Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 80 2 Solution Created 2026-10-03 Updated 2026-10-14
Use the steady axisymmetric top-hat plume model, with plume radius , uniform mean upward speed and reduced gravity . The Boussinesq approximation uses constant reference density in inertia and retains the small density difference only in buoyancy. Assume a quiescent hydrostatic ambient, a slender fully turbulent plume, negligible molecular transport and an entrainment velocity , with constant positive entrainment coefficient . Define the ambient buoyancy frequency by ; positive is stable stratification.
The integrated volume, momentum and buoyancy balances of a Boussinesq top-hat plume in a stratified ambient areThe first follows from side entrainment, the second from the integrated buoyancy force, and the third accounts for the changing ambient density. Common factors of are cancelled. Equivalently, the requested radius, speed and reduced-gravity equations areThese equations stop describing an upward light plume once vanishes, and their light-plume interpretation requires .
For nonzero , seek plume similarity in power-law stratification in the form , and . The first balance gives . Momentum gives and . The buoyancy balance then requires , soChoosing gives the exact power-law family, including its normalization:These are the amplitudes and physical branches of a power-law stratified plume. No source flux was specified, so an arbitrary finite-source solution is not determined by alone; the displayed family is an exact similarity solution, rather than the general solution for arbitrary initial fluxes.
With the specific flux convention , and , their values areThe physical volume flux is , mass flux is , momentum flux is , and kinematic buoyancy flux is . Thus the “specific mass flux” is the mass flux divided by reference density, with or without the explicitly stated factor .
The parameter restrictions follow directly from positivity. Positive radius needs . Positive and real positive further require and , givingFor stable , buoyancy flux decreases with height, while volume and momentum fluxes increase. However, diverges at throughout this interval. Also the ambient density gradient and plume speed are singular there. Thus this stable similarity branch can describe a region away from a virtual source, but cannot represent a finite-buoyancy-flux point source extending all the way to . For ordinary regular stable stratifications, including positive , a finite-source rising plume instead loses buoyancy and eventually reaches neutral buoyancy and a finite rise limit; the pure-power ansatz does not supply such a solution.
For unstable and , the buoyancy flux grows with height and tends to zero at the origin. The plume draws its increasing buoyancy from an unstable ambient rather than a finite positive source buoyancy flux. Such an ambient must be maintained or treated locally, since it is itself convectively unstable. If , its density profile is singular at the source, while for its increasing density contrast eventually violates the Boussinesq approximation at sufficiently large height. These are formal local similarity solutions with corresponding physical limits. For and , the algebra also gives a real positive speed but negative : this is a heavy upward flow, not the specified light plume.
At nonzero , the values do not give finite coefficients. Respectively they would require zero entrained volume-flux growth, zero momentum-flux growth despite nonzero buoyancy, or constant buoyancy flux despite nonzero ambient stratification. They are not omitted regular endpoints.
When , the ambient is unstratified regardless of the nominal value of . The nontrivial positive-buoyancy point-source solution is the Boussinesq point-source plume:Here is the constant specific buoyancy flux with removed. Its and vanish at the point source, while . Its near-source singularity is the usual ideal point-source limit; a finite vent replaces that limit by finite source data or a virtual origin.
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 71 1 c Solution Created 2026-10-03 Updated 2026-10-14
For the steady positive-velocity branch, conservation of buoyancy flux gives . Substituting into the other two balances givesUse a power-law ansatz , . The entrainment balance forces and . The momentum powers are , and , so the nonzero inertial branch has . Therefore the constant-speed entraining gravity current on a slope isIts positive speed requires for . In the shallow, strictly hyperbolic range this meansThe same steady integral balances have a formal vertical-plume limit at , although their time-dependent system then loses strict hyperbolicity. At the lower threshold the assumed positive-flux, finite-speed solution fails.
The line source is singular: and the momentum flux tend to zero, but diverges as . Thus this is a downstream ideal-source solution, valid only far enough away that and the Boussinesq approximation and top-hat plume model hold. It does not describe the immediate source region.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 69 4 Solution Created 2026-10-03 Updated 2026-10-07
Write for the positive reduced gravity of dense fluid. With upwards, places denser fluid above lighter fluid and permits convective turbulence. A displacement over the radius scale samples a buoyancy difference of order ; balancing kinetic energy per mass with buoyant work over givesThe buoyancy-gradient mixing-length closure then gives an eddy diffusivity . We use the paper's unit-prefactor convention for this dimensional closure and its effective cross-sectional area ; order-one mixing constants and the cylinder's geometric factor are suppressed in its displayed evolution equation.
