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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.

Integrating the radial velocity over the thickness gives the radial volume flux per unit width:The full volumetric flow rate across a cylindrical surface, positive outward, isAn annulus of radial thickness $dr$ contains volume $2\pi rhdr$ to leading order. Its rate of gain equals incoming minus outgoing volume flux, soDividing and taking $dr\to 0$ yields the axisymmetric thin-film equation:The equation assumes no evaporation or injection through the boundaries. Integrating it over the whole drop also verifies conservation of mass when the edge carries zero volume flux.

Take $x$ along a horizontal gap and $s$ perpendicular to its plates, with $0<s<b$. Assume a Newtonian fluid, stationary plates, a steady flow, a laminar flow and a fully developed flow away from the entrances and side edges. The no-slip boundary condition holds on both plates. Any transverse gravity is absorbed in the hydrostatic pressure; there is no imposed transverse through-flow. Translational symmetry and the no-penetration boundary condition then give $\mathbf{u}=(u(s),0,0)$. The Navier-Stokes equation reduces toTwice integrating gives the plane Poiseuille flow:The volumetric flow rate through one gap isThere are $n-1$ internal gaps. Their total cross-sectional area is $(n-1)wb$, because the plate thickness is negligible, so the Darcy velocity isThus the permeability of a porous medium has the dimensions of squared length. If finite plate thickness were included, the Darcy velocity averaged over the whole manifold would also contain its porosity.

Both the local acceleration and the convective acceleration vanish: ${\partial }_t\mathbf{u}=0$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=u_x{\partial }_x\mathbf{u}=0$. The dominant balance is therefore between the pressure gradient and viscous stress, with no inertia contribution in this fully developed solution. This exact cancellation does not guarantee that arbitrary large-Reynolds number realizations remain laminar flows; entrances, disturbances and turbulence can invalidate the assumptions.

Under spherical symmetry, the continuity equation and Darcy's law giveHence $q_r=C/r^2$, and integrating the pressure gradient from infinity givesThe boundary condition at $r=a$ fixes $C={\Pi }_{1}a\Delta p/\eta$. Therefore spherical Darcy injection hasHere $q_r$ is the superficial Darcy velocity, not the pore velocity $q_r/\phi$.

For the comparison, assume geometrically similar pore networks, equal porosity, the same source radius and far boundary geometry, and equal pump efficiency. Dimensional analysis then gives ${\Pi }_1=C_{\mathrm{geom}}{d}^2$ with the same dimensionless factor. ConsequentlyAt a fixed desired volumetric flow rate, the required excess pressure and pumping power areAquifer A requires half the pumping power at equal throughput, or half the energy to inject the same volume over the same time. This compares the excess pressure-volume work dissipated by the Darcy flow; any different baseline compression or other plant costs need separate accounting.

The imposed comparison matters. At fixed $\Delta p$, A takes twice the volumetric flow rate and twice the instantaneous power, but the same excess energy per injected volume. The pore spacing alone does not determine permeability of a porous medium if network geometry or porosity changes; without the geometric assumptions the preference cannot be inferred from $d$ alone.

For the common constant mass density, the background hydrostatic pressure satisfies $d{p}_0/dz=-\rho g$. Define the excess pressure $P=p-p_0(z)$. In Darcy's law, the background pressure gradient cancels gravity:If the source imposes the same excess pressure relative to the local hydrostatic pressure everywhere on its spherical surface, $P(a)=\Delta p$, the spherical Darcy injection result survives exactly:It is important to specify what is maintained on the source. A literally uniform absolute pressure $p_s=p_0(0)+\Delta p$ on a finite sphere centered at $z=0$ instead gives $P(a,\theta )=\Delta p+\rho ga\cos \theta$, where $\theta$ is measured from the upward vertical. The decaying harmonic function outside the sphere is thenand thereforeThis supplies a concrete exception to an unqualified claim that gravity leaves the radial flow unchanged. The additional dipolar contribution has zero net volumetric flow rate, so $Q=4\pi {\Pi }_{1}a\Delta p/\eta$ still holds. Equal injected and resident mass densities are essential to the background cancellation.

