5.1. Hanging, intermediate and wrapping droplets

We perform three-dimensional numerical simulations for the above-mentioned configuration and find that the heavier droplet hangs from the lighter liquid interface for certain diameters and impact velocity of the droplet. Figure 4(a) shows the evolution of a hanging droplet (see supplementary movie 1 available at https://doi.org/10.1017/jfm.2023.137) on the impact of a drop of diameter $D=2$ mm, released from height $h = 4.98$ mm (impact velocity $\sqrt {2gh} = 0.313\ {\rm m}\ {\rm s}^{-1}$). Here, $t^*=t/\tau _c$, where capillary time $\tau _c$ is defined as $\tau _c=\sqrt {\rho _1 D^3/\sigma _{12}}$. As the droplet makes a transition from hanging to sinking, an intermediate case (see supplementary movie 4) is also observed, as shown in figure 4(b) for a droplet of diameter $0.6$ mm released from height of $h = 61.67$ mm (impact velocity $\sqrt {2gh} = 1.1\ {\rm m}\ {\rm s}^{-1}$). Figure 4(c) shows the case for a $2$ mm diameter droplet released from height $h = 50.97$ mm (impact velocity $\sqrt {2gh} = 1\ {\rm m}\ {\rm s}^{-1}$), in which the droplet sinks into the pool upon impact (see supplementary movie 7). It is observed that the droplet begins to slow down even before it makes contact with the pool. As the drop moves closer to the pool, a thin film of air separates the droplet (Duchemin & Josserand Reference Duchemin and Josserand2020) from the pool. This acts as a cushion and is responsible for the decrease in the impact velocity of the droplet. As the droplet makes contact with the pool, the TPCL diameter expands rapidly. After the impact, the droplet loses its kinetic energy drastically by displacing a portion of the pool towards the pool surface. This creates a crater in the pool, shrinking the TPCL diameter. The droplet sits in this crater and hangs from the surface. After the droplet loses all its kinetic energy, it starts moving upwards and keeps oscillating with a very small amplitude until it reaches an equilibrium height. The TPCL diameter again increases during this process. The evolution of the non-dimensional TPCL diameter with respect to the non-dimensional time for different hanging droplet cases is shown in figure 5(a). A capillary wave (Che & Matar Reference Che and Matar2018) is formed upon the impact of the droplet. It is also observed that the equilibrium height and the shape of the hanging droplet are independent of the release height of the droplet as long as it hangs from the surface. This independence of the final shape of the droplet on the impact velocity or release height differs from the results presented by Xie et al. (Reference Xie, Forth, Zhu, Helms, Ashby, Shum and Russell2020) owing to the representation of the coacervate with a single surface.

Figure 4.

Evolution of (a) hanging droplet of 2 mm diameter released from 4.98 mm height, (b) intermediate droplet of 0.6 mm diameter released from 61.67 mm height, and (c) wrapping droplet of 2 mm diameter released from 50.97 mm height.

Figure 5.

(a) Variation of non-dimensional TPCL diameter with respect to the non-dimensional time for different hanging droplet cases. Here, $d^*$ is defined as $d^*=d/D$, where $d$ is the TPCL diameter. (b) Variation of the non-dimensional height of the droplet for different cases of hanging and wrapping droplets. State diagrams for hanging and wrapping droplets as functions of (c) height and diameter, (d) excess energy and diameter.

Figure 5(b) shows the variation of the height of the centre of mass of the droplet from the pool surface for various cases. It can be observed that there exists a critical depth upon crossing which the droplet gets wrapped. Droplets that do not cross this critical depth tend to hang from the pool surface. The critical depth is found to be $0.74$ times the diameter of the droplet. We note that the critical depth is greater than half the diameter of the droplet, i.e. for a brief moment, the droplet goes completely below the pool surface, displacing the pool fluid. Since the computational domain is taken as a closed container such that no fluid exits the domain, the displaced fluid increases the pool height. Increased pool height results in additional pressure head which pushes the droplet upwards. However, if the pool height increases significantly, then it covers the top surface of the droplet and wraps it completely. The droplet sinks when it gets wrapped by the pool fluid.

