An Extension of the Stefan-Type Solution Method Applicable to Multi-component, Multi-phase 1D Systems

Abstract

We present an extension of the Stefan-type solution method applicable to multi-component, multi-phase 1D porous flows, and illustrate the method by applying it to phase separation dynamics in an NaCl–$\hbox {H}_2\hbox {O}$-saturated hydrothermal heat pipe. For this example, three mathematical models are constructed. The first two models concern the rate of progression of two interfaces, one separating brine from two-phase fluid and another separating two-phase fluid from single-phase liquid at seawater salinity. The brine layer model shows that the layer may reach quasi-steady-state thickness even while the salt content of the layer continues to increase; the two-phase layer model shows how variable heat flux at the top of the layer leads to departure from the linear growth rate predicted by a simpler model. The third model concerns the temperature profile in the entire column. The governing advection–diffusion equation has highly variable coefficients, with no negligible terms in it in the region of parameter space considered. We present a method to solve this type of equation by constructing a propagator and a corresponding Green’s function. Finally, we show how to use the developed framework to test the internal consistency of numerical simulations, again using the 1D heat pipe as an example.

Appendices

Appendix 1: Alternative Derivation of Equation (12)

The equations expressing the balance of salt or heat can both be cast in the form

$$\begin{aligned} \frac{\partial \mathcal {C}}{\partial t}+\frac{\partial \mathcal {F}}{\partial y}=0, \end{aligned}$$

(88)

where $\mathcal {C}$ represents the content of salt or energy and $\mathcal {F}$ represents combined advective and diffusive fluxes. Integrating both sides from the bottom to the top of the expansion layer gives

$$\begin{aligned} \int _0^{y_{ij}}\frac{\hbox {d}\mathcal {C}}{\hbox {d}t}\hbox {d}y = \mathcal {F}_{0}-\mathcal {F}_{ij} \equiv \mathcal {F}. \end{aligned}$$

(89)

According to the Leibniz integral theorem,

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\int _0^{y_{ij}}\mathcal {C}\hbox {d}y = \int _0^{y_{ij}}\frac{\partial \mathcal {C}}{\partial t}\hbox {d}y + \mathcal {C}(y_{ij},t)\frac{\hbox {d}y_{ij}}{\hbox {d}t}. \end{aligned}$$

(90)

Solving for the first term on the right-hand side and substituting the result into (89) gives

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\int _0^{y_{ij}}\mathcal {C}\hbox {d}y -\mathcal {C}_{ij}\frac{\hbox {d}y_{ij}}{\hbox {d}t}=\mathcal {F}, \end{aligned}$$

(91)

where $\mathcal {C}_{ij}\equiv \mathcal {C}(y_{ij},t)$. Defining the average content density as in equation (10) and substituting into the above equation leads to equation (12).

Appendix 2: Derivation of the Stefan Condition

In the original Stefan problem that involves a half plane filled with melting ice, the position of the melt interface is found by imposing the Stefan condition [see (Carslaw and Jaeger 1959], pg. 284); in this section, we show how this condition may be derived from equation (12). Consider the upper half plane ($y \ge 0$) filled with ice and heated from below. In our notation, the Stefan condition reads

$$\begin{aligned} \lambda _2\frac{\partial T_2}{\partial y}-\lambda _1\frac{\partial T_1}{\partial y}=L\rho \frac{\hbox {d}y_{12}}{\hbox {d}t}, \end{aligned}$$

(92)

where the subscript 1 refers to liquid, the subscript 2 refers to ice, and L is the latent heat of fusion. In this scenario, equation (12) becomes

$$\begin{aligned} \rho \frac{\hbox {d}y_{12}}{\hbox {d}t}[\langle h \rangle - h_{12}]+\rho y_{12}\frac{\hbox {d}}{\hbox {d}t}\langle h \rangle = \mathcal {F}_0-\mathcal {F}_{12}, \end{aligned}$$

(93)

where $\rho$ is the liquid density, h is the liquid enthalpy, and we have made the replacement $\mathcal {C}=\rho h$. For simplicity, we have assumed that the liquid is raised only slightly above freezing so that variation of density within the liquid phase may be neglected. First, we note that the heat flux into the $y_{12}$ interface from below is equal to the flux at the bottom boundary of the system minus the portion of that heat absorbed by the layer of liquid between the bottom boundary and $y_{12}$; mathematically, this statement translates to

$$\begin{aligned} \mathcal {F}_{12,-} = \mathcal {F}_0 -\rho y_{12} \frac{\hbox {d}}{\hbox {d}t}\langle h \rangle . \end{aligned}$$

(94)

Second, we relabel the quantity $\mathcal {F}_{12}$ as $\mathcal {F}_{12,+}$, emphasizing that it is the flux of heat upward from $y_{12}$. Making these replacements and setting

$$\begin{aligned} \langle h \rangle - h_{12} = L \end{aligned}$$

(95)

in equation (93) yields

$$\begin{aligned} L\rho \frac{\hbox {d}y_{12}}{\hbox {d}t} = \mathcal {F}_{12,-}-\mathcal {F}_{12,+}. \end{aligned}$$

(96)

Using Fourier’s law of heat conduction, we write

$$\begin{aligned} \mathcal {F}_{12,-} = -\lambda _{1}\frac{\partial T_1}{\partial y}, \end{aligned}$$

(97)

and

$$\begin{aligned} \mathcal {F}_{12,+} = -\lambda _{2}\frac{\partial T_2}{\partial y}, \end{aligned}$$

(98)

where both derivatives are evaluated at $y_{12}$. Substituting these expressions into (96) gives equation (92).

In the problem of the receding ice sheet, the classical Stefan condition can be used to find the form of $y_{12}(t)$ together with the temperature profile; however, this technique only works under very restricted conditions. For example, imposing a heat flux condition at the bottom boundary of the system instead of a constant temperature condition already leads to problems (see (Carslaw and Jaeger 1959)) for details). The generalized condition (12) avoids this difficulty by representing $y_{ij}(t)$ as the solution of its own differential equation instead of as an auxiliary quantity to be determined simultaneously with T(y, t).

Appendix 3: Derivation of Condition (20)

Using the definitions

$$\begin{aligned} T = Fy_{ij}\theta + T_{ij} + F(y_{ij}-y), \end{aligned}$$

(99)

and

$$\begin{aligned} \tau = \frac{\kappa _i t}{y_{ij}^2}, \end{aligned}$$

(100)

we find that

$$\begin{aligned} \frac{1}{F}\frac{\partial T}{\partial t} = \dot{y}_{ij}(\theta +1) + \frac{\kappa _i}{y_{ij}}\frac{\partial \theta }{\partial \tau } - 2\tau \dot{y}_{ij}\frac{\partial \theta }{\partial \tau }. \end{aligned}$$

(101)

We seek the condition under which the second term on the right-hand side of (101) dominates the first and the third terms. The first and third terms are both of the order $\dot{y}_{ij}$, while the second term is of order $\kappa _i/y_{ij}$. Hence, the desired condition is that

$$\begin{aligned} \dot{y}_{ij} \ll \frac{\kappa _i}{y_{ij}}, \end{aligned}$$

(102)

which is identical with (20). This condition is sufficient but not necessary, because the first and third terms on the right-hand side of (101) are of opposite sign, and so may approximately cancel one another even if (102) is not satisfied.

Appendix 4: Derivation of Propagator and Green’s Function

To solve for $\theta _{\mathrm{H}}(\zeta ,\tau )$ and construct the propagator, we begin by finding the eigenfunctions $\psi _n(\zeta )$ and eigenvalues $\lambda _n$ satisfying

$$\begin{aligned} -\frac{\hbox {d}^2}{\hbox {d}\theta ^2}\psi _n(\zeta ) = \lambda _n \psi _n(\zeta ), \end{aligned}$$

(103)

where the eigenfunctions are required to satisfy the homogeneous boundary conditions (22) and (23). The solutions, normalized on the interval $0 \le \zeta \le 1$, are

$$\begin{aligned} \psi _n(\zeta ) = \sqrt{2}\cos (\sqrt{\lambda _n}\zeta ), \end{aligned}$$

(104)

with

$$\begin{aligned} \lambda _n = \left[ \frac{(2n+1)\pi }{2} \right] ^2, \end{aligned}$$

(105)

and $n\in \mathbb {N}$. Now we search for a solution of the form

$$\begin{aligned} \theta _{\mathrm{H}}(\zeta ,\tau ) = \sum _{n=0}^{\infty } a_n(\tau )\psi _n(\zeta ), \end{aligned}$$

(106)

where the time-dependent coefficients $a_n(\tau )$ are to be determined. Substituting (106) into (21), setting $\tilde{f}(\zeta )=0$, and keeping in mind that the $\psi _n$ satisfy (103), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } \psi _n\left( \frac{\hbox {d}a_n}{\hbox {d}\tau }+\lambda _n a_n\right) = 0. \end{aligned}$$

(107)

The eigenfunctions are all linearly independent, so the only way the above equation can be satisfied is if each coefficient of $\psi _n$ vanishes; the resulting differential equation in $a_n(\tau )$ has the solution

$$\begin{aligned} a_n(\tau ) = A_n e^{-\lambda _n \tau }, \end{aligned}$$

(108)

where $A_n$ is a constant for each n. Equation (106) now takes the form

$$\begin{aligned} \theta _{\mathrm{H}}(\zeta ,\tau ) = \sum _{n=0}^{\infty } A_n e^{-\lambda \tau }\psi _n(\zeta ). \end{aligned}$$

(109)

It remains to choose the constants $A_n$ so that the initial condition (24) is satisfied. Applying the initial condition to (109) gives

$$\begin{aligned} \theta _{\mathrm{H}}(\zeta ,0) = \sum _{n=0}^{\infty } A_n \psi _n(\zeta ). \end{aligned}$$

(110)

Multiplying both sides of this equation by $\psi _m(\zeta )$, integrating from 0 to 1, and keeping in mind the orthonormality of the eigenfunctions results in

$$\begin{aligned} A_m = \int _0^1\theta (\zeta ',0)\psi _m(\zeta ')\hbox {d}\zeta '. \end{aligned}$$

(111)

Changing m back to n, substituting the above expression into (109), and rearranging give the solution to the homogeneous problem as

$$\begin{aligned} \theta _{\mathrm{H}}(\zeta ,\tau ) = \int _0^1 \left[ \sum _{n=0}^{n=\infty }\psi _n(\zeta ')\psi _n(\zeta )e^{-\lambda _n \tau } \right] \theta (\zeta ',0)\hbox {d}\zeta '. \end{aligned}$$

(112)

The expression in brackets is the kernel of the desired integral operator, which propagates the solution at $\tau = 0$ to that at a later time $\tau$. Therefore, the kernel that propagates a solution at $\tau =\tau '$ to a later time $\tau$ is

$$\begin{aligned} \mathcal {P}(\zeta ,\zeta ',\tau ,\tau ') = \sum _{n=0}^{\infty } \psi _n(\zeta ')\psi _n(\zeta )e^{-\lambda _n(\tau -\tau ')}. \end{aligned}$$

(113)

It can be shown that multiplying the propagator with the Heaviside step function $H(\tau -\tau ')$ produces the Green function required for solution of the associated inhomogeneous problem (see Bayin 2006); hence, the desired Green function is

$$\begin{aligned} \mathcal {G}(\zeta ,\zeta ',\tau ,\tau ') = \mathcal {P}(\zeta ,\zeta ',\tau ,\tau ')H(\tau -\tau '). \end{aligned}$$

(114)

The solution to the inhomogeneous problem is then

$$\begin{aligned} \theta _{\mathrm{I}}(\zeta ,\tau )=\int _0^{\infty }\int _0^1 \mathcal {G}(\zeta ,\zeta ',\tau ,\tau ')\tilde{g}(\zeta ')\hbox {d}\zeta ' d\tau '. \end{aligned}$$

(115)

Carrying out the integral as far as possible with general $\tilde{g}(\zeta ')$ results in

$$\begin{aligned} \theta _{\mathrm{I}}(\zeta ,\tau )= & {} \sum _{n=0}^{\infty } \psi _n(\zeta ')\int _0^1 \psi _n(\zeta ')\tilde{g}(\zeta ')\hbox {d}\zeta '\int _0^{\tau }e^{-\lambda _n (\tau -\tau ')} d\tau ' \nonumber \\= & {} \sum _{n=0}^{\infty } \frac{1-e^{-\lambda _n \tau }}{\lambda _n}\psi _n(\zeta )\int _0^1 \psi _n(\zeta ')\tilde{g}(\zeta ')\hbox {d}\zeta ', \end{aligned}$$

(116)

which is the same as equation (27). Using equation (24) and completing the integral in (112) yields

$$\begin{aligned} \theta _{\mathrm{H}}(\zeta ,\tau ) = \sum _{n=0}^{\infty }\frac{\sqrt{2}}{\lambda _n}e^{-\lambda _n \tau } \psi _n(\zeta ), \end{aligned}$$

(117)

which is the same as equation (26). The full dimensionless solution $\theta (\zeta ,\tau )$ is now given by (25).