Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 65 (Fluid Dynamics of Energy)

Take the boundary to be horizontal, put $z=0$ on it, and let $z=h(x,t)$ be the interface. Write $\Delta \rho =\rho -{\rho }_2>0$. Under the long-thin hydrostatic approximation, the deep ambient fluid has pressure $p_2=p_{\ast }-{\rho }_{2}gz$ to leading order, with negligible horizontal pressure gradient. Matching pressure at the interface gives, inside the dense porous gravity current,The Darcy velocity is superficial volume flux per total area, not pore speed. The horizontal Darcy law and depth-integrated flux therefore giveThe porosity $\varphi$ multiplies storage, so mass conservation is $\varphi h_t+q_x=0$. ConsequentlyThe deep-ambient and long-thin assumptions are what permit neglecting the ambient return-flow pressure variation, vertical inertia, and nonhydrostatic pressure corrections.

For a localized finite release, first use the reflecting half-line $x\ge 0$, so $h_x(0,t)=0$. Let $S=V/\varphi$ be the conserved cross-sectional area of the current; $V$ here is actual pore-fluid volume per unit cross-flow distance. A similarity solution of the porous medium equation has thickness proportional to $t^{-\alpha }$ and extent proportional to $t^{\beta }$. Conservation gives $\alpha =\beta$, and balancing $h_t$ against $(h{h}_x{)}_x$ gives $\alpha +2\beta =1$. Thus $\alpha =\beta =1/3$. PutSubstitution gives $-(F+\xi F^{\prime })/3=(F{F}^{\prime }{)}^{\prime }$. Zero flux at $\xi =0$ allows one integration: $F{F}^{\prime }=-\xi F/3$. In the positive-current region, $F^{\prime }=-\xi /3$, so $F=({\xi }_f^2-{\xi }^2)/6$. The normalization ${\int }_0^{{\xi }_f}Fd\xi =1$ gives ${\xi }_f^3=9$. The one-sided constant-volume porous gravity current is thereforeHere $[a{]}_{+}=\max (a,0)$. Its Darcy flux vanishes at the moving front because $h=0$ there, and its physical volume is $\varphi {\int }_0^{L}hdx=V$. This is the exact concentrated-release Barenblatt solution of the depth equation, physically used after the short initial adjustment to a long-thin current; an arbitrary finite initial patch requires its initial profile as well as its volume, although the same profile describes the usual late spreading regime. If a line release spreads symmetrically in both directions and $V$ denotes the total volume, use the same parabola on $-L<x<L$ with $L^3=9\lambda Vt/(2\varphi )$. If instead volume has already been divided by porosity, replace $V/\varphi$ by that geometric volume.

Drainage at superficial rate $\varphi \Gamma h$ changes mass conservation to $\varphi h_t+q_x=-\varphi \Gamma h$, soFor constant $\Gamma >0$, use the exponential drainage transform for porous-medium diffusionIndeed, $h_t+\Gamma h=e^{-\Gamma t}{\tau }^{\prime }{H}_{\tau }$ and $(h{h}_x{)}_x=e^{-2\Gamma t}(H{H}_x{)}_x$. Choosing ${\tau }^{\prime }=e^{-\Gamma t}$ leaves $H_{\tau }=\lambda (H{H}_x{)}_x$. Substituting the undrained Barenblatt solution givesUse the factor $1/2$ in $L_d^3$ for a total-volume symmetric release, as above. For general initial data the same transform is exact, but $H$ is the corresponding undrained solution rather than automatically a source parabola.

The remaining physical volume is $V(t)=V{e}^{-\Gamma t}$. Early times have $\tau \sim t$, recovering undrained spreading. Late times have $\tau \to 1/\Gamma$, so the front approaches a finite runout while the current drains exponentially:The front slows because drainage reduces the hydrostatic driving head. The finite transformed time also explains why drainage does not guarantee that an arbitrary initial profile reaches its undrained universal large-time shape.

