Article pubs.acs.org/JPCC
Model for Rolling Angle Ran Minghao,† Yang Chengxia,† Fang Yuan,† Zhao Kaikai,† Ruan Yaqin,† Wu Jie,† Yang Hong,‡ and Liu Yinfeng*,† †
Department of Polymer Materials, School of Materials Science and Engineering, Shanghai University, Shanghai 201800, China Boping Aerospace Materials Company, Ltd., Jiading District, Shanghai 201800, China
‡
ABSTRACT: A new method was proposed to estimate the rolling angle of a droplet on a superhydrophobic surface. Based on the relationship between the adhesion work and the force moment of the spherical droplet, a rolling angle model was established. For smooth surfaces, the equation we obtained is sin α ≥ [(4γl)/(g·δ)]·[(3π2)/(ρm2)]1/3[(1 + cos θ)2/ (3 + cos θ)][(2 + cos θ)/(1 − cos θ)]2/3, and for rough surfaces, the equation is sin α ≥ [(4fγl)/(g·δ)]·[(3π2)/(ρm2)]1/3{[(1 + q cos θY)(1 + cos θ)]/(3 + cos θ)}[(2 + cos θ)/(1 − cos θ)]2/3. The relationship between the rolling angle, the contact angle, and the surface structure was investigated. The equations we obtained revealed the factors that have important influences on the rolling angle, including the surface roughness, the trapped air, the mass of droplet, and the chemical composition of the surface. The simulation results showed that the rolling angle decreases with the increases of the contact angle, the surface roughness, the proportion of trapped air, the weight of the droplet, and the hydrophobicity of the surface material. These results agreed well with the most widely used explanation of other researchers.
1. INTRODUCTION Superhydrophobic materials play a crucial role in many industrial applications, such as waterproofing, self-cleaning, lubricating, and anti-icing.1 Interest in this field has expanded greatly during the past decade. Many efforts had been made to investigate the design, preparation, application, and theoretical models for superhydrophobic systems. A superhydrophobic surface is defined as having a water contact angle greater than 150° and a rolling angle lower than 10°, which is recognized by most researchers. In nature, one of the plants that possesses such a property is the lotus, and it also shows a self-cleaning phenomenon. Any materials with these characteristics are said to be having the “lotus effect”.2 It is found from the natural lotus that the superhydrophobic phenomenon must consist of two dominant factors: surface roughness and low surface energy.3,4 Many different approaches have been taken to prepare superhydrophobic surfaces, including sol−gel method,5,6 particle filling method, 7−9 template method, 10,11 CVD method, 12 phase separation method, 13 and embossing method.14 While the traditional static contact angle measurement has been widely used as a criterion for the evaluation of hydrophobicity of a surface, this alone is insufficient for the evaluation of the sliding properties of droplets on surfaces. In fact, to a superhydrophobic surface, the rolling angle is a parameter which can directly reflect whether a liquid drop would stay on the surface or roll off immediately. But many researchers neglected the rolling angle and put their emphasis only on the contact angle. A commonly accepted wettability theory of solid surface is the famous Young’s equation, eq 115, which was created in the early 19th century © 2012 American Chemical Society
cos θ Y =
γs − γsl γl
(1)
where γl, γs, and γsl are the surface tensions of liquid−gas, solid−gas, and liquid−solid, respectively. However, it is suitable only for flat smooth surfaces. In the middle of the 20th century, Wenzel modified Young’s equation for a rough-surface model:16
cos θ = q cos θ Y
(2)
where q is the roughness factor. For more accurate consideration, Cassie supposed that the liquid−solid interface is composed of two fractions:17 one is the area of the solid surface touched with the liquid, and the other is the part touched with the trapped air. The formula is cos θ = f cos θ Y + f − 1
(3)
where f represents the remaining area fraction which is defined as the ratio of liquid−solid interface area to the total area of the solid surface. In 1956, an experiential equation describing the sliding angle on a smooth surface was proposed by Wolfram18,19 2r π sin α = k mg (4) where α is the sliding angle, g is the gravitational acceleration, m is the mass of the droplet, r is the radius of the contact circle Received: October 25, 2011 Revised: February 4, 2012 Published: February 16, 2012 8449
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between the liquid and the solid, and k is a proportionality constant. About two years later, Wolfram explained the constant k via the force equilibrium of the droplet20 (Figure 1).
