2. Experiment set-up and measurements

The schematic of the experimental set-up is shown in figure 2. Impact experiments were performed at room temperature with 40 % relative humidity in the ambient environment. The droplets were injected from a needle with an inner diameter of 2.18 mm connected to a syringe pump at a flow rate of 50 μL min⁻¹ (Pico Plus, Harvard Apparatus). Another needle was implemented inside the main needle to inject air into the liquid droplet, where the tips of both needles were located in a line. When a droplet started to form at the tip of the liquid needle, the second needle was used to inject an air bubble into the droplet. The gas injection process was done with constant flow rate, manually. Liquid and air injection were continuous until further increment in the droplet weight caused the hollow droplet detachment from the needle tip. It should be mentioned that the proposed set-up is capable of producing hollow droplets, but the size of hollow droplets produced vary significantly and the experiments had to be repeated several times to produce a hollow droplet with almost exact sizes.

Figure 2. Schematic of the experimental set-up of hollow droplet impact.

The height of the needle above the surface was varied between 50 and 700 mm, resulting in impact velocities ranging from 0.5 to 3.6 m s⁻¹. A high-speed camera (Phantom v711, Vision Research) recorded the impacting droplets. The process was illuminated by a lamp enabling filming the process at a rate of 5000 fps and an exposure time of 30 μs. The video signal was recorded in the memory of a PC, and the images were analysed using ImageJ software (version 1.46, National Institutes of Health, Bethesda, MD).

The apparent, advancing and receding contact angles of water droplet on the aluminium surface were measured as 85°±2°, 94°±2° and 65°±2°, respectively. The fall of hollow droplets from 50 mm height was recorded on the video camera, and the spread droplets were weighed on a balance accurate to 0.05 mg. From the taken images, the outer diameter of the droplet $({D_h})$ and the diameter of the bubble $({D_{b - image}})$ were measured. Considering the density of water and calculating the total volume of the hollow droplet, the actual diameter of the bubble $({D_b})$ was evaluated. Measurements show that, due to the fisheye effect, ${D_{b - image}}$ is not equal to ${D_b}$ and it needs to be corrected. These measurements were repeated for ten droplets and at the end, the following correlation was derived to calculate the actual size of the bubble from token images (Nasiri et al. Reference Nasiri, Amini, Moreau and Dolatabadi2021):

(2.1)\begin{equation}{D_b} = \; 0.86\times {D_{b - image}}.\end{equation}

The details of the hollow droplet size calculations and experimental set-up are presented in the previously published article of Nasiri et al. (Reference Nasiri, Amini, Moreau and Dolatabadi2021). To report the experimental data, ten similar hollow droplets with less than 10 % standard deviation in the size of droplet and bubble were measured.

3. Numerical analysis

To explain the experiments and provide a better understanding of the reasoning behind the formation and detachment of liquid counter-jet, a numerical model has been used to solve compressible continuum equations for the drop flattening on a surface. Figure 3 shows schematics of a hollow droplet impacting a flat surface. Several parameters are defined to characterise this process. The main geometrical parameters are the initial diameter of the hollow droplet $({D_h})$ and the initial diameter of the bubble inside the hollow droplet $({D_b})$, which are shown in figure 3(a). The flattening of a hollow droplet is schematically shown in figure 3(c) showing the related parameters of counter-jet diameter $({D_{cj}})$, counter-jet height $({h_{cj}})$, counter-jet velocity $({U_{cj}})$, spreading diameter (D), splat height $(h)$ and spreading velocity $({U_R})$. Equivalent diameter $({D_{eq}})$ of a hollow droplet is defined as the diameter of a dense droplet with the same weight (figure 3b). These diameters are related to each other by the definition of $\alpha ,\beta $ parameters as

(3.1a,b)\begin{equation}{D_h} = \alpha {D_{eq}},\quad {D_b} = \beta {D_{eq}}.\end{equation}

From (3.1), it can easily be concluded that ${\alpha ^3} - {\beta ^3} = 1$. To study the impact process, a hollow droplet with a size of ${D_h} = 5.6\ \textrm{mm},\;{D_b} = 4.5\ \textrm{mm}$ and ${D_{eq}} = 4.4\ \textrm{mm}$ is simulated at different velocities. The hollow droplet vertically impacts a flat aluminium surface with an initial temperature of 300 K. The ambient pressure and temperature are 1 atm and 300, respectively. The properties of fluids used in this simulation are presented in table 1.