Take positive volumetric flow rate in the direction of increasing $x$. For an arterial pressure wave, the characteristic admittance of an arterial wave givesThe backward wave contributes the opposite flow sign. Write $p_j$ for each junction's complex pressure amplitude, suppressing the common $e^{i\omega t}$ factor, and put $r=RY$, $C_{\beta }=\cos \beta$, $S_{\beta }=\sin \beta$. The topology is the two paths $K\to L\to N$ and $K\to M\to N$; their intermediate and common distal junctions each have a hydraulic resistance. The coordinate convention in the source puts the upstream end at $-\ell$ and the downstream end at zero in each segment. Consequently the wave transfer matrix of an arterial segment isEquivalently $q_u=Y(C_{\beta }{p}_u-p_d)/(i{S}_{\beta })$ and $q_d=Y(p_u-C_{\beta }{p}_d)/(i{S}_{\beta })$, with their continuous limits understood when $S_{\beta }=0$.

Symmetry gives $p_L=p_M=p_s$. Distal mass conservation and the resistive relation $Q_N=p_N/R$ giveAt $L$, the incoming flow supplies both its bed and segment $D$, soChoose the incident amplitude $P_A$ at $K$. If $P_{AR}$ is the reflected amplitude in vessel $A$, pressure continuity equation gives $p_K=P_A+P_{AR}$, while inlet flow is $2Y(P_A-P_{AR})=2Y(2P_A-p_K)$. Equating this to the two outgoing segment flows yieldsEliminating $p_s,p_K$ from these three relations gives the wave transmission through a resistively loaded arterial loop resultFor large $r$ at fixed phase, $p_N=2P_A{e}^{-2i\beta }+O(r^{-1})$ and $p_M=C_{\beta }{p}_N+O(r^{-1})$. HenceThese leading terms must not be used as nonzero denominators near a phase at which the normal intermediate flow vanishes.

In the travelling coordinate $X=x-ct$, mass conservation becomes $[\alpha (u-c){]}^{\prime }=0$. Thus the conserved relative flux is $Q$ andFor a periodic peristaltic pumping solution, the neglected-inertia momentum equation isTake ${\alpha }_0$ to be the spatial mean area, so $⟨{\alpha }_1⟩=⟨{\alpha }_2⟩=0$. This normalization, and zero net pressure drop over a period, are implicit in the specified periodic forcing. The constant leading state has no pressure gradient, giving $Q_0=-{\alpha }_{0}c$. Put $B={\tilde{P}}_0^{\prime }$ and $d=R_{0}c/{\alpha }_0$. At first order,Period averaging gives $Q_1=0$, and the zero-mean solution isFor the second-order mean balance, expand the velocity and friction:Averaging the exact pressure derivative gives zero, so $⟨R_0{u}_2+R_1{u}_1⟩=0$. Hence the weak peristaltic pumping of a collapsible tube result isThe mean laboratory volumetric flow rate is $c⟨\alpha ⟩+Q={ϵ}^2{Q}_2+O({ϵ}^3)$, forward with the wave when $c>0$. The negative leading relative flux is therefore consistent with positive mean pumping.

Use the ideal one-dimensional pipe model: constant density, steady inviscid flow, approximately uniform endpoint speeds $u_1,u_2$, and equal body-force potential at the two ends. Conservation of mass gives the constant volumetric flow rate $q=u_1{S}_1=u_2{S}_2$, and the mass flow rate is $m=\rho q$. The steady Bernoulli equation givesFor flow from a wider to a narrower section, $S_1>S_2$ and $\Delta p>0$. Solving for the positive mass throughput gives the Venturi effect formulaThe formula requires these hydraulic assumptions; steady incompressibility alone does not impose uniform endpoint profiles or equal potential. If $V_1\ne V_2$, the same derivation replaces $\Delta p$ by $\Delta p+\rho (V_1-V_2)$. Viscous losses would also modify the pressure balance. The equal-area case is singular in this pressure-based determination: in the ideal equal-potential model its pressure difference must vanish and does not determine the throughput.

