On the modelling of joint formation in dissolutive brazing processes

Abstract

In this paper, earlier dissolutive wetting models describing the dynamics of an axisymmetric alloy drop spreading on pure metal substrate are extended to describe reactive wetting and subsequent joint formation in brazing processes. A two-dimensional time-dependent problem is formulated, and the model equations are nondimensionalized, revealing the possibilities for asymptotic model reduction. Whilst the numerical solution of the time-dependent problem, which contains two moving contact lines and would not in general be amenable to lubrication theory, is relegated to future work, the steady-state problem is analyzed in detail. The analysis offers an arguably more transparent alternative to an earlier energy minimization approach for finding the location of the meniscus, which ultimately constitutes the joint. The results of the present model are found to compare favourably to those of earlier experimental and theoretical work.

Appendices

Appendix A: Determining the velocity scale, $\left[ u\right]$

In addition to $\left[ u\right] =D/l,$ another possibility is to take $\left[ u\right] =D\varGamma /l^{2}.$ In this case, we have

$$\begin{aligned} \left[ t\right] =\frac{l}{\left[ u\right] }=\frac{l^{3}}{D\varGamma }, \end{aligned}$$

(A.1)

and (4.19) and (4.20) become

$$\begin{aligned} \dot{X}_{1}&=\left. -\left\{ \frac{\tan \theta _{L}}{c\tan \theta _{b,x_{1}} }\left( \frac{F_{XX}}{\left( 1+F_{X}^{2}\right) ^{3/2}}\right) _{X}\right\} \right| _{\left( X_{1}\left( \tau \right) ,0\right) }, \end{aligned}$$

(A.2)

$$\begin{aligned} \dot{Y}_{1}&=\left. -\left\{ \frac{\tan \theta _{L}}{c\tan \theta _{b,y_{1}} }\left( \frac{F_{XX}}{\left( 1+F_{X}^{2}\right) ^{3/2}}\right) _{X}\right\} \right| _{\left( 0,Y_{1}\left( \tau \right) \right) }, \end{aligned}$$

(A.3)

respectively. Also, (4.5) and (4.17) become

$$\begin{aligned}&\frac{\partial c}{\partial \tau }+U\frac{\partial c}{\partial X}+V\frac{\partial c}{\partial Y}=\frac{1}{\delta }\left( \frac{\partial ^{2}c}{\partial X^{2} }+\frac{\partial ^{2}c}{\partial Y^{2}}\right) , \end{aligned}$$

(A.4)

$$\begin{aligned}&\frac{\delta c}{\sqrt{1+F_{X}^{2}}}\frac{\partial F}{\partial \tau } =\mathbf {n}_{F}\cdot \left( c\mathbf {u}-D\nabla c\right) \quad \text {at }Y=F\left( X,\tau \right) , \end{aligned}$$

(A.5)

respectively. It is now convenient to refer to Table 1, which contains the model parameters. Now, since $\delta \ll 1,$ (A.4) and (A.5) reduce to

$$\begin{aligned}&\frac{\partial ^{2}c}{\partial X^{2}}+\frac{\partial ^{2}c}{\partial Y^{2}} \approx 0, \end{aligned}$$

(A.6)

$$\begin{aligned}&\mathbf {n}_{F}\cdot \nabla c \approx 0\quad \text {at }Y=F\left( X,\tau \right) ; \end{aligned}$$

(A.7)

i.e. this is already the steady state. Also, from (A.1), we would obtain $\left[ t\right] \sim 10^{10}$ s, and it is clear that we must have taken the wrong choice for $\left[ u\right] .$

Appendix B: $\varepsilon \ll 1$

In this case, the F and H problems decouple in an interesting way. From (5.31), and assuming that $\mathcal {A}\sim O\left( 1\right) ,$ we now expect that

$$\begin{aligned} \int _{0}^{X_{*}}F\left( X\right) \mathrm {d}X\sim O\left( \varepsilon \right) . \end{aligned}$$

(B.1)

For $Y<0,$ this should mean that $X\sim O\left( 1\right) ,F\sim O\left( \varepsilon \right) ,$ whereas for $Y>0,$ we expect $F\sim O\left( 1\right)$ when $X\sim O\left( \varepsilon \right)$.

