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deuterium proton capture is much faster than the weak reaction that produces deuterium. Its abundance therefore adjusts so that ${\lambda }_{21}{n}_2{n}_1\simeq {\lambda }_{11}{n}_1^2/2$. Substituting this balance eliminates the fast intermediate from the slower proton and helium-3 evolution equations. This is a quasi-steady approximation, not an assumption that hydrogen itself is in nuclear equilibrium.

Take a reservoir with constant horizontal cross-sectional area $A$, one small orifice of area $a$, no inflow, and constant stipulated discharge coefficient $C_d>0$. Let the initial head be $H>0$. For constant mass density $\rho$, the stored mass above the outlet level is $\rho Ah$ and outgoing mass flow rate is $\rho Q$. Mass conservation cancels $\rho$ and gives the initial value problemThe downward speed of the free surface is $Q/A$, smaller than the ideal jet speed by the factor $C_{d}a/A$. Thus a small opening relative to the reservoir supports the negligible-surface-velocity assumption.

For $h>0$, separation of variables or the chain rule gives $d\sqrt{h}/dt=-k/2$. Integrating yieldsThe model reaches zero head in finite time even though volumetric flow rate tends to zero. Extending the squared expression beyond $T$ would make head grow again and violate the original ordinary differential equation. The zero-head continuation enforces no further outflow. For an elevated side opening, zero head means the surface has reached that opening, leaving fluid below it; $T$ is not the time to remove every drop from such a tank.

For an original synthetic example, stipulateThe initial volume above outlet level is $AH=0.0250{\mathrm{m}}^3=25.0\mathrm{L}$. The initial ideal jet speed is $2.215\mathrm{m}{\mathrm{s}}^{-1}$, the initial volumetric flow rate is $Q_0=0.1373\mathrm{L}{\mathrm{s}}^{-1}$, and the predicted zero-head time is $T=364.1\mathrm{s}$. These are consequences of stipulated parameters, not experimental measurements. At $t=T/2=182.1\mathrm{s}$, head is $H/4=0.0625\mathrm{m}$, stored volume is $6.25\mathrm{L}$, and volumetric flow rate is $Q_0/2=0.06866\mathrm{L}{\mathrm{s}}^{-1}$. Half the initial volume has already left at $t=T(1-1/\sqrt{2})=106.7\mathrm{s}$.

An independent mass conservation check integrates the outgoing volumetric flow rate:Multiply both sides by mass density to obtain discharged mass. A dimensional analysis check gives $[k]={\mathrm{m}}^{1/2}{\mathrm{s}}^{-1}$, $[Q]={\mathrm{m}}^3{\mathrm{s}}^{-1}$ and $[T]=\mathrm{s}$. The triangular area under $Q(t)$ explains why the initial-rate estimate $AH/Q_0$ is only half the actual model time. Doubling $A$ doubles $T$; doubling $a$ or $C_d$ halves $T$, within the same assumptions.

This finite-time result is an idealized prediction. The constant $C_d$ and quasi-steady approximation need not survive very small head or the initial startup. Appreciable viscosity in a narrow outlet tube requires a different pressure balance, as in Poiseuille flow; appreciable surface tension can prevent zero head altogether.

Balances and meanings. The first equation is conservation of energy per unit area. Here $c$ is effective areal heat capacity, $Q$ is the solar constant, and $Q/4$ accounts for the ratio of the Earth's intercepting disc area to its total surface area. The albedo $a$ reflects a fraction of that input. The outgoing thermal radiation is the Stefan–Boltzmann law modified by $\gamma (p)$, an effective transmission factor for the greenhouse effect; it normally decreases as atmospheric carbon dioxide increases.

The albedo relaxes towards its temperature-dependent equilibrium on the ice-adjustment time $t_i$. The atmospheric mass inventory per unit pressure is $M_{{\mathrm{CO}}_2}A/(M_{a}g)$, using the molar mass ratio of carbon dioxide to air and hydrostatic atmospheric mass. Its sources and sinks are volcanic input $v$, removal by chemical weathering over land area $A_L$ at rate $W$, and the atmosphere-to-ocean exchange $h(p-p_s)$. The ocean equation is a molar mass conservation balance: ${\rho }_{{\mathrm{H}}_2\mathrm{O}}{V}_{\mathrm{oc}}$ is ocean water mass, division by $M_{{\mathrm{CO}}_2}$ converts the incoming carbon-dioxide mass to moles, and $bC$ removes dissolved carbon by burial/export. With $C$ in moles per kilogram, $b$ has units of kilograms of water-equivalent per year.

