 Solution

At a fixed height, first determine the stored quantities per unit vertical height and per unit length along the line source. With $\Delta \rho =\rho -{\rho }_0$, the triangular-profile line plume has the useful profile integralThe geometric volume is $\mathcal{A}=2b$; the mass is $\mathcal{D}={\rho }_0{I}_0+\Delta \rho I_1=b(\rho +{\rho }_0)$. The upward momentum is $\mathcal{P}=W({\rho }_0{I}_1+\Delta \rho I_2)$, while the buoyancy force is $\mathcal{B}=-g\Delta \rho I_1$. ThusHere positive buoyancy corresponds to a rising light plume. If instead buoyancy is defined kinematically, divide $\mathcal{B}$ by ${\rho }_0$; the density-weighted convention here is the one matching the printed equations.

Multiply each local density by the vertical velocity to obtain the fluxes through a horizontal section, still per unit source length:In particular, the stored momentum equals the mass flux $Q$, and the buoyancy flux is $F=g({\rho }_{0}V-Q)$. The different triangular-profile factors in storage and flux must not be replaced by top-hat coefficients.