5.2 Surface Tension Capillary Pressure: The Young-Laplace Equation. Figure 5.1 Origin of surface tension at liquid-vapor interface.

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1 5.2.1 Capillary Pressure: The Young-Laplace Equation Vapor Fo Fs Fs Fi Figure 5.1 Origin of surface tension at liquid-vapor interface. Liquid 1

2 5.2.1 Capillary Pressure: The Young-Laplace Equation Figure 5.2 Arbitrarily-curved surface with two radii of curvature 2

3 Displace the surface, the change in area is A = ( x + dx ) ( y + dy ) xy (5.1) If dxdy 0, then A = y dx + x dy (5.2) The energy required to displace the surface δ W = σ ( x dy + y dx) (5.3) Energy attributed to generating this pressure δ W = p xy dz = pcap xy dz (5.4) 3

4 From the geometry of Fig. 1 x + dx x = RI + dz RI or Similarly (5.5) x dz dx = RI (5.6) y dz dy = RII (5.7) 4

5 For equilibrium, the two expressions for energy must be equal σ ( x dy + y dx) = p xy dz (5.8) xy dz xy dz = p xy dz σ + RII RI (5.9) The pressure difference Δp=(ρv-ρl) between the two 1 phases becomes 1 pcap = p = σ + (5.10) RI RII This expression is called the Young-Laplace equation, and it is the fundamental equation for capillary pressure. 5

6 5.2.2 Interface Shapes at Equilibrium Figure 5.3 Shape of the liquid-vapor interface near a vertical wall. 6

7 The first principal radius of curvature RI 1 d 2 z / dy 2 = RI 1 + (dz / dy ) 2 [ ] (5.11) 3/ 2 The Young-Lapace equation becomes pv pl = σ / RI (5.12) At a point on the interface that is far away from the wall, the liquid and vapor pressures are the same, i.e., pv (,0) = pl (,0) (5.13) The liquid and vapor pressures near the wall are pv ( z ) = pv (,0) ρ v gz (5.14) pl ( z ) = pl (,0) ρ l gz (5.15) The relationship between the pressures in two phases can be obtained by combining eqs. (5.13) (5.15), i.e., (5.16) pv pl = g ρ l ρ v z ( ) 7

8 Combining eqs. ( ), the equation for the interface shape becomes 1+ g (ρ l ρ v ) z σ dz dy 3/ 2 d2z = 0 2 dy (5.17) Multiplying this equation by dz/dy and integrating gives g (ρ l ρ v ) z + 1+ σ 2 2 dz dy 2 1/ 2 = C1 (5.18) Since at y both z and dz/dy equal zero, C1 = 1. The boundary condition for equation (5.18) is (5.19) (dz / dy ) y = 0 8

9 Equations (5.18) and (5.19) can then be 1/ 2 solved for z(0) 2σ (5.20) z0 = z (0) = (ρ ρ v )g Using (5.20) as a boundary condition, integrating eq. (5.18) 2L z z (5.21) 2L y = cos h c Lc z0 where 1 1 cos h c z c + 4+ L σ Lc = (ρ ρ v ) g 1/ 2 2 1/ Lc 1/ 2 (5.22) 9

10 Example 5.1 Two parallel plates are put into bulk water at the bottom end (see Fig. 5.4). Estimate the minimal distance between the plates at which the central point of the liquid-air interface between the plates is not elevated from the bulk water level at equilibrium condition. Assume that water completely wets the material of the plates. The system temperature is 20 C, σ = N/ m, and ρ=l 999 kg/m. 10

11 2y z y water Figure 5.4 Two parallel plates in bulk water. 11

12 Solution: Since the density of vapor is much less than that of the liquid eq. (5.22) can be simplified to obtain σ Lc = ( ρ ρ ) g v l 1/ 2 σ B ρ g l 1/ = / 2 ρv =ρl = m The rise of the liquid surface at y=0 is obtained from eq. (5.20): 2σ z0 = z (0) = ( ρ ρ ) g v l = / 2 1/ 2 2σ = ρ g l 1/ 2 = m 12

13 The value of y for z = 0 can be found by using eq. (5.21), i.e., 2L y = cos h c Lc z0 1 2L cos h c z = cos h = / 2 1 z Lc 1/ 2 z Lc cos h / / 2 i.e., y = Lc = = m = 2.48mm This value of y is one-half the minimal distance between the plates. Therefore, the minimal distance is about 5.96 mm. 13

