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An eddy diffusivity is a transport closure representing unresolved turbulence by a mean scalar flux proportional to a negative mean concentration gradient. A local estimate is $K\sim U\ell$, where $U$ and $\ell$ are turbulent speed and length scales. In a line plume with constant characteristic speed and width proportional to $z$, this gives $K\propto z$. Its coefficient is an independent closure parameter; fluid entrainment conservation alone does not determine it.

For a plume of full width $w(z)$, its horizontally averaged concentration is $\overline{C}(z,t)=c(z,t)/w(z)$, where $c$ is horizontally integrated plume concentration. In a line plume with $w=\mu z$, division by $z$ generally changes the location of maximum concentration. A gradient closure for $c$ differs from one for $\overline{C}$: $w{\partial }_z\overline{C}=c_z-c/z$, so integrated diffusive flux $-Kw{\partial }_z\overline{C}$ equals $-K(c_z-c/z)$.

Use $G=g\beta (T-T_0(z))$ for the plume reduced gravity, and take $z$ upwards. For a symmetric line plume define fluxes per unit source length,The integrals span the plume; its edges can move horizontally with $z$. The fluid entrainment rate at one moving edge is the inward normal transport relative to that edge. For an edge $x=b(z)$ it is $E=w{b}^{\prime }-u$; symmetry supplies equal transport through the other edge. The Batchelor entrainment hypothesis closes its mean value as $E=\alpha w_c$, where $w_c$ is a representative vertical velocity and $\alpha$ is the entrainment coefficient. The integrated volume conservation equation is therefore $q^{\prime }=2\alpha w_c$.

Using volume conservation, the vertical acceleration equation has conservative form ${\partial }_x(uw)+{\partial }_z(w^2)=G$. Incoming ambient has zero vertical momentum; the integrated momentum conservation equation is consequently $m^{\prime }=\int Gdx$. For the temperature anomaly, $T_0$ depends on height andAmbient fluid enters with $G=0$, giving the buoyancy flux balance. The integral model isHere $N$ is the buoyancy frequency when $N^2>0$; the same algebra permits unstable temperature gradients with $N^2<0$.

These are mean integral balances. The advective equations printed in the question omit explicit turbulent stresses and heat fluxes; a turbulent plume interpretation requires the usual entrainment closure for that unresolved transverse transport. Assuming vanishing edge stresses and heat fluxes, and negligible additional axial turbulent-flux contributions, gives the displayed bulk model. It would be inconsistent to treat a discontinuous top-hat profile as a smooth exact pointwise solution of the inviscid temperature-advection equation.

In a homogeneous ambient fluid, $N=0$ and $R={\rho }_0$ is constant. Use kinematic fluxes or, equivalently, the primitive form of the unsteady top-hat line-plume balances:Here $g^{\prime }=g({\rho }_0-\rho )/R>0$ is the positive buoyant reduced gravity. Seek a nonsteady separable power-law ansatz with both time differentiation and vertical transport retained. Matching powers in the volume equation forces $b\propto z$ with no time dependence. Matching the two momentum terms then gives $w\propto z/\tau$, and the buoyancy force has $g^{\prime }\propto z/{\tau }^2$, where $\tau =t+t_{\ast }>0$ is a shifted age. Thus writeSubstitution determines the constants rather than just the exponents. The volume equation gives $2BA=\alpha A$, so for a nonzero buoyant plume $B=\alpha /2$. The momentum equation gives $C=-A+3A^2$, while the buoyancy equation gives $C(-2+3A)=0$. The buoyant nontrivial branch has $A=2/3$ and $C=2/3$. Consequently the separable decaying top-hat line plume isThe Boussinesq approximation requires $2z/(3g{\tau }^2)\ll 1$. The corresponding mass density-weighted fluxes, to this order, areFor a direct check, $Q^2/M=R\alpha z/2$ is time independent and $FQ/M=R\alpha z^2/(3{\tau }^2)$. Hence $(Q^2/M{)}_t+Q_z=2R\alpha z/(3\tau )=R\alpha M/Q$, $Q_t+M_z=R\alpha z^2/(3{\tau }^2)=FQ/M$, and $(FQ/M{)}_t+F_z=0$. This verifies all three balances and the required width coefficient.

