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5.4: A floating non-wetting body is supported by a combination of buoyancy and curvature forces, whose relative magnitude is prescribed by the ratio of displaced fluid volumes Vb and Vm. Equations (5.27-5.31) thus indicate that the curvature force acting on the floating body is expressible in terms of the fluid volume d...
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19 Prof. John W. M. Bush 5.3. Fluid Statics Chapter 5. Stress Boundary Conditions Figure 5.5: a) Water strider legs are covered with hair, rendering them effectively non-wetting. The tarsal segment of its legs rest on the free surface. The free surface makes an angle θ with the horizontal, resulting in an upward ...
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indicates the dependence of contact length on body weight for over 300 species of water-striders, the most common water walking insect. Note that the solid line corresponds to the requirement (5.35) for static equilibrium. Smaller insects maintain a considerable margin of safety, while the larger striders live close to...
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y). We define a functional f (x, y, z) = z − h(x, y) that necessarily vanishes on the surface. The normal to the surface is defined by n = ∇f |∇f | = and the local curvature may thus be computed: zˆ − hxxˆ − hyyˆ 1/2 1 + h2 + h2 y x ) ( − (hxx + hyy) − hxxhy ∇ · n = 2 + hyyhx 3/2 2 + 2hxhyhxy ) ( 1 + h2 + h2...
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h2 r2 θ , ( ) −hθθ − h2hθθ + hrhtheta − rhr r ∇ · n = − 2 hrh2 − r2hrr − hrrh2 1/2 r 1 + h2 +r θ 1 h2 r2 θ θ + hrhθhrθ r2 ( ) In the case of an axisymmetric interface, z = h(r), these reduce to: n = zˆ − hrrˆ 1/2 (1 + h2) r , ∇ · n = −rhr − r2hrr 3/2 r2 (1 + h2) r 5.5 Appendix C : Frenet-Serret Equations ...
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42) (5.43) C I MIT OCW: 18.357 Interfacial Phenomena 21 Prof. John W. M. Bush 6. More on Fluid statics Last time, we saw that the balance of curvature and hydrostatic pressures requires −ρgη = σ∇ · n = σ We linearized, assuming ηx ≪ 1, to find η(x). Note: we can integrate directly −ηxx (1+η2 )3/2 . x ρgηηx = σ 1...
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�e, gives σ cos θe = weight of fluid d| isplaced above z = 0. σ cos θ = ρgzdx vert. proj. of T | {z } 1 Z x wei ght id of f lu } {z (6.1) (6.2) (6.3) (6.4) Note: σ cos θe = weight of displaced fluid is +/− according to whether θe is smaller or larger than π .2 Floating Bodies Without considering interfacial effects, one a...
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interaction of the menisci of floating bodies • attractive or repulsive depending on whether the menisci are of the same or opposite sense • explains the formation of bubble rafts on champagne • explains the mutual attraction of Cheerios and their attraction to the walls • utilized in technology for self-assembly on...
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S: 1 2 2 ΔP + ΔρΩ2 r = σ∇ · n curvature " centrif ugal " Nondimensionalize: Δp + 4B0 ′ = ∇ · n, , Σ = r a ( = 2 ′ aΔp ) σ 3 a ΔρΩ2 8σ = where Δp = = Rotational Bond number = const. Define surface functional: f (r, θ) = z − h(r) ⇒ vanishes on the surface. Thus ∇f n = |∇| and ∇ · n = zˆ−hr (r)ˆr...
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raindrop scaling, we expect ℓc ∼ but both σ, Δρ higher by a factor of 10 ⇒ large tektite size suggests they are not equilibrium forms, but froze into shape during flight. Q2: Why are their shapes so different from those of raindrops? Owing to high ρ of tektites, the internal dynamics (esp. rotation) dominates the ...
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in a denser fluid, spinning with angular speed Ω. The energy per unit drop volume is thus E = 1 ∆ρΩ2r2 + Minimizing with respect to r: V 4 2γ . r 2 2 (cid:1) = 1 ∆ρΩ2r − 2γ = 0, which occurs when r = d E dr V r Vonnegut’s Formula: γ = 1 ∆ρΩ2 V L as it avoids difficulties associated with fluid-solid contact. (cid:1) Note: r...
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porous media such as soil or sand. Historical Notes: • Leonar do da Vinci (1452 - 1519) recorded the effect in his notes and proposed that mountain streams may result from capillary rise through a fine network of cracks • Jacques Rohault (1620-1675): erroneously suggested that capillary rise is due to suppresion of a...
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ition Condition: I > 0. Note: since I = S + γLV , the imbibition condition I > 0 is always more easily met than the spreading condition, S > 0 ⇒ most liquids soak sponges and other porous me- values of the imbibition parameter I: dia, while complete spreading is far less common. I > 0 (left) and I < 0 (right). F...
