Transport Phenomena II

Transcription

1 Transport Phenomena II Andrew Rosen April 25, 2014 Contents 1 Temperature Distributions with More Than One Independent Variable The Microscopic Energy Balance Unsteady State Heat Conduction: Separation of Variables Technique Introduction The Steady State Solution The Transient Solution both steady and unsteady state Heating a Semi-Infinite Solid Heating a Finite Slab Complexification Background Unsteady Heat Conduction near a Wall with Sinusoidal Heat Flux Laplace Transforms General Definitions Common Transforms Using the Laplace Transform to Solve Initial-Value Problems Reworking the Semi-Infinite Slab Boundary Layer Theory for Nonisothermal Flow Velocity Boundary Layer Thermal Boundary Layer Laminar and Turbulent Velocity Boundary Layers The Momentum and Energy Balance The Boundary Layer Equations Boundary Layer Similarity External Flow Flat Plate in Parallel Flow Hydrodynamic Solutions Laminar Flow Equations Heat Transfer Solution Laminar Average Boundary Layer Parameters Laminar Turbulent Flow Cylindrical and Spherical Systems in Cross Flow Potential Flow Sphere in a Water Bath Tube Flows Area Average Quantities Laminar Flow in Circular Tubes Heat Flow

2 5 Diffusivity and the Mechanisms of Mass Transport Kinetic Theory and Lennard-Jones Potential Fick s Law of Binary Diffusion Molecular Mass Transport Temperature and Pressure Dependence of Diffusivities Theory of Diffusion in Gases at Low Density Mass Average and Molar Average Velocity Summary of Mass and Molar Fluxes Concentration Distributions in Solids and in Laminar Flows Shell Mass Balances and Boundary Conditions Diffusion Through a Stagnant Gas Film Diffusion through a Spherical Film Diffusion with a Heterogeneous Chemical Reaction e.g. Gas Reacting on Solid Catalyst Diffusion with an Instantaneous Heterogeneous Reaction Diffusion with a Slow Heterogeneous Reaction Diffusion with a Homogeneous Chemical Reaction e.g. Gas Dissolving in Liquid Gas Absorption with Chemical Reaction in Agitated Tank Diffusion into a Falling Liquid Film Gas Absorption Gas Absorption from Rising Bubbles Diffusion into a Falling Liquid Film Solid Dissolution Diffusion and Chemical Reaction Inside a Porous Catalyst Equations of Change for Multicomponent Systems The Equations of Continuity for a Multicomponent Mixture Summary of the Multicomponent Equations of Change Simultaneous Diffusion, Convection, and Reaction with a Porous Plug Concentration Profile in Tubular Reactors Concentration Distributiuons with More than One Independent Variable Time-Dependent Diffusion Gas Adsorption with Rapid Reaction Mass Transfer at an Interface with a Semi-Infinite Body Diffusion and Chemical Reaction in Isothermal Laminar Flow Along a Soluble Flat Plate Taylor Dispersions Unsteady-State Interphase Diffusion Interphase Transport in Nonisothermal Mixtures Rotating Disks Correlation of Binary Transfer Coefficients in One Phase Interaction of Phase Resistances i.e. Leeching Appendix Gradient Divergence Curl Laplacian

3 1 Temperature Distributions with More Than One Independent Variable 1.1 The Microscopic Energy Balance The microscopic energy balance states that where v and q are vector quantities ρĉp [ ] T t + v T = q In the limit of no convection and applying Fourier s law, where Fourier s law is T ρĉp t = k T q = k T This can be occasionally rewritten as q = k T L Occasionally, k can be assumed to be independent of temperature, which then makes the energy balance where α is the thermal diffusivity defined as T t = α 2 T α = k ρĉp The units of α are length-squared per unit time The units of k are frequently watts per meter-kelvin= The heat flow is given as ˆ Q = A which simplifies to the following if q is assumed to be uniform over the area e.g. if you use an average heat transfer coefficient, h to find q The energy over a period of time can be given by q da Q = qa = Ak T E = A 1.2 Unsteady State Heat Conduction: Separation of Variables Technique Before I begin this section, I d like to point out that Section 1.2 is not explore in full rigor, and you are not advised to try to understand the poor explanations I ve included here. The takeaway points are how to use the technique of separation of variables and that we add the unsteady state term to the steady state term to get a true equation for the temperature profile. Problem: Consider a rod of length l, initially at T 0, that is perfectly insulated except at the ends such that there is 1-D flow of temperature. Find the steady-state temperature profile if the left side is held at a constant T L and the right side at a constant T R ˆ t 0 q dt 3

4 1.2.1 Introduction 1. Since we are not worrying about radial temperature, we shall use Cartesian coordinates 2. The boundary and initial conditions are T 0, t = T L T l, t = T R T x, 0 = T 0 3. Try a solution of T x, t = X x ˆT t 4. The microscopic energy balance is 5. Therefore, plugging in the trial solution yields which can be rewritten as 6. Now, we analyze the possible conditions ˆT T t = T α 2 x 2 X ˆT = α ˆT X ˆT = αx X = Constant a If the expression equals zero, it is the steady-state solution b If the expression equals a positive quantity, then the temperature would increase up to positive or negative infinity, so it is unphysical. This is shown below where C 2 is some arbitrary positive constant: ˆT ˆT = C2 1ˆT d ˆT dt = C2 now rewriting this temporarily using the differential operator D yields D ˆT C 2 ˆT = 0 ˆT D C 2 = 0 which has roots of such that the solution to the ODE is D = C 2 ˆT = C 1 exp C 2 t which approaches infinity if C 1 is positive and negative infinity if C 1 is negative c If the expression equals a negative quantity, then the temperature would be something that s feasible. The same mathematical progression would show that for a negative constant C 2 : ˆT = C 1 exp C 2 t which decays to zero at infinite time. Since t is steady-state, the ˆT t term should drop out i.e. the transient term goes to zero, which it does. Therefore, a negative constant value is the unsteady state solution 7. Let s tackle this problem by addressing the steady-state solution, then the unsteady state solution, and finally adding them together 4