Considering the first case, if $F=\varepsilon \bar{F},$ with the expectation that $\bar{F}$ is an $O\left( 1\right)$ function, implies that (5.24) becomes, at leading order in $\varepsilon ,$

$$\begin{aligned} \frac{\mathrm {d}^{2}\bar{F}}{\mathrm {d}X^{2}}=\mathcal {\bar{C}}_{2}, \end{aligned}$$

(B.2)

where $\mathcal {C}_{2}=\mathcal {\bar{C}}_{2}/\varepsilon ,$ with $\mathcal {\bar{C}}_{2}$ an $O\left( 1\right)$ constant, whereas (5.28) and (5.29) reduce to

$$\begin{aligned}&\bar{F}\left( X_{*}\right) =0, \end{aligned}$$

(B.3)

$$\begin{aligned}&H_{X}\left( X_{*}\right) =-\tan \theta _{L}, \end{aligned}$$

(B.4)

respectively. For the second case, we set $X=\varepsilon \bar{X},$ so that (5.24) becomes

$$\begin{aligned} \frac{\mathrm {d}^{2}F}{\mathrm {d}\bar{X}^{2}}=\mathcal {\bar{C}}_{2}\left( \frac{\mathrm {d}F}{\mathrm {d}\bar{X}}\right) ^{3}, \end{aligned}$$

(B.5)

with (5.26) reducing to

$$\begin{aligned} H_{X}\left( 0\right) =-\cot \theta _{L}. \end{aligned}$$

(B.6)

Note that, at this stage, (B.2) and (B.5) are consistent with each other, and the implication is that the radius of curvature of F,  which is in effect $1/\mathcal {C}_{2},$ is small. Also, we can note that the problem for H has now decoupled from that for F. More specifically, we have, for H,  Eq. (5.23) subject to

$$\begin{aligned}&H_{X}\left( 0\right) =-\cot \theta _{L}, \end{aligned}$$

(B.7)

$$\begin{aligned}&H\left( X_{*}\right) =0,\end{aligned}$$

(B.8)

$$\begin{aligned}&H_{X}\left( X_{*}\right) =-\tan \theta _{L},\end{aligned}$$

(B.9)

$$\begin{aligned}&\int _{0}^{X_{*}}H\left( X\right) \mathrm {d}X =\mathcal {A}; \end{aligned}$$

(B.10)

so, there are enough constraints to determine $H,\mathcal {C}_{1}$ and $X_{*},$ and thus $H\left( 0\right) .$

Returning to the problem for F,  we have

$$\begin{aligned} \frac{\mathrm {d}^{2}\bar{F}}{\mathrm {d}X^{2}}=\mathcal {\bar{C}}_{2}, \end{aligned}$$

(B.11)

subject to

$$\begin{aligned} \bar{F}\left( X_{*}\right) =0, \end{aligned}$$

(B.12)

and

$$\begin{aligned} \frac{\mathrm {d}^{2}F}{\mathrm {d}\bar{X}^{2}}=\mathcal {\bar{C}}_{2}\left( \frac{\mathrm {d}F}{\mathrm {d}\bar{X}}\right) ^{3}, \end{aligned}$$

(B.13)

subject to

$$\begin{aligned} F\left( 0\right) =H\left( 0\right) . \end{aligned}$$

(B.14)

