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Charge conservation says that the total signed electric charge of an isolated collection stays constant. More generally, a collection closed to electric charge exchange has constant total electric charge, even if energy crosses its boundary. This is the charge bookkeeping relevant to a closed system or an isolated system; isolation from energy exchange is not required. Positive and negative electric charge count with opposite signs. Creating or removing an electron and a Positron together is compatible with charge conservation, because their combined electric charge is zero. ^[ref].

Imagine a fixed box initially containing electric charge $Q(0)=10\mu \mathrm{C}$. Stipulate incoming conventional current $I_{\mathrm{in}}=3\mu \mathrm{A}$ and outgoing conventional current $I_{\mathrm{out}}=1\mu \mathrm{A}$ throughout $0\le t\le 2\mathrm{s}$, with no other boundary transport. Since one ampere is one coulomb per second, the stored electric charge grows at $2\mu C/s$:If instead the outgoing electric current also equals $3\mu \mathrm{A}$, the stored electric charge remains $10\mu \mathrm{C}$ while electric charge passes through. These are stipulated transport examples; constant terminal electric currents alone do not guarantee a time-independent interior charge density.

Locally, charge conservation is the charge continuity equation, an instance of the general continuity equation:Here $\rho (\mathbf{r},t)$ is signed volume charge density and $\mathbf{J}(\mathbf{r},t)$ is current density. A positive divergence of current density means net outward transport from a small neighborhood, so its charge density decreases. For a fixed control volume $V$ with closed surface $\partial V$ and outward unit normal $\mathbf{n}$, the equivalent integrated balance isThe surface integral is signed: incoming positive electric charge contributes negatively to outward electric current, and outgoing positive electric charge contributes positively. Integrate the charge continuity equation over fixed $V$ and apply the divergence theorem to obtain this balance. Zero net boundary transport gives constant $Q_V$; the global and local statements describe the same conservation law at different scales. ^[ref].

In a steady transport state, charge density and current density are time-independent, so the charge continuity equation reduces to $\nabla \cdot \mathbf{J}=0$. At an electrical circuit junction, choose each terminal electric current $I_k$ positive outward and assume the terminals account for all boundary transport. ThenKirchhoff's current law, ${\sum }_k{I}_k=0$, follows when accumulation of electric charge at the junction is negligible. During storage, retain the same balance with its nonzero derivative; charge conservation continues to hold. This makes charge conservation useful for checking electrical circuit models and for tracking the charging of a capacitor. ^[ref].

The beautiful consistency with Maxwell equations is that the electric field and magnetic field already enforce this transport balance. For sufficiently smooth fields in vacuum SI, take the divergence of the Ampère-Maxwell equation, use that the divergence of a curl is zero, and substitute Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Constant vacuum permittivity ${ϵ}_0$ givesThe displacement current term supplies the part needed when electric charge accumulates. Historically, Maxwell added that term to extend the steady-current equation; see the existing charge conservation from Maxwell equations for this connection. ^[ref].

A control volume is a chosen region over which fluid mass conservation and momentum balance are integrated. It may be fixed or move with a prescribed velocity. Boundary fluxes must use velocity relative to the moving boundary. In Stokes flow, negligible inertia reduces the momentum balance to the sum of body forces and boundary tractions, making a carefully chosen control volume useful for finding drag without integrating directly over a curved body.

Take a fluid control volume spanning a nearly occluding sphere. The pressure difference contributes $\pi a^2\Delta P$ upwards, while the outer-wall shear in the sphere frame contributes $2\pi a\mu \int u_y(h)dz$. Since sphere-frame flux in a tube gives $u_y(h)=4U/h-6q/h^2$, the drag on the sphere is $F=\pi a^2\Delta P+2\pi a\mu (4U{I}_1-6q{I}_2)$. Using the wall shear avoids integrating traction over the strongly curved sphere directly. The remaining tube pressure drop is fixed by Hagen-Poiseuille flow.