5 1.2.2 The Steady State Solution 1. If the expression equals zero the steady state solution, then we can evaluate this as follows: such that and creating a new constant will yield ˆT ˆT = 0 ˆT = C 1 α X X = 0 X = C 2X + C 3 T x, t = C 1 C 2 X + C 3 T x, t = C 4 X + C 5 which can be solved using the boundary and initial conditions to yield T 0, t = T L = C 5 T l, t = T R = C 4 l + C 5 so for the steady-state solution to heat conduction in a rod that is perfectly insulated except at the ends: TR T L T x, t = x + T L l The Transient Solution both steady and unsteady state 1. If the expression equals a negative constant i.e. C 2, let s try the differential equation of X X = C2 Rewriting this temporarily using the D differential operator yields which has roots of such that the solution of the ODE is D 2 X + C 2 X = 0 X D 2 + C 2 = 0 D = ± C 2 = ±i C 2 = ±ic X = C 1,x,i cos Cx + C 2,x,i sin Cx 2. Let s now try the other differential equation for a negative constant, which can be solved, as earlier with C 2, to yield 3. Finally, we can write the solution as ˆT ˆT = C2 ˆT = C 3,x exp C 2 αt T = X ˆT = C 3,x exp C 2 αt [C 1,x,i cos Cx + C 2,x,i sin Cx] which can be rewritten absorbing C 3,x into the other constants to yield T uss = exp C 2 αt [C 1,x cos Cx + C 2,x sin Cx] 5

6 4. Since we defined the problem to be at steady state at x = 0 and x = l, it is clear that T uss = 0 at these values. Therefore, C 1,x = 0 and C 2,x = exp C 2 αt B sin Cl 5. Since we just stated that T uss = 0 at x = l, we have an issue. We can have B = 0, but that means T uss = 0 for all x and t. Clearly that s not possible. There are no parameters to alter in exp C 2 αt either. This leaves us with forcing the sin Cl term to zero. To do this, we can state that C n = nπ l where n is an integer greater than or equal to 1. Rewriting our expression for the unsteady state temperature yields T uss = exp n2 π 2 αt nπx l 2 B n sin l 6. After a lot of math and applying the half-range sine expansion in addition to the initial conditions, we find that for the case of T R = 100, T L = 50, and T x, 0 = 100 that B n = 100 nπ and x T x, l = 50 l π for both steady and unsteady state 1.3 Heating a Semi-Infinite Solid [ 1 nπx π 2 n sin n 2 ] αt exp l l 2 n=1 Problem: Consider a solid material occupying the space from y = 0 to y = that is initially at temperature T 0. At time t = 0, the surface at y = 0 is suddenly raised to temperature T 1 and maintained at that temperature for t > 0. Find the time-dependent temperature profiles T y, t. Assume a constant k. Solution: 1. The microscopic energy balance in the y direction states that 2. We can introduce a dimensionless variable of T t = T α 2 y 2 Θ = T T 0 T 1 T 0 to simplify the calculations such that Θ t = Θ α 2 y 3. The boundary and initial conditions state that: 4. Since Θ is dimensionless, it must be related to Θ y, 0 = 0 Θ 0, t = 1 Θ, t = 0 the only possible dimensionless group from the given variables. Therefore, y αt since this or multiplicative scale factors of it is Θ = Θ η where η = y 4αt a The 4 term in the denominator is included for mathematical simplicity during the derivation 6

7 5. The differential equation in Step 2 can be broken down from a PDE to an ODE a First, The value for η t b Next, The value for η y Θ t = Θ η η t can be found from taking the derivative of η with respect to t. This yields Θ t = dθ 1 η dη 2 t Θ y = Θ η η y can be found from taking the derivative of η with respect to y. This yields i. We want 2 Θ y 2 though, so 2 Θ y 2 = y c Therefore, 6. A new set of boundary conditions are needed for η Θ y = dθ 1 dη 4αt Θ y d 2 Θ dθ + 2η dη2 dη = 0 = d2 Θ dη 2 1 4αt a At η = 0, Θ = 1 since this is when y = 0, and it was stated earlier that Θ0, t = 1 b At η =, Θ = 0 since this is when y =, and it was stated earlier that Θ, t = 0 7. To solve this differential equation, introduce ψ = dθ dη to make the equation dψ dη + 2ηψ = 0 a This yields, ψ = C 1 exp η 2 8. Integrating ψ yields, ˆ η Θ = C 1 exp η 2 d η + C The boundary conditions of Θ = 0 and Θ = 1 can be used here to find C 1 and C 2, which produces the equation Θ η = 1 erf η which corresponds to the following solution for the heating of a semi-infinite slab 1 : Θ = T y, t T 0 T 1 T 0 y = 1 erf 4αt 1 T 0 is the initial temperature, T 1 is the surface temperature, and T is the temperature at a point in space and time afterwards 7

8 a Important Point: The above expression tells you about the temperature at some point along the semi-infinite axis. If you have a semi-infinite object, but you want to find the temperature somewhere on the finite axis, then this is not the equation for you. 10. If the heat flux is desired at the wall i.e. surface 2, q s = q y y=0 = k T y = y=0 k παt T 1 T 0 since d 2 erf u = exp u 2 du dx π dx As a side-note, the thermal penetration thickness, δ T, is defined as the value of η when erf η = 0.99 such that there is a 1% change in Θ. This occurs close to η = 2, so 1.4 Heating a Finite Slab δ T = 4 αt Problem: A solid slab occupying the space between y = b and y = b is initially at temperature T 0. At time t = 0, the surfaces at y = ±b are suddenly raised to T 1 and maintained there. Find T y, t. Assume a constant k. Solution: 1. The microscopic energy balance in the y-direction states that T t = T α 2 y 2 2. We can introduce the following dimensionless variables to simplify the math: Θ = T 1 T T 1 T 0 η = y b τ = αt b 2 a It should be noted that the steady state solution goes to zero because at t =, T = T 1 such that Θ = 0 for all τ > 0 3. This makes the microscopic energy balance now read 4. The initial and boundary conditions are now: a At τ = 0, Θ = 1 b At η = ±1, Θ = 0 for τ > 0 2 Additional expressions: q s = period of time is E = t 0 qs dt = 2k t T 1 T 0 πα. Θ τ = 2 Θ η 2 kρcp T 1 T 0. At constant T and t,.q s,2 k2 ρ 2 C P,2 =. The energy J/m 2 over a πt q s,1 k1 ρ 1 C P,1 8

