Dimensions of k in Nusselt Number

AI Thread Summary
The discussion focuses on determining the dimensions of thermal conductivity (k) in the context of the Nusselt number equation, Nu = hD/k. The heat transfer coefficient (h) is correctly identified as J/(m² s K) or W/(m² K) in SI units. The participant clarifies that the hour (hr) is not a standard SI unit, which affects the dimensional analysis. Consequently, thermal conductivity (k) is concluded to be expressed as W/(m K), and in MLTΘ notation, it is represented as MLT⁻³Θ⁻¹. This analysis aids in understanding the relationships in heat transfer calculations.
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Homework Statement


A common dimensionless group used in heat transfer calculations is defined as:

Nu=\frac{hD}{k}

where h is the heat transfer coefficient, and D is the diameter of the pipe in which heat transfer takes place. Please determine the dimensions of the quantity k in terms of mass (M), length (L), time (T) and temperature (\theta) and in terms of SI units.

Homework Equations



Nu=\frac{hD}{k}

The Attempt at a Solution



So far, I've worked out that the units for h are J/(m2 hr C), but I'm having difficulty expressing k in terms of what is given.

Am I wrong is thinking that J/(m2 hr C) can be expressed as W/(m2 K)? I'm thinking that this is relevant to the end units of k, but I am unsure if the dimensions are correct.
 
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Well, in SI units h, the heat transfer coefficient is given in J/(m² s K), or W/(m² K) as 1 W = 1 J/s. Hour (hr) is not one of the SI system's base units.

Therefore, k, the thermal conductivity, should be expressed as W/(m K) and in MLTΘ notation it is expressed as MLT⁻³Θ⁻¹.
 
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