Semester S1 2023 Assessment Notice Faculty of Engineering ASSESSMENT DETAILS Unit code(s) Title of assessment Assessment type Assessment duration Materials required CHE3167 Transport Phenomena & Numerical Methods – Final Assessment Moodle Assessment 10 minutes reading time + 3 hours writing + 30 minutes upload time Pen & paper, calculator, scanning device ACADEMIC INTEGRITY Intentional plagiarism or collusion amounts to cheating under Monash University Statute & Regulations. Plagiarism: Plagiarism means to take and use another person’s ideas and or manner of expressing them and to pass these off as one’s own by failing to give appropriate acknowledgement, including the use of material from any source, staff, students or the internet, published and unpublished works. Collusion: Collusion means unauthorised collaboration on assessable written, oral or practical work with another person. 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No part of this assessment has been previously submitted as part of another unit/course. ● I acknowledge and agree that the assessor of this assessment may for the purposes of assessment, reproduce the assessment and: i. provide to another member of faculty and any external marker; and/or ii. submit it to a text matching/originality checking software; and/or iii. submit it to a text matching/originality checking software which may then retain a copy of the assignment on its database for the purpose of future plagiarism checking. ● I certify that in completing this assessment I have not plagiarised the work of others, participated in unauthorised collaboration or otherwise breached the academic integrity requirements in the Student Academic Integrity Policy. In completing this assessment task you agree to the statements above. If you do not agree to the Student Statement, please submit directly to your Unit Coordinator by the due date, providing a written explanation of which aspect of the Student Statement you do not agree with and why. USE OF GENERATIVE ARTIFICIAL INTELLIGENCE Generative AI tools cannot be used in this assessment task: In this assessment, you must not use generative artificial intelligence (AI) to generate any materials or content in relation to the assessment task. Answering Instructions for Candidates 1. Answer all four questions. 2. Marks for each question are indicated. The total marks for the exam are 100. 3. A list of formulae used in the course is appended at the end (pages 9 to 23). 2/23 CHE 3167 —Transport Phenomena and Numerical Methods FINAL ASSESSMENT 2023 The exam is open book and notes 100 marks Q1. Diffusive Mass Transfer in a Reactive Tank (25 marks) Consider a tank with radius R with a source of species a that diffuses from the top (z = 0) to the bottom of the tank (z = L). The mole fraction of a at the entrance is x = xa0. The bottom of the tank is coated with a catalyst that converts a to b with a first order reaction rate, which means that the reaction rate (in [moles m−3s−1]) is dependent on the concentration of a: −ra = ctksxa where ct is the total molar concentration of a and ks is the rate