Some physico-chemical data can be found at the web page (E-Tables):

Transcription

1 Reminiscences 1

2 Physical data have been supplied to Problem_#4. Some physico-chemical data can be found at the web page (E-Tables): A lot of data can be found at: 2

3 THREE-STAGE PLANT 50 MMscfd 30% CO psia 14.6% CO m m MMscfd 7.9% CO psia 30% CO m % CO % CO % CO 2 8 streams: 8 x 3 = 24 unks. 2 amounts specified + 7 concentrations 7 concentrations can be easily evaluated so 2+14 = 16 data values are known There is 8 unknown values and we can formulate 4x2 balances + 1x sum of concentrations (stream 2) = the problem is overdetermined!!! 3

4 Batch process: In a batch process after filling the feed vessel with the starting volume, concentration and/or diafiltration (washing) starts until the final volume or desired washing is reached. Product is either in the vessel or the collected permeate. Q w, c w =0 V 0,c 0 c R V f, c f c F c P 4

5 5 ˆ f W P R f c V c V Q Q volumetric flow rate of washing solution volumetric flow rate of permeate f final state 0 initial state

6 Multicomponent gas separation: stirred to stirred system n, z nr, x F i J i M, Pi i n P, y i 6

7 P i Ji xi pr yi pp M y i Pi x i pr yi pp P x P x J / i i i pr yi pp i i p M red yi Pk Sik xi pred yi nc nc nc nc n P C j J x j p y p P x p y p P x p y / P S x p e y j R j P j j R j P j j red j k j1 j1 M j1 j1 j1 jk j r d j z y 1 x i 1,, nc i i i We have 2n C equations for x i and y i supposing feed composition (z i ) is given together with the cut () value and permeabilities or selectivities (P i or S ik ). 7

8 8

9 9

10 ULTRAFILTRATION (UF) 10

11 HISTORY 11

12 Pressure driven membrane separation process for removal of macromolecular (or colloidal) solutes ( Daltons) Membranes: asymmetric microporous with pore size (approx.) nm Separation: due to molecule size (MW) & shape vs. pore size & structure = solute rejection Applications: separation of proteins, polysaccharides, nucleic acids, thickening of solutions (dairy industry, fruit juices ), waste water treatment 12

13 Membrane rejection/retention fundamental factor affecting separation Solute rejection is a complex process (size shape)! 13

14 Membrane fouling main problem of UF separations Fouling due to small/colloidal particles present in the feed generally reversible (turbulence, backflushing ) generally irreversible (membrane blocking) 14

15 Turbulence? UF device configuration, channel size, stirring Re lu feed/retentate stream velocity Spacers/obstacles 15

16 BACK FLUSHING reversal of flow direction Due to membrane fouling the permeate flux declines with time of operation membrane cleaning is necessary BACK FLUSHINGS? 16

17 Concentration polarization and gel layer formation hydrodynamics of UF modules and from it resulting mass transfer rate is of great importance Development of concentration polarization layer and (possibly) of gel layer at the membrane feed side with operational pressure 17

18 polarization modulus 18

19 Permeate flux vs. applied transmembrane pressure membrane permeability J max k c ln c c i, gel i, bulk 19

20 MASS TRANSFER COEFFICIENT? k c Di k d D c h Sh i definition Sherwood number Re udh Reynolds number Sc D i D i Schmidt number 4S dh o dh d hydraulic diameter of retentate channel dh 2h tubes slits 20

21 Laminar flow of retentate stream: d Sh 1.62 ReSc h L Turbulent flow of retentate stream: d h 0.1 ReSc 10 L 4 Re 10 Sh 0.04 Re 3/4 Sc 1/3 4 1/3 21

22 Permeate flux in ultrafiltration? 22

23 J driving force V p R tot difference in osmotic pressures permeate flux density (m/s) R R R M tot M g PM g P g total, membrane, gel layer resistances pure water flux density (m/s) J V p R M p P M M evaluation of membrane resistance or permeability using pure water permeation data

