Solid Processing.pdf

SOLIDS PROCESSING4200:461/561 Fall 2004 TABLE OF CONTENTS 1. INTRODUCTION 2. A CHEMICAL PROCESS INDUSTRY PERSPECTIVE 3. PROPERTIES OF PARTICULATE SOLIDS 4. BULK PROPERTIES OF SOLIDS 5. FLUIDIZATION 6. ELUTRIATION 7. SOLID/LIQUID SEPARATIONS 8. PRETREATMENT OF S/L MIXTURES 9. SEGREGATION MECHANISMS 10. HOPPER DESIGN 11. GRADE EFFICIENCY 12. CYCLONES 13. CONVEYING 14. DUST EXPLOSIONS SOLIDS NOTES 1, George G. Chase, The University of Akron 1. INTRODUCTION 1.1 Organization of the Course Chapters 1 and 2 of this course give an introduction to this course with an emphasis on industrial applications. Chapters 3 and 4 introduce you to fundamental properties of particles and particulate systems. The remaining chapters cover specific topics such as conveying, hopper design, and separations. For undergraduates this is a design course, hence you are expected to apply design methodology. You will be assigned laboratory projects to work on in which you will apply engineering judgment with design skills to produce a final product. Graduate students are required to write a paper concerning a particular aspect of solids processing. There are two labs in this course. The labs are well defined and structured. A few class periods are dedicated to the labs, though you will probably have to spend some time out of class to complete the labs. Depending on schedules and availablility, there may be a short field trip to tour a facility that handles bulk solids. Also, I may invite guest engineers to teach selected topics. 1.2 Acknowledgement I acknowledge engineer Karl Jacob, at the Dow Chemical Company at Midland Michigan, for his enthusiastic support in helping me prepare and organize this course. Karl introduced me to many of the topics used in pneumatic conveying and hopper design and helped write many of the notes. I acknowledge the National Science Foundation GOALI grant CTS 9613904 for its financial support that made it possible for Karl Jacob and I to collaborate in developing this course. Finally, I acknowledge the American Filtration and Separations Society and its members’ knowledge on a number of topics including particle size characterization, surface science and effects of surfactants, and various methods of fluid-particle separations. Many discussions with members of AFS over the past several years have helped me to refine my course notes and to focus on essential aspects of fluid/particle separations. The collective knowledge is vast and one course can only attempt to introduce students to selected topics. 1-2 SOLIDS NOTES 1, George G. Chase, The University of Akron 1.3 Overview Solids processing is a topic area that can cover a very wide range of processes. Processes could include: particle sizing & shaping flocculation pastes packing & compaction absorption/desorption mixing Brownian motion leaching rheological applications crushing/grinding particle classification (separation by size) caking crystallization separations fluidization filtration slurry flow catalytic reactors settling agglomeration drying digestion floatation surface phenomena ion exchange packed beds These topics and many more are covered in Perry’s Handbook. Perry’s Handbook has several chapters devoted specifically to topics in solids processing: • • • • • • Handling of Bulk Solids and Package of Solids Size Reduction and Size Enlargement Adsorption and Ion Exchange Gas-Solid Systems Solid-Solid Systems Solids Drying and Gas-Solid Systems as well as many topics that are buried within other chapters. Solids processing is an important part of industrial operations. In the chemical process industry roughly 60% of the products are particulate in form. When you add in products that at some intermediate step are in particulate form then 80 to 90% of all chemical processes used in industry require application of solids processing either directly or indirectly. (These items are listed in HANDOUT 1.1). 1-3 HANDOUT 1.1 Typical Solids Processing operations particle sizing & shaping flocculation packing & compaction absorption/desorption mixing Brownian motion leaching pastes rheological applications crushing/grinding particle classification (separation by size) caking crystallization separations fluidization filtration slurry flow catalytic reactors agglomeration settling drying digestion floatation surface phenomena ion exchange packed beds Sections in Perry’s Handbook • • • • • • Handling of Bulk Solids and Package of Solids Size Reduction and Size Enlargement Adsorption and Ion Exchange Gas-Solid Systems Solid-Solid Systems Solids Drying and Gas-Solid Systems . SOLIDS NOTES 2. New York.A. 1977. George G. Shreve and J. The intent of showing you these flow charts is to make you aware of the important role that solids handling and fluid/solids separations have in the industrial operations. Some examples from Shreve and Brink (see HANDOUT 2. Chase. However. Industries include: water conditioning coal chemicals industrial carbon ceramics paints explosives and propellants agriculture fermentation pulp and paper synthetic fibers petrochemicals environmental cleanup glass industry phosphorous production potassium production nuclear industries food and food processing sugar and starch wood chemicals plastics rubber industries pharmaceuticals A sampling of the process flow charts are in the 1977 publication.. Brink. In today's industries we could add other flow charts such as for terephthalic acid production for the polymer industry. The University of Akron 2. many of the unit operations steps would be the same as shown in these flow charts. A CHEMICAL PROCESS INDUSTRY PERSPECTIVE The chemical process industry makes heavy use of solids materials handling equipment and separations. include process flow diagrams showing the materials handling and fluid/particle separations steps.N. Chemical Process Industries.1) R. 2-1 . McGraw-Hill. but for the most part these processes have not been altered significantly. We are not going to go over each process in detail. 