Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 52 / 6

Use the Boussinesq approximation and a top-hat plume model. Let $v(z)<0$ be the uniform ambient return velocity outside the plume. Negligible source volume flux and an impermeable cylindrical wall require zero net transport through every horizontal section:The entrainment speed must be specified relative to that moving ambient. Taking it to be $\alpha (w-v)$ givesThe surrounding unmodified fluid has zero buoyancy relative to the initial fluid. Neglecting thermal or compositional loss therefore givesLet $B_s$ denote the conserved specific source buoyancy flux: $B_s=B_o/\pi$ if $B_o$ is the actual integrated source flux, or $B_s=B_o$ if the source strength has already been normalized by $\pi$.

To obtain the usual integral-model singularity one must also state a pressure closure. Here take the whole-section mean pressure to have only its reference ambient hydrostatic pressure, and neglect wall force and extra vertical turbulent-stress fluxes in the integral budget. Internal momentum transfers between plume and ambient cancel in that whole-section balance. The remaining upward force is the plume buoyancy, givingThe two momentum fluxes add, even though $v$ is negative: vertical momentum transported downward also contributes $v^2$ to the flux through an upward-oriented horizontal section. These three flux balances, with the algebraic return-velocity constraint, govern $b,w,g^{\prime }$ in the confined plume with compensating return flow.

The pressure-neglecting closure is an integral approximation, not a claim that the return fluid has zero acceleration. If a mean pressure departure $p_{\ast }(z)$ is retained, the whole-section momentum balance instead contains $-(R^2/{\rho }_0)p_{\ast }^{\prime }(z)$, along with any wall-force or stress contribution. The entrainment hypothesis alone does not specify that function. The singularity calculated below belongs to the stated pressure-neglecting model; it is not forced by continuity alone or asserted to be an infinite velocity in the real fluid.

Normalize all fluxes by $\pi$, rather than including that factor in only some of them. The specific plume volume flux, momentum flux and buoyancy flux areThe ambient transport is $-Q$ and its specific momentum flux isPut $s=b^2/R^2=Q^2/(R^{2}M)$, with $0<s<1$. The total specific momentum flux isThe first differential equation follows immediately from relative-speed fluid entrainment:For the second, differentiate $J=M/(1-s)$ and use $s^{\prime }=2s{Q}^{\prime }/Q-s{M}^{\prime }/M$. Since $J^{\prime }=b^2{g}^{\prime }=B_{s}Q/M$, this yieldsThis is the requested two-equation system, with $B=B_s$ and zero total volume flux as the additional conservation conditions. Near the pure source, $Q,M\to 0$ and $s\to 0$. Eliminating height in that limit givesThe zero integration constant selects a pure plume; a nonzero imposed source momentum would select a different branch and generally a different turning height. The leading radius is $b\sim 6\alpha z/5$, which recovers the unconfined pure-plume limit near the source.

It is helpful to solve using the nonsingular total flux $J$ and volume flux $Q$ before recovering $M$. The relation $J=M/(1-Q^2/(R^{2}M))$ is quadratic:The plus sign selects the small-area pure-source branch. Its discriminant vanishes when $J=4Q^2/R^2$, givingThereforeThe specific upward plume and ambient momentum-flux contributions are equal, not opposite. The ambient's transported momentum has the opposite sign from its transported volume, so their product remains positive.

To show that the pure branch actually reaches this fold in finite height and that it is a genuine pole, introduceThen $s=q^2/m$ andThe pure-source condition is $j\sim (5/8{)}^{2/5}{q}^{4/5}$. Before the fold, $s<1/2$ and $m>2q^2$. Thus for every fixed positive $q_1$, the derivative $dj/dq$ is bounded above by a constant times $q^{-2}$ on $q\ge q_1$. If the branch continued for arbitrarily large $q$, $j$ would remain bounded, contradicting $j\ge 4q^2$. Hence it reaches $j=4q^2$ at a finite $q_o>0$. Its height is finite because $d\zeta /dq=(1-s)/(2\sqrt{m})$ is finite away from the source and behaves like a constant times $q^{-2/5}$ at the source, an integrable singularity.

For the numerator, write $m=j(1-s)$. ThenAt the fold $j=4q_o^2$, so $q_o^3\ge 5/256$. On the other hand the dimensionless numerator in the $m$ equation at $s=1/2$ isIt is strictly negative because $5/256>1/(64\sqrt{2})$. The denominator $1-2s$ approaches zero from above. Consequently $dM/dz\to -\infty$ at finite $z_o$, although the fluxes and velocities themselves remain finite. This proves the confined-plume momentum fold rather than merely noticing a possible zero denominator.

The dimensionless pure-source branch uniquely determines the remaining amplitudes. They can be specified without an elementary antiderivative by integrating the displayed scalar equation for $j(q)$ until $j=4q^2$, and settingA direct numerical integration gives $q_o\simeq 0.281610$, ${\zeta }_o\simeq 0.400049$, so the dimensional results for this specified entrainment convention areA convention using $\alpha w$ instead of $\alpha (w-v)$ would produce different numerical amplitudes; the relative-speed convention was stated in part (a).

Physically $z_o$ is a turnover or breakdown height for a slender steady plume surrounded by uniform counterflow. The opposing streams then occupy equal areas and have equal speeds; the single-valued small-area branch cannot continue smoothly. Above it one expects lateral spreading, stronger recirculation, mixing and unsteady buoyant structures rather than a continuation of the same slender top-hat column. A detailed flow or upper-layer evolution requires restoring pressure and stress dynamics and, eventually, the modified ambient stratification. The model does not predict an infinite physical velocity, nor does it require that all buoyant fluid forever remain below $z_o$.