The finite volume method follows the amount of a conserved quantity inside each control volume. Its unknowns are cell averages, rather than values at isolated points: a cell's width times its stored average approximates that cell's total. A numerical flux transfers an amount across a face. Using one shared transfer for both neighboring cells makes the loss from one exactly the gain of the other, as explained in shared face flux cancellation.^[ref]

For a one-dimensional scalar conservation law without sources, let $u_t+{\partial }_{x}f(u)=0$, with conservation law flux $f$. Fix cells $I_j=[x_{j-1/2},x_{j+1/2}]$ of widths $h_j>0$. Define the exact cell average and its numerical approximation byThe integral balance over one cell and one time interval of length $k>0$ isFor a smooth solution this follows by integration of the differential equation; with jump discontinuities, it expresses the conservation law in integral form, using appropriate face traces when they exist. Replacing each time-averaged face conservation law flux by a numerical flux givesThus the finite volume method implements a balance directly, without needing a pointwise derivative at a jump. The tube balance in Riemann Problems and Jupyter Solutions by David I. Ketcheson, Randall J. LeVeque, and Mauricio J. del Razo (Advection, §Conservation of mass) motivates this construction from conservation of mass.

This is useful for shock waves: their speed is constrained by the Rankine-Hugoniot condition, itself an integral balance. A nonconservative approximation can give a visibly plausible jump with the wrong speed. Conservation is one requirement; a nonlinear scalar conservation law also needs the correct entropy solution, and the chosen numerical flux and update need their own stability and accuracy analysis.^[ref] The original periodic upwind finite volume advection example isolates the shared-transfer mechanism in a problem with a known translated solution.

For dipole pressure $\delta p\sim Cx/(x^2+y^2)$, choose a rectangular control volume whose transverse size tends to infinity faster than its streamwise size. Each end face contributes $\pi C$ to $\int \delta p{n}_{x}ds$, so fixed-throughflow boundary work gives $\Delta \mathcal{D}=-2\pi UC$, the displayed result. This control-surface choice removes the integrated far-field velocity correction. A circular cutoff of the pressure-only functional gives $-\pi UC$ instead and does not implement the same throughflow normalization.

A heat exchanger transfers heat between a fluid and another fluid or its surroundings. In measuring the Joule-Thomson effect with a porous plug, an upstream coil restores the inlet temperature before the measurement throttle; it belongs outside the approximately insulated measurement control volume.

Apply the steady-flow energy balance to an entire liquefaction apparatus, allowing a warm gas outlet and a separate cold liquid outlet. With incoming heat $q$ per unit feed mass, no shaft work, and negligible changes in bulk kinetic energy and gravitational potential energy, conservation of mass and conservation of energy give $h_{\mathrm{in}}+q=y{h}_l+(1-y)h_g$. Solving this equation gives the displayed liquid mass fraction.

The two outlet streams need not share a temperature. For example, a counterflow heat exchanger can warm returning gas before it crosses the overall control volume boundary, while collected liquid leaves at the cold end. When the outlet temperatures differ, $h_g-h_l$ includes sensible heating as well as phase change; it is not simply a latent heat. A colder two-phase separator nested inside the apparatus has its own boundary states and a different local enthalpy balance.

Supersonic communication. A point source moving at speed $V>c_s$ outruns its small acoustic waves. After a time $t$, the source has moved $Vt$, while a disturbance emitted at its old position has radius $c_{s}t$. The envelope of these spherical wavefronts is the Mach cone, with half-angle $\mu$ satisfyingInformation is restricted to the downstream cone, so upstream gas cannot smoothly anticipate a supersonic obstacle. Finite compressions can also steepen because their characteristics converge. Such disturbances form a shock wave, across which the fluid variables change rapidly. A perfectly uniform supersonic flow need not contain a shock: the cone expresses the communication constraint, while a disturbance or compression supplies the reason for a discontinuity.

