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For a horizontal forced plume, the plume virtual origin lies downstream and below the physical source on the far-field vertical asymptote. The origin shift in centreline arc length is distinct from its vertical height: if $z=s-D+o(1)$ and $\hat{s}=s+s_0$, the virtual height is $z_v=-D-s_0$. Both geometric shifts scale with the jet length at fixed entrainment coefficient.

A line plume is invariant along one horizontal direction, so its integral fluxes are measured per unit span. A wall line plume entrains through one exposed edge and has width proportional to distance from its plume virtual origin.

For a rising light plume use $g^{\prime }=g(\rho -{\rho }_p)/{\rho }_0>0$ in the Boussinesq approximation, with ambient density as the inertial reference. The full cross-sectional kinematic volume flux, momentum flux and buoyancy flux of a top-hat plume areHere momentum flux per unit reference density is meant, not momentum per unit volume. The Batchelor entrainment hypothesis says ambient fluid crosses the plume edge at mean inward speed $\alpha w$, where $\alpha$ is the dimensionless entrainment coefficient. Volume added per unit height is $2\pi b\alpha w$. With $a_e=2\alpha \sqrt{\pi }$, the integrated plume balances in uniform ambient density areSeek a pure plume with no persistent source volume or momentum scale. Equating powers of $z$ gives $Q\propto z^{5/3}$ and $M\propto z^{4/3}$; constant buoyancy gives $B=B_0$. Equating coefficients yields, relative to the ideal source or a fitted plume virtual origin,Thus $Q\sim B_0^{1/3}{z}^{5/3}$, $M\sim B_0^{2/3}{z}^{4/3}$ and $b=6\alpha z/5$. A finite nozzle or nonzero source momentum requires a near-source transition rather than this singular point-source solution at arbitrarily small $z$.

Let ${\mathcal{B}}_0>0$ be the source buoyancy flux per unit length. The conversion from the heat-source strength depends on its units: if $F$ is a kinematic temperature-volume flux, ${\mathcal{B}}_0=g\beta F$; if it is a physical heat power per unit length, ${\mathcal{B}}_0=g\beta F/({\rho }_0{c}_p)$, where $c_p$ is specific heat capacity at constant pressure. This explicit convention avoids treating heat power as a buoyancy flux without conversion.

For uniform ambient $N^2=0$, so $\mathcal{B}={\mathcal{B}}_0$. The ideal pure line-source limit has negligible initial volume and momentum transport. The flux equations admit $w=w_{\ast }$ constant, $q=2\alpha w_{\ast }z$ and $b=\alpha z$. Substituting into the momentum equation gives $2\alpha w_{\ast }^2=2bG$, and the conserved buoyancy flux then fixes $2\alpha w_{\ast }^3={\mathcal{B}}_0$. ThusThis verifies all three differential equations, not just their dimensional exponents. The plume grows in width while maintaining constant upward velocity; its temperature anomaly decreases like $z^{-1}$. The zero-height singularity is the ideal point-source limit. A finite source or nonzero momentum input introduces a plume virtual origin or a forced near-source region; the heat strength alone cannot uniquely determine that finite-source region.

Use the flux normalization of the supplied equations, with the common factor $\pi$ suppressed. Put $q=w{b}^2$, $m=w^2{b}^2$ and $B_0=g^{\prime }q$. The top-hat plume model reduces toEliminate height to obtain $dm/dq=B_{0}q/(2\alpha m^{3/2})$. Integrating with zero source volume and momentum flux gives the pure plume balanceEquivalently, seek $b=dz$ and $w=s{z}^{-1/3}$. The volume balance gives $d=6\alpha /5$, and the momentum and buoyancy balances give $g^{\prime }=(4s^2/3)z^{-5/3}$ and $B_0=4d^2{s}^3/3$. ThusA finite source replaces the singular point origin by a plume virtual origin and source conditions. The Boussinesq approximation requires the fractional density difference to be small, so density can be replaced by ambient density in inertia and entrainment while retaining the small difference in buoyancy. In this solution it requires $g^{\prime }/g\ll 1$. The approximation necessarily fails sufficiently near the ideal point source, where $g^{\prime }$ diverges, and can also fail for strongly heated or otherwise large-density-contrast sources. The similarity solution is appropriate farther away once entrainment has diluted the anomaly, assuming a uniform ambient, constant entrainment coefficient and negligible confinement, heat loss and stratification.