24 Membrane area (module size)? A M V J P V mass balance (m 3 /s) kinetics (m/s) 24

25 How gel layer concentration can be estimated? Experiments, experiments. 25

26 Continuous ultrafiltration SINGLE STAGE 26

27 TWO STAGES 27

28 THREE STAGES 28

29 Batch ultrafiltration c b 29

30 R c c 1 P b module retention cb () t c b0 V0 V() t, when R 1 c () t V cb0 V() t b 0 ln Rln, when R1 30

31 DIALYSIS 31

32 Concentration difference driven membrane separation process for separation of low molecules from large molecules in a solution Membranes: symmetric, dense (dialysis tubes) Separation: due to different rates of diffusion in membrane Applications: desalting of protein solutions, polysaccharides, nucleic acids, recovery of spent acids or alkalis, drinking water 32

33 Description solution 1 membrane solution 2 c 1 c 1m m, D Partitioning of the component between solution and membrane (equilibrium): c 1m S1c1, ip c2m S2c2, ip S i distribution/partition coefficients) FLUX OF A COMPONENT c 1,ip c 2m j k c c n L1 1 1, ip c 2,ip c 2 j k c c n m 1m 2m j n j k c c n L2 2, ip 2 k m D m k m is (a formal) mass transfer coefficient in the membrane used to describe flux in membrane formally in the same way as mass transfer in liquid films (D is diffusivity) 33

34 j n S1c1 S2c2 S 1 S k k k when S 1 = S 2 =S 1 2 L1 m L2 c c c c jn K c1 c m 1 K k S k k k S D k L1 m L2 L1 L k S k k L1 m L2 k m D and k L1, k L2 are evaluated from Sherwood number m P n S total component flux: k m D S n K c c A m 1 2 m permeability of the membrane 34

35 BATCH DIALYSIS dn d d dn2 d d d Vc d Vc K c c A K c c A when V 1 =const. and V 2 =const. V, c V1, c1 2 2 dc1 A K c c d V dc2 d A K c c V

36 c 10 K A ( V1 V2) c1 ( ) V1 V2 exp V1 V2 VV 1 2 c 10 K A ( V1 V2) c2( ) V1 V1 exp V1 V2 VV c1 t, c2 t t sec. 36

37 CONTINUOUS DIALYSIS V c V1 c1, 1, 1, in n A, out V2, c2, out V2, c2, in c c c c 1, in 2, out 1, out 2, in n jna K A K c A c mean 1, in c2, out ln c c 1, out 2, in 37

38 STRIPPING SOLUTION Possible (historical) arrangement of a semi-batch dialysis DIALYSIS TUBE FEED SOLUTION PERMEATE VESSEL 38

39 LABORATORY DIALYSIS 39

40 An Arrangement of Industrial Dialysis 40

41 Acid Recovery by Diffusion Dialysis 41

42 MEDICAL DIALYSIS 42

43 1) An experiment is being conducted to determine the suitability of a cellophane membrane for using it in hemodialysis process. The membrane is mm thick. In the experiment, conducted at 36 C using common salt (NaCl) as the diffusing solute, the membrane separates two compartments containing stirred aqueous solutions of NaCl. The bulk liquid phase concentration of diffusing solute in the upstream is mol cm -3 and mol cm -3 in the downstream. The mass transfer coefficients on either side of the membrane is same which calculated as m s -1. Experimental data obtained gave a flux of mol NaCl s -1 m -2 at quasi steady state condition. Calculate: The membrane permeability in m s -1 and (S D) in m 2 s -1. The percentage resistance to diffusion in liquid film. 2) Calculate the rate of removal of urea in grams per hour from blood in a cellophane membrane dialyzer at 37 C. The membrane is mm thick and having an area of 2.0 m 2. The mass transfer coefficient on the blood side is m s -1 and that on aqueous side is m s -1. The membrane permeability is m s -1. The concentration of urea in blood is 0.02 g per 100 cm 3 and that in the dialyzing fluid can be assumed zero. 3) A liquid containing dilute solute at a concentration of kmol m -3 is flowing rapidly by a membrane of thickness m. The distribution coefficient of the solute equals 1.5 and diffusivity is m 2 s -1 in the membrane. The solute diffuses through the membrane and its concentration on the other side is kmol m -3. The mass transfer coefficient k L1 is large and can be considered as infinite and k L2 = m s -1. Calculate the flux and the concentrations at the membrane interfaces. Sketch the concentration profile also. 43

