Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 90 (Environmental Fluid Dynamics)

Assume the shallow-water approximation: the horizontal scale is long compared with the depth, the hydrostatic approximation applies, and the cross-section-averaged velocity is approximately uniform. Neglect viscosity and drag for this part. Volume conservation gives the constant discharge $Q=bhu$. The steady horizontal momentum equation isso its integral is the Bernoulli equation head $\mathcal{H}=H+h+u^2/(2g)$. The shallow-water specific energy measured above the bed is thereforeThe Froude number compares the flow velocity with the long-wave speed $\sqrt{gh}$:At fixed $Q,b$, the shallow-water specific energy tends to infinity as $h$ tends to zero or infinity. Its unique minimum occurs at the critical depthThe two depth branches above that minimum are subcritical flow, $F<1$ with $h>h_c$, and supercritical flow, $F>1$ with $h<h_c$. The long-wave characteristic speeds relative to the bed are $u\pm \sqrt{gh}$. Thus a subcritical flow can receive information from downstream, whereas both characteristic curves travel downstream in a positive supercritical flow.

A hydraulic control is a critical section where one characteristic speed vanishes and a smooth flow can pass between the two branches. Regularity there supplies a condition relating $Q$, the depth and the geometry; together with reservoir or inlet data, it can determine the discharge. Merely finding $F=1$ is not enough: a finite depth gradient also requires the geometric forcing to vanish in the critical momentum equation.

Differentiate the Bernoulli equation at fixed discharge, using $u_x/u=-b_x/b-h_x/h$. This givesAt a smooth hydraulic control, $F=1$ and the right-hand side must vanish. For the specified geometry,For the intended positive parameters $\beta ,\eta ,a>0$, positive finite depth requiresThe control lies between the bed summit and the width minimum, rather than necessarily at either one. For $a<0$ the corresponding interval is $a<x_c<0$. If $a=0$, the coincident summit and throat admit $x_c=0$ without imposing a depth through this first-derivative relation. The limiting cases $\beta =0$ or $\eta =0$ must be handled before division: their control is at the bed summit or throat, respectively.

At $x_c=a/2$, the critical-depth formula simplifies toThis establishes the requested depth without assuming that the bed summit and throat coincide.

There is a further regularity qualification if a critical candidate is to be a simple, smooth transcritical hydraulic control. At fixed $Q$, define the critical-head envelopeReal depths require $\mathcal{H}\ge T(x)$. At a control $\mathcal{H}=T(x_c)$, so $T$ must have a local maximum, not a local minimum. The first condition $T_x=0$ is precisely the compatibility equation above, andFor the positive-parameter case, a nondegenerate control therefore additionally satisfiesIndeed $T_{xx}=-2\eta [3a+\beta (a-x_c{)}^2(3a-10x_c)]/[3(a-x_c)b(x_c)]$. The interval $0<x_c<a$ gives all positive-depth candidates, while this inequality excludes candidates with no smooth critical crossing. At $a/2$ it becomes $\beta a^2<6$ for a simple control; the requested depth relation remains necessary whenever such a control exists. Equality is a degenerate critical point requiring higher-order analysis. For example, $a=1$, $\beta =16$ makes the midpoint a local minimum of $T$, so positivity of the depth alone would incorrectly identify it as a regular control.

Model the specified drag by a prescribed head loss, $(E+H{)}_x=-\gamma$. A modified conserved head is consequentlywith an arbitrary reference constant absorbed into ${\mathcal{E}}_{\gamma }$. More generally the last term is ${\int }_0^x\gamma (s)ds$. The effective bed elevation for the shallow-water specific energy calculation is $H+\gamma x$. The critical momentum equation becomesThis is the head-loss correction to hydraulic control.

When $\beta =0$, $b=1$, and regularity at $F=1$ gives $-2\eta x_c+\gamma =0$. Therefore, for $\eta ,\gamma >0$,The effective bed $-\eta x^2+\gamma x$ has a strict maximum there, with elevation increment $\Delta ={\gamma }^2/(4\eta )$ relative to $x=0$.

Let $d=h(0)$ and require $h_c=2d$. At the control $Q^2=g{h}_c^3=8g{d}^3$. At $x=0$ the shallow-water specific energy is $d+Q^2/(2g{d}^2)=5d$, while at the control it is $3h_c/2=3d$. Equality of the modified head gives $5d=3d+\Delta$. HenceThe inlet has $F(0{)}^2=8$ and is supercritical flow; the branch whose depth increases into the control makes the required smooth transition. At the control the increasing branch has slope $h_x=\sqrt{2\eta h_c/3}$, as follows by expanding the modified head to second order.

