Diffusional Growth of Liquid Phase Hydrometeros. I. Diffusional Growth of Liquid Phase Hydrometeors A. Basic concepts of diffusional growth. 1. To understand the diffusional growth of a droplet, we must consider two important processes a. Water vapor is transferred to the droplet by vapor diffusion i. A gradient of vapor develops around a drop that is growing or evaporating - the droplet is not in equilibrium b. Condensation (evaporation) results in latent heat release (absorption) i. The net effect is that the droplet is warmed (cooled) ii. The warming (cooling) slows the growth (evaporation ) of the droplet as the saturation vapor density is directly related to the temperature iii. If the drop is to grow, the excess heat must be removed. This can occur by conduction with the environment. B. In summary, we must be concerned with 1. The mass flux of water molecules toward/away from the droplet 2. The conduction of heat away/toward the droplet 3. We will solve these individually as well as simultaneously with the help of the Clausius - Clapeyron equation. a. Alternative methods are also possible; however, we will not look at these. C. Assumptions: 1. Initial work will be based on the following assumptions a. Critical radius droplet b. Stationary droplet c. Isolated droplet d. Spherical (symmetric) Area = 4pa2, where a is droplet radius D. Mass flux of vapor Now the flux of vapor molecules to / from the surface of a droplet is 1. The mass rate of change a. The mass flux toward the drop through radius r from the drop is proportional to i. the gradient of vapor density through any radial ii. diffusivity of water vapor iii. area of a shell with radius r b. with r = mn dM / dt = 4p D a[r v,¥ - r v,a] i. The droplet mass change is a function of i. radius a ii. diffusivity iii. [r v,¥ - r v,a] • if [r v,¥ - r v,a] > 1; dM/dt > 0 • if [r v,¥ - r v,a] < 1; dM/dt < 0 2. Growth due to mass flux in terms of radius 1 M = r w V = r w p 4/3 a3 (r w is density of water) dM / dt = 4p D a [r v,¥ - r v,a] r w 4/3 p da3 / dt = 4p D a [r v,¥ - r v,a] r w (4/3) p 3a2 da / dt = 4p D a [r v,¥ - r v,a] r w 4p a2 da / dt = 4p D a [r v,¥ - r v,a] r w a da / dt = D [r v,¥ - r v,a] a da / dt = D [r v,¥ - r v,a] / r w a. We can integrate this to get a as a function of time t for mass flux of water vapor... [a2(t1) - a2(t0)]/ 2 = D [r v,¥ - r v,a] (t1-t0) / r w t1 = t0 + r w [a2(t1) - a2(t0)] {2 D [r v,¥ - r v,a] } b. The time for a droplet to grow by vapor diffusion given SS, T, and initial radius i. Initial conditions D = 2.11x10-5 m2/s r w = 1000 kg / m3 SS = 1.01 r v,¥ = 4.895x10-3 kg/m3 T = 273.15 K p = 101325 Pa r(t0) = 5x10-6 m radius dt t+dt (m) (sec) (accumulated) 5x10-6 m 0 0 10x10-6 m 20x10-6 m 30x10-6 m 40x10-6 m 50x10-6 m 100x10-6 m 400x10-6 m 1000x10-6 m 5000x10-6 m E. The conduction of heat away / toward the droplet As a droplet grows, the droplet heats. This increases its saturation vapor pressure for equilibrium over the drop surface. For a steady state drop growth, we find the steady state droplet temperature. 1. This derivation is analogous to that for mass flux of water vapor 2. The heat flux equation necessary to keep droplet T constant a. The conduction (diffusion) of heat toward the droplet at radius r from the drop is proportional to i. the gradient of temperature through any radial 2 ii. thermal conductivity iii. area of a shell with radius r (see picture from before) -dQ / dt = 4p a2 K ¶{a[Ta - T¥ ]/r + T¥ }/¶r (r=a) d. ¶T¥ /¶r = 0, so -dQ / dt = 4p a2 K ¶{a[Ta - T¥ ]/r}/¶r (r=a) e. or, noting ¶(1/r)/¶r = -1/r2 ¶r/¶r -dQ / dt = 4p a2 K {-a[Ta - T¥ ]/r2¶r/¶r} (r=a) f. thus, with r = a (droplet surface) -dQ / dt = 4p a2 K {-a[Ta - T¥ ]/a2} -dQ / dt = 4p K a[T¥ - Ta] g. The droplet heat change is a function of i. radius a ii. conductivity iii. Thermal gradient • if [T¥ - Ta]< 1; -dQ/dt > 0 • if [T¥ - Ta]> 1; -dQ/dt < 0 3 This equation can be re-written in terms of mass change noting that -dQ/dt = -L dM/dt where L is the latent heat of vaporization -L dM / dt = 4p K a[T¥ - Ta] F. Now we need to discuss the balance conditions for the heats associated with condensation and conduction. 