Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 90 / 4

The entrainment coefficient $\alpha$ is the ratio of the inward ambient-fluid velocity normal to a buoyant plume edge to the characteristic buoyant plume velocity. For an upward buoyant plume, take $u_e=\alpha w>0$. There are two edges at half-width $b$, so the full-section volume source is $2u_e$. Dividing every integral balance by two gives consistent half-section equations.

For top-hat plume profiles, volume conservation and mass conservation areThe mass entering the buoyant plume is ambient mass with local mass density ${\rho }_0(z)$. The Boussinesq approximation uses a common reference mass density in inertia while retaining the mass density deficit in buoyancy. The mass equation may be kept to the first order in that deficit to derive its transport.

Multiply the volume equation by the time-independent ambient mass density. The product rule givesSubtract the mass equation and multiply by $g$. The entrainment sources cancel exactly, leavingwhereFor stable stratification, ${\rho }_0^{\prime }<0$ and the buoyancy frequency $N$ is real. Rising fluid moves into lighter ambient fluid and loses its mass density deficit relative to that local ambient. Thus environmental stratification reduces the buoyant plume's buoyancy flux, even though the ambient fluid entering by entrainment contributes no mass density deficit at its own height.

Keep the half-section convention of part (a). Define the mass density-weighted mass flux, buoyancy flux and momentum flux byThe full-section fluxes are $2Q,2F,2M$. These factors must be changed consistently if full fluxes are used; mixing a full-width flux with a half-width entrainment source would introduce a spurious factor two.

For positive $Q,M$, $w=M/Q$, $\rho b=Q^2/M$, and $({\rho }_0-\rho )gb=FQ/M$. The mass equation in part (a), the momentum equation, and the buoyancy equation therefore becomeThese are direct algebraic rewritings with the mass density factors retained. In particular, $FQ/M$ is the buoyancy inventory per unit height as well as the integrated buoyancy force. If desired, $\rho$ can be reconstructed from $F/(gQ)={\rho }_0/\rho -1$.

At leading Boussinesq order use a constant reference mass density $R$ in the inertial fluxes, $Q=Rbw$, $M=Rb{w}^2$, and write $g^{\prime }=g({\rho }_0-\rho )/R$ so $F=R{g}^{\prime }bw$. Set ${\rho }_0/\rho =1$ in the stratification source and use $N^2\simeq -g{\rho }_0^{\prime }/R$. The unsteady top-hat line-plume balances are thenEquivalently the kinematic fluxes $q=Q/R=bw$, $m=M/R=b{w}^2$ and $f=F/R=g^{\prime }bw$ obey the same equations with $R$ removed from the first source. This explicit distinction between mass density-weighted and kinematic fluxes fixes the normalization for the power-law solution.

In a homogeneous ambient fluid, $N=0$ and $R={\rho }_0$ is constant. Use kinematic fluxes or, equivalently, the primitive form of the unsteady top-hat line-plume balances:Here $g^{\prime }=g({\rho }_0-\rho )/R>0$ is the positive buoyant reduced gravity. Seek a nonsteady separable power-law ansatz with both time differentiation and vertical transport retained. Matching powers in the volume equation forces $b\propto z$ with no time dependence. Matching the two momentum terms then gives $w\propto z/\tau$, and the buoyancy force has $g^{\prime }\propto z/{\tau }^2$, where $\tau =t+t_{\ast }>0$ is a shifted age. Thus writeSubstitution determines the constants rather than just the exponents. The volume equation gives $2BA=\alpha A$, so for a nonzero buoyant plume $B=\alpha /2$. The momentum equation gives $C=-A+3A^2$, while the buoyancy equation gives $C(-2+3A)=0$. The buoyant nontrivial branch has $A=2/3$ and $C=2/3$. Consequently the separable decaying top-hat line plume isThe Boussinesq approximation requires $2z/(3g{\tau }^2)\ll 1$. The corresponding mass density-weighted fluxes, to this order, areFor a direct check, $Q^2/M=R\alpha z/2$ is time independent and $FQ/M=R\alpha z^2/(3{\tau }^2)$. Hence $(Q^2/M{)}_t+Q_z=2R\alpha z/(3\tau )=R\alpha M/Q$, $Q_t+M_z=R\alpha z^2/(3{\tau }^2)=FQ/M$, and $(FQ/M{)}_t+F_z=0$. This verifies all three balances and the required width coefficient.

This nonsteady branch is not the steady, continuously supplied, constant-flux line plume. The latter has constant $w$ and half-width $b=\alpha z$ in the same top-hat convention. Also, the unshifted power law has $Q,F\to 0$ as $z\to 0$ and is singular at $\tau =0$. It is therefore an interior separable solution, not by itself a solution with a nonzero continuously operating point-source flux at the origin. A finite-width source boundary, a virtual origin, or a matching region is needed to connect it to physical source and initial data. For example, replacing $z$ by $z+z_{\ast }$ supplies a finite base width and time-varying positive base fluxes when $z_{\ast }>0$. No source or initial flux history is specified in the question to select that matching; the requested power-law width follows from the local balances above.