9 5. We can implement the method of separation of variables by stating Θ η, τ = f η g τ 6. Substituting the trial solution for Θ yields fg = gf 7. Dividing by fg yields 1 dg g dτ = 1 d 2 f f dη 2 8. If the left side is a function of only τ, and the right side is a function of only η, then both must be equal to some constant. The constant will be chosen as c 2 solely to simplify the math. Therefore, dg dτ = c2 g d 2 f dη 2 = c2 f 9. Integrating the above equations yields g = A exp c 2 τ f = B sin cη + C cos cη 10. The integration constant B must equal zero since Θ η, τ must equal Θ η, τ due to symmetry about the xz-plane. Applying the boundary conditions then yields C cos c = 0 a If C = 0, then the temperature profile is always zero, which cannot be the case. Other solutions exist as follows: c = n + 1 π 2 where 11. Therefore, Θ n = A n exp n = 0, ±1, ±2,..., ± n π 2 τ C n cos n + 1 πη Since Θ is the sum of all Θ n, Θ = D n exp n π 2 2 τ cos n + 1 πη 2 n=0 where D n = A n C n + A n+1 C n Applying the initial condition yields 1 = n=0 D n cos n + 1 πη 2 9

10 14. The solution for D n is the following see BSL for more details D n = 2 1n n + 1 π Plugging D n into the equation for Θ in Step 12 yields T 1 T 1 n = 2 exp T 1 T 0 n=0 n π n π 2 αt b 2 cos n + 1 π y 2 b a The solution to this equation as well as the analogous equations for cylindrical and spherical systems is graphically shown in BSL Figures , , and For a cube-like solid, T 1 T x = Θ T 1 T 0 a, αt y a 2 Θ b, αt z b 2 Θ c, αt c 2 = Θ x Θ y Θ z where each side is 2a, 2b, and 2c a Important Note: You multiply the Θ of every finite dimension you have. For a semi-infinite cylinder, you would just useθ r, but for a finite cylinder, you d have to do Θ r Θ z. b Important Note: The middle is 0 since the top is a and the bottom is a 1.5 Complexification Background Recall that e iθ = cos θ + i sin θ The approach to complexification is to translate a real system to a complex one, solve the system, and extract the real part of the solution frequently used with periodic conditions Unsteady Heat Conduction near a Wall with Sinusoidal Heat Flux Problem: A solid body occupying the space from y = 0 to y = is initially at temperature T 0. Beginning at time t = 0, a periodic heat flux is given by q y = q 0 cos ωt = q 0 R [ e iωt] is imposed at y = 0. Here, q 0 is the amplitude of the heat flux oscillations, and ω is the frequency. It is desired to find the temperature in this system, T y, t, in the periodic steady steady state. Assume a constant k. Solution: 1. The microscopic energy balance in the y-direction states that T t = T α 2 y 2 2. If both sides are multiplied by k and both sides are acted on by the operator y rearrangement, k T = α 2 t y y 2 k T y then with a little 10

11 3. The definition of heat flux can then be used such that 4. The boundary conditions are: a At y = 0, q y = q 0 R [ e iωt] b At y =, q y = 0 5. We postulate an oscillatory solution of the form where q of a complex function of y q y t = q y α 2 y 2 q y = R [ q e iωt] 6. Substituting the trial solution into the equation in Step 3 yields R [ q iωe iωt] [ d 2 q ] = αr where q means complex q dy 2 eiωt 7. The above expression is mathematically equivalent to d 2 q iω dy 2 q = 0 α 8. The new boundary conditions are: a At y = 0, q = q y b At y =, q = 0 9. The differential equation has the solution of iω iω q = C 1 exp y + C 2 exp 1 + i y α 2α 10. Since 1 i = ± i, ω ω q = C 1 exp 1 + i y + C 2 exp 1 + i y 2α 2α 11. The second boundary condition requires that C 1 = 0, and the first boundary condition requires that C 2 = q 0. Therefore, ω q = q 0 exp 1 + i y 2α 12. As such, q y = R [ ] [ ] ω ω ω q 0 exp 1 + i y e iωt = q 0 exp y R exp i y 2α 2α 2α ωt 13. The above expression is equivalent to q y = q 0 exp ω ω y cos ωt y 2α 2α 11

12 14. Integrating Fourier s law yields 15. The above expression simplifies to 1.6 Laplace Transforms General Definitions T T 0 = q 0 k ˆ T0 k d T = T The Laplace Transform, a linear operator, is defined as ˆ y qȳdȳ α ω ω ω exp y cos ωt y 2α 2α π 4 F s = ˆ 0 e st ft dt We write the Laplace transform as F s = L [ft] If F s = L [ft], then we say that ft is the inverse Laplace transform, written as ft = L 1 [F s] The First Shift Formula states that L [ e αt ft ] = F s a where F s = L [ft] Equivalently, L 1 [F s] = e αt L 1 [F s + a] The Second Differentiation Formula states that L [t n ft] = 1 n dn ds n L [ft] While the Laplace and inverse Laplace operators are linear, they have analogous properties with integrals, so L [A B] L [A] L [B] Common Transforms The following is a list of basic transforms and inverse transforms: L [ e λt] = 1 [ ] 1 1 and L = e λt s λ s λ L [1] = 1 [ ] 1 and L 1 = 1 s s L [t n ] = n! [ ] 1 s n+1 and L 1 s n = tn 1 n 1! [ ] s L [cos βt] = s 2 + β 2 and L 1 s s 2 + β 2 = cos βt [ ] β L [sin βt] = s 2 + β 2 and L 1 1 s 2 + β 2 = 1 sin βt β 12

13 1.6.3 Using the Laplace Transform to Solve Initial-Value Problems 1. Transform both sides of the differential equation, incorporating the initial data by means of the first differentiation formula, L [ D k x ] = s k L [x] s k 1 x0 s k 2 x 0... x k 1 0 where D d dt 2. Solve algebraically for L [x] in terms of s 3. Obtain x as the inverse Laplace transform of the equation found in Step Reworking the Semi-Infinite Slab First, realize that Laplace transforms convert the t to s such that L [Θ x, t] Θ x, s Recall that Θ x, 0 = 0 Θ 0, t = 1 Θ, t = 0 The differential equation that applies for this system if we consider the x direction is Θ t = Θ α 2 x 2 Taking the Laplace transform of both sides yields the following once the first differentiation formula is used, [ 2 ] Θ sl Θ 0 = αl x 2 Since we said that L [Θ] = Θ, Rewriting this with more familiar notation yields s Θ = α 2 Θ x 2 Θ = s α Θ The boundary conditions can be rewritten using the Laplace transform as Θ 0, s = 1 s Θ, s = 0 Recall that this second-order homogeneous differential equation can be solved as D 2 s Θ α Θ = 0 Θ D 2 s = 0 α such that the roots are which means that the solution is Θ = C 1 exp s D = ± α x s α + C 2 exp x s α 13

