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Cooling element optimizing -advanced help needed

  1. Sep 19, 2006 #1
    Here is the situation: We have 100w heatload coming from little die which has only 1cm^2 of surface area. We cool it with Ln2 as cold as possible (in normal atmospheric pressure) by using simple round container which is made of 99.99% copper. Lets assume that system is perfectly isolated so all the energy for Ln2 boiling comes from the 100w 1cm^2 surface area.

    Lets easen the situation by choosing cooling elements wall thickness 5mm so that we have to optimize only the bottom thickness. The inside bottom surface area is roughly 3,25*3,25*3,14=33cm^2 and the wall surface area inside 6,5*3,14*20-x (around 390cm^2).

    There are many holes in the task.

    First of all we should know the amount of wattage 1cm^2 of copper surface area can transfer to Ln2. This also depends on the distance from the heat source so the core surface area above the core moves energy much more efficiently to Ln2 than in upper parts of the pipe due the temperature difference.

    Lets assume the cooling element has smooth shining inside surface (bad for boiling but it makes sense for this task since the bottom thickness will be this way something else than "thinnest possible" since the extra surface area from upper section of the cooling element is needed for most cold result).

    I dont know what level of knoweledge this forum has but im thinking this as an "breakfast challenge" for some physics teacher or professor.

    I will give more information if there is anyone who has slightest idea how to calculate this (i know this goes 3-dimensional stuff when done right). Here is picture from the system:

    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Sep 19, 2006 #2
    The heat transfer of an annular pipe is:

    [tex] q_r = \frac {2 \pi L k (T_{s,1} - T_{s,2} ) } {ln (\frac{r_2}{ r_2})} [/tex]

    k- conduction coefficient ~= 401 W/m-C
    Ts,1 = Temperature at hot surface (C)
    Ts,2 = Temperature at cold surface (C)
    L = length of rod
    r = radius
    r2 = radius outter (M)
    r1 = radius inner (M)
    Last edited: Sep 19, 2006
  4. Sep 20, 2006 #3
    I think that would help us calculating the heatloss (heat coming trough the pipe from ambient to ln2) but isnt what im looking for here (im pretty sure there isnt ready formula for this one).

    I figured that this system might need to be divided in two parts: bottom and pipe. Then it might be easier to solve.
  5. Sep 20, 2006 #4


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    Hi Ilkka,
    It's not obvious to me what you're trying to determine. Are you trying to determine the total amount of heat that is going into the LN2 or are you trying to calculate the temperature of the 100 W heat source? If you're trying to determine what thickness the bottom of the container should be, what criteria do you have for the temperature that's allowed for the 100 W heat source? Also, are we to assume the supply of LN2 is to be replenished? If not, are you trying to determine how long the LN2 will last?
  6. Sep 20, 2006 #5
    Keyword: newton's law of cooing.
  7. Sep 21, 2006 #6
    We are trying to find out optimal thickness of the base (presented X in the picture) of this copper cooling element to get 100w heat source as cold as possible. The cooling element is kept full of Ln2 as it boils away.

    The picture i put in my first post should help a bit.
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