Each of these problems is clearly missing a boundary condition, and hence two more conditions would be required. One would hope for continuity of F and its derivative, so that the two problems would “join up”, in the sense that F and $\mathrm {d}F/\mathrm {d}X$ as computed by the two problems should be continuous at some value of X that would have to be determined. However, it is impossible to achieve this, whilst at the same time ensuring that the curvature of F is constant. Thus, the only apparent possibility is that

$$\begin{aligned} F\left( 0\right) =0, \end{aligned}$$

(B.15)

and hence that

$$\begin{aligned} \bar{F}\left( 0\right) =0, \end{aligned}$$

(B.16)

which abandons the requirement that $\mathrm {d}F/\mathrm {d}X$ is continuous. In this case, dissolution does not start at all from the point (0,0). Note that (B.15) will mean that F is multivalued at $X=0,$ and indeed for $X<0.$

The remaining problem for $\bar{F}$ is (B.11), subject to (B.12) and (B.16), and we obtain

$$\begin{aligned} \bar{F}=\frac{1}{2}\mathcal {\bar{C}}_{2}X\left( X-X_{*}\right) . \end{aligned}$$

(B.17)

The remaining problem for F is (B.13), subject to (B.14) and (B.15). To solve, it is easier to rewrite the problem as

$$\begin{aligned} \frac{\mathrm {d}^{2}\bar{X}}{\mathrm {d}F^{2}}=\mathcal {\bar{C}}_{2}, \end{aligned}$$

(B.18)

subject to

$$\begin{aligned}&\bar{X}\left( 0\right) =0, \end{aligned}$$

(B.19)

$$\begin{aligned}&\bar{X}\left( Y_{*}\right) =0, \end{aligned}$$

(B.20)

which leads to the solution

$$\begin{aligned} \bar{X}=-\frac{1}{2}\mathcal {\bar{C}}_{2}F\left( F-Y_{*}\right) . \end{aligned}$$

(B.21)

Finally, $\mathcal {\bar{C}}_{2}$ will be given from (5.31) by

$$\begin{aligned} \frac{1}{2}\mathcal {\bar{C}}_{2}\int _{0}^{X_{*}}X\left( X_{*}-X\right) \mathrm {d}X+\frac{1}{2}\mathcal {\bar{C}}_{2}\int _{0}^{Y_{*}}F\left( Y_{*}-F\right) \mathrm {d}F=\mathcal {A}, \end{aligned}$$

(B.22)

leading to

$$\begin{aligned} \mathcal {\bar{C}}_{2}=\frac{12\mathcal {A}}{X_{*}^{3}+Y_{*}^{3}}. \end{aligned}$$

(B.23)

So,

$$\begin{aligned} F=\left\{ \begin{array} [c]{ll} \frac{1}{2}\varepsilon \mathcal {\bar{C}}_{2}X\left( X-X_{*}\right) &{}\quad X>0,\\ \frac{1}{2}\left\{ Y_{*}\pm \sqrt{Y_{*}^{2}+\frac{8\varepsilon }{\mathcal {\bar{C}}_{2}}X}\right\} &{}\quad X<0. \end{array} \right. \end{aligned}$$

(B.24)

Fig. 9

f and h ($\theta _{L}=0.1,\varepsilon =0.445$), obtained using the assumption that $\varepsilon \ll 1$

A preliminary result for f and h is given in Fig. 9 using $\theta _{L}=0.1$$\left( 5.71^{\text {o}}\right)$ and $T=615\,^{\circ }$C, corresponding to $\varepsilon =0.445.$ Note that the sum of the area in $y<0$ and $x<0$ should be 0.445 that in $x,y,>0.$

Furthermore, Fig. 10 shows the extent of the dissolution zone for $T=595\,{^\circ }$C, 615 ${^\circ }$C and $625\,{^\circ }$C. Note that the curve for $T=595\,{^\circ }$C is literally on top of the x- and y-axes. For this $\varepsilon \ll 1$ analysis, these curves are calculated using the results from the $O\left( 1\right)$ problem for H;  thus, $X_{*}$ and $Y_{*}$ are the same for all of these curves.

Fig. 10

f ($\theta _{L}=0.1, T=595\,^{\circ }$C, 615 $^{\circ }$C, 625 ${^\circ }$C), obtained using the assumption that $\varepsilon \ll 1$

Appendix C: A one-circle dissolution zone for $\varepsilon \ll 1$?

