# Thermal Ckt Parallel Configuration & 1-D HeatFlow Contradictory?

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1. Apr 30, 2015

### ltkach2015

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

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. May 1, 2015

### BvU

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

Q2: see Q1

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 ($\vec {\dot Q}_y = 0$).

3. May 1, 2015

### ltkach2015

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
4. May 1, 2015

### BvU

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 :: 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 ${d\dot Q\over dz} = 0$