14 5.2.3 Effects of Interfacial Tension Gradients Since surface tension depends on temperature, permanent nonuniformity of temperature at a liquid-vapor interface causes a surface tension gradient. This may in turn establish a steady flow pattern in the liquid. 14

15 The surface tension of a multicomponent liquid that is in equilibrium with the vapor is a function of temperature and composition of the mixture, i.e., σ = σ (T, x1, x2,l, xn 1 ) (5.23) The change of surface tension can be caused by either change of temperature or composition, i.e., σ dσ = dt + T xi N 1 σ dxi i = 1 xi T, x j i (5.24) 15

16 Curve-fit equations for surface tension are almost linear σ = C0 C1T (5.25) This motion of a liquid caused by a surface tension gradient at the interface is referred to as the Marangoni effect. The most well-known example of surfacetension-driven flow is Bernard cellular flow, which occurs in a thin horizontal liquid layer heated from below. 16

17 hδ Figure 5.5: Cellular flow driven by surface tension gradient. 17

18 Boundary condition at the interface τ yx x = δ u σ T = µl = x T x x= δ x= δ (5.26) Local temperature Velocities T = T + T = (Tw ζ y ) + T (5.27) v = v + v = v (5.28) u = u + u = u (5.29) 18

19 Substituting eqs. (5.27)-(5.29) into the continuity, momentum and energy equations, and subtracting the corresponding base flow equations (5.30) T = θ ( y )eiα x + β t v = V ( y )eiα x + β t The case of marginal stability 8αˆ (αˆ cosh αˆ + Bi sinh αˆ )(αˆ sinh αˆ cosh αˆ ) Ma = 3 αˆ cosh αˆ sinh 3 αˆ (8Crαˆ 5 cosh αˆ ) /( Bo + αˆ 2 ) (5.31) (5.32) where Ma is the Marangoni number, where α=2π/λ ζ (dσ / dt )δ Ma = α lµ l 2 (5.33) 19

20 Figure 5.6 Stability plane for the onset of cellular motion 20

21 Wave number Biot number Bond number Bo = Crispation number αˆ = α δ (5.34) hδ δ Bi = kl (5.35) ( ρ l ρ g ) gδ 2 σ µ lα Cr = σδ l (5.36) (5.37) 21

22 The liquid film is stable if its Marangoni number is below that predicted by eq. (5.32). The marginal stability predicted by eq. (5.32) for some typical combinations of parameters is illustrated in Fig Since the Fourier components of all wavelengths can be contained in a random disturbance, the system becomes unstable when it is unstable at any wavelength. For a system with Cr < 10 4 and Bi 0, Bo 0, Fig 5.6 indicates that the critical Marangoni number is about 80 and the associated dimensionless wave number αˆ = 2. The Marangoni effect can have an important influence on the evaporation of a falling film, and cause vapor bubbles in a liquid with a temperature gradient to move toward the high temperature region during boiling. 22

23 Example 5.2 A 0.3-mm-thick water film sits on a surface held at a temperature of Tw = 80 C. The top of the liquid film is exposed to air at a bulk temperature of TG = 20 C, and the convective heat transfer coefficient between the liquid film and the air is hδ = 10 W/m2K. Determine whether the water film is stable. 23

24 Solution: Since the liquid film is very thin, the temperature drop across the liquid film will be very small. The properties of the liquid film can be determined at the wall temperature of 80 C, i.e., kl = 0.67 W/m-K, ρ l = kg/m3, α l = 1.64x10-7 m2/s, µ l= 351.1x10-6 N-s/m2, σ = N/m, and dσ/dt = -1.7x10-4 N/m C. The Biot number is hδ δ Bi = = = kl 0.67 At steady-state, the surface temperature of the interface, Tδ, satisfies Tw Tδ kl = hδ (Tδ TG ) δ 24

25 i.e., hδ δ Tδ = Tw + TG kl hδ δ Tw + BiTG 1+ = kl 1 + Bi o = = C which demonstrates that the temperature drop across the liquid film is minimal. The Bond number is obtained by using eq. (5.37), i.e., Bo = ( ρ l ρ g ) gδ σ 2 ρ gδ B l σ The Crispation number Cr is = = µ lα l Cr = = = σδ

26 which is less than The critical Marangoni number, Mac, below which the liquid film is stable, can be obtained from Fig. 5.4; its value is 80 at αˆ = 2. The Marangoni number of the system can be obtained from eq. (5.33), i.e., ζ ( dσ / dt )δ 2 Tδ Tw ( dσ / dt )δ 2 Ma = = α lµ l δ α lµ l ( ) = = > Ma c Therefore, the system is unstable and Marangoni convection will occur. 26