Write $\theta =T-T_e$ and use the Boussinesq approximation for air, ${\rho }_e-\rho \simeq \rho \theta /T_e$. A uniformly heated, well-mixed room has the buoyancy pressure difference $\Delta p_b=\rho gH\theta /T_e$ between its openings. Let $C_o$ denote the combined hydraulic coefficient of the two equal openings, so the effective opening area for pressure-driven ventilation givesFor example, if each orifice has conventional discharge coefficient $C_d$ in $V=C_{d}A\sqrt{2\Delta p/\rho }$, the two pressure losses add and $C_o=C_d$. The question's coefficient is the prefactor in the final heat-load law; define $c=(2C_o^2{)}^{1/3}$. This explicitly states the numerical loss-coefficient convention. Steady heat balance is $Q=\rho C_{p}V\theta$. Eliminating $\theta$ givesHere $Q$ is total heating power and $C_p$ is specific heat capacity at constant pressure. Writing $c=C_d$ without this conversion would incorrectly change the power of the discharge coefficient.

After the reduction, cooler newly heated air occupies the bottom and displaces the old warmer air upward. Use the ideal two-layer displacement ventilation model: each layer is well mixed, the upper layer remains at ${\theta }_0$, walls have negligible heat capacity, and the sharp interface is advected without exchange between the layers. Let the lower layer have depth $h$ and excess temperature $\theta$. Integrating the hydrostatic buoyancy over the full height givesThe room's excess thermal energy is $E_T=\rho C_p{A}_r[h\theta +(H-h){\theta }_0]$. The floor supplies $\lambda Q$, incoming air has zero excess temperature, and the ceiling removes $\rho C_{p}V{\theta }_0$. Thus the required heat balance isThis moving-interface energy term must include the loss of old upper-layer volume; balancing only the lower-layer temperature would miss it.

For a genuine reduction assume $0<\lambda <1$. SetThe integrated buoyancy factor is $H{\theta }_0(1-x)$, so $V/V_0=\sqrt{1-x}$. Since $Q=\rho C_p{V}_0{\theta }_0$, the thermal balance becomes the ventilation after a heating reduction equationThe material interface also moves with the upward displacement velocity, $A_r\dot{h}=V$. Therefore ${\eta }_s=\sqrt{1-x}$, and subtracting the two equations givesThe second relation is also the lower-layer energy balance: its total excess energy increases at rate $\lambda Q$. It does not assume its temperature stays constant. Since $x_s>0$ until $x=1-{\lambda }^2$, the solution remains $0\le x<1-{\lambda }^2$, and $V/V_0\ge \lambda$. Hence ${\eta }_s\ge \lambda$, so the lower layer reaches $\eta =1$ in finite time $s_H\le 1/\lambda$. Also ${\eta }_s\le 1$, so $s_H\ge 1$.

An explicit implicit solution is useful. Put $y=\sqrt{1-x}$, which decreases from $1$ toward $\lambda$. Separation givesFilling occurs when $1=1-y_H^2+\lambda s_H$, so $\lambda s_H=y_H^2$ and ${\theta }_H/{\theta }_0=y_H^2$. These relations determine the exit time and the continuous temperature at that time. For complete switch-off $\lambda =0$, the limiting solution is $y=1-s/2$, $x=\eta =s-s^2/4$ for $0\le s\le 2$, so the cold lower layer fills the room at $s_H=2$ even though its final ventilation rate is zero.

After the interface exits, the whole room is well mixed. For $z=\theta /{\theta }_0$, the natural ventilation rate is $V=V_0\sqrt{z}$, and the well-mixed ventilation temperature balance becomesEquivalently, $\rho C_p{A}_{r}H\dot{\theta }=\lambda Q-\rho C_p{C}_{o}A\sqrt{gH/T_e}{\theta }^{3/2}$. The heat-removal term is strictly increasing in $\theta \ge 0$, so every nonnegative solution tends to the unique stable equilibriumFor $\lambda =0$ the final temperature is $T_e$. The unchanged-load case $\lambda =1$ has no distinct new temperature layer and is outside the reduction scenario.