It was seen in Appendix B that if $\varepsilon \ll 1,$ then it is not possible that a single circular arc will join the two contact lines. However, it is not yet clear that this is the case if $\varepsilon$ is larger, as will be the case when the difference between $c_{0}$ and $c_{E}\left( T\right)$ is large, or $c_{E}\left( T\right)$ is sufficiently small. Thus, we now consider this situation.

Suppose that such a circular arc exists and that it is part of the circle

$$\begin{aligned} \left( X-X_{0}\right) ^{2}+\left( Y-Y_{0}\right) ^{2}=R^{2}, \end{aligned}$$

(C.1)

where $X_{0},Y_{0}$ and R are to be determined; then we would need

$$\begin{aligned} X_{0}^{2}+\left( Y_{*}-Y_{0}\right) ^{2}&=R^{2}, \end{aligned}$$

(C.2)

$$\begin{aligned} \left( X_{*}-X_{0}\right) ^{2}+Y_{0}^{2}&=R^{2}. \end{aligned}$$

(C.3)

Also, since

$$\begin{aligned} \left( X-X_{0}\right) +\left( Y-Y_{0}\right) \frac{\mathrm {d}Y}{\mathrm {d}X}=0, \end{aligned}$$

(C.4)

we have

$$\begin{aligned} F_{X}\left( 0\right) =\frac{X_{0}}{Y_{*}-Y_{0}},\quad F_{X}\left( X_{*}\right) =\frac{X_{*}-X_{0}}{Y_{0}}. \end{aligned}$$

(C.5)

We observe that

$$\begin{aligned} X_{0}^{2}+\left( Y_{*}-Y_{0}\right) ^{2}=\left( X_{*}-X_{0}\right) ^{2}+Y_{0}^{2}, \end{aligned}$$

(C.6)

which gives

$$\begin{aligned} Y_{*}^{2}-2Y_{*}Y_{0}=X_{*}^{2}-2X_{*}X_{0}, \end{aligned}$$

(C.7)

whence

$$\begin{aligned} Y_{0}=\frac{Y_{*}^{2}-X_{*}^{2}+2X_{*}X_{0}}{2Y_{*}}. \end{aligned}$$

(C.8)

Also,

$$\begin{aligned} \frac{1}{2}R^{2}\theta -X_{0}Y_{0}-\frac{1}{2}X_{0}\left( Y_{*} -Y_{0}\right) -\frac{1}{2}Y_{0}\left( X_{*}-X_{0}\right) =\varepsilon \mathcal {A}, \end{aligned}$$

(C.9)

where

$$\begin{aligned} \theta =\frac{\pi }{2}+\tan ^{-1}\left( \frac{Y_{*}-Y_{0}}{X_{0}}\right) +\tan ^{-1}\left( \frac{X_{*}-X_{0}}{Y_{0}}\right) ; \end{aligned}$$

(C.10)

here, (C.9) comes from Eq. (5.31), i.e. the conservation of solute condition.

Numerical experimentation to determine f and h indicated that there are not only great difficulties in finding a numerical solution, but also difficulties in finding a good initial guess; it is necessary to find this because the governing equations are nonlinear. Moreover, it is not even known if a solution exists. Thus, it turns out to be more worthwhile to consider what would be the properties of such a solution, if it exists. We know that we must have

$$\begin{aligned} H_{X}\left( 0\right)&=\frac{X_{0}+\left( Y_{*}-Y_{0}\right) \tan \theta _{L}}{Y_{*}-Y_{0}-X_{0}\tan \theta _{L}}, \end{aligned}$$

(C.11)

$$\begin{aligned} H_{X}\left( X_{*}\right)&=\frac{X_{*}-X_{0}-Y_{0}\tan \theta _{L} }{Y_{0}+\left( X_{*}-X_{0}\right) \tan \theta _{L}}, \end{aligned}$$

