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In a cross-section normal to the cylinder axes, measure $y$ from the line of contact. The two circular boundaries have the parabolic approximationfor their separation. The contact point of the meniscus is at $y=w+o(w)$, so the leading cross-sectional area isThe area of the small meniscus cap is $O(w^4/a^2)$ and is lower order. At its upper end the gap has width $h(w)=w^2/a$. A tangent semicircle therefore has radius $w^2/(2a)$ and curvature

The Young–Laplace equation makes the liquid pressure, relative to the nearly uniform gas pressure,At a fixed $y$, lubrication theory gives a planar Poiseuille flow through a gap of width $h(y)$, with axial flux per unit $y$Integrating across the cusp,The continuity equation $A_t+Q_z=0$ now gives

Set $q=w^3$. Since $w^4{w}_z=\frac{1}{5}(q^{5/3}{)}_z$, this is the porous medium equationConservation of the fixed volumeand dimensional analysis give the self-similar solution $q=t^{-3/8}F(z/t^{3/8})$. One integration of the resulting ordinary differential equation yields the compactly supported Barenblatt solutionEquivalently,Its tip is $z_N=(C/B{)}^{1/2}{t}^{3/8}$. Usingin the volume constraint gives $C^2=8aV\sqrt{B}/\pi$. Therefore

For vertical cylinders at equilibrium, hydrostatic pressure gives $p-p_{\mathrm{atm}}=-\rho gz$. Balancing this with the capillary pressure $-2\gamma a/w^2$ gives the large-height profile

With ${\mathbf{u}}_0=0$ and ${\theta }_0=-z$, the incompressibility condition and temperature equation hold because ${\nabla }^{2}z=0$. The momentum equation is satisfied by the hydrostatic pressurebecause $\nabla p_0=\sigma Ra{\theta }_0\hat{\mathbf{z}}$. Thus this is the conductive basic state.

Take the rock at $z=0$, the till–ice interface at $z=h$, and the ice surface at $z=h+H$. The common leading hydrostatic pressure has horizontal gradientThe thin-film momentum equations areImpose no slip at the rock, continuity of velocity and shear stress at $z=h$, and zero shear at the ice surface. Expanding for $h/H\ll 1$ gives the till flux and ice fluxThe basal shear stress isApplying mass conservation separately to the ice and till, with erosion source $E{\tau }_0$, givesand

In a steady state $q_i=q_0$. Balancing the two terms in the till equation over length $L$ givesand henceindependently of $q_0$. Without substantial lubrication, the usual shallow-ice balance gives $H^4\sim \mu q_{0}L/(\rho g)$. ThereforeThe sheet is essentially unlubricated when $\boxed{\mathcal{E}\ll 1}$.

For $\mathcal{E}\gg 1$, the sliding term dominates the ice flux:Integrating the steady till equation from $x=0$, where $H=H_0$ and $h=0$, givesEliminating $H_x$ between these equations yieldsWith $F=H/H_0$ and $\xi =x/L$, a second integration giveswhere the physical branch decreases from $F(0)=1$ to $F(1)=0$. The length condition determinesUsing this relation in the till thickness givesThus $H$ decreases monotonically from $H_0$ to zero. The till starts at zero, rises to one interior maximum at $F=1/\sqrt{3}$, and returns to zero at the margin.

The shallow-water approximation requires $h/L\ll 1$, a nearly hydrostatic pressure, negligible vertical acceleration, and approximately depth-independent horizontal velocity, concentration, and density. The large Reynolds number allows viscous stress to be neglected away from thin boundary layers, while the deep ambient is taken to remain stationary.

For unit channel width, volume conservation, chemical conservation, and mass conservation are respectivelyandEntrained ambient fluid contains no chemical and enters with density ${\rho }_0$; detrained fluid carries the local concentration and density. The ambient has no horizontal momentum, whereas detrained fluid carries horizontal momentum $\rho u$ per unit volume. The depth-integrated momentum conservation law is thereforeso $M=-\rho u{w}_d$.

Let $z$ measure height above the sponge. Under the long-wave approximation, hydrostatic pressure gives $p_x=\rho g{H}_x$, and the horizontal balance for viscous fluid flow isThe no-slip boundary condition $u(0)=0$ and stress-free boundary condition $u_z(H)=0$ giveIntegrating this lubrication theory profile gives the horizontal volume flux per unit span

Local mass conservation includes the downward loss found in part i:Consequently the required nonlinear partial differential equation isIf the imposed inlet flux is $Q_0{t}^{\alpha }$ and the moving front is $x=x_N(t)$, sufficient boundary conditions areFor a current released onto a dry substrate one also takes the initial condition $H(x,0)=0$ away from the source. The front position is part of this moving-boundary problem.