Simulations for different droplet diameters and release heights are performed (see supplementary movies 1–9). It is observed that the tendency of a droplet to hang from the pool surface increases as the droplet radius or the release height is reduced. A droplet of diameter $2$ mm released from height $10$ mm gets wrapped and sinks into the pool. In contrast, a droplet of diameter $0.4$ mm released from height $60$ mm hangs from the pool surface. Figure 5(c) shows both the hanging and wrapping states as functions of the droplet diameter and the release height. A nonlinear curve divides both states. There are also a few cases that lie very close to the curve dividing the two states. These are the cases where a droplet, upon impact with the pool, briefly gets stuck at the pool surface and slowly moves downwards, eventually sinking into the pool. The cases close to the curve dividing the two states are the intermediate cases.

5.2. Energy balance for hanging droplets

Empirical energy calculations are performed to justify hanging and wrapping states for different cases. It is assumed that the droplet is released from height $h$, and the entire potential energy is converted into kinetic energy at the time of impact:

(5.1)\begin{equation} E_{KE}=\frac{\rm \pi}{6}\,D^3\rho_1 gh. \end{equation}

This is the entire energy available with the droplet that is used to overcome different forms of energy requirements. Three different forms of energy loss are considered here for energy balance. First, a part of the available energy is spent to displace the pool fluid to the pool surface to create a crater for the droplet. It is observed from figure 5(b) that if a droplet is getting wrapped, then it needs to attain a critical depth. From this, the displaced volume $\Delta \mathcal {V}$ is approximated as

(5.2)\begin{equation} \Delta \mathcal{V} \approx \frac{\rm \pi}{4}\,D^2\times 0.74D + \frac{\rm \pi}{12}\,D^3. \end{equation}

Here, it is assumed that the crater is cylindrically shaped with a hemisphere at one of its ends. The centre of mass of the crater is given as

(5.3)\begin{equation} z_c = \frac{\dfrac{\rm \pi}{4}\,D^2\times 0.74D \times \dfrac{0.74}{2}\,D + \dfrac{\rm \pi}{12}\,D^3\times \left(0.74+\dfrac{3}{16}\right)D}{\dfrac{\rm \pi}{4}\,D^2\times 0.74D + \dfrac{\rm \pi}{12}\,D^3}. \end{equation}

This $z_c = 0.5431D$ is the height by which the crater needs to be lifted, and the energy required for this is calculated as potential energy loss. Taking a correction factor $k_{PE}$, the potential energy loss is evaluated as

(5.4)\begin{equation} E_{PE} \approx k_{PE} \times \rho_2\,\Delta \mathcal{V}\,gz_c = k_{PE} \times 4.4916\rho_2D^4. \end{equation}

The correction factor for the potential energy loss is taken as unity. By taking the shape of the crater as defined above, change in surface area can be evaluated for different surfaces. Taking the product of these surface changes with their respective surface tension values gives an estimate of the energy required for the destruction and creation of new surfaces:

(5.5)\begin{equation} E_{S} \approx \left( 0.74{\rm \pi} D^2\sigma_{23} + \frac{\rm \pi}{2}\,D^2\sigma_{12}\right)-\left(\frac{\rm \pi}{2}\,D^2\sigma_{13} + \frac{\rm \pi}{4}\,D^2\sigma_{23} \right). \end{equation}

Using the values of $\sigma _{12}$, $\sigma _{23}$ and $\sigma _{13}$, this energy is further approximated as

(5.6)\begin{equation} E_{S} \approx k_S \times 0.005{\rm \pi} D^2. \end{equation}

The actual crater is not exactly cylindrical, but rather has curved edges and capillary waves. Therefore, the actual surface generated should be bigger than estimated. As a result, the adjustment factor $k_S$ should be greater than $1$. The liquid in the pool is highly viscous, hence large viscous losses are also expected due to the motion of the droplet into the pool. Since the force experienced by a droplet when moving through another fluid is not known exactly, the following assumptions are made to approximate this energy loss: (a) the droplet is assumed to be a rigid sphere moving through the pool; (b) flow speed past the droplet is taken as constant. Under these assumptions, the drag force experienced by the droplet is given by

(5.7)\begin{equation} F_d = \tfrac{1}{2}C_d\rho_2 V^2 \times A, \end{equation}

where $A={\rm \pi} D^2/4$ is the frontal area. The distance travelled by the droplet $z_d$ is the sum of the height of the centre of mass of the drop at the time of impact above the interface ($0.5D$) and the critical depth ($0.74D$). We also have to include a correction factor $k_V$ to accommodate the aforementioned assumptions. This gives the viscous losses as

(5.8)\begin{equation} E_V \approx k_V \times \tfrac{1}{2}C_d\rho_2 V^2 A z_d = 0.155 k_V C_d\rho_2 V^2{\rm \pi} D^3. \end{equation}

When the droplet descends from the interface, its velocity decreases, and its shape changes, resulting in a decrease in the drag coefficient. Therefore, the overall viscous losses will be lower than the estimated value and should also be accounted for by the correction factor. Here, $C_d$ is the drag coefficient for flow past a sphere and is dependent on the droplet diameter $D$, impact velocity $V$, density $\rho _2$, and viscosity $\mu _2$ of the pool. The Reynolds number $Re = {\rho _2 V D}/{\mu _2}$ satisfies $1< Re<1000$. Hence the drag coefficient is defined using the relation given by Schiller and Naumann (Flemmer & Banks Reference Flemmer and Banks1986). Considering the above-mentioned four energies, it is determined whether a droplet, if it crosses the critical depth, still has additional energy to move further downwards. Excess energy is calculated as

(5.9)\begin{equation} E_{{excess}} = E_{KE}-E_{PE}-E_S-E_V. \end{equation}

The values of the correction factors $k_S$ and $k_V$ are tuned such that the available data set for the final state of droplet impact gives a distinct distribution in terms of the excess energy. Figure 5(d) shows the state diagram for hanging and wrapping droplets as functions of excess energy and the diameter of the droplet. It can be seen that the droplets with sufficient energy to spend on different losses tend to get wrapped and sink into the pool, whereas the droplets that have less energy to begin with, such that they have negative excess energy, tend to hang from the pool surface. There are also intermediate cases where the available energy is nearly equal to the energy required and thus has close to zero excess energy. These droplets initially lose their entire kinetic energy upon impact and then gradually sink into the pool. It is observed that the majority portion of the available energy is spent to overcome the viscous loss, and the remaining energy is spent for surface energy. A very small part of the available energy is spent on potential energy loss. Thus a larger droplet with higher initial energy can still hang from the surface if either the viscosity of the pool fluid is increased or the interfacial tension value used for the coacervate is increased. We note here that the excess energy is just a function of $D$ and $h$, and it converts the nonlinear distribution of hanging and wrapping droplets in figure 5(c) into a linear distribution in figure 5(d).

5.3. Force balance for hanging droplets

The viscous force has a significant impact on the droplet's rate of descent; nevertheless, after it has reached equilibrium, it is the surface tension force and the buoyant forces that are responsible for maintaining the droplet's attachment to the surface by balancing its weight. We present the calculations for the force balance based on an analytical approach and the outcomes of the numerical simulations. The weight of the droplet is calculated as

(5.10)\begin{equation} F_{w} = \mathcal{V}_1 \times \rho_1 \times g. \end{equation}

Here, $\mathcal {V}_1$ is the total volume of the droplet. The buoyant force is defined as

(5.11)\begin{equation} F_{b} = \mathcal{V}_2 \times \rho_2 \times g. \end{equation}

Here, $\mathcal {V}_2$ is the volume of pool fluid displaced by the droplet below the TPCL. It is worth noting that the droplet at equilibrium is not completely immersed beneath the pool's surface. A little portion of the droplet remains above the TPCL line. There are also pockets of air bubbles trapped between the droplet and pool interfaces. The volume of these air bubbles is also included in $\mathcal {V}_2$ to calculate the buoyant forces. Now consider a system with a droplet including the droplet–air interface and the droplet–pool interface. The forces acting on the system in the vertical direction are only the gravitational force, the buoyant force, and the surface tension due to the air–drop interface and drop–pool interface. Since the hanging droplet system consists of two different surfaces wrapped around a common ring, i.e. the TPCL, the surface tension forces due to each of the two interfaces can be evaluated by taking the product of the pressure jump across the interface and the projected area at their boundary. The pressure jump at the interface can be calculated using Young's Laplace equation:

(5.12)\begin{equation} \Delta p_{ij} = \frac{2\sigma_{ij}}{R_{ij}}, \end{equation}

where $R_{ij}$ is the radius of curvature of the interface between the phases $i$ and $j$. Therefore, the vertical force on the hanging droplet owing to the surface tension computed from the values of $R_{12}$ and $R_{13}$ obtained from the simulations is given by

(5.13)\begin{equation} F_{\sigma} = \left(\frac{2\sigma_{12}}{R_{12}}-\frac{2\sigma_{13}}{R_{13}}\right) \times \frac{{\rm \pi} d^2}{4}. \end{equation}

Based on the values of $\mathcal {V}_1$ and $\mathcal {V}_2$ obtained from our simulations, in order to satisfy the force balance on the hanging droplet in the vertical direction, the ideal value of the vertical component of surface tension force should be

(5.14)\begin{equation} F_{\sigma0} = F_w-F_b. \end{equation}

The $F_{\sigma }$ values computed from the simulations are in fact very close to $F_{\sigma 0}$, as demonstrated in table 4 for four cases of hanging droplets signifying the dynamical balance.

Table 4.

Calculation for surface tension force balance for the hanging droplet.

Our simulations show that the shape of a hanging droplet at equilibrium resembles a combination of two spherical caps with varying radii. Hence we model the droplet with two spherical sections of radii $R_{12}$ and $R_{13}$, respectively, as shown in figure 6(a). Considering the two interfaces as part of purely spherical sections, the radius of curvature and the TPCL diameter can be related as

(5.15)\begin{equation} d = 2R_{12}\sin\beta = 2R_{13}\sin\phi. \end{equation}

Using (5.15), the volumes of the upper and lower spherical sections, $V_{13}$ and $V_{12}$, can be evaluated as

(5.16)\begin{gather} V_{13} = \frac{\rm \pi}{24\sin^3\phi}\,d^3(2+\cos\phi)(1-\cos\phi)^2,\end{gather}

(5.17)\begin{gather}V_{12} = \frac{\rm \pi}{24\sin^3\beta}\,d^3(2-\cos\beta)(1+\cos\beta)^2. \end{gather}

The droplet and the pool fluids are taken as immiscible because there is no chemical reaction taking place at the interface. Therefore, the volume of the droplet must be conserved:

(5.18)\begin{equation} V_{12}+V_{13} = \frac{\rm \pi}{6}\,D^3. \end{equation}

Again using (5.15) in (5.13), the vertical surface tension force on the droplet is calculated analytically as

(5.19)\begin{equation} F_{\sigma,model} = {\rm \pi}d (\sigma_{12}\sin\beta-\sigma_{13}\sin\phi). \end{equation}

However, considering the vertical force balance on the droplet, i.e. using (5.14), (5.17) and (5.16), we get

(5.20)\begin{equation} F_{\sigma,model} = \left((\rho_1-\rho_2)V_{12}+(\rho_1-\rho_3)V_{13}\right)g. \end{equation}

Apart from the force balance on the droplet, the interfaces between the three phases, air, droplet and pool, are also considered to be massless. Therefore, at the junction of the three phases, i.e. at the TPCL, the vertical and the horizontal surface tension forces due to the three interfaces must balance each other (Phan et al. Reference Phan, Allen, Peters, Le and Tade2012; Xie et al. Reference Xie, Forth, Zhu, Helms, Ashby, Shum and Russell2020). Therefore,

(5.21)\begin{gather} \sigma_{23}\sin\alpha-\sigma_{12}\sin\beta+\sigma_{13}\sin\phi = 0, \end{gather}

(5.22)\begin{gather}\sigma_{23}\cos\alpha+\sigma_{12}\cos\beta-\sigma_{13}\cos\phi = 0. \end{gather}

Eliminating $\alpha$ (see figure 3b) from (5.21) and (5.22), we get

(5.23)\begin{equation} (\sigma_{13}\sin\phi-\sigma_{12}\sin\beta)^2 + (\sigma_{13}\cos\phi-\sigma_{12}\cos\beta)^2 = \sigma^2_{23}. \end{equation}

Equations (5.18), (5.19), (5.20) and (5.23) can be solved simultaneously to obtain the values for $\beta$, $\phi$, $d$ and $F_{\sigma,model}$. Figures 6(b), 6(c) and 6(d) show an excellent match of the contact angles ($\beta$ and $\phi$), radii of curvature for the two interfaces, and the surface tension forces respectively between our simulations and the analytical solution obtained by solving the force balance equations for the hanging drops.

Figure 6.

(a) Shape of droplet at equilibrium. (b) Contact angles $\beta$ and $\phi$ versus diameter of the droplet at equilibrium. (c) Radius of curvature for the interface versus diameter of the droplet at equilibrium. (d) Surface tension force on the droplet versus diameter of the droplet at equilibrium.