Conduction - finite difference method

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yecko
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Homework Statement


Textbook problem figure with a cross-section of a heat sink showing device width and sink dimensions, a 5 by 5 finite difference nodal grid with symmetry line and adiabatic plane, and answer fields for total heat transfer per unit depth and thermal resistance.


Homework Equations


Finite difference method

The Attempt at a Solution


Handwritten finite difference heat conduction derivation for a computer project, showing the total heat flow equation with terms for qa, qb and qc, numeric substitutions, a resistance result of 1.588 x 10^-3 m K/W, and a small node diagram labelled T1 to T6 with heat flows qa, qb and qc.

Handwritten derivation titled computer project: total heat flow q_tot' as a sum of three terms, a finite-difference expression with 300 times a sum of three fractions equal to 1.263 x 10^4 W/m, and R' = (Td - Ts)/(2 q_tot') = 2.38 x 10^-3 m K/W. A small node grid labelled Td, T1, T4, T5, T6 is at the top right.

I have tried two different approaches, but still i am wrong in the question. Can anyone guide me how to attempt this question?
Thank you
 
Attachments
  • Textbook problem figure with a cross-section of a heat sink showing device width and sink dimensions, a 5 by 5 finite difference nodal grid with symmetry line and adiabatic plane, and answer fields for total heat transfer per unit depth and thermal resistance.
    6BB605E5-3BBB-4CA0-AADE-414022EA84B9.png
    23.5 KB · Views: 921
  • Handwritten finite difference heat conduction derivation for a computer project, showing the total heat flow equation with terms for qa, qb and qc, numeric substitutions, a resistance result of 1.588 x 10^-3 m K/W, and a small node diagram labelled T1 to T6 with heat flows qa, qb and qc.
    36D8F330-C4B5-451E-92CE-8059FD0FD165.png
    38.7 KB · Views: 787
  • Handwritten derivation titled computer project: total heat flow q_tot' as a sum of three terms, a finite-difference expression with 300 times a sum of three fractions equal to 1.263 x 10^4 W/m, and R' = (Td - Ts)/(2 q_tot') = 2.38 x 10^-3 m K/W. A small node grid labelled Td, T1, T4, T5, T6 is at the top right.
    6C9774F5-7878-48B0-9A0D-F03FEEF0C72B.jpeg
    23 KB · Views: 719
Chestermiller said:
The first approach gets the right answer. Why are you saying that it's wrong?
The system says I am wrong (ref: first screenshot in #1), and the following is what my instructor says:
"Are these temperatures computed by you, by solving the finite difference equations ? If these are given in the problem, these are for input data corresponding to the case of delta x = delta y."
I have no clue in understanding how should I do it...
Thank you.