4th ed. The University of Akron You should note the variety of equipment that is used: crushing and screening conveyor vibrating feeders flotation cells thickeners dryers grinding mills rotary kiln classifiers screw conveyors leaching digesters fourdrinier oven bin storage bins classifiers and screens pebble mills filters pneumatic conveying slurry mixers dust collector cyclone separators cake washing crystallizers beaters All of these operations have in common the handling and processing of particulate solids. Chase.SOLIDS NOTES 2. George G. 2-2 . . 4th ed. PARTIAL LIST OF INDUSTRIES THAT USE FLUID/PARTICLE PROCESSES: water conditioning environmental cleanup coal chemicals glass industry industrial carbon phosphorous production ceramics potassium production paints nuclear industries explosives and propellants food and food processing agriculture sugar and starch fermentation wood chemicals pulp and paper plastics synthetic fibers rubber industries petrochemicals pharmaceuticals LIST OF OPERATIONS FOUND IN THE EXAMPLE INDUSTRIAL PROCESSES: crushing and screening oven bin conveyor storage bins vibrating feeders classifiers and screens flotation cells pebble mills thickeners filters dryers pneumatic conveying grinding mills slurry mixers rotary kiln dust collector classifiers cyclone separators screw conveyors cake washing leaching crystallizers digesters beaters fourdrinier . Chemical Process Industries. 1977. New York.HANDOUT 2.A.N. McGraw-Hill. Brink. Shreve and J.1 R. b. A fractal length can be determined which characterizes the size of the particle and its dimensionality somewhere between linear and two-dimensional. The second method is through the use of Fractals. The sphere of the same projected area as viewed from above when lying in a position of maximum stability (as with a microscope). The values of the polynomial coefficients characterize the particle shape. The sphere of the same surface area per unit volume. Based on the measurement techniques the particle sizes are typically related to equivalent sphere diameters by a. The sphere of the same surface area as the particle. f. The sphere of the same volume of the particle. If you are buying a sleeping bag I suggest the last one.SOLIDS NOTES 3. There are two other methods that I know of for sizing particles that are not based upon comparison to a standard (sphere) shape. For example. The first method is to fit the particle area projected shape to a polynomial type of relation. The sphere with the same settling velocity as the particle in a specified fluid. b. Length of your longest chord (height)? As you can deduce. We will not be spending any time with these latter two methods though they would be interesting topics for a term paper.1. 3-1 . PROPERTIES OF PARTICULATE SOLIDS Before we can discuss operations for handling and separating fluid/particle systems we must understand the properties of the particles. d.1 Individual particle characteristics In your assigned reading is a discussion on the characterization of particles. The sphere of the same area when projected on a plane normal to the direction of motion. The University of Akron 3. e. Circumference around your waist? 2. The sphere which will just pass through the same size of square aperture as the particle (as on a screen). 3. The way that we characterize the particles largely depends on the technique used to measure them. Chase. If you are sizing a life jacket belt you would interested in the first size. Diameter of a sphere of the same displacement volume as your body? 3. g. how would you quantify yourself if measured by 1. the measured values have different meanings and will be important relative to those meanings. Sizes of common materials are listed in HANDOUT 3. George G. c. a. The way that we measure a particle size is as important as the value of the measured size. There are a number of properties of particles that are of interest besides its size and shape. “Filtration Spectrum” which compares. Chase. You can normalize the plot by dividing the masses of each size by the total mass. if we add the mass fractions cumulatively we get the Cumulative Mass Fraction plot. A number of suppliers now sell small spherical particles of nearly uniform size distributions for calibration purposes. Distributions can be reported either in terms of frequency (differential form) or by cumulative (integral form) as shown below. Finally.2 lists the Tyler and the US standard mesh nominal sizes as well as the screen opening sizes in mm and inches. Some of these methods depend upon calibration with known particle sizes. Except for the extreme case of long thin fibers. chemically react with each other. lets suppose that you measure the mass of particles by size by some unspecificed process.4. Several sieve standards exist which classify particles according to the size hole through which the particles can pass.3 and 3. shown in Figure 3-3. As an example your measured data may be plotted as shown in Figure 3-1. Particles can repel or attract each other due to static charge build up. One of the better known instruments for this is the Coulter Counter. The University of Akron Probably among the earliest forms of particle classification (sizing) to be developed is sieving. Also in this handout is the Osmotics Inc. Many of these techniques are listed in HANDOUTS 3. break up. To explain how we mathematically represent the distribution data. (2) by length. For a given material. the particle mean size will be of the same order of magnitude of the dimensions of the particle no matter which method is used. the relative sizes of common materials. among other things. they can stick. Class HANDOUT 3. A brief description of the electronic particle counter principle is given in HANDOUT 3. 