Write the given pressure-dependent ratio of permeability of a porous medium to sponge thickness aswhere $K$ is constant. The fluid pressure drop across the sponge is the hydrostatic pressure of the current,because the air pressures above and below are equal. Therefore Darcy's law gives the downward volume flux per unit horizontal area asThus the sponge drainage obeys a power law in the current thickness, as described by pressure-dependent Darcy drainage.

At latitude ${\theta }_0$, resolve the planetary angular velocity asFor velocity $(u,v,w)$, the Coriolis acceleration isThe traditional approximation drops the terms involving the horizontal rotation component $\Omega \cos {\theta }_0$. The retained horizontal force is thereforeIts direct velocity-scale requirement isFor an incompressible flow, $W/U=O(H/L)$, so a sufficient condition istogether with the small aspect ratio that makes hydrostatic pressure the leading vertical momentum balance. The approximation consequently becomes delicate near the equator.

The beta plane is the local Taylor expansionwhereand $a$ is the planetary radius. It requires a local Cartesian region,and $H/a\ll 1$. Away from the equator, treating the variation as a perturbation of an f-plane also requires $\beta L/|f_0|\ll 1$; near the equator one instead retains the linear term as the leading Coriolis parameter.

Begin with the Boussinesq approximation primitive equations on a beta plane, decompose every field into a zonal mean and a disturbance, and average over longitude. The zonal momentum equation then contains the divergence of the eddy momentum flux $\overline{u^{\prime }{v}^{\prime }}$, while the mean density equation contains the divergence of the eddy density flux $\overline{{\rho }^{\prime }{v}^{\prime }}$. At small Rossby number, use geostrophic balance, hydrostatic pressure, thermal-wind balance, and the leading eddy equations to combine those fluxes.

Defineand introduce the residual mean circulationThe added eddy-induced velocity is nondivergent, soIt absorbs the eddy density-flux divergence into advection by the transformed circulation, giving

The mean zonal momentum equation becomeswhere the zonally averaged Eliassen–Palm flux in the meridional-vertical plane isThus the transformed Eulerian mean gathers the wave forcing into one flux divergence and makes density evolve under one residual circulation.

Define the residual-circulation stream function byThis satisfies residual mass conservation identically. LetThe momentum equation is

Differentiate geostrophic balance vertically and hydrostatic pressure meridionally to obtain thermal-wind balanceAfter a time derivative, the transformed density equation givesOn the other hand, the $z$ derivative of momentum givesEliminating ${\overline{u}}_{tz}$ yields the Eliassen equation for residual circulation

Put $l=\pi /L$. For the specified quasi-geostrophic streamfunction, the complex amplitudes of the disturbance fields areThe product of these two amplitudes is purely imaginary after one is conjugated, so its zonal mean vanishes:Consequently

Geostrophic balance and hydrostatic pressure giveThereforeSince $d{\rho }_s/dz=-{\rho }_0{N}^2/g$, the vertical Eliassen–Palm flux isThus it has the stated formwith

Define the indoor-to-outdoor reduced gravityBecause the whole opening lies above the interface, its indoor side contains upper-layer fluid of density ${\rho }_H$. Under hydrostatic pressure, the pressure difference varies linearly about a neutral pressure level. Equal opening geometry and equal discharge coefficients for inflow and outflow put that level at the vent midpoint. At vertical distance $y$ from it, the ideal Bernoulli equation gives speed $\sqrt{2g_V^{\prime }|y|}$.

Writing $Q_V$ for either one-way volumetric flow rate, integration over one half of the opening givesThe total unsigned exchange is $2Q_V$. The ideal sharp-edged inviscid model has $C_d=1$; an empirical $C_d<1$ represents contraction and losses.

