The Nusselt number is defined as $$\text{Nu}=\frac{h\delta}{k},$$ with

  • $h$ as convection heat transfer coefficient,$\text{W}/(\text{m}^2\text{K})$,
  • $\delta$ as characteristic length, $\text{m}$, and
  • $k$ as thermal conductivity, $\text{W}/(\text{m}\text{K})$.

In words: "Therefore, the Nusselt number represents the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction across the same fluid layer." from [1]. This is theoretically illustrated in e.g. Wikipedia.

In other words: “The ratio of the actual heat transfer to that by conduction.” from Coulson & Richardson Volume 1

The above descriptions lead to the conclusion that $\text{Nu} \geq 1$ must hold, where $\text{Nu}=1$ corresponds to pure conduction (no convection).

However, when choosing a small enough characteristic length in empirical correlation equations, $\text{Nu} < 1$ is possible. For example, when calculating the natural convection around a melting aluminum wire in air with a diameter of 0.3mm as the characteristic length $\delta$, various empirical correlations from e.g. Cieśliński, 2021 provide Nusselt numbers significantly smaller than one. For example, "one of the most famous" correlations, the one from Churchill and Chu (here used to model the wire as a horizontal cylinder), yields $\text{Nu} = 0.75$, while even making sure that the Rayleigh number is well within the valid range. This is confusing. Should such numbers be rounded up to 1?

[1] Ҫengel and Turner, 2001, Fundamentals of Fluid Sciences, IE


1 Answer 1


The problem is in the interpretation of the Nusselt number as the ratio of convective heat transfer to conductive heat transfer. The Nusselt number is, more correctly, the convective heat transfer coefficient reduced to dimensionless form, or, in other words, the dimensionless convective heat transfer coefficient.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.