For turbulent flow inside a plain tube, the working correlation is $\mathrm{Nu} = 0.023 \, \mathrm{Re}^{0.8} \, \mathrm{Pr}^{n}$. Read aloud: Nusselt equals nought-point-nought-two-three, Reynolds to the nought-point-eight, Prandtl to the n. Every symbol has already been introduced except $n$, the Prandtl exponent, and it is the whole exam: $n = 0.4$ when the tube fluid is being heated, $n = 0.3$ when it is being cooled. One digit, and a systematic error in every rating you write afterwards.

The correlation has a licence, and outside it the number it returns is decoration: $\mathrm{Re}$ above 10 000, $\mathrm{Pr}$ between 0.6 and 160, and a tube at least ten diameters long. It is also only accurate to about ±25 % — which is why plant engineers add fouling margin rather than extra decimal places.

And the last step is the one students skip. Nu is a bare number; nobody buys a bare number. Since $\mathrm{Nu} = \dfrac{hL}{k}$, the film coefficient comes back out as $h = \dfrac{\mathrm{Nu} \, k}{L}$, where $k$ is the FLUID's conductivity and $L$ is the tube's inside diameter. Push the units through: $\mathrm{W/(m \cdot K)}$ divided by metres is $\mathrm{W/(m^2 \cdot K)}$, exactly what a film coefficient wears. If they had not cancelled, the rearrangement would be wrong, no appeal — units guide, but the check runs one way only.