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An inclined forced plume curves upward under buoyancy. In a stationary homogeneous ambient, its horizontal momentum flux is conserved and its vertical momentum flux increases. For kinematic plume fluxes, $m_x^{\prime }=0$ and $m_z^{\prime }=fq/m$, where $m=\sqrt{m_x^2+m_z^2}$ and the derivative is with respect to centreline arc length.

Use the conventional kinematic plume fluxesThen $\pi Q$ is the volume flux, $\pi {\rho }_{r}Q$ the dimensional mass flux, $\pi {\rho }_{r}M$ the dimensional momentum flux, and $\pi F$ the kinematic buoyancy flux. Calling $\pi Q$ a mass flux uses a density-normalized convention: the reference density has been suppressed. If density-weighted momentum and buoyancy fluxes are desired, multiply $M$ and $F$ by the same constant ${\rho }_r$ throughout. Here the conventional kinematic definitions make the Boussinesq approximation explicit.

For an upward plume with $Q,M>0$, the inverse relations areThe volume conservation, vertical momentum conservation, and buoyancy balances from part a becomeHere the vertical momentum conservation equation uses ${\rho }_r$ in the inertia and retains $({\rho }_0-\rho )gA={\rho }_{r}B$ in the buoyancy force. The ambient has no imposed vertical momentum at entrainment.

To classify the resulting partial differential equations, work in the nonsingular primitive variables $(A,w,g^{\prime })$ with $A>0$. Subtract $w$ times the volume equation from the momentum equation and $g^{\prime }$ times the volume equation from the buoyancy equation. This givesThe principal matrix is thereforeAll three characteristic speeds are real and equal to $w$, so the model is a hyperbolic system if that term means only reality of the characteristic roots. However, for $A>0$, $(K-wI)v=0$ requires $v_w=0$, leaving only two independent eigenvectors. HenceThis weak hyperbolicity of the unsteady top-hat plume prevents a complete set of independent characteristic fields. For example, freezing $A,w$ in the homogeneous principal part gives $D_t\delta w=0$ and $D_t\delta A=-A{\partial }_z\delta w$. A Fourier perturbation of wave number $k$ therefore produces a term proportional to $kt$ in $\delta A$. Real characteristic roots alone do not provide uniform control of high-frequency perturbations. The flux-variable transformation above is invertible in the stated region, so changing to $(Q,M,F)$ does not remove this degeneracy.

For a steady upward plume with prescribed $N^2=-G^2<0$, the kinematic plume fluxes satisfyThe positive sign in the last equation expresses the gain of buoyancy flux as the plume rises through the prescribed unstable ambient. Seek a positive power law solution $Q=q{z}^p$, $M=m{z}^r$, $F=f{z}^s$, with $z>0$. Matching powers givesThe last relation gives $s=p+1$ and the first $r=2p-2$. Substitution into the middle relation yields $4p-5=2p+1$, henceThe amplitude equations areFrom the first relation $m=9q^2/(4{\alpha }^2)$, while the last two give $m^2=G^2{q}^2/16$. Taking positive amplitudes gives $m=Gq/4$, and consequentlyReconstructing the radius, speed and reduced gravity gives the top-hat plume in constant unstable stratification:As a direct check, $M_z={\alpha }^2{G}^2{z}^3/9=FQ/M$ and $F_z={\alpha }^2{G}^3{z}^3/9=G^{2}Q$, while $Q_z={\alpha }^{2}G{z}^2/3=2\alpha \sqrt{M}$.

The corresponding ambient gradient is ${\rho }_{0z}={\rho }_r{G}^2/g$. Thus $\rho (z)={\rho }_0(z)-{\rho }_r{G}^{2}z/(4g)$; for a linear ambient this is ${\rho }_0(0)+3{\rho }_r{G}^{2}z/(4g)$. The plume is locally lighter than the ambient even though both densities increase upward in this prescribed unstable stratification.

The coordinate $z$ can be measured from a virtual origin. This exact power law branch has fluxes tending to zero there and describes buoyancy generated by the maintained unstable ambient; it does not impose arbitrary nonzero source fluxes at that origin. The given idealization holds the ambient quiescent despite its static instability. Finally, the Boussinesq approximation limits the modeled height range to small relative density changes, here $G^{2}z/g\ll 1$. Within these assumptions, the displayed positive branch proves the requested existence.

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.

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.

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.