The downward reduced-gravity transport magnitude and its convergence are different quantities:The printed expression labelled a flux is actually the second quantity, a flux convergence per unit height. The dimensions distinguish them: has units , whereas has units . A constant positive gradient has nonzero transport but zero convergence, providing a direct counterexample to identifying the derivative with the transport itself.
The upward total buoyancy flux is . Conservation therefore yieldsThis flux convergence in turbulent buoyancy mixing obtains the displayed evolution equation with the correct transport interpretation.
For the arrested cubic buoyancy profile, an undisturbed region below has and no turbulent buoyancy flux. The joining condition is andIn a steady state, integrating the conservation equation from that front gives . For a nonzero mixed region,and henceThe associated gradient is , so it matches the required zero-gradient state continuously. We take the mixed interval to start immediately above ; adding a further zero interval would merely relocate the arrest front.
The source buoyancy flux is , with in the Boussinesq approximation. The stated top-source convention means that this entire flux enters the downward turbulent transport there: . Since steady transport also satisfies ,This is the extent in the paper's dimensional-closure normalization. It presupposes positive ; with no upflow there is no finite steady arrest depth. For a mixture of the two original fluids, is necessary, so an extrapolation to cannot represent a physical concentration profile under this top-flux boundary model. A dilute-source regime avoids appreciable source-volume corrections.
Assuming the same scalar-mixing coefficient for a passive tracer, its eddy diffusivity in the mixed region isIt vanishes at the arrest front and increases linearly towards the source. A different turbulent scalar diffusivity ratio would multiply this result by its corresponding constant.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 70 2 Solution Created 2026-10-03 Updated 2026-10-06
Use and constant entrainment parameter . The printed integrals are normalized by , so a top-hat plume model has , and ; the physical area is still . Keeping this normalization avoids spurious factors of in the flux equations.
Eliminate height between the two flux balances. The plume flux-balance invariant follows fromsoFor positive fluxes, both and increase. In particular , so as . Therefore , whatever the finite source imbalance, and . Integrating gives the attracting similarity solutionThis proves attraction to pure plume similarity for physical positive source fluxes. Sources with a vanishing initial volume or momentum flux enter the positive regime immediately and can be obtained by the corresponding limiting solutions; a point buoyancy source has the same similarity form with zero virtual-origin offset. The statement assumes positive buoyancy flux: a jet with is a different asymptotic problem.
Define the plume balance parameterPure plume balance means at each height, or equivalently : momentum, buoyancy and entrainment have their similarity relation, even when the source has finite radius. A forced plume has , meaning excess momentum relative to this balance; a lazy plume has , meaning insufficient momentum. The invariant ensures that the sign of the imbalance is preserved while . In particular, a strictly positive source remains pure at every height precisely whenFor this source the exact solution is the same similarity form with replaced by . Writing ,Thus the radius growth of an axisymmetric pure plume is and the plume virtual origin lies at , with . Equivalently, in terms of top-hat source properties, pure plume balance is .
For the stated source-to-source geometry, the relevant horizontal distance is . Consequently the edge criterion is , rather than the radius at which two expanding edges first touch. Since ,Here . A positive merging height requires ; equality puts it at the source, while an already wider source satisfies the geometric condition from the start. The no-interaction premise extrapolates independent plumes even though their edges would first touch earlier, when each radius is .
Use with and to get . The individual plume properties at the specified height are thereforeThe source volume flux affects the height through the virtual origin, but it does not affect these properties at a prescribed radius in an exactly pure plume.
For the stipulated merger of two equal pure plumes, the new radius is . This factor is outside , as required by the area. Since speed and reduced gravity are held fixed, the merged fluxes are , and . Their balance parameter isThe immediate combined plume is lazy. The same result follows from : enlarging the radius at fixed increases this parameter.
At fixed , the radius required for pure plume balance isThat is half the stipulated combined area . This fixed-speed pure-plume merger and flux conservation comparison exposes the limitation of demanding immediate pure balance with unchanged speed and reduced gravity: reducing the area to this value would also halve all three merged fluxes. A conservative merger must instead adjust its properties and begin out of pure balance. Under the given constant-entrainment equations, the resulting lazy plume subsequently approaches the attracting similarity solution with total buoyancy flux .