Let the Eulerian time average commute with space and time derivatives and let the statistically steady mean have no time derivative. Define $U=\overline{u}$, $V=\overline{v}$, $\hat{u}=u-U$ and $\hat{v}=v-V$, so $\overline{\hat{u}}=\overline{\hat{v}}=0$. Averaging instantaneous incompressibility givesand subtracting gives ${\hat{u}}_x+{\hat{v}}_y=0$ for the fluctuations. Start from the instantaneous streamwise Navier-Stokes equationUse incompressibility to write $u{u}_x+v{u}_y={\partial }_x(u^2)+{\partial }_y(uv)$. The averaged products are $\overline{u^2}=U^2+\overline{{\hat{u}}^2}$ and $\overline{uv}=UV+\overline{\hat{u}\hat{v}}$. The mean conservative terms reduce back to $U{U}_x+V{U}_y$ because $U_x+V_y=0$. The Reynolds-averaged momentum equation is consequentlyThe signs use the positive covariance convention for the Reynolds stress. In particular, both quadratic averages involve the fluctuations, not products of the mean $U$ with a fluctuation; the converted TeX loses the overbar/hat structure here.

Let $L$ be the streamwise evolution scale, $\delta$ the jet width, and $U_c$ its velocity scale. A slender jet has $\epsilon =\delta /L\ll 1$; mean incompressibility gives $V=O(\epsilon U_c)$. Both mean advection terms are then $O(U_c^2/L)$. For turbulent parcels crossing a transverse distance $l$, retention of their original streamwise momentum gives a fluctuation $\hat{u}\sim -l{U}_y$; a comparable transverse fluctuating speed gives the down-gradient mixing length closureThe mixing length is an effective distance of momentum retention, with statistical coefficients absorbed into $l$; it is not the net transverse displacement of a mean streamline. This closure models the Reynolds stress rather than deriving it exactly.

In a slender equilibrium jet, the fluctuation magnitude inferred from momentum balance is $u^{\prime 2}\sim U_c^2\delta /L=\epsilon U_c^2$. The transverse Reynolds stress gradient then has size $u^{\prime 2}/\delta \sim U_c^2/L$, whereas the streamwise normal-stress gradient is $u^{\prime 2}/L\sim \epsilon U_c^2/L$. At sufficiently high Reynolds number, molecular transverse diffusion is smaller providedStreamwise molecular diffusion is smaller still by ${\epsilon }^2$. A free jet has ambient mean pressure constant to leading order; transverse momentum corrections imply only a higher-order mean streamwise pressure force. Thus high speed must be accompanied by the slender-jet ordering, not used alone to discard every other term.

For a symmetric jet decreasing away from the centre, $U_y<0$ on $y>0$ and $U_y>0$ on $y<0$. Accordingly $\overline{\hat{u}\hat{v}}=\mathrm{sgn}(y)l^2(U_y{)}^2$ off the centreline, and the leading equation isAt the centreline the signed form requires the stress to tend continuously to zero; otherwise differentiation would add a distributional term. The undifferentiated Reynolds stress form is the appropriate global equation.

The proposed mixing length can be motivated as a local phenomenological estimate: on either half of a self-similar jet, the centreline supplies a distinguished reference plane, and an eddy at distance $|y|$ might retain momentum over a fixed fraction $C_1|y|$ of that transverse distance. This givesIt is compatible with the dilation $x\mapsto ax$, $y\mapsto ay$ and introduces no independent outer length. In the turbulent plane jet similarity regime its typical value at $|y|\sim \delta$ is $C_1\delta$, so the scaling from part (b) has $\delta /L\sim C_1^2$.

