1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fin/Extended surface differential equation for temperature

  1. Nov 15, 2015 #1
    I'm trying to deduce the differential equation for temperature for a triangular fin:
    phpjpZDsu.png

    I know that for a rectangular fin, such as:
    k1WrY.png

    I can do:
    Energy entering the left:
    [tex]q_x= -kA\frac{dT(x)}{dx} [/tex]

    Energy leaving the right:
    [tex]q_{x+dx} = -kA\frac{dT(x)}{dx} - kA\frac{d² T(x)}{dx²}dx [/tex]

    Energy lost by convection:
    [tex]dq_{conv} = h_eA(T-T_e)[/tex]
    [tex]dq_{conv} = h_eP(T-T_e)dx [/tex]

    [tex]q_x - q_{x+dx} - dq_{conv} = 0 [/tex]

    [tex]q_x = q_{x+dx} + dq_{conv}[/tex]

    [tex]-kA\frac{dT(x)}{dx} = -kA\frac{dT(x)}{dx} - kA\frac{d²T(x)}{dx²}dx + h_eP(T-T_e)dx[/tex]

    [tex]kA\frac{d²T(x)}{dx²}dx = h_eP(T-T_e)dx[/tex]

    [tex] \frac{d²T(x)}{dx²} = \frac{h_eP(T-T_e)}{kA}[/tex]

    But I don't understand why A = Pdx on the energy lost by convection, so I don't know how to adapt that for a triangular fin.

    Seems to me that T(x,y,z) still changes predominantly on the x axis, so I assume the other two equations remain unchanged.
     
  2. jcsd
  3. Nov 15, 2015 #2
    There are two areas involved here, one perpendicular to the flow of heat in the x direction (which you call A), and the other perpendicular to the flow of heat from the fin surface into the surrounding air (which is Pdx). For a fin with triangular shape, A is decreasing with x, and P is about constant. Also, since A is a function of x, when you do your heat balance on a differential section of fin, the area A must stay inside the heat flux derivative.

    Chet
     
  4. Nov 15, 2015 #3
    So this is correct
    121rdht.png
    ?

    [tex]q_x = -kA(x)\frac{dT(x) }{dx}[/tex]
    [tex]q_{x+dx} = -kA(x)\frac{dT(x) }{dx} - k\frac{dA(x)}{dx}\frac{d²T(x)}{dx²}[/tex]
    [tex]dq_{conv} = h_eP(T(x)-T_e)dx[/tex]
    [tex]-kA(x)\frac{dT(x) }{dx} = -kA(x)\frac{dT(x) }{dx} - k\frac{dA(x)}{dx}\frac{d²T(x)}{dx²} + h_eP(T(x)-T_e)dx [/tex]
    [tex] k\frac{dA(x)}{dx}\frac{d²T(x)}{dx²} = h_eP(T(x)-T_e)dx [/tex]
    [tex]\frac{dA(x)}{dx}\frac{d²T(x) }{dx²} = \frac{h_eP(T(x)-T_e)}{k}[/tex]
    [tex]T(x)-T_e = θ(x)[/tex]
    [tex]m² = \frac{h_eP}{k}[/tex]
    [tex]\frac{d²T(x)}{dx²} = \frac{d²θ(x)}{dx²}[/tex]
    [tex]\frac{dA(x)}{dx}\frac{d²θ(x) }{dx²} = m²θ(x)[/tex] (1)

    What I've posted here is just a portion of an exercise I was doing, I would then have to solve the ODE to find an equation for T(x). So now I need to get rid of this derivative of A(x). To be honest the only thing I can think of to proceed from here is to find an equation for A(x) and derivate it.

    otg6bk.png
    [tex]A(x) = 2h(x)P[/tex]
    [tex]c² = h² + x²[/tex]
    [tex]h(x) = sqrt(c²-x²)[/tex]
    [tex]A(x) = 2sqrt(c²-x²)P[/tex]
    [tex]\frac{dA(x)}{dx} = \frac{-2Px}{sqrt(c²-x²)}[/tex]
    [tex]\frac{d²θ(x) }{dx²} = -\frac{m²θ(x)sqrt(c²-x²)}{2Px}[/tex]

    And this is feeling too awkward to be correct, lol. Can you integrate back at (1)?
    ------
    Just thought of
    [tex]n²(x) = -\frac{m²sqrt(c²-x²)}{2Px}[/tex]
    [tex]\frac{d²θ(x)}{dx²} = n²(x)θ(x)[/tex]
    [tex]\frac{d²θ(x)}{dx²} - n²(x)θ(x) = 0[/tex]
    [tex]θ(x) = c_1e^{nx} + c_2e^{-nx}[/tex]
    I know how to continue from here, but I'm having the sensation that this is wrong.
     