This nonsteady branch is not the steady, continuously supplied, constant-flux line plume. The latter has constant $w$ and half-width $b=\alpha z$ in the same top-hat convention. Also, the unshifted power law has $Q,F\to 0$ as $z\to 0$ and is singular at $\tau =0$. It is therefore an interior separable solution, not by itself a solution with a nonzero continuously operating point-source flux at the origin. A finite-width source boundary, a virtual origin, or a matching region is needed to connect it to physical source and initial data. For example, replacing $z$ by $z+z_{\ast }$ supplies a finite base width and time-varying positive base fluxes when $z_{\ast }>0$. No source or initial flux history is specified in the question to select that matching; the requested power-law width follows from the local balances above.

The Batchelor entrainment hypothesis takes the inflow speed through the plume's exposed outer edge to be $\alpha W$, where $\alpha$ is the entrainment coefficient. A wall plume has only one such edge, so $dV/dz=\alpha W$.

The Boussinesq approximation replaces density by a constant reference value ${\rho }_0$ in inertia and mass flux while retaining the small density deficit in buoyancy. It requires $|\hat{\rho }-\rho |/{\rho }_0\ll 1$. A sufficiently hot radiator can violate this near the source, where thermal expansion is large and the developed-plume description may also fail.

Put $B_0=F_0/{\rho }_0$ for the kinematic buoyancy flux per unit length. Dimensional analysis for a line plume givesThe plume rise time is therefore $t_{\mathrm{rise}}\sim H/B_0^{1/3}$. Changing the room stratification requires a plume volume comparable with $XH$, so $t_{\mathrm{room}}\sim XH/V(H)\sim X/B_0^{1/3}$. HenceThe plume consequently follows the slowly changing ambient through a quasi-steady approximation when $X\gg H$.

The source supplies the kinematic buoyancy flux $B_0=F_0/{\rho }_0$. The unstratified line plume velocity scale is $B_0^{1/3}$. Comparing this with the stratification time scale $N_0^{-1}$ gives the stratified line-plume height scaleIf $z_s\gg H$, the initial plume is only weakly affected by the stable density stratification and reaches the ceiling much like an unstratified wall plume. If $z_s\sim H$, its buoyancy flux falls substantially during the rise; it approaches neutral buoyancy near the upper room, overshoots because of its momentum flux, and spreads as a horizontal buoyant intrusion. If $z_s\ll H$, the plume reaches neutral buoyancy low in the room and forms a low intrusion after a modest overshoot. In each sketch the plume widens by entrainment, while increasing $N_0$ lowers the neutral-buoyancy and overshoot heights.

An ideal two-sided line plume has half-width $b=\alpha z$, constant velocity $w_{\ast }=(\mathcal{B}/(2\alpha ){)}^{1/3}$ and volume flux $q=2\alpha w_{\ast }z$. In a closed box of width $2R$, neglecting the plume's occupied area gives $2R\dot{h}=-q(h)$ for the filling-box first front, initially at ceiling height $H$. With negligible diffusion the front is a material interface; its ambient buoyancy jump retains the initial ceiling-discharge value $\mathcal{B}/q(H)=w_{\ast }^2/H$. The contemporaneous plume reduced gravity $w_{\ast }^2/h(t)$ is a different quantity. The upper ambient generally becomes stratified rather than remaining one uniformly mixed layer.

A two-sided line plume whose axial velocity and density anomaly both have triangular transverse profiles. For half-width $b$, ${\int }_{-b}^{b}f(x/b{)}^{n}dx=2b/(n+1)$. These profile moments set the different coefficients of stored mass, momentum, and their fluxes. It differs from a wall plume because both edges entrain ambient fluid.

A wall line plume is a one-sided line plume adjoining a solid wall. Under a top-hat pure-plume model, $d(bW)/ds=\alpha W$ because only its outer edge entrains ambient fluid.