SEPARATION PROCESSES 2018

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1 SEPARATION PROCESSES

2 Contact details: Pavel Hasal Building B, 1 st floor, room: B143 Pavel.Hasal@vscht.cz, tel

3 AIM OF THE COURSE: i. The aim of the course is to teach students principles of chosen separation processes most frequently used in chemical and similar technologies. The arrangements of separation processes will be generally described and then their mathematical description will be presented. ii. iii. iv. The students would acquire knowledge necessary for both verbal and mathematical description of selected separation processes. Students will know principles of mass and energy transport phenomema, phase equilibria and material and energy balancing of separation processes. The students will be able to describe principles of separation processes, will know their applications and will be able to set-up their mathematical models. The students will know how to solve numerically the problems concerning separation processes and to design basic dimensions of a separation equipment. 3

4 COURSE OUTLINE: 1) Introduction to separation processes. Material and energy balances. 2) Basic numerical methods in chemical engineering 3) Basics of fluid mechanics 4) Heat and mass transfer fundamentals 5) Separation processes rate and equilibria 6) Adsorption processes 7) Membrane separation processes 8) Ionic exchange and electrophoretic separation 9) Membrane process design 10) Gas permeation, reverse osmosis 11) Ultrafiltration, dialysis 12) Pervaporation 13) Crystallization 4

5 ASSESSMENT CRITERIA Homework: solution of individual problems (max 100 points). Final test in week 14 (the last one): a simple problem solution (max. 100 points). Repetitions possible, only the best result is considered. Oral examination max. 100 pts. Weights: Individual problems 1/3 Final test 1/3 Oral exam 1/3 Points (total) Grade A B C D E F 5

6 LITERATURE: (SHR) J. D. Seader, E. J. Henley, D. K. Roper: SEPARATION PROCESS PRINCIPLES. Chemical and Biochemical Operations. (Third Ed.), John Wiley & Sons, Inc., (ISBN ) 6

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8 On May 3 rd there will be Tuesday schedule! No Separation Processes! 8

9 Material and energy balances in chemical engineering applications 9

10 Outline Balances principles and nomenclature Mass and molar balances without chemical reactions How to solve balance problems recommended procedure Energy balances Examples 10

11 System Part of universe with real or fictitious boundaries that can communicate with surroundings systems (pot, membrane module, chemical reactor, chemical plant, bank account ) Systems o o o o Open mass and energy exchange Closed energy exchange Isolated no exchange Adiabatic no heat exchange Systems o Homogeneous single phase o Heterogeneous two or more phases Systems o Steady state no change in time o Dynamical state properties of the system change in time System communicates with surroundings by streams o Streams = flows of mass, energy, volume, cash or other extensive quantities 11

12 Chemical engineering systems - examples a) Open complex system with fictitious boundaries b) Fictitious differential cube a part of chemical reactor c) Chemical reactor. The system boundaries coincide with the reactor walls real boundaries. 12

13 Nodes parts of complex systems Process node Mixer Bypass Reflux Complex system consisting of 5 nodes 13

14 Processes taking place in systems Mechanical o Grinding, crushing, milling, sieving, sorting... Hydromechanical o Gas or liquid flow, filtration, stirring, sedimentation, fluidization... Heat transfer o Heat exchangers, evaporators, cooling towers... Mass transfer o Distillation, extraction, rectification, membrane processes, drying, adsorption, absorption... Chemical processes (reactions) o Batch, PFR, CSTR, and other reactors with chemical or biochemical processes Biological processes o Systems containing living organisms 14

15 Physical quantities Extensive quantities (EQ) o EQ can be summed over parts of a system o EQ express amount of some quantity o EQ may be balanced examples: mass, molar amount, volume, energy, enthalpy, momentum, money Intensive quantities (IQ) o IQ cannot be summed over parts of a system o IQ describe physical properties of systems o IQ cannot be balanced examples: temperature, pressure, velocity, radioactivity, color... 15

16 Composition of a mass within a system To be balanced, all components have to be identified in the system The contents of components are expressed as concentrations or fractions Mass (ρ i ), molar (c i )or volume (φ i ) concentrations m i n i V i V - mass of a component in the system - molar amount of a component in the system - volume of a component in the system - total volume of the system 16