Watanabe et al. introduced the surface roughness factor into eq 8 to describe the relationship between sliding angle and contact angle on a rough surface.23 In eq 9, the constant k related to the interaction energy is assumed to be proportional to the true contact area. sin α =
2qk·sin θ(cos θ + 1) g (q·cos θ Y + 1) ⎤1/3 ⎡ 3π 2 ⎥ ⎢ 2 ⎣ m ρ(2 − 3cos θ + cos3 θ) ⎦
where θ is the apparent contact angle of the rough surface and θY is the Young’s contact angle on the flat surface. These theories revealed the importance of some factors such as the surface roughness, the trapped air, and the hysteresis of contact angle. However, the big difference between the sliding behavior on the hydrophobic surface and the rolling behavior on the superhydrophobic surface was neglected. So there always exist some errors when simulating a superhydrophobic surface with large contact angle and small rolling angle by these equations.23 We know that rolling behavior is also a common phenomenon in nature. Droplets would form spheres and could roll off easily on the surfaces with the lotus effect. If there is any dust on the surface, the rolling droplet can also take it away. The so-called “self-cleaning effect”2,4 not only proves that the droplet on a superhydrophobic surface can roll like a ball but also proves that no water will be left on the surface after the droplet rolls off. Previous researchers, however, agreed on the formation of a thin water film that partially wets the solid surface left behind.21,25 It is reasonable for hydrophobic surfaces and can explain the phenomenon of the hysteresis of contact angle, but it does not agree well to some extent with the characteristics of superhydrophobic surfaces. There are so far precious few literatures aimed at the theoretical research about the rolling angle other than the sliding angle of the superhydrophobic surface. This paper attempts to understand the rolling behavior and attempts to establish a relationship between rolling angle and contact angle by analysis of force moment and energy equilibrium of rolling water droplet.
Figure 1. Profile view of a tilted droplet and the force on it.
r πγl cos θa ≤ r πγl cos θr + mg ·sin α
(5)
where θa and θr are the advancing and receding contact angle, respectively. Then k is given by k=
1 γ|cos θa − cos θr| 2 l
(6)
In 1962, Furmidge et al. derived another equation describing the relationship between the sliding angle and the hysteresis of the contact angle.21 When the droplet is descending a virtual displacement dx along the inclined surface, the work done by gravity must be equal to the work done by the surface tension in wetting a unit area of the surface mg ·sin α = γl (cos θr − cos θa)w
(7)
where w is the width of the droplet (w = 2r) in Figure 2.
2. ROLLING ANGLE MODEL FOR SMOOTH SURFACE 2.1. Droplet Shape and the Mass Center Dependence. The droplet on a surface is considered as a solid hemisphere with a small lack of shape at the bottom, and it is small enough so that the distortion due to the gravity can be neglected. An ideal case is considered where no hysteretic phenomenon of contact angle is assumed; i.e., advancing angle and receding angle are the same. It is noticeable that for a uniform sphere, the center of sphere (point O) and the center of mass (point O′) are coincident. But if there is a flattening on the bottom of the droplet, the sphere center and the mass center have different positions. Figure 3 shows the alternative cases of a droplet on a hydrophilic solid surface (θ ≤ 90°) or a hydrophobic one (θ > 90°), respectively. In Figure 3, R is the radium of the sphere, R′ is the height of the mass center to the solid surface, H is the total height of the droplet, and θ is the apparent contact angle. h is the distance from the sphere center to the solid surface which is defined as h = R·cos θ. It is obvious that h ≥ 0 when θ ≤ 90° and h < 0 when θ > 90°.
Figure 2. Plane view of a droplet during sliding.