44 DIALYSIS - 1 An experiment is being conducted to determine the suitability of a cellophane membrane for using it in hemodialysis process. The membrane is mm thick. In the experiment, conducted at 36 C using common salt (NaCl) as the diffusing solute, the membrane separates two compartments containing stirred aqueous solutions of NaCl. The bulk liquid phase concentration of diffusing solute in the upstream is mol cm -3 and mol cm -3 in the downstream. The mass transfer coefficients on either side of the membrane is same which calculated as m s -1. Experimental data obtained gave a flux of mol NaCl s -1 m -2 at quasi steady state condition. Calculate: 1) The membrane permeability in m s -1 and D AB K in m 2 s -1. 2) The percentage resistance to diffusion in liquid film. 44

45 DIALYSIS - 2 Calculate the rate of removal of urea in grams per hour from blood in a cellophane membrane dialyser at 37 C. The membrane is mm thick and having an area of 2.0 m 2. The mass transfer coefficient on the blood side is m s -1 and that on aqueous side is m s -1. The membrane permeability is m s -1. The concentration of urea in blood is 0.02 g per 100 cm 3 and that in the dialyzing fluid can be assumed zero. 45

46 DIALYSIS - 3 A liquid containing dilute solute at a concentration kmol m -3 is flowing rapidly by a membrane of thickness m. The distribution coefficient is 1.5 and diffusivity m 2 s -1 in the membrane. The solute diffuses through the membrane and its concentration on the other side is kmol m -3. The mass transfer coefficient k c1 is large and can be considered as infinite and k c2 = m s -1. Calculate the flux and the concentrations at the membrane interfaces. Sketch the concentration profile also. 46

47 U3.3 An aqueous solution of nonpurified proteins was filtered using an experimental ultrafiltration membrane. A dependence of the intensity of volumetric permeate flux (J V ) on applied transmembrane pressure difference (p) was measured. The results are given in the table: p [kpa] J V [10-3 m s -1 ] Using these data judge whether a gel layer was formed at the membrane surface and, eventually, estimate limiting value of the permeate flux intensity. Results: The gel layer has formed, limiting intensity of permeate flux is about m s

48 U3.4 The retentate stream in an ultrafiltration membrane module flows inside membrane fibers with inner diameter of 0.1 cm and length of 100 cm. The mean velocity of the flow in fibers is 3 m s -1. The retentate stream contains a dissolved protein. Its diffusivity equals m 2 s -1. The viscosity of the protein solution is Pa s and its density is 1100 kg m -3. The permeate flux intensity equals m h -1. Determine a value of the polarization modulus during the ultrafiltration and discuss significance of membrane polarization. How would situation change if fibers of inner diameter of mm are used? Results: The polarization modulus in 0.1 cm fibers is and the membrane polarization is therefore very strong. The modulus equals in larger fibers and the membrane polarization is therefore insignificant. 48

49 U3.6 It is known that the proteins of the blood plasma form a gel layer at the membrane surface during their ultrafiltration. The mass fraction of proteins in the gel layer equals 0.2. The molecules of the proteins can be considered as spherical particles of 30 nm in diameter. A filtration experiment with pure water yielded value of the intensity of volumetric flux of the permeate m s -1 at the transmembrane pressure difference p = kpa. When the blood plasma was filtered at p = kpa the intensity of volumetric permeate flux equaled m s -1. The filtration is performed at 20 o C. Auxiliary experiments yielded permeability of the gel layer of m 2. What is the thickness of the gel layer at the steady state, i.e., when limiting intensity of the permeate flux is achieved? Consider (for simplification) that the plasma viscosity and the permeate viscosity values are the same and equal to water viscosity at the temperature of the ultrafiltration. Result: The gel layer thickness is 68.5 m. 49

