totally obsorbs Δq(Δt)

    Then at 3Δt, another heat unit arrives. Point A obsorbs Δq(3Δt) and the neighbor point B (2Δl) obsorbs Δq(2Δt) and Point C obsorbs Δq(Δt).

    At this time, A totally obsorbs Δq(Δt)+Δq(2Δt)+Δq(3Δt), and B totally obsorbs Δq(Δt)+Δq(2Δt) and C totally obsorbs Δq(Δt).

    Now, you will find the rule that at the time t, if the distance between point A(Δl) and Q is ΔL
    Then the heat that point A obsorbs will be ƩΔq(NΔt)=Δq(Δt)+Δq(2Δt)+...+Δq(t)
    The heat that point P obsorbs will be ƩΔq((N-ΔL/Δl)Δt)
    Because the velocity is a constant, then ΔL/t=Δl/Δt=v
    Then the heat that point P obsorbs will be ƩΔq((NΔt-ΔL/v))=Δq(Δt)+Δq(2Δt)+...+Δq(t-ΔL/v)
    Then for A, it will be q(A)= ∫(dq/dt)dt. The region will be (0,t)
    For P, it will be q(P)=∫(dq/dt)dt. The region will be (0,t-ΔL/v)
    If you agree, just letΔT=(q(A)-q(P))/Cv, and then let lim P→A, you will get the fourier's law of heat conduction. Do you think this way can correctly desribe the heat conduction?
    Thank you Simon!
    Last edited: Jun 15, 2012
  11. Jun 15, 2012 #10
    My Point, in a word, is that the position determines the total heat it obsorbs and then determines the temperature it increases.
    And the heat going through each single interval has the same rate.
    And using this method, you can theoretically proove Fourier's law of conduction.
    Thank you Simon
    Last edited: Jun 15, 2012
  12. Jun 16, 2012 #11
    @Simon Bridge

    Any ideas?
  13. Jun 23, 2012 #12

    Simon Bridge

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    About what? You seem to have reached a conclusion.

    I don't think you need different heat absorption in different parts of the conductor to prove Fouriers law for heat conduction.

    Without an experiment, though, you can only demonstrate the internal logic of a proposition. Since Fourier's Law amounts to a synthetic proposition, you need empirical verification.
  14. Jun 24, 2012 #13
    Thanks Simon Bridege. If I have other problem, I will contact you!
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