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Diffusion processes Introduction Typical scenario of catastrophic pollution Accidental release of contaminant Transport of conatminant by underground water, streams, rivers, lake or ocean currents, atmosphere Diliution of contaminant by water or air (diffusion) Mechanism of contaminant transport and dilution crucial for understanding and control of polution phenomena Management of environmental pollution resources industry goods Wastes: Accidental realease of Transformation contaminants: Accumulation Dispersion in environment Transport Environmental Quality Effects on Effects on Effects on health climate ecosystem Possible strategies for pollution control Control sources of contaminants Prevention – the most recommended Waste treatment Less recommended choice, only if unavoidable Dispersion in environment Impact of waste storage Accidental release of contaminants Technical necessity e.g. heat, flue gas emission Related problems: Conservation of mass Point sources/distributed sources Continuous/accidental release Basic definitions Concentration: Mass M of contaminant in volume V gives concentration M kg c lim m3 (1) V 0 V in general: c c x , t (2) Relative concentration: c %; ppm (3) of solution Averages: Ensemble average: c x , t (4) t 0 T Time average: c x , t0 1 T c dt t0 (5) cV x , t 1 Space average: V c dV V (6) Basic definitions 1 1 Q Flux average: cf cU dA q dA (7) A Q A Where: Q U dA - volumeflux over surfaceA A kg q 2 specific mass flux m s Dilution: 1 (8) S p ion where: p relative volume concentrat volume of contaminant p (9) total volume Mass conservation equation for contaminant Conservation of mass of conatminant in volume V with no sources within the volume: c dV t V q n dA 0 A (10) chaneg of mass mass flux contained within volume Mass conservation equation for contaminant Due to Gauss-Ostrogradski transformation c t xi qi dV 0 V (11) Since V is an arbitrary volume: c (12) qi 0 t xi Flux of contaminant: qi U i c qDi (13) advection diffusion Substitution of Eq.(13) into Eq. (12) leads to c U i c qDi (14) t xi xi Mass conservation equation for contaminant If fluid is at rest: Ui 0 c qDi (15) t xi U i If fluid is incompressible: 0 xi c c Ui qDi (16) t xi xi If flow is turbulent (Reynolds decomposition): U i U i ui C C c (17) U i c U i C ui c Then mass conservation equation becomes: C t Ui C xi xi qDi ui c (18) Molecular diffusion Turbulent flux – closure problem Recommended strategies for turbulent transport analysis Basic mechanisms involved into transport phenomena: • Advection by flow • Naturl convection • Evaporation/condensation • Entrainment/deposition • Molecular/turbulent diffusion • Disperision in shear flows Separation of dominant mechanisms Splitting of the area of analysis into subdomians Recommended strategies for turbulent transport analysis Experimental versus computer modeling Type of analysis accuracy Time required Cost In-situ ivestigations + -- -- (full scale) Experimental (model) +- -- - investigations Computer modeling +- ++ + Dimensional analysis Order of magnitude +++ +++ Conclusions: •Numerical modeling is nowadays the most recommended, however keep in mind that: garbage computer garbage •Accuracy depends on physical features assumptions but their validity is often doubtful e.g. diffusivity in ocean varies in the range 10-9 (molecular) to 105 m2/s (turbulent) •Dimensional analysis often useful in rough approximations Molecular diffusion Molecular diffuison plays minor role in environmental transport processes but is a basis for understanding other types of diffusion According to Fourier’s (1822) law of heat transfer Fick’s law-thermal analogy (19) T heat flux k x where : k thermal diffusivit y coeffient Fick’s (1855) analogy for mass flux c qi D (20) xi where: q mass flux per unit surfaceand unit time D- diffusion constant[L2 /T] kg m2 kg Note on the units: q 2 ; D ; c 3 m s s m Fick’s law General form of diffusion equation: gradient of transported flux quantity i.e. it is gradient/flux relationsh ip for mass ( D), momentum ( ) or heat (k ) diffusion process D - physical constant depending on the properties of fluid and contaminant Example: For water: D~10-9 [m2/s] Schmidt number Sc D ν ~ 10-6 [m2/s] for water Sc 10 3 k ~ 10-5 [m2/s] for air Sc 1 For air: D ~ ν ~k ~10-5 [m2/s] A simple model for gradient/flux relationship Suppose a 1D pipe with concentration gradient in x-direction, two boxes created in a pipe i.e. left (L) and right (R) containing different numbers of molecules NL=10 and NR=20 L R L R ΔNL Δx Δx Δx Δx ΔNR (A) (B) Concentration of molecules in the initial (A) and consecutive moment (B) for 1D diffusion process A simple model for gradient/flux relationship If each molecule has propability p to pass from one box to the another, then if (say) p=1/5 after time t+Δt, because 1 1 N R 20 4; N L 10 2 5 5 in a new time step N L 10 2 4 12; N R 20 4 2 18 After each time step N Ln 1) N Ln ) ; N Rn 1) N Rn ) ( ( ( ( We obtain N R n 1 N Ln 1 N Rn N Ln And the process continues until the uniform distribution of the molecules is approached – the concentration difference dimnishes with time and the process tends toward uniform concentration Analytical description of gradient/flux relationship Each molecule has mass m M L N L m; M R N R m The mass flux during time Δt in positive x-direction is mass m p N L m pN R p M L M R negative mass flux flux Concentration in each box: ML MR cL ; cR 1 x 1 x Mass flux per unit area and unit time x x 2 c x 2c qx p cL cR p t t x 2 x 2 Analytical description of gradient/flux relationship Mass transfer should not be dependent on the size of the box x 2 lim p const D diffusivity x 0 t t 0 And hence neglecting higher order terms c qx D x We obtained a simple 1D model of Fick’s law Conclusions: •Boxes must be large enough, so that we can apply probability analysis N L ; N R 1 •Boxes must be small enough to apply first order approximation in Taylor series 2c c x 2 x x •Both the above conditions are always satisfied in normal conditions for molecular diffusion because the mass of a single molecule is small an their number is very large even in small volumes Similarity solutions and properties of the 1D diffusion equation Assume 1D diffusion in fluid at rest: c 2c D 2 t x (21) And uniform concentration along y and z axes c c 0 (22) y z Coordinates for 1D diffusion problem Similarity solutions and properties of the 1D diffusion equation Total mass of the contaminant: M c dV A c dx V (23) That means c will be sought in the form M 1 M 1 c (24) A Length A L Let us introduce: L referencelength scale T - referencetime scale Dimensional analysis yields: c c D 2 L2 DT T L (25) Time scale and the length scale in the diffusion process are naturally related Similarity solutions and properties of the 1D diffusion equation Example: Consider a diffusion process after initial injection of the the contaminant. The time scale is time t elapsed after injection. After substitution into Eq.(25) the concentration can be evaluated as: M 1 c (26) A Dt i.e. concentration may be obtained for a similarity solution f x, t M 1 c (27) A Dt The only possibility for non-dimensional function f is: x m f f 12 Dt m 2 (28) s s That finally yields: (29) f ; M 1 x c A Dt Dt Analytical solution of 1D diffusion equation for initial impulse injection of contaminant Introducing Eq.(29) into the r.h.s of diffusion equation: df d2 f 2c M 1 f ; f f (30) d d 2 x 2 A Dt 32 and introducing Eq.(29) into the l.h.s. of the diffusion equation c M 1 1 1 f f (31) t A 2 D t 3 Dt 2t Hence the diffusion equation takes the following form 1 f f f (32) 2 And finally: f 2 f (33) Note that similarity solution allowed to transform the p.d.e into the ordinary differential equation Analytical solution of 1D diffusion equation for initial impulse injection of contaminant After integration of Eq.