Transport processes 7. Semester Chemical Engineering Civil Engineering 1
Course plan 1. Elementary Fluid Dynamics 2. Fluid Kinematics 3. Finite Control Volume nalysis 4. Differential nalysis of Fluid Flow 5. Viscous Flow and Turbulence 6. Turbulent Boundary Layer Flow 7. Principles of Heat Transfer 8. Internal Forces Convection 9. Unsteady Heat Transfer 10. Boiling and Condensation 11. Mass Transfer 12. Porous Media Flow 13. Non-Newtonian Flow 2
Today's lecture Mass transfer Ficks law Steady mass diffusion through a wall Reynolds analogy 3
Molecular diffusion 4
Fick s 1 st law Rate of diffusion J dc = D dx 2 2 (kg/s m or kmol/s m ) or dc mdiff = DB dx (kg/s) D B is the binary diffusion coefficient (m 2 /s) of species in medium B D B is dependent on temperature and pressure D B T P 32 DB,1 P 2 T 1 = DB,2 P1 T2 32 5
Some diffusion coefficients Gasses 6
Some more diffusion coefficients Liquids and solids 7
Similarity with Heat transfer Thermal diffusion coefficient Thermal Gradient Q dt = k dx (J/s) dc mdiff = DB dx (kg/s) Mass diffusion coefficient Concentration Gradient 8
Example Calculate the Helium flux in a pipe (no bulk fluid motion) ssume the ideal gas law holds: PV = nrt n V P = = c RT Insert in Ficks 1 st law, rearrange, integrate: dc C C J = D J dx D dc J He N DHe N dx = = x x J He N 2 1 ( ) He,1 He,2 2 1 4 5 D phe,1 p,2 0.687 10 0.6 0.2 10 He N He 6 2 = = = 5.63 10 kmol m s RT x x 8314 298 0.2 0 9
Concentrations given by mol fractions y C [ kmol/kmol mixture] C + C = C total concentration of ( + B) = C y B also y + y = 1.0 J z = CDB dz B dy 10
Concentrations given by partial pressures p and Ideal gas law z also p p and y =, yb = p B Ru T p p = n = CRT u C = V RT J = D B C R u T dz B dp p + p = p total pressure p B p u 11
Concentrations given by mass fractions ω ρ M C [ kg / kg mixture] 3 kg mixture / m mixture [ kg / kmol ] 1 = ρ ω ω + ω = 1 M B J z = ρ M D B dω dz 12
Concentrations given by mass concentration ρ ρ M C 3 kg m mixture / 3 kg mixture m mixture [ kg / kmol ] / 1 = ρ ρ + ρ = ρ M B J z = 1 M D B d ρ dz 13
Fick s 2 nd law Transient mass diffusion C t = D 2 C 2 x Steady state concentration distribution 0 = D 2 C x 2 dolf Eugen Fick (1829-1901) 14
Boundary conditions Concentration is not necessarily a continuous function ppropriate boundary conditions can be found from equilibrium considerations: y H 0, surface, gas 2 = P sat @15 C 1.705kPa 0.0185 1.87% P = 100kPa = = y H 1 100% 2 0, surface, liquid = = 15
Mass diffusion through a wall Ficks law (mass basis): J m dω = = ρdb (kg/m s) dx Integrating over the length, L, of the wall: m ω ω ω ω = = L ρ D R 1 2 1 2 B diff, w 16
Example Calculate the Hydrogen flux through the walls of the pressure vessel C C N = π r r D = kmol s diff,1,2 10 4 1 2 B 1.228 10 / r2 r1 Converting to mass basis: m = MN = kg kmol kmol s = kg s diff diff Converting to yearly basis: m diff 10 10 2 / 1.228 10 / 2.46 10 / ( g kg ) 10 2.46 10 60 60 24 365 1000 / 7.8g/year = = 12 2 DB = 1.2 10 m / s 17
Equimolar diffusion For each molecule of species diffusing in one direction a molecule of species B is diffusing in the other direction N J dc dx D DB = B = N = = J B D B D B B dc dx B
Diffusion in a moving medium Superposition of diffusion velocity and bulk flow velocity Note, turbulence, simultaneous heat transfer, etc. may complicate things! V = bulk flow velocity ρ dω Vdiff, = DB ρ dx (m/s) 19
Total mass flux The total mass flux of species dω J = V + V = V D dx 2 ρ ρ diff, ρ ρ B (kg/m s) m m = ρ V = ρ V B B B m = ρ V = m + m tot mix B 20
Mass convection Prandtl number Molecular diffusivity of momentum ν Pr = = = Molecular diffusivity of heat α µ c p k Schmidt number Molecular diffusivity of momentum Sc = = Molecular diffusivity of mass Lewis number ν D B Sc Molecular diffusivity of heat α Le = = = Pr Molecular diffusivity of mass DB 21
Relative boundary layer thickness For laminar flow δ δ velocity thermal = Pr n δ δ velocity concentration δ δ n = Sc thermal n concentration = Le n 1/3 in all cases 22
Convective mass transfer Newtons law of cooling conv s s ( ) [ ] Q = h T T W Convective mass transfer mconv = hmass s ρs ρ ( ) [ kg/s] 23
Sherwood number Nusselt number hl Nu = k Sherwood number hmassl Sh = D B Stanton number St thermal = Nu Re Pr St mass = Sh ReSc 24
Reynolds analogy Pr=Sc=Le Reynolds analogi: the special case where the non-dimensional velocity, thermal and concentration boundary layer is the same Pr = Sc = Le also f f Nu Sh Re = Nu = Sh = = 2 2 RePr ReSc 25
General case Pr Sc Le Chilton-Colburn analogy: f = St Pr = St Sc 2 23 23 mass 26
Empirical correlations 27
Excercises 28