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Consider the reacting zinc and copper species in the zinc-copper displacement experiment. Split the net chemical equation into two half-reactions:

The zinc oxidation state rises from $0$ to $+2$; the copper oxidation state falls from $+2$ to $0$. Add the half-reactions: two released electrons match two accepted electrons. Cancelling $2e^{-}$ from both sides gives^[ref]Thus the closed net chemical reaction has zero net production of free electrons. Cancellation does not destroy electrons: it removes a matched internal exchange from the written chemical equation. The half-reactions alone are incomplete descriptions of the whole reacting system. In a galvanic cell, the exchange can pass through a wire; in the zinc-copper displacement experiment, the reacting materials meet at the same metal surface. The equations do not require a reservoir of free electrons in the bulk chemical solution.

Check both kinds of balance. The oxidation line has net electric charge $0$ on the left and $(+2-2)e=0$ on the right; the reduction line has $(+2-2)e=0$ on the left and $0$ on the right. The net chemical equation contains one zinc and one copper chemical atom on each side and has net electric charge $+2e$ on each side, where $e$ is the positive elementary charge. The written reacting species need not be electrically neutral as a group: the bulk chemical solution also contains compensating anions.

There is an even more literal check. Zinc has atomic number $30$ and copper $29$, so the initial pair contains $30+(29-2)=57$ electrons. The final pair contains $(30-2)+29=57$ electrons. Only their distribution has changed. Conserved chemical element counts give the usual chemical mass conservation inventory; changing the counts of neutral and charged species does not create or remove their atomic nuclei.

Matching electrons is necessary but does not replace balancing every chemical element. The acidic permanganate-iron balance illustrates unequal one- and five-electron exchanges, with water and protons also needed. Conversely, a balanced chemical equation alone establishes neither a reaction rate nor a detailed mechanism. Its usefulness is that it fixes the stoichiometry even when those additional questions remain open.

A layer of half-thickness $h_0$, dynamic viscosity $\mu$ and density $\rho$, between semi-infinite fluids of viscosity $\lambda \mu$ and densities $\rho \mp \Delta \rho$, relaxes a symmetric thickness Fourier mode. In the plug-like extensional viscosity regime, mass conservation and $4\mu (h{u}_x{)}_x=h\Delta \rho g{h}_x-{\sigma }_{xz}^{+}$ together with the Fourier traction map for a viscous half-space give the displayed rate, where $\kappa =|k|h_0$. Inner longitudinal extension dominates for $\lambda \ll \kappa$; outer tangential resistance dominates for $\kappa \ll \lambda$ while the plug approximation remains valid.

Incompressible mass conservation in a moving gap givesCombined with the local Couette-Poiseuille flux, this is the Reynolds lubrication equation.

For a thin gap $h$ with tangential wall velocities ${\mathbf{v}}_0,{\mathbf{v}}_h$, lubrication theory and the no-slip boundary condition give the integrated volume fluxConservation of mass gives $h_t+{\nabla }_{\parallel }\cdot \mathbf{q}=0$. This is the Reynolds equation; it determines the lubrication pressure from wall motion and gap geometry.

Let $q=M_1/M_2$, hold total binary mass fixed, use $R_L\propto a{q}^{3/10}$, and impose $d\log J=fd\log M_1$. The circular-binary orbital angular momentum then gives $d\log a/d\log M_1=2(q-1+f)$, whereas mass conservation gives $d\log q/d\log M_1=1+q$. Thus ${\zeta }_L=(23q-17+20f)/10$. If the donor obeys $R_1=K{M}_1^{\beta }$ with slowly varying $K$, loss of donor mass increases its overfill when ${\zeta }_L>\beta$, or $q>(17-20f+10\beta )/23$. This is a feedback criterion on the time scale of the assumed donor response, not automatically a dynamical instability for every choice of $\beta$. For $\beta =1/2$, the threshold is $(22-20f)/23$; if $f\ge 11/10$, every positive mass ratio satisfies it.

Particle mass conservation gives $h_c=[{\varphi }_1(H-z_m)+{\varphi }_2{z}_m]/{\varphi }_{\max }$. Averaging concentration over the entire original container gives $\bar{\varphi }={\varphi }_{\max }{h}_c/H$. This average differs from the concentration inside the deposit itself.

Sediment transport is movement of grains by a flowing fluid, through motion close to a bed or within a particle suspension. A flow's shear stress and fluctuating forces compete with grain weight, contacts and friction. The Shields parameter compares a bed-stress scale with the submerged grain-weight scale and helps organize the threshold for movement.

