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An ideal fluid issuing from a small hole a vertical depth $h$ below a free surface has speed

Here $h$ is the vertical distance from the free surface to the centre of the orifice, and $g$ is the positive magnitude of gravitational acceleration. The ideal prediction assumes incompressible flow, the same ambient pressure at the free surface and emerging jet, negligible velocity of the reservoir's free surface, and negligible viscous dissipation and surface tension effects. For a falling free surface, a quasi-steady approximation additionally assumes that the local outflow adjusts much faster than $h$ changes. It applies the Bernoulli equation at each instant, rather than claiming that the whole drainage is steady flow.^[ref]

A short proof uses a streamline connecting the reservoir's free surface at elevation $z_s$ to the emerging jet at elevation $z_o$. With constant mass density $\rho$, negligible surface speed and no mechanical energy losses, the Bernoulli equation givesEqual pressures cancel, leaving $v^2/2=gh$: a unit mass gains the same kinetic energy as a body in free fall through height $h$. This makes Torricelli's law a useful first estimate before the detailed velocity field is known. At one quarter of the original head, ideal jet speed is one half as large; it is not proportional to head.

Far from the accelerating outlet region, the reservoir has approximately hydrostatic pressure $p_{\mathrm{ambient}}+\rho gh$ at the outlet's elevation. The emerging jet instead has approximately ambient pressure when surface tension is negligible. Do not assign that bulk hydrostatic pressure to the emerging jet as well as adding the height difference: that would count the available mechanical energy twice. A sealed reservoir, a submerged outlet or significant surface tension changes the pressure conditions and requires a different balance.

Actual volumetric flow rate is commonly represented by a discharge coefficient:where $a$ is the geometric area of the orifice. This definition compares volumetric flow rates, not just speeds. If a nearly uniform jet at its vena contracta has area $a_j$ and speed $v_j$, define the two separate ratios $C_c=a_j/a$ and $C_v=v_j/\sqrt{2gh}$. Then $Q=a_j{v}_j$ gives $C_d=C_c{C}_v$. A contracted jet can consequently have a speed close to $\sqrt{2gh}$ while $C_d$ is appreciably below one. No particular $C_d$ is universal: calibrate it for the opening and operating range.

A triple point is a temperature–pressure pair at which three distinct bulk thermodynamic phases of a one-component substance coexist in thermodynamic equilibrium. In the familiar solid–liquid–gas case, their three phase coexistence curves meet. Other combinations, such as two crystal polymorphs and a liquid, also qualify: “triple” counts thermodynamic phases, rather than states-of-matter categories.

For the same substance, the chemical potential in each thermodynamic phase must agree:There are two independent equality conditions for two controls, temperature and pressure. Generically they select an isolated pair $(T_t,p_t)$; the third pairwise equality follows from the first two. Equivalently, the phase rule gives $F=1-3+2=0$. This counts independent intensive controls, not phase fractions: changing the amounts of the three thermodynamic phases need not change the ideal bulk temperature and pressure, provided all three remain present. Different sets of crystal polymorphs can give different triple points for the same substance.^[ref]

At a liquid-gas critical point, the two fluid thermodynamic phases instead merge, and their mass density difference and latent heat vanish. It is the endpoint of the liquid–gas phase coexistence curve, rather than a meeting of three distinct thermodynamic phases. The regular independent-equality count used for a triple point cannot explain this merger. A triple point can retain finite mass density and entropy differences between all three thermodynamic phases.^[ref]

Turgor pressure is the internal hydrostatic pressure above the external pressure in a biological cell. When water enters a walled biological cell by osmosis, expansion loads the cell wall. Rising pressure can oppose further net entry while the cell membrane retains its selective permeability. Mechanical support therefore does not require the cell wall to block water.

As an original mechanical check, idealize a small flat patch of cell wall of surface area $A$ under uniform turgor pressure $\Delta P=P_{\mathrm{in}}-P_{\mathrm{out}}$. Its outward force is $F=\Delta PA$. Doubling $A$ at fixed pressure doubles the load; equal internal and external pressures give zero net pressure load. This is a force balance, not a measured wall strength or a complete model of a curved, growing biological cell. It says nothing about which solutes cross the cell membrane.

For inviscid planar incompressible flow, write velocity as $(-{\psi }_y,{\psi }_x)$ and vorticity as $\zeta ={\nabla }^2\psi$. Taking curl of the Euler equations for an inviscid fluid removes pressure and gives the displayed transport equation, with $J(\psi ,\zeta )={\psi }_x{\zeta }_y-{\psi }_y{\zeta }_x$. The Poisson equation then recovers the streamfunction at each instant. A velocity tangent to the plane cannot stretch its perpendicular vorticity vector.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.

In a connected inviscid uniformly rotating barotropic fluid, momentum balance makes the displayed sum constant. The pressure term is the barotropic enthalpy function. Perturbations experience Coriolis acceleration, and their rotating-frame frequency depends on the chosen exponential sign convention.