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 74 4 a Solution Created 2026-10-03 Updated 2026-10-06
At a fixed height, first determine the stored quantities per unit vertical height and per unit length along the line source. With , the triangular-profile line plume has the useful profile integralThe geometric volume is ; the mass is . The upward momentum is , while the buoyancy force is . ThusHere positive buoyancy corresponds to a rising light plume. If instead buoyancy is defined kinematically, divide by ; 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 , and the buoyancy flux is . The different triangular-profile factors in storage and flux must not be replaced by top-hat coefficients.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 330 2 a Solution Created 2026-10-03 Updated 2026-10-06
Use the steady Boussinesq top-hat plume in a stratified ambient, with negligible ambient vertical momentum and no externally added heat or particles. The Batchelor entrainment hypothesis supplies inward velocity across circumference . Volume conservation gives , and , henceThe momentum flux increases by the integrated buoyancy force. Writing , the force per reference density and per is . Since and , this givesFor the buoyancy flux, entrained material has the local environmental mass density. Let be the plume's averaged density. Its scalar mass conservation is . Since , differentiating cancels both entrainment terms and givesIn stable density stratification, and a rising plume loses buoyancy flux. These are kinematic fluxes with the common factor removed, so no additional factor of belongs in the three differential equations.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 330 2 b Solution Created 2026-10-03 Updated 2026-10-06
Define . The buoyancy frequency definition gives . The powers in a plume similarity in power-law stratification satisfySolving givesA rising positive-buoyancy turbulent plume entrains, so ; its momentum flux increases, so ; its buoyancy flux decreases in the stable ambient, so . With positive coefficients, this requires , and . The strongest restrictions areIn this interval , and .
Coefficient matching makes the positive similarity solution explicit:soIt is an ideal scaling for , often a far-field description. As , the prescribed ambient stratification and this model's buoyancy flux are singular; a finite physical source would require a regularized near-source region.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 330 2 d Solution Created 2026-10-03 Updated 2026-10-06
Expand the square root and quotient in the logarithmic-height plume equations:The linear stability analysis givesIn particular, the second row contains , since the reciprocal of contributes another .
For the growing mode of power-law plume similarity, the characteristic polynomial isWithin the physical interval and , while as . There is a positive real eigenvalue, so the similarity equilibrium is unstable with increasing height. Perturbations behave as ; the result is not stability in physical time. Thus this exact positive-buoyancy similarity trajectory is not a generic attracting far field.
At the upper endpoint,The limiting polynomial is , reproducing the three stated eigenvalues. The positive eigenvalue tends to zero, with . The two negative modes decay as and ; the remaining mode is neutral at the limiting linear system. Its eigenvector is in relative-flux coordinates.
The neutral buoyancy-flux mode of a pure plume has a simple interpretation: a constant surviving buoyancy flux fixes the amplitudes of a pure plume, with and . A small fractional change in produces changes in in the ratio . It is retained rather than damped out.
There is a qualification at the exact endpoint. For fixed nonzero and , taking gives , not zero. The flux coefficients above also diverge as . Thus the constant-flux powers are the formal limiting scaling, not an exact constant-buoyancy solution in a nonzero endpoint stratification; cumulative ambient effects can require logarithmic corrections.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 330 3 a Solution Created 2026-10-03 Updated 2026-10-06
Seek a steady downslope solution with constant. The volume conservation equation reduces to , so the constant-speed entraining gravity current on a slope hasThe conserved buoyancy flux per unit width is , givingSubstitute these expressions into the momentum equation. Its advective term is and its hydrostatic term is . The balance is thereforewhich givesThe assumed slope inequality makes . The increasing thickness dilutes reduced gravity while leaving and the buoyancy flux constant. The denominator includes both bottom drag and the momentum needed to accelerate newly entrained ambient fluid.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 330 3 d Solution Created 2026-10-03 Updated 2026-10-06
For a steady no-source particle-laden gravity current, . Simply entraining particle-free ambient fluid increases volume and dilutes particle volume fraction; it cannot create particle buoyancy flux.
A steady increase can occur if sediment resuspension or bed erosion adds more suspended particle buoyancy than settling removes, if a tributary or lateral particle source supplies a positive flux, or if the buoyancy per unit particle amount increases through a change of the surrounding fluid's density. Changes of particle properties or another buoyancy source can also alter the flux, but they go beyond the fixed-particle, fixed-reference model. In a model with a source ,Thus the buoyancy production in a particle-laden current requires in a steady current. In an unsteady current, can be positive even without external addition: local depletion has , so more buoyancy leaves a section than enters it while stored particles are being released. An instantaneous spatial flux increase is not by itself proof of particle production.