Denote the four terms byso the radial or poloidal momentum equation is $I=P+G+C$.

i) In a thin Keplerian accretion disk, radial inertia and pressure are higher-order in $H/R$, leaving $G+C\simeq 0$.

ii) In a nearly static stellar atmosphere, $I=C=0$ and hydrostatic pressure balance gives $P+G=0$.

iii) In pressureless gravitational collapse, rotation and pressure are negligible, so $I=G$; this is free fall.

iv) A slim accretion disk retains radial inertia and radial pressure together with gravity and centrifugal support, so all four terms generally survive: $I=P+G+C$.

v) A stationary geometrically thick disk or torus has negligible poloidal inertia but order-one pressure support, giving $P+G+C=0$.

vi) Nonrotating Bondi accretion has $C=0$ and $I=P+G$. In a highly supersonic Bondi--Hoyle limit the pressure term is also negligible, reducing this to ballistic $I=G$.

vii) A sub-Keplerian advection-dominated accretion flow has significant pressure support and radial inflow as well as rotation, so again $I=P+G+C$, with $C$ smaller than the Keplerian value and the remaining inward gravity balanced by $P$ and $I$.

Let $K=k\Delta \rho g/\mu$. Hydrostatic pressure continuity at the lower boundary of the light current givesinside the current, so $p_x=\Delta \rho g{h}_x$. Darcy law gives the depth-integrated horizontal flux per unit transverse widthThus the porous gravity current satisfiesaway from the fracture. The boundary and front conditions areAt $x=L$, the pressure excess at the base of the fracture is $\Delta \rho g{h}_L$. Taking upward leakage as positive,This jump is the local mass conservation law for a leaky porous gravity current.

At late times, $q=Q$ to leading order on $0<x<L$, while $Q_l\simeq Q$. Integrating $q=-K(h^2{)}_x/2$ givesLeakage balance and the pressure-driven drop determine

In the far field, $h(L,t)\simeq h_L$ andBalancing the two sides with $h=O(h_L)$ givesMore precisely, setThe self-similar solution is determined byand $x_N=L+{\eta }_N(K{h}_{L}t/\varphi {)}^{1/2}$.

Let $\mathbf{u}=(u,v)$ and let the layer thickness be $H=H_0+\eta$. The inviscid shallow water equations areThey follow from a homogeneous incompressible fluid with small aspect ratio, hydrostatic pressure, negligible vertical acceleration, horizontal velocity nearly uniform through the depth, a material free surface, and a rigid stationary bottom. Rotation may be represented by an f-plane or beta plane. The model is useful because many atmospheric and oceanic motions are horizontally much broader than their depth, while the free surface or an internal density interface still supports waves and potential vorticity dynamics.

Let $\zeta =v_x-u_y$. Taking the vertical curl of momentum giveswhere $Df/Dt=\beta v$ supplies the planetary-vorticity term when $f$ varies. Continuity gives $DH/Dt=-H\nabla \cdot \mathbf{u}$. Combining the two equations yields material conservation of shallow-water potential vorticity:

At the final state of the slump test for yield stress, an axisymmetric deposit of height $h(r)$ is just able to support itself. Hydrostatic pressure gives radial pressure gradient $\rho g{h}^{\prime }(r)$, while lubrication theory makes the magnitude of the basal shear stressThe marginally yielded final profile therefore satisfiesIntegration givesUsing mass conservation, the known volume isSolving for the yield stress produces the estimateThus one measures the final radius $R$ and inserts it with $V$ and $\rho$. The estimate assumes a thin deposit, negligible surface tension, complete initial yielding, and a spatially uniform yield stress.

Dry sand is a granular material governed primarily by frictional stability. Its final free surface reaches the angle of repose ${\theta }_r$, so the deposit is approximately a cone,This constant-slope profile differs from the square-root edge of the yield-stress-fluid model.

The shallow-ice approximation makes the pressure hydrostatic pressure,Horizontal Stokes flow balance and the stress-free boundary condition at $z=h$ givesoThe basal shear stress and the stated basal sliding law are thereforeIntegrating once more and imposing $u(0)=u_b$ yieldsFor the right half of an ice cap, $h_x<0$, so both basal sliding and internal deformation carry ice away from the ice divide.

The horizontal velocities $u$ and $v$ point east and north, $\eta$ is the displacement of the free surface from its mean level, $H_0$ is the undisturbed depth, $g$ is gravitational acceleration, and $f$ is the constant Coriolis parameter on an f-plane. The three linearized shallow water equations are horizontal momentum balance and mass conservation:They follow from the rotating Navier-Stokes equation by assuming an inviscid homogeneous layer, hydrostatic pressure, horizontal scales much larger than $H_0$, depth-independent horizontal velocity, a flat impermeable bottom, constant $f$, and small surface displacement and velocity so that nonlinear products are neglected.

