Otto-von-Guerice University Magdeburg PROCESS SYSTEMS ENGINEERING Dr.-Ing. Richard Hane-Rauschenbach Project wor No. 1, Winter term 2011/2012 Sample Solution Delivery of the project handout: Wednesday, 16.11. 2011 Due date for the solved project: Wednesday, 30.11. 2011 1:15 p.m. Important 1. The project must be written in computer, or must be perfectly readable if it is prepared by hand. 2. The names of all the members of the team must be written in the project. The teams may have a maximum of three members. 3. The project must be delivered as hard copy, with all the sheets numbered, bound together, e.g. stapled i.e. no spare sheets will be accepted, and perforated in the left side for their storage in a ring binder. 4. No projects will be accepted after the due date and time. 5. Try to answer the questions as short and concrete as possible. 6. Indicate clearly the assumptions made for the derivations of the equations. Cross out the terms that you neglected, and clearly specify the assumption considered for neglecting such terms. 1 General Aspects 1.1 Tensor Notation 1.1.1 Write down completely i.e. expand the following equation! 2 P ϱ = ϱ v + j, + σ, = 1, 2, 3 1 - expanding for =1,2,3 results in ϱ = z 1 ϱ v 1 + j 1, z 2 ϱ v 2 + j 2, z 3 ϱ v 3 + j 3, + σ 2 1.1.2 Write down completely i.e. expand the following equation! 3 P ϱv j = ϱv z j v + P j + ϱ f j,, j, = 1, 2, 3 3 - considering first =1,2,3 results in ϱv j = z 1 ϱvj v 1 + P j1 z 2 ϱvj v 2 + P j2 ϱvj v 3 + P z j3 + 3 ϱ f j, 4 1
- or considering first j=1,2,3 results in ϱv 1 ϱv 2 ϱv 3 = ϱv z 1 v + P 1 + ϱ f 1, 5 = ϱv 2 v + P 2 + ϱ f 2, 6 = ϱv 3 v + P 3 + ϱ f 3, 7 - then, expanding for j=1,2,3, =1,2,3, respectively ϱv 1 ϱv 2 ϱv 3 = z 1 ϱv 1 v 1 + P 11 = z 1 ϱv 2 v 1 + P 21 = z 1 ϱv 3 v 1 + P 31 z 2 ϱv 1 v 2 + P 12 z 2 ϱv 2 v 2 + P 22 z 2 ϱv 3 v 2 + P 32 ϱv z 1 v 3 + P 13 + 3 ϱv 2 v 3 + P 23 + z 3 z 3 ϱv 3 v 3 + P 33 + ϱ f 1, 8 ϱ f 2, 9 ϱ f 3, 10 1.1.3 Extra tas. Show, that the following expression is valid! 2.5 P extra p v j z j = p v j δ j, j, = 1, 2, 3 11 - Expanding the right side for =1,2,3 results in p v j vj = p δ z j z j1 + v j δ 1 z j2 + v j δ 2 z j3 3 12 - Expanding the right side of the results also for j=1,2,3 results in p v j v1 = p δ z j z 11 + v 1 δ 1 z 12 + v 1 δ 2 z 13 13 3 + v 1 δ z 21 + v 2 δ 22 + v 2 δ 23 14 2 z 2 z 3 + v 1 z 3 δ 31 + v 3 z 2 δ 32 + v 3 z 3 δ 33 15 - with the definiton of δ j δ j = { 1, if j = 0, if j = 16 - follows 1 0 0 δ = 0 1 0 17 0 0 1 - and the equation simplifies to p v j v1 = p + v 2 + v 3 z j z 1 z 2 z 3 18 2
- application of Einstein summation convention gives p v j z j = p v j z j 19 1.2 Given the state variables internal energy U, chemical potential µ, density ϱ, temperature T and mass m! Which of the quantities are intensive, which extensive variables? Justify your answer! 2.5 P - intensive, changes not with system size: temperature T,chemical potential µ,density ϱ - extensive, changes with system size: mass m, internal energy U Remar: Specific internal energy u is an intensive variable 1.3 Name the three basic components of a process model and give an example each! 1.5 P - balance equations: e.g. mass balance or energy balance - state equations of thermodynamics: e.g. thermal or caloric equation of state - inetic equations constituative equations: e.g. empirical transport inetics as Fic s law or Fourier s equation of heat conduction 2 Balances 2.1 The thermodynamic-mechanical state of a pure fluid is completely characterized by its density field ϱt, z, its velocity field v j t, z and its internal energy field ut, z. From which balances can these fields be determined? 1.5 P - density field: mass balance - velocity field: momentum balance - internal energy field: energy balance 2.2 Material balances 2.2.1 Write down the component mass balance in local formulation and explain the meaning of each single term. 3 P - component mass balance in local formulation for component reads ϱ = ϱ v + j, + σ 20 - meaning of the terms left hand side: accumulation of first on right hand side: convective mass transport of second on right hand side: diffusive mass transport of third on right hand side: sin/source for 2.2.2 Give the definitions for the mass average velocity v and the diffusive flux j,! Show that holds! 2 P j, = 0 21 3
