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Heat transfer in a finite rod

  1. Nov 24, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A rod with length [tex]L[/tex] and section [tex]A[/tex] has its extremities in contact with 2 springs of heat whose temperatures are [tex]T_A[/tex] and [tex]T_B[/tex] such that [tex]T_A>T_B[/tex]. The rod is an environment where the temperature is worth [tex]T_0[/tex] (constantly).

    a)Determine the function [tex]T(x)[/tex] that describes the temperature of the rod in the steady state, for the case in which the rod is covered by a insulating material.

    b)Calculate the value of [tex]\vec q[/tex] in the 2 extremities of the rod.

    c)Determine [tex]T'(x)[/tex] while in steady state, that describes the temperature of the rod if we remove the insulating material and supposing that [tex]T_0=T_B[/tex].

    d)What is the value of [tex]\vec q[/tex] in each of the extremities in this new situation?

    e)Calculate the flux by unit of time that is emitted by the surface of the rod.
    2. Relevant equations
    None but the thermal conductivity coefficient is worth [tex]h[/tex]. (conductivity between the rod and the environment).
    The conduction coefficient is [tex]K[/tex] and the perimeter of the rod is [tex]P[/tex].


    3. The attempt at a solution
    a)In the steady state, [tex]T(x)=ax+b[/tex].
    I've found out that [tex]T(x)= \left ( \frac{T_A-T_B}{L}\right ) x+T_A [/tex]. At least it satisfies the initial conditions.

    b)[tex]\vec q =-KA \frac{\partial T}{\partial x}=-\frac{KA}{L}(T_B-T_A)[/tex]. I'm sure I made an error here. It seems that the q vector is constant in all the rod, sounds like possible though... I'm unsure I must say. If it wasn't constant then the temperature would change.

    c)Heat from the rod will be transfered to the environment at a rate [tex]\frac{dQ}{dt}=\bold q = hA(T_0-T)[/tex] where [tex]T[/tex] is the temperature of the rod.
    I'm a bit lost here. Should I replace [tex]T[/tex] by the [tex]T(x)[/tex] I've found in part a) ?

    Once I get help on this, I'll look for the rest. Thank you in advance.
     
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  3. Nov 25, 2009 #2

    Mapes

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    Looks to me like part (c) requires you to develop a differential equation describing the heat transfer. Can you perform an energy balance on a slice of the rod showing the inputs and outputs from conduction and convection?
     
  4. Nov 25, 2009 #3

    fluidistic

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    I'm trying to do so.
    I'm sorry, I forgot to give a data : the perimeter of the rod is P. (I think I don't forget anything now).

    I've sketch the slice of length [tex]dl[/tex]. Let [tex]\bold q_1[/tex] be the input from conduction and [tex]\bold q_2[/tex] be the output from convection. The area of the slice that is exposed to convection is [tex]Pdl[/tex]. Hence [tex]q_2=hPdl(T_0-T)[/tex] where [tex]T[/tex] is the temperature of the slice.
    [tex]\bold q_1=K\frac{\partial T}{\partial x} \big |_{dl}[/tex].
    So [tex]\bold{q_{\text{total}}}=q_1-q_2[/tex]...

    I don't have any confidence in myself on this. Could you correct me?
     
  5. Nov 25, 2009 #4

    Mapes

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    It's a very useful technique. When you get the hang of it, it'll be easy to construct differential equations describing fluxes (heat, matter, charge, etc.)

    Let's consider an infinitesimal slice of the rod with length dx, cross-sectional area A, perimeter P dx/L. The slice is so small that we can approximate its temperature with a single value T(x), but there is a finite temperature gradient.

    The heat energy coming in by conduction is, by Fourier's Law, [itex]-kA(dT/dx)[/itex]. The heat energy leaving the other end isn't exactly the same because of the temperature gradient. We approximate it with the first two terms of a Taylor series expansion:

    [tex]-kA\left(\frac{dT}{dx}\right)+\frac{d}{dx}\left[-kA\left(\frac{dT}{dx}\right)\right]dx=-kA\left(\frac{dT}{dx}\right)-kA\left(\frac{d^2T}{dx^2}\right)dx[/tex]

    The heat energy leaving by convection is simply [itex]hP(dx/L)[T(x)-T_B][/itex]. Sum all these energy terms and divide by dx and you'll have your differential equation. And hopefully you can see how you could handle variations like a position-dependent thermal conductivity or cross-sectional area, by not taking them outside of d/dx. Make sense?