17 Composition of mixtures Mass (w i ) and mole (x i ) fractions Relative mass (W i ) and relative (X i ) mole fractions o are used in processes with inert components (amount of inert component does not change within the system) Each identified component in the system can be balanced we can formulate n linearly independent balances for n components This is not valid in simple dividers and mixers (compositions of all inlets and outlets are identical) m total mass of the system n - total molar amount in the system m i - mass of an inert component n i - molar amount of an inert component 17

18 Principles of balancing Inputs + Sources = Outputs + Accumulation salary rente Bank account flat etc. shopping Inputs BQ that comes into the system through boundaries Outputs - BQ that leaves the system through boundaries Sources internal formation or extinction of BQ o o Positive interests, formation of a component by chemical reaction Negative transaction fees, decay of a chemical component Accumulation the difference between final and initial amounts of BQ within the system 18

19 Balancing Inputs + Fictitious Inputs = Outputs + Fictitious Outputs initial value interests salary rente Bank account rent shopping final value transaction fees Sources and accumulation can be treated as fictitious inputs or outputs o o o Positive sources = fictitious inputs Negative sources = fictitious outputs Accumulation Initial amount of a quantity in the system fictitious input Final amount of a quantity in the system fictitious output 19

20 Material balances One node Four streams Stream #2 Three chemical components: A, B, C Stream #1 Process node Stream #4 Stream #3 Mass balance of a component m A1 + m A2 = m A3 +m A4 Can be rewritten by means of mass fractions Sum of all mass fractions in a stream is always 1 w A1 + w B1 + w C1 = 1 w A2 + w B2 + w C2 = 1 w A3 + w B3 + w C3 = 1 w A4 + w B4 + w C4 = 1 m 1 w A1 + m 2 w A2 = m 3 w A3 + m 4 w A4 m 1 w B1 + m 2 w B2 = m 3 w B3 + m 4 w B4 m 1 w C1 + m 2 w C2 = m 3 w C3 + m 4 w C4 m 1 + m 2 = m 3 + m 4 Four balances obtained three of them are linearly independent 20

21 Material balances recommended procedure 1) Draw flowsheet nodes, streams. 2) Identify all (chemical) components in the system to be balanced. 3) Choose an extensive quantity to be balanced (BQ) mass, molar amount, volume, 4) Evaluate/identify compositions of streams (whenever possible). 5) Compose an input matrix. 6) When no extensive quantity is given (e.g., mass of a stream or mass of a component in a stream), then an extensive quantity is chosen/defined for one chosen stream. 7) Determine number of unknowns in the input matrix = Degrees of Freedom Analysis (DOF). 8) Set-up balance equations. 9) Set-up auxiliary equations, number of independent equations is equal to the number of unknowns (DOF). 10) Solve the set of equations = algebraic problém. 21

22 DoF Analysis (Degrees of Freedom) SHR: useful tool to assess solvability of a balance problem assessment of number of unknowns to be evaluated vs. number of equations and data N N N DoF vars eqns number of Degrees of Freedom = number of data that have be specified number of all variables describing the problem number of all independent equations Ex.: A single-phase stream with C components. N DoF = C = C + 3 concentrations T P amount 22

23 Xylene, styrene, toluene and benzene are to be separated with the array of distillation columns shown in figure (right), where F, D, B, D 1, B 1, D 2 and B 2 are the molar flow rates (mol/min). Concentrations are given in molar %. Evaluate all unknown component concentrations and fluxes. There is: 7 streams 7 (4+1) = 35 variables 6 unknown flow rates (B, D, B 1, D 1, B 2, D 2 ) 8 unknown concentrations (streams B and D) ===================================== 14 unknowns = 21 d.o.f. we have to provide 21 values 21 values specified in the schema 23

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29 Example: Natural gas composed of CH 4 and C 2 H 6 is combusted with excess amount of the air. The amounts of H 2 O and CO 2 in combustion gasses are in molar ratio 9:1. After removing of these two components the composition of resulting gas is: 8.3 mol.% of CO, 6.5 mol.% of 0 2 and 85,2 mol.% of N 2. Calculate: a) composition of the natural gas (in mol. %), b) amount of the combustion gasses evolved by combustion of 1 mole of the natural gas, c) what was the per cent excess of the oxygen? Hint: Consider these three reactions taking place within the burner: 2CH 4 + 3O 2 2CO + 4H 2 O 2C 2 H 6 + 5O 2 4CO + 6H 2 O 2CO + O 2 2CO 2 Results: a) Composition of natural gas: 85.5 mol.% of CH 4 and 14.5 mol.% of C 2 H 6, b) 13,3 mole of combustion gasses per 1 mole of natural gas, c) Excess of the oxygen: 11.7 %

