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In the shock frame, integrate mass conservation, momentum conservation, and total-energy conservation across a thin interval containing the stationary normal shock wave. No mass, momentum, or energy is stored in the vanishingly thin interval, so each flux has the same value on both sides. For a perfect gas, the specific enthalpy is $h=\gamma p/[(\gamma -1)\rho ]$. The resulting Rankine-Hugoniot conditions for a perfect gas areThe momentum flux includes both transported momentum and pressure force. The energy flux includes kinetic energy and enthalpy, the latter accounting for internal energy and the work needed to push gas through the interval. Entropy production can occur inside the normal shock wave even though total energy is conserved.

Take $u(r)>0$ to be the inward radial speed and select the regular critical solution for a gas reservoir of density ${\rho }_{\infty }$ and adiabatic sound speed $c_{\infty }$. Other mathematically subsonic steady solutions need a different inner boundary condition. For the polytropic equation of state, let $\gamma =1+1/n$, soMass conservation and the Bernoulli equation giveThe gas is at rest at infinity, so its kinetic and gravitational terms vanish there.

Differentiate these two conserved quantities, using $d\log \rho /dr=-2/r-u^{\prime }/u$. Eliminating the density derivative yields the polytropic Bondi accretion equationAt a regular sonic point, both sides vanish:The Bernoulli equation at that point reduces to $n{c}_{\infty }^2=(n-3/2)c_c^2$. For $n>3/2$, equivalently $1<\gamma <5/3$,Since $c_s^2\propto {\rho }^{1/n}$, the critical density is ${\rho }_c={\rho }_{\infty }[2n/(2n-3){]}^n$. Evaluate the conserved mass accretion rate at the sonic point:whereThe isothermal limit $\gamma \to 1$ gives $\lambda =e^{3/2}/4$, agreeing with Isothermal Bondi accretion.

For $\gamma =5/3$, $n=3/2$ and the Newtonian critical radius tends to zero while $c_c$ diverges. The limiting critical solution approaches Mach number one only as $r\to 0$; it has no finite-radius Newtonian sonic transition. This is a failure of the Newtonian inner approximation, not an obstruction to accretion onto a black hole. A physical Schwarzschild black hole has a finite event horizon, and the regular relativistic inflow becomes supersonic outside it.

The singular-looking factor in the rate has a finite limit. Writing $ϵ=n-3/2$, its logarithm is $ϵ\log [n/ϵ]\to 0$, so the critical Bondi accretion rate for gamma equals five thirds isThis is also the leading cold-reservoir limit of relativistic spherical accretion; an exact rate for relativistically hot gas requires the relativistic equations, not this Newtonian expression.

For completeness, the finite relativistic sonic point can be seen without assigning it an arbitrary inner radius. Let $b=c_s^2/c^2$ denote the relativistic squared sound speed and $g=\gamma -1$. For an isentropic relativistic gas with $p=K{\rho }_{\mathrm{rest}}^{\gamma }$, its dimensionless specific enthalpy is $h=g/(g-b)$. Conservation of rest-mass flux and relativistic Bernoulli energy gives the critical relationsFor $g=2/3$ and $c_{\infty }\ll c$, expansion gives $b_{\infty }\simeq (9/4)b_c^2$, henceHere the asymptotic relativistic and Newtonian sound speeds agree to leading order. This finite-radius transition explains why the gas can reach the horizon despite the Newtonian $r_c=0$ result.

In the synchronously rotating reference frame, stationary fluid obeyswhere $\Psi$ is the sum of the two gravitational potentials and the centrifugal potential. The circular binary's Kepler third law gives ${\Omega }^2=G(M_1+M_2)/a^3=GM/a^3$, hence

Uniform composition and the assumed central condensation make star 1 approximately barotropic, so one may define the specific enthalpyHydrostatic balance becomes $\nabla (h+\Psi )=0$, or $h+\Psi =\text{constant}$ throughout the star. Since $h$ is a monotone function of $P$ and $\rho$, constant-$P$ and constant-$\rho$ surfaces are equipotential surfaces of $\Psi$.