constant. It is assumed that the total concentration of species does not change and that b diffuses back through the fluid. Figure 1.1: Tank with heterogeneous reaction at the bottom surface. (a) What is a reasonable postulate for the steady state molar concentration profile in the tank? Justify your answers. (3 marks) (b) Using an appropriate shell control volume with a thickness of ∆z, set up the differential equation that governs the mole fraction xa using a species shell balance. You need to provide a second order ordinary differential equation. (7 marks) (c) Integrate the differential equation twice to find an expression for xa in terms of the integration constants E1 and E2, respectively. (2 marks) (d) Identify the two boundary conditions and use these to find expressions for E1 and E2. At the bottom surface, the flux of a that arrives at the surface gets consumed by the first order reaction. Substitute the expressions for E1 and E2 into the equation for xa and express it in terms of the non-dimensional concentration xa/xa0. (8 marks) (e) Find an expression for the diffusive molar flux of species a which is independent of the position z. When ks goes to ∞ or zero, what happens to the flux and what does that mean physically for these two cases? (5 marks) Page 3 of 23 CHE 3167 —Transport Phenomena and Numerical Methods Q2. Free convection in a slot (25 marks) A fluid of constant viscosity, with density given by, ρ ( T ) = ρ ̄ − ρ ̄ β ̄ ? T − T ̄ ? (in which β ̄ = −(1/ρ)(∂ρ/∂T)p evaluated at T = T ̄; you do not need to substitute the equation for β ̄), is confined in a rectangular slot as shown in Fig. 2.1. The slot has vertical walls at x=±B,y=±W,andatopandbottomatz=±H,withH≫W≫B. Thewallsare non-isothermal, with temperature distribution, Tw =T ̄+Ay so that the fluid circulates by free convection. The velocity profiles are to be predicted, for steady laminar flow conditions and small deviations from the mean density ρ ̄. Figure 2.1: Laminar free convection flow in a slot with non-isothermal walls. (a) Simplify the equations of continuity, motion, and energy according to the postulates: v=vz(x,y)δz; ∂2vz≪∂2vz; T=T(y) ∂y2 ∂x2 These postulates are reasonable for slow flow, except near the edges, y = ±W and z = ±H (i.e., within the blue shaded region in Fig. 2.1). You can also assume that the weight of the fluid is balanced by the pressure gradient, i.e., +W +H Tw (+W) -H -W Tw (-W) -B +B vz (x,y) z y x – p = −ρ ̄gz + constant (b) List the boundary conditions to be used with the problem after the simplification specified in part (a). (4.5 marks) (c) Assuming negligible viscous dissipation, solve for the temperature and velocity profiles. Note that since the net mass flow in the z-direction is zero, the velocity is an odd function of y (as shown in Fig. 2.1). (8.5 marks) (12 marks) Page 4 of 23 CHE 3167 —Transport Phenomena and Numerical Methods Q3. Time dependent flow in a pipe (25 marks) Industrial chemical and biological engineering processes include fluid flows in a pipe. For ex- ample, blood that flows through our arteries and delivery of reagents in a chemical