Two source scales obtained from dimensional analysis areThe jet length $L_M$ measures how far source momentum remains important relative to buoyancy-generated momentum. The source-volume length of a turbulent plume $L_Q$ measures how far imposed source volume remains important relative to entrained volume. Numerical constants and entrainment coefficient factors depend on the chosen normalization. From the equations here the useful exact volume-growth scale is ${\ell }_Q=[Q_0^3/(20{\alpha }^4{B}_0){]}^{1/5}$; the pure-similarity height whose flux equals $Q_0$ is $5{\ell }_Q/3$. A momentum crossover likewise contains a factor of order ${\alpha }^{-1/2}{L}_M$. The source radius $Q_0/\sqrt{M_0}$ is another useful geometrical combination, but it does not replace the two buoyancy/source crossover comparisons.

Integrate the differential relation from part (a) with finite source values:DefineThis plume flux-balance invariant rearranges toUse it to express $\sqrt{M}=[5B_0{Q}_0^2/(8\alpha ){]}^{1/5}[(Q/Q_0{)}^2-c{]}^{1/5}$, and substitute into $Q^{\prime }=2\alpha \sqrt{M}$. The coefficient satisfies $(2\alpha /Q_0{)}^5[5B_0{Q}_0^2/(8\alpha )]=20{\alpha }^4{B}_0/Q_0^3$, givingWith positive source fluxes, $c<1$. For $c<0$ the source has excess momentum relative to pure plume balance, so it is a momentum-dominated forced plume. For $c=0$ it is already in pure plume balance, and the exact solution is the pure similarity translated to a finite plume virtual origin. For $0<c<1$ the source has deficient momentum relative to its buoyancy and volume fluxes, a lazy plume. Here “forced” in the broad sense merely means finite source fluxes; this sign classification distinguishes its jet-like and lazy branches. At large height, finite $c{Q}_0^2/Q^2$ vanishes and every positive branch approaches pure plume balance.

Let ${\rho }_r$ be a constant reference density close to the ambient density, and let ${\rho }_p$ denote the mean plume density. The Boussinesq approximation permits full-cross-section kinematic fluxesThus $Q$ is the specific mass flux, equal to volume flux at this order; $M$ is the specific vertical momentum flux; and $F$ is the buoyancy flux. Physical mass and momentum fluxes are ${\rho }_{r}Q$ and ${\rho }_{r}M$. These definitions include the full circular area. If factors of $\pi$ are removed in a course's flux convention, divide $Q,M,F$ by $\pi$ throughout rather than mixing normalizations. The PDF's source density is ${\rho }_0$, distinct from ambient ${\rho }_a(z)$; the TeX changes the source symbol incorrectly.

For the top-hat plume model, let the plume radius be $r(z)$ and let $w,b$ be uniform inside it. ThenThe Batchelor entrainment hypothesis supplies the necessary closure: the inward edge velocity is $v_e={\alpha }_{e}w$, with constant entrainment coefficient ${\alpha }_e>0$. Entrainment across a cylindrical slice gives $Q^{\prime }=2\pi r{\alpha }_{e}w=2{\alpha }_e\sqrt{\pi }\sqrt{M}$. The entrained ambient has negligible upward momentum. The vertical buoyancy force on the slice is $\pi r^{2}b=F/w$, hence $M^{\prime }=FQ/M$.

For the buoyancy balance, let $R={\int }_A{\rho }_{p}wdA$ be the transported density scalar. Entrainment gives $R^{\prime }={\rho }_a{Q}^{\prime }$. Differentiate $F=g({\rho }_{a}Q-R)/{\rho }_r$ to get $F^{\prime }=g{\rho }_a^{\prime }Q/{\rho }_r$. Therefore the closed flux equations areHere $N$ is the buoyancy frequency when the ambient is stably stratified. These are the Boussinesq top-hat plume in a stratified ambient equations in the explicitly stated full-area convention. Stable stratification consumes positive buoyancy flux as the plume rises. The top-hat assumption alone would not determine $Q^{\prime }$ without an entrainment law.

