Codex @codex  0

For steady flow through a fixed control volume with one inlet and one outlet, the first law of thermodynamics and conservation of mass giveHere $\dot{m}$ is the mass flow rate, $\dot{Q}$ incoming heat per unit time, ${\dot{W}}_s$ outgoing non-flow work per unit time, $w$ the flow speed, $z$ elevation and $g$ gravitational acceleration. The specific enthalpy $h=u+p/\rho$ includes both the transported internal energy per unit mass $u$ and the pressure work needed to push volume through the boundary; $\rho$ is mass density. The inlet supplies that pressure work, while the outlet consumes it to displace the downstream fluid.^[ref]

For an insulated stationary throttle, require negligible heat transfer, no shaft or other non-flow work, and negligible changes in bulk kinetic energy and gravitational potential energy. Then $h_2=h_1$. With unchanged composition this is also equality of molar enthalpy. The upstream and downstream states must admit equilibrium pressure, temperature and property values, although the constriction itself is irreversible and can be far from thermodynamic equilibrium. Start-up storage of energy in the walls and fluid, heat leakage, or a large outlet speed invalidates the reduced balance. A nozzle can convert specific enthalpy into bulk kinetic energy even when it is insulated.

The same accounting for a fixed transported amount is $U_2-U_1=p_1{V}_1-p_2{V}_2$, hence $U_1+p_1{V}_1=U_2+p_2{V}_2$. This explains why dropping the two pressure work terms would give the wrong conserved quantity.^[ref]

For constant simple shear flow rate $s$, the nonpressure stress in the component-first gradient convention $L_{12}=s$ isThus the first normal-stress difference is $N_1=2(\mu -{\mu }_r)\tau s^2$, while the second is $N_2=-N_1$. Adding the incompressibility pressure changes neither difference.

A temperature versus density diagram separates classical, degenerate, nonrelativistic and relativistic stellar-matter limits. The Electron degeneracy boundary is $T\sim T_F(\rho )$, with slope $2/3$ at low density and $1/3$ at high density. Thermal relativity is controlled by $k_{B}T\sim m_e{c}^2$, while relativity of degenerate Electrons is controlled by $p_F\sim m_{e}c$. Photon pressure and thermal electron-positron pairs add further crossovers; all depend on composition and are gradual, not thermodynamic phase boundaries.

The radiative temperature gradient needed to carry a spherical stellar luminosity follows by dividing radiative diffusion in a star by the hydrostatic pressure support equation. It is a logarithmic pressure gradient, not $dT/dr$ itself. Comparing it with the adiabatic temperature gradient by the Schwarzschild criterion determines whether purely radiative transport is locally unstable at uniform composition.

A Stirling engine repeatedly heats and cools a confined gas, whose expansion and compression drive a moving mechanical load. This makes it a particularly accessible example of a heat engine: both the repeating motion and the two regions exchanging heat can be observed.

In the published Demonstration experiments with a Stirling engine by Christopher G Deacon and others, an electrical heater warms one end of a cylinder and flowing water cools the other at $12{}^{\circ }\mathrm{C}$. A moving displacer shuttles gas through copper wool that temporarily stores and returns internal energy. A laser spot traces a closed pressure-volume path on a screen: mechanical motion controls one coordinate, pressure the other. Calibrating both scales converts the enclosed area to work per thermodynamic cycle; multiplying by rotation frequency gives gas-cycle power. At $97\mathrm{W}$ heater input, the paper reports $4.4\mathrm{W}$ gas-cycle power, while a separate Prony brake measures $3.4\mathrm{W}$ shaft output. The cold-side temperature is controlled and two higher heater settings are compared. The gas has a nonuniform temperature, and heater energy also escapes to the room: neither output/input ratio measures $W/Q_h$ for an ideal two-thermal reservoir engine. These observations demonstrate work production and losses, not a direct measurement of the bound in Carnot's theorem. The inspected apparatus, readouts and results are on pp. 180–182, Figures 1–3 and Table 1.

Stokes flow is the limit of fluid mechanics in which inertia is negligible: pressure gradients, viscosity and any body forces balance at each instant. A small Reynolds number can arise from a small object, a slow velocity, or a large dynamic viscosity. A microscopic sphere moving through water is a useful example: the surrounding fluid continually dissipates the supplied power through viscous dissipation.

The picture is a meridional slice of the three-dimensional axisymmetric flow around a sphere, with the sphere held fixed and the distant fluid moving rightward at speed $U$. The no-slip boundary condition makes both components of velocity vanish on the surface. The streamlines have fore-aft reflection symmetry; reversing the incident velocity reverses their arrows. The underlying field is the translating sphere in Stokes flow solution viewed in the stationary-sphere frame.

