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So, I was studying some fin design in a heat transfer course , and then came the part where the efficiency is to be calculated, then I noticed that when he calculated the surface area ,enter image description here the sides of a rectangular fin weren't included, so I searched and I found out that it was neglected because it's a one dimensional system, I understand how will that add an extra dimension, but I don't understand how is this a one dimensional system to begin with, we have conduction in $x$ direction and convection in $y$ direction so that's $2$ dimensional.

Edit: I just watched a video where he said he neglected the sides because they have negligible area...makes sense but I am not sure that this is $100\%$ right because he also ignored the tip and we don't ignore it in our course

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This is treated as a quasi-one dimensional system because, if we label the horizontal direction as $x$ and the vertical one as $y$, then it is assumed that there are no temperature gradients in the $y$ or $z$ direction, so:

$$\frac{\partial T}{\partial y}=0, \frac{\partial T}{\partial z}=0$$

So the temperature in the $y$ and $z$-directions is considered uniform. To put it even more clearly, temperature is dependent only on $x$. Not doing so would require solving the 3D Fourier PDE:

$$\frac{\partial T}{\partial t}=\kappa\Big(\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}\Big)+\frac{\dot{Q}(x,y)}{c_p \rho}$$

Which is mathematically far more challenging (for stationary state: $\frac{\partial T}{\partial t}=0$)

Using the quasi-one dimensional approach is a good approximation as long as the fin is thin (compared to its length). The DE becomes:

$$\frac{d^2T}{dx^2}-\frac{ph}{kA}(T-T_{\infty})=0$$

(Where $p$ is the perimeter of the fin and $A$ the cross-section)

Differential equations need boundary conditions and for this problem we have two choices:

  1. Assume the tip temperature is:

$$T(L)=T_{\infty}$$ This tends to be true for very long fins.

  1. Assume that due to the small area of the tip, no convection takes place there and of course no conduction can take place either (it's the end of the fin!), which mathematically means:

$$\Big(\frac{dT}{dx}\Big)_{x=L}=0$$

Neither assumptions are really 100 % correct but here your teacher chose the second one.

Here's my own derivation of the fin problem.

I just watched a video where he said he neglected the sides because they have negligible area...makes sense but I am not sure that this is 100% right because he also ignored the tip and we don't ignore it in our course.

There's not need to ignore the sides. Read my derivation.

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  • $\begingroup$ ok, i will just tell you what i understood and tell me if i am right , what i thought he meant by dimension ( i was obviously wrong) is that the heat flows out in one direction either x or y, what he meant by 1 dimension though is that all our equations depend on the distance on 1 axis only ( say the x-axis). i have 1 last question though, he assumes that no temperature gradient in either y or z direction, yet there is convection in the y direction but not the z, why ? ( assuming that the area of the sides is not negligible) $\endgroup$ – user28324 Jul 28 '16 at 15:03
  • $\begingroup$ Convection only depends on surface area, heat transfer coefficient and temperature difference, $y$ and $z$ coordinates aren't needed for that. Please read my own derivation, the link at the end of my answer. Thank you. Remember also we're talking about a fin with constant cross-section! $\endgroup$ – Gert Jul 28 '16 at 15:09
  • $\begingroup$ yes, that's my main problem , why ignore convection from the sides if the directions doesn't matter, i asking this because when i asked my professor he said that's because it's 1 dimension, the weird thing was his derivation was like yours, the area of is P*dx, which actually includes the sides $\endgroup$ – user28324 Jul 28 '16 at 17:31
  • $\begingroup$ i first thought maybe my professor is wrong, but every single textbook ignores the sides but not the tip so it's not a matter of area, why? $\endgroup$ – user28324 Jul 28 '16 at 17:38
  • $\begingroup$ physics.stackexchange.com/questions/21448/… in this thread he discusses my problem which is why do textbooks use the second expression (with the corrected length), but i didn't quite understand him $\endgroup$ – user28324 Jul 28 '16 at 17:41

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