For uniform ambient density, $F=F_0$ is constant. The source volume $Q_0$ becomes negligible when accumulated entrainment ${\int }_0^{z}c\sqrt{M(s)}ds$ is much greater than $Q_0$. In a jet-like region, $M\simeq M_0$, so the criterion is $z\gg Q_0/(c\sqrt{M_0})$. In a buoyancy-dominated far field with $Q\simeq C_Q{z}^{5/3}$, it is $z\gg (Q_0/C_Q{)}^{3/5}$, with $C_Q$ determined below. These are the two source-volume lengths of a turbulent plume. They are not the jet length, which measures the persistence of imposed source momentum. A nonzero source volume can also appear as a plume virtual origin; it is not valid to discard it at every distance simply because the source is geometrically small.

For $Q_0=M_0=0$, $F_0>0$, seek $Q=C_Q{z}^{5/3}$, $M=C_M{z}^{4/3}$. The two flux equations give $C_Q=(3c/5)\sqrt{C_M}$ and $C_M^{3/2}=9c{F}_0/20$. Thus the pure-plume solution isFor completeness its directly measurable top-hat fields areThis is an axisymmetric pure plume. In this regime the explicit source-volume criterion isFor $Q_0=F_0=0$, $M_0>0$, the buoyancy vanishes and momentum is constant. The pure-jet solution isThis is the jet limit of the top-hat plume model. Both ideal zero-volume sources have singular speeds at $z=0$; they model regions outside a finite source where entrainment has already made the source-volume correction small.

When $M_0$ and $F_0$ are both positive, use the jet solution near the source and the pure-plume solution far above it, with an appropriate virtual-origin correction. To quantify this transition, eliminate $z$:The conserved quantity is the plume flux-balance invariant. With negligible $Q_0$, comparison of its two terms and $Q\simeq c\sqrt{M_0}z$ gives a transition height of order $M_0^{3/4}/(\sqrt{c}\sqrt{F_0})$. The exact connection, including finite $Q_0$, is the finite-source forced-plume quadraturewith $M$ then recovered from the invariant. A positive integrand gives the physical increasing-volume branch. As $Q$ grows, the source constant is negligible relative to $Q^2$, proving attraction to pure plume similarity. One matches the two asymptotic regimes or uses this quadrature; the jet and plume solutions cannot be linearly superposed because their governing flux equations are nonlinear.

Use pi-normalized kinematic plume fluxes: physical volume, momentum and buoyancy fluxes are $\pi Q$, $\pi M$ and $\pi F$. For a top-hat plume model in a homogeneous ambient, constant entrainment coefficient $\alpha$ givesEliminating height between the first two equations yields $d(M^{5/2})/dQ=5FQ/(4\alpha )$. Hence the plume flux-balance invariant is $M^{5/2}-5F{Q}^2/(8\alpha )$. Pure plume balance means that this invariant vanishes, including at a finite-volume source:It does not mean that $Q_s$ and $M_s$ must individually be zero. The balanced solution has a plume virtual origin below the physical source:where $z_v=[Q_s/(C{F}_s^{1/3}){]}^{3/5}>0$. For given nonzero source fluxes, the displayed balance is a compatibility condition; they cannot be chosen independently and still describe an exactly balanced source.

Use $F=F_s>0$ and constant entrainment parameter $\alpha >0$. The printed integrals are normalized by $\pi$, so a top-hat plume model has $Q=b^{2}w$, $M=b^2{w}^2$ and $F=b^{2}w{g}^{\prime }$; the physical area is still $\pi b^2$. Keeping this normalization avoids spurious factors of $\pi$ in the flux equations.

Eliminate height between the two flux balances. The plume flux-balance invariant follows fromsoFor positive fluxes, both $M$ and $Q$ increase. In particular $Q^{\prime }\ge 2\alpha \sqrt{M_s}>0$, so $Q\to \infty$ as $z\to \infty$. Therefore $C/(\lambda Q^2)\to 0$, whatever the finite source imbalance, and $M\sim {\lambda }^{2/5}{Q}^{4/5}$. Integrating $Q^{\prime }\sim 2\alpha {\lambda }^{1/5}{Q}^{2/5}$ 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 $F=0$ is a different asymptotic problem.

