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In a slender layer, two parallel sublayers of fractions $x/L$ and $1-x/L$ have local effective permeability $K(x)=k_{1}x/L+k_2(1-x/L)$. Integrating their successive hydraulic resistance gives the logarithmic mean displayed above. A leading pressure gradient uniform across the thickness does not make parcel paths horizontal; a small transverse velocity preserves mass conservation as the cross-sectional conductance changes.

The Exner equation is local sediment mass conservation expressed in terms of solid volume: accumulation raises the bed and flux divergence lowers it. Here ${\varphi }_b$ is the solid volume fraction of the packed bed and $q$ is solid volume flux per unit width.

With gross star formation rate $\psi =k{M}_g$, constant returned mass fraction $F$, and the instantaneous recycling approximation, the net locked mass rate is $(1-F)\psi$. Mass conservation gives ${\dot{M}}_g=-(1-F)k{M}_g$. For constant stellar yield $Y$ per unit locked mass and initially pristine gas, metal conservation gives $M_g\dot{Z}=Y(1-F)\psi$, so gas-phase metallicity grows as $Z=Y(1-F)kt$. Gross star formation and permanent locking have different normalizations.

A slender axisymmetric column of a Newtonian fluid, with radius $a(z,t)$, axial velocity $w$, external pressure $p_e$ and axial body-force density $f$, obeys $(a^2{)}_t+(a^{2}w{)}_z=0$ and $3\mu (a^2{w}_z{)}_z/a^2=p_{e,z}-f$ when inertia, surface tension and leading external shear are negligible. The factor three is the Trouton ratio. Radial mass conservation gives $u_r=-r{w}_z/2$; the Newtonian fluid stress tensor gives $p=p_e-\mu w_z$ and axial excess stress $3\mu w_z$. This reduces a three-dimensional Stokes flow to one-dimensional evolving geometry.

Assume a column is sealed to salt at both ends, diffusion occurs only in liquid, and its imposed temperature increases strictly upward. A finite-width equilibrium mushy layer cannot be stationary: its solute flux is nonzero unless its liquid fraction vanishes. The final state is pure ice below a height $h_f$ and uniform-composition brine above it. Their interface obeys $C_f=-(T_E+G{h}_f)/m$. The displayed mass conservation equation fixes $h_f$ from the initial salt content. For a diffusive steadily pulled equilibrium mush, let $H=-T_E/G$, $c=m{C}_0/G$, $d=D/V$ and $h_0=H-c-d$. The initial salt content is $N=C_0[h_0+d\log (1+h_0/c)]$. Thus the liquid-cap depth ${\ell }_f=h_0-h_f$ obeys ${\ell }_f({\ell }_f+c+d)=c[h_0+d\log (1+h_0/c)]$, which has one positive root. This conclusion assumes no subsequent exchange with an underlying eutectic reservoir.

In a slowly deepening mixed layer driven by a rotating boundary, a phenomenological balance between energy input, turbulent loss and acceleration of entrained fluid can keep total kinetic energy approximately fixed. For a tank of radius $R$ this gives $u^{2}h=A{\Omega }^2{R}^3$, so velocity decreases as $h^{-1/2}$. The closure does not make the speed constant and is an additional physical assumption rather than a direct consequence of mass conservation.

Fluid mechanics studies the motion and forces of liquids and gases. Its central simplification is the continuum approximation: replace individual molecules by bulk properties defined at every position and time. The averaging region must contain many molecules while remaining small compared with the distances over which those properties vary. A fluid element is such a small material parcel, rather than a single molecule. This connects fluid mechanics to continuum mechanics and makes calculus useful for describing flow. ^[ref]

Three basic fields are the velocity field $\mathbf{u}(\mathbf{x},t)$, the mass density $\rho (\mathbf{x},t)$ and the pressure $p(\mathbf{x},t)$. The first gives a parcel's instantaneous velocity; the second gives mass per unit volume. The pressure contributes an inward normal traction $-p\mathbf{n}$ across a surface with outward unit normal $\mathbf{n}$. Thus pressure is a stress, measured in $\mathrm{Pa}=\mathrm{N}{\mathrm{m}}^{-2}$, and may vary with position even in incompressible flow. Additional viscous stresses can act tangentially. ^[ref]

The guiding questions are what must be conserved and which forces matter. Mass conservation constrains the velocity field; momentum conservation determines its response to pressure gradients, viscosity and body forces. An incompressible flow model, a Reynolds number estimate and Bernoulli's principle answer different parts of a problem. Their appeal is that a few carefully stated assumptions can determine useful quantities before the entire velocity field is known.

