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Extended Surfaces (Fins)

  1. Mar 17, 2016 #1
    I am working on a project and I am having difficulty understanding a concept

    I have to analyze a rectangular fin in 2 cases (Adiabatic tip AND Convective tip) and I am having difficulty understanding which surface area to use.

    For the first case, I want to find Afin which according to my book is:
    Afin = 2wL + wt -> why does it not include the lateral sides (2*L*t)? It's just the top, bottom and front surface.

    Schematic: http://imgur.com/YgfHmUw

    Same thing with the corrected length Lc.
    The area with the corrected length is Afin=2wLc -> but this now only includes the top and bottom surface (i get that I don't include the tip because it's adiabatic) but still no lateral sides.

    My professor taught me this formula: Afin = P*L
    where P is perimeter. This formula includes all 4 sides and then we can add the tip cross section if needed.

    Someone please explain the difference.
  2. jcsd
  3. Mar 18, 2016 #2
    When you Say fin you Mean something that is Not finite along w director. This allow You to considerazione The problem as mono dimensional.
  4. Mar 18, 2016 #3
    I know that it is assumed in my case that heat transfer is 1 directional (normal to the cross section) however that does not explain why the lateral sides are not included.

    please see this http://physics.stackexchange.com/questions/244253/finned-surface-corrected-length/ [Broken], it explains everything I am confused about.
    Last edited by a moderator: May 7, 2017
  5. Mar 18, 2016 #4
    They are Not included because those sides are Not suppose to Loss any heat. Normaly The heat losses for a fin are espressed per unit length. If you consider a finite fin on the directIon w you got to do some aproximations to use The same relations you Will use for an infinite fin. U also have to do An aproximation to use The same relation u use for An adiabatic tip fin with a convertive tip fin .
    Maybe you have to consider The differenze between a fin And a pin..
  6. Mar 18, 2016 #5
    In the simplest heat transfer model of a fin, only the convective heat transfer on the top and bottom surfaces are included, and the convective heat transfer on the sides is neglected (i.e., the sides that are L x t). This problem is asking you to determine how the answer changes when you include the convective heat on the sides (using the same convective heat transfer coefficient as on the top and bottom). They want you to show that the answer you get is the same as if you neglected the convective heat transfer on the sides, and increased the length of the fin by t/2.
  7. Mar 18, 2016 #6
    Thank you for the explanation. However my professor taught us that the "easiest" way to calculate Afin=PL - where P is perimeter of cross section.
    If I am neglecting the sides, then that formula should not apply. Yet, I've seen it used with rectangular fins (not just circular) hence it's confusing. One side is telling me to exclude the side area, while the formula taught does include it.
  8. Mar 18, 2016 #7
    Let's see the results of your analysis when you include the sides. Then I'll show you how to get the results that the problem statement is looking for.

  9. Mar 18, 2016 #8
    Here's the calcs I did, and the diagrams I was referring to ("book" and "table")

    Attached Files:

  10. Mar 18, 2016 #9
    OK. Your analysis had numbers plugged in and is not done algebraically. So it is impossible to relate to your problem statement. So here is my differential equation:

    $$ktW\frac{d^2T}{dx^2}-2h(W+t)(T-T_0)=0$$where ##T_0## is the outside bulk temperature.

    Is this consistent with your differential equation? Once we get agreement on the differential equation, we can proceed further.

  11. Mar 18, 2016 #10
    I've never seen that formula before....and to be honest we never really start with a differential equation. The only time we've derived is when we start with Fourier's Law of Conduction (Qx = kAc dT/dx) and the definition of a derivative, in order to find the temperature distribution equation T(x)

    We jump straight into Area cals and then go from there
  12. Mar 18, 2016 #11
    Here's the scope of the project and my initial calcs, this is the same way we do it in class.

