Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 84 (Bouyancy Effects in Fluid)

The Boussinesq approximation assumes that the fractional mass density variations are small. Replace density by a constant reference value in inertia, pressure scaling and mass conservation, but retain its small anomaly in the gravitational force: the large hydrostatic force otherwise cancels, leaving precisely the buoyancy-driving term. For temperature-driven motion, write $\rho ={\rho }_r[1-\alpha (T-T_r)]$ with $|\alpha (T-T_r)|\ll 1$. The resulting velocity is an incompressible flow; compressional heating and acoustic effects are omitted in the shallow, low-compressibility regime.

For this convection problem, use the strong Boussinesq approximation in the constant-property sense: the thermal expansion coefficient, kinematic viscosity, thermal diffusivity and heat capacity are also independent of temperature over the layer. This additional assumption makes the conduction profile linear and the perturbation coefficients constant. This usage of “strong” is the convection convention described in courses.physics.ucsd.edu/2021/Spring/physics218c/Spiegel%20Convection.pdf . Some stratified-flow literature uses “strong” instead to emphasize a globally constant reference density; that condition is also satisfied here. The equations below specify precisely which assumptions are being used.

Introduce the thermal diffusivity $\kappa$ and positive thermal expansion coefficient $\alpha$, which are necessary material data even though the question names only $\nu$ explicitly. Measure $z$ upward from the bottom. The conductive state of Rayleigh-Bénard convection is $T_b=T_0+\Delta T(1-z/d)$ with zero velocity. Let $\theta =T-T_b$ and let $\pi$ be perturbation pressure divided by the reference density, after subtracting the hydrostatic base pressure. Linearizing givesThe sign in the last equation is positive because upward motion carries warmer base-state fluid into cooler surroundings. Taking the divergence of momentum gives ${\nabla }^2\pi =g\alpha {\theta }_z$. Apply a Laplacian to its vertical component and subtract the vertical derivative of that pressure identity. This eliminates pressure:Apply the temperature operator to obtain the vertical-velocity equationThis is the pressure- and temperature-eliminated linear stability analysis equation for Rayleigh-Bénard convection.

At each boundary, impermeability gives $w=0$, the stress-free boundary condition gives $w_{zz}=0$, and perfect thermal conduction gives $\theta =0$. To obtain the second condition, zero tangential traction gives ${\partial }_z{u}_h+{\nabla }_{h}w=0$; since $w$ vanishes identically along the boundary, differentiate horizontal incompressibility to get $w_{zz}=0$. For the fully eliminated velocity equation, $\theta =0$ and the preceding momentum relation also imply $w_{zzzz}=0$ there. Thus suitable formulations areor, for the sixth-order vertical equation alone, $w=w_{zz}=w_{zzzz}=0$ at both boundaries. Fixed boundary temperature means a zero perturbation there, not zero perturbation heat flux.

Use normal modes $w=W\sin (n\pi z/d)e^{\sigma t+i\mathbf{k}\cdot {\mathbf{x}}_h}$, $n\ge 1$, $k=|\mathbf{k}|>0$. Put $K^2=k^2+n^2{\pi }^2/d^2$. Substitution yields the stress-free convection growth-rate polynomialAt stationary marginal stability, $\sigma =0$. In terms of the Rayleigh number $\mathrm{Ra}=g\alpha \Delta T{d}^3/(\nu \kappa )$ and dimensionless horizontal wave number $a=kd$, the free-slip convection neutral curve isWriting $y=a^2$ and differentiating gives $d\log {\mathrm{Ra}}_n/dy=3/(y+n^2{\pi }^2)-1/y$. Its zero is $a^2=n^2{\pi }^2/2$, with minimum $27n^4{\pi }^4/4$. The smallest vertical mode is $n=1$, so the critical temperature difference isThe first unstable horizontal wavelength is $2\sqrt{2}d$. To check that an oscillatory mode cannot destabilize earlier, expand the growth-rate equation: its linear coefficient is $(\nu +\kappa )K^2>0$, and its constant term changes sign exactly at the stationary neutral curve. For positive temperature difference the discriminant is $(\nu -\kappa {)}^2{K}^4+4g\alpha \Delta T{k}^2/(d{K}^2)>0$, so the crossing root is real. Below the global critical value every convection mode decays; at criticality the selected mode is marginal; above it some modes grow and eventually require nonlinear advection, producing convective heat transport. This is exchange of stabilities in stress-free convection.

Strictly, the linearized equations remain the correct small-amplitude equations on either side of onset. What fails above $\Delta T_{\max }$ is stability of the conductive state and long-time validity of a small-perturbation description, not the algebra of linearization. The threshold also presumes that the required small density variations and constant-property approximation hold at that temperature difference.

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 the steady slender-plume model implicit in the requested height-only similarity. Put $\theta =T-T_0$ and introduce permeability $K$, dynamic viscosity $\mu ={\rho }_0\nu$, fluid heat capacity per volume $C_h={\rho }_0{c}_p$, and effective transverse thermal diffusivity $\kappa$. These material constants are needed to give dimensional fluxes. Take $u,w$ to be Darcy velocities, so the thermal transport coefficient for the steady equation is $\kappa =k_e/C_h$ when $k_e$ is the effective conductivity of the saturated medium.

