Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 345 / 3

It is important to fix the mass flux convention. With literal mass units and the corresponding density-weighted buoyancy flux, the Boussinesq approximation and circular top-hat plume model giveReplacing $\rho$ by ${\rho }_0$ in inertial fluxes is the Boussinesq approximation; the density difference is retained in the buoyancy flux. Define the kinematic plume fluxesIf the symbols $Q,\mathbf{M},F$ instead denote the customary kinematic fluxes, simply omit the factors ${\rho }_0$ in all conversions below. Keeping the two conventions consistent is essential for the jet length and dimensional prefactors.

The Batchelor entrainment hypothesis prescribes inward edge velocityThe constant entrainment coefficient relates ambient inflow to the local axial velocity. A centreline segment of arc length $ds$ entrains volume $2\pi b{u}_{e}ds$. Hence, setting $E=2\alpha \sqrt{\pi }$,The buoyancy flux is conserved because the homogeneous entrained ambient has zero density deficit, and mixing has no buoyancy source or sink. The imposed zero source mass flux with nonzero source momentum flux is a singular point-source idealization, not a finite-radius nozzle with zero emitted fluid. Indeed $V=m/q$ and the plume reduced gravity $f/q$ diverge as $q\to 0$ with nonzero source fluxes. The Boussinesq approximation cannot remain valid arbitrarily close to that point. The top-hat plume model is applied outside the unresolved source region.

For a neutrally buoyant horizontal turbulent round jet, a segment's horizontal momentum flux changes only if an external horizontal force or ambient momentum input exists. Neither is present, soPositive buoyancy adds a vertical force, not a horizontal one. Therefore $d{M}_x/ds=0$ remains true for an inclined forced plume even when $F_0\ne 0$; in general $M_x=M_0\cos {\theta }_0$.

The net upward force on a segment is $g({\rho }_0-\rho )\pi b^{2}ds$. Since $F=g({\rho }_0-\rho )\pi b^{2}V$ and $V=|\mathbf{M}|/Q$, vertical momentum balance givesIn kinematic plume fluxes, the equations are $m_x^{\prime }=0$, $m_z^{\prime }=fq/m$, where $m=(m_x^2+m_z^2{)}^{1/2}$. The centreline geometry satisfiesFor $F_0>0$, $Q>0$ beyond the source, and an initially nonvertical turbulent round jet pointing right, $M_z$ increases and the centreline turns toward the vertical. The magnitude of momentum flux is not conserved once the tangent has a vertical component.

Put $m_0=M_0/{\rho }_0$ and $f_0=F_0/{\rho }_0>0$ under the literal mass convention. Their dimensions are $[m_0]=L^4{T}^{-2}$ and $[f_0]=L^4{T}^{-3}$. Dimensional analysis givesThe jet length compares source momentum with buoyancy. The region near the source behaves as a turbulent round jet; sufficiently far away, the forced plume behaves like a pure plume. With the constant entrainment coefficient retained explicitly, substantial turning occurs over a length of order ${\ell }_J=L_J/\sqrt{E}$, where $E=2\alpha \sqrt{\pi }$. Statements involving $s\ll L_J$ or $s\gg L_J$ normally hold this dimensionless coefficient fixed.

Near the point source, $m\sim m_0$ and the Batchelor entrainment hypothesis gives $q\sim E\sqrt{m_0}s$. ThereforeIntegrating the centreline tangent givesTo leading order the centreline is a straight turbulent round jet at its source inclination. For the horizontal source, the first curvature is the horizontal forced-plume trajectoryFor a vertical source, $\cos {\theta }_0=0$, the trajectory stays vertical rather than developing this cubic transverse displacement. If $F_0=0$, there is no finite buoyancy-induced jet length and the turbulent round jet remains straight.

For a horizontal source, $m_x=m_0$ throughout. Far downstream $m_z\gg m_0$, so the kinematic plume fluxes obey the vertical pure plume equations to leading order,Set $q=q_{\ast }{f}_0^{1/3}{\hat{s}}^{5/3}$ and $m_z=m_{\ast }{f}_0^{2/3}{\hat{s}}^{4/3}$ with $\hat{s}=s+s_0$. Matching exponents and coefficients gives $(5/3)q_{\ast }=E\sqrt{m_{\ast }}$ and $(4/3)m_{\ast }^2=q_{\ast }$. Consequentlyand the requested literal mass and density-weighted buoyancy fluxes areto leading order. The radius is $b\sim (6\alpha /5)\hat{s}$ and the axial velocity decreases as ${\hat{s}}^{-1/3}$.

The centreline becomes nearly vertical but retains finite horizontal drift:Since $dz/ds=1-O({\hat{s}}^{-8/3})$, there is a finite positive geometric deficit $D={\lim }_{s\to \infty }(s-z)$, and $z=s-D+o(1)$. The equivalent vertical plume virtual origin is therefore located atFor the horizontal point-source forced plume, the virtual source is downstream and below the actual source. The horizontal turbulent round jet first entrains substantial ambient fluid while rising only a little. Thus at the height where it turns upward it already has a finite radius and volume flux; an equivalent pure plume must have begun rising from below to acquire these. The turning displacement, entrained radius divided by its far-field spreading angle, and virtual vertical depth all scale asAt fixed entrainment coefficient this is simply $O(L_J)$. The arclength origin $s=-s_0$ should not be interpreted as a physical height: the curved near field supplies the additional shift $D$. The numerical constants depend on the chosen top-hat plume model and entrainment coefficient; scaling does not fix them.

The inclined forced plume bends from a cubic near-field trajectory toward a vertical asymptote. Dashed far-field spreading lines extrapolate to the equivalent plume virtual origin; the marked location is obtained from the integral model for this illustration, not a universal numerical prediction.