This argument is a modeling assumption, not a physical necessity: a jet centreline is not a wall that prevents turbulent eddies from crossing it. Thus a mixing length vanishing there cannot automatically be justified throughout the core. Parts (e) and (f) exhibit the precise difficulty: the resulting finite-velocity similarity profile has singular centreline shear and zero shear at a finite free edge, contradicting the printed edge definition. The local dimensional motivation does not make all the stipulated global jet properties consistent.

Use mean incompressibility to convert advection into momentum-flux divergence:Therefore the reduced Reynolds-averaged momentum equation givesFor a localized free jet, $UV$ and the Reynolds stress tend to zero in the quiescent ambient as $y\to \pm \infty$. Assume sufficient decay to differentiate its momentum integral. Integrating across the jet givesThus $M_0=\int U^{2}dy$ is constant. It is a specific momentum flux per unit span: multiplication by density gives the momentum flux. A clipped profile with nonzero stress just inside its artificial edge would fail this boundary-flux assumption unless an additional external momentum source were supplied.

First derive the formal turbulent plane jet similarity equations, then check the printed edge conditions. Define a streamfunction by $U={\psi }_y$, $V=-{\psi }_x$, so mean incompressibility is automatic. Momentum conservation gives $U_c^2\delta \sim M_0$. With $l=C_1|y|$, the closure balance $U_c^2/x\sim C_1^2{U}_c^2/\delta$ gives $\delta \propto x$. Hence $U_c\propto M_0^{1/2}{x}^{-1/2}$ and the consistent self-similar ansatz isAn odd $f$ gives symmetric $U$ and antisymmetric $V$. Direct differentiation givesThe advection and Reynolds stress terms becomeThus the two required consistency equations, the first for $\eta >0$ and the second for a symmetric finite-width profile, areThe second is momentum normalization; it is the equation referenced as (2) in the PDF, although its display in part (d) has no printed number. Use an infinite upper limit for a decaying profile of infinite support. Symmetry sets $f(0)=0$. A finite centreline velocity and continuous zero centreline stress give the integrated first equationThese equations establish the similarity reduction. They also give a direct finite-edge stress condition for a mixing-length jet. At a finite edge ${\eta }_w>0$, $U=0$ requires $f^{\prime }({\eta }_w)=0$. For finite $f({\eta }_w)$, the integrated equation forces $f^{\prime \prime }({\eta }_w)=0$. A nontrivial decreasing velocity profile has nonzero shear somewhere inside, so zero edge shear cannot be its maximum. Allowing a one-sided nonzero edge shear instead produces nonzero edge stress and an unbalanced jump in the conservative momentum equation.

For a concrete diagnostic, take $f^{\prime }=1-{\eta }^2$ on $0\le \eta \le 1$, so $f=\eta -{\eta }^3/3$. It has zero edge velocity and its largest one-sided shear magnitude at the edge, matching that descriptive condition. Yet at $\eta =1$ the integrated equation has left-hand side zero and right-hand side $8C_1^2>0$: such a profile is not a solution of the stated momentum closure.

The centreline is also problematic for a smooth interpretation. If $f^{\prime }(0)=a>0$, then $f\sim a\eta$, and the decreasing branch of the integrated equation givesA continuous formal profile can have this cusp, but it is not a differentiable symmetric velocity field, and molecular diffusion cannot be neglected uniformly in that core. Therefore there is no nontrivial regular similarity solution satisfying all the printed edge requirements. The boxed consistency equations are valid; the additional coincidence of a zero-velocity edge with maximum shear is not. A consistent jet model must relax that edge definition and regularize the core or use a different mixing length.

The eddy viscosity in the prescribed mixing length model isThus at a fixed similarity coordinate $\eta$, every nonzero value grows as $x^{1/2}$. Its maximum occurs at a fixed coordinate maximizing ${\eta }^2|f^{\prime \prime }|$, so the maximum itself has the same downstream scaling.