Conserved fluxes. Work in the shock frame, take velocities normal to the front, and integrate the conservation laws through a thin steady control volume of unit area. Mass conservation gives constant flux $j=\rho u$, soThe normal momentum flux is the advected part $\rho u^2$ plus the pressure stress $p$. With negligible body-force impulse across the thin front,For an ideal gas with constant adiabatic index $\gamma$, the specific internal energy is $e=p/[(\gamma -1)\rho ]$. The energy flux iswhere $h_{\mathrm{sp}}$ is the specific enthalpy. No radiation or other heat flux escapes in this adiabatic shock model. Conservation of the energy flux, divided by the nonzero mass flux, givesThese are the Rankine-Hugoniot conditions for a perfect gas. The shock is adiabatic but irreversible: its entropy increases, so an isentropic relation cannot be imposed across the discontinuity.

Strong-shock compression. Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and neglect $p_1$ and the upstream enthalpy compared with ${\rho }_1{u}_1^2$ and $u_1^2$. Momentum conservation gives $p_2={\rho }_1{u}_1^2(1-1/\chi )$. Substitute this into energy conservation:ThusThe root $\chi =1$ describes no shock. The compressive branch of the Strong-shock Rankine-Hugoniot conditions isIt also gives $p_2=2{\rho }_1{u}_1^2/(\gamma +1)$. A monatomic gas has $\gamma =5/3$, hence $\chi =4$.

The finite slab in its initial rest frame. Denote the shock's lab speed by $V_s$. The upstream gas is initially stationary, so $u_1=V_s$ and the downstream shock-frame speed is $u_2=V_s/4$. Transforming to the initial slab frame givesThis is planar shock compression of a finite slab. In the uniform post-shock approximation, breakout occurs at $t_b=L/V_s$. The original front face has then moved $(3V_s/4)t_b=3L/4$, while the shock reaches the originally stationary far face at $L$, leaving thickness $L/4$. Equivalently, the fixed mass per area $\rho L$ becomes $4\rho L_2$. The result assumes a maintained approximately constant shock speed and no important boundary rarefaction during the crossing; an unconfined finite cloud need not remain uniformly compressed.

Escape estimate and the sound-speed convention. Let $M_c$ be the slab mass, $W>0$ its binding-energy magnitude in the star's field, and $E_{\mathrm{th}}$ its initial internal energy. Treat the potential as nearly uniform across the slab. For the bulk-ballistic escape estimate, requireThe Mach-number convention in shock unbinding estimates matters here. With the usual upstream adiabatic sound speed, $c_{\mathrm{ad}}^2=\gamma p/\rho =5p/(3\rho )$, define ${\mathcal{M}}_{\mathrm{ad}}=V_s/c_{\mathrm{ad}}$. ThenIf instead the shock speed is normalized by the isothermal sound speed, $c_T^2=p/\rho$, then ${\mathcal{M}}_T^2=(5/3){\mathcal{M}}_{\mathrm{ad}}^2$ andThe original PDF's “around five” is consistent with the latter convention or an order-of-magnitude sound-speed estimate. It is not the exact coefficient obtained with the conventional adiabatic Mach number and the stated bulk kinetic-energy criterion. Here $W=-E_{\mathrm{grav}}$ if the potential is zero at infinity; using negative gravitational energy directly on the right would be incorrect. Subsequent conversion of shock-generated internal energy to expansion, external pressure, or radiative losses would change the physical escape threshold, so this numerical factor is not universal.

The first argument mixes a material fluid element with a fixed spatial volume. For a material volume element following the fluid, its volume is generally time dependent:Therefore the material derivative cannot be taken outside the product with $\delta V$ as in the proposed cancellation. Instead, mass conservation statesThe momentum balance for a material fluid element, using Newton's second law, is consequentlyFor a fixed control volume, fluid crosses its boundary and carries momentum; a momentum flux term must be included. Its local conservative form is ${\partial }_t(\rho \mathbf{u})+\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=-\nabla p$, equivalent to the displayed Euler equations after using the continuity equation. The erroneous equation omits the term $\rho \mathbf{u}\nabla \cdot \mathbf{u}$ needed to convert a material derivative of momentum density into this conservative form.

