Boiling Heat Transfer and Two-Phase Flow Fall 2012 Rayleigh Bubble Dynamics Derivation of Rayleigh and Rayleigh-Plesset Equations: Let us derive the Rayleigh-Plesset Equation as the Rayleigh equation can be otained from this equation by neglecting viscous effects and surface tension. Let us first begin the derivation of the R-P equation by considering the velocity of some point in space outside the bubble. The velocity at this point is denoted u(r,t) and is a function of the radial coordinate extending from the center of the radius as well as time. By making use of inverse square law to the conserved quantity of mass, we may re-cast the the form of the velocity as a function that varies only with time divided by the radial coordinate squared: u(r,t) = F(t) r 2 Let us now consider a mass balance at the interface of the bubble and surrounding fluid. The mass rate of evaporation of vapor in the bubble can be expressed as: ṁ v = ρ r 4πR 2 dr where ρ v is the vapor density at the saturation temperature within the bubble and R is the radius of the bubble. This mass flow rate must balance with the rate of liquid entering the bubble from the surrounding fluid. Considering the relative velocity between the bubble wall and the surrounding fluid, u l, the mass flow rate of liquid entering the bubble may be expressed as: ṁ l = ρ l 4πR 2 u l The equation is negative as the mass is flowing into the bubble, i.e. v n is negative. Summing the mass flow rates due to evaporation and liquid entering the bubble and re-arranging results in the following expression for the velocity of liquid, u l. u l = ρ v dr ρ l At the bubble surface, the velocity u(r = R,t is the difference between the bubble wall speed and the velocity of liquid moving toward, i.e. into the bubble. u(r = R,t) = dr u l Substituting the previously found expression for u l results in: u(r = R,t) = dr ( 1 ρ ) v ρ l Substituting our assumed expression for the velocity u(r,t) = F(t) results in the following: r 2 F(t) = R 2 dr ( 1 ρ ) v ρ l 1
But if the density of the vapor much less than that of the liquid ρ v << ρ l, the term in the parentheses is nearly equal to one and the unknown function F(t) is: F(t) = R 2 dr Now if we consider the momentum balance in the radial direction (no momentum transfer in other two coordinates): u t + u u r = 1 ( ( p 1 ρ l r + ν l r 2 r 2 u ) 2u ) r r r 2 Substitution of u = F(t) r 2 into the radial momentum equation results in: 1 df r 2 2F2 r 5 = 1 p ρ l r Notice that the viscous terms in the equation have vanished. Integrating from r to results in the following: 1 df r F2 2r 4 = P r P ρ l Substitution of F(t) = R 2 dr into this expression and reducing results in: R R + 3 2Ṙ2 = P R P ρ l (1) We now remain with one more step. In the above equation, we have the driving pressure differnential expressed in terms of P R and P. However, P R is not readily known, rather, we would like to have the driving pressure expressed in terms of the bubble pressure P B and the pressure at infinity P. In order to accomplish this task, we consider a force balance between the bubble surface tension and the pressure within as well as external to the bubble. 2S R = P B P e If we consider an element of fluid just outside/adjacent to the bubble surface, we can see that the external pressure acting on the bubble must balance with the normal stress of the fluid, i.e. equal but oppostie in sign. 2S R = P B + σ rr If the liquid is considered to be incompressible, the normal stress will be comprised of a normal component due to pressure and an additional component due to rate of strain. The pressure is equal in magnitude to the pressure at the bubble wall, i.e. P = P R but opposite in sign. σ rr = P + 2µ u r = P R + 2µ u r 2