For a certain droplet with a given mass m, the radius of the contact circle changes with the contact angle. In 1998, Murase et al. modified Wolfram’s empirical eq 4 and established a relationship between sliding angle and contact angle via geometric calculation.22 Substitution of (3/4)πR′3ρg = mg, R′ = [(1/4)(2 − 3 cos θ + cos3 θ)]1/3R, and r = R·sin θ into eq 4 leads to eq 8 ⎡ 9m2(2 − 3 cos θ + cos3 θ) ⎤1/3 g ρ1/3 sin α k=⎢ ⎥ 6 sin θ ⎦ ⎣ π2
(9)
(8)
where ρ is the density of the liquid and R′ and R are the radius of the droplet before and after attachment to the surface. By using this equation, the constant k for the surface can be obtained from measurable values: α, θ, and m. 8450
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The gravity of the inclined droplet is mg; the support force N is equal to the vertical component of the gravity F⊥, and the friction force f is equal to the parallel component of the gravity F∥. Considering a condition without rolling, the droplet would start moving while the parallel component of the gravity F∥ is larger than the friction force f, and this is the basic for sliding models. 2.3. Force Moment Analysis. Then we can try to analyze the force moment of the droplet. When a tilted drop starts rolling, the instantaneous contact point P can be regarded as the rolling axis of the rolling drop shown in Figure 4. The force moment gives the droplet a trend to roll. Among these forces that act on the droplet, the friction force acts through the contact point P and thus no force moment is created. On the other hand, gravity has two components. One of them is the component which is vertical to the slope F⊥; its value is equal to the support force N. Their arms of force are the same, but these two forces have opposite directions. So the force moments they created counteract each other. The parallel component of the gravity F∥ acts on the mass center of the droplet and thus creates a force moment whose rolling axis is the instantaneous contact point P. The moment should be calculated via the height of the mass center rather than the height of the center. As we described before, only gravity created a force moment, and this makes the droplet have the trend to roll down. If there are no other forces acting on the droplet, the droplet must begin to roll. But in fact, many experiences tell us that the droplet cannot roll in any case. Hence, there must exist another force moment which prevents the droplet from rolling in this system, and obviously, it comes from the adhesion force between liquid and solid. As assumed in Figure 5, once there is some water in two glass sheets, they are difficult to separate. We can know from the
Figure 3. Centers of sphere and mass (a) droplet on a hydrophilic surface (θ < 90°, h > 0); (b) droplet on a hydrophobic surface (θ > 90°, h < 0).
The distance between two centers can be easily deduced by a mathematical method as follows: |O′O| =
3(2R − H )2 4(3R − H )
=
3[2R − (R − h)]2 4[3R − (R − h)]
=
3(R + h)2 4(2R + h)
(10)
Therefore, the height of mass center could be obtained: R′ = |OO′| − h = = R·
3(R + h)2 −h 4(2R + h) 4 − (1 + cos θ)2 4(2 + cos θ)
(11)
Equation 11 gives the dependence of R′ on R and the contact angle θ. 2.2. Force Analysis. Considering a water droplet on an inclined slope shown as Figure 4, the inclined angle of the slope Figure 5. Capillary pressure between the glass sheets.
Young−Laplace equation that the pressure inside the water is less than the atmospheric pressure. The pressure difference makes the water membrane become thinner, and it presses these two glass sheets together tightly. This is the so-called “capillary pressure”. When we push the glass sheet along it, the capillary pressure will not prevent us. But if we try to separate these two glass sheets vertically, the capillary pressure will. The value of this kind of capillary adhesion force is always equal to the force which acts on the glass sheets vertically before they are greater than the critical force that can separate the two glass sheets vertically. The condition described before is similar to that of rolling. The adhesion force which exists at the interface of liquid and solid prevents the droplet from rolling. Before achieving the critical rolling condition, the force moment due to the gravity M is not big enough to conquer the preventing force moment due to the adhesion M′. They maintain a balance. While the condition changes and the balance is broken, such as reducing the surface tension or increasing the tilt angle and the mass of the droplet, three results may exist: One is the condition that
Figure 4. Force analysis of a droplet on solid slope with a slant angle α.
