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Heat losses from pipes

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  1. Sep 16, 2018 at 5:46 AM #1
    Hi,

    I wanted to do some rough "back of the envelope"-calculations of heat losses from pipes, with water circulating in them, to compare to the results of various programs' simulations.
    I was not very successful as I was perhaps hoping to get rough estimates in the range 20-200% off, but right now I'm in the magnitude of 1000s times off from the simulations.

    I'm not entirely sure of if I simply don't know what I'm doing and I have the completely wrong approach, or if there is something I fail to see along the way.

    As a rough example a 15 mm pipe with an inner diameter of 13 mms:

    from:
    https://www.engineeringtoolbox.com/conductive-heat-loss-cylinder-pipe-d_1487.html
    Conductive heat loss through the wall of a cylinder or pipe can be expressed as

    Q = 2 π L (ti - to) / [ln(ro / ri) / k]

    I'm calculating the heat loss as W/m so I have:

    Q = 2 π * k * (T_inside - T_outside) / ln(ro / ri)

    The thermal conductivity of copper ~ 400 W/mK ( https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html)

    dT = 22 degrees Celsius
    ro = 7.5 mm
    ri = 6.5 mm

    Q = (2 π * 400 W/(m K) * 22 C) / ln(7.5 mm / 6.5 mm)
    = 386 385 W/m

    To get an idea of the magnitude I look at the following data to get an approximation. These are close to the various simulations I've seen, even if there are some larger differences at times.
    As an example the value estimated for 1/2" pipe with the same temperature difference is 26 W/m
    https://www.engineeringtoolbox.com/copper-pipe-heat-loss-d_19.html

    So I'm not even close and I'm stuck and don't know what the approach I should take.

    My questions therefore are:
    -What am I doing wrong and how should I do this the right way?
    -If I'm not directly doing anything wrong then how should I interpret these results?
     
  2. jcsd
  3. Sep 16, 2018 at 8:57 AM #2

    russ_watters

    User Avatar

    Staff: Mentor

    Welcome to PF!

    Where does the heat go when it leaves the pipes? What's that mechanism called? Did you do any calculations on that...?
     
  4. Sep 16, 2018 at 1:13 PM #3
    (Thank you! Been checking out the forum for years, but the questions are mounting so I decided to join =D )

    Sorry, now I didn't fully understand, but what I attempted to do was a sort of general estimation of the values I could expect by making a simple assumption of just conduction and steady-state. (An example attached)
    That example is quite a lot from the ones in my old schoolbooks, but the issue is now that it seems like this leads to some very, very different values to what would be seen in reality. Therefore I'm a bit confused.
     

    Attached Files:

  5. Sep 16, 2018 at 1:33 PM #4
    The convective resistance is going to be a lot higher than the conductive resistance. So with the convective resistance included, the estimated heat transfer rate is going to be much lower than with pure conduction. See some heat transfer texts for some typical values of the convective heat transfer coefficient for natural convection from a pipe.
     
  6. Sep 16, 2018 at 2:48 PM #5
    To expand on what @Chestermiller said, the right way is to calculate three heat transfer coefficients:
    1) The film coefficient from the fluid inside the pipe to the pipe inside wall.
    2) The conductive resistance of the pipe itself.
    3) The film coefficient from the outside wall of the pipe to atmosphere.

    Then add them up to get the total heat flow resistance, then bring in the temperature difference to get heat flow. Normally the heat flow resistance of copper pipe is so low that it can be neglected.
     
  7. Sep 16, 2018 at 7:23 PM #6

    Tom.G

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    Science Advisor

    The linked page shows the heat loss to free air.

    Your calculation shows heat loss thru the pipe wall ONLY; as if the the pipe was embedded in an infinite heat sink of zero thermal resistance. You must take into account the heat transfer from the pipe outer wall to the air, once you do that you will find the pipe outer wall is very nearly the same temperature as its inner wall.

    Cheers,
    Tom
     
  8. Sep 17, 2018 at 4:17 AM #7
    Thanks guys. Now I think I grasp it.

    I'll return later on with my new calculations. :)
     
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