Lecture2_ICE_2024_2025.pdf course year 10
- 1. Teaching by: Vassily Hatzimanikatis (vassily.hatzimanikatis@epfl.ch) Fridays, 14h15 - 17h00 2024-2025 Assistants: Denis Joly (denis.joly@epfl.ch) Konrad Lagoda (konrad.lagoda@epfl.ch) Zi Xuan Ng (zixuan.ng@epfl.ch) Introduction to Chemical Engineering Office hours: Mondays 16h-19h (CH H4 625) or schedule by email
- 2. Example 3: Distillation In a distillation process working at steady state, 2,000 kg/h of a binary mixture of benzene (B) and toluene (T) containing 70% B by mass is separated. There are two output streams in this process: a top stream, which carries 1100kg B/h, and a bottom stream, which contains 450kg T/h. What is the flow rate and composition of all streams in this process? 0) Understand the problem! Any new concept? Type of process? 1) Translate the problem into flowchart: draw the operation unit and the streams 2) Label the flowchart: total flow rates and individual flow rates 3) Define basis: time AND mass units 3) Basis: kg/h F1,B = 1400 kg B/h F1,T = 600 kg T/h F2 ? F2,B = 1100 kg B/h F2,T (kg T/h) ? F3 ? F3,T = 450 kg T/h F3,B (kg B/h) ? distillation F1? 1)
- 3. Example 3: Distillation 4) Mass balances: How many equations can we write? • One for each molecular species • One for each element • One for total mass Total : 3 equations F1,B = 1400 kg B/h F1,T = 600 kg T/h F2 ? F2,B = 1100 kg B/h F2,T (kg T/h) ? F3 ? F3,T = 450 kg T/h F3,B (kg B/h) ? distillation F1? How should we modify the mother of all equations to describe our process? (consider the process classification) General: Input + generation – output – consumption = accumulation Specific to our problem? Input – output = 0
- 4. Example 3: Distillation 1. For benzene: 1400 – (1100 + F3,B) = 0 (steady state/st. st.) → F3,B = 300 kg/h 2. For toluene: 600 – (450 + F2,T) = 0 (st. st.) → F2,T = 150 kg/h 3. Total mass balance : In – out = 0 2000 – (1100 + 150 +450 + 300) = 0 F1,B = 1400 kg B/h F1,T = 600 kg T/h F2 ? F2,B = 1100 kg B/h F2,T (kg T/h) ? F3 ? F3,T = 450 kg T/h F3,B (kg B/h) ? distillation F1? Which mass balances shall we choose? Any combination if they are linearly independent We have defined 3 mass balances, but … How many mass balances do we need to solve the problem? Clues: In this problem, what is the… Number of molecular species? Number of unknowns? Number of linearly independent equations?
- 5. REMINDER: A linear independent system of equations Two equations are linearly dependent, if: • one can be obtained from the other by scaling it by a factor • both equations produce identical graphs • both equations provide the same information Is the following system of equations linearly independent? 2x + 2y = 4 x + y = 2 x – y = 0 And the following one? x + y = 5 2x + y = 7
- 6. Example 4: Mixing We are working with two different binary mixtures of methanol/water. The first one contains 50 %-kg/kg (weight) methanol and the second one 80 %-kg/kg (weight) methanol. We mix 200 g of the first one and 150 g of the second one. What is the composition of final mixture? 0) Understand the problem! Any new concept? Type of process? 1) Translate the problem into flowchart: draw the process 2) Label the flowchart: mass and compositions 3) Define basis: time? mass units? 3)Basis: total time t0 → tf 50% M 200 g 80% M 150 g tf t0 ? 6
- 7. Example 4: Mixing 4) Mass balances General: Input + generation – output – consumption = accumulation Specific to our problem: ? Input = Accumulation Methanol: (0,5 . 200 + 0,8 . 150) = Mmethanol 100 + 120 = 220 g methanol Water: (0,5 . 200 + 0,2 . 150) = Mwater 100 + 30 = 130 g water Total: 220 + 130 = 350 g 50% M 200 g 80% M 150 g tf t0 ? Composition of final mixture: methanol: 220/350 = 63% kg/kg, water: 130/350 = 37% kg/kg 7
