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Transient solution for a semi-infinate solid from a pipe

  1. Oct 20, 2009 #1

    cgw

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    Does anyone have a solution for the heat transfer from a pipe into a solid. I am looking for the temperature in the solid based on radius and time and also the heat transfer based on time. The temperature in the pipe (surface of the solid) is constant. The solid has a constant lower temperature at t=0.
    (I have the functions for a flat wallbut can not determine how it would work radially)
     
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  3. Oct 20, 2009 #2

    minger

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    So, the pipe has a constant temperature and you want a the temp of the surrounding material? Treat it as a circular cross section. Give the size of your material a large "radius" and impose the pipe temperature at the materials ID. Then just give it an outer ambient temperature at r>>ID. It's a straight forward conduction problem from there.

    I can look up the solution if you want, but since its hollow, I'm pretty sure there are no Bessel functions involved. I recall it being straight forward.
     
  4. Oct 21, 2009 #3
    if it's semi infinite can you model the pipe as the intersection between a semi infinite cuboid and semi infinite cylinder (using the outer rad of the pipe) and the solid as the intersection between two semi infinite cuboids?

    this is coming from a class i did over a year ago but i hope it's some help. if the problem wasn't transient you could just use shape factors.
     
  5. Oct 21, 2009 #4
    This should be physically similar to the plane case, except you are solving the heat equation in cylindrical polar coordinates. Solve this for the area outside the pipe and use the two BCs and an IC.

    I would give this a go as I can't see it being too complicated, but before I try I want to clear up your use of the word "semi-infinite", which makes no sense to me in this geometry- to me this phrase would mean "from 0 to infinity" as opposed to "from - infinity to infinity", which is what I'd assume your pipe is. Is the pipe divided cross-sectionally into two halves??
     
  6. Oct 21, 2009 #5

    minger

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    Semi infinite is a term used to describe some heat transfer problems. For introductory transient heat transfer, you're typically taught two methods, lumped-capacity and semi-infinite solid. Applicability of these two methods are based on the Biot number:
    [tex]
    Bi = \frac{hL}{k}
    [/tex]
    In the case of lumped capacity, the Biot number is less than 0.1. This means that you approximately treat the entire solid as uniform temperature (i.e. thermal conductivity overwhelms the convection).

    The other case is where the Biot number is large. This approximately means that due to the size and heat transfer properties, one side of the body is essentially unaffected by the fluid and stays at constant temperature.

    In this case, yes you could treat this as a semi-infinite cylinder, where the outer radius is held at the freestream, or ambient temperature of the surrounding fluid, and the inner radius is suddenly changed from ambient to whatever temperature you're testing.

    As mentioned, if the pipe thickness is much less than the diameter, then you can treat this system as a plane wall. In this case, we can write the solution to the problem in non-dimensional form as:
    [tex]
    \theta = \frac{T-T_{\infty}}{T_i - T_{\infty}} = \sum_{n=1}^\infty C_n e^{\left[(-\zeta^2_n Fo)\cos(\zeta_n r^*)\right]
    [/tex]
    Where [tex]r^* = r/r_o[/tex] and [tex]Fo = \alpha t/r^2_o[/tex] and the coefficient:
    [tex]
    C_n = \frac{4\sin\zeta_n}{2\zeta_n + \sin(2\zeta_n)}
    [/tex]
    Where the zeta values are the positive roots of the transcedental equation:
    [tex]
    \zeta_n \tan\zeta_n = Bi
    [/tex]

    Or, you can treat it as a semi-infinite solid in which case you end up with things that look like:
    [tex]
    \theta = \mbox{erf} \left(\frac{r}{2\sqrt{\alpha t}}\right)
    [/tex]
    Where erf is the error function.
     
  7. Oct 21, 2009 #6
    I see... I was trained in mathematics so my method would be to take the heat equation in plane polars, solve it in the region r>a, the radius of the pipe, and use the constraints:

    T = T0 + Ti at r = a, for all theta, and t
    T -> T0 as r -> infinite, for all theta and t

    T at r>a = t0 for t = 0.


    so du/dt = del^2 u = (1/r)*d/dr (r*du/dr) + (1/r^2)*d^2u/dθ^2
    Not sure about the solution to this though, might take a lot of messing around.

    edit - doh, BCs are temperature, heat equation function is heat.
     
  8. Oct 21, 2009 #7

    cgw

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    [tex]
    \theta = \mbox{erf} \left(\frac{x}{2\sqrt{\alpha t}}\right)
    [/tex]
    is the solution to the one dimensional semi infinite solid. The radius of the pipe is too small to approximate it as a plane.
    The pipe wall thickness can be ignored for now (it is << than the solid).

    The function is
    [tex] \frac{1}{r} \frac{\partial}{\partial r} (kr \frac{\partial T}{\partial r} ) = \rho c \frac{\partial T}{\partial t} [/tex]

    I can barely follow how the one dimensional is solved (which is in any heat transfer text). I can not figure how to solve the radial. Someone must have done it.
     