Figure 3. Geometrical model of a droplet impacting a flat surface: (a) hollow droplet before and after impact; (b) dense droplet before and after impact; (c) hollow droplet spreading after impact.

Table 1. Properties of water and air at T = 300 K.

Due to the symmetry of the process, a 2-D-axisymmetric domain was used to simulate the droplet impact on a flat surface. The computational domain and boundary conditions are shown in figure 4. The behaviour of a hollow droplet, including spreading and bubble ruptures, was captured in the simulations with grid numbers from 208 000 to 1 200 000. The results were validated with extracted experimental data. Compromising the accuracy and time, a structured grid with total grid numbers of 340 000 was chosen to simulate all the cases. The results for mesh independency of the solution have been reported in the previous study of the authors (Nasiri et al. Reference Nasiri, Amini, Moreau and Dolatabadi2021). With the current grid distribution, approximately 100 cells per diameter exist inside the hollow droplet before the impact. It should be mentioned that cells were more concentrated near the symmetry axis and along the bottom boundary, the cells are fine and uniform at the impact location (around 0.125 mm) and the grid distribution is constant at different time steps.

Figure 4. Computational domain and boundary conditions of the specified problem.

The no-slip boundary condition is applied on the bottom surface and the pressure boundary condition is zero gradient. The dynamic contact angle is used to simulate the liquid motion on the bottom wall. The apparent, advancing and receding contact angles for water on an aluminium surface are reported in table 2. For the other side of the domain, velocity boundary conditions are assumed to be zero gradient while the pressure is atmospheric pressure.

Table 2. Measured contact angles of water on different surfaces.

The mass, momentum and energy equations are the governing equations to be solved to simulate hollow droplet flattening with the VOF method as

(3.2)\begin{gather}\frac{{\partial \rho }}{{\partial t}} + {\nabla }\boldsymbol{\cdot }(\rho \; \boldsymbol{U}) = 0,\end{gather}

(3.3)\begin{gather}\frac{{\partial (\rho \boldsymbol{U})}}{{\partial t}} + {\nabla }\boldsymbol{\cdot }(\rho \boldsymbol{UU}) = \; - \boldsymbol{\nabla }p + \rho \boldsymbol{g} + \boldsymbol{\nabla }\boldsymbol{\cdot }\{ \mu [{\nabla }\boldsymbol{U} + {({\nabla }\boldsymbol{U})^T}]\} + {\boldsymbol{F}_{vol}}.\end{gather}

(3.4)\begin{gather}\frac{{\partial (\rho {C_p}T)}}{{\partial t}} + \boldsymbol{\nabla }\boldsymbol{\cdot }(\rho {C_p}\boldsymbol{U}T) = {\nabla }\boldsymbol{\cdot }(k{\nabla }T) + \frac{{\partial p}}{{\partial t}} - \left( {\frac{{\partial \rho K}}{{\partial t}} + \boldsymbol{\nabla }\boldsymbol{\cdot }(\rho \boldsymbol{U}K)} \right),\end{gather}

where $\rho $, $\boldsymbol{U}$, P, $\boldsymbol{g}$, K and ${C_p}$ are mixture density, velocity field, pressure, gravitational acceleration, the kinetic energy of fluid, and the specific heat, respectively. Surface tension is considered as a source term $({\boldsymbol{F}_{vol}})$ and can be calculated using the Brackbill method (Brackbill, Kothe & Zemach Reference Brackbill, Kothe and Zemach1992). To solve momentum equations correctly, the pressure field can be calculated from the following equation (Miller et al. Reference Miller, Jasak, Boger, Paterson and Nedungadi2013):

(3.5) \begin{equation}\left( {\gamma \frac{{{\varPsi_l}}}{{{\rho_l}}} + (1 - \gamma )\frac{{{\varPsi_g}}}{{{\rho_g}}}} \right)\left[ {\frac{{\partial p}}{{\partial t}} + \boldsymbol{U}\boldsymbol{\cdot }\boldsymbol{\nabla }p} \right] + \boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{U} = 0,\end{equation}

where ${\varPsi _l}$, ${\varPsi _g}$, ${\rho _l}$, ${\rho _g}$ and $\gamma $ are the compressibility and the density of liquid and gas, and the liquid–gas volume fraction, respectively. Defining compressibility as $\varPsi = \partial \rho /\partial P$ based on the ideal gas equation of state, gas compressibility and liquid compressibility take the forms of