50 U3.7 An aqueous colloidal solution is concentrated in an ultrafiltration membrane module. The inet concentration of the dissolved substance is 50 kg m -3, the outlet concentration would increase to 200 kg m -3. The retention factor of the membrane used for filtration equals one (an ideal membrane). The volumetric flow rate of the feed is 3.6 m 3 h -1. The dissolved substance creates a gel layer at the membrane surface where its concentration equals 300 kg m -3. The module consists of tubes with the inner diameter of 5 mm and length of 1250 mm. The retentate stream flows inside the tubes. Evaluate the intensity of the permeate volumetric flux and the total membrane area needed for the filtration at mean velocity of the retentate stream within the tubes of 1 m s -1. The density of the retentate stream is 1260 kgm -3, dynamic viscosity is Pa s and the diffusivity of the dissolved substance in water is m 2 s -1. Results: The intensity of permeate volumetric flux is m s -1 and the membrane area of 26 m 2 is necessary. 50

51 U8.10 A salt is removed from an aqueous solution of a protein by dialysis. The protein solution is tightly closed within a dialysis tube and its volume is 50 cm 3. The tube surface area available for dialysis equals 110 cm 2. The dialysis tube is made of a membrane whose thickness is 120 m. The diffusivity of the salt within the membrane is m 2 s -1. The initial salt concentration was set to 0.37 mol dm -3. The mass transfer coefficient at the inner (retentate) side of the membrane is m s -1. The mass transfer coefficient at the outer (permeate) side of the tube equals m s -1. The tube is immersed in pure water whose volume is many times larger than the volume of the protein solution. Determine at what time the salt concentration in the solution within the tube will reach 10% of its initial value. Result: 2530 s. 51

52 U8.11 The phenol contaminates a solution of macromolecular substances. A continuous dialysis is suggested to remove the phenol from the solution. The dialysis equipment consists of a glass pipe (inner diameter: 8 cm) within which 20 membrane dialysis tubes made of a copolymer of a polycarbonate and polyether are placed. The outer diameter of the dialysis tubes is 10 mm, the tube wall thickness is 50 m. The phenol diffusivity in the tubes walls equals m 2 s -1. The contaminated solution flows at flow rate of 0.25 m 3 h - 1 within the dialysis tubes, the inlet phenol concentration is mol dm -3. The outer surfaces of the dialysis tubes are washed with water that enters the pipe at zero phenol concentration. The water flow rate is m 3 h -1. Determine the length of the dialysis tubes necessary to reduce the phenol concentration in the treated solution to 10% of its inlet value. Suppose counter-current operation of the device. The mass transfer coefficient at the inner surface of dialysis tubes equals m s -1 and at the outer surface equals m s -1. Result: The length of tubes is 14.2 m. 52

53 U8.1 A permeation experiment using pure hydrogen has been performed to determine permeability of a paladium based membrane for the hydrogen moles of H 2 permeated the membrane within 50 seconds when the steady state was reached with the membrane of the thickness of 1 mm, of total surface area of 9 cm 2 and at 30 C. The pressure difference across the membrane was kept constant at 1.44 MPa. Determine the membrane permeability. Result: Permeability of the membrane for hydrogen equals mol s -1 Pa -1 m -1. U8.2 Permeability of a polyethylene foil for oxygen equals kg m-1s-1pa-1 (at 30 C). The foil of thickness of 150 m is used to wrap perforated paper box with fruits. The metabolic activities of the fruits in the box result to oxygen consumption of 2.1 g of oxygen per hour. The box dimensions are m. The resistance of the box itself to the oxygen transport can be neglected. What will be the oxygen partial pressure within the box in a steady state if the box is surrounded with the air at 98 kpa and 30 C? Result: The oxygen partial pressure value in the box is kpa. 53

