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Thermal Ckt Parallel Configuration & 1-D HeatFlow Contradictory?

  1. Apr 30, 2015 #1

    1.
    My Conceptual Questions (5) is at 3.

    CONSIDER:

    Case 1: two dissimilar slabs of material (say slab 1 and slab 2) connected in series (bonded at their interface). There is a temperature difference: T1 @ slab1 and T2 @ slab2.

    Case 2: two dissimilar slabs of material bonded together, i.e. parallel configuration. There is a temperature difference: T1 @ the midpoint of the entering interfaces configuration and likewise T2 @ end.


    CONSTRAINTS:
    -1-Dimensional Heat Flow (x-direction)
    -thin yet long materials
    -steady state

    2. None. There isn't really any equations other than parallel and series addition for thermal conductivity and resistivity for materials arranged longitunditally or traverse-ly relative to direction of heat flow.



    3. My own conceptual questions posed try to use reason to determine/understand. So, I considered it an attempt.

    Question 1) Despite that I say that I have constrained Case 2 to be a 1-Dimensional Heat Flow the heat will still move in a direction other than 'x', say 'y', even if just infinitesimally small because of the parallel configuration thereby violating the 1-Dimensional Heat Flow. Is this true? If so can you show this approximation mathematically?
    Question 2) Also, would this necessitate for something like a '2-Dimensional Heat Flow'?
    Question 3) Source 1 (below) talks about non-planar geometry (axisymmetric configurations?), however they constrain heat conduction along the single coordinate 'r', thus there is not a variation of heat flow in any other direction; could I not just constrain the heat to flow through series connected materials and drop the Quasi? Because would the Quasi part imply there is a slight variation in another dimension? -Also, the MIT source calls this Quasi-One-Dimensional-Heat-Flow-in-Non-Planar-Geometry.
    Question 4) If this is just an assumption for non-planar geometry fine. But why? Empirically accurate?
    Question 5) Wouldn't Case 2 be considered planar geometry? If so, how does that situation not contradict the 1-Dimensional Heat Flow Constraint? Or this just another assumption?



    -Thank you.


    MISC. Similar Question Asked by student in the MIT course questioning the accuracy of the 1-Dimensional Heat Flow:

    MP 16..2How specific do we need to be about when the one-dimensional assumption is valid? Is it enough to say that dA/dx is small?

    The answer really is ``be specific enough to enable one to have confidence in the analysis or at least some idea of how good the answer is.'' This is a challenge that comes up a great deal. For now, if we say that A is an area defined per unit depth normal to the blackboard then saying dA/dx is small, which is a statement involving a non-dimensional parameter, is a reasonable criterion.




    SOURCES:
    1) http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node119.html
    1
    2) http://www.me.umn.edu/courses/old_me_course_pages/me3333/essays/essay 5.pdf
     
  2. jcsd
  3. May 1, 2015 #2

    BvU

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    Q1: No. ##{dT\over dy}=0## so no heat flow.

    Q2: see Q1 :smile:

    Q3: Quasi means you can move in 2 out of 3 dimensions (a thin cylindrical shell in this case) and not experience a change in heat flow: ##\dot Q(r, \theta, z) = \dot Q(r)##. By the way, this only works if ##{ dT\over dz }= 0##.

    Q4: mathematically it's perfectly Ok to assume ##{ dT\over dz} = 0## and ## {dT\over d\theta} = 0##, and assuming ## {dT\over d\theta} = 0## is physically completely acceptable too.

    Q5: see Q1. No, it's 1-D. They tell you that it is. Mathematically. Physically: wide slabs, so to speak -- either that, or perfectly isolated at the edges (##\vec {\dot Q}_y = 0##).
     
  4. May 1, 2015 #3
    Wow that makes a lot of sense! Though I still have some questions and responses to your answers.


    Answer 1&2: dT/dy = 0; no temperature difference between top and bottom! It's the temperature difference that drives the heat flow.

    Answer 3: Although the heat can still move in that additional dimension θ, there is not a change in heat flow in θ; because of the mathematical assumption Quasi-1-Dimensional Heat Flow Assumption? Doesn't this mathematical assumption also imply that dT/dθ = 0 too(as in Answer 4)? How does just setting dT/dz = 0 allow for this and not both dT/dθ and dT/dz?

    So as opposed to a 1-Dimensional and/or Quasi-1-Dimensional Heat Flows, a 2-Dimensional Heat Flow would indicate heat flow in two dimensions and with changes in those dimensions, e.g. dT/dr ≠ 0, dT/dθ ≠ 0? I am sure this would extend to the third dimension z too.....?

    Answer 4: Ok mathematically (theoretically) this makes sense. But experimentally this would be an approximation right?

    Answer 5: Aww yes I did forget to write the constraint insulated at top and bottom.

    Lastly, a parallel configuration would imply a quasi- 1-dimensional heat flow right? if dT/dy, and dT/dz are set equal to zero.


    -edit1: added the dT/dy and dT/dz part ==0
    -edit2: fixed everything that said quasi-static to just quasi.....sorry brain is used to doing mechanics.
     
    Last edited: May 1, 2015
  5. May 1, 2015 #4

    BvU

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    Correct. ##dT/dy = 0## so no heat flow in the y direction. In contrast ##dT/dx \ne 0## so plenty heat flow in the x direction. I thought I had picked up on your notation and choice of coordinates. Clarify if you think I am mistaken.

    It's not a mathematical assumption, it's a physical assumption that arises from seeing the symmetry (no physical difference if ##\theta \rightarrow \theta + \Delta \theta##). That translates into a mathematical condition ##{dT\over d\theta}=0##. Similarly translation symmetry (changing ##z\rightarrow z+ \Delta z##) leads to ##{dT\over dz}=0##. So a three dimensional problem is reduced to a 1 dimensional one. It's still a three dimensional configuration, so that's why it is called quasi-1d.

    Note that not only there is no change in heat flow in the ##\theta## direction: there is no heat flow in that direction.

    As I tried to indicate carefully ("By the way, this only works if ## { dT\over dz }= 0 ##".) the outward radial heat flow in source 1 -- where they mention "situations where fluids are pumped and heat is transferred" -- results in a temperature decrease (##dQ = c_pdT##), making the problem quasi 2-d !
    Yes. See above. It works for calculations on short or well-insulated pipes where the heat 'loss' doesn't lower the temperature too much.
    You can call it that, yes. Or you can call it two quasi- 1-dimensional heat flows ::smile:: because the heat flow in the slab that conducts better is not equal to the heat flow in the other slab. But within each of the slabs ##{d\dot Q\over dz} = 0##
     
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