14 Applying the second boundary condition yields C 1 = 0 such that s Θ = C 2 exp x α The second boundary condition yields C 2 = 1 such that s Θ = 1 s x s exp a This Laplace transform is tabulated and can be readily found to be x Θ = 1 erf 4αt 2 Boundary Layer Theory for Nonisothermal Flow 2.1 Velocity Boundary Layer The local friction coefficient is defined as Don t forget that τ s is evaluated at y = 0 C f 2τ s ρu 2 The boundary layer thickness for a velocity boundary is where u = 0.99u Recall that the shear stress is given by τ s = µ u y y=0 2.2 Thermal Boundary Layer The thermal boundary layer looks like the following for T s > T For a thermal boundary layer, the boundary layer thickness is the position where The local surface heat flux can be obtained by q s y=0 = k f T y y=0 = h T s T T s T T s T = 0.99 such that h = k f T y y=0 T s T 14

15 Also, ˆ Q = ˆ q da s = T s T h da s which is equivalent to Q = ha s T s T where h is the average convection coefficient for the entire surface For a sphere, A s = 4πr 2 = πd 2 For a cylinder, A s = 2πrL = πdl If it s a 1-D characteristic length such that Q has units of W/m for a cylinder or sphere, then it s L = 2πr = πd It follows that ˆ h = 1 A h da s For 1D heat transfer of a flat plate 3, h = 1 L ˆ L 0 h dx 2.3 Laminar and Turbulent Velocity Boundary Layers The term x c marks the transition between the laminar and turbulent regions The turbulence brings in fluid from the undisturbed main stream closer to the plate 3 hturb = 1 x [ xc 0 h lam x dx + ] x x c h turb x dx 15

16 The Reynolds number is defined as 4 Re x = ρu x µ = u x ν where ν µ ρ The critical Reynolds number that marks the transition from the laminar to turbulent regions is approximated as the following for a flat-plate Re x,c = Differences in the thickness of the velocity and thermal boundary layers tend to be much smaller in turbulent flow than in laminar flow since turbulence causing mixing that reduces the importance of conduction in determining the thermal boundary layer thickness 2.4 The Momentum and Energy Balance The assumptions that shall be used here include: steady state, incompressible fluid, no body forces ignore g, constant properties e.g. µ, ρ, and continuity i.e. no mass accumulation Recall that the momentum balance states that v ρ t + v v = P + µ 2 v + ρ g For the system we are describing, this simplifies to ρ v v = P + µ 2 v If we state that u = v x and v = v y, the x-momentum equation is ρ u u x + v u 2 u = µ y x u y 2 P x The full energy balance is given by where µφ is the viscous dissipation term ρc p v T = k 2 T + µφ 4 Additional expression: Re x,c Re L = xc L 16

17 2.5 The Boundary Layer Equations The additional assumptions for boundary layer theory is the gradients perpendicular to the surface are much greater than the gradients parallel to the surface and that P changes slowly with x such that P x dp dx where P is the pressure gradient within the boundary layer and P is the pressure gradient of the free stream The x-momentum equation simplifies to the following inside the boundary layer u u x + v u y = 1 dp ρ dx + ν 2 u y 2 since 2 u x 2 2 u in the boundary layer i.e. gradients normal to the object s surface are much larger y2 than those along the surface The energy equation simplifies to the following as well u T x + v T y = α 2 T y 2 + ν C P 2 u y since With no viscous dissipation 2 T x 2 2 T y 2 u T x + v T y = T α 2 y Boundary Layer Similarity To non-dimensionalize our equations, we introduce the following variables: x x L y y L u u V T v v V The shear stress can be expressed equivalently as τ s = µ u y = y=0 T T s T T s P P ρv 2 µv L u y y =0 such that C f = 2τ s ρv 2 = 2 u Re L y y =0 17

18 The non-dimensionalized x-momentum equation states that u u u + v x y = dp dx P Re L y 2 The boundary condition at the wall is u x, 0 = 0 and for the free stream is u x, = u x V The similarity parameter used is Re L = ul ν The non-dimensionalized x-thermal equation states that u T T + v x y = 1 2 T Re L P r y 2 The boundary condition at the wall is T x, 0 = 0 and for the free stream is T x, = 1 The similarity parameter used is P r = ν α = momentum thermal We expect that u = f x, y, dp dx, Re L and T = f x, y, dp dx, Re L, P r The heat transfer coefficient can be expressed similarly as h = k f T L y The Nusselt number is defined as Nu hl k f = T y y =0 = fx, Re L, P r y =0 Nu hl k f = fre L, P r It is also true that so the following plots are true Nu = CRe m L P r n 18

19 Thermophysical quantities are frequently measured at the film temperature, which is defined as T f = T s + T 2 Unless otherwise stated, evaluate thermophysical quantities at the film temperature. An exception to this would be evaluating the heat flux at a surface. Since you re evaluating it at the surface, you d use the thermophysical properties at the surface temperature and not the film temperature 3 External Flow 3.1 Flat Plate in Parallel Flow Hydrodynamic Solutions The Blasius solution states the following it ignores P u u x + v u y = ν 2 u y 2 The thermal boundary layer distance scales as νx δ v and The stream function is defined as and u 2 x u νu δ 2 v u = ψ y v = ψ x Note that the signs are reversed for u and v in BSL We will then define and It turns out that 5 5 I am now writing fη as simply f fη = ψ u νx u νx ψ = u fη u u η = y νx u = df u dη u x = u 2x η d2 f dη 2 u y = u u νx 2 u y 2 = u2 d 3 f νx dη 3 d 2 f dη 2 19

20 These equations can be combined with the momentum balance to yield see BSL 4.3, d3 f dη 3 + f d2 f dη 2 = 0 whose values are tabulated in Table 7.1 of Incropera Laminar Flow Equations See Incropera Section 7.2 for the math, but since u u = 0.99 for η = 5 δ v = 5 u νx = 5x Rex The shear stress at the wall may be expressed as τ s = µ u u d 2 f y = µu y=0 νx dη 2 so τ s = 0.332u ρµu x η=0 = 0.664ρu2 2 Re x Note that the unit of shear stress is N/m 2 or, equivalently, At the boundary layer i.e. η = 5, the local friction coefficient is Heat Transfer Solution Laminar C f,x = 2τ s,x ρu 2 = Rex kg m s 2 The energy equation in the boundary layer without viscous dissipation can be rewritten as the following when the dimensionless temperature is introduced and a similarity solution of the form T = T η is assumed: d 2 T dη 2 + P r 2 f dt dη = 0 with boundary conditions of T 0 = 0 and T = 1 For P r > 0.6, the surface temperature gradient is given as dt dη = 0.332P r 1/3 η=0 The local convection coefficient can be expressed as q s h x = = T T s k T u 1/2 dt T s T T s T y = k y=0 νx dη From this, we can state that for P r > 0.6 η=0 which means that Nu x = h xx k = 0.332Re1/2 x P r 1/3 δ v δ t = P r 1/3 20