54 U8.3 A gas releasing from a bioreactor is transferred to an analyzer by means of a tubing made of the silicone rubber; the tubing outer diameter is 5 mm and wall thickness is 0.25 mm. The length of the tubing is 5.2 m. The gas pressure in the tubing is 125 kpa, the outer pressure is 99 kpa. The gas from the bioreactor contains 4.5 vol.% of carbon dioxide, 16.5 vol.% of oxygen and 4.3 vol.% of water vapour; the rest up to 100 % is nitrogen. The permeabilities of the silicone rubber are: m3(m s Pa)-1 for oxygen, m3(m s Pa)-1 for carbon dioxide, m3(m s Pa)-1 for water vapour the permeabilities were measured under the same conditions as are those of the permeation from the tubing. What amounts of O2, CO2 and H2O vapour will permeate through the tubing wall? Will it cause significant change in gas composition prior the gas enters the analyzer? Results: Oxygen permeates into the tubing at the rate m3s-1, m3s-1 of carbon dioxide escape from the tubing together with m3s-1 of water vapour. The composition of the gas will change significantly. U8.4 In a single-stage membrane separator the air is to be enriched with oxygen in such a manner that oxygen concentration in the permeate is 30 vol.%. The retentate side pressure is kept at 1 MPa, the permeate side is kept at 0.2 MPa. The permeate is leaving the separator at 40 C. The permeability of the membrane for the oxygen is m2s-1pa-1 (at normal temperature and pressure). The thickness of the active layer of the membrane is 10 m. The selectivity of the membrane for oxygen with respect to nitrogen equals 2.2. The separator would produce 10 m3h-1 of the enriched air (at temperature and pressure of the permeate). Determine composition of the retentate stream, the area of the membrane necessary for the enrichment and the stage cut value. Consider ideal mixing of the gas at both sides of the membrane. Results: The oxygen concentration in the retentate will be 19 mol.%. The necessary membrane area is 2.5 m2. The stage cut equals

55 U8.5 An experimental reverse osmosis membrane has been used for treatment of a waste water containing dissolved sodium carbonate (Na2CO3). The intensity of pure water flux through the membrane was 1.5 m3 per 1 m2 of the membrane area per 1 day (laboratory experiment value) at the pressure difference of 3 MPa at 30 C. The full scale waste water treatment will be performed at the same temperature but at the pressure difference of 5 MPa. The required sodium carbonate concentration in the retentate is 0.05 kmol m-3. The streams at both membrane sides can be considered as ideally mixed. Determine the intensity of volumetric flux of the permeate through the ideal membrane supposing that concentration polarization of the membrane can be neglected. Determine also the membrane area required for treatment of 125 m3 of waste water per day; the water contains kmol m-3 Na2CO3. Results: The intensity of volumetric flux of the permeate is m s-1. The membrane area is 24.9 m2. U8.6 The reverse osmosis (RO) is used (RO) to treat an aqueous solution of low molecular substances that comes to the RO module at the rate of 26 m3 per day. The concentration of the substances equals 0.1 kmol m-3 and the osmotic pressure of the solution equals 0.46 MPa. We can suppose that the osmotic pressure of the solutions is directly proportional to concentration of dissolved substances concentration. The concentration polarization of the membrane can be neglected. The RO is performed at the pressure difference of 3 MPa and the stage cut equals 0.4. The retention factor value of the module is 0.9. An experimentally determined permeance of the membrane (Pm=P/m) for pure water equals m s-1 Pa-1. Determine volumetric flow rates of the retentate and the permeate and concentrations of dissolved substances in both streams. Determine also the required membrane area. Results: The permeate volumetric flow rate is 10.4 m3/day, the retentate flow rate is 15.6 m3/day. The retentate contains kmol m-3 of dissolved substances, the permeate contains kmol m-3 of the dissolved substances. The membrane area is 103 m2. 55