  9. Oct 22, 2009 #8

    minger

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    Here is a 1D transient example in cartesian coordiantes. The equation of heat is:
    [tex]\frac{\partial^2 T}{\partial x^2} = \frac{1}{\alpha}\frac{\partial T}{\partial t}[/tex]
    The boundary conditions are:
    [tex]
    \begin{equation}
    \begin{split}
    T(x,0) &= f(x) \\
    T(0,t) &= 0 \\
    T(L,t) &= 0
    \end{split}
    \end{equation}
    [/tex]
    These are homogeneous boundary conditions, which are important. They will make life easier. Typically you can at the very least make one boundary condition homogeneous by imposing T* = T-T(0,t), etc.

    The key to solving these problems is separation of variables. We write the solution to the problem as a combination of functions:
    [tex]
    T(x,t) = \Psi(x)\cdot\Gamma(t) = \frac{\Psi''}{\Psi} = \frac{1}{\alpha}\frac{\Gamma'}{\Gamma} = \lambda^2
    [/tex]
    Where we are homoegenous in the x direction. We separate the equation into two:
    [tex]
    \begin{equation}
    \begin{split}
    {\Psi''} + \lambda^2\Psi = 0 \\
    {\Gamma'} + \alpha\lambda^2\Gamma = 0
    \end{split}
    \end{equation}
    [/tex]
    The solution to these two equations are:
    [tex]
    \begin{equation}
    \begin{split}
    \Psi &= A\sin\lambda x + B\cos \lambda x \\
    \Gamma &= C e^{-\alpha \lambda^2 t}
    \end{split}
    \end{equation}
    [/tex]
    We impose our boundary conditions
    [tex]
    \begin{equation}
    \begin{split}
    x&=0\rightarrow \Psi = B cos\lambda x = 0 \rightarrow B = 0 \\
    x &= L \rightarrow A\sin\lambda L = 0
    \end{split}
    \end{equation}
    [/tex]
    We combine the solutions for:
    [tex]
    T(x,t) = \sum_{n=1}^\infty A_n e^{-\alpha \lambda_n^2 t} \sin\lambda_n x
    [/tex]
    Where
    [tex]
    \lambda_n = \frac{n\pi}{L}\,\,\, n=1,2,3...
    [/tex]
    Imposing the initial condition:
    [tex]
    \begin{equation}
    \begin{split}
    &f(x) = T(x,0) \rightarrow f(x) = \sum_{n=1}^\infty A_n \sin\lambda_n x \\
    &A_n = \frac{2}{L} \int_0^L f(x) \sin\frac{n\pi}{L}x
    \end{split}
    \end{equation}
    [/tex]
    Which makes the final solution:
    [tex]
    T(x,t) = \frac{2}{L}\sum_{n=1}^\infty e^{-\alpha\left(\frac{n\pi}{L}\right)^2 t} \sin\frac{n\pi}{L}x \int_0^L f(x)\sin\frac{n\pi}{L}x\,dx
    [/tex]

    This may look difficult because it is. It's very difficult. This is graduate level mathematics. The class Conduction is merely a coverup; a way to teach partial differential equations without scaring people to dropping the class.

    Aside from that, this equation was homogeneous, and only 1D. Additional dimensions don't add too much difficulty, but imposing non-homoegenous boundary conditions really confuses things. I can post an example with non-homoegenous boundary conditions, but as said, I would draw your control volume such that you can just assume a semi-infinite solid. Use the error function.
     
  10. Oct 22, 2009 #9
    I don't think it is valid to use separation for the heat equation, is it? That suggests the fractional change in temperature over a time interval is the same at every point, where in real life, some initially cool parts may warm, and some initially warm parts may cool as heat is dissipated along the spacial dimension.

    Also, I don't see the validation for using a finite domain in your solution, which you've implicitly done by defining the L value? The initial temperature profile f(x) is not nessecarily periodic.


    The way I remember solving the heat equation was through a horrible non-dimensionalising process, and made use of Green's functions to eventually get a solution similar to the normal distribution. :\
     
  11. Oct 22, 2009 #10

    minger

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    I think perhaps you're thinking about the same problem, but with non-homoegeneous boundary conditions. In that case, you have Bessel functions. For spherical coordinates you end up with Legendre polynomials.