Susbtitution of this expression back into the surface tension force balance results in: 2S R = P B P R + 2µ u r Recalling that u = F(t) r 2 P R = 2S R + P B 4µ l r and F = R 2 dr, the expression for the P R becomes: dr Substituting Equation 2 into Equation 1 results in the Rayleigh-Plesset equation: (2) R R + 3 2Ṙ2 + 4ν lṙ R + 2S ρ l R = P B P l ρ l Rayleigh-Plesset Equation Neglecting viscous and surface tension forces, the R-P equation becomes the Rayleigh equation: R R + 3 2Ṙ2 = P B P l ρ l Rayleigh Equation Problem posed: Determine the variation in bubble radius as a bubble moves up through a column of water. Assume that the bubble is composed entirely of a perfect gas and there is no mass diffusion in or out of the bubble. Utilize both the Rayleigh and the Rayleigh-Plesset equations to describe the time-rate-of-change of the bubble radius and compare the two results. A separate dynamics equation will be needed for the time-varying change in bubble height as it moves up through the water column. Solution Assumptions: Dynamics Equation for the Motion Bubble Height: The dynamic equation for bubble height may be derived from Newton s Second Law of Motion accounting for the external forces due to gravity, buoyancy, and drag. Taking a coordinate system where h = 0 at the bottom of the water column and is positive in the upward direction, i.e. as the bubble moves toward the liquid surface, the governing differntial equation is: d 2 h 2 = ρ liquidv disp g g C DρA m 2m ( ) dh 2 The displaced volume of the bubble may be related to the bubble radius assuming that the bubble is a sphere: d 2 h 2 = 4πgρ liquid 3m R 3 g C DρA 2m ( ) dh 2 Numerical Discretization of Governing Differntial Equations: 3
Given that the Rayleigh-Plesset and dynamic equation for bubble height are of second-order, we will transform both into two first-order differential equations through a change of variables. This is possible given that the dynamics problems at hand are initial-value type problems of the general form: d 2 ( y dx 2 = G x,y, dy ) dx y(x 0 ) = y 0, dy dx (x 0) = y 0 Making a change of variables transforms the equation into a system of two, coupled, first-order differential equations: dy dx = u du dx = G(x,y,u) y(x 0 ) = y 0 u(x 0 ) = y 0 Performing this transformation for the Rayleigh-Plesset Equation: dr = α dα = 3 2R α2 4ν liquid R 2 α 2S ρ liquid R 2 + P bubble P liquid Rρ liquid R(0) = R 0 α(0) = R 0 In a similar fashion for the Rayleigh Equation: dr = β dβ = 3 2R β 2 + P bubble P liquid Rρ liquid R(0) = R 0 β(0) = R 0 For the dynamic height equation: dh = ζ dζ = 4πgρ liquid 3m R 3 g C DρA 2m ζ 2 h(0) = h 0 ζ (0) = h 0 The equations can be solved numerically using finite differences with a constant step size, t. For example, discretization the dynamic height ODEs using Euler s method results in the following (assuming the index i starts at 1): h i+1 = h i + tζ i ζ i+1 = ζ i + t ( 4πgρliquid 3m R 3 i g C ) DρA 2m ζ i 2 h i=1 = h 0 ζ i=1 = h 0 However, the Euler method (simplest form of Runge-Kutta method) is not very accurate with a global truncation error t. As a result, the more accurate classical 4th-order Runge-Kutta method 4
will be employed for numerical solution of the differential equations. The 4th-order Runge-Kutta method has a global truncation error ( t) 4 and can be written for the solution to y = f (t,y) as: y i+1 = y i + 1 6 (K 1 + 2K 2 + 2K 3 + K 4 ) K 1 = t f (t i,y i ) K 2 = t f (t i + 0.5 t,y i + 0.5K 1 ) K 3 = t f (t i + 0.5 t,y i + 0.5K 2 ) K 4 = t f (t i + t,y i + K 3 ) Numerical Solution Procedure: The numerical solution proceeds as follows: 1. The dynamic equation for the bubble radius is solved, R i R i+1. 2. The dynamic equation for the height is solved, h i h i+1. 3. The pressure in the bubble is updated (P bubble,i P bubble,i+1 ) based on the newly found bubble radius, R i+1, using the perfect gas relation. P bubble,i+1 = 4m RT 3πMR 3 i+1 4. The surrounding pressure of the bubble is updated (P liquid,i P liquid,i+1 ) based on the newly computed height of the bubble in the water column, h i+1. P liquid,i+1 = P bottom ρgh i+1 In this equation, m is the mass of fluid (assumed constant), R is the universal gas constant, and M is the molar mass of the gas within the bubble. 5. Repeat step 1 with updated pressures. Computational Parameters: 1. The time-step sizes chosen for the Euler and 4th-order Runge-Kutta methods were t = 1.0E 04 and t = 1.0E 03 respectively. 2. The Euler solutions were run for 44 iterations and the RK4 solutions were run for 375 iterations with water as the working fluid. When the working fluid was switched to oil, the RK4 solution was run for 250 iterations. Physical Parameters: Physical parameters remaining constant with changes in working fluid: 1. Gravitational acceleration, g = 9.8 m s 2. 5
