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A source volume $Q$ makes roof outflow exceed floor inflow by $Q$. At zero floor inflow, an entraining plume cannot support a steady cold lower layer; the warm layer reaches the floor. With reduced gravity $g^{\prime }=B/Q$, the entire hydrostatic head drives roof discharge $Q=A_e\sqrt{g^{\prime }H}$. This gives the stated threshold. Equal splitting of pressure head is inappropriate at blocking. The same result in physical-area notation is $Q_c=(2C_d^2{A}_{\mathrm{geom}}^{2}BH{)}^{1/3}$.

Specific enthalpy is enthalpy per unit mass. For a simple compressible fluid of fixed composition, the first law of thermodynamics giveswhere $T$ is temperature, $s$ is specific entropy, $p$ is pressure, and $\rho$ is mass density. Along an isentropic flow, $dh=dp/\rho$. For a perfect gas with constant specific-heat ratio $\gamma$, $h=\gamma p/[(\gamma -1)\rho ]$.

For number density $n$, rest-frame energy density $\rho$ and pressure $p$, the enthalpy per particle is $ϵ=(\rho +p)/n$. In an isentropic barotropic fluid, $d\rho =ϵdn$ and $dp=ndϵ$. It differs from enthalpy per unit mass by the particle mass when that mass is fixed. Using $ϵ$ as thermodynamic variable gives $n=p^{\prime }(ϵ)$ and $\rho =ϵp^{\prime }(ϵ)-p(ϵ)$.

Specific entropy is entropy per unit mass. For a perfect gas with constant specific heat capacity $c_v$ at constant volume and specific-heat ratio $\gamma$, differences obeywhere $p$ is pressure and $\rho$ is mass density. Thus a homentropic flow has constant $p/{\rho }^{\gamma }$.

Specific heat capacity is heat capacity per unit mass:Its units are $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and the constraint remains part of the definition. In particular, specific heat capacity at constant volume and specific heat capacity at constant pressure are different responses. For homogeneous material, doubling the mass doubles its heat capacity while leaving its specific heat capacity unchanged. ^[ref]

As an illustrative material model, give a fixed-volume body mass $m=2\mathrm{kg}$ and constant specific heat capacity at constant volume $c_V=400\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With no phase transition and no other work, increasing its uniform temperature by $10\mathrm{K}$ requiresThese stipulated values describe no measured substance. The right panel of the diagram in heat capacity shows the straight line $Q=m{c}_V\Delta T$: a temperature increase of $5\mathrm{K}$ takes $4000\mathrm{J}$ of heat. If the specific heat capacity at constant volume depends on temperature, use $Q=m{\int }_{T_1}^{T_2}{c}_V(T)dT$ instead. The corresponding fixed-pressure calculation uses specific heat capacity at constant pressure and gives $Q=\Delta H$ under the assumptions in heat capacity.

Choosing between specific heat capacity, molar heat capacity and volumetric heat capacity is choosing whether to compare equal mass, equal amount of substance or equal volume. A small sample with large specific heat capacity can still have less total heat capacity than a much larger sample.

The specific heat capacity at fixed pressure and composition is $c_P=(\partial h/\partial T{)}_P$, where $h$ is specific enthalpy. Equivalently, it is heat capacity at constant pressure divided by mass. Under the process assumptions in heat capacity, it includes expansion work as well as the internal energy change. The sensible heat example uses this response separately for ice and water.

For a sphere falling with speed $U$ through a tube of radius $a$, let $Q$ be the downward laboratory volume flux. In the sphere frame the wall moves upwards and the narrow annulus carries upward flux per unit circumference $q=(\pi a^{2}U-Q)/(2\pi a)$. Couette-Poiseuille flow in a thin gap gives $q=Uh/2-h^3{p}_z/(12\mu )$. Thus the pressure below minus above the sphere is $\Delta P=6\mu (2q{I}_3-U{I}_2)$, using lubrication resistance integrals for a nearly occluding sphere. A nearly piston-like laboratory flux can coexist with a nonzero relative gap flux.

Use radius-deficit $ϵ$, $F^{\ast }=F/(6\pi \mu aU)$ and $\delta =3a/(4L)$. The control-volume drag formula for a sphere in a tube and the radius-deficit normalization of sphere-tube lubrication integrals give three leading resistances: $c_3{ϵ}^{-5/2}$ for $\delta \ll {ϵ}^{5/2}$, $1/\delta$ for ${ϵ}^{5/2}\ll \delta \ll {ϵ}^{1/2}$, and $2\pi \sqrt{2}/(3\sqrt{ϵ})$ for $\delta \gg {ϵ}^{1/2}$. They correspond to gap bypass with almost no tube throughflow, pressure resistance of tube throughflow, and local annular resistance, respectively.

For an energy-conserving strong spherical shock wave expanding into mass density $C{r}^{-\beta }$, $0\le \beta <3$, dimensional analysis gives the displayed radius law. Neglect ambient pressure, radiation losses and gravity in this blast-dominated approximation. Write $u=\dot{R}U(\xi )$, $\rho =C{R}^{-\beta }D(\xi )$ and $p=C{R}^{-\beta }{\dot{R}}^{2}P(\xi )$, with $\xi =r/R$. The radial continuity equation, momentum equation and adiabatic pressure equation reduce toThe Strong-shock Rankine-Hugoniot conditions give $U(1)=P(1)=2/(\gamma +1)$ and $D(1)=(\gamma +1)/(\gamma -1)$. The explosion energy fixes the radius coefficient through $E=4\pi C{R}^{3-\beta }{\dot{R}}^2{\int }_0^1[D{U}^2/2+P/(\gamma -1)]{\xi }^{2}d\xi$.