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π/2 2. H increases as θ decreases. Hmax for θ = 0 3. we’ve implicitly assumed R ≪ H & R ≪ lC. The same result may be deduced via pressure or force arguments. By Pressure Argument Provided a ≪ ℓc, the meniscus will take the form of a spherical cap with radius R = . Therefore cos θ pA = pB − 2σ a 2σ cos θ ⇒ H =...
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side is the total momentum flux, which evaluates to πa2ρz˙ = m˙ z˙, so the force balance on the column may be expressed as dt I S 2   m + ma Inertia ; Added mass ;  z¨ = 2πaσ cos θ − mg −  1 πa2 ρz˙ 2 2 − 2πaz · τv (8.2) capillary f orce ; weight ; dynamic pressure ; viscous f orce ; 2 r 2 2 ...
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radial inflow extends to ∞, but u = U . In the spherical cap, m0 = u(r) decays. Volume conservation requires: πa2U = 2πa2ur(a) ⇒ ur(a) = U/2. Continuity thus gives: 2πa2ur(a) = 2πr2ur(r) ⇒ ur(r) = r2 Thus, the K.E. in the far field: 1 m a a ur(a) = 2r dm, where dm = ρ2πr2dr. |r=a = ef f U 2 (r)2 2 U . 1 2 2π 3 ...
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m = πa2zρ, ma = πa3ρ (added mass) and τv = 7 6 − 4µ a z˙ into (8.2) we arrive at z + a z¨ = 7 6 ) ( 2σ cos θ ρa 1 − z˙2 − 2 8µzz˙ ρa2 − gz (8.3) The static balance clearly yields the rise height, i.e. Jurin’s Law. But how do we get there? MIT OCW: 18.357 Interfacial Phenomena 28 Prof. John W. M. Bush ...
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and the force balance becomes: 1 z˙ = 2 6 σ cos θ 7 ρa2 7 6 2σ cos θ ρa az¨ = t2 . 7 6 ) ( 2 2σ cos θ ρa 7 We thus infer . As the col­ 2σ cos θ ⇒ ρa 2 4σ cos θ ρa 1/2 z˙ = U = ( 4σ cos θ ρa ) z = ( is independent of g, µ. 1/2 ) t. ρa Viscous Regime (t ≫ τ ∗) Here, inertial effects become negligible, ...
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= −z ∗ − 1 z ∗2and so infer z = 2t∗ . ∗ ∗ ∗ 2 √ Redimensionalizing thus yields Washburn’s Law: z = Note that z˙ is independent of g. [ 1/2 σa cos θ 2µ t ] Late Viscous Regime: As z approaches H, z ≈ 1. Thus, we consider t∗ = [−z ∗ −ln(1−z ∗)] = ln(1−z ∗) and so infer z = 1 − exp(−t ∗). ∗ Redimensionalizing yie...
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2 2 dt t (Washburn’s Law). 1/2 √ ∼ ( Note: Front slows down, not due to g, but owing to increasing viscous dissipation with increasing col- umn length. σa cos θ 2µ t ) Figure 8.5: Horizontal flow in a small tube. MIT OCW: 18.357 Interfacial Phenomena 29 Prof. John W. M. Bush 9. Marangoni Fl...
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normal stress jump across a fluid interface, they do not contribute to the tangential stress jump. Consequently, tangential surface stresses can only be balanced by viscous stresses associated with fluid motion. Thermocapillary flows: Marangoni flows induced by temperature gradients σ(T ). Note that in general surface...
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The Prandtl number Pr = O(1) for many common (e.g. aqeous) fluids. E.g.1 Thermocapillary flow in a slot (Fig.9.2a) Surface Tangential BCs τ = · ≪ 1 if Re ≪ 1, so one has ∇2T = 0. ν κ viscous stress U ∼ 1 H Δσ. ≈ µ U a ν = Δσ L dσ ΔT dT L µ L U H Figure 9.2: a) Thermocapillary flow in a slot b) Thermal convect...
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d U ρ 2 2 2 d .κ Criterion for Instability: τdif f τrise 3 ∼ gαΔT d κν ≡ Ra > Rac ∼ 103, where Ra is the Rayleigh number. Note: for Ra < Rac, heat is transported solely through diffusion, so the layer remains static. For Ra > Rac, convection arises. The subsequent behaviour depends on Ra and Pr. General...
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due to buoyancy ⇒ not recognized until Block (Nature 1956). 3. Pearson (1958) performed stability analysis with flat surface ⇒ deduced Mac = 80 . 4. Scriven & Sternling (1964) considered a deformable interface, which renders the system unstable at all Ma. Downwelling beneath peaks in Marangoni convection, upwelling ...