2. Drag coefficient of a sphere, C D = 0.47. 3. Temperature (bubble growth assumed isothermal), T = 300 K. 4. Universal gas constant, R = 8.31446 J mol K. 5. Molar mass of dry air (what is assumed to be inside bubble), M = 0.02896 kg mol. 6. Mass of air, m = 1.0 kg. Physical parameters with water as working fluid: 1. Approximate density of water at 300K, ρ = 999 kg m 3. 2. Kinematic viscosity of water at 300K, ν = 1.4E 06 m2 s. 3. Surface tension of water at 300 K, S = 7.197 N m. Physical parameters with mineral oil as working fluid: 1. Approximate density of oil at 300K, ρ = 870 kg m 3. 2. Kinematic viscosity of oil at 300K, ν = 1.26E 04 m2 s. 3. Surface tension of water at 300 K, S = 2.4E 02 N m. Numerical Results: It is clear from the results in Figure 1 for the variation in bubble radius over time, that the Euler method does not perform well in capturing the non-linear behavior of the governing differential equation and the produces highly erroneous results if the RK4 method is considered to be correct or nearly so. Figure 1: Bubble Radius vs. time, RK4 and Euler 6
The inaccuracy of the Euler method is also visible in Figure 2 with the change in bubble height over time. The Euler method does not even appear to be a linearization of the RK4, but rather, completely misses critical physics involved in the problem. The Euler solution extrapolates too quickly with a total elapsed time two orders of magnitude smaller than the RK4 method. Figure 2: Bubble Height vs. time, RK4 and Euler It is clear in Figure 3 that the bubble radius and the differential pressure across the bubble surface are inversely related. This is readily apparent both in the form of the Rayleigh and Rayleigh-Plesset equations but may also be reasoned from a simply physical argument. Figure 3: Bubble Pressure vs. time 7
Figure 4: Bubble Radius vs. time Figure 5: Bubble Height vs. time 8
Thermal Effects: Figure 6: Bubble Pressure vs. time Thus far, the bubble contents have been considered to be purely gas that obeys the perfect gas relation. We will now consider the case in which the bubble contents are comprised both of a quantity of perfect gas as well as vapor resulting from evaporation of liquid at the bubble boundary. Using Dalton s Law of partial pressures, we may conclude that the total pressure in the bubble is the sum of the vapor pressure at the bubble temperature and the pressure of the gas, given that both pressures are considered to be if the vapor or gas occupied the entire volume of the bubble. P B (t) = P v (T B ) + P gas = P v (T B ) + 4m RT B 3πMR 3 Now if we replace P B (t) in the R-P equation and add and subtract P v (T ), the resulting equation becomes: R R + 3 2Ṙ2 + 4ν lṙ R + 2S ρ l R = P v (T ) P (t) ρ }{{ L } 1 + P v (T B ) P v (T ) ρ L } {{ } 2 + 4m RT B ρ L 3πMR 3 }{{} 3 (3) In Equation 3, term 1 is an instantaneous driving term determined by conditions far from the bubble, 2 is a thermal term, and 3 is the partial pressure arising from the perfect gas present in the bubble. In our original isothermal analysis, we considered the bubble to be at the same temperature as the surrounding fluid, i.e. T B = T, however, we now are considering the case in which these two temperatures are not necessarily equal. In order to determine the bubble temperature, T B, we must consider a model of heat transfer between the bubble and the surroundings. 9