- 8. Course Schedule Recommended textbook: Elementary Principles of Chemical Processes Richard M. Felder & Ronald W. Rousseau Date Subject 13-Sep 1. Fundamentals of Material Balances 1.1. Process definition and classification 1.2. Material balance calculations 20-Sep 1.3. Balances on multiple-unit processes 1.4. Chemical reaction stoichiometry 27-Sep 1.5. Balances on reactive processes 04-Oct Review on Mass Balances 11-Oct 1.5. Balances on multiple unit reactive processes 18-Oct 2. Energy and Energy Balances 2.1. Energy balances on closed systems 2.2. Open systems at steady state 01-Nov 3. Balances on Non-Reactive Processes 3.1. Energy balance calculation 3.2. Changes in Pressure, Temperature, Phases 3.3. Mixing and Solution 08-Nov 4. Balances on Non-Reactive Processes Problems: Mass and Energy Balances on non-Reactive Systems 15-Nov Midterm Exam: Mass & Energy Balances non-Reactive Systems 22-Nov Review Midterm 29-Nov 5. Balances on Reactive Processes 5.1. Heats of reaction/combustion 5.2. Combustion reactions 5.3. Enthalpy of reaction 5.4. Energy balance calculation 06-Dec 6. Energy balances on mixing processes Review 13-Dec Review and Study Session
- 9. Session II: Friday 20 September 2024 After studying this session, you will be able to: 1. Perform Degree of Freedom analysis on a unit operation 2. Perform a Material Balance on a single unit process (following the general procedure) 3. Perform Mass Balances on multiple-unit processes 9
- 10. 1. Degree of freedom 10
- 11. What is degree of freedom? In chemical engineering, the number of independent equations that can be obtained from a non-reactive system is equal to the number of unique molecular species in that system E.g.: (from Session I) Number of equations: ➢ One for each molecular specie : B, T ➢ One for total mass: Total = B+T Are they independent? Only 2 (B and T) or (B and Total) or (T and Total) • Not all problems are solvable: problems can have too much information (unsolvable) or give not enough data (infinite solutions) • A degree-of-freedom analysis is used to determine whether a problem is solvable with the information given F1? F1,B = 1400 kg B/h F1,T = 600 kg T/h F2 ? F2,B = 1100 kg B/h F2,T (kg T/h) ? F3 ? F3,T = 450 kg T/h F3,B (kg B/h) ? 11
- 12. How to perform a degree-of-freedom analysis? 1. Draw a properly labeled flowchart: what we have and what we don’t have to define the mass balance for all the species. Each stream must have information about its total flow rate and its composition (this is equivalent to having the flow rate of each species in the stream) 2. Count the number of unknowns: what information is missing ? 3. Count the number of independent equations: For any system, we can write exactly as many independent material balances as unique chemical species we have. If we write a balance equation for total mass, we should omit one of the chemical species, because a balance equation for total mass is not independent and can be obtained by summing all the equations for unique chemical species 12
- 13. How to perform a degree-of-freedom analysis? 4. Calculate the Degrees of freedom (DOF) DOF = Number of unknowns – Number of independent equations ➢ If the DOF = zero, the problem is solvable, one solution (determined) ➢ If the DOF > zero, the problem has infinite solutions (underdetermined) ➢ If the DOF < zero, the problem is unsolvable (we have too many constraints) (???) 13
- 14. We have two different mixtures of methanol/water 1: 50 % W methanol 2: 60 % W methanol and we want a final solution 56% W of methanol? General: Input + generation – output – consumption = accumulation Specific to our problem: ? Input = Accumulation Example 1: Mixing 14 50% M 60% M tf t0 56% M 50% M 50% W 60% M 40% W tf t0 56% M 44% W S1 S2 Sf 1. Draw the flow chart… and label it!!!! The mass balance equations are well defined with the information from the flowchart
- 15. We have two different mixtures of methanol/water 1: 50 % W methanol 2: 60 % W methanol and we want a final solution 56% W of methanol? Input = Accumulation Mass balance equations: Methanol: 0,5 . S1 + 0,6 . S2 + 0 – 0 = Sf . Xf = Sf . 0,56 Water: 0,5 . S1 + 0,4 . S2 = Sf . 0,44 Total: S1 + S2 = Sf DOF analysis: 2. Number of unknowns: 3 (S1, S2 , Sf ) 3. Number of independent equations: 2 4. DOF = 3 – 2 = 1 50% M 60% M tf t0 56% M S1 S2 Sf Example 1: Mixing 15 The problem is missing required information, it has infinitely many solutions (we can’t solve it*) !
