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For a Newtonian fluid and a flat wall with $0<b<\infty$, the Navier slip boundary condition is equivalent to a tangential wall-on-fluid tractionThe sign expresses resisting friction: ${\partial }_{n}u$ uses the inward unit normal, whereas the wall-on-fluid traction uses the outward unit normal of the fluid domain. The irreversible loss of mechanical energy per unit wall area and time isFor two tangential directions replace $u_{\mathrm{rel}}^2$ by the squared Euclidean norm of relative tangential velocity. At a moving wall this relative-velocity loss must be distinguished from the power supplied by the wall's motion. Bulk viscous dissipation also remains present. A finite slip length models reduced interfacial resistance, without declaring every boundary ideal or removing dynamic viscosity.

The limiting loss depends on how the flow is driven. In Couette flow with a slipping lower wall,It tends to zero for both $b\to 0$ and $b\to \infty$ at fixed $H$ and $U$. At $b=0$ impose the no-slip boundary condition directly, rather than dividing by zero. In planar Poiseuille flow with equal wall slip at fixed $G$, each wall instead dissipates $G^{2}b{H}^2/(4\mu )$, which grows with $b$ because relative velocity grows. The bulk loss per unit width is $G^2{H}^3/(12\mu )$; adding both wall losses gives $Gq$, the pressure work per unit channel length and width. This energy balance shows why small friction coefficients do not guarantee small losses under every driving constraint.

NASA's Wind Tunnel Experiment Details, slide 27, describes a practical manometer measurement: partially fill a Tygon tube with water and connect its two ends to total pressure and static pressure ports in a wind tunnel. The total pressure connection pushes its liquid meniscus downward; the vertical length difference supplies the pressure difference. The slide predicts increasing separation as the tunnel speed increases. It supplies an apparatus and method, rather than a table of observed readings.^[ref]

A control for this practical is to expose both connections to the same atmospheric pressure before the run: settled liquid meniscus heights should agree in a symmetric column. Read each liquid meniscus at eye level, reconnect sealed tubing to the identified ports, and wait for settling. Recheck the equal-pressure zero afterwards; reverse connections to check the sign. These checks establish neither mass density nor ruler accuracy.^[ref]

The inference is a differential pressure from hydrostatic pressure, with stagnation pressure on one side and static pressure on the other in NASA's practical. Converting that difference to a flow speed additionally needs the hypotheses discussed under dynamic pressure and Pitot-static tube. A changing pressure can leave the column oscillating; leaks, blocked pressure tappings or trapped liquid in a gas connection can alter or delay the readout. The column should be read after settling, and temperature should be recorded when choosing the measuring liquid's mass density.

Two corrections clarify the limits. If both gas ports are at the same height and carry the same gas of constant mass density ${\rho }_g$, keeping its hydrostatic pressure gives $p_L-p_R=(\rho -{\rho }_g)g\Delta h$. Different port heights require their separate gas column heights. Also, a curved liquid meniscus can sustain a capillary pressure jump through surface tension. Write $J_L,J_R$ for liquid pressure minus gas pressure at the two surfaces. Neglecting gas weight, the balance becomes $p_L-p_R=\rho g\Delta h+J_R-J_L$. Identical meniscus shapes in matching clean bores cancel those jumps; unequal bores or different wetting conditions need not. Thus the useful simplicity of $\rho g\Delta h$ rests on identifiable physical conditions.

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.

A white dwarf is a compact stellar remnant supported mainly by electron degeneracy pressure. Its ions carry most of the mass, while its Electrons provide most of the supporting pressure. The Pauli exclusion principle makes compression fill quantum states with larger momenta; support therefore persists even when thermal motion becomes small. Cooling and hydrostatic equilibrium are separate questions: a white dwarf can radiate stored energy while retaining degeneracy support.

The surprising trend is that a more massive cold white dwarf is smaller. Its higher mass density raises the Fermi momentum, eventually requiring special relativity. The degenerate-electron pressure crossover explains why this response approaches a limiting stiffness and why the Chandrasekhar limit exists. Subrahmanyan Chandrasekhar's 1983 lecture develops this connection between quantum statistics, stellar structure and stellar evolution.^[ref]

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.