21 3.1.4 Average Boundary Layer Parameters Laminar The average boundary layer parameters are C f,x = Rex and for P r > 0.6 since Nu x = h x x k = 0.664Re1/2 x P r 1/3 h x = 2h x Turbulent Flow The equations for turbulent flow over a flat plate, the local friction coefficient is approximately C f,x = Re 1.5 x for Reynolds numbers between R x,c and 10 8 Also, the velocity boundary layer thickness is δ = 0.37xRe 1/5 x The local Nusselt number for a Prandtl number between 0.6 and 60 is Nu x = Re 4/5 x P r 1/3 3.2 Cylindrical and Spherical Systems in Cross Flow At the leading edge θ = 0, For a cylinder, Hilpert s relation states that for P r > 0.7 Nu D = 1.15Re 1/2 D P r1/3 Nu D = hd k = CRem DP r 1/3 Note that the transition for cylindrical cross flow is R x,c

22 Technically, there is a more accurate equation for the average Nusselt number for cylindrical cross-flow given by the following expression, that is valid for all Re D and P r > 0.2 [ Nu D = Re1/2 D [ P r1/3 ] 2/3 1/ P r ] 5/8 4/5 ReD For a sphere, Nu D = Re 1/2 D P r1/3 Technically, there is a more accurate equation for the average Nusselt number for spherical cross-flow given by, 1/4 Nu D = Re 1/2 D Re2/3 D P r 0.4 µ µ s where all properties except µ s are evaluated at T and is applicable for 0.71 < P r < 380, 3.5 < Re D < , and 1 < µ µ s < Potential Flow There really isn t much to say here. description of potential flow The stream functions are defined as and See my Transport Phenomena I review packet for a detailed u = ψ y v = ψ x As streamlines are more compact closer to the sphere or cylinder in potential flow, it means the velocity of the fluid is increasing 3.4 Sphere in a Water Bath Problem: Consider a hot sphere in a cold bath such that it is quiescent i.e. no flow - only conduction and is at steady state Always start with boundary conditions. They are, for T r: T = T T R = T R There are two ways to approach this both will yield the same answer. approach as opposed to a microscopic energy balance where We can try a differential 4πr 2 q r r r=r 4πr 2 q r r=r+ r = 0 such that it reads conduction in minus conduction out equals zero. Note that it is area times flux in minus area times flux out. The area here is 4πr 2 for a sphere This expression can be rewritten by dividing through 4π and dividing by dr such that d r 2 q r = 0 dr 22

23 Using Fourier s Law, once constants i.e. k are factored out d r 2 dt = 0 dr dr Note that this equation could also have been obtained using the spherical equation of the microscopic energy balance such that 2 T = 0 1 d r 2 r 2 dt = 0 d r 2 dt = 0 dr dr dr dr Anyhow, we can integrate the above expression twice to yield T = C 1 r + C 2 which turns to the following when boundary conditions are applied: T = R r T R T + T T T T R T = R r The radial flux from the sphere surface can be found by q r r=r = k dt dr = k T R T r=r R We know that Newton s law of cooling applies at the solid-liquid interface, so k T R T R = h T R T h = k R = 2k D Recall that the Nusselt number is defined as Nu = hd k, so for solely conduction in a sphere 4 Tube Flows Nu = Area Average Quantities An area average quantity is defined as f = f da da For the area average shear stress around a sphere, τ r, z r dr dθ τ = 2π r r dr dθ 0 0 The mixing cup temperature is defined as T mc = v z T v z At a constant heat flux, T r, z = dt mc z dz where both quantities are for fully developed thermal regions and laminar flow in a tube. Therefore, the axial temperature gradient is independent of the radial location 23

24 4.2 Laminar Flow in Circular Tubes Heat Flow The log-mean temperature difference is defined as T lm T o T i To ln T i where the subscript i represents the fluid flow through the tube, and the subscript o represents fluid flow over the tube For internal flow in a circular cylinder, Q = hπdl T lm From Transport Phenomena I, we know that for laminar flow in a circular tube, v z,max = 2v z,mean and v z = P 0 P L R 2 [ r ] 2 1 4µL R It will simply be stated that Nu D = hd k = 4.36 for laminar flow in a circular tube with constant surface heat flux To see how the local Nusselt number changes with different conditions, consult Figure in Bird, αz which shows a plot of Nu loc vs. v z D 2 5 Diffusivity and the Mechanisms of Mass Transport 5.1 Kinetic Theory and Lennard-Jones Potential The Kinetic Theory of Gases makes the assumption that all atoms are hard spheres that collide elastically and there are no intermoelcular forces. From this: The mean free path is given as λ = 1 2πd2 n where d is the diameter and n is the number density The mean molecular speed is ū = 8kB T πm where k B is Boltzmann s constant i.e m 2 kg s 2 K 1, or, equivalently, J/K The collision frequency is The dynamic viscosity is Z = 1 4 nū µ = 1 3 nmūλ = 1 3 ρūλ = 2 πmkb T 3π πd 2 24

25 The thermal conductivity is given as and has units of K/m The Lennard-Jones potential states that that k = 1 2 nk Būλ = mkb T/π πd 2 [ σ 12 σ ] 6 φr = 4ε r r where φr is the potential energy, σ is the characteristic diameter of the molecules, and ε is the maximum energy of attraction between a pair of molecules The values for σ are tabulated in Table E.1 of BSL The viscosity of a pure monatomic gas may be written as µ = 5 πmkb T 16 πσ 2 Ω µ MT = σ 2 Ω µ where the second equation has σ with units of angstrom, T with units of Kelvins, M is the molecular weight unitless, and µ with units of g/cm s Ω µ is called the collision integral for viscosity and is tabulated in Table E.2 of BSL The thermal conductivity using Lennard-Jones parameters is T M k = σ 2 Ω µ 5.2 Fick s Law of Binary Diffusion Molecular Mass Transport If we define w i has the mass flow of substance i, ω i as the mass fraction of substance i, D ij as the diffusivity of i in j, w i A = ρd ω i ij L where w i A is the mass flux of substance i The units of D ij are length-squared per unit time It is also important to recall that ω i = 1 This can be rewritten as i j i = ρd ij ω i where j i is the mass flux of substance i. This is Fick s Law. The mass flux is defined as j i ρω i v i 25