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ffects both the alcohol concentration c and temperature θ. The density ρ(c, θ) and surface tension σ(c, θ) are such that ∂ρ < 0, ∂ρ < 0, dσ < 0, dσ < 0. Evaporation results in surface cooling and so may generate either Rayleigh-B´enard or Marangoni thermal convection. Since it also induces a change in surface chemist...
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tears of wine phenomenon is shown in Fig. 10.1. Evaporation of alcohol occurs everywhere along the free surface. The alcohol concentration in the thin layer is thus reduced relative to that in the bulk owing to the enhanced surface area to volume ratio. As surface tension decreases with alcohol concentration, the su...
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by evaporation of alcohol from the free surface. 34 10.2. Surfactants Chapter 10. Marangoni Flows II Figure 10.2: a) A typical molecular structure of surfactants. b) The typical dependence of σ on surfactant concentration c. 10.2 Surfactants Surfactants are molecules that have an affinity for interfaces; common e...
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the bulk. An insoluble surfactant cannot dissolve into the bulk, must remain on the surface. • Volatility prescribes the ease with which surfactant sublimates. Theoretical Approach: because of the dependence of σ on the surfactant concentration, and the ap­ pearance of σ in the boundary conditions, N-S equations mus...
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0. Many surfactants have sufficiently small Ds that surface diffusivity may be safely neglected. MIT OCW: 18.357 Interfacial Phenomena 35 Prof. John W. M. Bush 10.2. Surfactants Chapter 10. Marangoni Flows II Figure 10.3: The footprint of a whale, caused by the whales sweeping biomaterial to the sea surface. The ...
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stress balance (through the reduction of σ), but also the tangential stress balance through the generation of Marangoni stresses. The presence of surfactants will act to suppress any fluid motion characterized by non-zero surface divergence. For example, consider a fluid motion characterized by a radially divergent su...
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, who would do similarly in an attempt to calm troubled seas. Finally, the suppression of capillary waves by surfactant is at least partially responsible for the ‘footprints of whales’ (see Fig. 10.3). In the wake of whales, even in turbulent roiling seas, one seas patches on the sea surface (of characteristic width...
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was ruffled by a slight breeze, a streak of calm such as, to use their own illustration, a cask of oil usually diffuses around it when in the water. The feverish anxiety about the missing woman suggested some strange connection between this singular calm and the mode of her disappearance. Two or three days after - why ...
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��shing ground, and he calculated, truly enough, that the fish would very soon destroy all means of identification; but it never entered into his head that as they did so their ravages, combined with the process of decomposition, set free the matter which was to write the traces of his crime on the surface of the wate...
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owing to the presence of soap, which is thus visible as a white streak. 10.3 Surfactant-induced Marangoni flows 1. Marangoni propulsion Consider a floating body with perimeter C in contact with the free surface, which we assume for the sake of simplicity to be flat. Recalling that σ may be though of as a force per un...
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��lm increases the surface area, decreases Γ and so increases σ. Fluid is thus drawn in and the film is stabilized by the Marangoni elasticity. Figure 10.5: Fluid is drawn to a pinched area of a soap film through induced Marangoni stresses. MIT OCW: 18.357 Interfacial Phenomena 38 Prof. John W. M. Bush 10.4. Bub...
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dynamics. The flow generated by a clean spherical bubble or radius a rising at low Re = U a/ν is in­ tuitively obvious. The interior flow is toroidal, while the surface motion is characterized by divergence and convergence at, respectively, the leading and trailing surfaces. The presence of surface contamination change...
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��uid jet Consider a circular orifice of a radius a ejecting a flux Q of fluid density ρ and kinematic viscosity ν (see Fig. 11.1). The resulting jet accelerates under the influence of gravity −gzˆ. We assume that the jet Reynolds number Re = Q/(aν) is sufficiently high that the influence of viscosity is negligible; furthe...
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finds 1/2 U (z) U0  1 + = 2 z Fr a acc. due to g + 1 − 2 a We r dec. due to σ (  )      � where we define the dimensionless groups � (11.3) (11.4) (11.5) Fr = We = = U 2 0 ga 2a ρU0 σ IN ERT IA GRAV IT Y = IN ERT IA CU RV AT U RE Now flux conservation requires that = F roude N umber (1...
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algebraically to yield the thread shape r(z)/a, then this result substituted into (11.5) to deduce the velocity profile U (z). In the limit of We → ∞, one obtains r a 1 + = ( 2gz 2 U0 ) −1/4 , U (z) U0 1 + = ( 2gz 2 U0 ) 1/2 (11.10) 40 11.2. The Plateau-Rayleigh Instability Chapter 1...
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equations. The perturbed columnar surface takes the form: , ˜ R0 + ǫeωt+ikz R = (11.12) where the perturbation amplitude ǫ ≪ R0, ω is the growth rate of the instability and k is the wave number of the disturbance in the z- direction. The corresponding wavelength of the varicose perturbations is necessarily 2π/k...
