Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 345 1 Solution Created 2026-10-03 Updated 2026-10-06
Write for the height of the filling-box first front above the floor and for the upward plume volume flux crossing it. Neglect the plume's horizontal area compared with , use the Boussinesq approximation, and measure buoyancy relative to the initially unmodified ambient. The first front separates fluid already modified by plume discharge from the unmodified fluid below; the entire upper region need not be uniformly mixed during an unventilated filling box model transient.
A finite injected volume flux cannot enter a hermetically sealed, fixed-volume room under the incompressible flow assumption. If , an equal overflow or exhaust is needed. Taking that exhaust at the ceiling, volume conservation of the region above the front givesThe source volume flux is subtracted because only entrained room fluid is removed from the initially unmodified region. The printed pure plume law has , so it cannot be an exact finite- source law down to the physical floor. In the far field where , or for a heat source whose injected volume flux is negligible, it givesHere is when the plume discharge first reaches the ceiling. Retaining while using the far-field law gives only while that law remains a valid approximation; it must not be extrapolated to predict negative plume entrainment. A plume virtual origin below the floor is one way to match a nonzero source volume flux.
For displacement ventilation, let be the total ceiling exhaust, the total buoyant-source volume flux, and the separate ambient inflow into the lower layer. Volume conservation and the upper-layer buoyancy flux balance giveThe ambient inflow has zero reference buoyancy. In the ideal two-layer closure the upper layer is uniformly mixed with reduced gravity , and the lower layer has zero reference buoyancy. The plume supplies the upper layer while lighter upper-layer air is removed at the ceiling. An interface is driven downward when and upward when ; since , its equilibrium is stable.
In the negligible-source-volume pure plume limit, identifying gives the intended one-source answerThe prescribed inequality makes , so both layers fit in the room. It does not, by itself, justify neglecting a finite source : the pure plume source region must also be small compared with , normally requiring . If denotes additional background ventilation rather than total exhaust, replace in both formulas by . If it is total exhaust, the separate ambient inlet carries . These conventions must not be mixed.
For independent equal axisymmetric pure plumes, adding their volume fluxes givesThus the conventional multiple-plume displacement ventilation calculation with negligible source volume flux and separate ambient ventilation givesSplitting the buoyancy flux increases total fluid entrainment, so the same ventilation balances the plumes at a lower interface. The global buoyancy flux is unchanged, which leaves the upper-layer reduced gravity unchanged. Independence requires sufficiently separated sources and plumes that do not merge below the interface.
There is, however, a substantive problem with the literal source-volume specification in this paragraph of the PDF. It puts all through the buoyant sources, so and . For every positive interface height, entraining plumes haveThere is no fresh ambient inflow to replace the lower-layer fluid being entrained. Consequently the stated positive steady two-layer configuration does not follow from those literal hypotheses. This is no steady displacement layer without ambient supply, not a small correction to the scaling. For example, a matched plume virtual origin givesWhen , this yields . The unshifted answer equals the omitted in this case, so neglecting the source region cannot be justified. Also each source has reference buoyancy , already equal to the proposed upper-layer value; a steady positive plume cannot entrain zero-buoyancy lower fluid and still deliver that same value. The conventional answer is valid for a corrected arrangement with negligible-volume buoyancy sources and a separate ambient inlet, or with source flux and the remaining supplied directly to the lower layer. A real room may have other circulation or continuous stratification, but that requires a different model.