- the mass average velocity, v, is defined as follows ϱv = N ϱ v, 22 =1 - the diffusion flux with reference to v is defined as j, = ϱ v, v 23 - summation of all the diffusive fluxes gives N j, = =1 N =1 ρ v, v N =1 ρ }{{} ϱ 24 - with Eq. 22, it finally follows N j, = 0 25 =1 2.2.3 Derive from the result of 2.2.1 a differential equation for the concentration c! 3 P - starting point is ϱ = ϱ v + j, + σ 26 - with ϱ = M c and nowing that M = const, we get M c = M c v j, + σ 27 - and finally c = c v + j, + σ, = 1, 2, 3 28 M M 2.2.4 Derive from the result of 2.2.3 a differential equation for the total pressure p, defined as p = c RT 29 Assume that the temperature T is constant. 3 P - starting point is c = c v + j, + σ 30 M M - summing up over all components, we have c = - considering sum rule in differentiation, we find c = j c v +, M c v + j, σ + M 31 M + σ M 32 4
- using equation 29 and assuming T = const we get p = j pv +, RT + RT M σ M, = 1, 2, 3 33 2.2.5 Extra tas. Derive from the result of 2.2.3 a differential equation for the total pressure p. Assume that T is not constant! 3 P extra - starting point is Eq. 32 c = c v + - using Eq. 29 we get p = p RT RT v + j, M j, M + σ M 34 + σ M 35 - with constant R and expanding the left hand side and the first term on right hand side we get 1 p RT + p R 1 = 1 T RT pv pv R 1 j, T + M σ M 36 - with multiplying by RT we get p + pt 1 = pv T pv T - finally, with application of chain rule we get [ p p T T = pv + pv T RT T 2.3 Given a single-phase tubular reactor, in which the reactions [ 1 RT T j, M j, M σ M ] σ M ], = 1, 2, 3 A B C 39 tae place in the presence of a solvent. 2.3.1 How many state variables are required in order to describe the thermodynamic-mechanical state of the system? From which relations can those variables be determined? 3 P - Gibbs phase rule F = N + 2 P gives the degree of freedom, i.e. how many intensive states can be chosen independently of each other - here: F=5 with N = 4 number of components and P = 1 number of phases, means 5 intensive state variables o e.g. and temperature T energy balance o e.g. 4 =N component densities ϱ from 4 component mass balances or o e.g. and temperature T energy balance 37 38 5
o e.g. 3 =N-1 component densities ϱ from 3 component mass balances o e.g. total density ϱ from continuity equation - in addition an extensive variable is needed o e.g. reactor length L - the mechanical state has to be characterized o mass average velocity v momentum balance o j, constituative relations 2.3.2 Assume, the system can be described with the following model. c = v c z + 1 M σ, {A, B, C} 40 Which assumptions have been incorporated into the model formulation? 2 P - mass transport is dominated by convection diffusion can be neglected - constant mass average velocity, no momentum equation needed - constant molar mass - isothermal, as there is no energy balance - 1D model, neglecting z 2 and z 3 direction 2.3.3 How many boundary and initial conditions are required to solve Eq. 40? Justify your answer and give appropriate boundary and initial conditions! 3 P - Three PDE s with first order in time have to be considered. Since for each order in time and each equation one IC is needed, three times one IC s have to be defined: e.g. c t = 0, z = c 0 z - Three PDE s with first order in space have to be considered. Since for each order in space and each equation one IC is needed, three times one BC s have to be defined: e.g. c t, z = 0 = c in t 2.4 Given the momentum balance in local formulation. ϱv j = ϱv z j v + P j + ϱ f j,, j, = 1, 2, 3 41 2.4.1 Comment on the meaning of the individual terms! Do not forget, that P j is composed of two terms! Give an example for f j,! 