The numerical discharge uses a constant loss rate on the entire segment from $0$ to $x_c$. The wording specifies that rate only near the control, which is sufficient to locate $x_c$ but does not by itself determine the inlet-to-control head loss. If the loss elsewhere is unspecified, write ${\Lambda }_c={\int }_0^{x_c}\gamma (s)ds$. The same calculation gives $h_c=H(x_c)-H(0)+{\Lambda }_c$ and $Q=\sqrt{g}{h}_c^{3/2}$; the boxed value is the constant-loss specialization.

The shallow-water approximation assumes a horizontal length scale $\ell$ much greater than the current depth, $h/\ell \ll 1$. Vertical acceleration is then small compared with gravity and pressure is given by the hydrostatic approximation. A depth-averaged model also assumes that the horizontal velocity and excess mass density can be represented by nearly uniform cross-sectional values. Here the ambient fluid is deep or sufficiently undisturbed that its motion and interface-pressure perturbations can be neglected.

The Boussinesq approximation assumes $|\rho -{\rho }_0|/{\rho }_0\ll 1$: mass density differences are retained in gravitational forces but a common reference mass density is used for inertia. A dense bottom current has the positive reduced gravityThus the pressure excess over the ambient hydrostatic value at height $z$ within the current is ${\rho }_0{g}^{\prime }(h-z)$. We neglect bed drag, molecular diffusion and horizontal turbulent stress in the leading balances, but retain the prescribed turbulent entrainment across the interface. Negligible bed drag does not imply that entrainment has no effect on the current's velocity or mass density.

The occupied cross-sectional area and interface width areThe incoming volume flux per unit channel length is $h{w}_e$. Volume conservation therefore gives $A_t+(Au{)}_x=h{w}_e$. Ambient fluid carries no excess mass density, so integrating the excess-mass density balance gives $(A{g}^{\prime }{)}_t+(Au{g}^{\prime }{)}_x=0$. These two equations imply dilution of the reduced gravity, even though total integrated buoyancy is conserved.

The integrated horizontal hydrostatic pressure force, divided by the reference mass density, isThe prismatic walls contribute no axial component of pressure force. Since ambient fluid enters with zero horizontal velocity, it supplies no horizontal momentum flux. Thus the entraining shallow-water current in a triangular channel obeys the conservative equationsTo put them in primitive form, let $D={\partial }_t+u{\partial }_x$ be the material derivative. The volume equation gives $Dh+(h/2)u_x=w_e$. Subtracting $u$ times the volume equation from the momentum equation, and $g^{\prime }$ times it from the buoyancy equation, yieldsThe momentum source is the acceleration of newly entrained ambient fluid, not bed drag. The source in the reduced gravity equation describes mixing with ambient fluid of zero excess mass density. In particular, the rectangular-channel coefficients cannot be used: the triangular area produces both $h/2$ in the depth equation and $h/3$ in the buoyancy-gradient force.

For the state vector $q=(h,u,g^{\prime }{)}^T$, the primitive equations are $q_t+\mathsf{A}{q}_x=S$, withThe characteristic polynomial isConsequently the three characteristic speeds areFor $h>0$ and $g^{\prime }>0$, the hyperbolic system has three distinct real eigenvalues. The long-wave speed relative to the current is $c$, while $u$ transports a buoyancy/contact disturbance.

A left eigenvector for the material characteristic is $(0,0,1)$, givingFor $u\pm c$, left eigenvectors can be chosen asMultiplying $q_t+\mathsf{A}{q}_x=S$ by these rows gives the characteristic compatibility for an entraining triangular-channel current. With all differentials evaluated along the indicated characteristic,The source follows directly from ${\ell }_{\pm }S=-2u{w}_e/h\pm 2c{w}_e/h\mp 2g^{\prime }{w}_e/(3c)$ and $g^{\prime }/c=2c/h$.

Since $4dc=(2c/h)dh+(h/c)d{g}^{\prime }$, an equivalent form isCombining it with the material buoyancy equation also givesOnly when entrainment vanishes and $g^{\prime }$ is spatially constant do $u\pm 4c$ become constant Riemann invariants on these characteristic curves. Dropping the buoyancy differential in the general problem would lose one of the required transport couplings. The characteristic formulas with $1/c$ apply to the nondegenerate region $g^{\prime }h>0$; dry or neutrally buoyant limits require their conservative form instead.