For the second task, take a homogeneous fluid with constant mass density $\rho$, prescribed gravitational acceleration $\mathbf{g}$, incompressible flow and inviscid flow. Its equations areUnder the replacement $\rho \mapsto \tilde{\rho }=C\rho$, the same velocity solves these equations if $p\mapsto \tilde{p}=Cp$, up to a spatially constant pressure offset. This is density scaling invariance of incompressible Euler flow:The comparison assumes the same geometry and compatible velocity or gravitational free surface conditions; an independently specified nonuniform boundary pressure would supply additional information. The gravitational field is taken as prescribed. If the fluid's own self-gravity must be included, its field changes with mass density, so the invariance need not hold. A single freely moving particle's density-independent gravitational acceleration does not justify neglecting that distinction.

Take outward radial volume flux per unit width as positive. The streamfunction in polar coordinates immediately givesFor spanwise width $W$, the volumetric flow rate is $Q=Wq$. The wedge inside radius $r$ has cross-sectional area $\alpha r^2/2$, whose rate of increase is $r^2\dot{\alpha }/2$. Opening therefore draws in exactly the volume needed to fill the expanding wedge; closing expels it. This is also a direct check of conservation of mass for the moving control volume. It follows from the wall boundary conditions and remains true even though the printed interior streamfunction is erroneous.

Measure pressure relative to its hydrostatic value and let $\Delta P$ be the lower-side excess pressure minus the upper-side excess pressure. The long unobstructed parts of the tube act as Poiseuille resistances in series. Their lengths add to $L$ at leading order because $l\ll L$; entrance and exterior resistances are smaller for $L\gg a$. ConsequentlyHere $Q$ is downward laboratory-frame volume flux.

Take $z$ upwards in the particle frame and let $s$ measure distance from the particle surface to the tube wall. In this frame the particle is stationary and the wall moves upwards at speed $U$. In the thin, slowly varying gap, pressure is uniform across $s$ and the leading axial Stokes equation is $\mu w_{ss}=p_z$. No slip givesThe gap's upward relative flux is constant, while conservation of displaced volume gives $2\pi aq=\pi a^{2}U-Q$. Thus $p_z=6\mu U/h^2-12\mu q/h^3$, and integration from bottom to top yieldsCurvature corrections to the gap circumference are relative order $h/a$. Rapid end regions do not contribute to the leading singular neck resistance.

For the force use a cylindrical fluid control volume containing the particle, with full-area end caps just beyond it and the particle surface as an inner boundary. The upward pressure force on the caps is $\pi a^2\Delta P$. The outer wall exerts axial traction $\mu w_s(h)$ on the fluid, whereas the particle's traction on the fluid integrates to minus its upward hydrodynamic resistance $F$. The zero-inertia momentum balance is thereforeThe velocity profile gives $w_s(h)=4U/h-6q/h^2$, provingUsing the outer-wall traction is essential; directly using the particle-side shear and a flat-end pressure force would omit the pressure on the sloping particle surface. This is control-volume drag formula for a sphere in a tube.

The four dimensionless relations areSolving them givesWithin the preceding lubrication and tube-resistance approximations, the force isSince $h/a\ll 1$, $I_3^{\ast }\gg I_2^{\ast }\gg I_1^{\ast }$. Dropping the smaller additive $-2I_2^{\ast }+4I_1^{\ast }/3$ terms in the numerator yields the requested leading formulaThe Cauchy-Schwarz inequality gives $(I_2^{\ast }{)}^2\le I_1^{\ast }{I}_3^{\ast }$, so the retained $\delta$ term is positive. The term multiplied by $\delta$ must be retained: it can compete with $I_3^{\ast }$ even though the unmultiplied lower integrals cannot.

For a sphere of radius $a(1-ϵ)$, the neck has $h=aϵ+z^2/(2a)$ at leading order. With $z=a\sqrt{2ϵ}t$, the integrals reduce to $\sqrt{2}{ϵ}^{1/2-n}{\int }_{-\infty }^{\infty }(1+t^2{)}^{-n}dt$. For $n=1,2,3$ these last integrals are $\pi ,\pi /2,3\pi /8$, giving the full two-sided integrals. PutThe printed sphere integrals are half these values: they correspond to integrating over one side of the neck rather than the stated interval spanning both sides. The substitution above establishes the radius-deficit normalization of sphere-tube lubrication integrals. Using the two-sided values, the force becomesso the three answers areThese are the corrected sphere-tube drag regimes with a radius-deficit gap. If the supplied half-integrals are treated as stipulated data instead, the same algebra gives $3\pi {ϵ}^{-5/2}/(8\sqrt{2})$, $1/\delta$ and $\pi \sqrt{2}{ϵ}^{-1/2}/3$ in the three regimes. Thus the first and third numerical drag coefficients double when the geometrically correct integrals are used; the regime boundaries and dominant flow patterns are unchanged.