3-2 .SOLIDS NOTES 3. bounce off of each other. they are affected by van der Waals forces (when they are small enough). to obtain the mass fractions as shown in Figure 3-2. George G. agglomerate.2 Measurements There are a number of methods for measuring particle sizes and size distributions. Some of the more advanced methods of particle size measurement not only measure the particle sizes but they will also provide the size distributions of the particles. and they are effected by the surrounding fluid phase due to drag an buoyant forces. there are four types of particle size distributions that are possible: (1) by number. 3. (3) by surface. and (4) by mass (or volume).5. 7 Mass Fraction 0.9 0. Example mass quantities of a imaginary sample of particles. mm 4 5 Figure 3-2. 3-3 .5 0.4 0.SOLIDS NOTES 3. x.5 1 0.1 0 1 2 3 Diameter.8 0.5 Mass.5 2 1. Mass fractions from data in Figure 3-1. The University of Akron 5 4. grams 3 2.3 0. 1 0. mm 4 5 Figure 3-1.6 0.5 4 3. Chase.5 0 1 2 3 Diameter.2 0. George G. x. x.9 0. Cumulative mass fraction plot of data from Figure 3-1. Furthermore. mm 4 5 Series5 Series4 Series3 Series2 Series1 Figure 3-3.3 0.6 0.SOLIDS NOTES 3. ∆Fm as ∆Fm ( xi ) = where ∆Fm ( xi ) ∆x ∆x (3-2) ∆Fm is the slope of the curve on the cumulative mass fraction plot. The University of Akron 1 0.8 Cumulative Mass Fraction 0.4 0.7 0. From these Figures we see that the cumulatve mass fraction can be written mathematically as Fm ( x n ) = ∑ ∆Fm ( xi ) i =0 n (3-1) as a function of the nth particle size.5 0. ∆x dx (3-3) 3-4 . we can write the increment in the cumulative mass. We define ∆x this slope to be the frequency distribution of the mass fraction. where f m ( xi ) = ∆Fm ( xi ) dFm = . Chase.1 0 1 2 3 Diameter.2 0. George G. f m . Svarovsky. etc. If the particle size distribution is determined on a microscope by measuring projected areas or by laser attenuation then the surface fraction or frequency distribution based on surface area is f s ( x)∆x = Area of all particles of size x . Butterworth. number. Total area of all particles (3-7) Since f is a fractional amount. we can relate the cumulative mass fraction to the frequency distribution by Fm ( x n ) = = = ∑ n i =0 n ∆Fm ( xi ) ∆x ∆x m ∑f i =0 xn 0 ( x i ) ∆x m . Chase. F ( x ) = ∫ f ( x )dx x 0 (3-9) which is the area under the f(x) curve from 0 to x. (3-4) ∫f dx Let the fractional amount of particles of size x be for any type of measurement (by mass. Solid-Liquid Separation. If the particle size distribution is determined as the number fraction then the number frequency distribution is given by f N ( x ) ∆x = Number of particles of size x . London. The University of Akron Hence. chapter 2). Total number of particles (3-6) where ∆x is the differential range above and below size x that the number count represents. then integrating over all particle sizes gives the whole. 1990. Plots of F and f have the general form 3-5 ..) be represented as f ( x ) = DISTRIBUTION FREQUENCY (3-5) (see L. or ∫ f ( x)dx = 1 0 ∞ (3-8) and if we integrate over only the range from zero to some size x we get the cumulative fraction. area.SOLIDS NOTES 3. 3rd ed. F(x). George G. F(x). Similarly. and are denoted with subscripts N or S. Number Distribution 2. Linear and volume (mass) basis for the distributions also exist and are denoted by subscript L or M. The University of Akron 1 F(x) f or F f(x) 0 x Figure 3-4.SOLIDS NOTES 3. 1. f(x). Typical f and F curves. x x 0 0 . Distribution by Length 3. and the cumulative fraction. and k3 are geometric shape factors. Distribution by Mass fN(x) fL(x) fS(x) fM(x) (Equivalent to distribution by volume) (Not used in practice) The several types of distributions are all related to the number distribution by f L ( x) = k1 x f N ( x) (3-11) (3-12) (3-13) f S ( x) = k 2 x 2 f N ( x) f M ( x) = k 3 x 3 f N ( x) where k1. may be based on numbers of particles or surface areas as described above. Distribution by Surface 4. where f and F are also related by f ( x) = dF ( x ) dx (3-10) The frequency distributions. the cumulative distributions can be related FL ( x) = ∫ k1 x f N ( x)dx = ∫ k1 x dFN ≅ ∑ k1 xf N ∆x x x 0 0 (3-14) (3-15) (3-16) 3-6 FS ( x) = ∫ k2 x 2 f N ( x)dx = ∫ k2 x 2 dFN ≅ ∑ k2 x 2 f N ∆x x x 0 0 FM ( x) = ∫ k3 x 3 f N ( x)dx = ∫ k3 x 3 dFN ≅ ∑ k3 x 3 f N ∆x . George G. Chase. k2. we substitute ln(x) for x and ln(σg) for σ. which is skewed to the left compared to the familiar bell curve. The most widely used function is called the log-normal distribution. and combine with Eq. Figure 3-5b. we start with f s ∆x = ∑ n j Aj n j Aj = ∑ N∆xf Nj Aj N∆xf Nj Aj = ∑ f Nj Aj f Nj Aj (assuming constant ∆x ). (3-12). The lognormal function is best described first by considering the normal distribution of the Gaussian (bell shaped) curve shown in Figure 3-5a: ⎛ ( x − x)2 ⎞ dF 1 ⎟ = exp⎜ − 2 ⎜ ⎟ dx σ 2π 2 σ ⎝ ⎠ (3-19) where F is the cumulative undersize fraction of particles.is the size increment range that n j represents. and ∆x j . The University of Akron Often. x is the particle size. experimental data are reported