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In this continuation $\lambda$ measures eccentricity, not the viscosity ratio used in Question 2. Take $ϵ\to 0$ with $0\le \lambda <1$ fixed. For the nearly occluding sphere in a cylindrical tube, the narrow gap has thickness $H\sim ϵa$ and axial length $\ell \sim a\sqrt{ϵ}$, obtained by balancing $h_0$ and $x^2/(2a)$. In the sphere frame the tangential velocity scale is $U$. The lubrication approximation then gives the gap scalesThe large local pressure excursion is distinct from the eventual net pressure jump. Away from the sphere the velocity varies across a radius $a$, so the far Hagen-Poiseuille flow hasThese scales also show why the gap pressure can be much larger than its shear stress while the leading net jump can be smaller than that pressure.

Use $y=a-r$, so $y=0$ is the moving tube wall and $y=h$ is the locally stationary sphere. The no-slip boundary conditions are $u(0)=-U$, $u(h)=0$. Integrating $\mu u_{yy}=p_x$, with pressure independent of $y$ to leading order, gives the Couette-Poiseuille flow in a thin gapDifferentiate at both surfaces and eliminate $p_x$ using this moving-boundary lubrication flux:The coordinate reversal means ${\sigma }_{rx}=-{\sigma }_{xy}$ at the tube wall.

At leading order $q$ is independent of $x$ in each angular strip, though it may depend on $\theta$. Both boundaries are stationary as shapes in the sphere frame and impermeable, so the integrated continuity equation is $q_x+a^{-1}(q_{\theta }^{\mathrm{circ}}{)}_{\theta }=0$. The pressure changes along $x$ over $a\sqrt{ϵ}$ but along the circumference over $a$. Thus circumferential velocity is only $O(U\sqrt{ϵ})$, and its flux is $O(Ua{ϵ}^{3/2})$. Circumferential flux divergence is smaller than the possible leading axial divergence by a factor $ϵ$. Consequently $q_x=0$ to the order needed. This is not an assumption of equal flux at different angular positions.

For fixed $\theta$ define the lubrication resistance integrals for a nearly occluding sphereIn particular, $J_1=\pi \sqrt{2a/h_0}$, $J_2=J_1/(2h_0)$ and $J_3=3J_1/(8h_0^2)$. The local pressure relation givesA putative generic value $q=O(U{h}_0)$ makes this jump $O(\mu U/(a{ϵ}^{3/2}))$. However, the integrated wall shear in part (ii) is only $O(\mu Ua/\sqrt{ϵ})$, so the force-free global balance permits a common net pressure jump only of order $\mu U/(a\sqrt{ϵ})$. The outer flow has circumferentially uniform pressures on both sides. Therefore the $O({ϵ}^{-3/2})$ pressure jump must cancel in every angular strip, not merely after angular averaging. The pressure recovery condition in lubrication flow determinesThis explains how the leading-order part of the global force balance selects the local flux.

The total sphere-frame volume flux is nowTransforming back to the laboratory adds $\pi a^{2}U$, so $Q_{\mathrm{lab}}=\pi a^{2}U(1-4ϵ/3)$ to this order. For physical pressure, Hagen-Poiseuille flow gives $Q_{\mathrm{lab}}=-\pi a^4{p}_x/(8\mu )$. ThusThe printed positive expression is the pressure-drop gradient in the direction of the laboratory flow; the Poiseuille hint omits the minus sign when $p_x$ is read as the derivative of physical pressure. With the specified wall velocity $-U$, the laboratory flow is positive and its physical pressure gradient is negative.

Next insert the leading flux into the wall shear stress:Let $\Delta p>0$ be the localized extra pressure drop, after subtracting the regular far Poiseuille gradient. Contributions outside the neck are only $O(\mu U/a)$ and are smaller than the $O({ϵ}^{-1/2})$ jump. Since ${\sigma }_{rx}=-{\sigma }_{xy}$, part (ii) gives $\pi a^2\Delta p=a\int \int {\sigma }_{xy}(0)dxd\theta$. Hence the leading pressure drop isFor example, the concentric limit is $\Delta p\sim 4\pi \mu U\sqrt{2/ϵ}/a$. The integral grows as the minimum gap shrinks; the fixed-$\lambda <1$ qualification prevents treating a further near-contact limit as uniformly covered.

On the sphere the leading shear stress is ${\sigma }_{xy}(h)=-2\mu U/h+4\mu U{h}_0/h^2$, soPositive and negative tangential stresses cancel within each angular strip. Thus the leading tangential resultant, and its leading torque contribution, vanish: negligible rotation is consistent with the torque-free condition. The shear is not zero pointwise, and the pressure forces are not zero. Pressure on a sphere cannot produce a torque about its centre because its normal is radial; its force contributions and the matched far pressure jump remain essential to the force-free transport of a nearly occluding sphere.

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 a sphere falling with speed $U$ through a tube of radius $a$, let $Q$ be the downward laboratory volume flux. In the sphere frame the wall moves upwards and the narrow annulus carries upward flux per unit circumference $q=(\pi a^{2}U-Q)/(2\pi a)$. Couette-Poiseuille flow in a thin gap gives $q=Uh/2-h^3{p}_z/(12\mu )$. Thus the pressure below minus above the sphere is $\Delta P=6\mu (2q{I}_3-U{I}_2)$, using lubrication resistance integrals for a nearly occluding sphere. A nearly piston-like laboratory flux can coexist with a nonzero relative gap flux.