- #1

Jasones

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So, in

Nu=[itex]\frac{hD}{k}[/itex]

..but what are the dimensions of

So far, I've worked out that the units for h are J/(m

Am I wrong is thinking that J/(m

Nu=[itex]\frac{hD}{k}[/itex]

*h*is the heat transfer coefficient, and*D*is the diameter of the pipe in which heat transfer takes place.....but what are the dimensions of

*k*in terms of mass (M), length (L), time (T) and temperature ([itex]\theta[/itex])?So far, I've worked out that the units for h are J/(m

^{2}hr C), but I'm having difficulty expressing k in terms of what is given.Am I wrong is thinking that J/(m

^{2}hr C) can be expressed as W/(m^{2}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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