In a steady state, geostrophic balance givesThe relative vorticity is $\zeta =v_x-u_y$. Expanding the shallow-water potential vorticity $(f+\zeta )/(H_0+\eta )$ to first order givesThus one convenient normalization of its disturbance isTaking the curl of momentum and using continuity shows $q_t=0$.

The variables $u,v,w$ are eastward, northward, and upward velocity, $\varphi =p^{\prime }/{\rho }_0$ is the pressure perturbation divided by reference density, $\sigma$ is buoyancy, $N$ is the constant buoyancy frequency, and $\beta y$ is the equatorial approximation to the Coriolis parameter. The equations express, respectively:

Together they are the hydrostatic Boussinesq approximation for long equatorial waves.

Observe each gas volume, temperature, pressure and moisture condition, and reduce both volumes to the same temperature, pressure and dry-gas basis. Equal scale readings alone do not establish equal amounts of gas.

Read the calibrated volume of each gas in a graduated tube or eudiometer, with the eye at the meniscus to avoid parallax. Allow the gas and the surrounding bath to reach thermal equilibrium, then read a thermometer close to the gas. Heating expands a fixed amount of gas, so a larger reading could reflect a higher temperature rather than more gas. The temperature used in a gas law must be an thermodynamic temperature.

Read the atmospheric pressure with a barometer. If a gas is confined over a liquid, also observe the difference between the inside and outside liquid levels. With $h$ the internal surface height above the external surface, the total internal pressure isHere ${\rho }_{\ell }$ is the confining density of the liquid and $g$ the gravitational acceleration; the sign reverses automatically when $h$ is negative. Equalizing the levels before taking the volume reading removes this hydrostatic pressure correction. A mercury barometer can itself require temperature and calibration corrections for accurate work.

If the gas is saturated with water vapor, subtract the vapor pressure of water at its measured temperature:The observed pressure includes both the test gas and water vapor; counting it all as test gas overestimates the amount. The saturation assumption needs time for equilibration. For a dry gas, no water vapor subtraction is made; another collecting liquid may require its own vapor pressure correction. A collecting liquid must also be chosen so that it neither appreciably dissolves nor reacts with the gas: water, for instance, is unsuitable for preserving the whole amount of freely soluble Ammonia.

For dilute gases obeying the ideal gas law, the volume $V_0$ at a common chosen $T_0,p_0$ isConsequently the corrected comparison isThis is also the ratio of amounts of substance in the ideal gas approximation. Check for leaks, trapped air, changing composition and incomplete transfer. At high pressure or near condensation, real gas behaviour may require compressibility factors; with $pV=ZnRT$, the inferred amount ratio contains the further factor $Z_2/Z_1$.

At the same temperature and pressure, one volume of HCl corresponds to $\frac{1}{2}$ volume of hydrogen and $\frac{1}{2}$ volume of chlorine; one of steam to one of hydrogen and $\frac{1}{2}$ of oxygen; one of ${\mathrm{N}}_2\mathrm{O}$ to one of nitrogen and $\frac{1}{2}$ of oxygen; one of NO to $\frac{1}{2}$ of each element; and one of ammonia to $\frac{1}{2}$ of nitrogen and $\frac{3}{2}$ of hydrogen. All elemental volumes here refer to their ordinary diatomic gases.

The simplest way to avoid an erroneous factor of two is to balance each chemical reaction using ${\mathrm{H}}_2$, $\mathrm{C}{\mathrm{l}}_2$, ${\mathrm{N}}_2$ and ${\mathrm{O}}_2$. Under Avogadro's law, their stoichiometric coefficients are also gas volume ratios, provided all gases are compared at the same temperature and pressure and ideal gas behaviour is a good approximation.

If the common chosen volume of each compound is $V$, the corresponding ordered elemental volumes are therefore $(V/2,V/2)$, $(V,V/2)$, $(V,V/2)$, $(V/2,V/2)$ and $(V/2,3V/2)$. The ratios concern molecular counts and hypothetical free elemental gases, not separate compartments of elemental gas still present inside a compound.

“Hydrochloric acid” must here mean gaseous hydrogen chloride, rather than its aqueous chemical solution; water must remain as steam. If water condenses or Ammonia dissolves in the collecting liquid, measured volumes no longer show these simple ratios. For an experimental comparison, dry-gas and hydrostatic pressure corrections are those discussed in 3.