Separation Processes: Filtration

Transcription

1 Separation Processes: Filtration ChE 4M3 Kevin Dunn, Overall revision number: 305 (September 2014) 1

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4 Filtration 4

5 Filtration section Filtration: a pressure difference that causes separation of solids from slurry by means of a porous medium (e.g. filter paper or cloth), which retains the solids and allows the filtrate to pass [Geankoplis, p 905] 5

6 References on filtration Geankoplis, Transport Processes and Separation Process Principles, 4th edition, chapter 14. Perry s Chemical Engineers Handbook, 8th edition, chapter 18. Seader, Henley and Roper, Separation Process Principles, 3rd edition, chapter 19. Uhlmann s Encyclopedia, Filtration 1. Fundamentals, DOI: / b02 10.pub3 6

7 Why filtration? Example: alkaline protease, used as an additive in laundry detergent [MIT OCW, Course , Separation Processes for Biochemical Products, 2005] 7

8 Commercial units: rotary drum filter [Seader, modified from fig 19-13] [YouTube video] 8

9 Commercial units: plate and frame 9

10 Commercial units: plate and frame [Flickr: cdeimages;] [also see YouTube video] 10

11 Commercial units: plate and frame (beer clarification) [Flickr: cizauskas] 11

12 Questions to discuss 1. What characteristics of a filtration system will you use to judge the unit s performance? 2. What factors can be used to adjust the units s performance? Example: a rotary drum filter 12

13 Poiseuille s law Recall from your fluid flow course that laminar flow in a pipe (considering no resistance): P L c = 32 µ v D 2 P = pressure drop from start (high P) to end of tube [Pa] L c = length being considered [m] µ = fluid viscosity [Pa.s] v = fluid s velocity in the pipe [m.s 1 ] D = pipe diameter [m] 13

14 Carmen-Kozeny equation through a bed of solids (cake) P L c = 32 µ v D 2 from which we derive the Carmen-Kozeny equation: P c L c = k 1 µ v ɛ ( (1 ɛ)s0 ɛ ) 2 P c = pressure drop through the cake [Pa] k 1 = 4.17, a constant [ ] ɛ = void fraction, or porosity typical values? [ ] S 0 = specific area per unit volume [m 2.m 3 = m 1 ] S 0 = specific surface area per unit volume is a property of the solids Prove in the next assignment, for spheres, S 0 = 6 d = f (d) 14

15 Solids balance Mass of solids in the filter cake = A L c (1 ɛ) ρ p Mass of fluid trapped in the filter cake = A L c ɛ ρ f small Mass of fluid in the filtrate = V ρ f What key assumptions are being made here? ρ p = solid particle density [kg.m 3 ] ρ f = fluid density [kg.m 3 ] V = volume of filtrate collected [m 3 ] A = cross sectional area for filtration [m 2 ] Then define slurry concentration: C S = mass of dry solids mass of dry solids volume of liquid in slurry volume of filtrate = AL c (1 ɛ) ρ p V 15

16 Exercise 1. Calculate the mass of solids in the cake for the case when ρp = 3000kg.m 3 A = 8m 2 L c = 10cm 2. Also calculate the mass of water in the cake. 16

17 Deriving the flow through the filter Our standard equation for fluid flow: 1 A dv dt = v = Q A for a given velocity v, and volumetric feed flow rate, Q. But, from the Carmen-Kozeny equation: P c = k 1µv (1 ɛ) 2 S0 2 L c ɛ 3 ( P c ) (ɛ) 3 v = (µ)(k 1 )(L c )(1 ɛ) 2 (S0 2) from our definition for C S we can solve for L c C S V L c = A (1 ɛ) ρ p 1 A dv dt = v = ( P c ) (A) (1 ɛ)(ɛ) 3 (ρ p ) (µ)(c S )(V ) (1 ɛ)(k 1 ) (1 ɛ) (S0 2) = P c µc S V α /A 17

18 The general filtration equation 1 A dv dt = P c µc S V α /A C S = slurry concentration [(kg dry solids)/(m 3 filtrate)] α = specific cake resistance [m.kg 1 ] All aspects of engineering obey this general law: transfer rate J = flux = transfer area including the filtration equation: 1 A dv dt 1 A dv dt = = driving force resistance P c µc S V α /A = P c µr c R c = resistance due to the cake = C SV α A [m 1 ] 18

19 Resistance due to the filter medium In a similar way, we can define the filter medium s resistance: 1 A dv dt = P m µr m Notes: P m = pressure drop across the medium [Pa] R m = resistance due to the filter medium [m 1 ] From a practical standpoint, R m is empirical for the given filter We wrap up all other minor resistances into R m also (e.g. pipe flow into/out of filter) The flux through the filter cake is exactly the same as through the medium After filtration gets started, we very often have R m R c 19

20 Bringing it all together As with resistances in series (you learned in Physics I), we have: 1 A dv ( dt = Pm add with P ) c = P tot µr m µr c µ (R m + R c ) this is called the general filtration equation. R c = resistance due to the cake [m 1 ] R m = resistance due to the medium [m 1 ] P tot = total pressure drop = ( P c + P m ) [Pa] 20