A buoyantly rising inviscid bubble in otherwise quiescent unbounded liquid has an exterior Stokeslet with $F=4\pi \mu aU=4\pi \rho g{a}^3/3$. Its tangential traction vanishes on the sphere. The dynamic normal traction is $-3F\cos \theta /(4\pi a^2)$, while the hydrostatic normal-traction variation is $\rho ga\cos \theta$. Their dipoles cancel at the terminal velocity, leaving a constant total normal stress compatible with uniform gas pressure even when surface tension is zero. This is an exact spherical solution in the stated Stokes model, not a proof of shape stability under arbitrary perturbations.

For $P_0>0$, $\alpha \ne 0$, $T\ge T_0$ and physical $T>0$, the atmospheric pressure-temperature profile $P=P_0\exp (\alpha \sqrt{T-T_0})$ is equivalent toThe branch restriction is essential: squaring does not create a second physical branch. Its temperature gradient is $dT/d\log P=2\sqrt{T-T_0}/\alpha$. Thus, in hydrostatic equilibrium, $\alpha <0$ gives an atmospheric thermal inversion, while $\alpha >0$ gives outward cooling. At $T=T_0$ the inverse branch has zero derivative; $\alpha =0$ gives constant pressure and cannot describe a nontrivial hydrostatic vertical interval.

Let the inner-centre displacement from a fixed outer centre have magnitude $r=\Delta e$, with $0\le e<1$. The leading full-film journal bearing resistance to translation in the displacement direction and perpendicular to it isFor $h=\Delta (1-e\cos \theta )$, the Reynolds lubrication equation gives radial squeezing flux $a{v}_r\sin \theta$ and tangential squeezing flux $a{v}_t[-\cos \theta +3e/(2+e^2)]$. The additive constant enforces periodic pressure. Integrating the pressure traction gives these coefficients; at $e=0$ they coincide, as required by rotational symmetry.

The pressure obtained by bringing a flow to rest without losses; in compressible flow the usual definition specifies an isentropic process for stopping. For the constant-mass density, same-height Bernoulli equation discussed under dynamic pressure, $p_0=p+q$.

A state function assigns a value to each equilibrium thermodynamic state of a specified thermodynamic system. Its change depends only on the initial and final states. Internal energy $U$, entropy $S$ and enthalpy $H=U+pV$ are state functions; here $p$ is pressure and $V$ is volume. Transferred heat $Q$ and work $W$ belong to a thermodynamic process, rather than to a state. A system stores internal energy; it does not store separate inventories of “heat received” and “work received”.^[ref]

Use the explicit convention $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work done by the thermodynamic system on its surroundings. For a closed system with unchanged endpoint bulk kinetic energy and external potential energy, the first law of thermodynamics readsThe notation $dU$ denotes an exact differential of a state function; $\delta Q$ and $\delta W_{\mathrm{by}}$ denote incremental transfers. Along a reversible thermodynamic process of a simple compressible system,where $T$ is absolute temperature. These are rules for particular thermodynamic processes, not definitions of stored heat or stored work. In particular, a quasistatic process with friction can remain near thermodynamic equilibrium while producing entropy.^[ref]

For any state function $F$ and two equilibrium states $A,B$, an integral of its exact differential givesThe proof is telescoping: add the changes $F(x_{i+1})-F(x_i)$ along a subdivided path, and all intermediate values cancel. The smooth limit gives the integral. Conversely, if an integrable differential form has zero integral around every accessible closed path in a connected region, define $F(B)$ by integrating from a reference state. Two routes differ by a closed path, so the definition is unambiguous. This argument needs every accessible loop, rather than a local test that might miss a hole in the domain; see path independence.

The same cancellation does not hold separately for heat and work. A thermodynamic cycle has $\Delta U=0$ but can have nonzero net transfers, with $\oint \delta Q=\oint \delta W_{\mathrm{by}}$. Nor does the fact that entropy is a state function make $Q/T$ its general change: the formula $dS=\delta Q_{\mathrm{rev}}/T$ uses a reversible thermodynamic process and its varying temperature.

This distinction makes endpoint calculations powerful. We can compute an internal energy, entropy or enthalpy difference along a convenient reversible reference path even when the actual thermodynamic process is irreversible. Its actual heat and work must still be obtained from its own boundary interactions. The appealing cancellation is visible in two reversible ideal-gas paths between the same states: a different mechanical output requires a matching difference in thermal input, while stored internal energy arrives at the same value.

A state of matter is a broad category of material behavior. A solid resists a sustained change of shape, a liquid flows while remaining difficult to compress, and a gas expands to fill its container. These distinctions are useful first descriptions, but they do not enumerate all thermodynamic phases. Plasma adds collective behavior of charged particles, while liquid crystals show that flow and structural order need not disappear together.

Distinguish three questions: what broad state of matter does the material resemble, which thermodynamic phase is present, and what are its temperature, pressure and other state variables? Two samples at different temperatures can be in the same liquid thermodynamic phase. Conversely, crystal polymorphism gives different solid thermodynamic phases of the same substance, distinguished by their crystal structures. A phase transition between them need not change the broad state-of-matter category.

For an imagined substance with crystalline forms A and B, melting A produces a solid–liquid phase transition, while converting A to B produces a solid–solid phase transition. A sample containing A and B has two thermodynamic phases even though all its material is solid. Cutting one uniform crystal into two pieces does not create a second thermodynamic phase: count distinct equilibrium forms, rather than pieces or grains. This distinction matters when applying the phase rule.

A phase diagram locates stable thermodynamic phases using temperature and pressure. Its liquid–gas phase coexistence curve ends at a liquid-gas critical point, around which a path connects liquid and gas without a first-order phase transition. Broad categories can therefore remain useful without a sharp boundary.^[ref]

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]