- 16. (we can’t solve it*) ? → We can define one more unknown: If we define that we want 370 g of the final 56% W solution… Mass balance equations: Methanol: 0,5 . S1 + 0,6 . S2 = 370 . 0,56 Water: 0,5 . S1 + 0,4 . S2 = 370 . 0,44 Total: S1 + S2 = 370 … we can calculate the mass of the stream 1 and 2: S1 ≈ 148g and S2 ≈ 222g • In this case, the mass of the final solution has been defined as the calculation basis 50% M 50% W 60% M 40% W tf t0 56% M 44% W S1 S2 Sf = 370 g Example 1: Mixing, continued… 16
- 17. General procedure for Single-Unit Process Material Balance Calculations (in non-reactive processes) 17
- 18. 0) Check given units. Make sure that you understand the system. 1) Draw flowchart and label: • Stream flow (rate): volumetric/mass (rate) • Stream composition: mass/mol fraction for each component • Identify the units (to be converted if they don’t match) • Identify information given or needed for each stream • Identify possible given relations between streams 2) Choose a basis: Unit, time or numerical convenience (m = 100 kg/min, t = 1 h…) → problem specific 3) Reformulate the problem as list of known/unknowns and check units (Remember not to apply a volumetric balance!) Problem set up How to solve mass balance questions 18
- 19. 4) Formulate equations. Can be : • Material balances: • Molecular species • Elemental balances • Process specifications: • Relation between streams • Distribution of species • Physical constrains: • Mole/mass fractions • Is it not a liquid? Is it a gas stream? • Physical properties How to solve mass balance questions 19
- 20. 5) Perform a degree-of-freedom analysis • Degrees of freedom (DOF) = Number of unknowns – Number of equations • Minimize/eliminate dependent equations • Identify equations with less unknowns • Highlight unknowns 6) Solve equations Always outline outline solution procedure 7) Calculate required/asked quantities 8) Rescale How to solve mass balance questions 20
- 21. A liquid mixture containing 45.0% benzene (B) and 55.0% toluene (T) by mass is fed to a distillation column. A product stream leaving the top of the column (the overhead product) contains 95.0 mole % B, and a bottom product stream contains 8.0 % of the benzene fed to the column (meaning that 92% of the benzene leaves with the overhead product). The volumetric flow rate of the feed stream is 2000 L/h and the density of the feed mixture is 0.872. Determine: • the mass flow rate of the overhead product stream ? • the mass flow rate and composition (mass fractions) of the bottom product stream? 0) Check the units mass/volume/mol??? 1) Draw and label the flowchart V1 = 2000 L/h ṁ1 (kg/h) ṁ2 (kg/h) 0.95mol B/mol 0.05mol T/mol 0.45 kg B/kg 0.55 kg T/kg ṁ3 (kg/h) ṁB3, ṁT3 . Example 2: Material Balance on a Distillation Column 21
- 22. 2) Choose a basis: Time= 1 hour 3) Reformulate the problem as a list of known/unknowns : Knowns: Unknowns: 4) Formulate equations: Example 2: Material Balance on a Distillation Column 22
- 23. 5) DOF analysis • 4 unknowns (ṁ1, ṁ2, ṁB3, ṁT3) Mass balances (for both species) One relation for the density (physical relation) Derived one relation for benzene (8% down, 92% up) (process specs) 6) Set up and solve equations The volumetric flow rate for the conversion: ṁ1 = (2000 L / h) × (0.872 kg / L) → ṁ1 = 1744 kg / h Relationship information for benzene: ṁB3 = 0.08 × (0.45 × ṁ1) → ṁB3 = 62.8 kg of B / h 4 equations DOF=0 Example 2: Material Balance on a Distillation Column 23
- 24. Benzene balance : 0.45 × ṁ1 = ṁ2 × yB2 + ṁB3 → ṁ2 ≈ 766 kg of benzene / h Another way to find ṁ2 : a bottom product stream contains 8.0 % of the benzene fed to the column (meaning that 92% of the benzene leaves with the overhead product). 0.92 × 0.45 × ṁ1 = ṁ2 × yB3 → ṁ2 ≈ 766 kg of benzene / h Toluene balance : 0.55 × ṁ1 = 0.058 × ṁ2 + ṁT3 → ṁT3 = 915 kg of toluene / h Total mass balance to verify the calculations ? The additional calculations: ṁ3 = ṁB3 + ṁT3 = 62.8 kg / h + 915 kg / h ≈ 978 kg / h yB3 = ṁB3 / ṁ3 = 62.8 kg de B / (978 kg / h) = 0.064 kg of B / kg yT3 = 1- yB3 = 0.936 kg T / kg Example 2: Material Balance on a Distillation Column 24
