A numerical example

If the vapour condenses at one of the walls, the vapour base enthalpy, $h_{A_{r}}$, is naturally taken as the enthalpy of the vapour at $T_{*r}$ relative to that of the condensed state at the same temperature. This can easily make $[1-\exp(-\varPhi _T)]\varLambda$ (the dimensionless `latent heat transfer') the dominant component of the overall energy transfer (4.42). This is the same conclusion as reached by Yan et al. (1989), from their numerical studies of the finite length channel. For points well up the channel from the entrance, their results are well described by the above formula.

For example, in their Case II ($T_{*r}=30$$\circ$C, $\Delta T_{*}=20$K) with both walls saturated with water vapour and the total pressure about 1 atm), the dimensionless parameters take on the values $\varLambda =66.0$ and $\varPhi _T =0.118$. Equation (4.42) gives $\mbox{\textit{Nu}}=8.349$. For a point 560 channel widths up from the entrance, Yan et al. (1989) give $\mbox{\textit{Nu}}=8.347$, which is excellent agreement. With saturated boundary conditions such as these, the vapour will certainly be supersaturated in the channel, so that gas-phase condensation is possible, but this was neglected in the cited work. Equation (4.42) is therefore applicable.

It is obvious from this example that the humidity difference has a large effect on the energy transfer, as will often be the case.

Numerous further numerical examples, due to Dr Harry Suehrcke, supporting this conclusion can be found in our recent joint paper (Suehrcke & McBain 1998).