(3.6a,b)\begin{equation}{\varPsi _g} = \; \frac{1}{{zRT}}\; ,\quad {\varPsi _l} = \; \frac{1}{{{a^2}}},\end{equation}

where z represents compressibility factor with the isothermal equation of state assumption. Parameter a represents sound velocity in the liquid. It should be noted that regarding the impact velocity of the hollow droplet (1–6 m s⁻¹), the compressibility does not play a role in the dynamic of hollow droplet flattening and can be ignored.

To track the interface of two fluids, the VOF equation is used (Miller et al. Reference Miller, Jasak, Boger, Paterson and Nedungadi2013):

(3.7) \begin{equation} \left. {\begin{array}{*{20}{c@{}}} {\dfrac{{\partial (\gamma )}}{{\partial t}} + \boldsymbol{\nabla }\boldsymbol{\cdot }(\gamma \boldsymbol{U}) = \dfrac{{ - \gamma {\varPsi_l}}}{{{\rho_l}}}\dfrac{{\textrm{D}p}}{{\textrm{D}t}} = \gamma (1 - \gamma )\zeta + \gamma \boldsymbol{\nabla }\boldsymbol{\cdot }(\boldsymbol{U})}\\ {\zeta = \left( {\dfrac{{{\varPsi_l}}}{{{\rho_l}}} + \dfrac{{{\varPsi_g}}}{{{\rho_g}}}} \right)\dfrac{{\textrm{D}p}}{{\textrm{D}t}}} \end{array}} \right\}.\end{equation}

By solving the above equation, the value of $\gamma $ can be obtained for each computational cell. The cell is occupied by liquid if $\gamma = 1$, whereas the cell is occupied with gas if $\gamma = 0$. Any value in the range of $0 < \gamma < 1$ indicates that the cell contains a liquid–gas interface. Once $\gamma $ has been determined, the surface tension force can be calculated as follows (Brackbill et al. Reference Brackbill, Kothe and Zemach1992):

(3.8) \begin{equation} {\boldsymbol{F}_{vol}} = \sigma \kappa ({\nabla }\gamma ),\quad \kappa =- {\nabla }\boldsymbol{\cdot }\left( {\frac{{{\nabla }\gamma }}{{|{\nabla }\gamma |}}} \right),\end{equation}

where $\sigma $ and $\kappa $ represent the surface tension of the liquid and the surface curvature, respectively, and $\nabla \gamma $ is a continuous function that is zero everywhere in the domain except for the transitional area at the interface. The mixture properties can be expressed as

(3.9)\begin{gather}\mu = \gamma {\mu _l} + (1 - \gamma ){\mu _g},\end{gather}

(3.10)\begin{gather}\rho = \gamma {\rho _l} + (1 - \gamma ){\rho _g},\end{gather}

(3.11)\begin{gather}k = \gamma {k_l} + (1 - \gamma ){k_g},\end{gather}

(3.12)\begin{gather}{c_p} = \gamma {c_{p,l}} + (1 - \gamma ){c_{p,g}}.\end{gather}

To optimise the computational time, an adaptive time step control feature was applied. This controller, which keeps the solution stable, was adjusted based on the Courant–Friedrichs–Lewy (CFL) number $(Co = {U_i}\Delta t/\Delta x)$ at the beginning of the time iteration loop. Using the values of the velocity of the phase fractions and $\Delta t$ from the previous time step, the maximum local CFL number $(C{o_0})$ was calculated, and a new time step was iteratively initiated as

(3.13)\begin{equation}\Delta t = min\left\{ {\frac{{C{o_{max}}}}{{C{o_0}}}\Delta {t_0};\left( {1 + {\lambda_1}\frac{{C{o_{max}}}}{{C{o_0}}}} \right)\Delta {t_0};{\lambda_2}\Delta {t_0};\Delta {t_{max}}} \right\},\end{equation}

where $C{o_{max}}$ and $\Delta {t_{max}}$ prescribe values for the CFL number and time step, respectively (Yeganehdoust et al. Reference Yeganehdoust, Attarzadeh, Karimfazli and Dolatabadi2020). It should be noted that codes of OpenFOAM open-source solver have been used to simulate the droplet flattening.