Times and reduction. Linearize the temperature equation at fixed $a,p$ about radiative equilibrium. Its exponential relaxation time isThe equality uses $\sigma \gamma T^4=Q(1-a)/4$. A stock-over-flux thermal time, sometimes used as a scale, is four times this, $0.381\mathrm{y}$; both describe a rapid thermal response relative to the other processes.

For pressure in atmospheres, write $I=M_{{\mathrm{CO}}_2}A10^5/(M_{a}g)$ for the atmospheric carbon-dioxide mass per atmosphere. The factor $10^5$ is essential because $h$ is given per atmosphere. The pressure and slow ocean exponential relaxation times areThus $t_T\ll t_p\ll t_i=10^4\mathrm{y}<t_C$. The leading quasi-steady approximation givesSubstitution in the two remaining equations yieldsThe coefficient of $\dot{a}$ remains $t_i$; the printed $t_T$ in the reduced equation is a typographical error. Eliminating a fast temperature variable does not replace the ice-response time. The ocean time here is the slow time after atmospheric balance cancels the exchange flux; holding pressure fixed after imposing a carbon-dependent $p_s$ would describe a different conditional relaxation.

The supplied volcanic input givesabout $41.1\%$ of the stated present value $2.0\times 10^{-3}$; that value is about $2.43C_0$.

Carbon-dioxide levels. Since $A_{L}W=3.0\times 10^{11}\mathrm{kg}{\mathrm{y}}^{-1}=v$, the atmospheric balance with $p_s=C/K$ givesThis uses the specified present ocean concentration, as requested, rather than replacing it by the model's smaller volcanic-only equilibrium.

After sustained additional input $v_i$, $C$ barely changes during the atmospheric adjustment. At this stageThis is an intermediate balance, approached over several centuries. On the slower ocean time,with pressure following $p\simeq C/K+v_i/h$. Since $C(0)<C_{\infty }$, pressure continues to rise slowly. Its eventual level in this stipulated constant-parameter model isThe two-reservoir carbon adjustment therefore has a rapid rise towards the frozen-ocean balance followed by a much longer increase. The figure solves the coupled linear pressure–ocean equations, retaining their small inventory correction to the reduced time scales.

Write $e=E^{\prime }(1)>0$, $q_1=q^{\prime }(1)>0$ and introduce perturbations $(H,U,C,S)e^{\lambda t+ikx}$ to water depth, velocity, concentration and bed height. Take the leading quasi-steady approximation $ϵ=0$ for the fast water and suspension response, retaining the small friction parameter $\delta$. For $k\ne 0$, the linearized equations areHere surface height is $H+S$, and linearization of $1-u^2/h$ gives $H-2U=-3U$. Eliminating $H,C,U$ gives the dispersion relationIn particular, putting $B=F^2-1\ne 0$ and expanding at large $|k|$,Thus the temporal growth rate tends to a finite constant, while the phase velocity isFor supercritical flow $F>1$, the waves propagate upstream. They grow if $\beta e(F^2-1)>3\delta q_1$; at leading order in small $\delta$, this holds for fixed positive $\beta ,e$ away from critical flow. They are antidunes, upstream-migrating bedforms coupled to the free surface. Indeed $H=-U$ and $S\simeq -(F^2-1)U$, so surface height $H+S\simeq -F^{2}U$ is in phase with the bed perturbation in this regime.

For clarity, neglecting $\delta$ completely gives the exact leading-order expressionsThese also identify the missing qualification in the source: the unstable upstream branch requires $F>1$. For $F<1$ this leading model gives decay and downstream propagation. The near-critical case $F^2-1\simeq 0$ needs a different balance; it is not covered by the expansion. Small nonzero $ϵ$ perturbs the slow bedform branch and adds fast water modes, so the stated result concerns the leading slow morphodynamic branch. This is the high-wavenumber antidune dispersion calculation.