Bedload transport carries grains along or near the bed, while suspended sediment balances settling against upward mixing and entrainment. The transport budget changes the bed itself: mass conservation in the Exner equation relates downstream flux divergence to erosion or deposition. A flow entering an erodible reach can take a finite distance to establish its transported grain population, described by sediment transport saturation. These couplings make a bed an evolving part of the fluid problem rather than a fixed rough wall. Measuring both transported material and bed-height change tests whether the sediment budget closes.

Sediment transport saturation is relaxation of a transported grain population toward the equilibrium sediment flux supported by local flow conditions. A change in bed shear stress does not immediately give a fully adjusted grain speed and population. Close to equilibrium, a steady model isFor constant $q_{\mathrm{sat}}$ and saturation length $L_{\mathrm{sat}}>0$, this gives $q=q_{\mathrm{sat}}+[q(0)-q_{\mathrm{sat}}]e^{-x/L_{\mathrm{sat}}}$. The spatial lag matters for bedform growth.^[ref]

A published wind tunnel test used a $1\mathrm{m}$ wide, $0.5\mathrm{m}$ high and $4.5\mathrm{m}$ long tunnel. Repeated measurements of bed elevation after stopping the flow supplied an erosion rate; integrating that rate inferred downstream flux by mass conservation. The fitted near-equilibrium relaxation length was distinguished from the earlier grain-ejection stage, so the total development distance was not treated as the same parameter.^[ref]

On a horizontal bed, the shallow-ice approximation with a linear basal drag law gives $u=-(\rho g/\mu )h_x(h/\beta +hz-z^2/2)$ and $q=-(\rho g/\mu )(h^3/3+h^2/\beta )h_x$. With no ice accumulation or ice ablation, mass conservation is $h_t+q_x=0$. Internal shear supplies the cubic mobility and basal sliding the quadratic mobility.

A slender viscous thread has a cross-sectional scale much smaller than its axial variation scale. For an axisymmetric Newtonian thread of area $A$ and nearly uniform axial velocity $w$, mass conservation and the Trouton ratio givewhen inertia and surface tension are negligible and $z$ points downward.

For steady polytropic flow with $p=K{\rho }^{\gamma }$, $K>0$, $\gamma >1$, a sonic section has $u_{\ast }=c_{\ast }=(\gamma K{\rho }_{\ast }^{\gamma -1}{)}^{1/2}$. Substitution in mass conservation $\dot{M}=A_{\ast }{\rho }_{\ast }{u}_{\ast }$ gives the displayed density. These conditions fix the sonic state for a prescribed flux and throat area, subject to compatible reservoir and downstream boundary conditions.

When the velocity distribution is independent of position and there are no forces, the free transport equation contributes zero and only collisions evolve $f$. For a bounded cutoff kernel, $\Vert Q(f,g){\Vert }_1\le C\Vert f{\Vert }_1\Vert g{\Vert }_1$ gives local existence and uniqueness by the Banach fixed-point theorem. Positivity and mass conservation bound the norm and permit global continuation in this particular model. Stronger kernels require moment or weighted estimates.

For a sink $W=W_0(r/r_g{)}^{-5/2}$ outside $r_g$, mass conservation gives $d{F}_{\mathrm{acc}}/dr=2\pi rW$ in steady state. Remote feeding gives $F_{\mathrm{acc}}={\dot{M}}_d-{\dot{M}}_w$ inside $r_g$, and ${\dot{M}}_d-{\dot{M}}_w\sqrt{r_g/r}$ outside, where ${\dot{M}}_w=4\pi W_0{r}_g^2$. The inward flux decreases towards the star because mass is diverted into the wind. A finite feeding radius or outer truncation supplies an additional boundary condition.

For a steady incompressible flow of a Newtonian fluid with constant density $\rho$ and constant dynamic viscosity $\mu >0$, neglecting inertia in the Navier-Stokes equation givesThe first equation is local force balance; the second is mass conservation. The term $\mu {\nabla }^2\mathbf{u}$ is the divergence of the viscous part of the Newtonian fluid stress tensor. The body force $\mathbf{f}$ has units of force per unit volume, $\mathrm{N}{\mathrm{m}}^{-3}$, and equals $\rho \mathbf{g}$ for uniform gravity. Appropriate boundary data for velocity or traction, together with suitable distant conditions for an exterior flow, complete the problem; a rigid surface normally imposes the no-slip boundary condition.

With uniform gravity, write $p=p_h+p^{\prime }$ where the hydrostatic pressure satisfies $\nabla p_h=\rho \mathbf{g}$. Then the disturbance equation is $-\nabla p^{\prime }+\mu {\nabla }^2\mathbf{u}=0$. This removes gravity from the fluid disturbance equation, while the hydrostatic pressure still supplies buoyancy on an immersed body. Omitting it from the total force balance would give the wrong Stokes settling velocity.