5. Theoretical analysis

In characterising the hollow droplet impact, a void water sphere with a diameter of ${D_h}$ is considered in which the diameter of the entrapped bubble is ${D_b}$. The mass of this hollow droplet is equal to the mass of a dense droplet with a diameter of ${D_{eq}}$. After impact, the hollow droplet divides into two portions. One portion spread on the surface in the shape of a flat cylinder with a diameter of D and height of h, while the other portion forms a counter-jet in the shape of a vertical cylinder with a diameter of ${D_{cj}}$ and height of ${h_{cj}}$. It is assumed that no splashing occurs and the edge effects associated with the rim formation are negligible. The droplets are considered large enough that the capillary length could be neglected. (i.e. $\rho gD_{eq}^2/\gamma \approx 1$). Therefore, the effect of gravity is neglected and the only contribution to the potential energy of the drop arises from the surface tension (Amini Reference Amini2020). In addition, as impact velocity increases, the droplet kinetic energy will be much larger than surface energy (i.e. $We \gg 1$), and surface tension and contact angle effects will eventually become negligible (Pasandideh-Fard et al. Reference Pasandideh-Fard, Qiao, Chandra and Mostaghimi1996).

Before impact, the effective energies of the hollow droplet are kinetic energy (KE ₁), surface energy (SE ₁) and gravitational energy, and the energy conservation equation of a hollow droplet can be written as

(5.1)\begin{equation}K{E_1} + S{E_1} + G{E_1} = K{E_{2s}} + S{E_{2s}} + G{E_{2s}} + W + K{E_{2cj}} + S{E_{2cj}} + G{E_{2cj}}.\end{equation}

Considering the hollow droplet exactly before the impact, the centre of the droplet is located at ${h_D} = {D_{eq}}/2$. Assuming a constant surface tension, and the energy values on the left side are given by

(5.2) \begin{gather}K{E_1} = \Big({\frac{1}{2}\rho U_0^2} \Big) \left({\frac{\rm \pi}{6}D_{eq}^3}\right),\end{gather}

(5.3)\begin{gather}S{E_1} = {\rm \pi}D_h^2\varUpsilon + {\rm \pi}D_b^2\varUpsilon = {\rm \pi}{\alpha ^2}D_{eq}^2\varUpsilon + \; {\rm \pi}{\beta ^2}D_{eq}^2\varUpsilon = \; {\rm \pi}D_{eq}^2\varUpsilon ({\alpha ^2} + {\beta ^2}),\end{gather}

(5.4)\begin{gather}G{E_1} = (\rho g{h_D})\left( {\frac{{\rm \pi} }{6}D_{eq}^3} \right) = \Big({\; \rho g\frac{{{D_{eq}}}}{2}\; } \Big)\left( {\frac{{\rm \pi} }{6}D_{eq}^3} \right).\end{gather}

After the hollow droplet impact, a portion of the droplet spreads and the other portion forms a counter-jet. When the spreading part reaches the maximum spreading diameter, its kinetic energy is zero. The gravitational energy is also negligible and assumed to be zero. The only effective energy is the splat surface energy. The surface energy can be calculated by

(5.5)\begin{gather}K{E_{2s}} = \; 0,\end{gather}

(5.6)\begin{gather}G{E_{2s}} = \; 0,\end{gather}

(5.7)\begin{gather}S{E_{2S}} = \frac{{\rm \pi} }{4}D_{max}^2Y(1 - \cos {\theta _a}),\end{gather}

where ${\theta _a}$ is advancing contact angle (Pasandideh-Fard et al. Reference Pasandideh-Fard, Qiao, Chandra and Mostaghimi1996). In addition, the work done by viscous force can be calculated by

(5.8)\begin{equation}W = \int_0^{{t_c}} {\int_0^v {\phi \,\textrm{d}\varOmega {t_s}} } ,\end{equation}

where $\varOmega $ is the volume of viscous fluid, ${t_c}$ is time for the droplet to reach maximum spreading and $\phi $ is viscous dissipation function, which is estimated by

(5.9)\begin{equation}\phi \sim \mu {\left( {\frac{{{U_0}}}{L}} \right)^2}.\end{equation}

Here, $\mu $ is liquid viscosity and L is characteristic length estimated by the boundary layer thickness $(\delta )$ at a solid–liquid interface.