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ur, u˜z, p˜) = R(r), Z(r), P (r) e ) ( ωt+ikz . (11.16) Substituting (11.16) into equations (11.13-11.15) yields the linearized equations governing the per­ turbation fields: Momentum equations : ωR = − 1 dP ρ dr ik ρ P ωZ = − Continuity: dR R r dr + + ikZ = 0 . (11.17) (11.18) Figure 11.3: A cylindri...
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havedness of our solution requires that R(r) take the form ) R(r) = CI1(kr) , (11.21) where C is an as yet unspecified constant to be determined later by application of appropriate boundary conditions. The pressure may be obtained from (11.21) and (11.17), and by using the Bessel function identity I 0(ξ) = I1(ξ): ...
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�p = σ R0 − ǫσ R2 1 − k2R0 0 ( ωt+ikz 2 e ) Cancellation via (11.11) yields the equation for p˜ accurate to order ǫ: p˜ = − ǫσ R2 0 1 − k2R2 e ( ) 0 ωtikz . Combining (11.22), (11.24) and (11.29) yields the dispersion relation, that indicates the dependence of the growth rate ω on the wavenumber k: ω2 =...
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(11.32) By inverting the maximum growth rate ωmax one may estimate the characteristic break-up time: tbreakup ≈ 2.91 � 3 ρR0 σ (11.33) Figure 11.4: The dependence of the growth rate ω on the wavenumber k for the Rayleigh­ Plateau instability. Note: In general, pinch-off depends on Oh = σR . µν 1/2 2 ( ) , λ...
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waves that will travel at U and so appear to be stationary in the lab frame. For jets falling from a nozzle, the result (11.5) may be used to deduce the local jet speed. 11.3 Fluid Pipes The following system may be readily observed in a kitchen sink. When the volume flux exiting the tap is such that the falling stre...
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surfactant gradient are balanced by the viscous stresses generated within the jet. The quiescence of the jet surface may be simply demonstrated by sprinkling a small amount of talc or lycopodium powder onto the jet. The fluid jet thus enters a contaminated reservoir as if through a rigid pipe. A detailed theoretical...
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π hx π where R is the rim 1. for dependence on geometry and influence of µ, see Savva & Bush (JFM 2009). √ 2. form of sheet depends on Oh = √ µ 2hρσ . 3. The growing rim at low Oh is subject to Ra-Plateau ⇒ scalloping of the retracting rim ⇒ rim pinches off into drops 4. At very high speed, air-induced shear s...
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�uid coating on a cylindrical fiber. Define mean thickness ∗ = h λ 1 λ 0 h(x)dx (12.1) Local interfacial thickness h(x) = h ∗ + ǫ cos kx (12.2) Volume conservation requires: λ 0 π(r + h)2dx = λ 0 π(r + h0)2dx ⇒ (r + h ∗ + ǫ cos kx)2dx = (r + h0)2λ ⇒ λ (r + h ∗ )2λ + ǫ2 λ 2 = (r + h0)2λ ⇒ (r + h ∗ 0 ǫ2 )...
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(r + h∗)λ + 4 Substitute for h∗ from (12.3): J ( (r + h∗ + ǫ cos kx)(1 + ǫ2k2 sin (r + h)ds = 1/2 λ 0 λ 0 dh 1 2 f f ] 2π(r + h)ds < 2π(r + h0)λ? 1 (r + h∗)ǫ2k2λ. 4 ǫ2 1 − 4 r + h0 + 1 4 (r + h ∗ )ǫ2k2 < 0 We note that the result is independent of ǫ: k2 < (r + h0) −1(r + h ∗ −1 ≈ ) 1 (r + h0)2 i.e. unst...
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= 0 (12.7) where dp is the gradient in curvature pressure, which is independent of y ( a generic feature of lubrication problems), so we can integrate the above equation to obtain dx v(y) = 1 dp µ dx y2 2 (cid:18) − hy (cid:19) Flux per unit length: h Z Conservation of volume in lubrication problems requires that Q(x +...
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ous mode. Note: • Recall that for classic Ra-P on a cylindrical fluid thread λ∗ ∼ 9R. • We see here the timescale of instability: τ ∗ = 12µ(r+h )4 3 σh0 0 . • Scaling Argument for Pinch-off time. When h ≪ r, ∇ p ∼ σh0 1 ∼ µ v 2 ⇒ v ∼ r τ r h0 r 2 ∼ h 3 0 σh0 µ r3 ⇒ ⇒ (12.8) (12.9) (12.10) (12.11) (12.12) (12.13) τpinch ∼...
