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Homework Help: Newton's Law of Cooling for Multilayers

  1. Jul 9, 2012 #1
    1. A problem that I need to solve for class is asking me to solve Newton's Law of Cooling for a 4-layered system.

    The system is a cube with three layers (also cubic) around it. The inside cube and the three layers all start at the same temperature that is lower than the ambient temperature. Each layer has different thermal properties and the ambient temperature around the outside is held constant. We are not given any data about initial temperature etc. They just want us to submit an excel model.

    Even if you could explain a simplified solution for a two/three - layer system (so that I can see how to solve when there are two sources/drains of heat) that would be so helpful

    2. Newton's Law of Cooling

    T(t) = Tenv + [T(0) - Tenv]e^(-rt)

    T(t) is the temperature of the layer in question
    Tenv is the ambient temperature which is constant
    h is the heat transfer coefficient for two particular layers
    A is the area over which heat transfer occurs
    C is the heat capacity of the layer in question
    t is time

    3. I've attempted to modify a solution that I found regarding home insulation...

    Let the temperatures of the cube and the insulating layers (going outwards) be T1, T2, T3 and T4.

    I defined r values for each interface (with an intention to come back and work this through more precisely later): r1 for cube/layer 1, r2 for layer 1/layer 2, r3 for layer 2/layer 3, r4 for layer 3/environment

    I got some simultaneous equations:

    T1(t) = T2(t) + [T1(0) - T2(t)]e^(-r1t)
    T2(t) = T1(t) + [T2(0) - T1(t)]e^(-r1t) + T3(t) + [T2(0) - T3(t)]e^(-r2t)
    T3(t) = T2(t) + [T3(0) - T2(t)]e^(-r2t) + T4(t) + [T3(0) - T4(t)]e^(-r3t)
    T4(t) = T3(t) + [T4(0) - T3(t)]e^(-r3t) + Tenv + [T4(0) - Tenv]e^(-r4t)

    Is this the right direction to go in? Or am I mental?
  2. jcsd
  3. Jul 9, 2012 #2
    It may be useful for you to look at an energy balance. The total energy will be conserved in this system, with the heat transfer through one layer equal to the heat transfer through each subsequent layer. Look up Fourier's Law of Thermal Conduction.
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