Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 76 / 5 / d / Solution

The flux solution gives $w=M/Q=s{z}^{-1/3}$, while $b^2=Q^2/(\rho M)$. ThusAt $z=z_b$, $\rho ={\rho }_0/2$, $T=2T_0$ and $g^{\prime }=g/2$. This height marks the transition between a strongly non-Boussinesq near-source plume and the dilute far-field plume. For $z\gg z_b$, $b\sim dz$ and $g^{\prime }\sim g(z_b/z{)}^{5/3}=(4s^2/3)z^{-5/3}$, recovering the Boussinesq solution. For $z\ll z_b$, $b\sim d{z}_b^{5/6}{z}^{1/6}$, $g^{\prime }\to g$ and $\rho \propto z^{5/3}$.

The ideal point-source limit still has a divergent velocity and temperature; it is not a literal resolved finite burner. In particular, zero source mass flux does not imply zero physical hot-gas volume flux: $w{b}^2$ tends to $B/g$ in this limiting solution. Source size, compressibility and heat transfer impose additional near-source limits on the model.