Integrating the resistive induction equation once introduces a constant $C$:Because $b(L)=b(-L)=0$, integrating this relation again over the channel gives $C=B_{0}Q/(2L)$, where $Q$ is the volumetric flow rate per unit span. Put $m^2=B_0^2/({\mu }_0\rho \nu \eta )$ and $H=mL$, taking $B_0\ge 0$ without loss of generality. Reversing the imposed field reverses $b$ but leaves $u$ and $Q$ unchanged. Eliminating $b^{\prime }$ from the magnetohydrodynamic momentum equation givesIts two zero wall values remove the odd homogeneous solution. Write $u=U[1-\cosh (mz)/\cosh H]$. Integrating the first relation with the two magnetic wall values yieldsTherefore $U=(G{L}^2/\nu H^2)H\coth H$. Substitution and integration produce the velocity and induced field:Here $H=B_{0}L/\sqrt{{\mu }_0\rho \nu \eta }$ is the Hartmann number. The hyperbolic cosine makes $u$ even, while the hyperbolic sine makes $b$ odd. Direct differentiation verifies both coupled equations and all four wall values.

Integrating the velocity gives the fluxThe apparent singularity at $H=0$ is removable. The small-field limit is plane Poiseuille flow:The cubic coefficient of $b$ is its first-order response in $B_0$; $b$ itself vanishes at zero imposed field. For a fixed positive $G$, the flux decreases monotonically as $H>0$ increases. One exact way to see the sketch's monotonicity is the partial-fraction expansion of the hyperbolic cotangent:whose derivative is strictly negative for $H>0$. The flux has a horizontal tangent at $H=0$ and approaches zero asThe flux is even if signed $H$ is used.

For $H\gg 1$, put $\zeta =z/L$. Away from the walls,The velocity has a nearly uniform core, with thin Hartmann layers of thickness $\delta =L/H$ enforcing the no-slip boundary condition. Near the upper wall, with $s=H(1-\zeta )$ fixed, $u/U_c\sim 1-e^{-s}$ and $b/B_s\sim -1+s/H+e^{-s}$; the lower wall follows by even/odd symmetry. Thus $b$ is negative for $0<z<L$, positive for $-L<z<0$, and returns rapidly to zero at both walls. The required sketches show a decreasing flux, a flat velocity core, and an odd induced field with magnetic wall layers.

Write $\mathbf{g}=-g{\mathbf{e}}_z$. The vertical component of the balance gives hydrostatic pressure,Its horizontal gradient is ${\nabla }_{h}p=\rho g{\nabla }_{h}h$. The horizontal balance is consequentlyThus $\mathbf{u}\cdot {\nabla }_{h}h=0$: the velocity is tangent to every level set of $h$, so those contours are streamlines of the geostrophic flow.

Across a vertical section joining the contours $h_0$ and $h_0+\Delta h$, an element of horizontal distance normal to the contours is $dn=dh/|{\nabla }_{h}h|$, while $|\mathbf{u}|=(g/f)|{\nabla }_{h}h|$. The volumetric flow rate is thereforeFor dimensional analysis, $[g]=L{T}^{-2}$, $[f]=T^{-1}$ and $[h_0]=[\Delta h]=L$, so the result has dimension $L^3{T}^{-1}$, exactly that of a volume flux.

The volumetric flow rate is $Q=2\pi {\int }_0^{R}u(r)rdr$. Integration by parts, using finite $u(0)$ and $u(R)=0$, givesBecause the stress magnitude is $\tau ={\tau }_{R}r/R$, change variables from $r$ to $\tau$:Therefore the required power is $n=3$. The fundamental theorem of calculus gives the pipe form of the Weissenberg–Rabinowitsch equation:Finally $\eta ({\dot{\gamma }}_R)={\tau }_R/{\dot{\gamma }}_R$, soA measured pressure-drop--flow-rate curve therefore recovers the wall viscosity without assuming a constitutive form.

Let $W_f$ be the vertical velocity of the carrier fluid. Each species has its isolated Stokes settling velocity relative to that fluid, so $W_i=W_f+{\hat{W}}_i$. The mixture has no imposed vertical volumetric flow rate. Using the particle volume fractions, its zero-volume-flux condition isSubstitution of $W_i=W_f+{\hat{W}}_i$ gives the common carrier-fluid backflowConsequently the hindered settling velocities areThe same $W_f$ appears for both populations precisely because every particle is assumed to feel the same volume-averaged backflow.