Hindered settling is the reduction of the particle settling velocity by displaced-fluid backflow and interactions with other particles. In the specified model, the downward speed is $V_s$ at vanishing concentration and falls to zero at the maximum packing fraction ${\varphi }_{\max }$. The characteristic speed of a concentration disturbance will differ from this individual-particle speed.

Take $z$ positive upwards. For a locally uniform horizontal suspension without diffusion or bulk fluid motion, the upward particle flux isParticle volume conservation gives the scalar conservation lawThis is the quadratic hindered-settling flux. Along its characteristic curves,For a smooth initial profile ${\varphi }_0(\zeta )$, the map is $z=\zeta +c({\varphi }_0(\zeta ))t$. Its Jacobian is $1+(2V_s/{\varphi }_{\max }){\varphi }_0^{\prime }(\zeta )t$. A shock develops where this first reaches zero: the compressive condition is ${\varphi }_0^{\prime }<0$, andwhen the minimum is negative and no earlier boundary discontinuity is imposed. A discontinuous compressive initial state can instead contain a shock from $t=0$.

Let $z=s(t)$ separate states ${\varphi }_L$ below the shock and ${\varphi }_R$ above it. Integrating the conservation law across a moving thin interval gives the Rankine-Hugoniot conditionEquivalently, the flux relative to the shock agrees on both sides. For this quadratic flux,Since $j^{\prime \prime }=2V_s/{\varphi }_{\max }>0$, a physical compression shock has ${\varphi }_L>{\varphi }_R$ and satisfies $c({\varphi }_L)>\dot{s}>c({\varphi }_R)$: characteristic curves enter it from both sides. These conditions select the shock over a nonphysical expansive discontinuity.

Use the upward coordinate of the preceding kinematic sedimentation calculation, with the given smooth profile on $0\le \zeta \le h$ and clear fluid above. This asks for the first characteristic breaking of that profile; a separately imposed packed-bed discontinuity would be a different boundary problem.

Substitution of the initial concentration into the characteristic speed givesThe characteristic Jacobian isIt first vanishes at the upper edge $\zeta =h$. That edge has zero concentration and descends initially at $V_s$, soAlthough the derivative of the initial profile jumps to zero in the clear region, the concentration itself is continuous there. The shock starts with vanishing amplitude. Both limiting concentrations at onset are zero, and the Rankine-Hugoniot condition consequently givesThis is a downward speed $V_s$.

After formation the upper state remains clear, ${\varphi }_R=0$, and the lower state is positive. The parabolic-profile sedimentation shock hasIt therefore initially slows down in magnitude: its signed upward-coordinate speed increases from $-V_s$. One can also verify that its captured left state increases. Write $r={\zeta }_L/h$, where ${\zeta }_L$ labels the incident lower characteristic. Conservation of the particle volume initially above that characteristic givesThe derivative of the right-hand side with respect to $r$ is $-(r+3)/(r+1{)}^3<0$. Thus $r$ decreases as time increases, and ${\varphi }_L={\varphi }_{\max }(1-r^2)/3$ increases. The downward shock speed continues to decrease on this branch until a bed or another boundary interacts with it. No later boundary behavior is needed for the requested conclusion.

Assume the carrier fluid has ambient mass density ${\rho }_0$, the particles have mass density ${\rho }_p>{\rho }_0$, and volumes add. The well-mixed current then hasHere $g_0^{\prime }$ is the particle-fluid reference reduced gravity per unit volume fraction, which is the convention needed in the requested expansion. The initial current reduced gravity is $g_0^{\prime }{\varphi }_0$; it should not be confused with $g_0^{\prime }$ itself. The Boussinesq approximation requires $\varphi ({\rho }_p-{\rho }_0)/{\rho }_0\ll 1$.

For particles to remain well mixed, the turbulent vertical mixing time must be short compared with the settling time. In a turbulent diffusion description with diffusivity $K_z$, require $h^2/K_z\ll h/[V_s(1-\varphi )]$, or $K_z\gg h{V}_s(1-\varphi )$. Equivalently, characteristic turbulent vertical speeds should substantially exceed the particle settling velocity. Turbulence can maintain an approximately uniform concentration in the body while particles still settle relative to the fluid. At the impermeable bed, fluid has no normal transport into the solid boundary, but particles can reach and be captured by it. Assume no resuspension after capture.