- 25. 3. Mass balance on multiple-unit processes 25
- 26. Multiple-unit processes • In the vast majority of real chemical processes, there will be: • more than one raw material (feed) entering the system AND • more than one unit operation through which the product must pass in order to achieve the desired result • Examples: • mixing reactants before going to reactor • blending products • separating products • recycling • Universal icons to show the process units on the paper 26
- 27. How to do mass balance for multiple-unit systems? Multiple-unit processes are tougher to solve, but it is doable! 1) Label a flowchart completely with all the relevant unknowns 2) Perform a degree of freedom analysis on each unit operation 3) Find and extract with dashed line a unit operation or combination of unit operations (subsystems) for which the degrees of freedom is zero 4) Calculate all of the unknowns involved in this combination 5) Recalculate the degrees of freedom for each process, treating the calculated values as known rather than as variables 6) Repeat these steps until everything is calculated 27
- 28. Mass balance on multiple units E B C D A F2 F3 Prod. 1 F1 Prod. 3 Prod. 2 How many systems do we have? We have 5 systems: ▪ The overall system : A ▪ Mixing and splitting points are also systems: F1 and F2 are mixing (point B) and D is splitting (point D) ▪ The internal units: C, E We have to define the mass balances for each system and start with the unit with DOF=0 28
- 29. Example 3: Mass balance on 2 unit operations • Units • Flowchart • How many systems? S3 S2 S4 S1 P2=30 kg/h 0.6 kg A/kg 0.4 kg B/kg P1=40 kg/h 0.9 kg A/kg 0.1 kg B/kg 0.5 kg A/kg 0.5 kg B/kg m3 kg/h X3 A % 1-X3 B % U1 U2 m2 kg/h X2 A % Y2 m1 kg/h X1 A % Y1 F2=30 kg/h 0.3 kg A/kg 0.7 kg B/kg 29 Basis : 100kg/h feed
- 30. Example 3: Mass balance on 2 unit operations • S1 (global system): 2 mass balances // 2 unknowns (m3, x3) • S2 (U1): 2 mass balances // 2 unknowns (m1, x1) • S3 (U2): 2 mass balances // 4 unknowns (m2, x2 , m3, x3) • S4 (System mixing): 2 mass balances // 4 unknowns (m1, m2, x1, x2) S3 S2 S4 S1 P2=30 kg/h 0.6 kg A/kg 0.4 kg B/kg P1=40 kg/h 0.9 kg A/kg 0.1 kg B/kg 0.5 kg A/kg 0.5 kg B/kg m3 kg/h X3 A % 1-X3 B % U1 U2 m2 kg/h X2 A % Y2 m1 kg/h X1 A % Y1 F2=30 kg/h 0.3 kg A/kg 0.7 kg B/kg 30 Basis : 100 kg/h
- 31. Example 3: Mass balance on 2 unit operations Basis : 100kg feed Mass balance on S1 (global system): S1 : In – Out = 0 → 100 + 30 – 40 – 30 – m3 = 0 → m3 = 60 kg / h S1 (specie A): 50 – 36 + 9 – 18 – 60x3 = 0 → x3 = 0.0834, y3 = 0.9166 → m3,A = 5 kg A / h, m3,B = 55 kg B / h Mass balance on S2 (U1): S2 (U1): 100 – 40 – m1 = 0 → m1 = 60 kg / h S2 (specie A): 50 – 36 – 60x1 = 0 → x1 = 0.23, y1 = 0.77 → m1,A = 13.8 kg A / h, m1,B = 46.2 kg B / h 31 Unknowns left to find : m1,x1,m2, x2 Unknowns left to find : m2, x2
- 32. Example 3: Mass balance on 2 unit operations Mass balance on S3 (U2): S3 (U2): m2 – (30 + 60) = 0 → m2 = 90 kg / h S3 (specie A): 90x2 – 18 – 5 = 0 → x2 = 0.256, y2 = 0.744 → m2,A = 23.04 kg A / h, m2,B = 66.96 Kg B / h 32 Mass balance on S4 (Mixing): S4 : m1 + F2 – m2 = 0 S4 (specie A) : m1 x1 + 0.3 F2 – m2 x2 = 0 S4 (specie B) : m1 y1 + 0.7 F2 – m2 y2 = 0 We have found all the unknowns without having to use S4 -> not all units have to be used. But if we wanted to write the mass balances anyway:
- 33. Example 3: Mass balance on 2 unit operations S3 S2 S4 S1 P2=30 kg/h 0.6 kg A/kg 0.4 kg B/kg P1=40 kg/h 0.9 kg A/kg 0.1 kg B/kg 0.5 kg A/kg 0.5 kg B/kg m3 kg/h X3 A % 1-X3 B % U1 U2 m2 kg/h X2 A % Y2 m1 kg/h X1 A % Y1 F2=30 kg/h 0.3 kg A/kg 0.7 kg B/kg 33 100 kg/h DONE ☺ 60 kg/h 0.23 kg A/kg 0.77 kg B/kg 90 kg/h 0.256 kg A/kg 0.744 kg B/kg 60 kg/h 0.0834 kg A/kg 0.9166 kg B/kg