26 The mass flux is measured with respect to the motion of the center of mass, so j i = 0 In general, for a mixture, i v = i ω i v i which translates to the velocity is equal to the sum of the mass fraction of each substance times its respective velocity The Schmidt number is defined as The Lewis number is defined as Also, recall that the ideal gas law states and Sc = Le = ν D ij α D ij ρ = MP RT c = P RT where c is a molar density defined as c ρ M The mass fraction is most frequently written as when volume is constant In addition, The mole fraction is most frequently written as ω i = ρ i ρ c = ρω M when volume is constant x i = c i c 5.3 Temperature and Pressure Dependence of Diffusivities For a binary-gas mixture at low pressure, P D AB P ca P cb 1/3 T ca T cb 5/12 1 M A + 1 M B b T 1/2 = a TcA T cb where the subscript c represents a critical property, which can be obtained from Table E1 in BSL. In this equation, D AB has units of cm 2 /s, P has units of atm, and T has units of K For a non-polar gas pair, a = and b = excluding helium and hydrogen 26

27 For a non-polar gas and water, a = and b = Interdiffusion is synonymous with self-diffusion and is denoted by D AA Using reduced temperatures and pressures, one can find the reduced self-diffusivity from Figure in BSL if cd AA is given at a specific temperature and pressure. Recall that Regular quantity Reduced quantity= Critical Quantity To implement this method, realize that the cd AA r is constant for a given system even if temperature and pressure changes. Therefore, if you are given cd AA at one temperature and pressure and want it at another temperature and pressure, use Figure to find the critical diffusivity, and then use this value at the new reduced pressure and temperature to find the new not critical diffusivity Another way to find the self-diffusivity is to use the equation where c is a concentration given in mol/cm 3 1 cd AB c = /2 P 2/3 ca M A M B T 1/6 ca For a binary-gas mixture at high density and low pressure, one can use Figure by replacing the above formula with 1 cd AB c = /2 P ca P cb 1/3 M A M B T ca T cb 1/ Theory of Diffusion in Gases at Low Density The kinetic theory of gases states that, for rigid spheres, D AB = 2 kb T π 2 m A m 2 B da + d B n π 2 With the use of Lennard-Jones constants, the above equation can be rearranged to If the ideal gas law is assumed, cd AB = T D AB = M A M B σab 2 Ω D,AB T 3 1 M A + 1 M B 1 P σ 2 AB Ω D,AB where the variables are in the conventional units previously described. The values for the collision integral for diffusivity can be found in Tables E1 and E2 of BSL If not given, the values for σ AB and ε AB can be given by for nonpolar gas pairs and σ AB = σ A + σ B 2 ε AB = ε A ε B 27

28 5.5 Mass Average and Molar Average Velocity When we state v i, or the velocity of species i, this is not the velocity of an individual molecule of i. Instead, it is the mean of all the velocities of molecules of species i. In essence, is the mass average velocity v = N i=1 ρ i v i N i=1 ρ i N = ω i v i i=1 The molar average velocity is given as v = N i=1 c i v i N i=1 c i N = x i v i i=1 5.6 Summary of Mass and Molar Fluxes As stated earlier, Fick s Law states that the mass flux is j A = ρd AB ω A This can be stated as where this is now the mole flux J A = cd AB x A The combined mass flux for one species is The combined molar flux for one species is n A = j A + ρ A v N A = J A + c A v The mass flux for N species is The molar flux for N species is j A = n A ω A J A = N A x A N ni i=1 N N i i=1 For a binary system with one-dimensional diffusion, N A,z = cd AB x A z + x A N A,z + N B,z 6 Concentration Distributions in Solids and in Laminar Flows 6.1 Shell Mass Balances and Boundary Conditions The molar flux can be related to the concentration gradient by N A = cd AB x A + x A N A + N B Note that is acting as a gradient here since x A is a scalar 28

29 For small x A, the right-hand term drops out, and implementing yields x a = c A c N A = D AB c A when x A is small. This is frequently encountered at surfaces. In words, the shell mass balance states rate of mass of A in minus rate of mass of A out plus rate of production of mass of A by homogeneous reaction equals zero The boundary conditions include: The concentration at a surface can be specified e.g. x A = x A0 The mass flux at a surface can be specifies e.g. N A = N A0 If diffusion occurs in a solid, N A0 = k c c A0 c Ab may apply, where N A0 is the molar flux at the surface, c A0 is the surface concentration, c Ab is the concentration in the bulk fluid stream, and k c is the mass transfer coefficient The rate of chemical reaction at a surface can be specified. For an n-th order reaction, N A0 = k c c n A0 may apply 6.2 Diffusion Through a Stagnant Gas Film Problem: Consider the schematic shown below. Note that B is immiscible with A, so while B can be present in the system at steady state, there is no net flux of B down or out, just across such that N B.z = 0. For the full description of the problem, see Section 18.2 of BSL. 1. We write the mass balance in the z direction as 2. Solving for N A,z yields 3. A steady-state mass balance can be written as where S is a cross-sectional area N A,z = cd AB dx A dz + x AN A,z N A,z = cd AB dx A 1 x A dz SN A,z z SN A,z z+ z = 0 29

30 4. Dividing by S z and letting z 0 yields dn Az dz = 0 5. This can therefore be written as d cdab dx A = 0 dz 1 x A dz a For an ideal gas mixture, c is constant for a constant T and P. Also, for gases, D AB is usually independent of the composition such that d 1 dx A = 0 dz 1 x A dz which can be integrated to yield 6 ln 1 x A = C 1 z + C 2 6. Although this is not obvious, we can let C 1 = ln K 1 and C 2 = ln K 2 such that 7 1 x A = K z 1 K 2 7. The boundary conditions are: x A z 1 = x A1 and x A z 2 = x A2 8. Applying the boundary conditions yields 1 x A1 = K z1 1 K 2 and 1 x A2 = K z2 1 K 2, which can be combined to yield 1 x A2 = K z2 z1 1 1 x A1 a A little algebraic manipulation yields b We need an expression for K 2, so 1 xa2 K 1 = 1 x A1 1/z2 z 1 z1/z 1 2 z 1 xa2 1 xa2 1 x A1 = K 2 K 2 = 1 x A1 1 x A1 1 x A1 9. Plugging in the results for K 1 and K 2 yields z/z2 z 1 1 xa2 1 xa2 1 x A = 1 x A1 1 x A1 1 x A1 z1/z 2 z 1 z1/z 2 z 1 which can be rearranged to 1 x A 1 x A1 = 1 xa2 1 x A1 z z1/z 2 z To obtain the profile for x B, recognize that x A + x B 1 6 Note that a useful integral for these types of problems is 1 ax + b dx = 1 ln ax + b a 7 Generally speaking, for an equation of the form a ln 1 + bx A = C 1 z+c 2, you want to make C 1 = a ln K 1 and C 2 = a ln K 2. From this, the final equation will be of the form 1 + bx A = K1 zk 2. 30