Define the indoor-to-outdoor reduced gravityBecause the whole opening lies above the interface, its indoor side contains upper-layer fluid of density ${\rho }_H$. Under hydrostatic pressure, the pressure difference varies linearly about a neutral pressure level. Equal opening geometry and equal discharge coefficients for inflow and outflow put that level at the vent midpoint. At vertical distance $y$ from it, the ideal Bernoulli equation gives speed $\sqrt{2g_V^{\prime }|y|}$.

Writing $Q_V$ for either one-way volumetric flow rate, integration over one half of the opening givesThe total unsigned exchange is $2Q_V$. The ideal sharp-edged inviscid model has $C_d=1$; an empirical $C_d<1$ represents contraction and losses.

For plume radius $r$, top-hat speed $U$, and plume reduced gravity $g_B^{\prime }$, define the volumetric flow rate, momentum flux, and buoyancy fluxThe integral balances for a steady axisymmetric pure plume in a uniform lower layer areThe source is at the plume's virtual origin $z=0$. Solving these equations givesIt is useful to defineThen the remaining similarity laws take the compact formThe second relation also follows immediately from conservation of buoyancy flux, $B=Q_B{g}_B^{\prime }$.

Measure downward distance from the plume virtual origin byand let $\mathcal{B}$ be the magnitude of the cold plume's buoyancy flux per unit span. A one-sided wall line plume with width $b$, downward speed $W>0$, and reduced-gravity magnitude $g^{\prime }$ satisfiesThe pure plume solution isImposing $b(z_V)=H^{\prime }/2$ fixesThe constant speed is a special feature of a pure top-hat line plume; its width and volumetric flow rate grow linearly with downward distance while entrainment dilutes its density anomaly like $(z_o-z{)}^{-1}$.

Let $w(x)$ be the downward velocity between the vertical plates. A steady vertical momentum balance givesSymmetry requires ${\tau }_{xz}(0)=0$, so the stress magnitude isThe wall stress first reaches the yield stress when $\rho g{l}_c={\tau }_y$. The onset measurement therefore gives

For $l>l_c$, define $\Delta =l-l_c$. The central region $|x|\le l_c$ is a plug flow of a yield-stress fluid, while the layers $l_c<|x|\le l$ are yielded. The Herschel–Bulkley fluid law gives, for $x\ge l_c$,Integrating from the no-slip wall $w(l)=0$ givesin the yielded layer, and the plug moves atThe volumetric flow rate per unit span is consequently

Near onset, the plug term dominates andThe observed exponent two therefore corresponds to $n=1$, a linear post-yield constitutive law at small shear rate. Far above onset, the fully yielded contribution scales asThe observed exponent four corresponds to $n=1/2$, a shear-thinning square-root law at high shear rate.

A plausible stress curve is therefore odd in $\dot{\gamma }$, starts at the two yield-stress values $\tau =\pm {\tau }_y$, and has finite linear slope just after yield:It then crosses smoothly to the concave high-rate behaviorwith the negative-rate branch fixed by odd symmetry. This combines the two power laws inferred independently from the measured flux limits.

Arterial recoil, the muscle pump, respiratory movements, venous tone and the mechanics of cardiac filling assist blood circulation. The heart remains its principal continuing source of mechanical energy. These agencies differ in whether they store cardiac work, contribute new muscular work, or direct flow without supplying power.

During ventricular systole, the aorta and other large elastic arteries expand and store elastic potential energy. During diastole, their recoil continues to drive blood towards the peripheral vessels, while closure of the aortic valve limits return into the heart. This elastic reservoir smooths an intermittent cardiac input into a more continuous downstream flow. It does not manufacture extra energy: it releases part of the work performed during the preceding ejection.

Contraction of surrounding skeletal muscles compresses veins. Venous valves favour displacement towards the heart and limit backward flow when the compression is released. Alternating contraction and relaxation therefore produces a muscle pump, especially useful for return from dependent limbs. The added work comes from the contracting muscles; valves guide the displacement but supply no power themselves. Persistent compression without refilling is not equivalent to a series of pumping strokes.