in discrete form (such as from a sieve analysis). For these data it is easier to work with discrete forms of the integral equations: FN ( xi ) = ∑ f N ( x j )∆x j j =1 i (3-17) where fN (x j ) = nj N∆x j (3-18) where n j is the number of particles in the jth set.SOLIDS NOTES 3. Let A j = πx 2 j . To obtain the log-normal distribution. upon rearrangement we get k2 = 1 . George G. This gives ⎛ ln x − ln x 1 xdF g ⎜ exp⎜ − = 2 dx 2 ln σ g ln σ g 2π ⎜ ⎝ ( ) 2 ⎞ ⎟ ⎟ ⎟ ⎠ (3-20) 3-7 . N is the total number of particles. It is a two-parameter function that gives a curve. As an example. Chase. and x is the mean particle size. x f ∆ x ∑ Nj 2 j There are several equations that are typically fitted to the distribution. to find k2. σ is the standard deviation. This function is normally used because in most cases there are many more measured fine particles than larger particles. at xm). Figure 3-5a Normal Gaussian curve.5 Sigma = 3 ⎛ − 1⎞ exp⎜ ⎟ ⎝ 4b ⎠ 0. 7. (3-21) ⎛ (ln x − ln x ) 2 ⎞ 1 dF m 2 ⎟ exp − ln σ g / 2 exp⎜ − = 2 ⎟ ⎜ dx x m ln σ g 2π 2 ln σ g ⎠ ⎝ ( ) (3-22) in which xm represents the mode because it is the size at which dF/dx has its maximum (recall f(x) = dF/dx. 3-8 . 351-352. Powder Technology. Log-normal curve.3 0.5 (3-22b) (3-22c) f(x) 0. Chase. but the larger standard deviation causes the second curve to have a smaller peak and more spread.2 1 b= 2 ln 2 σ g to simplify the calculations. Svarovsky (L. Both curves have the same area. we get the more convenient form Figure 3-5b. (3-22) as ⎛ ⎞ dF 2⎛ x ⎞ ⎟ ⎟ ⎜ ln b = a exp⎜ − ⎜ x ⎟⎟ ⎜ dx m ⎝ ⎠⎠ ⎝ where 1 a= xm ⎛b⎞ ⎜ ⎟ ⎝π ⎠ 1/ 2 (3-22a) 1 0. hence f is maximum at its mode. George G. The University of Akron where x g is the geometric mean and is equal to the median size (where 50% of the particles are greater in size and 50% are smaller in size).(3-20) is rearranged and the substitution ln xm = ln x g − ln 2 σ g is applied.7 Sigma = 1.4 0. Comparison between log normal curves with σ g = 1.8 0.9 0. 0. 1973) recommends writing Eq.SOLIDS NOTES 3. When Eq. Svarovsky.6 0.1 0 0 -0.5 and σ g = 3 .1 5 10 15 20 25 30 X Figure 3-6. 63 1.99 1.71 1.725 1 Total 21 fdx f F 0. f is 0.904762 0.675 2 3.659 2.952381 0. 3-9 . fdx is determined by 7/21=0.2 0 1.5 1.575 9 2.42 1. the average candy diameter (assuming spherical shape) is calculated.047619 0.7 Diameter.590 2.58 1.857143 0.6 1.66667. Plot of frequency and cumulative frequency distributions for M&M’s.952381 0.06 1. The University of Akron EXAMPLE 3-1 (HANDOUT 3.6) A sample of M&M’s ™ with peanuts are weighed as listed in Table 3-1.508 2.7 1.55 cm and are assigned to the average size of 1.555 2. Plot the frequency distribution and the cumulative frequency Table 3-1.59 1.5 and 1.05 = 6.(3-13) and (3-14) the frequency and cumulative frequency distributions are calculated.6 1. Grams Dia.952381 1 1 Frequency Distribution of M&Ms Frequency Distribution 10 8 6 4 2 0 1.542 2.55 are in the second increment.94 1.5 1. all M&Ms with size between 1.37 1. The size ranges start with 1.01 1.05 cm. All M&Ms of size less than 1.35 1.36 1.65 1.33333/0.6 0.525 7 2.511 2.550 2.571429 0.586 2.4 0.5 to 1.SOLIDS NOTES 3.45 1 0.75 1.666667 0.570 2.525 cm.047619 0. there are 7 M&Ms in the size increment range of 1.4 1.809524 0.475 1 2. Using the formulas in Eqs. and so on.47 1.21 1. Mass and diameter distribution of M&M’s. F is determined by cumulative summing the values fdx.333333 6.75 f F 1. The particle sizes are added up in ∆x increments of 0.625 1 2.55 1.45 to 1.55 1.18 1.614 1.047619 0.473 1. cm Figure 3-7.672 1.047619 0.668 1.50 cm.50 are counted in the first increment.428571 8.540 2.65 1.18 1.380952 0.33333.501 2.8 0. distribution of the average diameter of the candies.501 2. Chase. George G. 2.952381 0.53 1. Using an average density of 1.57 1.598 1.544 1.49 1.565 2.095238 1. For example.588 2.23 grams per cubic centimeter. The values for nj are determined by counting the number of M&Ms that fall in a given size increment and are assigned to the average size in the increment. cm Size < Avg size No.22 1.578 2. 50. there is a bewildering number of different definitions of "mean" size for a particle. (HANDOUT 3. The median is the x value at which F(x) = 0. but all of the other means may be different. Chase.3 Choice of Mean Particle Size As shown in handout 3 and the previous discussions. The University of Akron The results of the summation are plotted in Figure 3-7. The mode is the x value at which f(x) is a maximum.8.7) MODE HARMONIC MEAN ARITHMETIC MEAN MEDIAN f QUADRATIC MEAN CUBIC MEAN f x Figure 3. 3. 3-10 . The choice of the most appropriate mean is vital in most applications.8 Comparison of size distributions. As can be seen in Figure 3. two different size distributions may have the same arithmetic mean.SOLIDS NOTES 3. George G. 3 or x c = ∑ xi3 3 ni = ∑ xi3 f Ni ∆x N where f N ∆x = ni N hence xc = 3 ∑x 3 i f Ni ∆x Suppose you have the mass distribution frequency of a set of particles and you want the geometric mean. How would you calculate the geometric mean from the given mass distribution frequency? log( x g ) = ∑ log( xi ) f M ( xi )∆x hence 3-11 . x h Example.SOLIDS NOTES 3. x a QUADRATIC MEAN. George G. The University of Akron The various means are defined by: g x = ∫ g ( x)dF 0 () 1 (3-23) or by the equivalent expressions g ( x) = ∫ g ( x) f ( x)dx = ∑ g ( x) f ( x)∆x 0 1 (3-24) g(x) = x x2 x3 log x 1/x NAME OF MEAN ARITHMETIC MEAN. CUBIC MEAN. suppose we want the cubic mean of a set of particles for which we know the number distribution. Chase. The mean is defined such that x c N = ∑ xi3 ni . xq xg HARMONIC MEAN. x c GEOMETRIC MEAN. 01 0.00 