30 natural gas CO 2 + H 2 O 3 2 air BURNER CO + O 2 + N CH 4 C 2 H 6 O 2 N 2 CO CO 2 H 2 O A B C D E F G Reaction # Imaginary streams 2 CH O 2 2 CO + 4 H 2 O (1) 5,6 2 C 2 H O 2 4 CO + 6 H 2 O (2) 7,8 2CO + O 2 2CO 2 (3) 9,10

31 #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 n [mol] 1????????? x A CH 4? / x B C 2 H 6? / x C O / / /3 x D N x E CO / / /3 x F CO ? x G H 2 O ? --- 4/ / There are 13 unknowns we need 13 equations We can formulate: 2x sum of molar fractions for streams #1 and #3 3x ratios of molar amounts of imaginary streams (given by reaction stoichiometries) 7x balance equations 1x ratio of molar fractions in stream #3: x G3 /x F3 = 9 The problem is solvable as there is 13 equations for 13 unknowns. NDoF Nvars Neqns

32 RESULTS #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 n [mol] x A CH / x B C 2 H / x C O / / /3 x D N x E CO / / /3 x F CO x G H 2 O / / Ratio of amount of combustion gases to amount of natural gas: (n 3 +n 4 )/n 1 = Excess of the oxygen: excess=(x C2 n 2 -(x C6 n 6 + x C8 n 8 + x C10 n 10 ))/(x C6 n 6 + x C8 n 8 + x C10 n 10 ) = 0.40 oxygen entering with stream #2 oxygen necessary to burn al methane and all ethane

33 Problem #1: Mass balance of a crystallizer 1000 kg/h of 10 w% sodium carbonate solution in water is mixed with a recycle stream. The obtained mixture is continuously introduced into an evaporator, where 30 w% sodium carbonate solution is produced. This solution is then led into a crystallizer where crystals of sodium carbonate dodecahydrate are produced together with 20 w% sodium carbonate solution. Certain part of this solution is recycled. The other part is withdrawn from the system. Yield of the process is 90 %, i.e. 90 % of sodium carbonate introduced into the system is obtained in the form of crystals. Determine mass flow rate of the reflux stream. 33

34 Problem #2: Pyrite calcination 1000 kg of pyrite composed of 85 mass % of FeS 2 and 15 mass % of rubbish is calcined with 100 % excess of the air. The calcination is described by chemical reaction 4 FeS O 2 = 2 Fe 2 O SO 2 The rubbish does not react with the oxygen and does not release any gasses.the solid material (waste) after calcination contains 2 mass % of FeS 2. Calculate: a) concentration of Fe in solid waste, b) volume of air (at 20 o C and MPa) required for the calcination, c) volume of the gas evolved by calcination (at 300 o C and MPa) and its composition in mole fractions. Results: a) the waste contains 54.9 mass % of Fe, b) 4610 m 3 of air, c) volume of evolved gasses is 8760 m 3 and its composition : 7.7 mol.% SO 2, 11.0 mol.% O 2 and 81.3 mol.% N 2. 34

35 HINTS: 5 2 Fe 2 O SO 2 1 pyrite gaseous products 3 2 air solid products FeS O 2 Components: A... pyrite B... rubbish C... O 2 D... N 2 E... Fe 2 O 3 F... SO 2 35

36 Problem #3: Sulphur is combusted with stoichiometric amount of air in a furnace and sulphur dioxide is produced. SO 2 is then partly oxidized with the excess amount of dry air to SO 3 in the first catalytic chamber. The sulphur trioxide is completely removed from the gas phase in the first absorber. Resulting gas containing 3.3 mol.% of SO 2 and 8.3 mol.% of O 2 is conveyed to the second catalytic chamber and the second absorber. In the second chamber all SO 2 is oxidized to SO 3. All SO 3 is removed from the gas in the second absorber. Calculate: a) what fraction of the sulphur entering the process was converted to SO 3 in the first chamber, b) b) what is the excess of oxygen used for sulphur combustion, c) c) composition of the gas phase leaving the second absorber (volumetric per cent). Results: a) 70.5 % of total sulphur, b) b) excess of oxygen is 39.5 %, c) c) gas phase consists of 7 vol.% of O 2 and 93 vol.% of N 2 36