(33) we obtain: f 2 f C (34) Determination of the constant C x odd C(x,t) C(x,t) (a) (b) Initial injection at t=0 x At t >0 x Expected evolution of the concentration. Concentration of contaminant at the initial injection (a) and after time elapsed t (b) c even Analytical solution of 1D diffusion equation for initial impulse injection of contaminant c f f even f odd f odd even odd And due to symmetry requirements C 0 (35) And finally: (36) f f 2 That yields the solution: 2 x2 f C1 exp cx, t C1 M 1 4 4 Dt exp (37) A Dt Note that C1 the only unknown quantity Analytical solution of 1D diffusion equation for initial impulse injection of contaminant From the contaminant mass conservation equation Eq. (23) x 2 dx C1 exp 4 Dt Dt 1 (38) From tables of definite integrals we find exp a 2 x 2 dx a (39) Substitution into eqaution (38) leads to: 1 C1 4 (40) And finally we obtain the equation for concentration evolution due to molecular diffusion x2 c x, t M 1 4 Dt exp A 4Dt (41) Statistical measures of concentration distribution establidhed due to diffusion processs First order moment – expected value: [E] x cx dx x odd , c even, x c odd , i.e. : E x cx dx 0 Second order moment – variance: x2 x cxdx 4 Dt exp 4Dt dx A 1 2 2 M Having known that the following definite integral is: 2 2 2 x exp a x dx 2a 3 ; and a 1 2 Dt Variance can be expressed as: 2 2Dt Variance is a measure of the contaminant displacement from the initial positiion i.e. is a measure of teh size of the cloud Diffusion and random walk model Definition of radnom walk process: Step ξ of the length λ randomly to the right or to the left (drunkard’s walk t (42) Probablity density function (pdf) of the random walk is: p 1 (43) 2 ∞ ∞ Δ – Dirac delta p(ξ) function ξ -λ +λ Diffusion and random walk model Consider n successive and independent (i.e. no memory of previous step) ξi steps; i=1…n and the random walk variable X(t): X t 1 2 n (44) Where: X(t) is the particle position after n steps Problem: what is the pdf of the process X(t)? Central Limit Theorem (CLT): The probability density function of the sum of n independent random variables tends towards the normal distribution with variance equal to the variance of the sum whatever the individual distributions of these variables provided n tends towards infinity Pdf of normal (Gaussian) distribution process is: x2 px 1 exp 2 2 (45) 2 Diffusion and random walk model Variance of the radnom walk process: 2 1 2 i n 2 12 22 i2 n2 2i j (46) Since the steps are independent: (47) i j 0 Then for random walk process: 2 n2 (48) If τ is teh time required for every step and V is velocity such that: V (49) Then the time t to make n steps is: t n (50) Adn the variance of random walk process: 2 t t2 2 n 2 nV 2 n V V V t (51) n n Diffusion and random walk model 2 V t (52) Close analogy of random walk process to molecular diffusion with λ analogous to mean free path and V analogous to molecular agitation (≈ temperature), which finally leads to the expression for diffusivity: V D (53) Example: N molecules of mass m injected into infinitely thin layer Diffusion and random walk model Total mass of injected molecules: N m M0 (54) If the molecules move according to random walk due to collisions then according to Central Limit Theorem probability p(x)dx that one given molecule is in the slice (x,x+dx) is the Gaussian pdf and the number of molecules in slice (x,x+dx) yields N px dx (55) Mass of the contaminant in volume Adx: N m px dx (56) Concentration of the contaminant: Nm p x dx M 0 c x, t px mass (57) volume Adx A Since the pdf is Gaussian then the concentration reads: x2 cx, t exp 2 , where 2 V t M0 1 (58) A 2 2 Diffusion and random walk model If diffusion process is described by Fick’s law: c 2c D 2 t x the solution for initial mass M0 injected at x=0: x2 cx, t M0 1 exp 2 , where 2 2 Dt 2 A 2 From comparison of Fick’s law and random walk: 1 D V 2 Conclusion: Any random walk process produces the diffusion with diffusivity 1 V 2 Diffusion and random walk model Example: Diffusion 1024 particles injected at x=0 and t=0 and then dispersed due to random walk Illustration of particles diffusion by random walk process Diffusion and random walk model In order to smooth the distribution let us calculate the average distribution for the intermediate step n=6.5 and compare it with the Gaussian distribution, i.e.: 1 x2 N N0 exp 2 2 (59) 2 for 1, 1 2 n x2 N X , n 6.5 160.2 exp (60) n 6.5; N 0 13 -7 -6 -5 -4 -3 -2 -1 0 1 2 x Random 4 8 28 48 84 128 140 160 140 128 walk Gaussian 3.7 10.1 23.4 46.8 80.1 117.8 148.3 160.2 148.3 117.8 process Conclusion: after only 6.5 steps the distribution obtained from random walk is already close to the Gaussian process Brownian motion