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1 − t2) = 2σxˆ (12.16) At time t = 0, planar sheet of thickness h punctured at x = 0, and retracts in xˆ direction owing to F c. Observation: The rim engulfs the film, and there is no upstream disturbance. Figure 12.4: Surface-tension-induced retraction of a planar sheet of uniform thickness h released at time t = 0. Ri...
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Rupture of a Soap Film (Culick 1960, Taylor 1960) Chapter 12. Instability Dynamics Note: Surface area of rim/ length: p = 2πR where m = ρhx = πρR2 ⇒ R = radius. Therefore the rim surface energy is σP = σ2π is σ 2x + 2(πhx)1/2 . ] [ hxπ Scale: 2x The rim surface area is thus safely neglected once the sheet has re...
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sheet and rim depend on the value of Oh. Figure 12.6: The typical evolution of a retracting sheet. As the rim retracts and engulfs fluid, it eventually becomes Rayleigh-Plateau unstable. Thus, it develops variations in radius along its length, and the retreating rim becomes scalloped. Filaments are eventually left by...
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curvature force is given by Using the first Frenet-Serret equation (Lecture 2, Appendix B), F c = 1 C σ (∇ · n) ndl thus yields (∇ · n) n = dt dl F c = σ 1 C dt dl dl = σ (t1 − t2) = 2σx (13.1) (13.2) (13.3) There is thus an effective force per unit length 2σ along the length of the sheet rim acting to ...
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Re = so inertia dominates gravity. The sheet radius is prescribed by a balance of radial forces; specifically, the inertial force must balance the curvature force: > 1000, the circular sheet is subject to the flapping instability. We thus consider U R ∼ 30·10 ∼ 30 ∼ 0.1 0.01 ν ∼ 3·104 ≫ 1. Fr = 2 103 ·10 ρU σ U ...
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a 50 Prof. John W. M. Bush 13.3. Lenticular sheets with unstable rims (Taylor 1960) Chapter 13. Fluid Sheets 13.3 Lenticular sheets with unstable rims (Taylor 1960) Figure 13.2: A sheet generated by the collision of water jets at left. The fluid streams radially outward in a thinning sheet; once the fluid ...
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(13.7) The sheet thickness is again prescribed by (13.5), but now Q = Q(θ), so the sheet radius R(θ) is given by the Taylor radius R(θ) = ρu2 Q(θ) n 4πσu (13.8) Computing sheet shapes thus relies on either experimental mea­ surement or theoretical prediction of the flux distribution Q(θ) within the sheet. 13.4 L...
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is defined to be the distance from the origin to the rim centreline, and un(θ) and ut(θ) the nor­ mal and tangential components of the fluid velocity in the sheet where it contacts the rim. v(θ) is de­ fined to be the velocity of flow in the rim, R(θ) the rim radius, and Ψ(θ) the angle between the po­ sition vector r an...
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flow into the rim from the fluid sheet and the centripetal force resulting from the flow along the curved rim: ρu2 h + n 2 ρπR2v rc = 2σ (13.10) Note that the force balance (13.7) appropriate for sheets with unstable rims is here augmented by the centripetal force. The tangential force balance at the rim requires ...
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rc = sin Ψ r ∂Ψ ∂θ ( + 1 ) (13.11) (13.12) (13.13) The system of equations (13.9-13.13) may be nondimensionalized, and reduce to a set of coupled ordinary equations in the four variables r(θ), v(θ), R(θ) and Ψ(θ). Given a flux distribution, Q(θ), the system may be integrated to deduce the sheet shape. MIT OC...
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13.5. A recent review of the dynamics of water bells has been written by Clanet (Ann.Rev.). We proceed by outlining the theory required to compute the shapes of water bells. We consider a fluid sheet extruded radially at a speed u0 and subsequently sagging under the influence of a gravitational field g = −gzˆ. The inn...
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orem indicates that 2 2 = u + 2gz 0 u (13.15) The total curvature force acting normal to the bell surface is given by 2σ∇ · n = 2σ 1 rc ( + cos φ r ) (13.16) Note that the factor of two results from there being two Figure 13.5: A water bell produced by the free surfaces. The force balance normal to the ...
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or, the bell with also close due to σ MIT OCW: 18.357 Interfacial Phenomena 53 Prof. John W. M. Bush 13.6. Swirling Water Bell Chapter 13. Fluid Sheets 13.6 Swirling Water Bell Consider the water bell formed with a swirling jet (Bark et al. 1979 ). Observation: Swirling bells don’t close. Why no...
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Wind over water: A layer of fluid of density ρ+ moving with relative velocity V over a layer of fluid of density ρ− . Define interface: h(x, y, z) = z − η(x, y) = 0 so that ∇h = (−ηx, −ηy, 1). The unit normal is given by nˆ = ∇h |∇h| = (−ηx, −ηy, 1) 2 + ηy ηx 2 + 1 ) ( 1/2 Describe the fluid as inviscid and irr...