    Attached Files:

  13. Mar 18, 2016 #12
    Well this equation is based on Fourier's law. I've never seen a fin analyzed without starting with the heat balance differential equation. So, I'll present the solution to the equation, and see if you can relate to it. Right now, gotta watch a Michigan basketball game. So, be back tomorrow.

  14. Mar 18, 2016 #13
    No worries, I appreciate the help.
  15. Mar 19, 2016 #14
    The differential equation in post #9 can readily be rewritten as:
    $$\frac{d^2(T-T_0)}{dx^2}-m^2(T-T_0)=0\tag{1}$$where $$m=\sqrt{\frac{2h}{kt}(1+t/W)}\tag{2}$$Note the factor of (1 + t/W) in Eqn. 2, which you were lamenting was omitted from the equation in your book. They simply neglected the t/W with respect to the 1. To make you happy, we will not neglect this term in our development.

    Eqn. 1 represents a 2nd order ordinary homogeneous differential equation with constant coefficients. For the adiabatic end case (at x = L), the boundary conditions are T = Tw @ x = 0 and dT/dx=0 at x = L, where Tw is the temperature at the base of the fin. The solution to Eqn. 1, subject to these boundary conditions is:

    From Eqn. 3, the temperature gradient at x = 0 is given by:
    So the rate of heat flow to the fin (and from the fin to the surroundings) is given by:
    The rate of heat flow from the fin to the surroundings is also given by:
    $$Q=h(2W+2t)L\eta(T_w-T_0)\tag{6}$$ where ##\eta## is defined as the fin efficiency. If we combine Eqns. 5 and 6, we obtain:
    $$h(2W+2t)L\eta=kWtm\tanh{(mL)}\tag{7}$$or equivalently,$$\eta=\frac{\tanh(mL)}{mL}\tag{8}$$

    This completes the analysis of the adiabatic tip case. To solve the problem where the fin has a convective tip, we need to modify the boundary condition at x = L to now read: ##-k\frac{dT}{dx}=h(T-T_0)## @ x = L. Do you think you can solve the differential equation for this modified boundary condition, and then determine the modified fin efficiency?
  16. Mar 19, 2016 #15
    Yeah....seems like I'm gonna fail this class because I have no idea what's going on. I am trying to follow your derivation and what my book has and nothing clicks.
    I get the idea of boundary conditions. When x=L my BC becomes Qcond = Qconv.

    My book doesn't offer any explanation at all. It just states the same about eq (1), it's a 2nd order ODE and the solution to it is:
    T(x) - To = (C1)e^(mx) + (C2)e^(-mx)
    Then it goes into different cases and their solutions (see second picture).

    My professor copies the text book, doesn't explain any of the steps and whenever we ask questions he just refers us to his notes or the book.

    I appreciate the help, but I'm at the point where I'm ready to say f*ck it, memorize the formulas and hope for the best.

    Attached Files:

  17. Mar 19, 2016 #16
    Hi. I'm sorry that you are struggling so much with this. It can be very frustrating.

    It isn't clear to me exactly where your difficulty lies yet. There are usually several steps to attacking a problem like this:

    1. Articulating in your own words the physical mechanisms involved
    2. Translating the description of the physical mechanisms into the language of mathematics (formulation of the equations).
    3. Solving the equations for the cases of specific interest.

    I am getting the sense that your main difficulty is item 3. This is because I noticed that, above, you were able to articulate the correct physical mechanism for the boundary condition at the tip.

    I looked over the solution to the differential equation that you attached to your post. It was exactly the same result for the temperature profile that I obtained, in exactly the same form. But, I feel that they left out several steps that would have been helpful to you.

    Please tell me where you want me to focus my efforts:

    (1) In elaborating on the derivation of the differential equation or
    (2) In elaborating on how to obtain the solution to the differential equation subject to the boundary conditions

    I can help you in both these areas.

    Sorry again for your frustration.

  18. Mar 19, 2016 #17
    Sir you have nothing to apologize for, the one who should be apologizing is my professor for simply feeding us the final formulas without any derivation or explanation for that matter. We did not cover the derivation in class, and as you mentioned, the book leaves out a lot of steps so it's not clear.

