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One Dimensional Slab Heat Transfer Taylor Expansion in Glasstone

  1. Mar 23, 2014 #1
    Hi There,

    I came across the following passage in Sam Glasstone's 'Nuclear Reactor Engineering'

    WqhHDFw.png

    See where I underlined in red that taylor series expansion? I don't understand how (dt/dx)_(x+dx) is equal to that.

    I know it's a Taylor series expansion, but where did the x+dx go?
     
  2. jcsd
  3. May 4, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
  4. May 5, 2014 #3
    Well,

    I know it's the incremental form of the Taylor series:
    http://geophysics.ou.edu/solid_earth/readings/taylor/taylor.html [Broken]

    But I have no clue how to derive this incremental form.
     
    Last edited by a moderator: May 6, 2017
  5. May 5, 2014 #4

    Astronuc

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    Staff: Mentor

  6. May 12, 2014 #5
    The differential is supposed to be small, so we can neglect higher order terms in the Taylor series expansion since squaring a small number will give you an even smaller number. So you have heat entering and exiting your differential control volume,

    That is Net Heat Flow:

    qx- qx+dx.

    See the heat enters at position x and leaves at a position x + a very small amount (dx). Now if we do a Taylor series expansion

    qx+dx= qx +(dqx/dx)*dx

    Take this expression and substitute into the above expression for net heat flow.

    qx-qx+dx=qx-qx+(dqx/dx)*dx

    Then:
    qx-qx+dx=(-dqx/dx)*dx

    No you can't cancel those two dx terms because one is a partial differential and the other is total. The one in parentheses is the partial differential.

    Now we can describe qx using Fourier's law

    qx=-kA(dT/dx).

    Substitute this into our above expression

    -d/dx*(-k *dydz*dT/dx)*dx

    Net heat flow in x direction is now

    d/dx*(kdT/dx)*dxdydz

    The area in the x direction is given by y times z, similarly in the y direction area will be given by x and z

    If you do those equations for the y and z directions and then divide the whole equation by dxdydz you will obtain the heat equation.

    I assume you were ok with the other terms in the heat equation. If not I'm happy to explain the derivation of those as well
     
  7. May 12, 2014 #6

    Morbius

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    terryphi,

    I looks like a Taylor series expansion for the first derivative instead of the function itself.

    The Taylor series expansion for function "f" about x = 0 is:

    f(x) = f(0) + df/dx dx + 1/2 d2f/dx2 (dx)^2....

    In this case, the function f is dT/dx; so all the derivative are one degree higher.

    Greg
     
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