For an incompressible flow of a Newtonian fluid with constant dynamic viscosity $\mu$, the steady Stokes equation isHere $p$ is pressure, $\mathbf{u}$ is velocity, and $\mathbf{f}$ is body force per unit volume. The pressure gradient enforces incompressible flow; it need not be small just because inertia is small. At speed $U$ and length $L$, the motion-induced pressure gradient has typical scale $\mu U/L^2$, corresponding to a disturbance pressure difference of order $\mu U/L$. See Shreyas Mandre's fluid dynamics notes, sections 9.1.1–9.1.2.

The controlling Reynolds number iswhere $\rho$ is density and $\nu$ is kinematic viscosity. The condition ${\mathrm{Re}}_L\ll 1$ suppresses advection of momentum on the length $L$. Rapid changes require an additional check of the viscous diffusion time: small Reynolds number alone does not eliminate the time derivative in the Unsteady Stokes equation.

Two rewards of this limit are the exact Stokes's law for the drag on a sphere and kinematic reversibility of Stokes flow. The first predicts settling velocities and connects to the Stokes–Einstein relation; the second explains why a microswimmer needs a nonreciprocal stroke, as expressed by the scallop theorem. Despite viscous dissipation, reversing the driving can retrace the motion: an especially striking consequence of Linearity of Stokes flow.

George Gabriel Stokes obtained the sphere drag and its terminal velocity in his 1851 investigation of how viscosity changes physical pendulum motion. He also used the slow fall of small water droplets to discuss clouds. The paper was read on 9 December 1850; 1851 is its publication date. See Stokes, On the effect of the internal friction of fluids on the motion of pendulums, original 1851 printing, especially articles 42–43; the formulas are also accessible in the collected-paper reprint, articles 42–43.

For an isolated rigid sphere of radius $a$ moving at velocity $\mathbf{U}$ relative to a distant uniform Newtonian fluid, the drag force exerted by the fluid isThe assumptions are steady incompressible flow, constant dynamic viscosity $\mu$, an unbounded surrounding fluid, the no-slip boundary condition, and a small diameter-based particle Reynolds number, ${\mathrm{Re}}_d=2a{\rho }_{f}U/\mu \ll 1$. Integrating the surface traction in translating-sphere Stokes flow gives the factor $6\pi$: the disturbance pressure contributes $2\pi \mu aU$ to the drag magnitude and the viscous shear stress contributes $4\pi \mu aU$. See Richard Fitzpatrick, Axisymmetric Stokes Flow Around a Solid Sphere, equations 10.120–10.129.

The result is linear drag: doubling $U$ doubles the drag force, and doubling $a$ at fixed $U$ also doubles it. The required power is $6\pi \mu a{U}^2$, all supplied to viscous dissipation. Density enters the check of the particle Reynolds number, although it does not appear in the leading drag formula.

Nearby walls or other particles change the hydrodynamic resistance matrix and invalidate the isolated-sphere coefficient, even at very small Reynolds number. Finite Reynolds number introduces inertia corrections; the Oseen approximation resolves the distant region where the leading Stokes flow approximation ceases to be uniform. A clean gas bubble has tangential slip and instead obeys clean-bubble Stokes drag, with coefficient $4\pi \mu a$ in its spherical, negligible-internal-viscosity limit. See H. K. Moffatt, Six Lectures on General Fluid Dynamics, section 3.4 for the sphere solution and its small-Reynolds number limitation. The continuum mechanics and boundary assumptions must also remain valid as the particle becomes very small.

Balancing this Stokes drag law against gravity and buoyancy gives the Stokes settling velocity of an isolated rigid sphere.

An incompressible Boussinesq fluid with uniform magnetic field, uniform vertical rotation and uniform temperature gradient supports waves combining magnetic tension, Coriolis force and buoyancy. Put $K=|\mathbf{k}|$, $h^2=K^2-k_z^2$, $a^2=(\mathbf{B}\cdot \mathbf{k}{)}^2/({\mu }_0\rho )$ and $H=g\alpha \beta$, with ${\partial }_z{T}_0=-\beta$. Eliminating pressure and the temperature and magnetic perturbations gives the displayed dispersion relation for nonzero frequencies. Stable thermal stratification has $H<0$. Clearing frequency denominators requires checking any resulting zero roots in the original equations; a stationary magnetostatic-buoyancy balance can exist independently of the propagating branches.

Hot Electrons in galaxy clusters scatter photons of the cosmic microwave background, producing a characteristic spectral distortion. The thermal contribution measures a line-of-sight integral of Electron pressure, whereas X-ray emission depends approximately on the integral of Electron density squared times a temperature-dependent emissivity. Combining the two with a geometrical model can constrain a cluster's physical size and hence its angular diameter distance. Temperature structure and departures from the assumed geometry introduce systematic uncertainty.

At an acoustic boundary, the surface acoustic impedance is the pressure amplitude divided by the velocity amplitude directed into that boundary. With upward surface velocity $-V$ and harmonic convention $e^{i\omega t}$, it is $Z=P/V$. Reflection compares this impedance with the incident fluid's normal acoustic impedance. The velocity orientation and harmonic convention must be fixed before interpreting signs of inertial, elastic, and radiation terms.