3.5 P - on the left hand side: accumulation of momentum - first on the right: convective transport of momentum - second on the right: surface forces: with P j as pressure tensor, consisting of viscous tension π j and hydrostatic pressure pδ j - third on the right: volume forces, with f j, e.g. Gravitation field, Centrifugal field, Electrostatic field 2.4.2 Modify the equation given above for the case of a friction free fluid! 1 P - for a friction free fluid, that means no viscous tension and π j = 0, we get ϱv j = ϱv z j v + pδ j + ϱ f j,, j, = 1, 2, 3 42 2.5 Given the balance of the internal energy in local formulation. ϱu = ϱuv + q v j + f, j, P j, j, = 1, 2, 3 43 6
Convert the equation into its substantial formulation! 1.5 P - By rearranging the left hand side ϱ u + u ϱ = ϱuv + q v j + f, j, P j, j, = 1, 2, 3 44 - the substantial formulation of the internal energy balance is obtained under usage of the continuity equation, that reads as: ϱ = ϱv 45 - inserting Eq. 45 into Eq. 44 gives: ϱ u + u ϱv = ϱuv + q v j + f, j, P j, j, = 1, 2, 3 46 - combining the left hand side with first term on the right results in: ϱ u ϱv u = q + - from applying operator equation to u, it follows: f, j, P j v j, j, = 1, 2, 3 47 Du Dt = u + v u 48 - by inserting u = Du Dt v u 49 - on the left hand side we finally get ϱ Du Dt = q v j + f, j, P j, j, = 1, 2, 3 50 3 Constitutive Equations 3.1 What is the purpose of state equations? Give an example for a state equation, name and equation! 2 P - state equations connect the density fields with the thermodynamic state fields - an example is ideal gas law, which is a thermal state equation p = ϱ w M RT 51 3.2 Given is the following reaction scheme: A r 1 B A + B r 2 C 2C r 3 D 52 53 54 3.2.1 Give the general relation between the source density for a component, σ, and the reaction rates r m of an arbitrary reaction networ! 1 P 7
- σ is defined as follows σ = M R m=1 ν m, r m 55 3.2.2 Write down the source terms for the species A, B and C. 3 P - we get σ A = M A r 1 + r 2 σ B = M B r 1 r 2 σ C = M C r 2 2r 3 3.2.3 Consider power law and elementary reactions. Write down the inetic equation for reaction rates r 2 and r 3. 3 P - the power law reads - we get 3.3 Transport inetics N r = + c m N,+ c m, 56 =1 =1 r 2 = 2 c A c B 57 r 3 = 3+ c 2 C 3 c D 58 3.3.1 Write down in words the general structure of empirical transport inetics. 1 P - flux functiondriving force, or more precisely flux = -inetic coefficient x gradient of driving force 3.3.2 Specify the general formulation given in question 3.3.1 for the case of diffusive mass transport. What are the units of each of the variables appearing in this formulation? 2 P - we get - the units are j, = ϱd w 59 [j, ] = g/m 2 /s [ϱ] = g/m 3 [D ] = m 2 /s [w ] = 1 [z ] = m 3.4 Thermodynamics of irreversible processes TIP 3.4.1 Explain the relevance of TIP in the context of process inetics! 1 P - TIP is an macroscopic theory that maes universal statements about transport coefficients independently of the structure of the matter. 8
- it provides generally valid qualitative relations between flux entities and their conjugated thermodynamic driving forces. 3.4.2 According to TIP basic effects and cross effects can be distinguished. Give an example for each of them, naming the flux and the conjugated driving force, respectively! 2 P - basic effects, e.g. Diffusion, diffusion flux j, with chemical potential gradient µ / as driving force - cross effects, e.g. Thermal diffusion Soret effect, diffusion flux j, with temperature gradient T/ as driving force 3.4.3 In the framewor of TIP, explain why a heat flux cannot be driven by the deformation velocity v j /! 