(C.12)

which we can rewrite as

$$\begin{aligned} X_{0}+\left( Y_{*}-Y_{0}\right) \tan \theta _{L}&=\left( Y_{*} -Y_{0}-X_{0}\tan \theta _{L}\right) H_{X}\left( 0\right) , \end{aligned}$$

(C.13)

$$\begin{aligned} X_{*}-X_{0}-Y_{0}\tan \theta _{L}&=\left( Y_{0}+\left( X_{*} -X_{0}\right) \tan \theta _{L}\right) H_{X}\left( X_{*}\right) , \end{aligned}$$

(C.14)

i.e.

$$\begin{aligned} \left( \begin{array} [c]{ll} a &{}\quad b\\ c &{}\quad d \end{array} \right) \left( \begin{array} [c]{l} X_{0}\\ Y_{0} \end{array} \right) =\left( \begin{array} [c]{l} e\\ f \end{array} \right) , \end{aligned}$$

(C.15)

where

$$\begin{aligned} a&=1+H_{X}\left( 0\right) \tan \theta _{L}, \end{aligned}$$

(C.16)

$$\begin{aligned} b&=H_{X}\left( 0\right) -\tan \theta _{L},\end{aligned}$$

(C.17)

$$\begin{aligned} c&=H_{X}\left( X_{*}\right) \tan \theta _{L}-1,\end{aligned}$$

(C.18)

$$\begin{aligned} d&=-H_{X}\left( X_{*}\right) -\tan \theta _{L},\end{aligned}$$

(C.19)

$$\begin{aligned} e&=Y_{*}\left( H_{X}\left( 0\right) -\tan \theta _{L}\right) ,\end{aligned}$$

(C.20)

$$\begin{aligned} f&=X_{*}\left( H_{X}\left( X_{*}\right) \tan \theta _{L}-1\right) , \end{aligned}$$

(C.21)

so that

$$\begin{aligned} \left( \begin{array} [c]{c} X_{0}\\ Y_{0} \end{array} \right) =\frac{1}{ad-bc}\left( \begin{array} [c]{c} de-bf\\ -ce+af \end{array} \right) . \end{aligned}$$

(C.22)

Now, since

$$\begin{aligned} 2Y_{*}Y_{0}=Y_{*}^{2}-X_{*}^{2}+2X_{*}X_{0}, \end{aligned}$$

(C.23)

from (C.8) we have

$$\begin{aligned} 2Y_{*}\left( af-ce\right) =\left( Y_{*}^{2}-X_{*}^{2}\right) \left( ad-bc\right) +2X_{*}\left( de-bf\right) , \end{aligned}$$

(C.24)

and we need to consider whether

$$\begin{aligned} \phi :=2Y_{*}\left( af-ce\right) -\left( Y_{*}^{2}-X_{*} ^{2}\right) \left( ad-bc\right) -2X_{*}\left( de-bf\right) \end{aligned}$$

(C.25)

is ever zero or not. Here, the idea is that we solve (5.10), subject to (5.12), (5.16) and

$$\begin{aligned}&H_{X}\left( 0\right) =-\tan \left( \pi /2-\alpha \right) , \end{aligned}$$

(C.26)

$$\begin{aligned}&H_{X}\left( X_{*}\right) =-\tan \beta , \end{aligned}$$

(C.27)

where $\alpha$ and $\beta$ are the angles that the curve $Y=H\left( X\right)$ makes with the Y- and X-axes, respectively; note that this computation is independent of $\theta _{L},$ and that we merely sweep over all values of $\alpha$ and $\beta ,$ where $0<\alpha ,\beta <\pi /2$. Each computation generates values for $X_{*}$ and $Y_{*},$ which are then used in (C.25). If it is found that $\phi$ is never zero, then this is an indication that $X_{0}$ and $Y_{0}$ cannot be found, and hence the sought-after circle does not exist. Numerical experimentation suggests that this indeed is the case. As an example, Fig. 11 shows $\phi$ vs. $\alpha$ for $\theta _{L}=0,\pi /4,\pi /2,$ for $\beta =\pi /3$, indicating that $\phi$ is never zero, no matter what value of $\theta _{L}$ is tried. Similar graphs were also obtained for other values of $\beta$.

Fig. 11

$\phi$ vs. $\alpha$ for $\theta _{L}=0,\pi /4,\pi /3$ ($\beta =\pi /3$)

Appendix D: $\theta _{L}=0,\varepsilon =0$

When $\varepsilon =0,$ we have $F\equiv 0,$ and Eqs. (5.23)–(5.31) reduce to just

$$\begin{aligned} \frac{H_{XX}}{\left( 1+H_{X}^{2}\right) ^{3/2}}=H+\mathcal {C}_{1}, \end{aligned}$$

(D.1)

subject to

$$\begin{aligned}&H_{X}\left( 0\right) =-\cot \theta _{L}, \end{aligned}$$

(D.2)

$$\begin{aligned}&H\left( X_{*}\right) =0,\end{aligned}$$

(D.3)

$$\begin{aligned}&H_{X}\left( X_{*}\right) =-\tan \theta _{L},\end{aligned}$$

(D.4)

$$\begin{aligned}&\int _{0}^{X_{*}}H\left( X\right) \mathrm {d}X =\mathcal {A}. \end{aligned}$$

(D.5)

Multiplying (D.1) by $H_{X}$, integrating with respect to X and applying (D.3) and (D.4)

$$\begin{aligned} -\frac{1}{\left( 1+H_{X}^{2}\right) ^{1/2}}=\frac{1}{2}H^{2}+\mathcal {C} _{1}H-\cos \theta _{L}. \end{aligned}$$

(D.6)

Alternatively, integrating (D.1) with respect to X and applying (D.2), (D.4) and (D.5) gives

$$\begin{aligned} \cos \theta _{L}-\sin \theta _{L}=\mathcal {A}+\mathcal {C}_{1}X_{*} \end{aligned}$$

(D.7)

Now, consider the behaviour of H near $X=X_{*}$ when $\theta _{L}=0.$ Since $H,H_{X}\ll 1,$ we have, from (D.6),

$$\begin{aligned} \frac{H_{X}^{2}}{2}\sim \frac{H^{2}}{2}+\mathcal {C}_{1}H. \end{aligned}$$

(D.8)

If $\mathcal {C}_{1}\ne 0,$ we have just

$$\begin{aligned} \frac{H_{X}^{2}}{2}\sim \mathcal {C}_{1}H, \end{aligned}$$

(D.9)

which leads to

$$\begin{aligned} H\sim \frac{1}{2}\mathcal {C}_{1}\left( X_{*}-X\right) ^{2}. \end{aligned}$$

(D.10)

However, note that if $\mathcal {C}_{1}<0,$ which is clearly possible from (D.7) if $\mathcal {A}>1,$ then $H<0,$ which would be physically unrealistic. On the other hand, if $\mathcal {C}_{1}=0,$ i.e. $\mathcal {A}=1,$ we have, instead of (D.9),

$$\begin{aligned} H_{X}^{2}\sim H^{2}, \end{aligned}$$

(D.11)

which would give $H\sim e^{\pm X}$. It is clear that only the behaviour with the minus sign is of relevance here. Taking this one, we see that an alternative to (D.10) is the possibility that the contact line extends to infinity with $H\sim e^{-X}$ as $X\rightarrow \infty .$ In fact, for this case, there is an analytical solution for the meniscus given by [30,31,32]

$$\begin{aligned} X=\sqrt{2}-\ln \left( 1+\sqrt{2}\right) -\sqrt{4-H^{2}}+\ln \left( \frac{2+\sqrt{4-H^{2}}}{H}\right) . \end{aligned}$$

(D.12)

Finally, it is worth noting that, from the data in Table 1 and the values for $h_{0}$ and $w_{0}$ used in this paper, $\mathcal {A}\approx 0.02,$ indicating that the relevant behaviour is that given by Eq. (D.10).