    Last edited: Nov 15, 2015
  5. Nov 15, 2015 #4
    If P is the dimension shown on your diagram, then the differential heat balance should read:
    $$k\frac{\partial}{\partial x}\left(A(x)\frac{\partial T}{\partial x}\right)-2Ph(T-T_e)=0$$
    Also, from the figure, A(x) = P t(x), where t is the thickness of the fin at location x. So:
    $$k\frac{\partial}{\partial x}\left(t(x)\frac{\partial T}{\partial x}\right)-2h(T-T_e)=0$$
     
  6. Nov 18, 2015 #5
    I don't think I expressed myself properly before but anyway, finally managed to get it in the way I wanted:
    http://i.share.pho.to/0a247f78_o.jpeg
    http://i.share.pho.to/57739467_o.jpeg
    [tex]T(x) - T_e = θ(x) [/tex]
    [tex]m = \frac{h_e(L/e)(1+(e/L)²)^{1/2}}{k} [/tex]
    [tex]xθ'' + θ' - mθ = 0[/tex]
    Just need to solve that now.
     
  7. Nov 18, 2015 #6
    This is an Euler differential equation.
     
  8. Nov 18, 2015 #7
    Wouldn't it have to be
    [tex]x²θ′′+xθ′−mθ=0[/tex]
    for it to be an Euler differential equation?

    I tried to do power series but I got stuck in the end:
    [tex]y=\sum_{n=0}^\infty a_k x^k[/tex]
    [tex]θ'=\sum_{n=1}^\infty ka_k x^{k-1}[/tex]
    [tex]θ''=\sum_{n=2}^\infty k(k-1)a_k x^{k-2}[/tex]
    [tex]\sum_{n=2}^\infty k(k-1)a_k x^{k-1} + \sum_{n=1}^\infty ka_k x^{k-1} - \sum_{n=0}^\infty m²a_k x^k = 0[/tex]
    [tex]\sum_{n=1}^\infty k(k+1)a_{k+1} x^{k} + \sum_{n=0}^\infty (k+1)a_{k+1} x^{k} - \sum_{n=0}^\infty m²a_k x^k = 0[/tex]
    [tex]a_1 - m²a_o + \sum_{n=1}^\infty k(k+1)a_{k+1} x^{k} + \sum_{n=1}^\infty (k+1)a_{k+1} x^{k} - \sum_{n=1}^\infty m²a_k x^k = 0[/tex]
    [tex]a_1 - m²a_o = 0[/tex]
    [tex]a_1 = m²a_o[/tex]
    [tex]\sum_{n=1}^\infty k(k+1)a_{k+1} x^{k} + \sum_{n=1}^\infty (k+1)a_{k+1} x^{k} - \sum_{n=1}^\infty m²a_k x^k = 0[/tex]
    [tex]\sum_{n=1}^\infty[k(k+1)a_{k+1} x^{k} + (k+1)a_{k+1} x^{k} - m²a_k x^k]= 0[/tex]
    [tex]\sum_{n=1}^\infty[k(k+1)a_{k+1} + (k+1)a_{k+1} - m²a_k]x^{k}= 0[/tex]
    [tex]\sum_{n=1}^\infty[(k+1)(ka_{k+1} + a_{k+1}) - m²a_k]x^{k}= 0[/tex]
    [tex]\sum_{n=1}^\infty[(k+1)(k+1)a_{k+1} - m²a_k]x^{k}= 0[/tex]
    [tex]\sum_{n=1}^\infty[(k+1)²a_{k+1} - m²a_k]x^{k}= 0[/tex]
    [tex](k+1)²a_{k+1} - m²a_k= 0[/tex]
    [tex]a_{k+1}= \frac{m²a_k}{(k+1)²}[/tex]
    [tex]k = 1, a_2 = \frac{m²a_1}{4} = \frac{m^{4}a_0}{2!}[/tex]
    [tex]k = 2, a_3 = \frac{m²a_2}{9} = \frac{m^{6}a_0}{(3!)²}[/tex]
    [tex]k = 3, a_4 = \frac{m²a_3}{16} = \frac{m^{8}a_0}{(4!)²}[/tex]
    [tex]k = 4, a_5 = \frac{m²a_4}{25} = \frac{m^{10}a_0}{(5!)²}[/tex]
    [tex]a_{2k}= \frac{m^{4k}a_0}{(2k!)²}[/tex]
    [tex]a_{2k+1}= \frac{m^{2(2k+1)}a_0}{((2k+1)!)²}[/tex]
    [tex]θ(x) = a_0\sum_{n=0}^\infty \frac{m^{4k}x^{2k}}{(2k!)²} + a_0\sum_{n=0}^\infty \frac{m^{2(2k+1)}x^{2k+1}}{((2k+1)!)²}[/tex]
     
  9. Nov 18, 2015 #8
    Ooops. Sorry. My mistake.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Fin/Extended surface differential equation for temperature
  1. Differential equation (Replies: 6)

  2. Differential Equation (Replies: 1)

  3. Differential equation (Replies: 5)

Loading...