Heat Transfer - Conduction/Convection Question - Which area?

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Homework Statement


Q2.jpg

Q2a.jpg

Its been assumed that the surfaces TL and TR of the same constant temperature.

Homework Equations


Tmax = TL/R + (qdoto*L2)/(8*k)

q = ΔT/R

Rconvection = 1/hA

The Attempt at a Solution


The problem I am having with this question is conceptualising which dimensions to use (I have no solutions to this question but I am trying to see what is realistic or not).

For part a, I have done the following:
Assuming - L = 0.01m, h = 0.25m and w (depth into page) = unit length of 1.

TL/R = Tsurface = 200°C - ((0.8*106)*(0.01)2)/(8*1) = 190°C

From a thermal network, I know that the temperature difference between the plate and the convective fluid is:
ΔT = Tsurface - Tfluid = 190°C - 90°C = 100°C

Here is my problem, assuming that my associations for the geometry are correct, the convective area (for the thermal resistance equation) should be:
A = p*L + 2*(h*w), where p = perimeter.
Hence, A = 2*(0.25 + 1)*0.01 +2*(0.025*1) = 0.525 m2

And the heat generated, q = V*qdoto = (0.01*0.25*1)*(0.8*106) = 2,000 W

Hence, from the thermal resistance equation qo = ΔT/R = ΔT/(1/hA) ⇒ h = qo / (ΔT*A). h = (2000)/(100*0.525) = 38.1 W/m2K

This seems reasonable for a fluid such as oil, but begs the question whether assuming a unit length into the page is the correct thing to do in this case (where only two dimensions are given)?

Any help would be appreciated!
 
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Answers and Replies

  • #2
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You were expected to neglect the heat transfer around the perimeter of the element, and include only the heat transfer on the larger faces. Certainly, your first relevant equation does this. Then the length in question would cancel.
 
  • #3
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You were expected to neglect the heat transfer around the perimeter of the element, and include only the heat transfer on the larger faces. Certainly, your first relevant equation does this. Then the length in question would cancel.
Hi Chester, thanks for your reply.

Do you mean that the heat transfer area should only be the two faces that are perpendicular to the heat conduction in Figure a, i.e. what were used to derive the Tmax equation, and not the entire surface area of the box? I'm not sure I follow what you mean.
 
  • #4
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4,394
Hi Chester, thanks for your reply.

Do you mean that the heat transfer area should only be the two faces that are perpendicular to the heat conduction in Figure a, i.e. what were used to derive the Tmax equation, and not the entire surface area of the box? I'm not sure I follow what you mean.
Yes. That's exactly what I mean.
 

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