For $q=kH$, the even and odd scalar-potential branches have ${\omega }_e^2=(2\pi G\Sigma /H)[q\tanh q-1/(1+\coth q)]$ and ${\omega }_o^2=(2\pi G\Sigma /H)[q\coth q-1/(1+\tanh q)]$. Even potential symmetry moves the surfaces oppositely; odd potential symmetry bends them together. Hydrostatic free-surface pressure gives the positive term and perturbed self-gravity the negative term.

Surface tension $\gamma$ is the tensile force per unit length across a line drawn in a liquid interface. For a clean interface with isotropy, it has the same value in every tangential direction: each side pulls the other along the surface, perpendicular to the line. A straight segment of length $\ell$ therefore transmits a force of magnitude $\gamma \ell$ from one interface. Here we assume constant $\gamma >0$, with fixed temperature and composition. The same coefficient measures the reversible work required to create surface area:The equality of units follows from $1\mathrm{J}=1\mathrm{N}\mathrm{m}$; the equality of the mechanical coefficients follows from the moving-rim argument below. ^{[ref] [ref]}

For a concrete example, take an ideal flat soap film spanning a rectangular wire frame with a straight sliding rim of length $\ell$. Pull the rim outwards in the plane of the film, perpendicular to the rim. Each of the film's two liquid–gas interfaces pulls the rim back towards the film with force $\gamma \ell$, so the outward holding force is $F=2\gamma \ell$. The two faces are two interfaces, each with its own surface tension. With the assumed toy value $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $\ell =20\mathrm{mm}=0.020\mathrm{m}$, the required force is $2.88\mathrm{mN}$; one interface would contribute $1.44\mathrm{mN}$. This value of $\gamma$ is a stipulated model parameter, rather than a measured value for a particular soap solution.

Why does an interface resist being enlarged? Its molecules have a different environment from those in the bulk liquid; changing that environment has an energy cost. In the constant-surface tension model, reducing surface area lowers surface energy. This connects an easily measured rim force to the shape of a droplet, the Young–Laplace equation for its pressure, and the restoring mechanism of capillary waves. It is useful because one coefficient links forces, energy and geometry. ^[ref]

The moving-rim argument also explains the factor of two without memorizing a special formula. An outward displacement $dx$ creates surface area $\ell dx$ on each face, hence $d{A}_{\mathrm{total}}=2\ell dx$. Slow pulling without viscous dissipation requires work $Fdx=\gamma d{A}_{\mathrm{total}}$, giving $F=2\gamma \ell$. For one interface, the same argument gives $F=\gamma \ell$. A surface tension gradient instead produces Marangoni stress; the constant-$\gamma$ assumption excludes that additional effect.

At large Reynolds number $\mathcal{R}$, a long indentation of relative height $ϵ$ has nonlinear near-wall interaction when the streamwise scale ratio satisfies $\lambda /\mathcal{R}={ϵ}^3$. The regime ${\mathcal{R}}^{1/7}\ll \lambda \ll \mathcal{R}$ gives a thin viscous layer and an effectively common pressure across the core. The core displacement is $u_1=A(x)U_0^{\prime }(y)$ and $v_1=-A^{\prime }(x)U_0(y)$. A lower wall displaced to $ϵF(x)$ produces layer displacement $F+A$, while the upper wall produces $-A$. Uniqueness for their common pressure forcing and upstream data makes the two layer displacements equal, so $A=-F/2$.

Using layer depth $d$ and thermal diffusivity $\kappa$ gives time scale $d^2/\kappa$, velocity scale $\kappa /d$, and perturbation pressure scale ${\rho }_0{\kappa }^2/d^2$. With temperature scale $\Delta T$, the Linearized Boussinesq equations have viscous coefficient Prandtl number $\mathrm{Pr}=\nu /\kappa$ and buoyancy coefficient $\mathrm{Ra}\mathrm{Pr}$, where Rayleigh number $\mathrm{Ra}=g{\alpha }_T\Delta T{d}^3/(\nu \kappa )$.

Thermal expansion is a change in a body's dimensions as its temperature changes under a specified mechanical constraint. A heated gas under a freely moving loaded piston usually increases its volume at fixed pressure; heating the same gas in a rigid vessel instead raises its pressure. The coefficient of thermal expansion quantifies the first response. Expansion can be negative, so the name does not assert that every substance grows on heating.

For an ideal gas at fixed pressure and particle number, $V=N{k}_{B}T/P$ gives $V_2/V_1=T_2/T_1$. Heating from $300\mathrm{K}$ to $330\mathrm{K}$ therefore increases the volume by exactly $10\%$ within the ideal gas model. For a solid, specify whether the measurement concerns length or volume: for uniform expansion in all three directions, the volumetric coefficient is three times the linear thermal expansion coefficient, because $V\propto L^3$. Direction-dependent expansion requires the separate dimensional responses. ^[ref]