In the constant-property convection convention, the strong Boussinesq approximation supplements small relative mass density changes with constant dynamic viscosity, thermal diffusivity, specific heat capacity and thermal expansion coefficient throughout the modeled temperature interval. A uniform reference density replaces density in inertia and mass conservation, while a linear density anomaly is retained in gravitational buoyancy. The temperature obeys an advection-diffusion equation with constant coefficients. The term “strong” is not uniform across all fluid-mechanics literature: some stratified-flow treatments instead emphasize the use of a globally constant reference density. The constant-property convection usage is explained in courses.physics.ucsd.edu/2021/Spring/physics218c/Spiegel%20Convection.pdf .

A thin-film equation combines a lubrication flux with mass conservation to evolve a free-surface thickness. Gravity, capillary pressure, wall motion, and intermolecular forces may contribute to the flux.

For a no-slip planar liquid film with an insoluble surfactant, lubrication theory gives $q=-h^3{p}_x/(3\mu )+h^2{\gamma }_x/(2\mu )$ and surface velocity $u_s=-h^2{p}_x/(2\mu )+h{\gamma }_x/\mu$. Mass conservation and conservation of insoluble surfactant on a moving interface give $h_t+q_x=0$ and $C_t+(C{u}_s{)}_x=0$. With $\gamma ={\gamma }_0-AC$ and small slopes, $p_x=\rho g{h}_x-(\gamma h_{xx}{)}_x$. The different coefficients in $q$ and $u_s$ are essential: Marangoni stress and hydrostatic return flow transport fluid and surfactant differently.

The thin-film mass flux is the mass transported along a thin layer per unit width and per unit time. For constant density $\rho$, thickness $h$ and depth-averaged streamwise velocity $\overline{u}$,The quantity $q=h\overline{u}$ is instead the volume flux per unit width. Their units differ by density: $j_m$ has units $\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$, whereas $q$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$.

For a constant-density incompressible film with no sources, mass conservation becomes $h_t+q_x=0$. In a steady region, $q$ is constant; in a profile moving at speed $c$, the conserved relative flux is $q-ch$. Thus a slowly varying or moving transition does not automatically have constant laboratory-frame flux. Keeping the density factor and the reference frame explicit prevents confusing a geometric thickness balance with the transported mass.

An arterial reservoir, venous reservoir and a left ventricle exchange flows $Q_a,Q_v,Q_c$. Their volume balances are $C_a{\dot{p}}_a=Q_a-Q_c$, $C_v{\dot{p}}_v=Q_c-Q_v$, and $d(p_{LV}/E_{LV})/dt=Q_v-Q_a$. Ideal one-way valves give $Q_a=[p_{LV}-p_a{]}_{+}/R_a$, $Q_v=[p_v-p_{LV}{]}_{+}/R_v$; systemic resistance gives $Q_c=(p_a-p_v)/R_c$. Summing the three equations verifies total mass conservation for incompressible blood.

In axisymmetric ideal magnetohydrodynamics, $\Omega =u_{\varphi }/R$ is the material angular velocity. Cylindrical basis-vector derivatives and mass conservation give the displayed winding equation. Thus $B_{\varphi }/(\rho R)$ is an advected scalar only when ${\mathbf{B}}_p\cdot \nabla \Omega =0$. Differential rotation along a field line otherwise produces toroidal field. This does not violate magnetic flux freezing, which concerns flux through a material surface rather than a cylindrical vector component.

Consider steady isentropic flow toward a Newtonian gravitational potential $-GM/r$ through a tube of cross-sectional area $A(r)=C{r}^n$. Write $v>0$ for the inward speed and use a polytropic equation of state with specific-heat ratio $\gamma >1$. Mass conservation and the Euler equations for an inviscid fluid givewhere $c_s$ is the adiabatic sound speed. A regular sonic point therefore has $v_s=c_s$ and $r_s=GM/(n{c}_s^2)$. The Bernoulli equation, matched to a nearly stationary reservoir with sound speed $c_0$, givesFor $n>1/2$, a finite positive sonic point requires $1<\gamma <(2n+1)/(2n-1)$. Differentiating the flow equation at that point, with $x=r_s{v}_s^{\prime }/c_s$, givesIts discriminant is $4n[(2n+1)-(2n-1)\gamma ]$, so the same bound permits real regular slopes. The branch on which the Mach number rises inward selects the negative sign inFor a spherical tube $n=2$, this recovers the $5/3$ threshold of Bondi accretion; for a dipolar flux-tube area, $n=3$ gives $7/5$. The tube approximation must remain valid between the reservoir matching region and the accretor.