    A friend explained it to me this way: If it's adiabatic then there is no need to include the front face (tW), and the lateral sides (tL) are not included either way because they are so small compared to the rest of the surface area. Another thing he thought of was this: the shape was cut in two arbitrary sections exposing the lateral sides. In practice, that may not be the case and those sides may never be exposed at all hence they're not included.

    I spent 3 days trying to understand this by going to see my prof and reading the book, neither helped. This explanation makes the most sense to me. So I will just leave it at that, practice as many examples as I can.

    I have a feeling I will have to retake this either way because my professor is one of those people that makes tests unreasonably long and difficult, with questions we've never seen or discussed in class. The average on his tests is 30-40% and to him that's high. The most frustrating thing for me isn't the lack of clarity in the material, it's the fact that even if I do grasp the concept he's still a ridiculously tough marker as if he does it on purpose.

    Again, I appreciate the help but I cannot afford to keep going in circles around this topic when I have exams in 2 weeks.
  19. Mar 19, 2016 #18
    I understand. My offer to help you still stands. I think you are very close to overcoming the barriers. If you find that you have some extra time later on and feel you want to continue, I am ready to assist.

  20. Mar 19, 2016 #19


    User Avatar
    Gold Member

    QUOTE="ksukhin, post: 5415108, member: 412043"]
    A friend explained it to me this way: If it's adiabatic then there is no need to include the front face (tW), and the lateral sides (tL) are not included either way because they are so small compared to the rest of the surface area. Another thing he thought of was this: the shape was cut in two arbitrary sections exposing the lateral sides. In practice, that may not be the case and those sides may never be exposed at all hence they're not included.

    Take a look at m, where m2=kP/kA
    P being the perimeter
    A is the area for conduction

    For a fin of length L, width w, and thickness t, extending L in the x-direction out from the body, we take a differential element of size dx at position x.

    then we have for this element, Hin, Hout are by conduction into and out of the element dx; Hconv is convection to the fluid (air) over the distance dx

    then adding up all the heat flows we get,
    Hin at x + Hout at x+dx Hconv over width dx

    Hin = -kA dT/dx
    Hout = -kA dT/dx + -kA d/dx ( dt/dx ) dx = -kA dT/dx - kA (d2T/dx2) dx
    Hconv = h(Pdx) (T-T∞[

    At steady state, there is no energy storage in the fin, so we can add up the terms to equal zero, and voila,
    we get the equation as shown in step (1)

    Perimeter P can include or not include the sides.
    If you look at equation (2) for m, there is the term t/w. When t is a small fraction of w, the amount of heat lost by the sides for any fin becomes negligable. Try putting some t/w in the square root and see how much error you might get. Try for example 1/10 ( a t that is 1/10 the of w ).

    If one has an infinite width of fin and does an analysis of of heat flow per unit width ( to be then exteded over the width of the fin to get the total heat flow ), for the unit width the heat flow into and out of the sides is zero due to the fact that there is now no temperature gradient in the z-direction along the width ( except at the extreme edges where it should then surely be negligable ).

    In post 8, you showed some calculation for P. P is just that - a perimeter. you seem to have added some extra "perimeters" in there ir error.

    The rest of the mathematical analysis calculus of finding the complimentay slotion and particular solution of the equation.
    One ends up with the equation in your book,
    T-T = C1 e-mx + C2 emx

    Your book has T and T (Tambiant ) switched around so it looks a liitle different, but it is the same thing.

    Now, one adds in the boundary conditions, for the three cases, and you get the final equations with the hyperbolic terms, which are the specific solutions for fin heat flow.

    What you book could have done is add an appendix showing the steps form the differential equation to the compilimentary and particular solution, to the specifc solutions.
    But, those steps should be covered in an engineering or math calculus course. Check your books there to see if you have derivations of second order differential equations.

    Does that help.
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