The time required for the hollow droplet to reach the maximum spreading diameter can be estimated by assuming the outcome of the hollow droplet impact as two different parts, one is a cylindrical disk with the diameter of D and thickness of h responsible for the droplet spreading, and the other one is a cylinder with a diameter of ${D_{cj}}$ and height of ${h_{cj}}$ responsible for counter-jet formation (figure 3). Liquid flows from the droplet into the film and counter-jet through an area of diameter d with velocity ${U_0}$. The velocity at the edge of the splat during spreading $({U_R})$ is given by the conservation of mass law:

(5.10)\begin{equation}{U_0}\frac{{{\rm \pi} {d^2}}}{4} = {U_R}{\rm \pi} Dh + {U_{cj}}\frac{{{\rm \pi} D_{cj}^2}}{4}.\end{equation}

Based on figures 10 and 14, velocity $({U_{cj}})$, diameter $({D_{cj}})$ and volume $({V_{cj}})$ associated with the counter-jet are estimated as ${U_{cj}} \simeq 0.4{U_0}$, ${D_{cj}} \simeq 0.5{D_b}$, while ${D_b} = \beta {D_{eq}}$ and ${V_{cj}} \simeq 0.3\alpha \beta {V_{eq}})$.

The splat thickness $(h)$ after impact can be calculated by equating the volume of hollow droplet before impact $({V_h})$, the volume of the spreading liquid $({V_s})$ in the shape of a disk with height h and diameter ${D_{max}}$, and the volume of counter-jet $({V_{cj}})$ as a cylinder with a height of ${h_{cj}}$ and diameter of ${D_{cj}}$,

(5.11)\begin{gather}{V_h} = {V_s} + {V_{cj}},\end{gather}

(5.12)\begin{gather}\frac{{4{\rm \pi} }}{3} \times \frac{{D_{eq}^3}}{8} = \frac{{{\rm \pi} D_{max}^2h}}{4} + (1 - 0.3\alpha \beta ) \times \Bigg( {\frac{{4{\rm \pi} }}{3} \times \frac{{D_{eq}^3}}{8}} \Bigg),\end{gather}

(5.13)\begin{gather}h = (1 - 0.3\alpha \beta ) \times \frac{2}{3} \times \frac{{D_{eq}^3}}{{\; D_{max}^2}}.\end{gather}

Since d varies between $0$ and ${D_{eq}}$, an average value is assumed for $d\sim {D_{eq}}/2$. Substituting h from (5.13) and the d parameter into (5.10) results in

(5.14)\begin{gather}\frac{{\textrm{d}D}}{{\textrm{d}t}} = 2{V_R} = \frac{3}{{32}}\frac{{D_{max}^2{U_0}}}{{{D_{eq}}}}\left( {\frac{{1 - 0.4{\beta^2}}}{{1 - 0.3\alpha \beta }}} \right),\end{gather}

(5.15)\begin{gather}\frac{{{D^2}}}{{D_{max}^2}} = \frac{3}{8}\frac{{\; {U_0}}}{{{D_{eq}}}}\left( {\frac{{1 - 0.4{\beta^2}}}{{1 - 0.3\alpha \beta }}} \right){t_s}.\end{gather}

At maximum spreading condition $D = {D_{max}}$, therefore,

(5.16)\begin{equation}{t_s} = \frac{8}{3}\frac{{{D_{eq}}}}{{\; {U_0}}}\left( {\frac{{1 - 0.3\alpha \beta }}{{1 - 0.4\; {\beta^2}}}} \right).\end{equation}

In addition, the volume of viscous fluid can be calculated by

(5.17)\begin{equation}\varOmega = \frac{{{\rm \pi} D_{max}^2\delta }}{4}.\end{equation}

Pasandideh-Fard et al. (Reference Pasandideh-Fard, Qiao, Chandra and Mostaghimi1996) calculated the boundary layer thickness of a dense liquid droplet after impact on a surface as

(5.18)\begin{equation}\delta = 2\left( {\frac{{{D_{eq}}}}{{\sqrt {Re} }}} \right).\end{equation}

Similarly, considering the assumption ${V_{cj}} \simeq 0.3\alpha \beta \; {V_{eq}}$, the volume of the liquid spreading on the surface is $(1 - 0.3\alpha \beta ){V_{eq}}$ and the boundary layer for the hollow droplet thickness can be written as

(5.19)\begin{equation}\delta = 2\left( {\frac{{\sqrt {(1 - 0.3\alpha \beta )} \times {D_{eq}}}}{{\sqrt {Re} }}} \right),\end{equation}

in which the Reynolds number is defined as $Re = \rho {U_0}{D_{eq}}/\mu $. Substituting (5.9), (5.18), (5.13), (5.16) and (5.17) into (5.8) yields

(5.20)\begin{equation}W = \frac{{\rm \pi} }{3}\rho U_0^2{D_{eq}}D_{max}^2\frac{1}{{\sqrt {(1 - 0.3\alpha \beta ) \times Re} }}\Big( {\frac{{1 - 0.3\alpha \beta }}{{1 - 0.4\; {\beta^2}}}} \Big).\end{equation}

Kinetic energy of the counter-jet at maximum spreading time can be calculated as

(5.21)\begin{equation}K{E_{2cj}} = {\textstyle{1 \over 2}}\rho {V_{cj}}U_{cj}^2.\end{equation}

Considering ${U_{cj}} \simeq 0.4{U_0}$ and ${V_{cj}} \simeq 0.3\alpha \beta \; {V_{eq}}$ (figures 10 and 14), one can obtain

(5.22)\begin{equation}K{E_{2cj}} = \frac{1}{2}\rho {(0.4 \times {U_0})^2}\left( {\frac{{\rm \pi} }{6} \times D_{eq}^3} \right) \times (0.3\alpha \beta ).\end{equation}

However, the counter-jet surface energy can be calculated by considering the counter-jet in the shape of a cylinder with a semi-sphere on the top:

(5.23)\begin{gather}S{E_{2cj}} = S{E_{2cj1}} + S{E_{2cj2}},\end{gather}

(5.24)\begin{gather}S{E_{2cj}} = \frac{{\rm \pi} }{4}D_{cj}^2Y + {\rm \pi}{D_{cj}}{h_{cj}}Y.\end{gather}

Assuming ${U_{cj}} = 0.4{U_0}$, ${D_{cj}} \simeq 0.5{D_b}$, ${D_b} = \beta {D_{eq}}$ and ${V_{cj}} = 0.3\alpha \beta {V_{eq}}$, we get ${h_{cj}} \simeq 0.8(\alpha /\beta ){D_{eq}}$ and

(5.25)\begin{gather}S{E_{2cj}} = \frac{{\rm \pi} }{2}D_{cj}^2Y + {\rm \pi}{D_{cj}}{h_{cj}}Y,\end{gather}

(5.26)\begin{gather}S{E_{2cj}} = {\rm \pi}YD_{eq}^2\left[ {\frac{{{\beta^2}}}{8} + 0.4\alpha } \right].\end{gather}

Gravitational energy of the counter-jet at maximum spreading time can be calculated as

(5.27)\begin{equation}G{E_{2cj}} = \rho {V_{cj}}g\frac{{{h_{cj}}}}{2}.\end{equation}

Assuming ${D_{cj}} \simeq 0.5{D_b}$, ${D_b} = \beta {D_{eq}}$ and ${V_{cj}} = 0.3\alpha \beta {V_{eq}}$, we get ${h_{cj}} \simeq 0.8(\alpha /\beta ){D_{eq}}$ and

(5.28)\begin{equation}G{E_{2cj}} = (\rho g)\left( {0.3\alpha \beta \times \frac{{\rm \pi} }{6}D_{eq}^3} \right)\Big({0.4\frac{\alpha }{\beta }{D_{eq}}}\Big).\end{equation}

Substituting (5.2)–(5.7), (5.22), (5.26) and (5.28) into (5.11), and noting that $\alpha = \sqrt[3]{{1 + {\beta ^3}}}$ (3.1), one obtains

(5.29)\begin{equation}\frac{{{D_{max}}}}{{{D_{eq}}}} = \sqrt {\frac{{M\,We + 12N + Q\,Bo}}{{3(1 - \cos {\theta _a}) + 4\frac{{We}}{{\sqrt {Re} }} \times I}}} ,\end{equation}

in which M, N and I are defined as

(5.30)\begin{gather}M = (1 - 0.048\beta \sqrt[3]{{1 + {\beta ^3}}}),\end{gather}

(5.31)\begin{gather}N = \sqrt[3]{{{{(1 + {\beta ^3})}^2}}} + {\beta ^2} - \frac{{{\beta ^2}}}{8} - 0.4\sqrt[3]{{1 + {\beta ^3}}},\end{gather}