Define the plume balance parameterPure plume balance means $\Gamma =1$ at each height, or equivalently $C=0$: momentum, buoyancy and entrainment have their similarity relation, even when the source has finite radius. A forced plume has $\Gamma <1$, meaning excess momentum relative to this balance; a lazy plume has $\Gamma >1$, meaning insufficient momentum. The invariant ensures that the sign of the imbalance is preserved while $\Gamma \to 1$. In particular, a strictly positive source remains pure at every height precisely whenFor this source the exact solution is the same similarity form with $z$ replaced by $Z=z+z_0$. Writing $a=(9\alpha F_s/10{)}^{1/3}$,Thus the radius growth of an axisymmetric pure plume is $b^{\prime }=6\alpha /5$ and the plume virtual origin lies at $-z_0$, with $z_0=5b_s/(6\alpha )$. Equivalently, in terms of top-hat source properties, pure plume balance is $g_s^{\prime }{b}_s/w_s^2=8\alpha /5$.

For the stated source-to-source geometry, the relevant horizontal distance is $L_s$. Consequently the edge criterion is $b_m=L_s$, rather than the radius at which two expanding edges first touch. Since $b(z)=b_s+6\alpha z/5$,Here $b_s=Q_s/\sqrt{M_s}={\lambda }^{-1/5}{Q}_s^{3/5}$. A positive merging height requires $L_s>b_s$; 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 $L_s/2$.

Use $M^{5/2}=\lambda Q^2$ with $Q=b^{2}w$ and $M=b^2{w}^2$ to get $F_s=(8\alpha /5)b{w}^3$. 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 $b_c=\sqrt{2}{b}_m$. This factor is outside $b_m$, as required by the area. Since speed and reduced gravity are held fixed, the merged fluxes are $Q_c=2Q_m$, $M_c=2M_m$ and $F_c=2F_s$. Their balance parameter isThe immediate combined plume is lazy. The same result follows from $\Gamma =5g^{\prime }b/(8\alpha w^2)$: enlarging the radius at fixed $g^{\prime },w$ increases this parameter.

At fixed $w_m,g_m^{\prime }$, the radius required for pure plume balance isThat is half the stipulated combined area $2\pi L_s^2$. 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 $2F_s$.

Seek a nonzero-buoyancy, separable similarity solution of the unsteady Boussinesq triangular-profile line-plume balances, using powers of distance and time. Let $\tau =t+t_{\ast }>0$, allowing a time origin, and initially take distance $z$ from the virtual origin. Balancing the powers in the momentum and buoyancy equations and the entrainment balance gives the formOne way to recover the exponents is to let $Q\propto z^a{\tau }^p$ and $M\propto z^b{\tau }^q$. Matching the spatial powers in ${\partial }_t(Q^2/M)$, $Q_z$ and $M/Q$ gives $a=2,b=3$; matching the time powers gives $p=-1,q=-2$. The momentum equation then gives the powers of $F$. In the resulting solution, $Q^2/M$ is time-independent, so the storage term in the first equation vanishes exactly.

Direct substitution yields the coefficient equationsFor a rising buoyant solution $C>0$, the last equation gives $B=2A/3$. The first then gives $A=\alpha {\rho }_0$, and the second gives $C=B$. The separable decaying line-plume similarity is thereforeReconstructing the physical fields confirmsFor example, ${\partial }_t(FQ/M)=-2\alpha {\rho }_0{z}^2/{\tau }^3$ and $F_z=2\alpha {\rho }_0{z}^2/{\tau }^3$ cancel; the momentum balance similarly gives $-\alpha {\rho }_0{z}^2/{\tau }^2+2\alpha {\rho }_0{z}^2/{\tau }^2=FQ/M$.

The unshifted form uses $t_{\ast }=0$ and is defined for $t>0$, with a singular zero-time limit. To give a finite solution for all $t\ge 0$, choose $t_{\ast }>0$. The equations also permit replacing $z$ by $z+z_{\ast }>0$, a plume virtual origin; at a finite physical source distance this family has flux histories proportional to ${\tau }^{-1},{\tau }^{-2},{\tau }^{-3}$. The Boussinesq approximation requires $(z+z_{\ast })/(g{\tau }^2)\ll 1$ in the region modelled. This is a decaying similarity family; its time and virtual origins require source or initial data. The PDF supplies neither, so a unique forced-startup history cannot be selected from it. In particular, the singular unshifted field should not be presented as a regular solution at $t=0$.