For the intended displacement ventilation model, reducing the exhaust to makes the new equilibriumThe interface initially descends since its old plume volume flux exceeds the new exhaust. In ventilated filling-box relaxation, writing and taking the linearization of the two balances above givesAs , the two decay times areA dimensional analysis estimate of the full adjustment time is thereforeThe interface adjusts on the shorter lower-layer scale; replacing the buoyancy stored throughout the upper layer is slower. Exact equilibrium is approached asymptotically, and a specified small relative tolerance adds a logarithmic factor. For the literal all-inflow-through-sources arrangement there is no positive two-layer equilibrium to return to; is only a flushing-scale estimate for a different, whole-room adjustment.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 345 2 Solution Created 2026-10-03 Updated 2026-10-14
Use a two-sided top-hat line plume with full width , upward speed , and reduced gravity uniform across the plume. Fluxes are measured per unit span along the line source: the volume flux is , the kinematic momentum flux is , and the buoyancy flux is . If the inward edge speed is , where is the entrainment coefficient, the two exposed edges giveThese are the volume conservation, momentum conservation, and buoyancy flux balances for a Boussinesq approximation plume in an unstratified ambient. Entrained ambient fluid supplies neither vertical momentum nor reference buoyancy.
A pure plume has no persistent source length scale. Since a line-source has dimensions , dimensional analysis gives constant , , and . Substituting constant into and the momentum balance determines the coefficients:Thus and when width is full width and velocity and buoyancy have top-hat profiles. If width means half-width, instead. Other prescribed profile shapes change these numerical factors; the linear growth laws remain the same. The point-source in Question 1 has different dimensions, so its law must not be used here.
Let be local contaminant concentration and let be horizontally integrated plume concentration. Its conserved amount per unit span is . The dilute contaminant is a passive scalar; the maintained plume is unaffected by its impulsive release. A one-dimensional effective transport closure takes the integrated scalar flux to beThe constant advective speed and the eddy diffusivity scaling follow from the line-plume scales. The entrainment coefficient alone does not determine the scalar dispersion coefficient: this Fickian closure for the integrated variable is an additional modelling assumption. Mass conservation then gives the displayed transport equation in the PDF. With this effective closure and advective speed chosen as , ; if , then , with independent dimensionless mixing coefficient .
There is a second possible closure convention. If one instead applies local diffusive flux to horizontally averaged plume concentration , its integrated diffusive flux is . For , the total scalar flux is then . The same printed equation is obtained with , rather than . Thus the printed and should be regarded as effective coefficients unless the averaging and closure conventions are specified; no equality between and the velocity prefactor is universal.
Assume , , an initial impulse at the origin, no further contaminant input, and zero endpoint scalar flux for . For a similarity solution, putSubstituting into the advection-diffusion equation givesDecay at infinity makes the integrated constant zero. Hence , and normalization with the Gamma function gives . The resulting gamma impulse solution for linearly increasing diffusivity isIts integral over is . Its flux is , which vanishes at both endpoints for . The scale shrinks to zero as , so the normalized solution has weak convergence of probability measures to the required unit source impulse. The mass at the boundary is a full unit impulse on the half-line, not half of a whole-line impulse.
For the integrated quantity, , and consequentlyThis verifies the printed location for horizontally integrated plume concentration. It is a maximum because the derivative changes from positive to negative there. The normalized profile is a gamma distribution with shape : its expected value is and its variance is , so neither the mean position nor the spread should be mistaken for the modal position.
The PDF changes from “integral” to “averaged” concentration in its final request. A genuine horizontally averaged plume concentration divides by the growing width , and is thereforeIt is normalized by , not by . When , its interior maximum isWhen it decreases from a finite boundary supremum; when it is singular at the ideal point source and has no positive interior maximum. A finite source regularizes that singularity. Thus the final printed maximum is correct for the integrated variable , or for an “average” using a fixed reference width, but is not generally the maximum of the local mean . Both quantities have been given rather than silently identifying them. If , the smooth similarity formula is replaced by the advected impulse ; positive eddy diffusivity is essential to the gamma profile.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 345 3 c Solution Created 2026-10-03 Updated 2026-10-05
Put and under the literal mass convention. Their dimensions are and . 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 , where . Statements involving or normally hold this dimensionless coefficient fixed.