56 U8.7 The air is enriched with the oxygen in a membrane separator with the membrane made of silicone rubber. The permeate containing 27 mol.% of the oxygen has to be produced. The membrane selectivity for oxygen with respect to the nitrogen under the conditions of the separator equals 2.1. The reduced pressure value equals Determine the stage cut value necessary to achieve goals of the separation. Result: The stage cut value has to be set to U8.8 The methane is purified in a membrane separator. The traces of carbon dioxide is removed from the methane at 35 C and 2 MPa at the retentate side of the membrane. The permeability of the membrane (at conditions of the separation) for carbon dioxide is m2s-1pa-1, for the methane is m2s-1pa-1 (the permeabilities were evaluated at conditions at the retentate side of the membrane). The gas entering the separator contains 90 mol.% of CH4 and 10 mol.% of CO2. The pressure at the permeate side is 0.11 MPa, the stage cut equals 0.5. The thickness of the membrane active layer is 1 m. Determine values of the membrane selectivity for CO2 against CH4, the compositions of both permeate and the retentate and the intensities of volumetric flow rates of both outlet streams. Determine separator production rate (measured by the permeate volumetric flow rate) if the membrane area equals 200 m2? Results: The membrane selectivity is The retentate contains 1.68 vol.% of the carbon dioxide, the permeate contains 18.3 vol.% of the carbon dioxide. The volumetric flow rate of the permeate equals 108 m3h-1 (at the conditions of the permeate side). 56

57 U7.1 Experiments yielded equilibrium adsorption data listed in table: X A [mg/g] c A [mg/cm 3 ] 0,02 3, ,08 3, ,10 1, ,50 3, ,70 1, ,00 1,00 Determine parameters of Freundlich and Langmuir isotherms using the above data. Results: Freundlich isotherm: X A 1,00 c. Langmuir isotherm does not fit the data. 0,20 A 57

58 U7.4 Clear fermentation broth after separation of cells contains mol dm -3 of immunoglobulin G. 90% of the immunoglobulin present in the solution has to be separated by an adsorption to a synthetic nonpolar adsorbent. Adsorption equilibrium can be described by 5 0,35 the isotherm: XA 5,5 10 ca, where X A is the equilibrium concentration in the adsorbent (mol cm -3 ) and c A is the equilibrium concentration in the liquid (mol dm -3 ). Compute minimum amount of the adsorbent required for treatment of 2 m 3 of the broth in a single stirred adsorption stage. Result: 0,16 m 3 of the adsorbent. U7.5 Into the vessel filled with a bed of activated carbon a solution of the acetic acid was poured in such a way that void volume between the particles has been just filled up. The acid adsorbs to the carbon particles at 60 C. The vessel performs as an equilibrium stage. The equilibrium is 0,4119 expressed by relation Xm A 3,019 cal, where X ma is equilibrium concentration of acetic acid in the adsorbent (moles CH 3 COOH per 1 kg of carbon) and c AL is equilibrium concentration of CH 3 COOH in liquid (mol dm -3 ). Total volume of the bed of the adsorbent is V b = 1,34 m 3, porosity of the empty bed is e = 0,434, inner porosity of the carbon particles is p = 0,570, density of nonporous adsorbent is C = 1820 kg m -3. Determine amount of carbon necessary for adsorption, volume of the acetic acid solution within the bed and concentration of acetic acid in the liquid after reaching equilibrium. Initial concentration of acetic acid in solution is 0,25 mol dm -3. Results: 594 kg of the adsorbent, volume of the solution 0,582 m 3, concentration of acid 0,0022 mol dm

59 U7.6 The same final concentration of the acetic acid as in the problem U7.5 has to be reached by the adsorption in two consecutive stages. Determine amount of the adsorbent necessary in each stage if volume of the liquid and other conditions will remain the same as in problem U7.5. Výsledek: The first stage: 156 kg of adsorbent, the second stage 87 kg. U7.7 Albumin in a solution is adsorbed in a stirred one-stage adsorber with ion-exchange resin. Adsorption vessel is filled with kg of the solution containing 0,0012 mass % of the albumin. 10 kg of dry clean ionex is used for the adsorption. the equilibrium is described by yA the Langmuir isotherm XA, where X A is relative mass fraction of the albumin ya within the adsorbent and y A is the mass fraction of the albumin in the solution. Determine values of albumin concentrations in both the liquid phase and in the adsorbent. What fraction of the albumin will be transferred from the solution to the adsorbent? Results: Relative mass fraction in the adsorbent: 0,07 and mass fraction in the solution: 2, ,6% of the albumin will be transferred to the ionex. 59