    EDIT: Typo, changed "conduction" to "convection"
     
    Last edited: Nov 25, 2009
  6. Nov 25, 2009 #5

    fluidistic

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    Thank you very much for your help.
    I understand that the heat entering the slice is greater than the heat that leaves it by conduction. However I don't follow you when you approximate the gradient of temperature (I think you mean [tex]\frac{dT}{dx}[/tex]) by a Taylor series. Could you precise a bit more on this please? I see that the heat leaving the slice differs from the heat entering it by approximately [tex]-kA\left(\frac{d^2T}{dx^2}\right)dx[/tex] which is the second term of a Taylor's "polynomial"?

    I understand I then have to add up these energy terms. But I don't understand why dividing by dx would reach the differential equation of the entire rod. Shouldn't I divide by dx and multiply by L?
     
  7. Nov 25, 2009 #6

    Mapes

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    The heat entering is [itex]f(x)[/itex], the heat leaving is [itex]f(x+dx)[/itex]. Since dx is small, we can use the Taylor series expansion

    [tex]f(x+dx)=f(x)+f'(x)(dx)+\frac{1}{2}f''(x)(dx)^2+\dots[/tex]

    from which we take the first two terms only. [itex]f(x)=-kA[dT(x)/dx][/itex]. I'm not sure if this answers your question.

    Just to check that we're on the same page, what differential equation do you come up with?
     
  8. Nov 25, 2009 #7

    fluidistic

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    I'm confused about the Taylor's expansion. Wikipedia states [tex]f(x+a)=f(a)+\frac {f'(a)}{1!} (x)+ \frac{f''(a)}{2!} (x)^2+\frac{f^{(3)}(a)}{3!}(x)^3+ \cdots[/tex] hence I guess [tex]f(x+dx)=f(dx)+\frac {f'(dx)}{1!} (x)+ \frac{f''(dx)}{2!} (x)^2+\frac{f^{(3)}(dx)}{3!}(x)^3+ \cdots[/tex] which is different from your expansion.


    Summing the energy terms and dividing by dx (I don't know why I have to divide all by dx), I reach [tex]-KA \left [ \left ( \frac{dT}{dx^2} \right ) + \left ( \frac{d^2T}{dx^2} \right ) \right ] + \frac{hP}{L} (T(x)-T_B)[/tex] which I know is wrong since I have a strange term: [tex]\left ( \frac{dT}{dx^2} \right )[/tex].

    I feel sorry not to catch up with all this, Mapes. But I'm really willing to understand this all. Thank you once again for your time.
     
  9. Nov 25, 2009 #8

    Mapes

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    This first equation assumes small x compared to a, the second small dx compared to x. Other than that, they look the same to me.

    You need to combine three components: one for heat entering by conduction, one for heat leaving by conduction, and one for heat leaving by convection. Some terms will cancel out. I think you missed the first component.
     
  10. Nov 25, 2009 #9

    fluidistic

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    If I understand well, it means that [tex]f'(x) \cdot dx+f''(x) \cdot (dx)^2=f'(dx)\cdot x + f''(dx) \cdot x^2[/tex]. I'll trust you if you tell me yes, but I'm ashamed it's not obvious to me.



    Oh you're absolutely right, I focused too much on the math, too few on the physics.
    So I get [tex]-KA\left ( \frac{d^2T}{dx^2} \right )+\frac{hP}{L}(T(x)-T_0)[/tex]. But it's not yet an equation... I should get T'(x), not the derivative of the temperature, but the temperature distribution of the bar in steady state when there's convection and conduction on it.
     
  11. Nov 25, 2009 #10

    Mapes

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    Whoa, that's not what I'm saying. The wikipedia article is using different variables. The smaller quantity always gets squared, cubed, etc. The larger quantity is acted on by the function and its derivatives. You can't interchange them.

    It's an energy balance; its sum is zero. That's the differential equation of the system.
     
  12. Nov 25, 2009 #11

    fluidistic

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    Ah, thanks a lot. I think I'm getting it.
    I wasn't aware of the term "energy balance" before. I guess I should learn about it in my Thermodynamics course (Still more than 1 year to wait).
    So part c) asks me to solve for [tex]T(x)[/tex] in the equation... [tex]-KA\left ( \frac{d^2T}{dx^2} \right )+\frac{hP}{L}(T(x)-T_0)=0[/tex]. Ok it's well over my head for now. It's an intro physics course. I don't know really know what to think of it. I have other similar exercises...
    Thank you very much for your time. I'll think about it during the weekend. I'm going to focus on the lab part for the next 2 days. My final exam is on the next Monday.
     
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