process. For a newtonian fluid we consider a scenario where the fluid is flowing in the z-direction with a uniform velocity v0, and suddenly a long pipe is placed in the stream. There is a uniform pressure gradient along the horizontal pipe, and we want to describe how the velocity profile in the pipe develops over time. The radius of the pipe is expressed as r0. (a) By carrying out the approrpiate simplifications of the equations of continuity and motion, and indicating your reasons for doing so, express the z-component of the equation of motion. (4 marks) (b) Write the boundary conditions for the velocity in space (at the walls and centre line) and the initial condition for the velocity. (c) Show that adopting the following non-dimensional variables, ξ=r; τ=μt; φ= vz where: vmax=−r02dp (3 marks) (3.1) (5 marks) r0 ρr02 vmax leads to the following expression for the non-dimensional velocity φ: ∂φ 1 ∂ ? ∂φ? ∂τ=4+ξ∂ξ ξ∂ξ 4μ dz (d) Write the initial and boundary conditions for the velocity in non-dimensional form. Note that φ0 = v0/vmax. (3 marks) (e) Equation (3.1) can be solved using the method of separation of variables by assuming that the velocity can be represented as the summation of two functions, one of which, f is a function of the position ξ, and the other which is also a function of time which is g(ξ,τ): φ(ξ,τ)=f(ξ) +g(ξ,τ) (3.2) Show that the steady state solution leads to f(ξ)=φ∞ =1−ξ2 (3.3) (5 marks) (f) Using Equations (3.2) to (3.3), simplify Eq. (3.1) which describes the transient part of the solution. This equation can then be solved using the method of separation of variables by assuming g(ξ,τ) = E(ξ)T(τ). Substitute this into the simplified partial differential equation and separate the PDE into two ODE’s in terms of the constant (C3). Next, find the general expression for T(τ) in terms of the integration constants (C3 and C4). You do not need to solve for E(ξ) or g(ξ,τ). Note: These equations can be shown to result in an exponential decay with time when substituting the initial condition, and the radial dependence of the velocity results in a Bessel function of the first kind (J). At short times the velocity profile is nearly linear, and transforms into a parabolic profile, Eq. (3.4), where the integration constants can be found using the boundary and initial conditions. g(ξ,τ) = φt(ξ,τ) = Cxexp(−Cyτ)J(Cy∗,ξ) (3.4) (5 marks) Page 5 of 23 CHE 3167 —Transport Phenomena and Numerical Methods Q4. Flow between parallel plates: Finite Difference Method (25 marks) A viscous fluid with temperature-independent physical properties is in fully developed lami- nar flow between two flat surfaces placed a distance 2B apart as shown in Fig. 4.1. For z < 0, the fluid temperature is uniform at T = T ̃. For z > 0, heat is added at a constant, uniform flux q ̃ at both walls. Due to the symmetry of the problem, the numerical solution for the temperature dis- tribution T (x, z) for the entire domain can be simplified by only solving for the domain 0 ≤ z ≤ L and 0 ≤ x ≤ B. Figure 4.1: Laminar, incompressible flow between parallel plates, both of which are being heated