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± ∂x ) (−ηx) + ∂φ± ∂y (−ηy) + ∂φ± ∂z (14.1) (14.2) (14.3) (14.4) Linearize: assume perturbation fields η, φ± and their derivatives are small and therefore can neglect their products. Thus ηˆ ≈ (−ηx, −ηy, 1) and ∂η = ± 1 V ηx + ⇒ 2 ∂φ± ∂z ∂t ∂φ± ∂z = ∂η ∂t 1 ∂η ∓ V 2 ∂x on z = 0 (14.5) 3. No...
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gη + (ρ+ ∂φ± ∂t − ρ− ∂φ− ∂t ) + V 2 (ρ− ∂φ− ∂x + ρ+ ∂φ+ ∂x ) = −σ(ηxx + ηyy) (14.8) is the linearized normal stress BC. Seek normal mode (wave) solutions of the form η = η0e iαx+iβy+ωt φ± = φ0±e ∓kz iαx+iβy+ωt e where ∇2φ± = 0 requires k2 = α2 + β2 . V ∂η Apply kinematic BC: ∂x ∂φ± ∂z ∓ 1 2 ...
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so ω2 + iαV ρ− − ρ+ ρ− + ρ+ ( ) ω − α2V + k2C0 2 2 = 0 1 4 where C 2 ≡ k. Dispersion relation: we now have the relation between ω and k σ ρ−+ρ+ ρ−−ρ+ ρ−+ρ+ + g k ( ) 0 ω = 1 2 i ( ρ+ − ρ− ρ− + ρ+ ) k · V ± [ ρ−ρ+ (ρ− + ρ+)2 2 (k · V ) − k2C 2 0 1/2 ] where k = (α, β), k2 = α2 + β2 . The system i...
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overlies light fluid: ρ+ > ρ−, V = 0. Via (14.15), the system is unstable if C 2 0 = ρ− − ρ+ g ρ+ρ− k + σ ρ− + ρ+ k < 0 (14.16) i.e. if ρ+ − ρ− > 2 4π σ σk g = gλ2 . 2 Thus, for instability, we require: λ > 2πλc where λc = σ Δρg J is the capillary length. Heuristic Argument: Change in Surface Energy: ΔE...
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In a finite container with width smaller than 2πλc, the system may be stabilized by σ. 4. System may be stabilized by temperature gradients since Marangoni flow acts to resist surface defor­ mation. E.g. a fluid layer on the ceiling may be stabilized by heating the ceiling. Figure 14.3: Rayleigh-Taylor instability may be...
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one can find a critical V that destabilizes the system. Marginal Stability Curve: Figure 14.4: Kelvin-Helmholtz instability: a gravi­ tationally stable base state is destabilized by shear. V (k) = ρ− − ρ+ g ρ−ρ+ k ( + 1 ρ−ρ+ σk ) 1/2 (14.19) dV dk = 0, implies − Δρ + σ = 0 ⇒ kc = V (k) has a minimum where ...
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is, Wetting of textured solids Recall: In Lecture 3, we defined the equilibrium contact angle θe, which is prescribed by Young’s Law: cos θe = (γSV − γSL) /γ as deduced from the horizontal force balance at the contact line. Work done by a contact line moving a distance dx: Figure 15.1: Calculating the work done by mov...
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which point the contact line advances • begin with a drop in equilibrium with θ = θe • drain drop slowly with a syringe • θ decreases progressively until attaining θr, at which point the contact line retreats Origins: Contact line pinning results from surface heterogeneities (either chemical or textural), that pr...
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balance requires: } weight {z | } 2σ R (cos θ −1 cos θ2) = ρgH (15.3) Thus, an equilibrium is possible only if 2σ (cos θ −r ρgH. R cos θa ) > Note: if θa = θr (no hysteresis), there can be no equilib- of gravity. rium. Figure 15.4: A heavy liquid column may be trapped in a capillary tube despite the effects MIT OCW: 18....
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Figure 15.5: A raindrop may be pinned on a window pane. r = Total Surface Area Projected Surf. Area > 1 φS = Area of islands Projected Area < 1 The change in surface energy associated with the fluid front advancing a distance dz: dE = (γSL − γSV ) (r − φS)dz + γ(1 − φS)ds (15.4) (15.5) Spontaneous Wetting ...
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� < θe when θe < π/2 (hydrophilic) θ∗ > θe when θe > π/2 (hydrophobic). The intrinsic hydrophobicity or hydrophilicity of a solid, as prescribed by θe, is enhanced by surface roughening. Figure 15.6: A drop wetting a rough solid has an effective contact angle θ∗ that is generally different from its equilibrium value θe....