In the first regime, $Q^{\ast }\sim \delta c_3{ϵ}^{-5/2}\ll 1$, $q^{\ast }\sim 1/4$, and $\Delta P^{\ast }\sim c_3{ϵ}^{-5/2}\sim F^{\ast }$. Almost no displaced fluid escapes through the long tube; it bypasses upwards through the thin annulus. The large pressure-driven bypass resistance sets the drag, with shear smaller.

In the second regime, $Q^{\ast }\sim 1$, $\Delta P^{\ast }\sim F^{\ast }\sim 1/\delta$, andAlmost all displaced volume travels through the unobstructed tube, whose pressure resistance dominates the force. Both terms in this small gap flux are worth displaying: their crossover at $\delta \sim {ϵ}^{3/2}$ changes the balance between pressure-driven and wall-driven gap transport but does not create a fourth leading force regime.

In the third regime, $1-Q^{\ast }\sim 4ϵ/3$, $q^{\ast }\sim ϵ/3$, and $\Delta P^{\ast }\sim 1/\delta \ll F^{\ast }$. Tube throughflow still carries almost all the displaced volume, but the dominant resistance is local annular lubrication rather than the net tube pressure drop. The relative gap flux is now of order $Uh$, so wall-driven and pressure-driven components both matter locally. The formulas describe hydrodynamic resistance after subtracting hydrostatic buoyancy; terminal settling would balance it against the particle's excess weight.

Work in the body frame and approximate the narrow annulus locally by a planar gap. Let $y=0$ be the cylindrical wall, moving at $-U$ in this frame, and $y=h(x)$ the stationary cell surface. The red blood cell is represented by the given compliant body; the surrounding fluid is modeled as incompressible and Newtonian. Assume $h\ll a$, small local gap slope, and negligible inertia. The lubrication equations are $p_y=0$ and $\mu w_{yy}=p_x$, with no-slip conditions $w(0)=-U$, $w(h)=0$. Integration givesMass conservation makes the body-frame flux constant. With the sign convention of the problem,so the lubrication model of a compliant translating plug givesThe circumferential multiplier is $2\pi a$ to leading order, giving the relative volume flux $-2\pi aQ$.

Let $p_d$ be the pressure in front of the cell. Match the film to the reservoirs by $p(L)=p_d$ and $p(-L)=p_d+\Delta p$. Use $h=a-r$ and symmetric end matching for the parabolic-gap reduction. The elastic law givesFor $r_0(-L)=r_0(L)$ this gives $h(-L)-h(L)=\Delta p/\alpha$. If the outer unstressed shape is not symmetric at the matching ends, the general relation instead includes $h(-L)-h(L)=\Delta p/\alpha -[r_0(-L)-r_0(L)]$. The proposed reduced end equation assumes the symmetric matching used here. The gap must remain positive. To determine the unknown flux and translating speed one also needs zero net axial force on the freely moving body, and a specified absolute downstream pressure relative to $p_0$, not merely a pressure difference. If an external force pulls the body, it must replace the force-free condition. The stated elasticity law fixes $p_0$; treating it as an unknown internal pressure would require a further cell-volume constraint.

The force-free condition is most easily obtained from a control volume containing both body and surrounding film. The pressure force is $\pi a^2\Delta p$, and the resisting wall traction is $2\pi a\mu {\int }_{-L}^L{w}_y(0)dx$. The local velocity solution gives $w_y(0)=4U/h-6Q/h^2$, soThis is the force-free pressure drop of an annular lubrication plug. It includes the film pressure and shear contributions without separately approximating the deformed body's surface normal.

