CHE412 Process Dynamics and Control BSc (Engg) Chemical Engineering (7th Semester) Week 3 (contd.) Mathematical Modeling (Contd.) Luyben (1996) Chapter 3 Stephanopoulos (1984) Chapter 5 Dr Waheed Afzal Associate Professor of Chemical Engineering Institute of Chemical Engineering and Technology University of the Punjab, Lahore wa.icet@pu.edu.pk 1 Modeling CSTRs in Series constant holdup, isothermal Basis and Assumptions A → B (first order reaction) Compositions are molar and flow rates are volumetric Constant V, ρ, T Overall Mass Balance πρπ ππ‘ = ρπΉ0 − ρπΉ1 = 0 i.e. at constant V, F3 =F2 =F1 =F0 ≡ F So overall mass balance is not required! F0 V1 K1 T1 Luyben (1996) F1 CA1 V2 K2 T2 F2 CA2 V3 K3 T3 F3 CA3 2 Modeling CSTRs in Series constant holdup, isothermal Component A mass balance on each tank (A is chosen arbitrarily) ππΆπ΄1 πΉ = πΆπ΄0 − πΆπ΄1 − π1πΆπ΄1 ππ‘ π1 ππΆπ΄2 πΉ = πΆπ΄1 − πΆπ΄2 − π2πΆπ΄2 ππ‘ π2 ππΆπ΄3 πΉ = πΆπ΄2 − πΆπ΄3 − π3πΆπ΄3 ππ‘ π3 kn depends upon temperature kn = k0 e-E/RTn where n = 1, 2, 3 Apply degree of freedom analysis! Parameters/ Constants (to be known): V1, V2, V3, k1, k2, k3 Specified variables (or forcing functions): F and CA0 (known but not constant) . Unknown variables are 3 (CA1, CA2, CA3) for 3 ODEs Simplify the above ODEs for constant V, T and putting τ = V/F 3 Modeling CSTRs in Series constant holdup, isothermal If throughput F, temperature T and holdup V are same in all tanks, then for τ = V/F (note its dimension is time) ππΆπ΄1 1 1 + πΆπ΄1 π + = πΆπ΄0 ππ‘ τ τ ππΆπ΄2 1 1 + πΆπ΄2 π + = πΆπ΄1 ππ‘ τ τ ππΆπ΄3 1 1 + πΆπ΄3 π + = πΆπ΄2 ππ‘ τ τ In this way, only forcing function (variable to be specified) is CA0. 4 Modeling CSTRs in Series Variable Holdups, nth order Mass Balances (Reactor 1) ππ1 = πΉ0 − πΉ1 ππ‘ π(π1πΆπ΄1 ) ππ‘ = πΉ0πΆπ΄0 − πΉ1πΆπ΄1 −π1π1(πΆπ΄1)n Mass Balances (Reactor 2) ππ2 = πΉ1 − πΉ2 ππ‘ π(π2πΆπ΄2) ππ‘ = πΉ1πΆπ΄1 − πΉ2πΆπ΄2 −π2π2(πΆπ΄2)n Mass Balances (Reactor 1) ππ3 = πΉ2 − πΉ3 ππ‘ π(π3πΆπ΄3) ππ‘ Changes from previous case: V of reactors (and F) varies with time, reaction is nth order Parameters to be known: k1, k2, k3, n Disturbances to be specified: CA0, F0 Unknown variables: CA1, CA2, CA3, V1, V2, V3, F1, F2, F3 CV Include MV Controller eqns V1 (or h1) F1 F1 = f(V1) = πΉ2πΆπ΄2 − πΉ3πΆπ΄3 −π3π3(πΆπ΄3)n V2 (or h2) F2 F2 = f(V2) V3 (or h3) F3 F3 = f(V3) 5 Modeling a Mixing Process Basis and Assumptions F (volumetric), CA (molar); Well Stirred Stephanopoulos (1984) Feed (1, 2) consists of components A and B Enthalpy of mixing is significant Process includes heating/ cooling H H ρ is constant 2 1 Overall Mass Balance π(ππ΄β) = π1πΉ1 + π2πΉ2 − ππ‘ π(β) π΄ = (πΉ1 + πΉ2 ) − πΉ3 ππ‘ Q π3πΉ3 in or out Component Mass Balance π(ππ΄ π) π΄ ππ‘ = (πΉ1 ππ΄1 + πΉ2 ππ΄2) − πΉ3 ππ΄3 H3 6 Modeling a Mixing Process Conservation of energy (recall first law of thermodynamics) βπΈ = βπ + βππ + βππ ± π + ππ π€ − πβπ£ H2 H1 βπ ≅ βπ» (for constant ρ/ liquid systems πβπ£ is zero) Energy Balance H3 enthalpy balance (h is energy/mass) π(ππππ) = π(πΉ1 ππ + πΉ2 ππ) − ππΉ3 ππ ± π ππ‘ We were familiar with energy ππΆπ βπ; how to characterize h (specific enthalpy) into familiar quantities (T, CA, parameters, …) H is enthalpy, h is specific enthalpy; CP is heat capacity, cP is specific heat capacity …. 7 Modeling a Mixing Process π(ππππ) = π(πΉ1 ππ + πΉ2 ππ) − ππΉ3 ππ ± π ππ‘ Since enthalpy depends upon temperature so lets replace h with h(T) β1 π1 = ππ π0 + ππ1 π1 − π0 β2 π2 = ππ(π0) + ππ2 π2 − π0 β3 π3 = ππ(π0) + ππ3 π3 − π0 enthalpy associated with ΔT was easy to obtain, how to obtain h(T0) πππ π0 = ππ΄1π»π΄ + ππ΅1π»π΅ + βπ»π1(π0) πππ π0 = ππ΄2π»π΄ + ππ΅2π»π΅ + βπ»π2(π0) πππ π0 = ππ΄3π»π΄ + ππ΅3π»π΅ + βπ»π3(π0) π»π΄ and π»π΅ are molar enthalpy of component A and B and βπ»ππ is heat of solution for stream i at T0. Put values of h in overall energy balance 8 Modeling a Mixing Process Re-arranging (and using component mass balance equations) ππ3 πππ3 π = ππ΄1πΉ1 βπ» π 1 − βπ» π 3 + ππ΄2πΉ2 βπ» π 2 − βπ» π 3 ππ‘ +ππΉ1 ππ1 π1 − π0 − ππ3 π3 − π0 + ππΉ2[ππ2 π2 − π0 − ππ3 π3 − π0 ] ± π If we assume cP1 = cP2 = cP3 = cP ππ3 πππ π = ππ΄1πΉ1 βπ» π 1 − βπ» π 3 + ππ΄2πΉ2 βπ» π 2 − βπ» π 3 ππ‘ +ππππΉ1(π1 − π3) + cpππΉ2(π2 − π3) ± π ο§ If heats of solutions are strong functions of concentrations then βπ» π 1 − βπ» π 3 and βπ» π 2 − βπ» π 3 are significant ο§ Mixing process is generally kept isothermal (how?) 9 Tips For Assessment (Exam) Introduction + Modeling (week 1-3) In exam, you may be asked short descriptive questions to check your understanding of process control and to prepare a mathematical model for a chemical process or processes and to make the system exactly specified (i.e. Nf = 0) 1. Consult your class notes, board proofs, discussions 2. Stephanopoulos (1984) chapters 1-5, examples and end-chapter problems 3. Luyben (1996) chapter 3 page 40 to 74. Practice examples and end-chapter problems for chapter 3. 10 Week 3 Weekly Take-Home Assignment 1. Follow all the example modeling exercises in Luyben (1996) chapter 3 page 40 to 74. Practice these example processes. 2. Solve at least 10 end-chapter problems from Luyben (1996) chapter 3 (Compulsory) Submit before Friday (Feb 7) Curriculum and handouts are posted at: http://faculty.waheed-afzal1.pu.edu.pk/ 11