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For a lossless polytropic flow with $\gamma >1$, negligible gravity and a sonic throat, the Bernoulli function is $u^2/2+c_s^2/(\gamma -1)=(\gamma +1)c_{\ast }^2/[2(\gamma -1)]$. On an expanding supersonic branch for which $\rho \to 0$, the specific enthalpy vanishes and the limiting velocity is the displayed value. If area grows as $C{r}^2$, mass conservation further gives $\rho \sim \dot{M}/(C{u}_{\infty }{r}^2)$. A subsonic pressure-confined branch need not have this asymptotic behavior.

Consider a steady radial axisymmetric magnetohydrodynamic wind with outward poloidal speed $u_p\to u_{\infty }>0$. First choose an outward-positive poloidal magnetic field $B_p>0$ and positive magnetohydrodynamic mass loading $k$, soIf the field-line angular velocity is $\omega$, ideal induction gives the large-radius toroidal fieldThe density is $\rho \sim kF/(u_{\infty }{r}^2)$, hence the toroidal Alfvén speed approachesThe absolute value of $\omega$ is necessary when quoting a nonnegative speed.

When specific enthalpy and gravitational potential vanish at infinity and the azimuthal material speed also decays, the Bernoulli invariant becomesThe second term is magnetic energy transport per unit mass flux, so terminal material motion need not carry all of the outward energy.

For a signed inward $B_p$, use signed $F_B=\lim r^2{B}_p$ and signed $k$ consistently. Then $F_B/k>0$ for an outward wind, $B_{\varphi }\sim -\omega F_B/(u_{\infty }r)$ and the speed formula uses $F_B/k$. Mixing an absolute magnetic flux with a signed mass-loading parameter would produce an unphysical sign.

A steady outward radial flow with negligible pressure force conserves $v^2/2-GM/r$. If its speed at a reference radius equals the mechanical escape velocity, this constant is zero and $v\propto r^{-1/2}$. Conservation of mass gives $\rho \propto r^{-3/2}$. An adiabatically evolving ideal gas carried along that leading flow has $p\propto r^{-3\gamma /2}$ and $T\propto r^{-3(\gamma -1)/2}$.

Finite specific enthalpy changes the dynamical problem: the full conserved Bernoulli equation is $v^2/2+h-GM/r=B$. At the same boundary mechanical escape speed, $B=h_e$ rather than zero. A warm wind therefore needs its enthalpy and equation of state specified before the explicit ballistic velocity can be replaced by a consistent pressure-driven solution.

This barotropic pressure potential obeys $\nabla h=\nabla p/\rho$ for a barotropic fluid. It gives the pressure contribution to a steady Bernoulli equation. In isentropic thermodynamics it agrees with specific enthalpy up to a constant; for an externally maintained isothermal ideal gas it is $c_s^2\log \rho$ and is not the constant thermodynamic enthalpy at fixed temperature.

In a steady adiabatic inviscid fluid with a conservative body force, the Bernoulli invariant is the displayed specific energy, constant along each streamline. Dotting the Euler equations with velocity and using the specific enthalpy identity gives $\mathbf{u}\cdot \nabla b=0$. In ideal magnetohydrodynamics, magnetic work need not vanish; the planar ideal-MHD Bernoulli invariant includes the corresponding electromagnetic correction.

For angular velocity depending only on cylindrical radius, the centrifugal acceleration is a gradient. Steady barotropic balance is $\nabla p/\rho =-\nabla \Psi$, hence specific enthalpy satisfies $h(\rho )+\Psi =\text{constant}$. On an invertible equation-of-state branch, mass density and pressure are functions of $\Psi$. Cylindrical rotation is essential to this scalar centrifugal-potential representation.

In a strictly cold flow governed by ideal magnetohydrodynamics the sound speed vanishes, so the slow magnetosonic wave speed is zero. A nonzero-speed outflow therefore has no ordinary finite interior slow crossing. For the cold radial magnetohydrodynamic wind integral, $f_y=-\alpha (x-x^{-1}{)}^2/(y-1{)}^3-\beta /(x^4{y}^3)<0$ on the sub-Alfvénic branch $y>1$, assuming $\beta >0$. A small retained specific enthalpy is needed to locate a finite slow point; the zero-temperature limit can move it to the launch boundary. The usual simultaneous numerator and denominator regularity conditions still apply to nondegenerate critical points.

The specific enthalpy is equal on both sides of an isothermal perfect-gas shock. The energy removed per unit area per unit time is $j(u_1^2-u_2^2)/2$, with $j>0$ the normal mass flux. Net cooling therefore selects compression shocks; an expansion jump requires heating.

For transonic accretion in a dipolar flux tube, the Bernoulli equation at the sonic point contains $c_s^2[1/(\gamma -1)-5/2]$. A positive reservoir energy requires $\gamma <7/5$. Compression in the $r^3$ dipolar flux-tube area raises the specific enthalpy more rapidly than in the $r^2$ geometry of spherical Bondi accretion, whose corresponding upper index is $5/3$.

For a fixed-composition inviscid fluid, specific enthalpy satisfies $dw=Tds+dp/\rho$. Combining this identity with the vorticity cross-product identity gives the displayed equation, with $\mathbf{\omega }=\nabla \times \mathbf{u}$ and $\epsilon =u^2/2+\Phi +w$. It exposes entropy gradients as the thermodynamic source of a mismatch between Bernoulli function gradients and the vorticity term. In a steady adiabatic process, its projection along a streamline makes $\epsilon$ constant there.

Let $h_c=\mathbf{u}\cdot \mathbf{B}$ be the cross-helicity density. The ideal magnetohydrodynamic induction equation implies $D\mathbf{B}/Dt=(\mathbf{B}\cdot \nabla )\mathbf{u}-\mathbf{B}\nabla \cdot \mathbf{u}$. Dotting the momentum equation with $\mathbf{B}$ eliminates the Lorentz force density, since it is perpendicular to the magnetic field. The material derivative of $h_c$ is thereforeUsing $dh=Tds+dp/\rho$ for specific enthalpy $h$ and specific entropy $s$, together with Gauss's law for magnetism, converts this toThe source vanishes when magnetic field lines lie in constant-specific entropy surfaces, in particular in a homentropic flow. Integrating over a volume conserves its total cross-helicity only if the boundary flux also vanishes. This local law is derived in Ogilvie's astrophysical fluid dynamics notes.

Cryogenics studies the production and use of very low temperatures. Its experimental challenge is a balance of rates: a refrigerator must remove energy from a cold region faster than unwanted heat transfer supplies it. A low temperature reading alone does not establish a useful cooling capacity. A maintained cold liquid bath is especially valuable because latent heat allows it to absorb heat while its phase fraction changes at nearly fixed temperature and pressure.

Helium liquefaction illustrates why staged cooling and recovery of a returning stream's cooling capacity matter. A counterflow heat exchanger transfers heat from the incoming stream to the colder returning stream without mixing them. This reduces the incoming specific enthalpy before expansion. The heat exchanger recovers cooling already produced; the surrounding refrigerator still requires input work and rejection of heat to a warmer environment.

In the rest frame of a strong planar shock wave, a perfect gas has ${\rho }_2/{\rho }_1=(\gamma +1)/(\gamma -1)$, $u_2/u_1=(\gamma -1)/(\gamma +1)$ and $p_2=2{\rho }_1{u}_1^2/(\gamma +1)$. Its downstream specific internal energy is ${ϵ}_2=p_2/[(\gamma -1){\rho }_2]=2u_1^2/(\gamma +1{)}^2$. Multiplying by mass flux $j={\rho }_1{u}_1$ gives the internal-energy flux in the title. The lost bulk kinetic energy flux instead equals $j(u_1^2-u_2^2)/2=\gamma F_{\mathrm{int}}$: this includes downstream pressure work and is the increase of specific enthalpy flux in the cold-upstream limit. Neither quantity is a temperature change per unit time of a fixed target mass without further information about that target.

For a spherical stellar polytrope of finite radius and zero surface pressure, the specific enthalpy is $(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =-GM/R$, hence $(n+1)\int PdV=-G{M}^2/R-2\Omega$. Eliminating the pressure integral with the stellar virial theorem proves the displayed formula. For a gas adiabatic stellar polytrope, $n=1/(\gamma -1)$ and $U=G{M}^2/[(5\gamma -6)R]$; a general polytropic index need not have this relation to the specific-heat ratio.

Helium liquefaction converts helium gas into a collectable liquid. It creates a bath for investigating material properties at very low temperatures, including superconductivity. The central question is how the incoming specific enthalpy compares with the specific enthalpies of the outgoing liquid and vapor. The liquefaction enthalpy budget makes that comparison explicit; a temperature decrease by itself does not specify a liquid yield.

In a Joule-Thomson effect cooling stage, a throttling pressure drop approximately preserves enthalpy under the assumptions of the steady-flow energy balance. For an ideal gas, specific enthalpy depends only on temperature: with positive specific heat capacity at constant pressure, equal specific enthalpy means equal temperature. Thus its Joule-Thomson coefficient is zero. An ideal gas also supplies no liquid branch. Cooling by throttling depends on real gas properties; expansion is not an automatic source of refrigeration.

For a local drop $dp<0$, $dT={\mu }_{\mathrm{JT}}dp$ cools only where the Joule-Thomson coefficient is positive. Precooling must bring the real gas into a useful cooling region, with enough cooling to overcome unwanted heat transfer. A finite expansion can cross a Joule-Thomson inversion curve; its final state requires the full enthalpy balance, including phase coexistence if liquid forms. Kamerlingh Onnes's 1908 helium liquefaction experiment combined external precooling with regeneration to reach that regime. The Joule-Thomson effect article develops the throttling identities and their assumptions.

For a fixed Newtonian gravitational potential, total energy density is $\mathcal{E}=\rho (u^2/2+\Phi )+p/(\gamma -1)+B^2/(2{\mu }_0)$. Its flux combines advected kinetic/potential energy, specific enthalpy and the Poynting vector. Dotting the ideal magnetohydrodynamic momentum equation with velocity, adding the adiabatic internal energy equation and the magnetic energy equation cancels the magnetic work. The pressure terms combine into $-\nabla \cdot (p\mathbf{u})$, proving the conservation law. A time-dependent imposed potential instead contributes $\rho {\partial }_t\Phi$.

An isenthalpic process preserves enthalpy for a fixed amount of material, or preserves specific enthalpy between inlet and outlet when comparing a flowing fluid of fixed composition. It need not preserve temperature or entropy. An insulated throttle satisfying the steady-flow energy balance assumptions is an important example.

The Joule-Thomson effect is the temperature response of a fluid flowing from higher to lower pressure through a valve or a plug of porous medium, when the throttle is insulated and the other steady-flow energy balance assumptions hold. Its equilibrium endpoints have equal specific enthalpy. The Joule-Thomson coefficient determines whether a small drop cools or warms; cooling is not universal.

A useful contrast is between three insulated expansions. In free expansion of a gas-only closed system into a vacuum, no external work occurs and $U_2=U_1$; an ideal gas returns to its initial equilibrium temperature. In a throttle, continuous flow carries pressure work across the boundary and preserves specific enthalpy under the stated assumptions; an ideal gas again has no endpoint temperature change. In a reversible adiabatic expansion of a loaded closed system, the gas supplies work, loses internal energy and cools, even if it is an ideal gas. That process preserves entropy; throttling generally produces entropy. For a small constant-enthalpy step, $dH=TdS+Vdp=0$ gives $dS=-Vdp/T>0$ when $dp<0$ and $T,V>0$. This equilibrium comparison computes endpoint entropy differences without supplying a reversible path through the plug.

The effect is useful in refrigerators because an appropriately prepared real gas can cool while passing through a simple valve. The inlet must lie in a suitable positive-Joule-Thomson coefficient region. The Joule-Thomson inversion curve makes that requirement precise; a pressure drop by itself is not a guarantee of refrigeration.

In an inviscid barotropic fluid subject to conservative body force, the circulation of a material loop is constant. For constant density, parameterize that loop by $\mathbf{r}(s,t)$ with ${\mathbf{r}}_t=\mathbf{u}$. Differentiating $\Gamma =\int \mathbf{u}\cdot {\mathbf{r}}_{s}ds$ gives an integral of $(D\mathbf{u}/Dt)\cdot d\mathbf{r}$ plus $\int {\partial }_s(|\mathbf{u}{|}^2/2)ds$. The latter vanishes on a closed loop, and the Euler equations make the former the integral of a gradient, also zero. For a barotropic density, $\nabla p/\rho$ is still a gradient of specific enthalpy. Constant circulation is consistent with changing local vorticity when a vortex tube changes its cross-sectional area.

For smooth inviscid barotropic fluid motion without gravity or magnetic forces, let $w$ be the specific enthalpy with $\nabla w={\rho }^{-1}\nabla p$, and set $q=\mathbf{u}\cdot \mathbf{\omega }$. ThenThis follows by dotting the barotropic vorticity transport equation with $\mathbf{u}$ and using the divergence of a cross product. A volume integral is constant when the resulting boundary flux vanishes.