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Apply volume conservation, mass conservation and vertical momentum flux balance to a slice of the triangular-profile line plume. Ambient ingestion contributes $2e$ of volume and $2{\rho }_{0}e$ of mass per unit height and source length. The ambient is quiescent, so it supplies no leading vertical momentum. The integrated driving force is the stored buoyancy $\mathcal{B}$. Using the quantities from part (a), the non-Boussinesq triangular-profile line-plume balances areTogether with Batchelor entrainment, these are three equations for $W,\rho ,b$. They neglect viscous boundary stresses and streamwise pressure-force corrections within the integral-plume approximation.

Multiplying volume conservation by $g{\rho }_0$ and subtracting $g$ times mass conservation cancels the ambient sources exactly, givingThis is buoyancy conservation in the homogeneous ambient; it is a consequence of the first two balances, not an additional independent equation.

Now take the Boussinesq approximation in inertia and entrainment, retaining the small density deficit in buoyancy. Then $Q={\rho }_{0}bW$, $M=(2/3){\rho }_{0}b{W}^2$ and $F=(2/3)gbW({\rho }_0-\rho )$. ConsequentlyThe mass source is $2\alpha {\rho }_{0}W=3\alpha {\rho }_{0}M/Q$. Substitution into the mass conservation, momentum and buoyancy balances gives the unsteady Boussinesq triangular-profile line-plume balancesThese use density-weighted mass and momentum fluxes. Removing ${\rho }_0$ from some flux definitions while retaining it in the source would mix incompatible conventions.

Seek a nonzero-buoyancy, separable similarity solution of the unsteady Boussinesq triangular-profile line-plume balances, using powers of distance and time. Let $\tau =t+t_{\ast }>0$, allowing a time origin, and initially take distance $z$ from the virtual origin. Balancing the powers in the momentum and buoyancy equations and the entrainment balance gives the formOne way to recover the exponents is to let $Q\propto z^a{\tau }^p$ and $M\propto z^b{\tau }^q$. Matching the spatial powers in ${\partial }_t(Q^2/M)$, $Q_z$ and $M/Q$ gives $a=2,b=3$; matching the time powers gives $p=-1,q=-2$. The momentum equation then gives the powers of $F$. In the resulting solution, $Q^2/M$ is time-independent, so the storage term in the first equation vanishes exactly.

Direct substitution yields the coefficient equationsFor a rising buoyant solution $C>0$, the last equation gives $B=2A/3$. The first then gives $A=\alpha {\rho }_0$, and the second gives $C=B$. The separable decaying line-plume similarity is thereforeReconstructing the physical fields confirmsFor example, ${\partial }_t(FQ/M)=-2\alpha {\rho }_0{z}^2/{\tau }^3$ and $F_z=2\alpha {\rho }_0{z}^2/{\tau }^3$ cancel; the momentum balance similarly gives $-\alpha {\rho }_0{z}^2/{\tau }^2+2\alpha {\rho }_0{z}^2/{\tau }^2=FQ/M$.

The unshifted form uses $t_{\ast }=0$ and is defined for $t>0$, with a singular zero-time limit. To give a finite solution for all $t\ge 0$, choose $t_{\ast }>0$. The equations also permit replacing $z$ by $z+z_{\ast }>0$, a plume virtual origin; at a finite physical source distance this family has flux histories proportional to ${\tau }^{-1},{\tau }^{-2},{\tau }^{-3}$. The Boussinesq approximation requires $(z+z_{\ast })/(g{\tau }^2)\ll 1$ in the region modelled. This is a decaying similarity family; its time and virtual origins require source or initial data. The PDF supplies neither, so a unique forced-startup history cannot be selected from it. In particular, the singular unshifted field should not be presented as a regular solution at $t=0$.

The unsteady Boussinesq triangular-profile line-plume balances admit a buoyant similarity solution $Q=\alpha {\rho }_0{z}^2/\tau$, $M=(2/3)\alpha {\rho }_0{z}^3/{\tau }^2$ and $F=(2/3)\alpha {\rho }_0{z}^3/{\tau }^3$, with $\tau =t+t_{\ast }>0$. It reconstructs $b=\alpha z$, $W=z/\tau$ and reduced gravity $z/{\tau }^2$. Origins can be shifted, but the zero-time unshifted solution is singular. Source and initial data are required to select the physical member and region of validity.