by a uniform wall heat flux q ̃, starting at z = 0. For a steady state heat transfer to this problem, by neglecting the viscous dissipation and axial heat conduction effects, the spatial distribution of the temperature profile T (x, z) can be described by the following partial differential equation, ρcˆv ∂T =k∂2T (4.1) pz∂z ∂x2 where k is the thermal conductivity, and ρ is the density of the fluid. The fully developed velocity profile for flow between two parallel plates is given by the expression, ? ?x?2? vz(x) = vz,max 1 − B (4.2) where vz,max is the maximum or centerline velocity of the fluid. Page 6 of 23 CHE 3167 —Transport Phenomena and Numerical Methods (b) It is desired to solve Eq. (4.1) numerically using the finite difference method for the compu- tational domain illustrated in Fig. 4.2. How many boundary conditions are required to solve Eq. (4.1)? Write down the boundary conditions. Figure 4.2: Finite difference grid for a laminar, incompressible flow between parallel plates. (3 marks) (c) Sketch the computational molecule for the second order finite difference (Crank-Nicolson) scheme that can be used to solve the partial differential equation (4.1) with guaranteed stability. (5 marks) (d) Using the computational molecule in (c), set up the O(h2) (where h is the step size in the x or z direction, which in general are not identical in magnitude) finite difference equations for a numerical solution of this problem to determine the temperature at a typical interior node. The grid numbering in the x and z directions is denoted by the indices i = 1, 2, …, m + 1 and l = 1, 2, …, n + 1, respectively, where m and n represent the number of intervals in each direction. By keeping all the z = l + 1 terms on the left hand side, and z = l terms on the right hand side, simplify your final form of difference equation with the help of the following expression for the parameter λi, ? ? ?xi?2??−1 k∆z λi = ρcˆpvz,max 1− B (∆x)2 (4.3) (5 marks) (e) For a computational mesh with m = 3, write the specific difference equations for all the interior nodes and boundary nodes that are required to compute the spatial temperature distribution at z = ∆z, i.e., T l=2, i = 1, 2, …, m + 1. i (5 marks) (f) Formulate the final set of equations in the matrix form [M][T] = [b] where [M] is the matrix of coefficients, [T ] is the column vector of unknown temperature at z = l + 1, and [b] is the column vector containing known constants or functions. Note that you are not required to solve the matrix equation for [T ] to answer this question. (4 marks) (3 marks) Page 7 of 23 CHE 3167 —Transport Phenomena and Numerical Methods END OF EXAM Page 8 of 23 2. Forward difference ∂T = 1 (−3Ti,j +4Ti+1,j −Ti+2,j)+O(∆x2) CHE 3167 TRANSPORT PHENOMENA & NUMERICAL METHODS CHE 3167 AIDE-ME ́MOIRE O(∆t) Finite Difference Equations: Forward difference ∂T = 1 ?Tl+1 −Tl ?