Put $s=2Q/U>0$, $\ell =\sqrt{s/\kappa }$, $h=sH$ and $x=\ell X$. The local parabolic elastic law is $p=p_0+\alpha [r_{00}-a-\kappa x^2/2+h]$, whence $p_x=\alpha \sqrt{\kappa s}(H_X-X)$. Substitution into the local lubrication equation givesThe derivative is with respect to the scaled coordinate $X$; a derivative with respect to dimensional $x$ would be inconsistent with this nondimensional equation.

Let $C=s/a$ and $\tilde{L}=L/\ell$. The elastic end condition gives $\Delta p=\alpha s[H(-\tilde{L})-H(\tilde{L})]$. Scaling the force integral and eliminating $\Delta p$ then givesThis completes the requested formulation and both dimensionless equations. The local parabolic approximation is used in the region contributing most strongly to the lubrication integrals; matching to the outer shape and reservoirs supplies the stated end conditions.

Take $z$ positive upwards and let $f$ denote axial body force per unit volume. With inertia and surface tension omitted, the extensional equations for a slender Newtonian column areMass conservation gives the first equation. The second uses the Trouton ratio three: the leading radial velocity is $u_r=-r{w}_z/2$, the radial normal traction gives $p=p_{\mathrm{ext}}-\mu w_z$, and the axial stress is $-p_{\mathrm{ext}}+3\mu w_z$. An axial control volume balance then gives the displayed equation. The approximation requires a slender column and slowly varying radius.

Let $h=R-a$. Since the container base is closed, the total volume flux at every section is zero. The column carries flux $\pi a^{2}w$, so the annulus carries $-\pi a^{2}w$ and has mean speed of order $a|w|/h$. Its interfacial shear stress is therefore of order $\lambda \mu a|w|/h^2$. Transmitting this stress through the core produces an axial velocity variation of order $\lambda a^2|w|/h^2$. The plug-flow criterion for a column in a low-viscosity annulus is consequently

Use modified pressure $P=p_{\mathrm{ext}}+\rho gz$, removing the annular fluid's hydrostatic pressure. In a local planar gap, with $y=0$ at the column and $y=h$ at the container wall, lubrication theory givesThe required return flux per unit circumference is $q_g=-aw/2$ to leading order in $h/a$. Its pressure-driven component is larger than the Couette component by $a/h$, so the annular return flow around a slumping column givesThe annular shear contribution to the integrated axial force is smaller than this pressure-gradient term by $h/a$. Retaining the leading terms and substituting $f=-(\rho +\Delta \rho )g$ gives the dimensional equationsThe boundary conditions are $w(0,t)=0$ at the base and $w_z(L(t),t)=0$ at the free upper end; the latter expresses zero excess axial extensional stress.

For $\lambda =0$ and initially constant $a=a_0$, put $G=\Delta \rho g/(3\mu )$. Then $w_{zz}=G$, so

For $\lambda >0$, defineThis function records the leading thin-annulus resistance; differences from curvature and the Couette term enter at the already neglected relative order $h/a$. Choose the extensional screening length of a slumping column and the associated scales asHere $h_0=R-a_0$ denotes the initial annular gap in this question. With $Z=z/\hat{z}$, $T=t/\hat{t}$ and $W=w/\hat{w}$, the dimensional equations reduce to

Initially $\varphi ={\varphi }_0$ and ${\Lambda }_0=L_0/\hat{z}$, so $W_{ZZ}=1+W$, with $W(0)=0$ and $W_Z({\Lambda }_0)=0$. The initial velocity profile of an annularly confined column isThe column thickens everywhere below its stress-free top, while that initial thickening rate is zero at the top.

For a tall column, ${\Lambda }_0\gg 1$, the limiting forms away from exponentially small end corrections areThe bulk falls at approximately $-\hat{w}$, where excess weight balances annular hydraulic resistance. Extensional stresses adjust this velocity to zero in a basal boundary layer of thickness $\hat{z}$; almost all thickening occurs there.

