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Water potential expresses the chemical potential of water relative to a stated pure-water reference in units of pressure. For a nearly incompressible liquid at fixed temperature, use a constant reference molar volume ${\bar{V}}_w$ and writeThe reference is pure water at the same temperature and a chosen pressure $P_0$. Both sides must use the same reference. Neglecting height differences and interactions with solid surfaces, the approximation isHere $a_w$ is the dimensionless thermodynamic activity of water, $R$ the molar gas constant, and $T$ the absolute temperature. The natural logarithm is therefore applied to a dimensionless quantity. The pressure contribution ${\Psi }_p=P-P_0$ increases with pressure; the solute potential ${\Psi }_s$ is negative when $a_w<1$. Across an available passive water route, net water transfer favors higher to lower water potential. The familiar plant description adds effects of height and interactions with solid surfaces when they matter; they are omitted in the two liquid compartments here.^[ref]

Adding solute and raising pressure can thus cancel in water potential while each contribution remains nonzero. Equality of water potential does not imply equality of pressure or of every solute concentration. In a supported plant cell, the cell wall permits a positive turgor pressure to balance a negative solute potential; plasmolysis illustrates the opposite loss of support after water leaves.

For time dependence $e^{i\omega t}$ and a segment of length $\ell$, let $\beta =\omega \ell /c$. Adding the forward and backward arterial pressure waves gives the displayed upstream-to-downstream matrix. The flow orientation is downstream at both ends. At a junction, pressure is continuous and incoming volumetric flow rates balance outgoing ones.

Consider two identical two-segment paths from an inlet to a distal junction, with equal resistive beds at the two intermediate junctions and the distal junction. The segment transfer matrix and junction flow balances determine all pressures. If one inlet branch is removed and the beds are weakly loading the waves, the remaining segments form a three-segment chain with an effectively closed far end. The altered reflection pattern can raise peripheral oscillatory flow amplitudes even though a branch was lost; the incident wave amplitude, rather than total inlet pressure or flow, is the prescribed forcing.

For negligible inertia, periodic forcing $p_e=ϵP_e\sin kX$, resistance $R\propto {\alpha }^{-n}$ and prescribed mean area ${\alpha }_0$, write $Q=-c{\alpha }_0+{ϵ}^2{Q}_2+\cdots$. First-order area obeys ${\tilde{P}}_0^{\prime }{\alpha }_1^{\prime }+R_{0}c{\alpha }_1/{\alpha }_0=-k{P}_e\cos kX$. Its mean square is $P_e^2/[2({\tilde{P}}_0^{\prime 2}+(R_{0}c/(k{\alpha }_0){)}^2)]$. Averaging the second-order friction balance gives $Q_2=c(n+1)⟨{\alpha }_1^2⟩/{\alpha }_0$, proving the displayed forward pumping rate. Periodicity entails zero mean pressure drop; the mean-area convention fixes the expansion's otherwise free constant.

For a distributed heat source, a well-mixed interior with temperature excess $\theta$ gains temperature-volume flux $\mathcal{H}$ and loses $|q|\theta$ by ventilation in either direction. With signed upward throughflow, $q|q|=K_A(G_T\theta \pm W)$, where $G_T=g{\alpha }_T{H}_b$ and $W$ is wind pressure divided by mass density. The plus sign is assisting wind. Steady temperatures satisfy ${\theta }^2(G_T\theta +W)={\mathcal{H}}^2/K_A$ for assistance, or ${\theta }^2|G_T\theta -W|={\mathcal{H}}^2/K_A$ for opposition. Direction restrictions must be applied to roots.

An apparatus that produces controlled fluid flow past a model for measuring quantities such as pressure, lift and drag.

The Young–Laplace equation relates the pressure jump across a stationary interface to its curvature and surface tension. Choose a unit normal vector $\mathbf{n}$ pointing from the inside region to the outside region, and define $\kappa ={\nabla }_s\cdot \mathbf{n}$ to be positive for a convex sphere. In mechanical equilibrium, with constant surface tension and no additional normal viscous stresses,Here $k_1,k_2$ are the signed principal curvatures in this convention. Their sum is twice the corresponding signed mean curvature; stating the convention prevents a factor-of-two ambiguity. The Young–Laplace equation is a local normal stress balance for one interface. ^[ref]

A spherical liquid droplet of radius $R$ in a gas has one interface, with $k_1=k_2=1/R$. An ideal thin soap bubble has two liquid–gas interfaces. Adding their pressure jumps givesFor the soap bubble, $\Delta p$ now means the inner gas pressure minus the outer gas pressure. The second formula assumes equal surface tensions on both faces and film thickness negligible relative to $R$, so their radii can be identified. Retaining distinct radii gives $2\gamma /R_{\mathrm{inner}}+2\gamma /R_{\mathrm{outer}}$. A gas bubble inside bulk liquid has only one interface and therefore uses $2\gamma /R$. ^[ref]

With the assumed constant toy surface tension $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, a droplet with $R=1\mathrm{mm}$ has excess pressure $144\mathrm{Pa}$; at $R=0.5\mathrm{mm}$ it has $288\mathrm{Pa}$. A thin soap bubble assigned the same $\gamma$ and $R=1\mathrm{mm}$ has excess pressure $288\mathrm{Pa}$. These are differences from the surrounding pressure, rather than absolute pressures. Halving the radius doubles the capillary pressure; doubling the interface count also doubles it at fixed radius and surface tension.

The spherical case has a short work argument. A virtual increase in radius gives $dV=4\pi R^{2}dR$ and $dA=8\pi RdR$. At mechanical equilibrium, the net pressure work balances the increase in surface energy:For a thin soap bubble, both interfaces gain approximately the same surface area, doubling the right-hand side and giving $4\gamma /R$. This is a virtual work balance between neighboring spherical configurations, rather than an assumption that a fixed-volume droplet actually expands. The general Young–Laplace equation replaces the spherical area-to-volume ratio by local curvature, making geometry directly measurable through pressure.

Besides the two propagating inertial-acoustic wave branches, a uniform barotropic Keplerian shearing sheet has a stationary branch balancing pressure against Coriolis acceleration. For radial wavenumber $k\ne 0$, setting $\omega =0$ in continuity gives $U_x=0$, and radial momentum gives the displayed azimuthal balance. The full determinant is proportional to $\omega ({\omega }^2-{\Omega }^2-c_s^2{k}^2)$. At $k=0$ the stationary branch is a uniform column-density change; it is not a travelling density wave.