31 11. Now, if the average concentration of B is desired, so x B = x B x B1 = z2 z 1 z2 z 1 x B dz dz z2 xb z 1 x B1 z2 z 1 dz dz 12. Define the non-dimensional height variable ξ = z z 1 z 2 z 1, such that dz = z 2 z 1 dξ and x B x B1 = 13. The integral table states that a x dx = ax ln x, so which yields x B x B1 = 1 xb 0 x B1 1 0 dξ ξ xb2 ln x B1 xb2 x B1 ξ dξ x B = x B2 x B1 x B ln xb2 ln x B1 14. The rate of evaporation is the rate of mass transfer at the liquid-gas interface and can be found by calculating N A,z at z = z 1. Therefore, N A,z z1 = cd AB dx A = cd AB dx B 1 x A1 dz x B1 dz z1 1 0 z1 a Implementing the dimensionless length, cd AB x B1 dx B dξ dξ ξ=0 dz = cd AB d x B /x B1 ξ z 2 z 1 dξ ξ=0 15. The final expression is N A,z z1 = cd AB z 2 z 1 ln xb2 x B1 cd AB = x A1 x A2 z 2 z 1 x B ln a This expression can be used to find the diffusivity constant of an evaporating substance 16. For diffusion with a moving interface, see Example in BSL 31

32 6.3 Diffusion through a Spherical Film Problem: Consider diffusion through a spherical shell with radii r 1 and r 2 where r 1 < r < r 2. For the full problem statement, see Example The shell balance states that in the r direction states that which can be rewritten as N A 4πr 2 r N A 4πr 2 r+ r = 0 d r 2 N A = 0 dr 2. We can state that N B,r = 0 since B is insoluble in A. Therefore, which can be solved for N A,r as 3. Therefore, N A,r = cd AB dx A dr + x AN A,r N A,r = cd AB dx A 1 x A dr d r 2 cd AB dx A = 0 dr 1 x A dr 4. Integrating this yields 5. Let C 1 = ln K 1 and C 2 = ln K 2 such that 6. As in the previous subsection, this comes out to 1 x A 1 x A1 = when the boundary conditions are applied ln 1 x A = C 1 r + C 2 1 x A = K 1/r 1 K 2 1 xa2 1 x A1 1/r1 1/r/1/r 1 1/r 2 7. The molar flow can be found as W A = 4πr1N 2 r1 A,r = 4πcD AB 1 xa2 1 r 1 1 ln r 2 1 x A1 which is applicable for any spherical surface of radius r between r 1 and r 2 32

33 6.4 Diffusion with a Heterogeneous Chemical Reaction e.g. Gas Reacting on Solid Catalyst Diffusion with an Instantaneous Heterogeneous Reaction Problem: Consider the heterogeneous chemical reaction of 2A B shown in the diagram below. For the full problem statement, see Section 18.3 in BSL. 1. From the stoichiometry, we know that 8 N B,z = 1 2 N A,z 2. We also know that which simplifies to N A,z = cd AB + x A N A,z 1 2 N A,z N A,z = cd AB dx A x A dz 3. The shell mass balance states that SN A,z z SN A,z z+ z = 0 which leads to 4. This yields 5. Integrating this yields dn A,z dz d cdab dz x A = 0 dx A = 0 dz 2 ln x A = C 1 z + C 2 for constant cd AB 6. Substituting C 1 = 2 ln K 1 and C 2 = 2 ln K 2 yields 1 x A 2 = Kz 1 K 2 7. The boundary conditions are x A 0 = x A0 and x A δ = 0 8 For a reaction aa bb, N B,z = b a N A,z 33

34 8. Applying the boundary conditions yields x A = z/δ 2 x A0 9. To get the molar flux, we need dx A. Since we want the molar flux at the film, and the film is at z = 0, dz we technically want dx A dz. The final result yields 9 so z=0 N A,z = 2cD AB δ Diffusion with a Slow Heterogeneous Reaction 1 ln x A0 Problem: Attempt the previous problem with a slow reaction i.e. not instantaneous. Assume that the rate A disappears at the catalyst surface is given as N A,z = k 1 c A = k 1 cx A, in which k 1 is a rate constant for the pseudo-first-order surface reaction. For the full problem statement, see Example in BSL. 1. The set-up is identical up until the boundary conditions at which point x A δ = N A,z x A δ = 0 2. Applying the boundary conditions yields 3. Evaluating dx A dz x A = and solving for N A,z yields, z=0 N A,z = 2cD AB δ z/δ N A,z k 1 c z/δ 2 x A0 NA,z k ln x A0 1 c k 1 c instead of 4. If k 1 is large note that this means the reaction is fast, but not so fast that it is instantaneous then N A,z = 2cD AB/δ D ln AB 1 1 k 1 δ 2 x A0 which can be obtained by a Taylor expansion on the logarithm term and keeping just the first term such that ln 1 + p p for small p 5. The Damkohler Number of the second order can be defined as Da II = k 1 δ D AB a In the limit of Da II, we obtain the expression for the instantaneous reaction b In words, the Damkohler number is the ratio of the chemical reaction rate compared to the diffusion rate i.e. mass transfer c A very fast reaction is governed by mass transfer, but a very slow reaction is governed by kinetics 9 The following is a helpful identity: d a bx = ba bx ln a dx 34

