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27 October, 16:20

By using the principle of dimensional consistency, what is the correct dimension for thermal conductivity in the SI?

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  1. 27 October, 16:40
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    from Fourier law,

    q = - K (∆T/L)

    but q = Q/A,

    hence,

    K = (QL) / (AΔT)

    Q = Joules/s = ML²/T³

    L = L

    A = L²

    ∆T = Kelvin = K

    since K = (QL) / (AΔT)

    substitute the fundamental units into the above equation we have

    K = (ML²/T³) (L) / (L²K) = ML³/T³L²K = ML/T³K

    the thermal conductivity, K dimensionality = ML/T³K

    or mass lenght per cube unit seconds per Kelvin
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