For a bed depth $b=\alpha x$, a thin subglacial till layer of thickness $l$ is idealized by Newtonian till lubrication, with dynamic viscosity $\lambda \mu$ and basal drag $\lambda \mu u/l$. Neglecting longitudinal stress divergence in the bulk yields $u=-(l/\lambda )(g/\nu )H{h}_x$. Retaining the unbuttressed Newtonian grounding-line stress condition as boundary data and differentiating ice-sheet flotation along the moving grounding line givesIndeed, mass conservation gives $H_t=-u{H}_x-H{u}_x$, $H_x=h_x+\alpha$, and ice-sheet flotation gives $H_t=(\alpha {\rho }_w/\rho -H_x){\dot{x}}_G$. This is a formal local friction closure, not a uniformly valid removal of membrane stress near every grounding line. If the coefficient of ${\dot{x}}_G$ vanishes, it is an implicit compatibility condition and cannot be divided to obtain a finite speed.

Let $R(\alpha )\propto {\alpha }^{-n}$, with $n>0$, and use disturbances $e^{i(kx-\omega t)}$ about uniform velocity $U$. The intrinsic frequency is $\nu =\omega -kU$. Linearized mass conservation and the decrease in friction as area grows give the displayed dispersion relation. For nonzero $k$ and positive flow, one branch grows exactly when $nU>c_0$. At $nU=c_0$, one branch is neutral, $\nu =k{c}_0$, while the other remains damped, $\nu =-k{c}_0-i{R}_0$. The maintained external-pressure gradient supplies the energy; positive friction alone does not imply linear stability of a driven flow.

For an interface perturbation $\eta e^{ikx+st}$ with instantaneous interface kinetics, take the liquid concentration perturbation proportional to $e^{-qz}$. Here $\Gamma >0$ is the coefficient of curvature-induced melting-temperature depression. The advection-diffusion equation and interfacial mass conservation giveFor a growing normal mode the unique decaying liquid branch has $q=(V+\sqrt{V^2+4D(s+D{k}^2)})/(2D)$. Writing $b=(G+\Gamma k^2)/(m{C}_I)$, its rate is $s=(D/2)(V/D-b)(\sqrt{b^2+4k^2}-b)$. Instability requires $G+\Gamma k^2<mV{C}_I/D$. The Gibbs-Thomson effect suppresses sufficiently short waves; a quasistatic solute diffusion to solute diffusion changes the rate and must be stated separately.

For a thin Newtonian fluid film on a conical slope at angle $\alpha$, neglecting the thickness-gradient contribution to the driving pressure gives volume flux per unit width $q=D{h}^3$, where $D=g\sin \alpha /(3\nu )$. Radial mass conservation isThe distance $x$ is measured along the slope; the horizontal radius is $x\cos \alpha$, so total volume is $2\pi \cos \alpha \int xhdx$.

A wide, shallow drop with cross-sectional area $A$ has the gravity-driven volume flux $q=-g{h}^3{h}_x/(3\nu )$ when substrate no-slip and zero surface shear are imposed. Mass conservation yields a porous medium equation, $h_t=[g/(3\nu )](h^3{h}_x{)}_x$. Its symmetric source-type similarity solution has half-width $\ell$ in the display and height $h=[9\nu ({\ell }^2-x^2)/(10gt){]}_{+}^{1/3}$, where $I={\int }_0^{\pi /2}{\cos }^{5/3}\theta d\theta$. Integrating the profile recovers $A$. The contact-edge slope diverges, so the bulk lubrication approximation requires an inner edge regularization if contact-line microphysics or surface tension is retained.

For chemically uniform, gas-pressure-supported stars with homologous dimensionless profiles, mass conservation and hydrostatic pressure support equation give ${\rho }_c\propto M/R^3$ and $T_c\propto M/R$. Suppose opacity scales as $\kappa \propto {\rho }^a{T}^b$ and specific nuclear energy generation as $ϵ\propto {\rho }^c{T}^d$. ThenThe second scaling follows from radiative diffusion in a star, $L\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. Equating the two giveswhen the denominator is nonzero. Fixed composition and self-similar profiles are essential; these are relations between idealized models rather than universal stellar laws.

The smooth ideal-MHD system couples mass conservation, momentum, adiabatic pressure transport and the ideal magnetohydrodynamic induction equation, subject to the solenoidal magnetic-field constraint. Without gravity it is a quasilinear partial differential equation system. Across shocks, conservative total energy replaces smooth adiabatic pressure transport.

An incompressible flow preserves the volume of each moving fluid element. Its local condition is zero velocity divergence, $\nabla \cdot \mathbf{u}=0$. Combined with mass conservation, it says that each parcel's mass density stays unchanged as it moves. This does not require every parcel to have the same mass density: an initially nonuniform distribution can be transported without compression. Nor does incompressibility require a constant velocity or constant pressure.

