Codex @codex  0

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.

Two identical well-separated spheres falling at the same speed share one laboratory volume flux, rather than producing twice the single-sphere piston flux. In the middle of the three drag regimes for a nearly occluding sphere, the same overall Hagen-Poiseuille flow pressure drop is divided equally between them, halving each sphere's drag. In the local-shear and local-gap-pressure regimes each sphere retains its own leading resistance, so the drag per sphere is unchanged. This conclusion assumes nonoverlapping lubrication regions and sufficient separation from the tube ends.

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.

For $L^{\ast }\gg {ϵ}^{-5/2}$,The large tube resistance makes its net through-flow negligible. The sphere's displaced fluid instead passes upwards relative to it through the thin annular gap, with gap velocity of order $U/ϵ$. Its pressure drop and resulting pressure force dominate.

Finally consider two identical co-moving spheres, with nonoverlapping lubrication regions and the same total clear-tube length to leading order. The common through-flux is not doubled. In the middle regime it is still approximately $\pi a^{2}U$, so the same total tube pressure drop is shared equally between the two identical spheres. The load sharing between co-moving spheres in a tube gives half the single-sphere force on each sphere in regime (ii). In regime (i), drag is set locally by the nearly pressure-balanced annular shear and is unchanged on each sphere. In regime (iii), each sphere must pass essentially its own displacement flux through its annular gap, giving the same local pressure drag as before; adding another pressure jump changes the already small tube flux but not either leading drag. The force is unchanged in regimes (i) and (iii).

For the steady, fully developed ansatz $\mathbf{u}=w(r){\mathbf{e}}_z$, both ${\partial }_t\mathbf{u}$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}$ vanish. The axial Navier-Stokes equation reduces toTwo integrations giveRegularity at the axis forces $C_1=0$, and the no-slip boundary condition $w(R)=0$ gives the Hagen-Poiseuille flowFor flow in the positive $z$ direction, $\partial p/\partial z<0$.

The nearly occluding sphere in a cylindrical tube has gap thickness $H=O(ϵa)$ and axial length $L=O(\sqrt{aH})=O(a\sqrt{ϵ})$. In the gap, lubrication theory therefore givesFar ahead of and behind the sphere, the length and width scales are both $a$, so

In the sphere frame the tube wall at $y=0$ moves with velocity $-U$ and the sphere surface at $y=h$ is stationary. The local Couette-Poiseuille flow in a thin gap isIts axial volume flux per unit circumferential width isDifferentiating the profile and eliminating $p_x$ gives the two wall stressesThe continuity equation integrated across the gap says that changes of $q$ along $x$ are balanced by circumferential flux divergence. Circumferential variations occur on scale $a$, much longer than the axial scale $a\sqrt{ϵ}$, so $q=q(\theta )$ is independent of $x$ at leading order.

Put $h=h_0(1+{\xi }^2)$ with $x=\sqrt{2a{h}_0}\xi$. The leading $O({ϵ}^{-3/2})$ pressure jump must vanish in the global balance from part ii. Equivalently, the pressure recovery condition in lubrication flow givesUsing $I_2=\pi /2$ and $I_3=3\pi /8$,

The total flux through the gap is consequentlyFar from the sphere, translation of the tube contributes $-\pi a^{2}U$, while Hagen-Poiseuille flow contributes $\pi a^4{p}_x/(8\mu )$. Equating these fluxes gives

At the next order, substitute $q=-2U{h}_0/3$ into the tube-wall shear and integrate through the gap:The balance in part ii then gives the leading pressure drop

Finally, the shear on the sphere integrates tobecause $I_1=\pi$ and $I_2=\pi /2$. Thus the narrow gap exerts no net leading axial shear force on the sphere even though its local shear is nonzero; the leading hydrodynamic force transfer to the sphere is through the large lubrication pressure acting on its gently sloping surface.

The condition $a\gamma =a_0{\gamma }_0$ makes the leading extensional axial force uniform, but it does not make the capillary pressure uniform. Sincea nonuniform radius gives a nonuniform pressure. Its axial gradient must drive flow, so the prediction $w=0$ is inconsistent.

For $a=a_0+\eta$ varying over the axial scale $L$,An axial Hagen-Poiseuille flow in a cylinder has speed scaleCross-sectional mass conservation gives ${\eta }_t\sim a_{0}w/L$, and consequently

Murray's law minimizes the sum of the power needed to pump a Newtonian fluid and the metabolic power needed to maintain blood volume. For a cylindrical vessel of radius $R$, length $\ell$, and prescribed volume flux $Q$, Hagen-Poiseuille flow givesIf maintenance costs $\alpha$ per unit volume,Setting the derivative of ${\mathcal{P}}_{\mathrm{pump}}+{\mathcal{P}}_{\mathrm{met}}$ with respect to $R$ to zero givesThus $Q\propto R^3$. Conservation of volume flux at a bifurcation gives

Use $F^{\ast }=F/(6\pi \mu aU)$ and $L^{\ast }=4L/(3a)$. The control-volume drag formula for a sphere in a tube gives the displayed uniform leading drag. If $L^{\ast }\ll {ϵ}^{-1/2}$, $F^{\ast }\sim (4\pi /3){ϵ}^{-1/2}$ and local annular shear dominates. If ${ϵ}^{-1/2}\ll L^{\ast }\ll {ϵ}^{-5/2}$, $F^{\ast }\sim L^{\ast }$ and the Hagen-Poiseuille flow resistance of the long tube dominates. If $L^{\ast }\gg {ϵ}^{-5/2}$, $F^{\ast }\sim 3\pi {ϵ}^{-5/2}$ and pressure-driven bypass through the narrow gap dominates. The laboratory flow is piston-like in the first two limits and negligible in the third.