A thermodynamic potential packages the equilibrium state in variables suited to the imposed constraints. For a simple fluid, $U(S,V,N)$ has differential $dU=TdS-pdV+\mu dN$. Legendre transforms give $H=U+pV$, $F=U-TS$ and $G=U+pV-TS$, with natural variables of a thermodynamic potential $(S,p,N)$, $(T,V,N)$ and $(T,p,N)$ respectively. At fixed particle number, Helmholtz free energy is minimized at fixed $T,V$; Gibbs free energy at fixed $T,p$.

A flow vessel is an open control volume, even if its walls are enclosed. For each inlet, pressure forces carry flow work $pdV$ in addition to internal energy. The steady first-law balance isWith no heat exchange, shaft work, or significant kinetic and gravitational energy differences, the enthalpy fluxes are equal. Mass flow must also balance. This equality concerns mass-flow-weighted specific enthalpies, not an unweighted sum over pipes.

Use the liquefaction yield with separate warm-gas and liquid outlets. The wording specifies that the outgoing gas is at 14 K; it does not require the outgoing liquid to have that temperature. The two outlet streams need not be in mutual thermal equilibrium. Taking $H$ as the printed specific enthalpy, the steady-flow energy balance givesFor fixed outlet enthalpies the greatest yield occurs at the smallest inlet enthalpy. Among the tabulated pressures this is 30 atm. Approximating the atmospheric-pressure gas enthalpy by the tabulated zero-pressure limiting value givesThis uses a low-pressure approximation: $p_0=1$ atm is not literally the zero-pressure table entry. Linear interpolation to 1 atm would instead give $H(p_0)\simeq 8.651\times 10^4\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ and $\alpha \simeq 0.193$. The sparse table does not determine an exact continuous optimum between entries.

Atmospheric-pressure liquid helium must be much colder than the 14 K gas outlet: its normal boiling point is about 4.224 K.^[ref] A counterflow heat exchanger can return cold gas through the incoming stream before that gas leaves the overall control volume, while the collected liquid leaves separately at the cold end. This is a possible interpretation of the whole apparatus, rather than an extra diagram specified by the question. The enthalpy calculation needs only the stated inlet and outlet properties; it does not assume one equilibrium vessel containing both phases at 14 K. In particular, $H(p_0)-H_{\ell }$ is an outlet-stream enthalpy difference, not the latent heat of vaporization at one common temperature.

Entropy at absolute zero. The third law of thermodynamics concerns the low-temperature limit of equilibrium states. In the zero-entropy formulation, the entropy of a perfect crystal with a nondegenerate ground state tends to zero as $T\to 0$. The statistical interpretation is $S=k_B\ln \Omega$: a unique accessible ground state has $\Omega =1$. The entropy-difference formulation of the third law of thermodynamics states that entropy differences associated with regular changes of mechanical parameters tend to zero as absolute zero is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate ground state can carry residual entropy.

For a phase of fixed composition and pressure,A finite limiting entropy constrains this integral. Ordinary equilibrium heat capacities therefore vanish as temperature approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic phonons and $C\propto T$ for a normal degenerate Fermi liquid. A constant positive heat capacity extrapolated to zero would give a logarithmically divergent entropy difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.

Gibbs free energy. For fixed particle numbers,Thus, when $S\to 0$, the Gibbs free energy approaches its zero-temperature value with a horizontal tangent:The equality $G(0,p)=H(0,p)$ uses the enthalpy $H=U+pV$; neither limiting value needs to be zero. The low-temperature Gibbs free-energy expansion makes the approach quantitative. For $C_p=a{T}^r$ with $r>0$ and $S(0)=0$,The second derivative is ${\partial }_T^{2}G=-C_p/T\le 0$, although it need not tend to zero. For a reaction between such equilibrium phases, ${\partial }_T\Delta G=-\Delta S\to 0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine residual entropy $S_0$ is retained instead, $G=G(0,p)-S_{0}T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.

Why helium remains a liquid. Helium atoms attract each other weakly and are exceptionally light. Their zero-point energy is large compared with the binding that would localize them in a solid. Both helium-4 and helium-3 can therefore remain liquid down to the lowest temperatures at low pressure; substantial applied pressure is needed to solidify them. This is not a violation of the third law. A quantum liquid can have a unique low-entropy ground state, and zero-point motion does not itself imply thermal entropy. The third law does not require every substance to crystallize at atmospheric pressure.

A helium-4 atom has integer spin zero and is a boson. Below the lambda point near $2.17\mathrm{K}$ at saturated vapor pressure, its liquid is superfluid. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the condensate fraction need not equal the superfluid fraction. The two-fluid model of superfluidity separates an entropy-carrying normal component of a superfluid from the superfluid component, whose entropy is zero. At sufficiently low temperature the dominant thermal excitations are acoustic phonons: their linear dispersion in three dimensions gives $U-U_0\propto V{T}^4$, $C_V\propto V{T}^3$ and $S\propto V{T}^3$. Higher-energy rotons are exponentially suppressed. Thus the normal component of a superfluid and entropy disappear as $T\to 0$, while superfluid flow and quantum motion can persist.

A helium-3 atom has spin $1/2$ and is a fermion. Well below its Fermi temperature, but above its superfluid transition, liquid helium-3 is a degenerate interacting Fermi liquid. Only quasiparticles close to the Fermi surface are thermally excited, giving $C_V\propto T$ and $S\propto T$. At temperatures of order a millikelvin, helium-3 also becomes superfluid through formation of Cooper pairs of fermions. This is not Bose-Einstein condensation of individual helium-3 atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic quasiparticle contribution to heat capacity is exponentially suppressed at temperatures well below the pairing gap. Gapless collective modes can still give thermal terms following a power law. The entropy nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.^[ref]

Helium-3 also illustrates the importance of nuclear spin entropy. Over an intermediate low-temperature range, localized spins in the solid can retain approximately $k_B\ln 2$ of entropy per atom, while the degenerate liquid has much less. Reversible conversion of liquid to the higher-entropy solid then absorbs heat; adiabatic compression along the melting curve can cool the sample. This is Pomeranchuk cooling. At still lower temperature, interactions between the nuclear spins remove this nearly free-spin entropy, so a constant $k_B\ln 2$ cannot be extrapolated to absolute zero.^[ref]

Liquefaction and cooling methods. A gas must be below its liquid-gas critical point to form a distinct liquid phase by compression. Compression above the critical temperature alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression heat, cools the compressed gas in a heat exchanger and uses expansion to lower its temperature further.

A throttling valve operates approximately at constant specific enthalpy when kinetic-energy changes and heat leakage are negligible. The Joule-Thomson coefficient isusing consistent molar quantities, or consistent total quantities for a fixed amount. A pressure drop cools the gas if ${\mu }_{\mathrm{JT}}>0$. An ideal gas has zero Joule-Thomson coefficient; real intermolecular interactions are essential. In particular, room-temperature helium warms on throttling, so it must first be brought into a region of positive Joule-Thomson coefficient. Expansion through an engine or turbine can instead remove work and produce approximately isentropic cooling.

A regenerative liquefier sends the cold returning gas through a counterflow heat exchanger to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces liquid. Modern helium liquefaction commonly combines work-producing expansion with a final throttle. ^[ref]

Once liquid exists, pumping away vapour lowers its saturation pressure and boiling temperature. The escaping vapour carries latent heat away, cooling the remaining liquid. At still lower temperatures, a dilution refrigerator exploits the different entropy carried by helium-3 in concentrated and dilute helium mixtures; adiabatic demagnetization and Pomeranchuk cooling provide other routes.

Approaching zero. The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach absolute zero in finitely many stages. As the entropy available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero entropy cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. Quantum liquids and refrigeration methods can approach $0\mathrm{K}$ while remaining consistent with vanishing equilibrium entropy; they do not provide a finite-operation route to absolute zero.

Work in the locally planar shock frame, taking $v_u,v_d>0$ towards the surface. For a stationary normal shock wave with negligible radiative loss inside the discontinuity, the Rankine-Hugoniot conditions for a perfect gas express conservation of mass, conservation of momentum, and conservation of energy per unit area asHere $h$ is specific enthalpy, including the pressure work; using only internal energy would omit that work. The gravitational potential is continuous across an infinitesimally thin front. For cold upstream gas, $P_u$ and $h_u$ are negligible, while a monatomic gas has $\gamma =5/3$ and $h=5P/(2\rho )$.

The hydrodynamic equations apply to flow parallel to the guiding magnetic field when its stress contributions cancel across the front. An oblique magnetohydrodynamic shock requires magnetic stresses and Poynting flux; a radiation-mediated front or appreciable escaping radiation requires the corresponding radiative momentum and energy flux terms. These are additional physics, not terms hidden in $h$ here.

For the stipulated Strong-shock Rankine-Hugoniot conditions, ${\rho }_d=4{\rho }_u$ and $v_d=v_u/4$. The ideal gas law is $P_d={\rho }_d{k}_B{T}_d/\mu$, with $\mu$ a dimensional mean particle mass. The specific enthalpy of a nonrelativistic monatomic gas is $5k_{B}T/(2\mu )$. Dividing the energy flux equation in d by the common mass flux givesThereforeAs an independent check, conservation of momentum gives $P_d={\rho }_u{v}_u^2-4{\rho }_u(v_u/4{)}^2=3{\rho }_u{v}_u^2/4$, and division by ${\rho }_d=4{\rho }_u$ gives exactly the same temperature. The numerical coefficient relies on $\gamma =5/3$ and the absence of dynamically important radiative or magnetic stresses.

An ideal Rankine cycle has four main steps: liquid water is pumped reversibly to high pressure; heat is added at high pressure to produce hot vapour or supercritical working fluid; the fluid expands approximately isentropically through a turbine to produce work; and low-pressure heat rejection in a condenser returns it to liquid. For steady flow, the main work and heat transfers are differences of specific enthalpy.

A steam implementation of the Carnot cycle would require compression of a wet liquid-vapour mixture and reversible isothermal heat transfer, both difficult to achieve with practical machinery. Complete condensation before the Rankine pump makes compression work small because liquid specific volume is small. High-pressure heating and superheating can then occur in a boiler without demanding isothermal addition of all the heat. Superheating also reduces turbine-exit moisture, although the calculated exit may still be wet. The pump, turbine, condenser and boiler can operate continuously as separate flow devices. The Rankine cycle sacrifices some ideal thermal efficiency in return for a practicable steam-plant configuration; no real engine can exceed the Carnot efficiency for the same extreme reservoir temperatures.

Assume an ideal isentropic turbine and use the supplied table consistently. Its inlet specific entropy is $s_3=6.282\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. At the condenser pressure the outlet mixture has $s_4=s_f+\chi (s_g-s_f)$, where $\chi$ is the vapour mass fraction. HenceThe outlet is approximately $71.5\%$ vapour and $28.5\%$ liquid by mass. Its specific enthalpy isThe source's common unit label omits the kelvin denominator for entropy: $s$ must be interpreted in $\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, while $h$ is in $\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. A real irreversible turbine would increase entropy and require an efficiency correction; no such correction is supplied.

Work in the rest frame of the planar shock, with nonzero mass flux $j=\rho u$. Conservation of mass gives $[j]=0$; the tangential and normal components of balance of linear momentum give $[jv]=0$ and $[p+ju]=0$. Thus $[v]=0$. For an ideal gas, specific internal energy is $e=c_{v}T=p/[(\gamma -1)\rho ]$, and specific enthalpy is $e+p/\rho =\gamma p/[(\gamma -1)\rho ]$. Total energy flux is $j[(u^2+v^2)/2+e+p/\rho ]$. Its continuity, with $v$ already continuous, givesPut $c^2=\gamma p/\rho$ and $M=u/c$. Since $p=ju/(\gamma M^2)$, the normal component of balance of linear momentum says $u[1+1/(\gamma M^2)]$ is constant. HenceThe energy equation says $u^2[1/2+1/((\gamma -1)M^2)]$ is constant, givingSet $a=M_{-}^2$, $b=M_{+}^2$, square the first relation and equate to the second. After clearing positive denominators the equation factors asThe first factor is the unchanged state. For a genuine shock the second factor vanishes, yieldingThe physical selection principle is entropy increase in the direction of flow. For $u_{\pm }>0$, the upstream side is $x<0$ and the downstream side is $x>0$. A compressive entropy-admissible shock has $\boxed{M_{-}>1,M_{+}<1}$; interpreting the printed “ahead” and “behind” spatial labels must respect these explicit side assignments. The expansion-shock branch would violate the second law.

Substitution gives the velocity ratioSince $v$ is unchanged while the normal component decreases, $|v|/u$ increases. Thus an oblique flow bends towards the shock plane, equivalently away from its normal. If $v=0$ there is no angular bending; the flow stays normal and simply slows. The tangential continuity, not a change in the tangential speed, causes the deflection.

Put $v=u-V$, the velocity in the shock frame. The mass jump condition gives a common mass flux $j=\rho v$. Subtracting $V$ times the mass balance from the momentum balance givesFor energy, transform the conserved energy to the shock frame by subtracting $V$ times momentum and adding $V^2/2$ times mass. The resulting energy flux isFor a genuine shock, $j\ne 0$; dividing its continuous energy flux by the same nonzero $j$ on both sides givesThe second term is the ideal-gas specific enthalpy. A contact discontinuity has zero relative mass flux and does not in general require equal specific enthalpies; the division here is appropriate for the shock specified in the question.

The thermal binding scale defined by equating heated gas sound speed to circular Keplerian orbit speed. It is $r_g=GM/c_s^2$. The circular-orbit binding energy is $GM/(2r)$ per unit mass, while the gas specific enthalpy is of order $c_s^2$. A heated disk surface at radii comparable to or larger than $r_g$ can therefore drive photoevaporation. The actual launching radius can differ by a factor of order unity; $c_s$ is not the ballistic escape speed.

Under the hypotheses of steady planar ideal-MHD field-line invariants, the specific enthalpy relation and the velocity projection of momentum give $\mathbf{B}\cdot \nabla Q=0$. The horizontal total energy flux, including the Poynting vector, is $k{\mathbf{B}}_{h}Q$. The magnetic correction accounts for work exchanged between the fluid and the field.

A polytropic flow uses a polytropic equation of state. When $K$ and the adiabatic exponent are constant, a smooth streamline has specific enthalpy $h=\gamma K{\rho }^{\gamma -1}/(\gamma -1)$ and adiabatic sound speed $c_s^2=\gamma K{\rho }^{\gamma -1}$. Spatially varying entropy can make $K$ depend on streamline; the constant-$K$ spherical wind is a special case.

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.

A stagnant mushy-layer model has no bulk advection, equal phase mass densities, pure solid, and local equilibrium on the liquidus. With liquid fraction $\varphi$, specific enthalpy $E=c_{p}T+L\varphi$ and bulk solute concentration $B=\varphi C$, neglecting solute diffusion inside the mushy layer gives $T_t+(L/c_p){\varphi }_t=\kappa T_{zz}$ and $(\varphi C{)}_t=0$. The latter fixes composition at each stationary material level; it does not automatically set every level to the initial liquid concentration when solute can diffuse ahead of the growing mush.

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]

Steam heating transports thermal energy using water vapor, commonly allowing it to condense in a heat-emitting component. At an approximately fixed local pressure, condensation releases latent heat while the phase-change temperature remains nearly fixed. The condensate can then return to the source for reheating.

For mass flow rate $\dot{m}$ through a steady emitter, the ideal transferred power is $\dot{Q}=\dot{m}(h_{\mathrm{in}}-h_{\mathrm{out}})$, where $h$ is specific enthalpy. That difference can include both latent heat from condensation and sensible heat from cooling the condensate. The emitter transfers this energy to its surroundings through convection and thermal radiation. Measuring only the condensate temperature change would miss the latent contribution. The mechanism illustrates why a large heat transfer need not be accompanied by a large change in the working fluid's temperature.

A steam turbine extracts mechanical work from flowing steam as it expands through stationary passages and moving blades. The moving blades change the flow's angular momentum and receive a corresponding torque. A shaft transfers the resulting power to a generator or another load. Unlike a reciprocating steam engine, the working stream passes continuously through the machine.^[ref]

For steady approximately adiabatic operation with negligible changes in inlet-to-outlet kinetic and gravitational energy, the first law of thermodynamics giveswhere $\dot{m}$ is mass flow rate and $h$ is specific enthalpy. An actual enthalpy drop of $500\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$ at $2\mathrm{kg}{\mathrm{s}}^{-1}$ therefore gives $1\mathrm{MW}$ of shaft power under these assumptions. If the reversible adiabatic drop to the same outlet pressure were $625\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, the turbine's isentropic efficiency would be $500/625=0.8$.^[ref]

A performance test measures inlet and outlet pressure and temperature, flow rate, shaft torque and rotational speed. If the exhaust is a liquid-vapor mixture, pressure and temperature alone do not determine its specific enthalpy; the vapor fraction or an additional energy balance is needed. Condensing exhaust and supplying useful process steam are different operating arrangements, so turbine efficiency and whole-plant fuel efficiency must also be distinguished.^[ref]