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Conduction and Convection across multilayers

  1. Apr 27, 2004 #1


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    I am working on a project that requires me to calculate the heat dissipation of a heatsink attached the a laser optic mirror. As the whole system is moving along a rail at 10m/s (average, 20m/s at fastest point) therefore forced convection takes place on the heatsink surface. I have modelled the heatsink as a solid block of pure aluminium and the mirror as pure silicon. The mirror absorbs 1% of a 300W laser, and is operating at a temperature that is similar to that of a CPU approx 100 deg C.

    Anything with Subscript 1 is the mirror, 2 is the mirror-heatsink boundary, 3 is the heatsink-air boundary

    Material Properties:
    Heatsink (Al) - thermal conductivity, k_2 = 247W/m-K
    Mirror (Si) - thermal conductivity, k_1= 2W/m-K

    Area of mirror, A_m = 1.9635e-3 m^2
    Length of mirror, L_1 = 10e-3m

    Area of Sink, A_h = 2.5e-3 m^2
    Length of Sink, L_2 = 25e-3m

    Velocity of Air = 10m/s

    The equations for a multilayer slab, are given below as 1 and 2.

    [tex] \dot Q=\frac{T_1-T_2} {L_1/(k_1A_m)} [/tex] ...... 1
    [tex] \dot Q=\frac{T_2-T_3} {L_2/(k_2A_m)} [/tex] ...... 2

    so by eliminating [tex] T_2 [/tex] we get,

    [tex] \dot Q=\frac{T_1-T_3} {L_1/(k_1A_m)+ L_2/(k_2A_m)}[/tex]

    Expressing in terms of T_3

    [tex] \dot Q+\left[ \frac{L_1} {k_1A_m} + \frac{L_2} {k_2A_m} \right] =T_1 - T_3[/tex]

    Why does rearranging for T_3 make Qdot +ve???? as on http://www.mayahtt.com/tmwiz/ [Broken] under the Conduction Calculator -> Conduction across a multilayer plate?

    I have worked out the temp by hand and with the calculator, i get 107.7 deg C but that is only if i change the sign of the Qdot

    For calculating the heat dissipation i have these equations,

    [tex] \dot Q=\frac{T_2-T_3} {L_2/(k_2A_h)} [/tex] ...... 3

    [tex] \dot Q=\frac{T_3-T_\infty} {1/(hA_h)} [/tex] ...... 4

    Eliminating T_3

    [tex] \dot Q=\frac{T_3-T_\infty} {L_2/(k_2A_h)+ 1/(hA_h)}[/tex]

    Standard Air assumptions for h @ 20 deg C

    I get a value of 41.4 W for dissipation.

    Im sure ive gone wrong somewhere but i cant for the life of me see where!!!
    Last edited by a moderator: May 1, 2017
  2. jcsd
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