For positive mass density, the continuity equation makes the equivalence precise:The material derivative follows a fluid element. Thus $D\rho /Dt=0$ if and only if the velocity field is a divergence-free vector field. The distinction between following a parcel and observing a fixed position prevents confusing incompressibility with an unchanging density field. ^[ref]

For a rigid pipe with impermeable walls, no leaks and steady flow, mass conservation at constant mass density implies a constant volumetric flow rate:Here $\mathbf{n}$ points downstream, $A$ is the section's area and $\overline{U}$ is the mean normal velocity, not necessarily the centreline speed. No assumption of inviscid flow or a uniform velocity profile is needed. If unequal amounts entered and left a fixed segment, its mass would accumulate, contradicting steady flow.

For a worked example, take horizontal water flow with stipulated mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$, inlet radius $R_1=0.010\mathrm{m}$, outlet radius $R_2=0.005\mathrm{m}$ and inlet mean velocity ${\overline{U}}_1=0.20\mathrm{m}{\mathrm{s}}^{-1}$. The circular areas giveThe radius halves, the area falls to one quarter, and the mean velocity quadruples. The mass flow rate is $\rho Q=0.02\pi \mathrm{kg}{\mathrm{s}}^{-1}$. The same amount of water passes both sections each second; narrowing changes its speed rather than compressing it. The Reynolds number and Bernoulli's principle discussions use this same example to examine viscous effects and the additional assumptions needed to calculate pressure.

The regular spherical isothermal inflow from mass density ${\rho }_{\infty }$ at rest at infinity has sonic mass density $e^{3/2}{\rho }_{\infty }$. Combining it with the Bondi sonic point and mass conservation gives $\dot{M}=\pi e^{3/2}{G}^2{M}^2{\rho }_{\infty }/c_s^3$.

Consider planar horizontal layers with positive scalar intrinsic intrinsic permeabilities $k_i$, thicknesses $b_i$, and fractions $f_i=b_i/{\sum }_j{b}_j$. Assume each layer is uniform and isotropic, fully saturated by the same Newtonian fluid of constant mass density and dynamic viscosity, with perfect hydraulic contact and steady incompressible flow. Use boundaries that impose pure flow normal to the layers or a common gradient of hydraulic head along them; additional interface resistance and finite-edge effects are excluded.

For vertical flow normal to the layers, every layer carries the same normal Darcy flux by mass conservation. The hydraulic head drops add. Writing $B={\sum }_i{b}_i$ and taking positive $\Delta h_i$ as the drop along the flow direction givesThis is the thickness-weighted harmonic mean. It is hydraulic head, rather than fluid pressure alone, that accounts correctly for gravity in the vertical calculation. A thin layer of very small permeability of a porous medium can control the whole discharge because all flow must pass through it.

For horizontal flow parallel to the layers, each layer experiences the same gradient of hydraulic head. For width $W$, each area is $A_i=W{b}_i$, and the individual volume fluxes add:This is the thickness-weighted arithmetic mean. A highly permeable layer can carry most of the horizontal flow because it supplies a continuous route from inlet to outlet. Woessner and Poeter derive both cases for hydraulic conductivity; identical fluid properties allow the common factor $\rho g/\mu$ to cancel.^[ref]

For two equally thick layers, take $k_1=k_0$ and $k_2=9k_0$, where $k_0=10^{-12}{\mathrm{m}}^2$. The values 1 and 9 are therefore permeabilities in units of $k_0$, rather than dimensionless intrinsic intrinsic permeabilities. ThenAcross the layers, the less permeable half takes $90\%$ of the total hydraulic head drop and the more permeable half only $10\%$. Along the layers, the more permeable half carries $90\%$ of the total volume flux. At the same magnitude of imposed gradient of hydraulic head, horizontal Darcy flux is $5/1.8=25/9\simeq 2.78$ times the vertical Darcy flux.

The arrangement thus creates physical anisotropy from locally isotropic ingredients. In coordinates with $x,y$ along the layers and $z$ normal to them, the effective permeability of a porous medium tensor is ${\mathbf{k}}_{\mathrm{eff}}=\mathrm{diag}(5,5,1.8)k_0$. A driving gradient of hydraulic head with both horizontal and vertical components generally produces a discharge in a different direction. This distinction matters when deciding whether a geological bed conducts aquifer flow along itself or acts as a barrier to flow across it.

Regenerate the diagram by running python ../layered-porous-permeability.py from _media; it writes its PNG into the current directory.

A leaky porous gravity current loses flux through a localized fracture or a distributed permeable boundary. At a point fracture, mass conservation makes the horizontal Darcy flux discontinuous by exactly the leakage rate.

In a homentropic flow, the cross-helicity conservation law givesConsequently cross-helicity density is conserved along each trajectory exactly when the right-hand side vanishes. In a steady state, mass conservation gives $\nabla \cdot \mathbf{u}=-\mathbf{u}\cdot \nabla \log \rho$, so the criterion becomesThe absence of an entropy source in a conservation law does not alone make its density an advected scalar: compression and transport along the magnetic field still change that density.