35 6.5 Diffusion with a Homogeneous Chemical Reaction e.g. Gas Dissolving in Liquid Problem: Consider a gas A diffusing into a liquid B. As it diffuses, the reaction A + B AB occurs. You can ignore the small amount of AB that is present this is the pseudobinary assumption. 1. Note that the reaction rate can be given as k 1 c A if we assume pseudo-first order. This makes the shell mass balance SN A,z z SN A,z z+ z k 1 c A S z = 0 2. This can be rewritten as dn A,z dz + k 1 c A = 0 3. If the concentration of A is small i.e. dilute, then we can state that x A goes to zero in such that 4. Combining this with the equation in step 2 yields N A,z = cd AB dx A dz + x A N A,z + N B,z N A,z = D AB dc A dz D AB d 2 c A dz 2 k 1 c A = 0 5. The boundary conditions are c A 0 = c A0 and N A,z L = dc A dz = 0. The first boundary condition L states that the concentration of A at the surface is fixed. The second boundary condition states that no A diffuses through the bottom of the container. 6. Multiply the equation in Step 4 by L 2 c A0 D AB L 2 d 2 c A c A0 dz 2 for later simplicity. This yields k 1 c A L 2 c A0 D AB = 0 7. Let s define the dimensionless variable known as the Thiele modulus: φ k 1 L2 /D AB 8. Let s also define the dimensionless length ξ z L such that dz = Ldξ 35

36 9. The concentration ratio ca be defined as 10. Using these variables, Γ c A c A0 d 2 Γ dξ 2 φ2 Γ = The general solution is given as Γ = C 1 cosh φξ + C 2 sinh φξ since and cosh p ep + e p 2 sinh p ep e p 12. The boundary conditions are at ξ = 0, c A = c A0, so Γ = 1 and ξ = 1, dγ dξ condition, 1 = C 1 cosh 0 + C 2 sinh 0 C 1 = 1 and and invoking the second boundary condition yields dγ dξ = φ sinh φξ + C 2φ cosh φξ 2 = 0. Applying boundary 0 = φ sinh φ + C 2 φ cosh φ C 2 = tanh φ 13. This yields which can be rearranged to 10 Γ = cosh φξ tanh φ sinh φξ Γ = cosh φ cosh φξ sinh φ sinh φξ cosh φ 14. Reverting to the original notation yields, [ c cosh k A 1 L2 /D AB 1 z ] = L c A0 cosh k 1 L2 /D AB 15. The average concentration in the liquid phase can be given by 16. The molar flux at the surface is c A c A0 = L 0 N A,z z=0 = D AB dc A dz c A c A0 dz = L 0 dz = tanh φ φ ca0 D AB = z=0 L cosh [φ 1 ξ] cosh φ φ tanh φ 10 Note that cosh x ± y = cosh x cosh y ± sinh x sinh y and sinh x ± y = sinh x cosh y ± cosh x sinh y 36

37 6.6 Gas Absorption with Chemical Reaction in Agitated Tank Problem: Consider the diagram shown below. Assume that each gas bubble is surround by a stagnant liquid film of thickness δ,which is small compared to the bubble diameter. Assume a quasi-steady concentration profile is quickly established in the liquid film after the bubble is formed. The gas A is only sparingly soluble in the liquid, so we can neglect the convection term. The liquid outside the stagnant film is at concentration c Aδ and is constant. Even though this is a spherical bubble, it is a thin shell, so you can treat it as a slab. For the full problem statement, see Example in BSL. 1. The setup is the same as before, but the boundary conditions are at z = 0, ξ = 0,c A = c A0, Γ = 1, and at z = δ, ξ = 1, c A = c Aδ, Γ = B if we state that B = C Aδ C A0. Note that the dimensionless length should be redefined accordingly as ξ = z δ and the Thiele modulus is redefined as φ = k 1 δ2 /D AB 2. From the previous problem, Γ = C 1 cosh φξ + C 2 sinh φξ 3. Applying boundary condition 1 yields C 1 = 1 4. Applying boundary conditions 2 yields C 2 = B cosh φ sinh φ 5. This means Γ = cosh φξ + B cosh φ sinh φ sinh φξ = sinh φ cosh φξ + B cosh φ sinh φξ sinh φ 6. Now equate A entering the liquid at z = δ to amount consumed in bulk: dc A SD AB dz = V k 1 c Aδ z=δ 7. We need the dc A dz term. This can be rewritten as z=δ dc A dz = dc A dξ dξ dz = dc A 1 dξ δ 37

38 8. Therefore, dc A dz = c A0 φ sinh 2 φ φ cosh 2 φ + Bφ cosh φ z=δ δ sinh φ using the identity cosh x 2 sinh x 2 = 1 yields dc A dz = c A0 Bφ cosh φ φ z=δ δ sinh φ 9. So, 10. This can be solved for B as SD AB c A0 δ Bφ cosh φ φ sinh φ 1 B = cosh φ + V Sδ φ sinh φ = V k 1 c Aδ 11. The total rate of absorption is N N A,z z=0 δ c A0 D AB = φ sinh φ which is plotted in Figure of BSL cosh φ 1 cosh φ + V Sδ φ sinh φ 6.7 Diffusion into a Falling Liquid Film Gas Absorption Problem: Consider the absorption of A into a falling film of liquid B. For the full problem, see Section 18.5 in BSL. 1. The velocity profile is found from Transport I as x ] 2 v z x = v max [1 δ 2. The concentration will change in the x and z direction, so at steady state N A,z z W x N A,z z+ z W x + N A,x x W z N A,x x+ x W z = 0 38

39 3. This then yields N A,z z 4. We now want expressions for the molar mass flux: + N A,x x = 0 N A,z = D AB c A z + x A N A,z + N B,z which reduces to the following because the transport of A in the z direction will be primarily by convection not diffusion N A,z = x A N A,z + N B,z c A v z x and in the x direction we have c A N A,x = D AB z since there is mostly diffusion in the x direction not convection 5. Therefore, 6. Inserting the velocity component yields c A v z z = D 2 c A AB x 2 x ] 2 ca v z,max [1 δ z = D 2 c A AB x 2 7. The boundary conditions are: at z = 0, c A = 0 and x = 0,c A = c A0 and x = δ, c A = 0 since there x is pure B at the top, the liquid-gas interface is determined by the solubility of A in B, and A can t diffuse through the wall 8. We shall use the limiting case of the Penetration Model, which states that there is only penetration in the outer layers of the film such that v z v z,max. This means, c A v z,max z = D 2 c A AB x 2 and the third boundary condition is changed to at x =, c A = 0 9. This looks like a semi-infinite solid problem, so c A x = 1 erf c A0 4DAB z/v z,max 10. The local mass flux at the gas-liquid interface may be found by N A,x x=0 = D AB c A x 11. The total molar flow across the surface at x = 0 is W A = ˆ W ˆ L 0 0 DAB v max = c A0 x=0 πz N A,x x=0 dz dy = W Lc A0 4DAB v max πl 39