In the first part of this section, we summarise the work of [25], in which an exact equation for the droplet’s evaporation rate is derived in terms of the droplets half-width, w, and contact angle, \(\theta \). We shall then investigate special cases where the evaporation rate can be written explicitly. We introduce the local inclination angle, \(\phi \in [-\pi ,\pi ]\), which is defined as the angle formed against the horizontal, to describe the surface of the droplet as
$$\begin{aligned} x&= -w \frac{\sin \phi }{\sin \theta }, \end{aligned}$$
(9a)
$$\begin{aligned} h&= -w\frac{\cos \theta -\cos \phi }{\sin \theta }. \end{aligned}$$
(9b)
We note that this formulation is applicable for all contact angles, unlike a direct formulation using h(x). Since the shape of the droplet is a circular segment, we use the conformal mapping
$$\begin{aligned} \frac{z-w}{z+w} = \left( \frac{\zeta -a}{\zeta +a}\right) ^\sigma , \end{aligned}$$
(10)
where \(z=x+\mathrm{i}y\) is the complex variable in the physical plane, \(\sigma =2(1-\theta /\pi )\), \(a = w/\sigma \), and \(\zeta \) is the complex variable in the transformed plane, see Fig. 2.
We note that (10) maps the z-plane onto the \(\zeta \)-plane, and in particular, a circular arc of width w and contact angle \(\theta \) is mapped to the perimeter of the upper half circle with radius a. In the transformed space, we can solve the Laplace system (7) to find the complex potential, \(\chi \), (which satisfies \(\mathrm {Re}(\chi )=c\)) given by
$$\begin{aligned} \chi (\zeta ) = 1-\frac{\ln {(\zeta /a)}}{\ln {(R/a)}}. \end{aligned}$$
(11)
Taking the derivative of \(\chi \) with respect to z, we find the mass flux across the surface
$$\begin{aligned} \dfrac{\mathrm {d}^{}{\chi }}{\mathrm {d}{z}^{}} = -\frac{1}{\zeta \ln {(R/a)}} \frac{4a^2 g^{\frac{1-\sigma }{\sigma }}}{(z+w)^2(1-g^\frac{1}{\sigma })^2}, \end{aligned}$$
(12)
where \(g(z) = (z-w)/(z+w)\). It follows that the normal derivative of the concentration is given by
$$\begin{aligned} \frac{\partial ^{}{c}}{\partial {n}^{}} = \hat{{\varvec{{n}}}}_x \mathrm {Re}\left( \dfrac{\mathrm {d}^{}{\chi }}{\mathrm {d}{z}^{}}\right) - \hat{{\varvec{{n}}}}_y \mathrm {Im}\left( \dfrac{\mathrm {d}^{}{\chi }}{\mathrm {d}{z}^{}}\right) , \end{aligned}$$
(13)
where \(\hat{{\varvec{{n}}}}_x,\hat{{\varvec{{n}}}}_y\) are the x, y components of the outward unit normal to the droplet’s interface. Hence, we can use (13) to find the evaporation rate, (8), which, in these transformed coordinates, is given as
$$\begin{aligned} \dfrac{\mathrm {d}^{}{V}}{\mathrm {d}{t}^{}} = \frac{w}{\sin \theta } \int ^{\theta }_{-\theta } \left. \frac{\partial ^{}{c}}{\partial {n}^{}}\right| _{(x(\phi ),h(\phi ))} \,\mathrm {d}{\phi }, \end{aligned}$$
(14)
with \(x(\phi )\) and \(h(\phi )\) given in (9). Equation (14) is the principle result of this section. It is an exact equation that predicts the evaporation rate of a droplet for a given width w and contact angle \(\theta \). In the following, we study two special cases for which it is possible to find asymptotic solutions of Eq. (14).
Limiting cases
We start by considering the case of nearly neutral wetting, i.e. the droplet interface is close to a semi-circle with the contact angle satisfying \(\left| {\theta -\pi /2}\right| \ll 1\). Observe that the surface of the droplet is then described, in polar coordinates \((r,\varphi )\), by
$$\begin{aligned} r = w(1+\varepsilon \sin \varphi ) +\mathcal {O}(\varepsilon ^2), \end{aligned}$$
(15)
where \(\varepsilon = \theta -\pi /2\). We consider a regular expansion for the concentration in terms of \(\varepsilon \). By solving the resulting leading and first-order systems, we find that the concentration is given by (see Appendix A for a derivation)
$$\begin{aligned} c = 1- \frac{\ln \frac{r}{w}}{\ln \frac{R}{w}} + \varepsilon \left[ \frac{2}{\pi \ln \frac{R}{w}}\left( 1- \frac{\ln \frac{r}{w}}{\ln \frac{R}{w}}\right) + \frac{4}{\pi \ln \frac{R}{w}}\sum _{n=1}^\infty \frac{\cos (2n\varphi )}{1-4n^2}\frac{\left( \frac{r}{R}\right) ^{2n}-\left( \frac{R}{r}\right) ^{2n}}{\left( \frac{w}{R}\right) ^{2n}-\left( \frac{R}{w}\right) ^{2n}}\right] +\mathcal {O}(\varepsilon ^2). \end{aligned}$$
(16)
By noting that the outward unit normal vector satisfies
$$\begin{aligned} \hat{{\varvec{{n}}}} = \hat{{\varvec{{r}}}} - (\varepsilon \cos \varphi ) \hat{{\varvec{{\varphi }}}} +\mathcal {O}(\varepsilon ^2), \end{aligned}$$
(17)
and applying Eq. (16) to (8) we find
$$\begin{aligned} \dfrac{\mathrm {d}^{}{V}}{\mathrm {d}{t}^{}} = -\frac{\pi }{\ln \frac{R}{w}} - \frac{2\varepsilon }{\left( \ln \frac{R}{w}\right) ^2} + \mathcal {O}(\varepsilon ^2), \end{aligned}$$
(18)
which gives the evaporation rate of a nearly neutral wetting setting.
The second special case we consider is the extremely hydrophilic droplet, i.e. a droplet with asymptotically small contact angle. This case has been examined in [24], and here, we give a brief summary of that method.
In this method, the far-field condition is not applied on the boundary of a semi-circle; instead, an upper semi-ellipse is used. This choice of fictitious infinity is to allow a sufficiently tractable conformal mapping solution. Using the upper semi-ellipse with major-axis \(\sqrt{R^2+1}\) that is aligned with the x-axis and minor-axis R as the fictitious infinity, i.e.
$$\begin{aligned} \left( \frac{x}{\sqrt{1+R^2}}\right) ^2+\left( \frac{y}{R}\right) ^2 = 1, \end{aligned}$$
(19)
we see that the domain is mapped to the rectangular domain \((u,v)\in (-1,1)\times (0,S)\) by the conformal mapping
$$\begin{aligned} z = -w\sin \left( \frac{\pi }{2}\zeta \right) , \end{aligned}$$
(20)
where \(\zeta = u+\mathrm{i}v\) and S satisfies \(R = 2\sinh \left( \frac{\pi S}{w}\right) \). In the transformed space, we obtain the complex potential
$$\begin{aligned} \chi = 1-\frac{v}{S}=1-\frac{\text {Im}(\zeta )}{S}. \end{aligned}$$
(21)
This result is then mapped back to the z-plane and applied to (8) to find
$$\begin{aligned} \left. \dfrac{\mathrm {d}^{}{V}}{\mathrm {d}{t}^{}}\right| _{\theta =0} = -\frac{\pi }{\mathrm {asinh}(R/w)}. \end{aligned}$$
(22)
Numerical results for homogeneous surfaces
We present here numerical results of Eq. (14) for a given and constant contact angle. In particular, in this section, we examine the instantaneous evaporation rate for a defined droplet, while the examination of a droplet lifetime is discussed in Sect. 5. We note that to fully determine the solution of the droplet, we need to either fix the volume V of the droplet or its width w. We consider each of these two cases in turn.
We start by fixing the width of the droplet to \(w=1\). As we vary the contact angle, the droplet’s volume is then given by Eq. (1). Figure 3a shows numerical results of Eq. (14) for different values of R. The blue and red lines represent the results from the asymptotic limiting cases given by Eq. (22) (nearly complete wetting) and (18) (nearly neutral wetting), respectively. We see that in both cases, there is excellent agreement between numerical and asymptotic results. We note that for the case of \(\theta =0\), the asymptotic result becomes more accurate as R increases (see inset panel of Fig. 3a). This is because in the analytical treatment of the zero contact angle case, the far-field conditions are applied on the elliptic curve (19) which tends to the circular arc of radius R as \(R\rightarrow \infty \). We also see that the asymptotic results derived here for nearly neutral wetting meet the exact results at a tangent to the point \(\theta =\pi /2\), as expected [24] .
Figure 3b shows the evaporation rate as a function of the contact angle when the volume of the droplet is fixed. Under this condition, the droplet’s width is given by
$$\begin{aligned} w =\sin \theta \left( \frac{2V}{2\theta -\sin 2\theta }\right) ^{1/2}. \end{aligned}$$
(23)
The red lines in Fig. 3b represent the asymptotic analysis given by Eq. (18), where we can see, again, that the two lines meet in tangent at \(\theta =\pi /2\). In order to conserve the volume as \(\theta \rightarrow 0\), we observe that \(w\rightarrow +\infty \), and therefore, the extremely hydrophilic case, given by (22), predicts that the volume change rate tends to negative infinity. This agrees with the numerical results shown in Fig. 3b.