Near the point source, and the Batchelor entrainment hypothesis gives . 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, , the trajectory stays vertical rather than developing this cubic transverse displacement. If , there is no finite buoyancy-induced jet length and the turbulent round jet remains straight.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 345 3 d Solution Created 2026-10-03 Updated 2026-10-14
For a horizontal source, throughout. Far downstream , so the kinematic plume fluxes obey the vertical pure plume equations to leading order,Set and with . Matching exponents and coefficients gives and . Consequentlyand the requested literal mass and density-weighted buoyancy fluxes areto leading order. The radius is and the axial velocity decreases as .
The centreline becomes nearly vertical but retains finite horizontal drift:Since , there is a finite positive geometric deficit , and . 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 . The arclength origin should not be interpreted as a physical height: the curved near field supplies the additional shift . 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.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 345 3 b Solution Created 2026-10-03 Updated 2026-10-05
Use the kinematic fluxes , and . In the steady Boussinesq approximation, the non-Boussinesq top-hat plume equations reduce toEliminating gives , where the integration constant vanishes for a pure plume from a point source. Write and . The two balances give and . The Boussinesq point-source plume is thereforeHere and tend to zero at the source, while remains positive. The plume Froude number is independent of height:The ideal point source is a far-field similarity idealization. Since , the Boussinesq approximation fails near and is valid only when . In terms of , this requires ; the density formula must not be extrapolated into its unphysical negative-density region.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 345 3 e Solution 2026-09-28
The final state is displacement ventilation: fresh air of density forms a cool lower layer, radiator-heated air forms a well-mixed upper layer, and the wall plume crosses their interface at height . Steady volume conservation requires the plume volume flux there to equal the imposed ventilation flux,
Below the interface the ambient is uniform, so is constant. Put . The triangular-profile wall line plume equations reduce toThe pure plume conditions at the radiator select the similarity solutionConsequently the displacement-ventilation interface height isThe upper-layer reduced gravity follows from its steady buoyancy balance as . The two-layer solution applies when the calculated interface satisfies .
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 345 3 d Solution 2026-09-28
Measure downward distance from the plume virtual origin byand let be the magnitude of the cold plume's buoyancy flux per unit span. A one-sided wall line plume with width , downward speed , and reduced-gravity magnitude satisfiesThe pure plume solution isImposing fixesThe constant speed is a special feature of a pure top-hat line plume; its width and volumetric flow rate grow linearly with downward distance while entrainment dilutes its density anomaly like .
Plume-fed doorway ventilation 2026-10-14
For a pure plume with , a steady sharp interface at has and warm-layer reduced gravity . A freely discharging upper doorway of height and width has available head . Equating plume supply to the broad-crested weir flux gives the displayed implicit height relation, with . Warm air above the doorway stores heat but adds no available head at its crest in this uniform-layer model.
Source-volume length of a turbulent plume 2026-10-14
Source volume flux is negligible only when entrained volume greatly exceeds it. In the jet limit of the top-hat plume model, this requires . In a pure plume with , it requires , where for full-cross-section fluxes and . These scales differ from the jet length which measures imposed momentum relative to buoyancy. Neglecting source volume does not automatically permit neglecting source momentum.
Starting plume 2026-10-05
A starting plume is the unsteady flow formed when a maintained buoyancy flux first begins. An idealized model joins a steady pure plume below its growing head to a buoyant thermal above. Turner's original starting-plume study develops this plume-and-thermal description.
Two independent point-source pure plumes with remove lower-layer volume . If plume 1 feeds the middle layer, its reduced gravity is . The upper outflow carries both sources, so , yielding the displayed ratio. Opening flow is driven by the integrated hydrostatic head . A plume crossing the lower interface has altered buoyancy relative to its new ambient and must be continued with nonzero incoming mass and momentum fluxes rather than reset as a zero-flux pure plume.
Two-sided top-hat line plume 2026-10-06
For a top-hat plume model with full width , upward speed , reduced gravity , and edge entrainment coefficient , define volume flux per unit span , kinematic momentum flux , and buoyancy flux . A uniform ambient and two exposed edges give , , and . A pure plume with hasIndeed with constant reduces the momentum balance to . If width means half-width, its growth coefficient is , not ; profile shape factors also alter the coefficients.