1 P - According to Curies symmetry principle only tensors with the same transformation characteristics can be lined to each other; v j / gives a second order tensor driving force, which cannot drive the vectorial flux q 4 Specific Problem Consider a long and flat meat sewer consisting of two parts see Fig. 1. z 1 =0 z 1 =i z 1 =L z 2 =H z 2 z 2 =0 T e T f z 3 z 1 Figure 1: Meat sewer consisting of two materials wood w and metal part m The material of the left part is wood. The right part is metallic. The top of the metallic part is held into a campfire with a homogeneous temperature T f. It can be assumed, that the part of the sewer that touches the fire also reaches T f dashed part. The environmental temperature field T e is continuously measured and can be assumed as nown at all distances from the fire. The outer shell of the wooden part is perfectly insulated from the environment in all directions no heat transfer occurs. 4.1 Neither the material nor the momentum balance is needed for the description of the thermodynamicmechanical state of the described system. Justify this statement for each named balance! 2 P - No material balance is needed, since only one component is considered and no changes in density and velocity appear - No momentum balance is needed, since no velocity has to be considered static system and therefore no friction and no volume forces appear 4.2 Given the general differential equation for the internal energy field in local formulation ϱu = ϱuv + q v j + f, j, P j, j, = 1, 2, 3 60 4.2.1 Discuss the meaning of the single terms appearing in Eq. 60 and consider which terms can be neglected for the given problem! Therefore discuss explicitly the influence of friction remind the meaning of P j, volume forces, heat convection and heat conduction remind the 9
meaning of q! Simplify the equation related to your considerations! Justify your answers! 4 P - term on the left: accumulation of internal energy simplification that is possible in the considered system: constant density ϱ = const - first term on right hand side: convective transport of internal energy simplification that is possible in the considered system: convection term will vanish, since velocity is zero, no movement - second term on right hand side: energy flux vector, consisting of heat flux q and enthalpy diffusion flux j, h simplification that is possible in the considered system: enthalpy diffusion flux will vanish since only one component considered no diffusion appears - third term on right hand side: occurs with multicomponent fluids where different forces acts upon the single components simplification that is possible in the considered system: will vanish since only one component considered no diffusion appears - last term on the right hand side: represents the conversion of one part of the performance of surface forces into internal energy through compression or friction, with P j, consisting of viscous tensionπ j and hydrostatic pressure pδ j simplification that is possible in the considered system: since no friction will appear, π j v j is equal to zero, and: since no velocity gradient will appear, pδ j v j will disappear - Including all considerations, it remains: ϱ u = q, = 1, 2, 3 61 4.2.2 Convert the result of tas 4.2.1 into a general differential equation for enthalpy h! 1.5 P - Using the definition of enthalpy h = u + p ϱ 62 - we get with the result from 4.2.1 ϱ h p = q, ϱ = 1, 2, 3 63 - since we now that pressure p and density ϱ are constant for the given problem, we get ϱ h = q, = 1, 2, 3 64 4.2.3 Convert the result of tas 4.2.2 into a general differential equation for temperature T! Hint: Use the total differential of enthalpy h! 3 P - the total differential of the enthalpy h reads Dh Dt = h DT T p,w =const Dt + h Dp h p T,w =const Dt + Dw w T,p=const Dt 65 - the second term at the right hand side can be neglected, because of constant pressure assumption 10