Write $h(t)$ for the height of the filling-box first front above the floor and $P(h)$ for the upward plume volume flux crossing it. Neglect the plume's horizontal area compared with $A$, 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 $Q$ cannot enter a hermetically sealed, fixed-volume room under the incompressible flow assumption. If $Q>0$, 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 $P(0)=0$, so it cannot be an exact finite-$Q$ source law down to the physical floor. In the far field where $Q\ll P(h)$, or for a heat source whose injected volume flux is negligible, it givesHere $t=0$ is when the plume discharge first reaches the ceiling. Retaining $Q$ while using the far-field law gives $-\dot{h}=[\lambda B^{1/3}{h}^{5/3}-Q]/A$ 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 $F$ be the total ceiling exhaust, $S$ the total buoyant-source volume flux, and $L=F-S$ 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 $b>0$, 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 $P(h)>F$ and upward when $P(h)<F$; since $P^{\prime }(h)>0$, its equilibrium is stable.

In the negligible-source-volume pure plume limit, identifying $F=Q_v$ gives the intended one-source answerThe prescribed inequality makes $h_1/H\ll 1$, so both layers fit in the room. It does not, by itself, justify neglecting a finite source $Q$: the pure plume source region must also be small compared with $h_1$, normally requiring $Q\ll Q_v$. If $Q_v$ denotes additional background ventilation rather than total exhaust, replace $Q_v$ in both formulas by $Q+Q_v$. If it is total exhaust, the separate ambient inlet carries $Q_v-Q$. These conventions must not be mixed.

For $n$ 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 $Q_v$ through the buoyant sources, so $S=F=Q_v$ and $L=0$. 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 $n^{-2/5}$ scaling. For example, a matched plume virtual origin givesWhen $S=F$, this yields $h_{\ast }=0$. The unshifted answer $h_n$ equals the omitted $z_0$ in this case, so neglecting the source region cannot be justified. Also each source has reference buoyancy $(B/n)/(Q_v/n)=B/Q_v$, 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 $S<Q_v$ and the remaining $Q_v-S$ 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 $F=Q_v/2$ makes the new equilibriumThe interface initially descends since its old plume volume flux exceeds the new exhaust. In ventilated filling-box relaxation, writing $V_{\ast }=A(H-h_{\ast })$ and taking the linearization of the two balances above givesAs $P_n^{\prime }(h_{\ast })=5F/(3h_{\ast })$, 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; $2AH/Q_v$ is only a flushing-scale estimate for a different, whole-room adjustment.

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.

Measure downward distance from the plume virtual origin byand let $\mathcal{B}$ be the magnitude of the cold plume's buoyancy flux per unit span. A one-sided wall line plume with width $b$, downward speed $W>0$, and reduced-gravity magnitude $g^{\prime }$ satisfiesThe pure plume solution isImposing $b(z_V)=H^{\prime }/2$ 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 $(z_o-z{)}^{-1}$.

A pure plume is a similarity regime in which the source supplies buoyancy flux but no dynamically persistent source volume or momentum flux. Its finite source region is represented by a plume virtual origin.

For the constant-entrainment axisymmetric plume equations $Q^{\prime }=2\alpha \sqrt{M}$, $M^{\prime }=FQ/M$ with $F>0$, pure balance means $M^{5/2}=5F{Q}^2/(8\alpha )$ at every height. This is a similarity relation between fluxes, not a requirement that the physical source have zero radius. A finite source on this curve is represented by a plume virtual origin. For fluxes normalized by pi and top-hat fields, the same relation is $g^{\prime }b/w^2=8\alpha /5$.

For pi-normalized top-hat fluxes, $b=Q/\sqrt{M}$. The pure similarity solution therefore has $b=(6\alpha /5)(z+z_0)$, so the radius grows linearly with slope $6\alpha /5$. The corresponding plume virtual origin is $-z_0$ and a finite source has $z_0=5b_s/(6\alpha )$. This law translates geometric plume-contact criteria into heights.