is α and there is a flattening on the bottom of the droplet. Point O is the center of sphere, R is the radium of the sphere, r is the radius of the contact circle of the liquid drop and solid surface, and θ is the contact angle and φ is the taper angle of the contact circle to the sphere center. Obviously, φ = 2(π − θ). The mass center of the droplet is O′, and the relationship between the height of mass center R′ and that of sphere center R is described in eq 11. 8451
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running angle. It is known that the running angle of at least δ could completely part from the solid surface. F∥ acts on the mass center of the drop, so the arm of force is the distance between the mass center and the solid surface R′ (Figure 4). Equation 11 gives the relationship between R′ and R; thus
the parallel component of gravity is greater than the friction force, but the force moment of gravity is less than the preventing force moment of adhesion. In this case, the droplet will slide down. Another condition is that the force moment of gravity can overcome that of the adhesion force, but the gravity component is less than the friction force; then the droplet will roll off rather than slide down. The last condition is that while the parallel component of gravity is greater than the friction force and the gravity force moment is larger than the preventing force moment too, the droplet will roll and slide simultaneously. It is obvious that the requirement of rolling behavior is that the force moment of gravity is greater than the preventing force moment of adhesion. Hence, the key point of this problem is the calculation of the biggest force moment that the adhesion force can make. But it is complex to calculate the exact value of the adhesion force moment. So we estimate the critical rolling condition via analyzing the energy relationship. 2.3. Energy Analysis. While the droplet is rolling, the force moment of gravity must be greater or at least equal to the force moment of adhesion. From analyzing the change of energy during the rolling process, the critical condition of rolling can be calculated. In our model, the rolling process could be divided to two steps as showed in Figure 6.
Em = M ·δ = F ·R′·δ = mg sin α·R ·
4 − (1 + cos θ)2 ·δ 4(2 + cos θ)
(14)
On the other hand, the calculation of the adhesion work can be understood from Figure 7.
Figure 7. Adhesion work of the droplet before and after contact with the smooth surface.
The total adhesive work between liquid and solid surface equals the change of energy before and after the droplet contact with the solid surface Ea = γlA a + γsA a − γslA a = (γl + γs − γsl)A a = γl(1 + cos θ Y )πr 2
Figure 6. Process of rolling: (a) original state, (b) intermediate state, and (c) final state.
= γl(1 + cos θ Y )π(R sin θ)2
In the first step, the droplet is raised up, and the liquid−solid contact changes to liquid−gas contact and solid−gas contact. The second step is the front part of the droplet rolling down to contact the slope surface and create a new liquid−solid interface. Then, the total energy taken in from this process should be the work of the moment of the parallel component of gravity and the wetting heat of the droplet contacting the front solid surface (step 2). The energy taken in would overcome the energy needed to separate the liquid−solid contact area (step 1) and would transfer to the rolling kinetic energy of the droplet:
Em + Er = Ea + E k
where γl, γs, γsl, and θY are same as in eq 1; Aa is the area of the liquid−solid contact circle (Aa = πr2); and r is the radius of liquid−solid contact circle. Substitution of eqs 14 and 15 into eq 13 leads to mg ·sin α·R ·
4 − (1 + cos θ)2 ·δ 4(2 + cos θ)
≥ γl ·(1 + cos θ Y ) ·π(R sin θ)2
(16)
Rearranging eq 16 gives
(12)
sin α ≥