Spontaneous inspiration supplies another assistance. Descent of the respiratory diaphragm raises abdominal pressure while expansion of the thorax lowers intrathoracic pressure. The resulting changes favour a pressure gradient towards the thoracic veins and right side of the heart. The effect depends on posture, the respiratory phase and whether vessels remain open. Positive-pressure inflation does not reproduce the same pressure changes as spontaneous inspiration.^[ref]

Venous smooth muscle can reduce the storage capacity of peripheral veins, transferring blood towards the central circulation. This changes the volume available for cardiac filling; it does not require a travelling wave of venous peristalsis. Relaxation and recoil of the heart, together with movement of the atrioventricular plane, also influence filling pressures. The older expression cardiac suction is therefore best interpreted through specific changes of chamber volume and pressure, rather than a separate force pulling uniformly through every vessel.

Gravity assists a downward column and opposes an upward one. For a complete circuit returning to its initial height it supplies no net gravitational work. Its important effects include hydrostatic pressure differences and dependent pooling. Likewise, the residual pressure gradient across the capillaries is chiefly the downstream consequence of the cardiac pump. Ordinary capillary action at a liquid meniscus is not a universal extra pump inside a closed, fluid-filled vascular network. Local vasodilation can improve perfusion by reducing vascular resistance, but it does not create the necessary pressure source.

With $Q$ the mean volumetric flow rate, $R$ an effective vascular resistance, and $\Delta P$ the driving pressure difference, the useful bookkeeping relation isIt separates an increase of driving pressure from a reduction of resistance. An instructive apparatus compares the outflow of a pulsatile pump through rigid tubing with its outflow through an elastic reservoir feeding the same resistance. Continued outflow between strokes demonstrates stored elastic potential energy. A separately compressed valved flexible tube demonstrates the additional muscle pump principle. Neither observation requires a mysterious attraction of blood by the tissues.

The lower wall has streamfunction zero. The streamline through the lip and along the upper fluid free boundary has streamfunction equal to the downstream volumetric flow rate, namely $a$ because the downstream speed is one and its depth is $a$. The complex potential of a two-dimensional potential flow therefore fillsThe chosen velocity potential origin gives $w(D)=ia$. The remote reservoir endpoints $A,E$ have velocity potential tending to $-\infty$; the downstream endpoints $B,C$ have velocity potential tending to $+\infty$.

Put $q=dw/dz=u-iv$. On the floor its values run from zero to one along the positive real axis. Above the plate the flow goes left toward the lip, so the plate velocities run along the negative real axis from zero at $E$ to $-1$ at $D$. On the fluid free boundary Bernoulli's equation and its constant pressure give $|q|=1$. Its downward velocity has $v<0$, hence $\mathrm{Im}q>0$. Consequently the hodograph plane domain isthe upper unit semicircle. Its marked boundary values are $q(A)=q(E)=0$, $q(D)=-1$, and $q(B)=q(C)=1$.

In steady state the incoming volumetric flow rate equals the discharge. Bernoulli's principle between a nearly stationary free surface and the atmospheric outlet gives the ideal Torricelli's law speed $\sqrt{2gh}$. The specified discharge coefficient multiplies the opening-flow rate, soThus, with $A=10^{-4}{\mathrm{m}}^2$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,Neglect surface-speed and capillary corrections. The coefficient includes contraction and loss effects; it should not be applied a second time to the calculated height.

Let $z$ increase down the pipe, $a=0.015\mathrm{m}$ be its radius and $r$ the distance from its axis. For steady, fully developed laminar flow of a Newtonian fluid, the axial Navier-Stokes equations reduce toWrite $K=-dp/dz+\rho g\sin 20^{\circ }$. Integrating twice, regularity at $r=0$ excludes a logarithmic term, and the no-slip boundary condition $u(a)=0$ gives the Hagen-Poiseuille flow profileThe volumetric flow rate follows by integrating over the circular cross-section:For the intended gravity-driven case with no imposed axial pressure gradient, $dp/dz=0$. Substitution givesandThe corresponding mean speed is about $0.1047\mathrm{m}{\mathrm{s}}^{-1}$ and the diameter-based Reynolds number is $\rho \overline{u}(2a)/\eta \simeq 3.49$, consistent with laminar flow. The original PDF does not specify inlet and outlet pressures. If they impose an additional pressure gradient, the general result with $K$ above applies; gravity alone does not determine that extra contribution.