+O(∆t) ∂t ∆t i,j i,j O(∆x) Finite Difference Equations: 1. Forward difference ∂T = 1 (Ti+1,j −Ti,j)+O(∆x) ∂x ∆x 2. Backward difference ∂T = 1 (Ti,j −Ti−1,j)+O(∆x) ∂x ∆x O(∆x2) Finite Difference Equations: 1. Central difference: ∂T = Ti+1,j − Ti−1,j + O(∆x2) ∂x 2∆x ∂2T = Ti+1,j − 2Ti,j + Ti−1,j + O(∆x2) ∂x2 ∆x2 ∂x 2∆x 3. Backward difference ∂T = 1 (3Ti,j −4Ti−1,j +Ti−2,j)+O(∆x2) ∂x 2∆x Page 9 of 23 Galerkin’sMethod: −x2−x1 Z x2 ̃ ?T1 T2? = dxNi(x) T 2 − T 1 − dx x1 dx dx; i = 1,2 CHE 3167 Finite Element Equations: Z Ni(x)[L{u}−f]dD=0 i=1,2,….,m N1=x2−x ; N2=x−x1 x2 − x1 −dN1 = dN2 = D x2 − x1 1 x2 − x1 x2 − x1 2 , i = 1, 2. T ̃ = N1(x)T1 + N2(x)T2 dx dx x1 Z x2 “d2T ̃ # dx2 Ni(x)dx x1 Z x 2 d 2 T ̃ x1 dx2 N1dx = − dx Z x 2 d 2 T ̃ d T ̃ x1 dx2N2dx= dx Z x2 Ni dx = “dT ̃ #x2 Z x2 dT ̃dNi(x) d T ̃ + x2 − x1 T 2 − T 1 x1 x1 TN1dx= 3+6 (x2−x1) x1 Z x2 ̃ ?T1 T2? TN2dx= 6+3 (x2−x1) x1 x2 Page 10 of 23 CHE 3167 Table 1: Constitutive laws for diffusive transport. TRANSPORT MECHANISM Law Relationship Flux Driving Force Proportionality Coefficient MOMENTUM Newton’s Law of Viscosity τ =−μ?∇u+∇uT? = −ρν ?∇u + ∇uT ? Shear stress τ (momentum flux) Velocity gradient ∇u Viscosity μ ENERGY Fourier’s Law q=−k∇T = −α∇ρcpT Energy or heat flux q Temperature gradient ∇T Thermal conductivity k MASS Fick’s Law ji =−Dij∇ρxi Mass or molar flux ji Concentration gradient ∇ρxi Diffusion coefficient Dij Page 11 of 23 CHE 3167 Table 2: Newton’s Law of Viscosity: τ = −μ?∇u+∇uT?. Note that ∇u = ∂uj/∂xi is the velocity gradient and ∇uT = ∂ui/∂xj is the transpose of the velocity gradient. Cartesian coordinates (x, y, z) τxx = −2μ ∂ux ∂x τyy = −2μ ∂uy ∂y τzz = −2μ ∂uz ∂z ∂u ∂u! τxy=τyx=−μ y+ x ∂u ∂u! τyz=τzy=−μ z+ y ∂u ∂u! τzx=τxz=−μ x+ z ∂z ∂x Cylindrical coordinates (r, θ, z) τrr = −2μ ∂ur ∂r 1∂u u! τθθ=−2μ θ+r τzz = −2μ ∂uz ∂z ∂x ∂y ∂y ∂z r∂θ r τrθ = τθr = −μ r τθz=τzθ=−μ θ + r ∂rr r∂θ 1∂u ∂u! z+ θ r∂θ ∂z “∂?u? 1∂u# ∂u ∂u! τzr=τrz=−μ r+ z Spherical coordinates (r, θ, φ) τrr = −2μ ∂ur ∂r ∂z ∂r 1∂u u! τθθ=−2μ θ+r r∂θ r τφφ = −2μ 1 ∂uφ + ur + uθ cot θ ! rsinθ ∂φ r “∂?u? 1∂u# τrθ = τθr = −μ r θ + r “sinθ∂?u? 1∂u# τθφ = τφθ = −μ φ + θ r ∂θ sinθ rsinθ ∂φ τφr=τrφ=−μ 1 ∂ur +r∂ ?uφ?# rsinθ∂φ ∂r r ∂rr r∂θ Page 12 of 23 CHE 3167 Table 3: Fourier’s Law: q = −k∇T Cartesian coordinates (x, y, z) qx = −k ∂T ∂x qy = −k ∂T ∂y qz = −k ∂T ∂z Cylindrical coordinates (r, θ, z) qr = −k ∂T ∂r qθ=−k ∂T r ∂θ qz = −k ∂T ∂z Spherical coordinates (r, θ, φ) qr = −k ∂T ∂r qθ=−k ∂T r ∂θ qφ=− k ∂T rsinθ ∂φ Page 13 of 23 Table 4: Fick’s Law: ji = −ρDij ∇yi Cartesian coordinates (x, y, z) = −ρD ∂yi ij ∂x = −ρD ∂yi ij ∂y j =−ρD ∂yi ir ij ∂r jiθ = − ρDij ∂yi r ∂θ j =−ρD ∂yi iz ij ∂z Spherical coordinates (r, θ, φ) j = −ρD ∂yi CHE 3167 j ix j iy j ∂yi ij ∂z = −ρD Cylindrical coordinates (r, θ, z) iz ir ij ∂r jiθ = − ρDij ∂yi r ∂θ jiφ = − ρDij ∂yi rsinθ ∂φ Page 14 of 23 Table 5: Analogy between momentum, energy and species molecular, convective and combined transport. CHE 3167 TRANSPORT MOMENTUM MOLECULAR TRANSPORT ENERGY Temperature Temperature gradient Fourier’s Law Heat flux q Temperature T Conductivity k Thermal diffusivity α Heat source Se Constant k Molecular work flux vector [σ · u] Molecular heat flux vector q Convective energy flux vector ρeu Combined energy flux vector E = q + [σ · u] +ρeu MASS Concentration Concentration gradient Fick’s Law Solute flux ji