4889. The University of Akron x g = 10 ∑ log( xi ) f M ( xi ) ∆x The mean particle size is rarely quoted in isolation.31 0.60 1.46 0.50 0.42 0.19 269.59 72. The results of a sieve analysis may give the size distribution as (HANDOUT 3.14 0.00 2.02 1243. It is usually related to some measurement technique and application and used as a single number to represent the full size distribution.2 Sieve analysis of a sample of particles.46 0.25 0.36 0.83 0.40 0.04 0.92 0. Usually enough is known about a process to identify some fundamentals.30 0.11 0.00 58141.65 0. Dividing the sample mass by the particle intrinsic density (assumed here to be 2.44 0. TRAY.00 0.70 0.16 0.00 TOTAL MASS 3.00 1.10 0.01 1.18 0. Consider measuring the size distribution by sieving.11 0.01 21045. Mass.25 0.01 1.38 0.62 1.06 1286.26 0. Note 1 Note 2 VOLUME AREA SIEVE AVG SIEVE MASS ON VOLUME V1 NUMBER NUMBER A1 TRAY AREA SIZE.42 0.06 1. The fundamental relations may be overly simple to describe the process fully.25 346.20 0.23 0.20 0.18 0. Comparison of mass versus number count.8) Table 3.00 1384.17 982. and area fractions are calculated. MM SIZE.14 192. Chase.71 0.01 518.00 The mass fraction is found simply by dividing the sample masses (sieve mass) by the sum of the masses.19 0.11 153. but it is better than randomly selecting mean definition.10 0. which can be used as a starting point. It is the application therefore which governs the selection of the most appropriate mean.10 0.85 0. number and area fractions are calculated.21 1266.92 0.19 0. EXAMPLE 3-2.01 0.23 0. George G.00 67293.67 0.14 0.64 27. If two size distributions have the same mean (as measured using the same methods) then the behavior of the two materials are likely to behave in the process in the same way.SOLIDS NOTES 3.13 0. number.03 3.40 504.14 0.24 0.06 5660.69 45.05 0.04 0. Sieve analysis of a sample of particles. gives 3-12 .05 0.58 0.03 38. Mass.70 0.00 0.59 0.60 0.19 269.03 38.90 0.98 0.63 10.50 0. MASS.01 0.88 0. Dividing the sample 3 volume by the volume of one particle ( 4 3 π R ) where R is the sieve size opening.20 0. The mean represents the particle size distribution by some property which is vital to the application or process under study.10 0.00 153782.60 320.10 0. MM g FRAC MM^3 FRAC MM^3 FRAC MM^2 MM^2 FRAC pan 0 0.15 0.06 76.79 254.03 0.06 0.82 1.08 0.29 0.6 g/cm3) gives the volume of the particles in the sample.00 0. mm Figure 3.30 0. hence the large number of fines will cause the fluid to be cloudy.00 Avg Particle Size. 0. The number and area fractions are found by dividing the sample values by the totals.SOLIDS NOTES 3. George G.45 0. The number distribution and surface area distribution are skewed greatly to the small particle size.60 0. Chase.25 0.50 0. The total surface area of the particles of a given size is obtained by multiplying the number of particles times the surface area of one particle ( 4π R 2 ).80 1.9.20 0.00 0.35 Fraction 0. This shows that a small mass of the fines contains a large number of particles. Comparison of the fractional distributions of the particle size distributions.05 0.20 0. 3-13 .00 Mass & Volume Frac Number Frac Area Frac 0.40 0.9 shows that the modes of the three distributions vary widely. The University of Akron the number of particles for that sample. A property such as turbidity is sensitive to the total number of particles.15 0. A process such as filtration is sensitive to the total surface area of the particles due to the drag resistance to flow across the surface.10 0. The plot in Figure 3.40 0. dewatering. George G. This surface area is related to the specific surface area.SOLIDS NOTES 3. Application: cake filtration. we apply the hydraulic radius. Chase. We can apply the concept of a friction factor and a Reynolds number. The University of Akron EXAMPLE 3-3.section area available for flow Rh = wetted perimeter volume available for flow = total wetted surface ⎛ volume of voids ⎞ (3-25) ⎜ ⎟ ⎝ volume of bed ⎠ = ⎛ wetted surface ⎞ ⎜ ⎟ ⎝ volume of bed ⎠ = ε a where ε is the bed porosity and a is a surface area. of the solids (total particle surface/volume of particles) by a = a s (1 − ε ) (3-26) The specific surface area in turn is related to the mean particle diameter (assuming the particle can be represented by a sphere) x = Dp = 6 6 ⋅ total volume of particles = as total surface of particles (3-27) For spheres the total volume of particles is given by 4 total volume = ∑ πR 3 3 1 2 = ∑ D pi πD pi 6 1 = ∑ xi S i 6 ( ) (3-28) and the total surface area of the particles is given by total surface area = ∑ 4π Ri = ∑ π Di = ∑ Si 2 2 (3-29) Hence. flow through packed beds and porous media. we get the mean particle diameter to be 3-14 . Rh. cake washing. cross . what definition of the mean should be used? In flows through a packed bed we can consider the pores to be conduits. Since the geometry of an arbitrary pore is not cylindrical. a s . If the particle size distribution is known. SOLIDS NOTES 3, George G. Chase, The University of Akron x= ∑x S ∑S = ∫ xdF i i 1 0 i where ∑S Si = dFs i (3-30) s where the latter expression is the analytical formulation. This latter expression defines the mean to be the arithmetic mean, x = ( x a ) s (from Eq. 3-23) of the distribution by surface. Next, we must relate this to a size distribution by mass (the usual way of measurement). The surface distributions by surface and mass can be related by f m ( x ) = kxf s ( x ) or dFs = dFm kx (3-31) where k is a constant that accounts for the geometric shape of the particles. It is assumed here that k is independent of x. Since the mean size is given in Eq. (3-30), then combining (3-30) and (331) we get (x ) a s = 1 1 dFm = ∫ k0 k 1 (3-32) where the integral is unity. If we go back to Eq.3-31, we can integrate to obtain ∫ dF = ∫ s 1 1 0 0 dFm kx (3-33) or, since k is not a function of x, k=∫ 0 1 dFm x (3-34) where the RHS of Eq.