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depends on surface texture) ⇒ demi-wicking / complete wetting 3. Wenzel state breaks down at large r ⇒ air trapped within the surface roughness ⇒ Cassie State 15.4 Cassie-Baxter State Figure 15.7: The wetting of a rough solid in a Wenzel state. In a Cassie state, the fluid does not impregnate the rough solid, lea...
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wetting of a rough solid in a Cassie-Baxter state. −1+φS For dEW > dEc, we require −r cos θe + cos θ∗ > −φs cos θe + (1 − φs) + cos θ∗, i.e. cos θe < r−φS = cos θc, i.e. one expects a Cassie state to emerge for cos θe > cos θc. Therefore, the criterion for a Wenzel State giving way to a Cassie state is identical t...
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0 (e.g. Lichen surface McCarthy) 3. MIT OCW: 18.357 Interfacial Phenomena 63 Prof. John W. M. Bush 16. More forced wetting Some clarification notes on Wetting. Figure 16.1: Three different wetting states. Last class, we discussed the Cassie state only in the context of drops in a Fakir state, i.e. suspended ...
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Figure 16.2: Wetting of a tiled (chem­ ically heterogeneous) surface. cos θ ∗ = θS cos θe − 1 + θS (16.2) as previously. As before, in this hydrophobic case, the Wenzel state is energetically favourable when dEW <dEC, i.e. cos θC < cos θe < 0 where cos θC = (θS − 1)/(r − θS), i.e. θE is between π/2 and θC. Howeve...
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: θe < π/2 Chapter 16. More forced wetting Figure 16.3: Relationship between cos θ∗ and cos θe for different wetting states. 16.2 Hydrophilic Case: θe < π/2 Here, the Cassie state corresponds to a tiled surface with 2 phases corresponding to the solid (θ1 = θe, f1 = φS) and the fluid (θ2 = 0, f2 = 1 − φS). Cassie-Bax...
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and system is described by “Wet Cassie” state. Johnson + Dettre (1964) examined water drops on wax, whose roughness they varied by baking. They showed an increase and then decrease of Δθ = θa − θr as the roughness increased, and system went from smooth to Wenzel to Cassie states. Water-repellency: important for cor...
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16. More forced wetting Figure 16.4: Contact angle as a function of surface roughness for water drops on wax. Figure 16.5: To remain in a Cassie state, the internal drop pressure P0 + 2σ/R must not exceed the curvature pressure induced by the roughness, roughly σ/ℓ. 16.3 Forced Wetting: the Landau-Levich-Derjaguin ...
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is this capillary suction inside the meniscus that resists the rise of thin films. √ √ √ MIT OCW: 18.357 Interfacial Phenomena 66 Prof. John W. M. Bush 16.3. Forced Wetting: the Landau-Levich-Derjaguin Problem Chapter 16. More forced wetting Thin film wetting We describe the flow in terms of two distinct re- gions...
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. /3 2 cCa where ℓc = Ca = Ca1/3 ≪ 1. Matched asymptotics give σ ,ρg µV .σ q 3 , or equivalently E.g.1 Jump out of pool at 1m/s: Ca ∼ 10−2 so h ∼ 0.1mm ⇒ ∼ 300g entrained. E.g.2 Drink water from a glass, V ∼ 1cm/s ⇒ Ca ∼ 10−4. Figure 16.7: Left: A static meniscus. Right: Meniscus next to a wall moving upwards with spee...
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: e 1. gravity prevents meniscus from extending to ∞ ⇒ h deduced by cutting it off at ℓc. 2. h is just a few times b (h ≪ ℓc) ⇒ lateral extent greatly exceeds its height. Forced wetting on fibers e.g. optical fiber coating. where h ≈ b ln(2ℓc/b). + 1 R2 z−h b = 0. ) Figure 17.2: Etching of the microtips of Atomic...
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a water-filled pipette, pumping oil out of rock with water. Figure 17.4: Left: Displacing a liquid with a vapour in a tube. Right: The dependence of the film thickness left by the intruding front as a function of Ca = µU/σ. In the limit of h ≪ r, the pressure gradient in the meniscus ∇p ∼ meniscus. As on a fiber, pre...
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low speeds, the contact line advances at the dynamic contact angle θd < θe • dynamic contact angle θd decreases progressively as U increases until U = UM . • at sufficiently high speed, the contact line cannot keep up with the imposed speed and a film is entrained onto the solid. Now consider a clean system free of hy...
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ussan 1979 ), ad­ vancing though a rolling motion. Flow near advancing contact line We now consider the flow near the contact line of a spreading liquid (θd > θe): • consider θd ≪ 1, so that slope tan θd = ≈ θd ⇒ z ≈ θdx. z x • velocity gradient: dU ≈dz U θdx • rate of viscous dissipation in the corner Φ = µ 1 1...
