High-Temperature Liquid Metal Infusion Considering Surface Tension-Viscosity Dissipation

Abstract

In considering the significant effect of the surface tension-viscosity dissipation driving the fluid flow within a capillary, high-temperature liquid metal infusion was analyzed for titanium, yttrium, hafnium, and zirconium penetrating into a packed bed. A model of the dissipation considers the momentum balance within the capillary to determine the rate of infusion, which is compared with the Semlak-Rhines model developed for liquid metal penetration into a packed bed assumed as a bundle of tubes mimicking the porosity of a packed bed. For liquid Ti, the penetration rate was calculated from 0.2 µs to 1 ms and rose to a maximum of 7 m/s at approximately 1 µs; after which, the rate decreased to 0.7 m/s at 1 ms. Beyond 10 µs, the decreasing trend of the rate of penetration determined by the model of dissipation compared favorably with the Semlak-Rhines equation.

Appendix

In this appendix, steps involved in solving the modified-SR equation are presented. We start with Eq. [3] with the appropriate initial conditions:

$$ \begin{aligned} \frac{{d}}{{{{d}}t}}\left( {\left( {\rho \pi R^{2} h + \rho V_{{L}} } \right)\frac{{{{d}}h}}{{{{d}}t}}} \right) = & 2\pi R\sigma \cos \left( \theta \right) - 8\pi \mu \left( {h + h_{0} } \right)\left( {\frac{{{{d}}h}}{{{{d}}t}}} \right) \\ - \rho \left( {\pi R^{2} h + \pi R^{2} h_{{L}} } \right)g - \frac{1}{4}\rho \pi R^{2} \left( {\frac{{{{d}}h}}{{{{d}}t}}} \right)^{2} \\ \end{aligned} $$

(3)

$$ \left. h \right|_{t = 0} = 0;\quad \left. {\frac{{{{d}}h}}{{{{d}}t}}} \right|_{t = 0} = 0. $$

(4)

We simplify Eq. [3] by dividing the left- and right-hand sides by πR ² and ignoring the gravitational and viscous force contributions resulting from an initial meniscus formation assumption (i.e., right-hand side terms containing h ₀ and h _L) as they become negligible shortly after the onset of the capillary rise when h >> h _L. This yields:

$$ \rho \frac{{d}}{{{{d}}t}}\left( {\{ h + R\tilde{V}_{L} \} \frac{{{{d}}h}}{{{{d}}t}}} \right) = 2\sigma \cos (\theta )/R - 8\mu hR^{ - 2} \frac{{{{d}}h}}{{{{d}}t}} - \rho gh - \frac{1}{4}\rho \left( {\frac{{{{d}}h}}{{{{d}}t}}} \right)^{2} $$

(5)

with $\tilde{V}_{\text{L}} = V_{\text{L}} /(\pi R^{3} ).$

The steady-state capillary height (h _s) may be obtained by setting all time derivatives to zero in Eq. [5], which then gives:

$$ h_{\text{s}} = \frac{2\sigma \cos (\theta )}{\rho gR}. $$

(6)

We may make use of Eq. [6] to rewrite Eq. [5] as:

$$ \rho \frac{{d}}{{{{d}}t}}\left( {\{ h + R\tilde{V}_{{L}} \} \frac{{{{d}}h}}{{{{d}}t}}} \right) = \rho g(h_{\text{s}} - h) - 8\mu hR^{ - 2} \frac{{{{d}}h}}{{{{d}}t}} - \frac{1}{4}\rho \left( {\frac{{{{d}}h}}{{{{d}}t}}} \right)^{2} . $$

(7)

Dividing Eq. [7] by the density ρ, we acquire:

$$ \frac{{d}}{{{{d}}t}}\left( {\{ h + R\tilde{V}_{{L}} \} \frac{{{{d}}h}}{{{{d}}t}}} \right) = g(h_{\text{s}} - h) - \frac{8\mu h}{{\rho R^{2} }}\frac{{{{d}}h}}{{{{d}}t}} - \frac{1}{4}\left( {\frac{{{{d}}h}}{{{{d}}t}}} \right)^{2} . $$

(8)

We introduce the following nondimensional capillary height and time:

$$ \tilde{h} = h/h_{\text{s}} ;\quad \tau = t\sqrt {g/h_{\text{s}} } $$

(9)

Then, Eq. [8] can be transformed into Eq. [10], which has the nondimensional capillary height and time:

$$ \frac{{d}}{{{{d}}\tau }}\left( {\{ \tilde{h} + (R/h_{\text{s}} )\tilde{V}_{\text{L}} \} \frac{{{{d}}\tilde{h}}}{{{{d}}\tau }}} \right) = (1 - \tilde{h}) - \frac{{8\mu h_{\text{s}} \tilde{h}}}{{\rho R^{2} \sqrt {gh_{\text{s}} } }}h\frac{{{{d}}\tilde{h}}}{d\tau } - \frac{1}{4}gh_{s} \left( {\frac{{{{d}}\tilde{h}}}{{{{d}}\tau }}} \right)^{2} . $$

(10)

Equation [10] can be written in nondimensionalized form by introducing a nondimensional viscosity ($\tilde{\mu }$):

$$ \frac{{d}}{{{{d}}\tau }}\left( {\{ \tilde{h} + (R/h_{\text{s}} )\tilde{V}_{\text{L}} \} \frac{{{{d}}\tilde{h}}}{{{{d}}\tau }}} \right) = (1 - \tilde{h}) - 8\tilde{\mu }\tilde{h}\frac{{{{d}}\tilde{h}}}{{{{d}}\tau }} - \frac{1}{4}\left( {\frac{{{{d}}\tilde{h}}}{{{{d}}\tau }}} \right)^{2} $$

(11)

with

$$ \tilde{\mu } = \frac{\mu }{{\rho R^{2} \sqrt {\frac{g}{{h_{\text{s}} }}} }}. $$

(12)

We define nondimensional rate of penetration $\tilde{v} = {{d}}\tilde{h}/{{d}}\tau$and use the fact that:

$$ \frac{{{{d}}^{2} \tilde{h}}}{{{{d}}\tau^{2} }} = \frac{{{{d}}\tilde{v}}}{{{{d}}\tau }} = \tilde{v}\frac{{{{d}}\tilde{v}}}{{{{d}}\tilde{h}}} = \frac{{d}}{{{{d}}\tilde{h}}}\left( {\frac{1}{2}\tilde{v}^{2} } \right). $$

(13)

Equation [11] can be written as:

$$ (\tilde{h} + \alpha )\frac{{d}}{{{{d}}\tilde{h}}}\left( {\frac{1}{2}\tilde{v}^{2} } \right) + \frac{5}{4}\tilde{v}^{2} = (1 - \tilde{h}) - 8\tilde{\mu }\tilde{h}\tilde{v}, $$

(14)

where $\alpha = R\tilde{V}_{\text{L}} /h_{\text{s}}$.

Further simplification of the left-hand side of Eq. [14] yields:

$$ \frac{{d}}{{{{d}}\tilde{h}}}\left\{ {(\tilde{h} + \alpha )^{5/2} f} \right\} = \left\{ {(1 - \tilde{h}) - 8\tilde{\mu }\tilde{h}\sqrt {2f} } \right\}(\tilde{h} + \alpha )^{3/2} $$

(15)

with

$$ f(\tilde{h}) = \frac{1}{2}\tilde{v}^{2} . $$

(16)

Equation [15] is subject to the initial condition:

$$ f(0) = 0 $$

(17)

Special Case $\tilde{\mu } = 0$

This case would be true during the initiation of the flow (for Ti, t < 0.1 µs). In this case Eq. [14] may be integrated explicitly by using Eq. [17] to obtain:

$$ (\tilde{h} + \alpha )^{5/2} f = \frac{2}{5}(1 + \alpha )\left\{ {(\tilde{h} + \alpha )^{5/2} - \alpha^{5/2} } \right\} - \frac{2}{7}\left\{ {(\tilde{h} + \alpha )^{7/2} - \alpha^{7/2} } \right\}. $$

(18)

This yields:

$$ f = \frac{2}{5}(1 + \alpha ) - \frac{2}{7}(\tilde{h} + \alpha ) - \beta /(\tilde{h} + \alpha )^{5/2} $$

(19)

with

$$ \beta = \frac{2}{5}\alpha^{5/2} (1 + \frac{2}{7}\alpha ). $$

(20)

Note that, as $\tilde{h}$ rises from zero, f increases from zero to a maximum and then falls back to zero again, at the maximum value of $\tilde{h}$, Since α is very small in comparison with unity, we have, approximately:

$$ \tilde{h}_{\hbox{max} } = \frac{7}{5} + \frac{2}{5}\alpha . $$

(21)

This expression is obtained by simply neglecting the last term in the right-hand side of Eq. [19] and setting it to zero. A better approximation is obtained by substituting Eq. [21] into the last term on the right-hand side of Eq. [19] before setting the equation to zero. The result is:

$$ \tilde{h}_{\hbox{max} } = \frac{7}{5} + \frac{2}{5}\alpha - \frac{7}{5}\alpha^{5/2} \frac{{1 + \frac{2}{7}\alpha }}{{\left\{ {\frac{7}{5}(1 + \alpha )} \right\}^{5/2} }} = \frac{7}{5} + \frac{2}{5}\alpha - \frac{5}{7}\sqrt {\frac{5}{7}} \alpha^{5/2} \frac{{1 + \frac{2}{7}\alpha }}{{(1 + \alpha )^{5/2} }}. $$

(22)

Noting that $\tilde{h}_{\hbox{max} }$ is larger than the steady-state value at $\tilde{h}_{\hbox{max} } = 1$, we see that the system is underdamped. We now compute the maximum value of f and, hence, the fluid velocity. Let $\tilde{h} = \tilde{h}_{m}$when f is maximized. The value of $\tilde{h}_{m}$is obtained by setting the derivative of $f({{d}}f/{{d}}\tilde{h})$ to zero:

$$ \frac{{{{d}}f}}{{{{d}}\tilde{h}}}_{{\tilde{h} = \tilde{h}_{m} }} = - \frac{2}{7} + \frac{5}{2}\beta /(\tilde{h}_{m} + \alpha )^{7/2} = 0. $$

(23)

Hence,

$$ \beta /(\tilde{h}_{m} + \alpha )^{5/2} = \frac{4}{35}(\tilde{h}_{m} + \alpha ). $$

(24)

Substituting Eq. [24] back into Eq. [19], with $\tilde{h}$ set to $\tilde{h}_{m}$, we acquire:

$$ f_{\hbox{max} } = \frac{2}{5}(1 - \tilde{h}_{m} ). $$

(25)

Now, substituting Eq. [20] into Eq. [24] and rearranging, we obtain:

$$ (\tilde{h}_{m} + \alpha )^{7/2} = \frac{7}{2}\alpha^{5/2} (1 + \frac{2}{7}\alpha ). $$

Then

$$ \tilde{h}_{m} = \left( {(1 + \frac{2}{7}\alpha )\frac{7}{2}\alpha^{5/2} } \right)^{2/7} - \alpha $$

(26)

General μ

In this case, Eq. [15] needs to be solved numerically, subject to the initial condition in Eq. [4]. By dividing the range of $\tilde{h}$ into intervals $[\tilde{h}_{k - 1} ,\tilde{h}_{k} ]$; k = 1, 2 … N, with $\tilde{h}_{0} = 0$, $\tilde{h}_{N} = \tilde{h}_{\text{s}}$. Let

$$ f_{k} = f(\tilde{h}_{k} ). $$

(27)

For

$$ \tilde{h}_{k} \le \tilde{h} < \tilde{h}_{k + 1} $$

(28)

we assume that $f = f_{k}$ is constant on the right-hand side of Eq. [15] during the integration process in the interval $\tilde{h}_{k} \le \tilde{h} < \tilde{h}_{k + 1}$. Equation [15] can therefore be approximated by:

$$ \frac{{d}}{{{{d}}\tilde{h}}}\left\{ {(\tilde{h} + \alpha )^{5/2} f} \right\} = (1 - \tilde{h})(\tilde{h} + \alpha )^{3/2} - 8\tilde{\mu }\tilde{h}(\tilde{h} + \alpha )^{3/2} \sqrt {2f_{k} } ;\quad \tilde{h}_{k} \le \tilde{h} < \tilde{h}_{k + 1} . $$

(29)

Equation [29] is solved with the condition:

$$ f(h_{k} ) = f_{k} . $$

(30)

Note that Eq. [29] may be written in abbreviated form as:

$$ \frac{{d}}{{{{d}}\tilde{h}}}\left\{ {(\tilde{h} + \alpha )^{5/2} f} \right\} = (1 - B_{k} \tilde{h})(\tilde{h} + \alpha )^{3/2} ;\quad \tilde{h}_{k} \le \tilde{h} < \tilde{h}_{k + 1} $$

(31)

with

$$ B_{k} = 1 + 8\tilde{\mu }\sqrt {2f_{k} } . $$

(32)

Integrating Eq. [31] over the interval $\left[ {\tilde{h}_{k} ,\tilde{h}_{k + 1} } \right]$ yields the following result:

$$ \begin{aligned} \left( {\tilde{h}_{k + 1} + \alpha } \right)^{5/2} f_{k + 1} - \left( {\tilde{h}_{k} + \alpha } \right)^{5/2} f_{k} = & \frac{2}{5}(1 + B_{k} \alpha )\left( {\tilde{h}_{k + 1} + \alpha } \right)^{5/2} - \frac{2}{7}B_{k} \left( {\tilde{h}_{k + 1} + \alpha } \right)^{7/2} \\ - \frac{2}{5}(1 + B_{k} \alpha )\left( {\tilde{h}_{k} + \alpha } \right)^{5/2} + \frac{2}{7}B_{k} \left( {\tilde{h}_{k} + \alpha } \right)^{7/2} . \\ \end{aligned} $$

(33)

Note that the only assumption made to arrive at the solution as in Eq. [33] is that B _k = constant inside the integral interval $\tilde{h}_{k} \le \tilde{h} < \tilde{h}_{k + 1}$. Also, since the nondimensional depth of penetration has been discretized, values of all $\tilde{h}_{k}$’s are known. Equation [33], therefore, can be used to solve for the nondimensional rate of penetration ($\tilde{v} = \sqrt {2f}$) by using the following equation:

$$ f_{k + 1} = \left( {\tilde{h}_{k + 1} + \alpha } \right)^{ - 5/2} \left[ {\left( {\tilde{h}_{k} + \alpha } \right)^{{\frac{5}{2}}} f_{k} + \frac{2}{5}\left( {1 + B_{k} \alpha } \right)\left( {\left( {\tilde{h}_{k + 1} + \alpha } \right)^{5/2} } \right) - \frac{2}{7}B_{k} \left( {\left( {\tilde{h}_{k + 1} + \alpha } \right)^{{\frac{7}{2}}} } \right) - \frac{2}{5}\left( {1 + B_{k} \alpha } \right)\left( {\left( {\tilde{h}_{k} + \alpha } \right)^{{\frac{5}{2}}} } \right) + \frac{2}{7}B_{k} \left( {\left( {\tilde{h}_{k} + \alpha } \right)^{7/2} } \right)} \right] . $$

(34)

The nondimensional time is evaluated by using Eq. [9]. By using the solution from Eq. [34] along with those from Eqs. [14] and [16], the relative importance of inertial $\left( {F_{\text{inertia}} } \right)$, gravitational ($F_{\text{g}}$), viscous ($F_{{{\mu }}}$), and end-drag ($F_{{\text{end}}{\text{-}}{\text{drag}}}$) forces with respect to the surface tension ($F_{\sigma }$) force can be computed as $\sqrt {2f} + \left( {\tilde{h} + {{\alpha }}} \right){{d}}f/{{d}}\tilde{h}$ (inertial/surface tension forces), $\tilde{h}$ (gravity/surface tension forces), $8\tilde{\mu }\tilde{h}\sqrt {2f}$ (viscous/surface tension forces), and $f/2$ (end-drag/surface tension forces), respectively.