The Darcy law with gravity isThe ambient is hydrostatic with $p_{a,z}=-{\rho }_{0}g$. A slender plume's pressure correction has vertical gradient smaller than the buoyancy term by the squared width-to-height ratio. Indeed, incompressibility gives $u/w$ of order width/height, and the horizontal Darcy relation makes the pressure correction of order $\mu w{\text{width}}^2/(K\text{height})$. Its vertical derivative has the additional inverse-height factor. Thus the leading vertical Darcy balance isThe proportionality is to temperature excess $T-T_0$, not absolute $T$; the velocity vanishes in the unheated ambient. This is the local buoyancy/drag balance of a porous thermal plume, rather than a free-fluid inertial momentum balance.

The governing continuity equation and steady advection-diffusion equation, with axial conduction neglected in the slender high-Rayleigh number limit, areChoose the streamfunction convention $u=-{\psi }_z$, $w={\psi }_x$. It enforces continuity identically and givesA change to the opposite streamfunction convention changes the sign of $\psi$, not the physical velocity or temperature.

The upward convected excess heat flux per unit source length isThis is the quantity denoted $F$ in this question; it differs from Question 2's buoyancy-flux symbol. In conservative form the heat equation is ${\partial }_x(u\theta )+{\partial }_z(w\theta )=\kappa {\theta }_{xx}$. Integrate across the plume:The temperature anomaly and its derivative vanish far from the plume; lateral entrainment brings in ambient-temperature fluid and thus no excess heat. There is no heat source or sink above the line source. Transverse conduction redistributes this heat without removing its total, while axial conduction is negligible in the stated approximation. In an unsteady plume a heat-storage term would have to be retained, so the convective flux alone would not in general be independent of height.

Let $\mathcal{J}=\beta F_h/C_h=\int w^{2}dx>0$. It is constant because $w=\beta \theta$. To obtain similarity exponents without guessing a profile, write the width as $\delta \propto z^a$ and central velocity as $W\propto z^b$. The fixed heat flux requires $W^2\delta$ constant, hence $2b+a=0$. Balancing vertical advection, including its continuity-required horizontal component, against transverse conduction gives $W{\delta }^2\propto \kappa z$, hence $b+2a=1$. Thus $a=2/3$ and $b=-1/3$, and the streamfunction scale $W\delta$ is proportional to $z^{1/3}$. A suitable similarity form isThe volume flux per unit span is $Q=\psi (+\infty )-\psi (-\infty )$ and therefore grows as $z^{1/3}$; it tends to zero at the ideal source. Thus the zero-source-volume requirement fixes the absence of an additive source-length scale, not a finite temperature at the singular point.

Substitution into the streamfunction equation givesChoose the scale normalization $AB=6\kappa$ and $f(+\infty )=1$. Reflection symmetry makes $f$ odd, so the relevant conditions are $f(0)=0$, $f^{\prime \prime }(0)=0$, $f^{\prime }(\infty )=0$, with finite $f(\infty )$; the heat input fixes the remaining dimensional scale. The nonlinear similarity equation becomesIntegrate once. Since $f^{\prime },f^{\prime \prime }\to 0$ in the ambient, the integration constant is zero, giving $f^{\prime \prime }+2f{f}^{\prime }=0$. A second integration gives $f^{\prime }+f^2=C$; the normalized far limit fixes $C=1$. Separation with $f(0)=0$ now yields the complete positive-plume profileIt satisfies the symmetry and decay conditions; the solution of the first-order initial-value equation is unique. The physical heat strength is determined using ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^4\eta d\eta =4/3$, obtained by substituting $t=\tanh \eta$:Equivalently, defineThe hyperbolic-secant porous plume profile and its velocity field are thenIn particular $W{b}^2=6\kappa z$. The far horizontal velocities are inward, $u(\pm \infty ,z)=\mp Wb/(3z)$, supplying the entrained ambient volume. They are consistent with $Q^{\prime }=2Wb/(3z)$.

Finally use specific Darcy-flux moments, per unit source length, for the requested flux relationships. The volume flux, kinematic momentum flux and buoyancy flux areThus the physical mass flux is ${\rho }_{0}Q(z)$, and the conventional density-weighted Darcy second moment is ${\rho }_{0}M$. If intrinsic pore-velocity momentum flux is required instead, a uniform porosity $\phi$ gives ${\rho }_{0}M/\phi$; keeping this convention distinct does not change the displayed Darcy moments. These are the conserved momentum flux of a porous thermal plume relationships. Constancy of $M$ follows here from proportionality of temperature and velocity plus heat conservation, not from ignoring the drag on the solid matrix in a total momentum balance.

The ideal source has zero limiting volume flux but divergent central speed and temperature. Consequently the Boussinesq approximation and slender-plume limit describe the region outside a finite source, where $|\alpha \theta |\ll 1$ and $b/z\ll 1$. Indeed $Wz/\kappa =6(z/b{)}^2\gg 1$ is the local high-Rayleigh condition. These limits improve downstream; they do not make the singular similarity field a literal finite-temperature solution at $x=z=0$.