37 HINTS: 1 2 furnace 3 catalytic 5 chamber I absorber I 7 catalytic 9 chamber II absorber II S + O2 SO2 2 SO2 + O2 2 SO3 2 SO2 + O2 2 SO3 Components: A... sulphur, B... O 2, C...N 2, D... SO 2, E... SO 3 37

38 Computer (algorithmic) solution of material balances example in Mathematica 38

39 ENTHALPY BALANCING 39

40 INTERNAL ENERGY The first law of thermodynamics o U internal energy o Q heat o W work dq System du dw du dq dw dq pdv The internal energy represents energy of the system that can be transformed into heat or work. Absolute/total value of internal energy of systems is unknown. We can, however, determine its changes. Change of the internal energy (du) is equal to the heat (dq) and work (dw) delivered into system 40

41 ENTHALPY There are volume, electric, magnetic and other works (W) that can be delivered to or produced by the system Volume work is given by a volume change - compression or expansion We can define a new extensive quantity enthalpy Total differential of enthalpy reads After substitution from the first law of thermodynamics, we obtain In isobaric systems (dp = 0), change of the enthalpy is equal to heat delivered to or removed from the system 41

42 ENTHALPY BALANCING Enthalpy of Material Inputs Enthalpy of Heat + = Material + Sources Outputs Enthalpy Accumulation within System Material Stream #2 Heat loss Q l Material Stream #1 Heat source Q s Process node - H Material Stream #4 Accumulation Material Stream #3 H 1 + H 2 + Q s Q l = H 3 + H 4 + H s 42

43 Specific/molar enthalpy Enthalpy of material stream H (heat contained in material) can be expressed as a product of mass/molar amount (m/n) and specific enthalpy/molar enthalpy (h m /h n ) H = mh m H = nh n h m or h n are given in [J/kg] or [J/mol] In the following slides, we will make use of single symbol h for both specific and molar enthalpies (h m /h n ) (the meaning of h will result from the context) 43

44 Enthalpy of mixtures Specific enthalpy of a mixture is given by mass contributions of all chemical components forming the mixture: h i is specific enthalpy of the component i h mix is specific enthalpy of mixing (values can be found in tables for particular mixtures) h mix =0 for an ideal mixture h = w i h i + h mix i 44

45 Enthalpy of a chemical component There is no absolute value of enthalpy Enthalpy value is always related to the reference state Reference state can be chosen arbitrarily We evaluate/study an increase or a decrease of enthalpy with respect to the reference state The enthalpy increase or decrease are equal to the amount of heat consumed by the system or released from the system Change of the enthalpy = How much heat I have to deliver to the system or to remove from the system 45

46 Reference state The reference state: enthalpy of a chemical component equals zero at the reference state The reference state is defined by stating: o o Temperature of the reference state State of matter (g,l,s) for each chemical component Proper selection of the reference state may significantly simplify the solutions of the problems: o The reference temperature would be chosen as a temperature in the studied system, e. g., temperature of a stream, o o The state of matter of a chemical component would be the same as one state of the component in the system, Availability of physical and chemical data has to be taken into account. 46

47 Enthalpy as function of temperature Change of specific/molar enthalpy with temperature is called specific/molar heat capacity Specific/molar heat capacity tells us how much heat has to be supplied in order to warm one kilo/mole of the component by 1 Kelvin Specific/molar heat capacity depends on temperature Change of enthalpy can be evaluated as the integral of a polynomial For engineering estimation of enthalpy, specific/molar heat capacity is considered to be a constant evaluated at the mean temperature T 47

48 Enthalpy as a function of the state of matter Change of state of matter is always accompanied with a large change of enthalpy value This enthalpy change depends on temperature and can usually be found in tables etc. The most common changes of state of the matter are: o o o h i,phase change = h i,new phase h i,old phase = h i,inverse phase change Vaporization condensation Sublimation desublimation Melting - freezing h i, vap = h i,g h i,l = h i,cond h i, sub = h i,g h i,s = h i,desub h i, melt = h i,g h i,s = h i,freez 48

49 Combined dependencies of enthalpy on temperature and state of matter (vaporization as an example) enthalpy is state variable Point A reference state, a liquid at temperature T ref Point B vapor at temperature T The total enthalpy change between A and B does not depend on the route from the point A to the point B. Solid line route boiling boil boil boil boil boil Dashed line route vap Dash-dotted line route vap 49

