Derivation process? (Heatsink Fin Heat Conduction Equations)

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EastWindBreaks
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Homework Statement


Slide titled Fin Equation showing a volume element of a fin at position x with length delta x, cross-sectional area Ac and perimeter p, with conduction and convection heat rates labelled, followed by the derivation of the fin differential equation and the excess temperature form.


Slide deriving fin heat transfer: general solution theta(x) = C1 e^(mx) + C2 e^(-mx), boundary conditions at fin base and tip, and the infinitely long fin heat rate Q = sqrt(hpkA_c)(Tb - Tinf) with a fin diagram.

I don't understand the derivation of the right side of the last equation.

Homework Equations

The Attempt at a Solution



Handwritten derivation of a fin temperature solution: T(x) minus T-infinity equals C1 e^(mx) plus C2 e^(-mx), with boundary conditions at x=0 and as L goes to infinity giving C1 = 0, leading to T(x) minus T-infinity equals C2 e^(-mx) and a ratio of temperature differences equal to e^(-mx).

I got to this point, I also don't understand why it did not include C_2 for the variation of temp. along the fin.
I am guessing the right side is the rate of heat transfer on the tip? which equals to the rate of heat transfer at x=0 under steady state. but i failed to see how they are equivalent, I don't understand where did sqrt( hpkA)*(Tb-T_inf) came from. any help would be appreciated!

 
Attachments
  • Slide titled Fin Equation showing a volume element of a fin at position x with length delta x, cross-sectional area Ac and perimeter p, with conduction and convection heat rates labelled, followed by the derivation of the fin differential equation and the excess temperature form.
    upload_2018-4-23_4-36-37.png
    49.6 KB · Views: 5,941
  • Slide deriving fin heat transfer: general solution theta(x) = C1 e^(mx) + C2 e^(-mx), boundary conditions at fin base and tip, and the infinitely long fin heat rate Q = sqrt(hpkA_c)(Tb - Tinf) with a fin diagram.
    upload_2018-4-23_4-37-2.png
    48.2 KB · Views: 2,331
  • Handwritten derivation of a fin temperature solution: T(x) minus T-infinity equals C1 e^(mx) plus C2 e^(-mx), with boundary conditions at x=0 and as L goes to infinity giving C1 = 0, leading to T(x) minus T-infinity equals C2 e^(-mx) and a ratio of temperature differences equal to e^(-mx).
    upload_2018-4-23_5-8-11.png
    12.1 KB · Views: 1,524

Discussion

Your second equation should be θb = θ(0) = C1 + C2, instead of 2. and since C1 = 0, C2 = θb. Now when you differentiate θ(x) wrt x and multiply both sides by -kA at x=0, it should work out OK.
 
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Dr Dr news said:
Your second equation should be θb = θ(0) = C1 + C2, instead of 2. and since C1 = 0, C2 = θb. Now when you differentiate θ(x) wrt x and multiply both sides by -kA at x=0, it should work out OK.
thank you, I got it now. the reason we differentiate θ(x) is because T_inf is treated as a constant right? therefore it is same as dT/dx.
 
Last edited:
That is correct.
 
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