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µ ∗ ≈ 30m/s. U = ∗ U 6ℓD 2 − θ2 e θd θd e ) (17.3) (17.4) (17.5) (17.6) MIT OCW: 18.357 Interfacial Phenomena 71 Prof. John W. M. Bush 17.1. Contact Line Dynamics Chapter 17. Coating: Dynamic Contact Lines Note: 1. rationalizes Hoffmann’s data (obtained for θe = 0) ⇒ U ∼ θ3 D 2. U = 0 for θd =...
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further reading: John E. Simpson - Gravity Currents: In the Environment and the Laboratory. Stage I: Re ≫ 1 Flow forced by gravity, and resisted by fluid iner­ tia: ∼ 2 ρ U R ⇒ U ∼ √ g ′h where g ′ = Δρgh R Δρ g. ρ Continuity: V = πR2(t)h(t) = const. ⇒ h(t) ∼ R2(t) g ′V 1 ⇒ RdR ∼ g ′V dt ⇒ R(t) ∼ ⇒ U ≡ ...
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∇pg ∼ R Continuity V = πR2(t)h(t) =const. Which dominates? important as the drop spreads !? Recall: ∼ ⇒ gravity becomes progressively more , Curvature: ∇pc ∼ R3 . = Bond number. Bo = Δpg Δpc ρgR γ ρgV γh ∼ 1 h γh 2 • drop behaviour depends on S = γSV − γSL − γ. • When S < 0: Partial wetting. Spreading...
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lie in the force imbalance at the contact line (S > 0) and its stability results from interactions between the fluid and solid (e.g. Van der Waals) Physical picture Force at Apparent Contact Line: γ cos θd − γSL = θd. γ(1 − cos θd) ≈ F = γ + γSL − for small 2 θ γ d 2 ∗ U 2 . Note: F ≪ S. Now recall from las...
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) (T anner ′ s Law) 1/10 ) , which is consistent with observation. (18.5) 18.2 Immiscible Drops at an Interface Pujado & Scriven 1972 Gravitationally unstable configurations can arise (ρa < ρb < ρc or ρc < ρa < ρb). • weight of drops supported by interfacial tensions. • if drop size R < lbc ∼ γbc (ρb−ρc)g , it...
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/4 SU µ 1/4 ( ) Phase IIIb Spreading driven by S, resisted by viscous dissipation in the underlying fluid. Blasius boundary layer grows on base of spreading current like δ ∼ νt. µ U πR2 ∼ S · 2πR where δ ∼ νt1/2 ⇒ R ∼ ν1/4t3/4 √ √ 1/2 √ . νt ⇒ R dR ∼ S µ dt δ S µ ( ) 18.4 Oil on water: A brief review When...
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Bush 18.5. The Beating Heart Stocker & Bush (JFM 2007) Chapter 18. Spreading 18.5 The Beating Heart Stocker & Bush (JFM 2007) . When a drop of mineral oil containing a small quantity of non-ionic, water-insoluble surfactant (Tergitol) Figure 18.5: An oil drop oscillates on the water surface. Note the ring of imp...
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Γ • for Γ = 0 no beating - stable sessile lens • for moderate Γ steady beating observed • for high Γ drop edges become unstable to fingers • for highest Γ, lens explodes into a series of smaller beating lenses. • beating marked by slow expansion, rapid retraction • odour of Tergitol always accompanies beating MIT...
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“beat­ ing heart”. Another surfactant-induced auto-oscillation: The Spitting Drop (Fernandez & Homsy 2004) • chemical reaction produces surfactant at drop surface • following release of first drop, periodic spitting • rationalized in terms of tip-streaming (Taylor 1934), which arises only in the presence of surfacta...
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∂φ on z = ζ. ∂z = 0 on z = −h ∂φ ∂ζ ∂x ∂x 1 2 + 2 ζxx (1+ζ2 x Recall: unsteady inviscid flows Navier-Stokes: )3/2 is the surface pressure. Figure 19.1: Waves on the surface of an inviscid ir­ rotational fluid. ∂u ∂t ρ 1 2 2 ) [ ( + ρ ∇ u − u × (∇ × u) = −∇ (p + Ψ) (19.1) ] 2 For irrotational flows, u = ∇...
https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf
∂φ + ρgζ + p0 − σζxx = f (t) on z = 0. Seek solutions: ζ(x, t) = i.e. travelling waves in x-direction with phase speed c and wavelength λ = 2π/k. Substitute φ into ∇2φ = 0 to obtain φˆzz − k2φˆ = 0 Solutions: φˆ(z) = e , e−kz or sinh(z), cosh(z). To satisfy B.C. 1: ∂ ˆ From B.C. 2: = 0 on z = −h so choose φˆ(z) = ...
https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf