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I Calculating thermal resistance across a metal flange

  1. Mar 8, 2016 #1
    I've been using the following equation for thermal resistance to make basic calculations for a few different engineering designs: $$R_t = \frac{L}{kA} $$
    Where L is the resistance path length, A is the cross-sectional area perpendicular to this path and k is the thermal conductivity of the material (in my case, it is 21.4 W/m⋅K). I ran into an issue with calculating resistance for heat traveling from the inside diameter of a metal flange to the outside. This flange is essentially a large, thin washer, with ID 1.13", OD 1.90" and a thickness of .060". With everything else I was doing, my L was linear, and my A was constant, or at least made up of a finite number of constant areas that could be added in series. If I was calculating the thermal resistance through the axial direction of this flange, it would be a simple task, but as I was trying to calculate it radially, I figured I needed to integrate Rt from the inside radius surface to the outside radius surface, but I think my difficulty has been in setting up the correct integral.

    I know that the cross sectional area at a given radius is essentially that of a cylinder: circumference x height, and I assumed that since I was trying to integrate from the inner radius (.565") to the outer (.950") that the integral would be in polar coordinates, giving me:
    $$R_t = \int_{r_0}^{r_1}\frac{L (rdr)}{k (2\pi rh)} $$
    where h is the thickness (.060") and L is the difference between my inner and outer radius (.385"). Everything moves to the outside of the integral except rdr/r, which leaves me simply with a factor of r after integration, which is evaluated between the inside and outside radius. This is the same thing as my L, so my resistance would seem to be ##R_t = L^2/2\pi kh ##, and this evaluates to 0.724 K/W (once I converted my k value to 0.543 W/in⋅K). The problem with this answer is twofold: for one, I was unsure how to analyze the dimensionality of my final R expression, since I had done integration, but on the surface, it doesn't lead to units of K/W. Secondly, while the final number seems more or less appropriate, a calculation of R assuming a constant cross sectional area of .358in2 (the area at the outermost radius) gave me a minimum possible R value of 1.98 K/W, even higher than what I got.

    I then considered that since my L was simply r1-r0, I may need to turn it into an additional r term in the integrand numerator, giving me $$R_t = \frac{1}{2\pi kh}\int_{r_0}^{r_1}\frac{r^2 dr}{r} $$
    Which evaluates to ##R_t = ({r_1}^2 -{r_0}^2)/4\pi kh ##, leaving me with 1.42 K/W--closer, but still below the minimum value.

    At this point, I am wondering if I am way off on my attempts to set up an integral, since my dimensions don't appear to match up. Additionally, I'm not sure I should be using polar coordinates in this case, or if my underlying R = L/kA formula is even applicable in this scenario. Would appreciate it if someone could point me in the right direction with this!

    [Update]: Interestingly, an R expression of ##[ln(r_1) -ln(r_0)]/2 \pi kh ## gives me the correct units and a realistic answer of 2.54 K/W, but that would require a 1/r term in my integrand, and I don't see why that would be appropriate.
    Last edited: Mar 8, 2016
  2. jcsd
  3. Mar 8, 2016 #2
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