The formal argument intended by the edge assumption is also clear. If one independently stipulates $|f^{\prime \prime }(\eta )|\le |f^{\prime \prime }({\eta }_w)|$ for $|\eta |\le {\eta }_w$, then ${\eta }^2|f^{\prime \prime }(\eta )|\le {\eta }_w^2|f^{\prime \prime }({\eta }_w)|$. This would put the maximum eddy viscosity at the edge and giveindependent of $x$, provided $f^{\prime \prime }({\eta }_w)\ne 0$. That is a conditional scaling calculation, not an admissible solution of the full momentum problem.

The printed edge claim fails for the reason established in part (e): $f^{\prime }({\eta }_w)=0$ forces $f^{\prime \prime }({\eta }_w)=0$, and thereforeA nontrivial profile has positive eddy viscosity in the interior; hence its maximum cannot occur at that finite free edge. Moreover $U(x,0)y_w/{\nu }_T(x,\pm y_w)$ has a zero denominator, so it is not a finite effective Reynolds number under the literal assumptions.

The intended downstream cancellation is valid if one uses a consistent interior reference point ${\eta }_{\ast }\in (0,{\eta }_w)$ with $f^{\prime \prime }({\eta }_{\ast })\ne 0$, for example the maximum of ${\nu }_T$. Since $U(x,0)=M_0^{1/2}{x}^{-1/2}{f}^{\prime }(0)$ and $y_w={\eta }_{w}x$,which is independent of $x$. This is a qualified correction of the reference point, not a proof of the false edge maximum.

Let $h_0=h(0)$ and $S=\pi R^2$. The closed tank has constant mass, soTherefore $({\rho }_l-{\rho }_u)h=({\rho }_l-{\rho }_u(0))h_0$. Under the Boussinesq approximation, define $g^{\prime }=g({\rho }_l-{\rho }_u)/{\rho }_l$ and obtainThe gravitational potential energy is especially easy to evaluate with the stipulated downward depth coordinate: the gravitational potential per unit mass is $-gz$. HenceThis proves the potential energy of two-layer entrainment relationThe positive sign expresses the lifting and homogenization of dense entrained fluid within the upper layer. The fixed thin interface is neglected in this leading two-layer energy integral.

To connect the prescribed interfacial drag coefficient to the energy supply, state the lumped interfacial power model for entrainment explicitly:It estimates the power transmitted by lid-driven flow to the mixing interface. Constant geometric and transfer factors are included in $c_D$ or $C_W$. It is the local stress-power interpretation required for the depth laws in this question, not an identity equating arbitrary lid shaft power to interfacial power. With the potential energy of two-layer entrainment from (a),In model S, $u_I=\Omega R$ and $B$ are constant, soThus the depth increases at a constant rate proportional to the inverse bulk Richardson number, with $h=h_0+2C_W{c}_D\Omega Rt/{\mathrm{Ri}}_B$, until the layer reaches $H$. Here the prescribed $u_I$ must be read as a characteristic speed. Literal pure solid-body rotation has zero radial velocity, so the PDF's parenthetical description of $u_I$ as radial cannot simultaneously be a strict component statement and the assumption $u_I=\Omega R$.

If the phrase about lid work is taken literally as shaft power $P_{\mathrm{lid}}=\Omega {\mathcal{T}}_{\mathrm{lid}}$, the interfacial stress alone does not specify the lid torque or its energy transfer to the interface. The preceding closure is therefore a visible model assumption, not a hidden consequence of solid-body rotation.