50 Problem #4: Enthalpy balance of a process mixer 10 kg of water at 25 C has been mixed with 3 kg of ice at 0 C with 5 kg of water at 100 C. What is temperature of the obtained mixture? Is there some ice in the resulting mixture? Consider specific heat capacity of water to be approximately constant within the relevant temperature interval. 50

51 Problem #4: Cooling a pipe by external flow A warm liquid is flowing through a horizontal pipe with a flow rate F (kg/s). The interal diameter of the pipe is D and its length is L (cf. the figure). A cooling fluid is flowing around the tube in a perpendicular direction. The cooling fluid has constant temperature T w. Warm liquid enters the tube with the temperature T ini. The density and heat capacity of the warm liquid are and c p, respectively. The overall heat transfer coefficient between the warm liquid and the cooling fluid is U (W/(m 2 K)). Find the temperature of the warm liquid as function of time and position within the tube, i.e., T(z, ). Suppose plug-flow within the pipe and neglect heat conduction resistances in radial direction. 51

52 INTRODUCTION TO SEPARATIONS 52

53 o Mixture: a basic form of an existence of matter. o Mixtures may form spontaneously but usually an energy input is needed and speeds-up the mixture formation. A SEPARATION of a mixture to individual components or less complex mixtures requires an energy input! Mixture C component i 1 i 53

54 Separation of a mixture to components (or to less complex mixtures) represents basic task in chemical and similar technologies (analytical separations?) PRODUCT 1 FEED MIXTURE SEPARATION PROCESS PRODUCT 2 PRODUCT N-1 PRODUCT N 54

55 MIXTURES AND THEIR PROPERTIES factors affecting separations Single phase mixtures multiphase mixtures Homogeneous mixtures heterogeneous mixtures Molecular, thermodynamic and transport properties Size and shape of particles 55

56 Classification of separation methods (a) phase creation (b) phase addition (c) solid barriers (d) solid agents (e) force field or gradient 56

57 (a,b) phase creation or phase addition When the feed to be separated is a single-phase then second separable phase must be developed before separation of the components can be achieved. Creation of the second phase? Energy separating agent (ESA) or a mass-separating agent (MSA) ESA heat transfer to/from mixture or shaft work to-from mixture MSA partially or fully immiscible with one or more constituents of the mixture 57

58 Full list of separations based on phase creation or addition gives Table 1.1 in Seader-Henley-Roper text-book: 58

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the barrier (filtration processes) or due to their different rates of permeation through the

61 (c) solid barriers Macro-, micro- and nano-porous barriers are used for separation according to mixture constituents properties Components are separated either by their selective retention by the barrier (filtration processes) or due to their different rates of permeation through the barrier (solutiondiffusion mechanism) Liquid membranes a modification of separations by solid barriers 61

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63 (d) solid agents A particulate/granular solid agent is mixed (in a proper way) with a gas, vapour or liquid mixture. Mixture constituents selectively bind to the solid agent. Phase equilibrium may be achieved (contact time vs. mass-transfer rate). Solid agent may be covered by a thin layer of selective agent (inert carrier + agent) or the solid agent itself binds the components. Adsorption chromatography ion exchange. 63

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65 (e) force field or gradient External fields are used to separate mixture constituents due to their various responses to applied field: 65

66 Typically, a sequence (a combination) of separation methods has to be used to achieve required separation and/or to meet legal regulations etc. Heuristics: 68

67 Selection of feasible separation method(s) Factors influencing selection of separation operation(s): A) Feed conditions 1. Composition, particularly of species to be recovered 2. Flow rate 3. Temperature 4. Pressure 5. Phase state (s, l or g) B) Product conditions 1. Required purities 2. Temperatures 3. Pressures 4. Phases 70

68 c) Property differences that may be exploited 1. Molecular 2. Thermodynamic 3. Transport D) Characteristics of of separation operation 1. Ease of scale-up 2. Ease of staging 3. Temperature, pressure and phase-state requirements 4. Physical size limitations 5. Energy requirements E) Economics 1. Capital costs 2. Operating costs 71

General Separation Techniques

ecture 2. Basic Separation Concepts (1) [Ch. 1] General Separation Techniques - Separation by phase creation - Separation by phase addition - Separation by barrier - Separation by solid agent - Separation