    What the hell I got time. For unsteady conduction in a cylinder with convection on the outside, the solution is:
    [tex]
    \theta(r,t) = T(r,t) - T_\infty = \frac{2}{r_o^2}\sum_{n=1}^\infty \frac{
    e^{-\lambda_n^2 \alpha t} J_0 (\lambda_n r) \int^{r_o}_0 F(r') J_0(\lambda_n r')r' \,dr'}
    {
    J_0^2(\lambda_n r_o) + J_1^2(\lambda_n r_l)}
    [/tex]
    Where the lambda values are found from the characteristic function:
    [tex]
    \lambda k J_1(\lambda r_o) - h J_0(\lambda r_o) = 0
    [/tex]
    and
    [tex]
    F(r) = f(r) - T_\infty
    [/tex]
    Where f(r) is the initial condition. If the initial condition is uniform temperature, then
    [tex]
    \int^{r_o}_0 F(r')J_0(\lambda_n r')r'\,dr' = \frac{r_o}{\lambda_n}J_1(\lambda_n r_o)(T_i - T_\infty)
    [/tex]
    Knowing that:
    [tex]
    \frac{d}{dr}\left[r J_1(\lambda r)\right] = \lambda r J_0(\lambda r)
    [/tex]
    Says that:
    [tex]
    \int^{r_o}_0 r J_0(\lambda r)\,dr = \left. \frac{1}{\lambda} r J_1(\lambda r) \right|^{r_o}_0 = \frac{r_o}{\lambda}J_1(\lambda r_o)
    [/tex]
    Or:
    [tex]
    \theta(r,t) = \frac{2}{r_o}\sum^\infty_{n=1}\frac{1}{\lambda_n}\frac{
    J_1(\lambda_n r_o) J_0(\lambda_n r)}{
    J_0^2(\lambda_n r_o) + J_1^2(\lambda_n r_o)}e^{-\alpha \lambda_n^2 t}
    [/tex]
    Using non-dimensional terms,
    [tex]
    \begin{equation}
    \begin{split}
    Bi &= \frac{r_o h}{k} \\
    Fo &= \frac{\alpha t}{r^2_o} \\
    \mu &= \lambda_n r_o
    \end{split}
    \end{equation}
    [/tex]
    The final solution can be elegantly written as:
    [tex]
    \theta(r,t) = 2\sum^\infty_{n=1} \frac{1}{\mu_n} \frac{
    J_1(\mu_n)J_0(\mu_n \frac{r}{r_o})}{J^2_0(\mu_n)+J_1^2(\mu_n)}e^{-\mu_n^2 Fo}
    [/tex]
    Where [tex]\mu_n[/tex] are the solutions of the characteristic equation:
    [tex]
    \mu J_1(\mu) - Bi J_0(\mu) = 0
    [/tex]
     
  12. Oct 23, 2009 #11

    cgw

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    My application is a pipe with a constant temperature fluid (water) buried in a solid (the ground) with a initial (lower) constant temperature at t=0. What is the temperature at time t and distance r? When will the heat transfer drop off to a certain rate? I though this would simple and the answer readily available. I guess I was wrong. After much looking I did find and expression (in a text by Ozisik). He used separation of variables and the expression involves Bessel functions and is too general for my practicle use.

    Minger - I wrote this without realy looking at your last reply. I will do so now.
     
    Last edited: Oct 23, 2009
  13. Oct 26, 2009 #12

    minger

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    If you can simplify, then your best bet is the error function.
    [tex]\theta = \mbox{erf} \left(\frac{r}{2\sqrt{\alpha t}}\right)[/tex]

    Assume that the pipe is the temperature of the fluid and then just calculate the thermal diffusivity (alpha) and there you go. From \theta (fractional change in temperature) you can calculate heat transfer.
     
  14. Oct 26, 2009 #13

    cgw

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    I am now thinking that the boundry layer makes a significant difference. I am trying a finite difference with small delta r and small delta t on a spread sheet. It is limited but at least the numbers seem right.
     
  15. Oct 27, 2009 #14

    minger

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    There is no moving fluid, there is no boundary layer.
     
  16. Oct 27, 2009 #15

    cgw

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    Any practical application I can think of would have a moving fluid in the pipe. Maybe a constant temperature at the pipe surface would approximate the situation close enough. One of the applications I am looking at is the heat transfer difference when using a copper pipe vs. a plastic pipe.
     
  17. Oct 27, 2009 #16

    minger

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    I thought you were interested in the temperature outside of the pipe?

    Are you familiar with the concept of thermal resistances? You can quickly apply thermal resistances to see the difference between materials.
     
  18. Oct 27, 2009 #17
    OT to minger: jeez, man do you think in LaTex or what? I imagine it would take me a couple of hours just to type that in, and that's after I figured it out in handwritten math...
     
  19. Oct 27, 2009 #18

    minger

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    Write one thesis (just one) and you'll be a master ;)
     
  20. Oct 27, 2009 #19
    Yeah someone told me its so much faster to type in latex than to use any other software, once you ave learnt it well enough... plus it looks 10x better than word!
     
  21. Oct 28, 2009 #20

    minger

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    It's very nice because you don't have to take your hands off the keyboard. Granted the backslash "\" is in a funky spot on the keyboard, so every now and then I need to take a break to give my pinky a break.
     
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