(3-34) is the definition for the Harmonic mean, 1 ( xh ) m , of the mass distribution given by Eq. (3-23). Hence this shows that the surface arithmetic mean is equal to the mass harmonic mean, ( xa ) s = ( xh ) m . Therefore, for flow through packed beds, filter cakes, etc., the appropriate mean particle size definition is the arithmetic average of the surface 3-15 SOLIDS NOTES 3, George G. Chase, The University of Akron distribution. This is shown to be equivalent to the mass distribution harmonic mean. EXAMPLE: 3-4. Mass recovery of solids in a dynamic separator such as a gravity settling tank. For a settling process in which mass recovery is to be optimized, which would be the most appropriate mean particle size? Total recovery of any separator can be obtained by combining the feed size cumulative distribution, F(x), with the operating grade efficiency curve, G(x). Mathematically, this is written as ET = ∫ G ( x)dF 0 1 (3-35) where Et is the recovery by mass. A simple plug flow model of the separation in a settling tank without flocculation gives the grade efficiency in the form G( x) = ut A Q (3-36) where A is the settling area, Q is the suspension flow rate, and ut is the terminal velocity of particle size x. Assuming Stoke's law for the terminal velocity ut = x 2 ∆ρ g 18µ A∆ρ g ⋅ ∫ x 2 dF 18Qµ 0 1 (3-37) then these three equations can be combined to obtain Et = (3-38) where the integral defines the quadratic mean of the particle size distribution by mass. We will discuss Grade Efficiency in further detail in a later section when we discuss separations processes. 3-16 SOLIDS NOTES 3, George G. Chase, The University of Akron 3.4 Drag Force on a Spherical Particle Probably the most significant force acting on particles in a fluid-particle medium is the drag force due to the relative motion between the fluid and the particles. A summary of the derivation of the governing equations is given here. Fb = mf g From a free body diagram, Figure 3.10, we can write a balance of forces acting on a spherical particle. The balance of forces shows that the accelerating force acting on the particle is given by Fa = Fg − Fb − Fk (3-39) PARTICLE MOVEMENT DIRECTION u Fk Initially, when a particle falls through a fluid the particle velocity accelerates. After a short distance the particle reaches its terminal velocity and its acceleration goes to zero. This means that the force of acceleration, Fa is zero. Hence, at terminal velocity the kinetic force acting on the particle is given by Fk = Fg − Fb (3-40) Fg = mp g Figure 3-10. Free body diagram on particle of diameter R. In Figure 3.10 mp is the mass of the particle and mf is the mass of the displaced fluid with the same volume as that of the particle. These masses are equal to the volume of the particle times the respective particle or fluid densities. The kinetic force becomes 3 Fk = 4 3 π R g ρp − ρ ( ) (3-41) We define the drag coefficient, Cd, by the expression Fk = Cd A KE (3-42) where A is the projected area normal to the flow and KE is the characteristic kinetic energy. When we substitute in the projected area of a sphere, πR2, and the kinetic energy, 1/2 ρu2, into Eq. (3-42) then we can derive a working equation for determining the drag coefficient as CD = 8 3 Rg (ρ p −ρ ) ρ u2 (3-43) In order to use this expression to determine values for CD we must run experiments. The experiments may be in the laboratory or they may be thought experiments for limiting case solutions. Lets consider the limiting case of creeping flow around the sphere as shown in Figure 3.8. This operation is discussed in some detail by Bird et.al. (1960). 3-17 The flow is in the positive z-direction such that there is symmetry in the φ-direction. The University of Akron Z r θ z φ rs in(θ Y ) x y X v∞ Figure 3-11.SOLIDS NOTES 3. For creeping flow the dominant term in the momentum balance is the viscous force term. the momentum balance in spherical coordinates becomes: ⎡∂ sin θ ∂ ⎛ 1 ∂ ⎞ ⎤ + 2 ⎜ ⎟⎥ ψ = 0 ⎢ r ∂θ ⎝ sin θ ∂θ ⎠ ⎦ ⎣∂ r 2 (3-47) where the [ ] term is a differential operator and where symmetry is assumed in the φ direction (hence no dependence on φ.and θ-directions. which at the continuum scale gives ∇ ⋅τ = 0 (3-44) where the stress tensor is related to the velocity by the Newtonian Fluid model. At distances far from the sphere the flow velocity is uniform at a value v∞. Equation (3-47) is solved with the boundary conditions 3-18 . Chase. it is mathematically easier to solve the resulting differential equations in terms of the stream function. This problem is equivalent to the particle falling in the negative zdirection through a stationary fluid. The stream function is related to the velocities in spherical coordinates by: vr = − 1 ∂ψ r sin θ ∂θ 2 (3-45) vθ = − 1 ∂ψ r sin θ ∂ r (3-46) In terms of the stream function. Since the fluid motion around the sphere varies in the r. ψ. Flow around a sphere of radius R. George G. we can evaluate the kinetic or drag force directly using ⎛ τ : ∇v ⎟ ⎞ r 2 dr sin θ dθ dφ v ∞ Fk = − ∫ ∫ ∫ ⎜ ⎝ ⎠ 0 0 R 2π π ∞ (3-56) Insertion of Newton’s Law of viscosity for the stress tensor in Eq.40) yields the kinetic force as 3-19 . George G. Chase. A more direct way is to recognize that the drag force on the sphere is directly related to the viscous dissipation. The last boundary condition suggests that ψ = f (r ) sin 2 θ (3-51) When this function is substituted into Eq. For a Newtonian fluid.2.B.SOLIDS NOTES 3. homogeneous fourthorder equation ⎛ d2 2 ⎞⎛ d 2 2⎞ ⎜ 2 − 2 ⎟ ⎜ 2 − 2 ⎟ f (r ) = 0 r ⎠ ⎝ dr r ⎠ ⎝ dr (3-52) Assuming a solution of the form f (r ) = Cr n shows that n may have the values of 1.(3.(3.41) and substitution of the velocity profiles in Eqs.39) and (3. (345)-(3-46) and (3-48)-(3-50)) gives the velocity profiles 3 vr ⎡ 3 R 1 ⎛ R ⎞ ⎤ = ⎢1 − 2 + 2 ⎜ ⎟ ⎥ cosθ v∞ ⎣ r ⎝r⎠ ⎦ ⎥ ⎢ 3 ⎡ vθ R 1⎛R⎞ ⎤ = − ⎢1 − 3 − ⎟ ⎥ sin θ 4 4⎜ v∞ r r ⎝ ⎠ ⎦ ⎥ ⎢ ⎣ (3-54) (3-55) We could derive an expression for the kinetic force on the sphere by using the momentum balances and an expression for the pressure distribution. The University of Akron vr = − vθ = − 1 ∂ψ =0 r sin θ ∂θ 2 at r = R at r = R for r → ∞ (3-48) (3-49) (3-50) 1 ∂ψ =0 r sin θ ∂ r 1 ψ →−2 v ∞ r 2 sin 2 θ The first two boundary conditions mathematically describe the contact of the fluid to the sphere surface. The third boundary condition shows that at distances far from the sphere the velocity becomes v∞.4 hence we get the functional form for f(r) as f (r ) = A + Br + Cr 2 + Dr 4 r (3-53) where A. Applying the boundary conditions and the definitions for the stream function (Eqs.C and D are constants.(3-47) we get the linear.1. and C D = 0. Cd. The University of Akron Fk = 6π µ v ∞ R This expression is known as Stoke's Law. Defining the Reynold's number as Re = d p v∞ ρ (3-57) µ (3-58) where d p is the particle diameter. Drag coefficient for spheres versus Reynolds number. The approximate curves are.1 0. (HANDOUT 3. C D = 24 / Rep (Stoke’s Law range for Rep<1).001 0.5 / Rep (Intermediate range for 1<Rep<1000). George G.12 for flow around spheres. defined by expression (3-42) may be combined with Eqs.(3-57) and (3-58) to derive Cd = 24 Rep (3-59) which is the Stoke's Law condition for the drag coefficient and holds for Rep less than one.SOLIDS NOTES 3. The drag coefficient.44 (Newton’s Law range 1000<Rep<100.01 0.1 1 10 Re 100 1000 10000 100000 Figure 3.12. For larger Reynolds numbers we need to use correlations obtained from experiments.000). we seek a correlation to relate the drag force to the Reynold's number. from left 3/ 5 to right. 3-20 . A correlation would allow us to extend our applications to flow conditions in which the creeping flow solution does not apply. A number of references give the familiar drag coefficient correlation as shown in Figure 3. Three approximate curves are overlayed onto the experimental curve. C D = 18.9) 100000 Expl curve 10000 Stokes Intermediate 1000 Newton Law Cd 100 10 1 0. Chase. Chase.SOLIDS NOTES 3. above Rep of about 105 there is a sudden decrease in the drag coefficient. size.44 yields ut = 173 . There are many other factors that may become significant in given situations..08.(3-60).4 of the text by Coulson and Richardson (Chemical Engineering. Reynold’s number). These correlations only relate the motion to a few of the important factors (density. where CD is a constant.3 in Coulson and Richardson (ibid) show the transition from smooth. The University of Akron For Reynold’s numbers less than 1 Stokes Law applies and this is known as Stoke’s Law range. In the book notation Re’=Rep and 2Cd=R’/ρu2. As can be seen in Figure 3. into the turbulent ranges and the formation of fluid eddies as the boundary layer separates from the particle surface. Figures 3. For Reynolds numbers greater than about 1000 and less than 105. Vol.(3-43) we can solve for the terminal velocity of the particle to be ut = 4 3 d p g ⎛ ρp − ρ ⎞ ⎟ ⎜ CD ⎝ ρ ⎠ (3-60) which applies to all flow regimes. 2. Eq. At the highest flow range new mechanisms can become important as the fluid separates away from the particle surface and cause the observed decrease in the drag coefficient. magnetic fields) sound waves rigid vs. 4th ed. Rep >105 we get Cd=0. Similarly in Newton’s Law range substitution of CD = 0. George G. Pergamon. Between these two ranges is known as the intermediate range. These include • • • • • • • • proximity to vessel walls particle surface roughness particle shape Brownian motion (for dp < 1 µm) external forces (electrical current. 1991). well behaved laminar flow (Stoke’s regime). deformable particles (ie. If we rearrange Eq. droplets) particle concentration The last topic in the list will be discussed further in a later section. d p g ρp − ρ ( ) ρ (3-62) Literature references have other correlations for representing these various ranges of Reynolds numbers. we get the terminal velocity to be ut = 2 gd p ρp − ρ ( ) 18µ (3-61) in Stoke’s Law range. When we substitute in Stoke’s Law.2 and 3. this is known and Newton’s Law range. 3-21 . John Wiley. 1950) and calculated the relationships plotted in Figure 3-10. pg 5-54).65 0.Y. George G.65 0. rounded Wilcox sand. John Wiley. natural (up to 3/8 inch) Glass.3.6 0. crushed Mica flakes Sand Average for various types Flint sand.SOLIDS NOTES 3. N.3 0. disks..28 0. especially when in the chemical process industry particles in settling operations tumble and rotate.83 0.8 Kunii and Levenspiel (Fluidization Engineering.Y.65 0. 