where Em is the work of the moment, Er is the energy released when creating a new liquid−solid interface, Ea is the energy required when the liquid−solid contact area changes to a liquid and a solid surface, and Ek is the kinetic energy of the droplet. In the unrolling state, Er does not occur and the kinetic energy Ek equals 0. Only if the work of the moment could furnish the energy necessary to make the droplet part from the surface could the drop start rolling. In other words, the energy that makes the droplet move must be greater than the energy of the adhesion force which clutches the droplet onto the surface: Em ≥ Ea
(15)
=
4πR γl(1 + cos θ Y ) (2 + cos θ)sin 2 θ · mg δ 4 − (1 + cos θ)2
4πR γl(1 + cos θ Y ) (1 + cos θ)(2 + cos θ) · mg δ (3 + cos θ)
(17)
For a hemispherical droplet with a given mass m, its radius R and its contact angle θ are not independent values. The relationship between R and θ can be gained via simple geometry calculation ⎤1/3 ⎡ 3m R=⎢ ⎥ ⎣ ρπ(2 − 3 cos θ + cos3 θ) ⎦
(13)
(18)
where ρ is the density of the liquid. Inserting eq 18 into eq 17 leads to
Here, the work of the moment Em equals the product of the moment of the parallel component of gravity M and the 8452
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4πγl(1 + cos θ Y ) (1 + cos θ)(2 + cos θ) · mg ·δ (3 + cos θ) ⎤1/3 ⎡ 3m ⎥ ⎢ ⎣ ρπ(2 − 3cos θ + cos3 θ) ⎦
A r = A w + Ad , 1 − f =
At A A − Aw = r = d Aa Ar Ar
(21)
As mentioned above, the total adhesion work can be calculated by estimating the change of energy before and after the droplet contact with the rough surface:
1/3 4γl ⎛ 3π 2 ⎞ = ·⎜ ⎟ g ·δ ⎝ ρm2 ⎠ (1 + cos θ Y )(1 + cos θ)(2 + cos θ)
Ea = (γlA a + γsA r ) − (γlA t + γsAd + γslA w ) = γl[A a − (1 − f )A a ] + γsqfA a − γslqfA a = fA a (γl + q γs − q γsl)
(3 + cos θ)[(1 − cos θ)2 (2 + cos θ)]1/3 1/3 4γl ⎛ 3π 2 ⎞ (1 + cos θ Y )(1 + cos θ) = ·⎜ ⎟ g ·δ ⎝ ρm2 ⎠ (3 + cos θ) ⎛ 2 + cos θ ⎞2/3 ⎜ ⎟ ⎝ 1 − cos θ ⎠
= fA a γl(1 + q cos θ Y )
(22)
Inserting eq 22 into eq 13 leads to eq 23: 1/3 4f γl ⎛ 3π 2 ⎞ (1 + q cos θ Y )(1 + cos θ) sin α ≥ ·⎜ ⎟ g ·δ ⎝ ρm2 ⎠ (3 + cos θ)
(19)
⎛ 2 + cos θ ⎞2/3 ⎜ ⎟ ⎝ 1 − cos θ ⎠
As the surface is a flat smooth surface, the apparent contact angle θ just is the Young’s contact angle θY. Then eq 19 can be simplified as follows:
(23)
With this model, the factors that influence the rolling angle can be known clearly, including measurable values such as the weight, density, surface tension, and contact angle. Moreover, the morphology parameters, q and f, also play an important role in the rolling angle.
1/3 4γl ⎛ 3π 2 ⎞ (1 + cos θ)2 ⎛ 2 + cos θ ⎞2/3 ⎜ ⎟ sin α ≥ ·⎜ ⎟ g ·δ ⎝ ρm2 ⎠ (3 + cos θ) ⎝ 1 − cos θ ⎠
(20)
By using this equation, the rolling angle of a droplet on a smooth surface can be estimated from measurable values: the weight, density, surface tension, and contact angle of the droplet.
4. RESULTS AND DISCUSSION Assume that the minimum running angle δ is equal to the taper angle φ; that is, δ = 2(π − θ). If a drop of distilled water (its surface tension is 72.75 × 10−3 N/m at 293 K and its density is 1 g/mL) with the volume of 10 μL is put on a smooth surface, the dependence of the rolling angle on the contact angle can be obtained by eq 20 as shown in Figure 9.
3. ROLLING ANGLE MODEL FOR ROUGH SURFACE Early at the middle of 20th century, scientists such as Wenzel and Cassie found that surface roughness has important influence on the wettability of a surface.16,17 After decades of research and experiments, one viewpoint that has been agreed widely upon is that surface roughness and trapped air are the main roles of a superhydrophobic surface.23,25 For a rough surface, the actual contact area of liquid and solid surface is not equal to πr2. The adhesive work should be modified. Figure 8 shows the adhesion work of the droplet before and after contact on a rough surface, where Ar is the real surface
Figure 9. Dependence of the rolling angle on the contact angle on a smooth surface.