Concentration Ci Diffusivity Dij Mass diffusivity Dij Chemical reaction Ri Constant Dij Molecular mass/ molar flux vector ji or Ji Convective mass/ molar flux vector ρiu or Ciu∗ Combined mass/ molar flux vector ni = ji + ρiu or Ni = Ji + Ciu∗ Quantity Driving Force Constitutive relationship First integration gives Second integration gives Transport coefficient Transport coefficient (in terms of diffusivity) Source term Assumption Viscous/work flux tensor/vector Diffusive flux tensor/vector Velocity Velocity gradient Newton’s Law of Viscosity Shear stress τ Velocity u Viscosity μ Kinematic viscosity ν Pressure gradient ∆p/L Constant μ Viscous momentum flux tensor τ Molecular momentum flux tensorσ=pI+τ CONVECTIVE TRANSPORT Convective flux Convective tensor/vector momentum flux tensor ρuu COMBINED TRANSPORT Combined flux tensor/vector Combined momentum flux tensor π = σ + ρuu Table 6: Continuity equation: ∂ρ/∂t + ∇ · ρu Cartesian coordinates (x, y, z) ∂ρ + ∂ (ρux)+ ∂ (ρuy)+ ∂ (ρuz)=0 ∂t∂x ∂y ∂z Cylindrical coordinates (r, θ, z) ∂ρ + 1 ∂ (ρrur)+ 1 ∂ (ρuθ)+ ∂ (ρuz)=0 Spherical coordinates (r, θ, φ) ∂tr∂r r∂θ ∂z ∂ρ + 1 ∂ (ρr2ur)+ 1 ∂ (ρuθsinθ)+ 1 ∂ (ρuφ)=0 ∂t r2 ∂r rsinθ ∂θ rsinθ ∂φ Page 15 of 23 Table 7: Navier-Stokes equation: ρDu/Dt = −∇p − [∇ · τ ] + ρg Cartesian coordinates (x, y, z) ρ ρ ρ ρ CHE 3167 ∂u ∂u ∂u ∂u! ∂p ∂ ∂ ∂ ! x+u x+u x+u x =− − τ+ τ+ τ +ρg ∂t x ∂x y ∂y z ∂z ∂x ∂x xx ∂y yx ∂z zx x ∂u ∂u ∂u ∂u! ∂p ∂ ∂ ∂ ! y+u y+u y+u y =− − τ+ τ+ τ +ρg ∂t x ∂x y ∂y z ∂z ∂y ∂x xy ∂y yy ∂z zy y ∂u ∂u ∂u ∂u! ∂p ∂ ∂ ∂ ! z+u z+u z+u z =− − τ+ τ+ τ +ρg ∂t x ∂x y ∂y z ∂z ∂z Cylindrical coordinates (r, θ, z) ∂x xz ∂p +ρgr ∂y yz ∂z zz z ∂ur ∂ur uθ ∂ur ∂ur u2 ! ∂t +ur ∂r + r ∂θ +uz ∂z − rθ r∂r r∂θ ∂z r = − ∂r 1∂ 1∂∂τ! − (rτrr)+ τθr + τzr − θθ ∂u ∂u u∂u ∂u uu! 1∂p ρθ+uθ+θθ+uθ+rθ=− ∂tr∂rr∂θz∂zr r∂θ “1∂ 1∂ ∂ τ−τ# − 2 (r2τrθ)+ τθθ + τzθ + θr rθ +ρgθ ∂u ∂u u∂u u ∂u u2+u2! ∂p r∂r r∂θ ∂z r ∂u ∂u u∂u ∂u! ∂p ρz+uz+θz+uz=− ∂t r ∂r r ∂θ z ∂z ∂z “1∂1∂∂# − r ∂r (rτrz)+ r ∂θτθz+ ∂zτzz +ρgz Spherical coordinates (r, θ, φ) ρr+ur+θr+φr−θφ=− ∂t r ∂r r ∂θ rsinθ ∂φ r ∂r “1∂1∂1∂τ+τ# − 2 (r2τrr)+ (τθr sinθ)+ r ∂r rsinθ ∂θ ∂uθ ∂uθ uθ ∂uθ uφ ∂uθ ρ ∂t+ur∂r+r∂θ+rsinθ∂φ+ θθ φφ +ρgr r τφr − uruθ−u2φcotθ! “1 ∂ 3 1 ∂ − r3 ∂r (r τrθ)+ rsinθ ∂θ (τθθ sinθ)+ rsinθ ∂φ τφθ + ∂uφ ∂uφ uθ ∂uφ uφ ∂uφ uφur +uθuφ cotθ ! ∂t +ur ∂r + r ∂θ + rsinθ ∂φ + r rsinθ ∂φ r 1 ∂p =−r∂θ 1 ∂ (τθr−τrθ)−τφφcotθ# r +ρgθ ρ 1 ∂p =−rsinθ ∂φ ” 1 ∂ 3 1 ∂ 1 ∂ − r3 ∂r (r τrφ)+ rsinθ ∂θ (τθφ sinθ)+ rsinθ ∂φ τφφ + (τφr −τrφ)+τφθ cotθ # r +ρgφ Page 16 of 23 ∂t x∂x y∂y z∂z ∂x ∂x2 ∂y2 ∂z2 x ∂u ∂u ∂u ∂u! ∂p ∂2u ∂2u ∂2u! y+u y+u y+u y =− +μ y+ y+ y +ρg CHE 3167 Table 8: Navier-Stokes equation: ρDu/Dt = −∇p + μ∇2u + ρg Cartesian coordinates (x, y, z) ρ ρ ρ ρ ∂u ∂u ∂u ∂u! ∂p ∂2u ∂2u ∂2u! x+u x+u x+u x =− +μ x+ x+ x +ρg ∂t x∂x y∂y z∂z ∂y ∂x2 ∂y2 ∂z2 y ∂u ∂u ∂u ∂u! ∂p ∂2u ∂2u ∂2u! z+u z+u z+u z =− +μ z+ z+ z +ρg ∂t x∂x y∂y z∂z ∂z ∂x2 ∂y2 ∂z2 Cylindrical coordinates (r, θ, z) z ∂ur ∂ur uθ ∂ur ∂ur u2! ∂t +ur ∂r + r ∂θ +uz ∂z − rθ ∂p =−∂r (∂”1∂ # 1∂2u ∂2u 2∂u) +μ (rur) + r + r − θ +ρgr ∂r r ∂r r2 ∂θ2 ∂z2 r2 ∂θ
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