Ask Question
27 October, 01:54

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

+1
Answers (1)
  1. 27 October, 03:50
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “By using the principle of dimensional consistency, what is the correct dimension for thermal conductivity in the SI? ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers