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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.

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.

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.

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.

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

Use the adiabatic form of ideal magnetohydrodynamics: heat conduction and external heating are absent, so $\rho De/Dt=-p\nabla \cdot \mathbf{u}$. Infinite electrical conductivity and zero viscosity alone would not remove the thermal source terms on the cover page. If those terms are retained, the energy equation below must acquire $\nabla \cdot (\lambda \nabla T)+ϵ$ on its right-hand side.

Expanding the continuity equation gives its conservative form ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Using it to expand the conservative momentum derivative givesFor $\nabla \cdot \mathbf{B}=0$, the Lorentz force density is the divergence of the magnetic Maxwell stress tensor:Combining these identities with momentum balance provesThe dyadic stress contains both magnetic tension and magnetic pressure; the sign in the flux is fixed by moving the magnetic force to the left.

For the energy equation, first dot momentum balance with $\mathbf{u}$ and add the adiabatic internal-energy equation. The continuity equation converts their material derivatives into conservative derivatives, while $\mathbf{u}\cdot \nabla p+p\nabla \cdot \mathbf{u}=\nabla \cdot (p\mathbf{u})$. The result isThe ideal magnetohydrodynamic induction equation gives the magnetic counterpart. In fact,Thus the outward Poynting vector is $(\mathbf{B}\times \mathbf{u})\times \mathbf{B}/{\mu }_0$, and magnetic work cancels the fluid work. Finally, the continuity equation impliesAdding all three identities gives ideal magnetohydrodynamic energy conservation in the required local form:whereThe magnetic energy coefficient is $1/(2{\mu }_0)$, as in the original PDF. For a self-consistent, time-dependent gravitational field this local balance should not be confused with a global self-gravitating energy integral: the latter uses gravitational field energy, or the familiar one-half factor in $\int \rho \Phi$ after the field is eliminated.

Work in the shock frame of a thin planar discontinuity, with normal speed $u>0$. A smooth gravitational potential has no finite jump through a vanishingly thin front; its advected contribution cancels using the common mass flux. Integrating the conservation laws gives constant normal mass, momentum and energy fluxes, together with the normal-field and tangential-electric-field conditions of the ideal magnetohydrodynamic shock conditions.

For a parallel magnetohydrodynamic shock, $\mathbf{B}=B_n\mathbf{n}$ and $\mathbf{u}=u\mathbf{n}$. The magnetic solenoidal constraint makes $B_n$ the same on both sides. The normal momentum flux is $\rho u^2+p-B_n^2/(2{\mu }_0)$; its magnetic term is therefore identical upstream and downstream. The Poynting vector is zero because $\mathbf{u}\times \mathbf{B}=0$. Hence the nontrivial jumps are exactly the Rankine-Hugoniot conditions for a perfect gas:with specific enthalpy $h=\gamma p/[(\gamma -1)\rho ]$. Physically, motion along the straight field neither bends it nor compresses its transverse flux. The field consequently provides no changed stress or energy flux in this purely parallel geometry. This conclusion does not apply to a shock that develops a transverse magnetic field.

For a perpendicular magnetohydrodynamic shock, the magnetic field is tangential and the velocity normal. Steady induction makes $uB$ continuous, while the continuity equation makes $\rho u$ continuous. Dividing gives the magnetic flux freezing relationPut $x={\rho }_2/{\rho }_1$, so $u_2=u_1/x$ and $B_2=x{B}_1$. Constant normal momentum flux now reads $\rho u^2+p+B^2/(2{\mu }_0)$. Substitution givesThe magnetic correction is negative for compression: part of the incoming ram pressure builds magnetic pressure instead of thermal pressure.

Here the Poynting vector has normal component $u{B}^2/{\mu }_0$. Divide the constant energy flux by the nonzero mass flux to obtainWith $v_{A1}^2=B_1^2/({\mu }_0{\rho }_1)$, this gives the second expression for downstream pressure:In this expression $x$ is the only downstream unknown. Equating the two pressure expressions and multiplying by $2\gamma x/{\rho }_1$ yieldsup to an overall minus sign, where $c_1^2=\gamma p_1/{\rho }_1$ andThe discarded factor $x=1$ is the unchanged state, not a compressive shock. This derives the polynomial for the compression ratio of a perpendicular magnetohydrodynamic shock without omitting its trivial branch.

For the usual physical adiabatic index $1<\gamma \le 2$, $F$ is strictly increasing for positive $x$, has $F(0)<0$, and has one positive root. Sincethat root is greater than one precisely when $u_1^2>c_1^2+v_{A1}^2$. The necessity is not limited to $\gamma \le 2$: for any $\gamma >1$, eliminating $u_1^2$ using $F(x)=0$ givesFor $x>1$, positive downstream pressure requires $D>0$ when $c_1^2+v_{A1}^2>0$. Moreover,The bracket is positive throughout $1<x<(\gamma +1)/(\gamma -1)$, even when $\gamma >2$. The exactly cold unmagnetized limit gives the familiar strong-shock upper compression ratio directly. Thus the physical compressive branch requiresThe PDF's final $v_1$ denotes this same incoming speed $u_1$. The right side is the squared speed of a perpendicular fast magnetosonic wave: the upstream flow must be super-fast-magnetosonic so that this compressive signal cannot run upstream against the incoming flow. At equality the shock amplitude vanishes into a characteristic disturbance.

Let $b=\frac{1}{2}{u}^2+\Phi +w$. Because the specific entropy is constant along the horizontal flow, the specific enthalpy identity gives $\mathbf{u}\cdot \nabla w={\rho }^{-1}\mathbf{u}\cdot \nabla p$. Dotting the ideal magnetohydrodynamic momentum equation with $\mathbf{u}$ therefore givesOn the field-aligned horizontal branch, $\mathbf{u}\times \mathbf{B}=U{\mathbf{e}}_z\times \mathbf{B}$. Directly evaluating the horizontal components of the curl givesThe scalar triple-product identity consequently makes the magnetic work $U{\mathbf{B}}_h\cdot \nabla B_z/{\mu }_0$. Meanwhile $\rho \mathbf{u}\cdot \nabla b=k{\mathbf{B}}_h\cdot \nabla b$. The previously established field-line constancy of $U,k$ now gives the planar ideal-MHD Bernoulli invariantThis is conservation of total energy transported along a magnetic field line, including electromagnetic work. To see its flux interpretation, the ideal magnetohydrodynamic energy conservation law has fluxThe second term is the Poynting vector, since $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. Substitution of $\mathbf{u}=k\mathbf{B}/\rho +U{\mathbf{e}}_z$ givesThe steady divergence of this horizontal energy flux is zero. Magnetic stress can change the ordinary hydrodynamic quantity $b$ in the Bernoulli equation by doing work; the term $-U{B}_z/({\mu }_{0}k)$ accounts for the accompanying Poynting flux. When $U=0$, the velocity is completely field-aligned, $\mathbf{E}=0$, and the usual hydrodynamic Bernoulli quantity is constant. These conclusions require the field-aligned branch and $k\ne 0$; the arbitrary constant-$E_z$ flow from the first part is not covered by this particular invariant.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

Write $v=u_x$ and $D={\partial }_t+v{\partial }_x$. The Euler equations and conservation of mass give $D\rho =-\rho v_x$ and $\rho Du=-p_x{e}_x$. The internal-energy density of the ideal gas is $e=p/(\gamma -1)$. The adiabatic pressure equation therefore givesMultiply the momentum equation by $u$ and use continuity to obtain the kinetic-energy balanceAdding the two equations moves ${\partial }_x(pv)$ into the flux. Thus the entire system is ${\partial }_{t}Q+{\partial }_{x}F=0$, whereThis derives the conservative energy flux of a polytropic ideal gas. In particular one-dimensional spatial dependence does not remove the transverse velocities from the total energy. The energy flux can equally be written $\rho v(u^2/2+w)$ with specific enthalpy $w=\gamma p/[(\gamma -1)\rho ]$.

Work in the frame of the steadily advancing mush, so the rigid primary-crystal skeleton moves with $-V{\mathbf{e}}_z$. Let $\varphi$ be solid fraction, $C$ the pore-liquid concentration, $\mathbf{q}=(u,w)$ the Darcy flux relative to that skeleton, and $K$ the common thermal conductivity. Set $\kappa =K/(\rho c_p)$, the thermal diffusivity. The actual pore velocity is $\mathbf{q}/(1-\varphi )-V{\mathbf{e}}_z$. Primary crystals are taken to exclude solute; the residual pore liquid freezes into a two-solid eutectic product at the bottom. These are the standard binary mushy layer assumptions implicit in using a single liquidus and an eutectic floor.

It is convenient to use hydraulic pressure $P=p+\rho gz$, absorbing the constant-density gravitational term. With equal phase properties, no solute diffusion, negligible fluid inertia and local liquidus equilibrium, a complete mush model isThe last equation follows from conservation of energy with mixture specific enthalpy $c_{p}T-L\varphi$. Sensible heat is transported by the total volume flux $\mathbf{q}-V{\mathbf{e}}_z$; the primary solid carries the latent-enthalpy flux $VL\varphi {\mathbf{e}}_z$. In particular the sign of the latent-heat term is fixed by the downward-moving skeleton.

At the liquid roof, imposeHere $P_0=p_0+\rho gh$. The gradient condition matches heat flux into the imposed liquid field, since there is no phase-fraction jump at this roof and the phase conductivities are equal. The pore velocity there equals the relative Darcy velocity. At the floor,There is no liquid penetration into fully solid material. The unspecified solid fraction at the floor is determined by the solute balance, not set to one: remaining eutectic liquid freezes as it crosses the floor. If $C_s$ denotes the bulk concentration of the final solid product, the floor solute balance and Stefan condition areMaintaining the floor at $T_E$ permits the solid-side bath to remove this final latent heat. Equivalently its required gradient is $T_z(0^{-})=T_z(0^{+})+\rho LV[1-\varphi (0)]/K$. No additional imposed solid-side gradient is specified. Laterally we take the quadratic-pressure strain solution, without sidewalls; finite sidewalls would require additional flow conditions and generally break this horizontal invariance.

For uniform permeability of a porous medium, incompressible flow gives ${\nabla }^{2}P=0$. The roof pressure and zero normal flow at the floor are satisfied byHorizontal outflow is balanced by downward inflow. The parameter $\alpha$ is an inverse time; $E^2$ in the printed pressure condition already has dimensions of pressure divided by length squared, so no extra density factor belongs in this definition.

With $T=T(z)$, $C=C(z)$ and $\varphi =\varphi (z)$, the $x$-dependence disappears from both scalar balances. Although $u$ is nonzero, its horizontal divergence cancels $w_z$ in $\nabla \cdot (\mathbf{q}C)$. The exact strain-driven steady mush equations becomeTogether with $T(0)=T_E$, $T(h)=T_0$, $T^{\prime }(h)=G$ and $\varphi (h)=0$, they determine the scalar fields and the unknown thickness. This reduction proves that the equations admit horizontally invariant scalar solutions despite the imposed straining flow. An equivalent useful form, obtained by differentiating the solute equation, isThis exact form also shows why one must not silently discard latent heat in an arbitrary strong-flow calculation.

Define $\Delta T=T_0-T_E>0$, $R=\alpha h/V$ and $S=L/(c_p\Delta T)$. For the stated regular ordering $R\gg S\gg 1$, strain transport dominates skeleton transport and the bounded phase-change enthalpy contribution. More precisely, at fixed finite thermal strain number $\alpha h^2/\kappa$, the dimensionless energy equation isso its skeleton and integrated latent-heat corrections are small in $1/R$ and $S/R$ respectively. The leading strain-dominated mush temperature profile obeys $\kappa T^{\prime \prime }+\alpha z{T}^{\prime }=0$. Put $\delta =\sqrt{2\kappa /\alpha }$ and $H=h/\delta$. Integrating first for $T^{\prime }$ and then for $T$ givesThe error function profile follows from integrating a Gaussian, with $\mathrm{erf}x=(2/\sqrt{\pi }){\int }_0^x{e}^{-s^2}ds$. Matching the roof gradient gives the requested transcendental thickness relationFor prescribed $G>0$ and $\Delta T>0$ the right side decreases continuously from infinity to zero as $H$ increases, so it fixes one positive leading-order thickness. The resulting $h$ must still satisfy the assumed ordering and the phase-fraction constraints.

To find the product composition, integrate the exact solute balance from the roof to an arbitrary depth. With $f=1-\varphi$,At $z=0$, the exact floor composition is thereforeThe minus sign has a direct physical meaning: the horizontal outflow removes more salt than arrives at the same depth from a uniform-$C_0$ liquid, leaving a depleted solid product. Using the leading error function profile andyieldsThe full solid fraction follows by substituting this same profile into the preceding integral formula for $f(z)$. It has $\varphi (h)=0$ and $1-\varphi (0)=C_s/C_E$. A physical branch requires $0\le C_s\le C_E$ and $0\le \varphi \le 1$. In this monotone profile, $C_s\ge 0$ ensures a nonnegative liquid fraction everywhere; it is not permissible to interpret a negative formal result as a material composition.

The error-function reduction is a regular leading approximation, not a uniform statement for every possible scaling of the thermal strain number. If $\alpha h^2/\kappa$ becomes singularly large along the ordering limit, a thin basal layer can amplify the discarded terms. The printed two inequalities alone do not specify that extra scaling. In such a regime the displayed exact scalar equations and boundary conditions, rather than an unqualified zero-latent-heat profile, provide the appropriate thickness and composition calculation.