21 Questions How would you use this equation? 1 A dv dt = P tot µ (R m + R c ) 1. to determine the medium resistance? 2. to determine the cake resistance? 3. to find the utility cost of operating the filter? 4. to predict the flow for a given filter when your boss wants a higher throughput? 21

22 Cake filtration most widely applied functions only when particles have been trapped on the medium, forming a bed/cake the cake is the filtering element, not the medium medium s pores are larger than particles often e.g. filter press, rotary vacuum drum filter 22

23 reduces cake resistance, R c 23 Cross-Flow Filtration (TFF) TFF = tangential flow filtration (parallel to medium) used with gel-like or compressible substances pores are smaller than particles used when absolute exclusion is essential inlet flow of suspension provides shear to limit cake build-up

24 Basic filtration equation recap 1 A dv dt = P tot µ (R m + C S V α /A) = P tot µ (R m + R c ) V = volume of filtrate collected [m 3 ] A = cross sectional area for filtration [m 2 ] P tot = total pressure drop = ( P c + P m ) [Pa] µ = fluid viscosity [Pa.s] C S = slurry concentration [(kg dry solids)/(m 3 filtrate)] R m = resistance due to the medium [m 1 ] R c = resistance due to the cake [m 1 ] α = specific cake resistance [m.kg 1 ] α = (k 1) (1 ɛ) (S0 2) (ɛ) 3 (ρ p ) which entries in the equation are a function of t? which entries in the equation are a function of V? 24

25 How do we determine cake resistance α? Recall: α = (k 1) (1 ɛ) (S 2 0 ) (ɛ) 3 (ρ p ) Measuring S 0 is difficult for irregular solids ɛ changes depending on many factors (surface chemistry, upstream processing) Is ɛ constant over time? What does it change with? α will also change as a function of P So we let α = α 0 ( P) f 25

26 Constant pressure (batch) filtration 1 A dv dt = Invert both sides, separate, divide by A B = P tot µ (R m + V C S α /A) = P tot µ (R m + R c ) dt A dv = µ (R m + V C S α /A) P tot dt dv = µ A ( P tot ) R µc S α m + A 2 ( P tot ) V dt dv = B + K pv t 0 dt = V 0 (B + K p V ) dv t = BV + K pv 2 2 µ A ( P tot ) R m [s.m 3 µc S α ] and K p = A 2 ( P tot ) [s.m 6 ] 26

27 Constant pressure (batch) filtration t = BV + K pv 2 2 Practical matters: how do we get constant P? which type of equipment has a constant pressure drop? what is the expected relationship between t and V? Plot it for a given batch of slurry. If P is constant, then ɛ is likely constant, and so is α 27

28 Example A water-based slurry of mineral is being filtered under vacuum, with a controlled pressure drop of 38 kpa, through a filter paper of 0.07 m 2. The slurry is at 24 kg solids per m 3 fluid. Use µ = Pa.s and the following data: V [L] t [s] t/v [s/l] plot a rough sketch of x = V against y = t/v 2. read off the intercept and slope 3. calculate medium resistance, R m [Ans: m 1 ] 4. calculate specific cake resistance, α [Ans: m.kg 1 ] 5. cake resistance R c at t = 280 s [Ans: m 1 ] [Flickr: debcha] 28

29 Pressure dependence of α Recall: α = α 0 ( P) f We can find tables of α 0 and f for various solids But they almost will never match our situation Finding the α value as a function of pressure drop: 1. Repeat the example above at several different P levels 2. Calculate α at each P (see example above) 3. Plot x = ( P) against y = α (expected shape?) 4. Take logs on both sides of equation: ln (α) = ln (α 0 ) + f ln ( P) Note: if f = 0 the cake is called an incompressible cake 29

30 Constant rate filtration At a constant rate, i.e. dv dt 1 dv A dt = 1 V A t = Q A = = constant We can solve the equations for P P µ (R m + C S V α /A) P = µr mv At P = µr mq A + µc sαv 2 t A 2 + µc sαq 2 A 2 t For many processes, the constant rate portion is fairly short Occurs at the start of filtration, where a plot of x = (t) against y = ( P) is linear Then we settle into constant pressure mode That s the part we have focused on earlier 30

31 Example continued t = BV + K pv 2 2 B = µ A ( P tot ) R m and K p = µc S α A 2 ( P tot ) 31

32 Example continued Suppose our prior lab experiments determined that α = ( P) 0.3, with P in Pa and α in SI units. Now we operate a plate and frame filter press at P of 67 kpa, kg dry solids with solids slurry content of C s = 300 m 3. The cycle filtrate time that the filter actually operates is 45 minutes (followed by 15 minutes for cleaning). Based on a simple mass-balance, the company can calculate that 8.5 m 3 of filtrate will be produced. 1. Calculate the area required. 2. If the slurry concentration had to double (still the same volume of filtrate), what would the required pressure drop have to be to maintain the same cycle time? 3. Create a plot of the volume of filtrate leaving the press as a function of time, t. 32

33 Pressure requirements change with concentration of slurry A = 81 m 2 stays fixed t = 2700 s stays fixed 33

34 Volume of filtrate produced against time A = 81 m 2 stays fixed 34

35 Extend your knowledge Research the following topics: Variable rate filtration: pressure and flow rate both vary along the characteristic centrifugal pump curve Filtering centrifuge (combine the topics from the last 2 weeks) 35