Under the Boussinesq approximation, total upper-layer kinetic energy is $K={\rho }_{l}Sh{u}_U^2/2$, where $u_U$ is its volume root-mean-square speed. For solid-body rotation of angular speed $\omega (t)$, direct radial averaging gives $u_U^2={\omega }^2{R}^2/2$, consistent with that energy expression. A constant $K$ implies $h{u}_U^2=h_0{u}_U(0{)}^2$. Define $C_K=u_U(0)/(\Omega R)$ to obtain the fixed-energy turbulent mixed layer relationWith $u_I=C_I{u}_U$, the interfacial power model for entrainment used in (b) givesMultiply by $h^{3/2}$ and integrate from $h_0$. This yieldsThe fixed-energy two-layer entrainment law is thereforeHere ${\mathrm{Ri}}_{B,0}=B/({\Omega }^2{R}^2)$ uses the same lid-speed normalization as part (b). All numerical geometry factors depend on the chosen characteristic stress/speed convention. The formula ceases to apply once $h=H$.

The exponent $2/5$ requires the local power to decrease as $u_I^3\propto h^{-3/2}$. It does not follow from constant upper-layer kinetic energy and a fixed fraction of unspecified lid power alone. For example, an imposed constant shaft power together with a constant fraction going to potential energy would give constant $\dot{h}$, not this exponent. The appropriate distinction between local interface power and global forcing appears explicitly in the primary study Entrainment and mixed layer dynamics of a surface-stress-driven stratified fluid, equations (3.6)–(3.7); the calculations above use that local interpretation.

For stable shear-driven turbulence, let $P$ be the rate of production of turbulent kinetic energy by mean shear and $B_m>0$ the rate consumed by buoyancy work. The Flux Richardson number isWith an upward coordinate and vertical turbulent buoyancy flux $\overline{w^{\prime }{b}^{\prime }}$, $B_m=-{\rho }_l\int \overline{w^{\prime }{b}^{\prime }}d\mathcal{V}$, while shear production is $P=-{\rho }_l\int \overline{u^{\prime }{w}^{\prime }}{\partial }_{z}Ud\mathcal{V}$. In a locally steady balance without net turbulent-energy transport or storage, $P=B_m+\mathcal{D}$, so $0\le {\mathrm{Ri}}_f\le 1$ and the dissipation-normalized mixing coefficient is $B_m/\mathcal{D}={\mathrm{Ri}}_f/(1-{\mathrm{Ri}}_f)$. A global ratio $C_W={\dot{E}}_P/P_I$ need not equal this local production-normalized Flux Richardson number when energy transport, storage, or bulk acceleration is important.

The interfacial Richardson number in this tank isIn model S,Thus the relative importance of interface stratification changes continually. If mixing efficiency depends on interfacial Richardson number, constant $C_W$ is not generally self-consistent over substantial deepening; it can at best describe a regime where that dependence is weak. In model K,A local efficiency depending only on this ratio can therefore remain constant, provided the fixed interface thickness and the other empirical coefficients remain appropriate.

The energetic reason to expect intermediate behavior is that entrainment must both lift and homogenize dense fluid and mobilize it. The whole upper-layer kinetic energy budget has the form $P_{\mathrm{lid}}=\dot{K}+{\dot{E}}_P+\mathcal{D}$, apart from additional boundary losses. Keeping the characteristic speed constant makes $K\propto h$ and requires a continuing power supply to accelerate new fluid, together with dissipation in an ever larger layer. Keeping $K$ constant instead forces the speed to decline as $h^{-1/2}$. A finite lid forcing plausibly produces an initial regime near the former and a deeper regime near the latter, rather than indefinite constant-speed entrainment.

One can make this expected bracket quantitative after matching the models' initial interface speed $u_0$. Suppose the actual speed lies between $u_0\sqrt{h_0/h}$ and $u_0$, and the same local coefficients are used. Set $D_0=2C_W{c}_D{u}_0^3/B$. ThenIntegrating these inequalities givesEquivalently, if $K\propto h^q$ with $0\le q\le 1$, the local power closure yields a late-time depth exponent $2/(5-3q)$ between $2/5$ and $1$. This is the intended energetic interpolation between K and S, not a universal bound following from the printed assumptions alone. Different efficiencies or unmatched initial coefficients can invalidate pointwise ordering; quantitative prediction requires the lid-to-interface power transfer and the actual energy budget.