(5.32)\begin{gather}I = \frac{{\sqrt {(1 - 0.3\beta \sqrt[3]{{1 + {\beta ^3}}})} }}{{(1 - 0.4{\beta ^2})}},\end{gather}

(5.33)\begin{gather}Q = (1 - 0.24\; \sqrt[3]{{{{(1 + {\beta ^3})}^2}}}).\end{gather}

The accuracy of the predicted maximum spreading from (5.29) was tested by comparison with experimental and numerical measurements for a variety of hollow droplets impacting a surface in a wide range of Weber and Reynolds numbers, and hollowness ratios. The results of theoretical prediction for hollow droplet flattening are compared with numerical data in figure 21.

Figure 21. Theoretical prediction of maximum spreading diameter of a hollow droplet: (a) with ${D_{eq}} = 4.4\ \textrm{mm}$ ${D_h} = 5.6\ \textrm{mm}$ and ${D_b} = 4.5\ \textrm{mm}$, impacting at different We and Re numbers; (b) with ${D_{eq}} = 4.4\ \textrm{mm}$, impacting on the surface with ${U_0} = 3.6\ \textrm{m}\ {\textrm{s}^{ - 1}}(We = 790,\;Ca = 0.0016)$ with different hollowness ratios.

The results show that the proposed model predicts maximum spreading diameter of a dense droplet of a hollow droplet with less than 10 % error. The hollowness ratio for the hollow droplets that have been tested varies between 0.25 and 0.85.

The magnitude of term $3(1 - \cos {\theta _a})$ in (5.29) can be at most 6. Therefore, in cases where $(We/\sqrt {Re} )I$ is large in comparison, the value of the contact angle will have little effect on the maximum spreading diameter of the hollow droplet. In addition, it can be concluded that if $We \gg I\sqrt {Re} $, the capillary effect can be neglected during the hollow droplet modelling. Furthermore, for the hollow droplets with $Bo \ll 1$, the effect of the gravity can be ignored during the modelling. It should be noted that this condition is applicable during the simulation of hollow droplet in a thermal spraying process in which the size of the particle is in the range of millimetres.

Considering the value for M and N parameters, if $We \gg 12$, (5.29) reduces to

(5.34)\begin{equation}\frac{{{D_{max}}}}{{{D_{eq}}}} = \sqrt {\frac{M}{{4I}}} R{e^{0.25}}.\end{equation}

Based on the previous studies (Gulyaev et al. Reference Gulyaev, Solonenko, Gulyaev and Smirnov2009; Gulyaev & Solonenko Reference Gulyaev and Solonenko2013), if $We \to \infty$,

(5.35)\begin{equation}\frac{{{D_{max}}}}{{{D_{eq}}}} = 0.7\left( {1 + \frac{4}{5}\chi } \right) + \left( {0.72 - \frac{2}{3}\chi } \right)R{e^{0.25(1 + \chi /5)}},\end{equation}

where $\chi = {(1 - ({D_h} - {D_b}))^3}$.

Also, if $Re \to \infty$, (5.29) reduces to

(5.36)\begin{equation}\frac{{{D_{max}}}}{{{D_{eq}}}} = \sqrt {M\,We + 12N + Q\,Bo} .\end{equation}

This is while based on the Gulyaev and Solonenko studies (Gulyaev et al. Reference Gulyaev, Solonenko, Gulyaev and Smirnov2009; Gulyaev & Solonenko Reference Gulyaev and Solonenko2013),

(5.37)\begin{equation}\frac{{{D_{max}}}}{{{D_{eq}}}} = 0.75(1 + 0.8\chi ) + (0.53 - 0.45\chi )W{e^{0.508 + 0.02\chi }}.\end{equation}

Based on the numerical and theoretical results for hollow droplet flattening, the maximum spreading of the hollow droplet can be between 80 and 100 % of the maximum spreading diameter of a dense droplet with the same mass. This percentage varies by the initial conditions before the impact including droplet velocity, bubblesize, etc.