Use a gravity-current box model of length $L(t)$, uniform depth $h(t)$ and uniform concentration $\varphi (t)$. Neglect ambient entrainment and the small volume change caused by depositing dilute solids, so $Lh=M$ remains constant. Take a constant front Froude number $F>0$ and the gravity-current front condition $\dot{L}=F\sqrt{g^{\prime }h}$. With ${\varphi }_{\max }=1$, the deposition flux is $V_s\varphi (1-\varphi )$ per unit bed area. The particle-volume budget isHence the integral model isThe model assumes a fixed rear boundary, one advancing front, a deep stationary ambient, negligible drag during the inertial spreading regime, and a perfectly absorbing bed. The supplied hindered-settling law is retained in the deposition closure.

For $0<\varphi <1$, eliminating time givesThe concentration integral is $2\mathrm{artanh}\sqrt{\varphi }$. ThereforeAs the particle concentration tends to zero, the current reduced gravity and front speed tend to zero. The hindered-settling runout in a rectangular channel isIt is a finite limiting distance; in this idealized model, the exponential tail of particle loss makes complete stopping asymptotic in time. For $L_{\infty }\gg L_0$, omit $L_0^{5/2}$. Using $\mathrm{artanh}x=x+x^3/3+O(x^5)$ givesThe original PDF has $V_s^2$ in this expression; the TeX aid's $V_s^3$ is a transcription error. Dimensions independently confirm the square: $M^3{g}_0^{\prime }/V_s^2$ has dimensions of length to the fifth power. If one instead names the initial current reduced gravity $G_{\mathrm{init}}=g_0^{\prime }{\varphi }_0$, the same expansion is $25F^2{M}^3{G}_{\mathrm{init}}[1+2{\varphi }_0/3+O({\varphi }_0^2)]/V_s^2$. Late loss of turbulent mixing, viscous resistance or bed resuspension can change the physical runout beyond this specified box model.

The entrainment coefficient $\alpha$ is the ratio of the inward ambient-fluid velocity normal to a buoyant plume edge to the characteristic buoyant plume velocity. For an upward buoyant plume, take $u_e=\alpha w>0$. There are two edges at half-width $b$, so the full-section volume source is $2u_e$. Dividing every integral balance by two gives consistent half-section equations.

For top-hat plume profiles, volume conservation and mass conservation areThe mass entering the buoyant plume is ambient mass with local mass density ${\rho }_0(z)$. The Boussinesq approximation uses a common reference mass density in inertia while retaining the mass density deficit in buoyancy. The mass equation may be kept to the first order in that deficit to derive its transport.

Multiply the volume equation by the time-independent ambient mass density. The product rule givesSubtract the mass equation and multiply by $g$. The entrainment sources cancel exactly, leavingwhereFor stable stratification, ${\rho }_0^{\prime }<0$ and the buoyancy frequency $N$ is real. Rising fluid moves into lighter ambient fluid and loses its mass density deficit relative to that local ambient. Thus environmental stratification reduces the buoyant plume's buoyancy flux, even though the ambient fluid entering by entrainment contributes no mass density deficit at its own height.

Keep the half-section convention of part (a). Define the mass density-weighted mass flux, buoyancy flux and momentum flux byThe full-section fluxes are $2Q,2F,2M$. These factors must be changed consistently if full fluxes are used; mixing a full-width flux with a half-width entrainment source would introduce a spurious factor two.

For positive $Q,M$, $w=M/Q$, $\rho b=Q^2/M$, and $({\rho }_0-\rho )gb=FQ/M$. The mass equation in part (a), the momentum equation, and the buoyancy equation therefore becomeThese are direct algebraic rewritings with the mass density factors retained. In particular, $FQ/M$ is the buoyancy inventory per unit height as well as the integrated buoyancy force. If desired, $\rho$ can be reconstructed from $F/(gQ)={\rho }_0/\rho -1$.

At leading Boussinesq order use a constant reference mass density $R$ in the inertial fluxes, $Q=Rbw$, $M=Rb{w}^2$, and write $g^{\prime }=g({\rho }_0-\rho )/R$ so $F=R{g}^{\prime }bw$. Set ${\rho }_0/\rho =1$ in the stratification source and use $N^2\simeq -g{\rho }_0^{\prime }/R$. The unsteady top-hat line-plume balances are thenEquivalently the kinematic fluxes $q=Q/R=bw$, $m=M/R=b{w}^2$ and $f=F/R=g^{\prime }bw$ obey the same equations with $R$ removed from the first source. This explicit distinction between mass density-weighted and kinematic fluxes fixes the normalization for the power-law solution.

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.