Introduce specific heat capacity $c_p$, latent heat per unit mass $L_f$, and $\Delta T=T_m-T_s>0$. Write the prescribed convective heat flux as $F=K(\overline{T}-T_m)$, whereThere is a dimensional convention to specify: with $Q$ measured in $\mathrm{W}\mathrm{k}{\mathrm{g}}^{-1}$ and dimensionless $\lambda$, this printed expression makes $K$ a heat transfer coefficient only if $\mu$ is kinematic viscosity. If $\mu$ denotes dynamic viscosity, the numerator needs ${\rho }_0^{2}g\alpha Q$ instead. The question says only “viscosity”; use the consistent kinematic interpretation, or equivalently the corrected dynamic formula. The following balances apply to either convention once $K$ has units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

For a thin lava lake crust with a linear temperature profile, Fourier's law gives the outward conductive flux $k\Delta T/a$. The Stefan condition balances it against upward thermal convection and freezing, while the well-mixed liquid loses the convective flux and gains radiogenic heating. To leading order in $a/H$ the radiogenically heated lava-lake crust model isThe fixed surface temperature already incorporates radiative cooling, so no extra radiative loss is added to the interior balance. The linear conductive profile is interpreted as a quasi-steady approximation, so sensible-heat storage within the crust is not evolved. Thinness alone does not establish this approximation: rapid thermal adjustment in the crust is also needed, for example a small Stefan number during the initial conduction-driven growth. Replacing the liquid depth by $H-a$ and retaining the moving liquid-volume enthalpy would introduce thin-crust corrections: the requested constant-temperature branch refers to the leading model above. A transient thermal profile in a thick or newly nucleating crust would require the full Stefan problem, rather than this linear-profile closure.

Using the thermal diffusion time $H^2/\kappa$, defineThe three independent dimensionless parameters are the convective-conductive coefficient $B$, the heating parameter $R$, and the Stefan number $S$. For positive $B,R,S$ the evolution equations becomewhere prime denotes $d/d\tau$. Temperature has the exact solutionConsequently the unique constant-temperature initial condition isAt this temperature, the convective flux equals the column's radiogenic heat production. Starting with no crust, the constant-temperature lava-crust growth law is obtained by separating $h^{\prime }=S(1/h-R)$:The crust grows towards $h_{\ast }=1/R$, or $a_{\ast }=k\Delta T/({\rho }_{0}QH)$, where conduction carries all the radiogenic input. The logarithm makes reaching this equilibrium take infinite time. For an initial thickness $h_0>0$, subtract $[-R{h}_0-\log (1-R{h}_0)]/(S{R}^2)$ on the same side of the equilibrium; the corresponding antiderivative uses $\log |1-Rh|$ on either branch. The thin-crust description through equilibrium requires $R\gg 1$.

For general initial temperature, the full early and intermediate history is specified by the exact temperature above and one scalar ordinary differential equation. A particularly convenient integral formulation for an initially bare surface isThis removes the infinite initial derivative of $h$. The right-hand side points into $y>0$ at zero, and is decreasing in $y$ when ${\theta }_0\ge 0$; two nonnegative solutions therefore cannot separate, which gives uniqueness. Its solution, together with the temperature formula, determines the entire model evolution and can be computed without a phase-boundary search. The early expansion follows by substitution:The dimensional leading crust thickness is $a\sim \sqrt{2k\Delta Tt/({\rho }_0{L}_f)}$. Thus hotter liquid reduces the first correction to conductive crust growth, whereas colder liquid permits more rapid thickening.

If ${\theta }_0>{\theta }_{\ast }$, the temperature decreases and the instantaneous stationary lava-crust thickness $h_e(\tau )=1/[B\theta (\tau )]$ increases towards $h_{\ast }$. Starting from zero, the crust always remains below this rising curve: at a first contact $h^{\prime }=0$ but $h_e^{\prime }>0$, so it cannot cross from below. Hence hot-start crust grows monotonically. It also stays thinner than the constant-temperature solution: the early expansion puts it below, and at a proposed first crossing their derivative difference is $-SB({\theta }_0-{\theta }_{\ast })e^{-B\tau }<0$, preventing crossing from below. If $0\le {\theta }_0<{\theta }_{\ast }$, the temperature increases and the instantaneous stationary lava-crust thickness decreases. The crust stays thicker than the constant-temperature solution by scalar comparison, but it may cross the decreasing $h_e$, reach a maximum and subsequently melt back towards $h_{\ast }$. Such overshoot is possible, not inevitable; the two relevant relaxation rates are $B$ and $S{R}^2$. The numerical plot uses $B=4$, $R=20$, $S=0.1$, so crust adjustment is faster than temperature relaxation and the cool-start example does overshoot:

The curves use ${\theta }_0/{\theta }_{\ast }=0.5,1,1.5$ and initially $h=0$. They solve the complete reduced time-dependent equations, rather than replacing the moving stationary thickness by the actual crust. At late times, set $h=h_{\ast }+v$, $\delta ={\theta }_0-{\theta }_{\ast }$; linearization gives $v^{\prime }=-S{R}^{2}v-SB\delta e^{-B\tau }$. For $S{R}^2\ne B$,At equal rates the forced term is $-SB\delta \tau e^{-B\tau }$. In particular, slow warming with $B<S{R}^2$ approaches equilibrium from a thicker crust, explaining the plotted remelting. These long-time expressions describe the linearized response; nonlinear corrections can have their own faster exponential rates. All predictions are restricted to $a/H\ll 1$, nonnegative liquid superheat, and the assumed vigorous-convection flux law. If crust growth violates those conditions, the reduced model must be replaced rather than extrapolated to total freezing.

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 $X\gg H$.

For a well-mixed lava lake of depth $H$ and a thin conductive crust of thickness $a$, put $\Delta T=T_m-T_s$, and use the convective flux $K(\overline{T}-T_m)$ with constant heat transfer coefficient $K>0$. The Stefan condition and liquid heat balance give, to leading order in $a/H$,Here latent heat $L_f$ is per unit mass and $k=\rho c_p\kappa$. Crust heating and changes of liquid depth are neglected. The linear conductive profile uses a quasi-steady approximation; thinness alone does not justify neglecting crust thermal storage. During initial growth, a small Stefan number supplies a sufficient rapid-adjustment regime. Define $h=a/H$, $\theta =(\overline{T}-T_m)/\Delta T$, and $\tau =\kappa t/H^2$. Scaling by the thermal diffusion time $H^2/\kappa$ and $\Delta T$ gives $h^{\prime }=S(1/h-B\theta )$ and ${\theta }^{\prime }=R-B\theta$, where $B=KH/k$, $R=Q{H}^2/(c_p\kappa \Delta T)$, and $S=c_p\Delta T/L_f$ is the Stefan number. Thus $\theta =R/B+({\theta }_0-R/B)e^{-B\tau }$ exactly. The reduction applies while $a/H\ll 1$, the conductive profile is approximately linear, and the thermal convection closure remains valid; it does not model complete freezing of a deep lake.

An ideal fluid issuing from a small hole a vertical depth $h$ below a free surface has speed

Here $h$ is the vertical distance from the free surface to the centre of the orifice, and $g$ is the positive magnitude of gravitational acceleration. The ideal prediction assumes incompressible flow, the same ambient pressure at the free surface and emerging jet, negligible velocity of the reservoir's free surface, and negligible viscous dissipation and surface tension effects. For a falling free surface, a quasi-steady approximation additionally assumes that the local outflow adjusts much faster than $h$ changes. It applies the Bernoulli equation at each instant, rather than claiming that the whole drainage is steady flow.^[ref]

A short proof uses a streamline connecting the reservoir's free surface at elevation $z_s$ to the emerging jet at elevation $z_o$. With constant mass density $\rho$, negligible surface speed and no mechanical energy losses, the Bernoulli equation givesEqual pressures cancel, leaving $v^2/2=gh$: a unit mass gains the same kinetic energy as a body in free fall through height $h$. This makes Torricelli's law a useful first estimate before the detailed velocity field is known. At one quarter of the original head, ideal jet speed is one half as large; it is not proportional to head.

Far from the accelerating outlet region, the reservoir has approximately hydrostatic pressure $p_{\mathrm{ambient}}+\rho gh$ at the outlet's elevation. The emerging jet instead has approximately ambient pressure when surface tension is negligible. Do not assign that bulk hydrostatic pressure to the emerging jet as well as adding the height difference: that would count the available mechanical energy twice. A sealed reservoir, a submerged outlet or significant surface tension changes the pressure conditions and requires a different balance.

Actual volumetric flow rate is commonly represented by a discharge coefficient:where $a$ is the geometric area of the orifice. This definition compares volumetric flow rates, not just speeds. If a nearly uniform jet at its vena contracta has area $a_j$ and speed $v_j$, define the two separate ratios $C_c=a_j/a$ and $C_v=v_j/\sqrt{2gh}$. Then $Q=a_j{v}_j$ gives $C_d=C_c{C}_v$. A contracted jet can consequently have a speed close to $\sqrt{2gh}$ while $C_d$ is appreciably below one. No particular $C_d$ is universal: calibrate it for the opening and operating range.