- the third term at the right hand side can be neglected, because the composition does not change one component only - due to the definition h c p = 66 T we stay with p,w =const Dh Dt = c DT p Dt - application of the operator equation 67 D Dt = + v 68 - for h Dh Dt = h + v h 69 - and T DT Dt = T + v T 70 - results in h + v h T = c p + v T 71 - since the velocity is equal to zero it remains h = c T p - inserting that into result of 4.2.2, we finally get 72 ϱc p T = q, = 1, 2, 3 73 4.2.4 Assume that changes in direction of the z 3 -coordinate can be neglected due to the flatness of the sewer. Insert appropriate transport inetics for q 1 and q 2 into the result of tas 4.2.3. Write down two energy balances, one for the wooden and one for the metallic part of the sewer. Keep in mind the different material properties of wood w and metal m. 2.5 P - Fourier s law of heat conduction reads: q = λ T, j, = 1, 2 74 - for the wooden part w with the assumtion of temperature independent coefficient λ w ϱcp w T w = λ w 2 T w z 2 1 + λ w 2 T w z 2 2 75 - for the metallic part m with the assumtion of temperature independent coefficient λ m ϱcp m T m = λ m 2 T m z 2 1 + λ m 2 T m z 2 2 76 11
4.2.5 Finally, for the metallic part results: ϱcp m T m = λ m 2 T m z 2 1 + 2 T m λ m z 2. 77 2 How many initial and boundary conditions are required for the solution of this partial differential equation PDE? Justify your answer! 2.5 P - Eq. 77 is a PDE with first order in t. Since for each order in time and each equation one IC is needed, one IC has to be defined here - Eq. 77 is a PDE with second order in z 1 and second order in z 2. Since each coordinate has to be separately accounted for and for each order in space two BC s are needed, two BC s for z 1 and two BC s for z 2, means four BC s have to be defined 4.2.6 Give appropriate initial and boundary conditions for Eq. 77! Keep in mind the information given above! Justify your answers! 3 P - suitable initial condition initial temperature profile T m z 1, z 2, t =0 = T m 0 z 1, z 2 78 - suitable boundary conditions for z 1 = L temperature fixed at the end to temperature of fire T m z 1 = L, z 2, t = T f t, z 2 79 - suitable boundary conditions for z 1 = i, specified heat flux at the boundary to wooden part λ m Tm = λ w Tw 80 z 1 z 1 - or z1 =i,z 2,t z1 =i,z 2,t T m z 1 = i, z 2, t = T w z 1 = i, z 2, t 81 - suitable boundary conditions for z 2, specified heat transfer at the upper and lower boundary with environment nown temperature T e λ m Tm z = β T m z 1, z 2 = H, t T e z 1, z 2 = H, t 82 2 z1,z 2 =H,t λ m Tm z = β T m z 1, z 2 =0, t T e z 1, z 2 =0, t 83 2 z1,z 2 =0,t 4.2.7 Consider the differential equation for the temperature of the wooden part of the sewer derived in tas 4.2.4. Does the required number of initial and boundary conditions change compared to the metallic part? Give appropriate boundary conditions for the wooden part of the sewer! Keep in mind the information given above! Justify your answers! 2.5 P - The required number of initial and boundary conditions does not change, same orders in t and z - suitable boundary conditions for z 1 =0, due to insulation λ w Tw z 1 = 0 z 2, t 84 z1 =0,z 2,t 12
- suitable boundary conditions for z 1 = i, specified heat flux at the boundary to metallic part see above λ m Tm = λ w Tw 85 z 1 z 1 z1 =i,z 2,t z1 =i,z 2,t - suitable boundary conditions for z 2, insulation at the upper and lower boundary λ w Tw z = 0 z 1, t 86 2 z1,z 2 =H,t λ w Tw z = 0 z 1, t 87 2 z1,z 2 =0,t 4.2.8 How are both energy balances coupled with each other? 1 P - the equation are coupled due to the flux about the same boundary by the boundary condition at z 1 = i λ m Tm z 2 z1 =i,z 2,t = λ w Tw z 2 z1 =i,z 2,t or 88 T m z 1 = i, z 2, t = T w z 1 = i, z 2, t 89 13