1. We have an equation for mass flux of vapor (which multiplied by L yields heat released during condensation) a. dM / dt = 4p D a [r v,¥ - r v,a] 2. We also have an equation for the heat conduction a. -dQ / dt = 4p K a [T¥ - Ta] 3. The rate of change of surface temperature of the droplet is a. 4/3 p r3 r w C dTr / dt = L dM/dt - dQ/dt i. C is the specific heat capacity ii. L dM/dt is the mass flux of vapor times L iii. dQ/dt is the heat conduction rate iv. Steady state growth; so dTr / dt = 0 4. The latent heat owing to mass flux and the heat conduction terms oppose each other. While latent heating warms it reduces the vapor density gradient, which reduces the vapor flux. To solve for the true mass growth, these need to be balanced (iterate) 4. Begin by assuming steady state growth, L dM/dt - dQ/dt = 0 5. From this we can state (the particle wet bulb relationship) 4p L D a [r v,¥ - r v,a] = -4p K a [T¥ - Ta] 6. Or, L D [r v,¥ - r v,a] = -K [T¥ - Ta] 7. Thus, the following must be balanced by the thermal and vapor fields [r v,¥ - r v,a] / [T¥ - Ta] = -K / (L D) 8. This equation can be solved by iterating as the vapor density at the droplet surface is a function of T at the surface. a. Others, such as Young (1974: JAS pg 1735) suggest that we can linearize the saturation vapor density relation to get r v,a = s Ta + b (where s and b are constant)to derive Ta 3 Ta = [D L (r v,¥-b) + KT¥ ] / [D L s + K] 9. Upon arriving at the equilibrium T and vapor density, we can solve the above mass flux equation for the mass growth. 10. An estimate of the influence of the of heating can be made as follows. a. Re-compute the drop temperature from 7 above. b. From this compute a new vapor density. c. The time for a droplet to grow based on the heat conduction equation can be computed given the following information: Compare this rate to that fromthe mass flux equation. i. Initial conditions D = 2.11x10-5 m2/s r w = 1000 kg / m3 SS = 1.01 r v,¥ = 4.895x10-3 kg/m3 T = 273.15 K p = 101325 Pa r(t0) = 5x10-6 m radius dt t+dt (m) (sec) (accumulated) 5x10-6 m 0 0 10x10-6 m 20x10-6 m 30x10-6 m 40x10-6 m 50x10-6 m 100x10-6 m 400x10-6 m 1000x10-6 m 5000x10-6 m G. The previous equations are tedious to solve. It is possible to solve simultaneously for the growth rate in terms of mass flux and heat conduction using the Clausius-Clapeyron equation. This is what is shown next (There are a number of ways of proceeding. I will choose that presented by Byers.) 1. We begin by writing the mass flux and thermal conduction equations a. L dM / dt = 4p L D a [r v,¥ - r v,a] b. -dQ / dt = 4p K a [T¥ - Ta] 2. Now from the particle wet-bulb relation, and the equation of state a. [r v,¥ - r v,a] / [T¥ - Ta] = -K / (L D) b. r v = ev / (Rw T) (Rw is the gas constant for water vapor) 3. We get, approximately (assuming T¥ » Ta in the equation of state; also note I moved a minus sign) [ev,¥ - ev,a] / {T¥ [Ta - T¥ ]} = K Rw / (L D) a. When the ambient vapor is at supersaturation with respect to the drop, ev,¥ > ev,a , then Ta > T¥ 4. Or, we get the following equation a da / dt = (S-1) / {r w L2 / (K Rw T¥ 2)] + r w Rw T¥ / (esv,¥D) } 5. Using dM/dt = r w 4 p a2 da/dt 4 a da/dt = [1/(r w 4 p a)] dM/dt 6. We get dM/dt = 4 p a (S-1) / {L2 / (K Rw T¥ 2)] + Rw T¥ / (esv,¥D) } 7. Notes a. This equation is entirely in terms of the ambient conditions b. It is very accurate for S near 1 (RH = 101%) c. It can be in error for |S-1| > 0.1 H. The influence of radius and solutions on diffusional growth. 1. In the previous section we computed the growth rate of a spherical liquid hydrometeor by steady state diffusion of water vapor and temperature assuming a balance between latent heat release and heat conduction.