It can be found from Figure 9 that the rolling angle increases while the contact angle decreases. It is because bigger contact angle implies a lower surface energy of the solid, which means a lesser interaction between the liquid and the solid surface. The lesser the interaction, the smaller the rolling angle. A lot of experiments also indicated this relationship between contact angle and rolling angle.23 It can be found from Figure 9 that when the contact angle is lower than 130°, the rolling angle increases rapidly and when the contact angle is lower than 120°, the rolling angle is greater than 90°. The most natural explanation is to consider that the droplet cannot move by the
Figure 8. Adhesion work of the droplet before and after contact with the rough surface.
area of the rough surface (the subscript r means rough), Aw is the contact area of liquid and solid under the droplet (the subscript w means wet), Ad is the contact area of gas and solid under the droplet (the subscript d means dry), and At is the contact area of gas and liquid under the droplet. Defining the wetted fraction f = Aw/Ar (0 ≤ f ≤ 1) and the surface roughness q = Ar/Aa (q ≥ 1), we have the following relationships: 8453
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way of rolling. In fact, to a smooth surface, its wettability depends mainly on the material of the surface. In other words, the surface energy plays a crucial role in its wettability. However, even a material with the lowest surface energy (6.7 mJ/m2 for a surface with regularly aligned closest-hexagonalpacked −CF3 groups) gives a water contact angle of only around 120°.1,24 This model indicates that a water droplet cannot roll down on a normal hydrophobic surface even of fluororesin without roughness. A superhydrophobic surface is, in fact, a rough surface with a large water contact angle. This condition is expressed by eq 23, and the influence of the surface structure and other factors on the rolling angle can be known clearly by varying these parameters. For the case that the surface is without any air traps, the wetted fraction f is to have a fixed value of 1, and thus the role of surface roughness q on the rolling angle can be studied by varying the value of q in Figure 10, where the volume of
Figure 11. Influence of the surface roughness on the critical Young’s contact angle.
can be known from Figure 12, where the volume of distilled water is 10 μL, the surface tension of water is taken 72.75 ×
Figure 12. Influence of the proportion of trapped air on the rolling angle.
Figure 10. Influence of surface roughness on the rolling angle.
distilled water is 10 μL, the surface tension of water is taken as 72.75 × 10−3 N/m at 293 K, and its density taken as 1 g/mL. The relationship between the apparent contact angle and the Young’s contact angle is cos θ = fq cos θY + f − 1, which was described in Watanabe’s work.23 While the value of q changes from 1 to 3, the curve moves down. This means the rolling angle decreases when the surface becomes more rough, which has been proved by many other researchers.9,23 In fact, it is similar to the condition described in Wenzel’s model, that the hydrophobicity enhances with the increase of its roughness if the surface is composed of hydrophobic materials. On the other hand, it is noticeable in Figure 10 that when the Young’s contact angle is larger than a certain value, the rolling angle gets the minimum value (0°). For example, the critical Young’s contact angle is 120° while q = 2. This means that for the solid surfaces with roughness q = 2, the material that composes the surface is hydrophobic enough to make the water droplet roll down if the contact angle of the smooth surface reaches 120°. Figure 11 gives the relationship between the surface roughness and the critical Young’s contact angle. It is obvious that the critical Young’s contact angle decreases with the increase of surface roughness. This also illuminates that surface roughness has positive effects on rolling. Another case is that the rough surface has some trapped air, and if the surface roughness is supposed to have a certain value of 1.5, the influence of the wetted fraction f on the rolling angle
10−3 N/m at 293 K, its density is taken as 1 g/mL, and cos θ = fq cos θY + f − 1. It can be found that when the value of f changes from 0.2 to 1, the curves become steeper. In other words, the rolling angle decreases as the proportion of the trapped air becomes larger. It is similar to the condition described in Cassie’s model: when there is more air trapped in the interface below the liquid drop, the hydrophobicity of the surface increases. Similarly, the influence of the volume of droplet on the rolling angle can be known if we assume q = 1.5 and f = 0.5, which is shown in Figure 13. The surface tension of the droplet is taken 72.75 × 10−3 N/m at 293 K, and its density is taken as 1 g/mL. The rolling angle decreases when the volume of the droplet changes from 1 to 40 μL. In fact, the weight of the droplet itself will not have any effect on the hydrophobicity of the surface material. The explanation for this phenomenon is that a heavier droplet can roll off easily because the parallel component of gravity is greater.