For a short column on this scale, ${\Lambda }_0\ll 1$, while still slender, the limits areThe annular-resistance term $W$ is small, so excess weight is balanced predominantly by extensional viscous stress. The top speed is $-\hat{w}{\Lambda }_0^2/2$, recovering the zero-annular-viscosity result. Thus $\hat{z}$ is the vertical distance over which extensional stress communicates the basal constraint before annular resistance screens it. The sketches below use separate natural normalizations for the two limits.

Use pressure with the ambient hydrostatic pressure removed, and let $\Delta P$ be the pressure below the falling sphere minus that above it. Away from the sphere the tube flow is Hagen-Poiseuille flow. The upper and lower clear lengths add to $L$ to leading order in $a/L$, so their pressure drops add:The large exterior reservoir returns this flux with negligible pressure loss compared with the long narrow tube. The sphere is assumed far enough from the ends that entrance effects and its occupied length give only lower-order corrections.

Now use a frame translating downwards with the sphere, and take $z$ positive upwards from its center. The sphere is stationary and the tube wall moves upwards with speed $U$. The narrow gap isIt has axial length scale $a\sqrt{ϵ}$, making lubrication theory appropriate. The sphere-frame flux in a tube is constant: the upward relative flux through the whole section is $\pi a^{2}U-Q$, so its flux per unit circumference isWith $y=0$ on the sphere and $y=h$ on the wall, Couette-Poiseuille flow in a thin gap givesSince $\Delta P=-\int p_{z}dz$, the pressure-jump relation isThe parabolic inner gap can be extended to infinite $z$ because these integrals are dominated by $|z|=O(a\sqrt{ϵ})$. The lubrication resistance integrals for a nearly occluding sphere areSetting $\xi =\tan \theta$ evaluates $J_1=\pi$, $J_2=\pi /2$, $J_3=3\pi /8$, hence

For the force, enclose the fluid between sections just above and below the sphere, including the annular layer. The net upward pressure force is $\pi a^2\Delta P$. The tube wall exerts upward shear traction $\mu u_y(h)$ on this fluid; the sphere exerts a downward force equal to the upward hydrodynamic force $F$ on itself. From the gap solution,The axial control volume force balance therefore gives the control-volume drag formula for a sphere in a tube:Using a control volume avoids having to integrate the curved-sphere normal pressure and tangential shear separately.

With $Q^{\ast }=Q/(\pi a^{2}U)$, $\Delta P^{\ast }=a\Delta P/(6\mu U)$, $q^{\ast }=2q/(aU)$, $F^{\ast }=F/(6\pi \mu aU)$, $I_n^{\ast }=a^{n-1}{I}_n$ and $L^{\ast }=4L/(3a)$, the three flux/pressure equations becomeSolving these linear relations givesThe force relation is $F^{\ast }=\Delta P^{\ast }+\frac{4}{3}{I}_1^{\ast }-q^{\ast }{I}_2^{\ast }$. Substitution first yieldsBecause $I_2^{\ast }/I_3^{\ast }=2ϵ/3$ and $I_1^{\ast }/I_3^{\ast }=2{ϵ}^2/3$, the terms $-2L^{\ast }{I}_2^{\ast }$ and $\frac{4}{3}{L}^{\ast }{I}_1^{\ast }$ are lower order than $L^{\ast }{I}_3^{\ast }$. Keeping the uniformly relevant leading terms givesThe squared $I_2^{\ast }$ is essential; replacing it by $I_2^{\ast }{I}_3^{\ast }$ would give the wrong sign and scaling. This establishes the three drag regimes for a nearly occluding sphere.

For $L^{\ast }\ll {ϵ}^{-1/2}$,The sphere nearly carries the tube's fluid down with it: the far-field flux is close to the piston displacement rate. The relative annular leakage is small in total volume, and the dominant force comes from the local annular shear term. The pressure force associated with the overall tube resistance is smaller.

For ${ϵ}^{-1/2}\ll L^{\ast }\ll {ϵ}^{-5/2}$,The flow remains approximately piston-like, but the dominant force is now the pressure needed to drive Hagen-Poiseuille flow through the long clear tube. Both displayed terms in $q^{\ast }$ are retained because their relative size changes inside this regime without producing an additional leading drag regime.