R. Brown (1826) – observation of motion of small particles A. Einstein (1905) – rigorous explanation practical application: motion of aerosol particles Starting point for analysis: macroscopic particle suspended in fluid, radius a ≈ 1 μm, particle subjected to molecular agitation i.e. collisions with molecules of fluid (much smaller than suspended particle) Numbers of collisions do not balance Finite probability to have more impulses on one side Particle will move Creation of impulse in brownian motion of macroscopic particles Brownian motion Suppose that the particle is spherical and obeys Stokes law: drag force 6 r U where: r radius of particle (59) viscosity U particle velocity Equation of motion for particle: dU (60 m 6 rU F dt Random impulse (+ or -) Inertia force Form fluid molecules Due to random impulse the motion of particle as in random walk Brownian motion Problem: what is the diffusivity of this process? 1 D V 2 To evaluate the diffusivity of random walk we need two scales linear scale time scale or V velocity scale V velocity scale From dimensional analysis the time scale can be estimated: F 6 rU U acceleration - a m m a U m U m a 6 rU 6 r (61) Brownian motion Between the impulses the motion equation is homogeneous: dU dU 6 r dt m 6 rU dt dt U m (62) t ln U ln U 0 U t U 0 exp t Velocity scale U0 is determined by the equipartition of energy 1 m U 02 k T (63) 2 where : k 1.38 10 23 J / K (Boltzman constant) Hence the velocity scale is: 2kT V U 2 2 0 (64) m And diffiusivity due to brownian motion can be evaluated as: 2kT m 1 kT Dbm V 2 m 6 r 3 r (65) Brownian motion Example: diffisuivity for spherical particles in air R [μm] 0.1 1 10 Dbm [m2/s] 1.3x10-10 1.3x10-11 1.3x10-12 Note: compare these values with kinematic viscosity for air equal ν= 1.5x10-5 [m2/s] i.e. diffusiivity of brownian motion is small compared with diffusivity of air Dispersion by turbulent motion Growth rate of the contaminant cloud due to moelcular diffusion: d D 2 2 Dt (66) dt Conclusion: the larger the cloud d The smaller growth rate: dt Extemely large time scales needed to achieve efficinet mixung by moldeular diffusion, for example in water: 1m 2 109 s 104 days D m 109 s Note: time constant for reaction of human nerves –diffusion of ions through the cellular membrane of the thickness h=10-6 m, τ=1ms Dispersion by turbulent motion How to intensify diffusion ? A possible way to speed up diffusion Break up the cloud into smaller fragments If the size of the cloud is σ2, then the diffusion time scale is: 2 2 , if the cloud is split into n smalleer parts then the time scale is t n n t0 D D 2 t 1 and the reduction of the diffuison time scale is : n t0 n Dispersion by turbulent motion What can be the role of turbulence in mixing enhancement ? Turbulence -> superposition of eddies Case 1: the size of the turbulent eddy L is much larger than the size of the cloud σ Cloud transported by turbulent eddy without any change of shape No speed-up of the molecular diffusion Dispersion by turbulent motion Case 2: L Cloud of contaminant distorted by turbulent eddy (random) Speed-up of the molecular diffusion by break-up of the mixing volume Dispersion by turbulent motion Case 3: L Small turbulent eddies distributed within the cloud of contaminant Molecular mixing augmented by turbulent eddies Dispersion by turbulent motion Possible mechanisms of turbulence interaction with molecular diffusion Vortex stretching Full range of eddies (from largest to smallest) appears in turbulent flow Even if initially only largest eddies exist (case 1) after a while eddies of sizes corresponding to cases 2 and 3 will be developed Vortex stretching Vortex stretching Steeper contration gradients inside the vortex Since {flux} ~ {concentration gradients} Enhanced diffusion in radial direction Summary of turbulent diffusion •Turbulence cannot directly enhance mixing and homogenisation at molecular level because fo disparity of scales smallest scales molecular of turbulent eddies scales •Turbulence can make molecular mixing more efficient by Reducing the local size of the contaminant volume cases 2+3 Making local concentration gradients steeper thus enhamcing the diffusive mass flux