5. CONCLUSION A rolling angle model was established by a grand new method, which aimed at the relationship between the adhesion work and the moment of the rolling droplet. For a smooth surface, the equation we obtained is the following: 8454
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(11) Lee, J. H.; He, B.; Patankar, N. A . J. Micromech. Microeng. 2005, 15 (3), 591−600. (12) Franco, J. A.; Kentish, S. E.; Perera, J. M.; Stevens, G. W. J. Membr. Sci. 2008, 318 (1−2), 107−113. (13) Fu, Q.; Rao, G. V.; Rama; Basame, S. B.; Keller, D. J.; Artyushkova, K.; Fulghum, J. E.; Lopez, G. P. J. Am. Chem. Soc. 2004, 126 (29), 8904−8905. (14) Lee, W.; Jin, M. K.; Yoo, W. C.; Lee, J. K. Langmuir 2004, 20 (18), 7665−7669. (15) Young, T. Roy. Soc. 1805, 95 (1), 65−87. (16) Wenzel, R. N. J. Phys. Colloid Chem. 1949, 53, 1466. (17) Cassie, A. B. D.; Baxter, S. Trans. Faraday Soc. 1944, 40, 546− 551. (18) Buzagh, A.; Wolfram, E. Kolloid Z. 1956, 149, 125. (19) Wolfram, E.; Faust, R. Wetting, Spreading, and Adhesion; Padday, J. F., Ed.; Society of chemical industry, Colloid and surface chemistry group, Academic Press: London, U.K., 1978; Chapter 10. (20) Buzagh, A.; Wolfram, E. Kolloid Z. 1958, 157, 50−53. (21) Furmidge, C. G. L. J. Colloid Sci. 1962, 17, 309−324. (22) Murase, H. Proceedings of the Fifth Interface Meeting of the Science Council of Japan; Tokyo, Japan, 1998 (in Japanese). (23) Masashi, M.; Akira, N.; Akira, F.; Kazuhito, H.; Toshiya, W. Langmuir 2000, 16, 5754−5760. (24) Nishino, T.; Meguro, M.; Nakamae, K.; Matsushita, M.; Ueda, Y. Langmuir 1999, 15 (13), 4321−4323. (25) Roura, P.; Fort, J. Langmuir 2002, 18, 566−569.
Figure 13. Influence of the weight of droplet on the rolling angle. 1/3 4γl ⎛ 3π 2 ⎞ (1 + cos θ)2 ⎛ 2 + cos θ ⎞2/3 ⎜ ⎟ sin α ≥ ·⎜ ⎟ g ·δ ⎝ ρm2 ⎠ (3 + cos θ) ⎝ 1 − cos θ ⎠
For rough surfaces, the roughness factor q and wetted fraction f were taken into account, and the equation is the following: 1/3 4f γl ⎛ 3π 2 ⎞ (1 + q cos θ Y )(1 + cos θ) sin α ≥ ·⎜ ⎟ g ·δ ⎝ ρm2 ⎠ (3 + cos θ)
⎛ 2 + cos θ ⎞2/3 ⎜ ⎟ ⎝ 1 − cos θ ⎠
Some factors, such as surface roughness, trapped air fraction, weight of the droplet, and hydrophobicity of the surface material, are important parameters affected on the rolling angle. The simulation results showed that the rolling angle of the droplet on a superhydrophobic surface was strongly affected by these factors. The rolling angle decreases with the increase of the contact angle, the surface roughness, the proportion of trapped air, the weight of the droplet, and the chemical composition of the surface. These results were reasonable and agreed well with the most widely used explanation of other researchers.
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AUTHOR INFORMATION
Notes
The authors declare no competing financial interest.
REFERENCES
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dx.doi.org/10.1021/jp210261b | J. Phys. Chem. C 2012, 116, 8449−8455