3-22 . and cylinders.Y. Chase. Kunii and Levenspeil studied this problem and developed a correlation based upon sphericity (1966).. jagged Most crushed materials SPHERICITY 1. McGraw-Hill.0 0. N. pg945.5 Drag Force on Non-Spherical Particles The shape and orientation of the particle has an important effect on the flow profiles around the particle. jagged Sand. 2001).. Figure 7. (6th ed. N. 6th ed. McCabe and Smith (Unit Operations of Chemical Engineering.G.. PARTICLE MATERIAL Sphere Cube Short Cylinder (Length=Diameter) Berl saddles Raschig rings Coal dust.75 0. Sphericity is a measure of how close a particle is to being a sphere defined as Φ= surface area of a sphere with same volume as the particle actual surface area of the particle (3-63) The sphericity of some common materials are given in Table 3.81 0.Y.11. pg 77) took data from Brown (G. (HANDOUT 3. N. HANDOUT 3. relating Cd to Rep.3 0. and Perry’s Chemical Engineer’s Handbook. Perry’s Handbook 6th ed.10).al. 1984) Figure 5-76) show the correlation for the drag coefficients for spheres..3. 1969. The University of Akron 3. 6th ed.6 to 0. Unit Operations. Table 3-3 Sphericity of Some Common Materials (McCabe & Smith. Brown et.87 0. McGraw-Hill. It is not practical to try to derive correlations for all particle shapes and orientations. (3-64) the terminal velocity can be calculated from the material properties and the sphericity.E+07 Plot to determine drag coefficients of irregularly shaped particles at terminal velocity. The particles are randomly oriented relative to the flow direction.0 1.E. George G. Chase.E+08 1.E+04 1.6 0. The particle diameter is the volume equivalent diameter.(3-43) and the definition of the Reynold’s Number we get Cd Rep 2 (ρ p − ρ ) ⎞⎛ d p uρ ⎞ ⎛8 =⎜ ⎜ 3 Rg ρu 2 ⎟ ⎟⎜ ⎜ µ ⎟ ⎟ ⎝ ⎠⎝ ⎠ 3 ( ) ρ ρ ρ gd − p p =4 3 2 =4 N GA 3 2 µ (3-64) where N GA = gd 3 ρ ( ρ s − ρ ) µ2 (3-65) is known as the Galileo number.E+00 1.E+05 1.E+09 1.E+06 1. SPHERICITY 0. and 3-61 are used with this chart. xv.4 0. With this chart and the correlation in Eq.E+01 1. Equations 3-54.E+02 1.E+02 1. The University of Akron It turns out that the product Cd Rep 2 is independent of velocity.E+03 1. Using Eq.E+01 1.E+10 1.01 1.E+03 1.E+04 Rep Figure 3-13. Shape is accounted for by the sphericity.8 1 . 1. 3-23 . Drag coefficient – Reynolds number relationship for non-spherical particles.2 0.E+00 1.SOLIDS NOTES 3. which makes it convenient for calculations. 3-60. of the sphere with the same volume as the particle. Φ⎤ ut* = ⎢ *2 + ⎥ *0. (3-66) and the dimensionless velocity and particle diameter are defined as ( ) (3-67) 1/ 3 and ⎡ρ ρ − ρ g⎤ p * ⎥ . The University of Akron Haider and Levenspeil (Powder Technology. Butterworth. Chase.0 0.SOLIDS NOTES 3.9 ..335 − 1744 . dp = dp⎢ ⎢ ⎥ µ2 ⎣ ⎦ 100 Sphericity = 1.1 0. George G. 63.. 0. Fluidization Engineering.23 0. Boston. Dimensionless terminal velocity and particle diameter are defined in Eqs.5 ( ) (3-68) 10 1 0.123 0.5 < Φ < 10 .12) where a curve fit of the plot gives ⎡ 18 2. 1989) also found a useful relationship for direct evaluation of terminal velocity of particles.043 0. 2nd.5 dp ⎢ ⎥ ⎣d p ⎦ ⎡ ⎤ ρ2 * ⎢ ⎥ u t = ut ⎢µ ρ − ρ g⎥ p ⎣ ⎦ 1/ 3 −1 for 0.026 Sphericity for Disks only ut* 0.01 1 10 100 dp* 1000 10000 Figure 3-14. 3-24 . 58. The correlation is shown in Figure 3-14 (HANDOUT 3.(3-67) and (3-68). Plot of data taken from Kunii and Levenspiel. 1991. SOLUTION: For a cube Φ = 0. water density = 1. syrup viscosity = 3000 cP.20 x10 − 3 m For the syrup: 2 6 ⎡ ⎤ ⎛ kg ⎞ ⎛ 100cm ⎞ 2 ⎢ (0.0 g/cc.07 = 0. 5mm on each side. 3-25 .95 g/cc. m ⎞ ⎛ −9 3 volume = l = (5mm) ⎜ ⎟ = 125 x10 m ⎝ 1000mm ⎠ 3 3 3 dp = 3 6 π 125x10 − 9 m 3 = 6. Properties: titanium density = 7. s/m and from the figure ut* = 17 hence ut = 0.20 x10 − 3 m⎢ 2 ⎥ ⎛ − 3 kg ⎞ 2 ⎢ ⎥ ⎟ (3000cP) ⎜ 10 ⎢ ⎥ m s cP ⎝ ⎠ ⎣ ⎦ . = 115 1/ 3 * dp ut* 3 ⎡ ⎤ ⎛ kg ⎞ ⎛ 100cm ⎞ 2 (0.07 hence ut = Similarly for the water ut* = ut [ 2551 . The University of Akron EXAMPLE 3-5. syrup density = 0.041m / s . s / m] 1/ 3 From the figure ut* = 0.67m / s . falling through maple syrup and falling through water. 170 .95g / cc) ⎜ ⎟ ⎟⎜ ⎢ ⎥ ⎝ 1000 g ⎠ ⎝ m ⎠ ⎢ ⎥ = ut ⎢ ⎥ ⎛ − 3 kg ⎞ ⎢ (3000cP)⎜ 10 ⎟ (7.81 . then the appropriate diameter is that of a sphere of the same volume.14 g/cc.95 g / cc)⎜ ⎟ (9. Chase.SOLIDS NOTES 3. Since settling depends upon mass average of the particle size.95g / cc)(9.807m / s 2 ) ⎥ ⎟ ⎜ ⎝ 1000 g ⎠ ⎝ m ⎠ ⎢ ⎥ = 6.95g / cc) (7. water viscosity = 1.807m / s 2 ) ⎥ m s cP ⎠ ⎝ ⎢ ⎥ ⎣ ⎦ = ut [170 .14 − 0. George G.0 cP. Compare the terminal velocity of a cube of titanium. s / m] * dp = 243 0. This shows that a change of 3000x in viscosity produces about a 10x change in the terminal velocity.14 − 0. 3-26 . The University of Akron More on Sphericity: We represent a bed of non-spherical particles by a bed of spheres of diameter Deff such that a bed of spheres and a bed of non-spheres have • • The same total surface area. The same fractional voidage.SOLIDS NOTES 3. The sphecific surface area of particles in either bed is found to be as ⎛ Surface area of one particle ⎞ =⎜ ⎟ ⎝ volme of one particle ⎠ π D2 sph / Φ = 3 π Dsph /6 6 = ΦDsph (3-69) For the whole bed ⎛ ⎞ 6(1 − ε ) Surface of all particles a=⎜ ⎟= ΦDsph ⎝ Total volume of particles in the bed ⎠ (3-70) Since there is no general relationship between Deff and d p (particle diameter corresponding to a sphere of the same volume). Deff = ΦDsph . Chase. the effective diameter of the particle is replaced with the sphericity times the defined diameter based on sphericity. but with a length ratio not less than 1:2 (peanut. εbed . estimate the relationship between d p and Deff from Φ values for corresponding disks and cylinders. a in a given volume of the bed. for example) Deff = ΦDsph = Φ 2 d p (3-73) • For very flat or needlelike particles. the best we can do without running experiments is as follows: • For irregular particles with no seemingly longer or shorter dimensions (hence isotropic in irregular shape) Deff = ΦDsph = Φd p • (3-71) For irregular particles with one longer direction . George G. In typical use of the Ergun Equation (McCabe & Smith). but with a length ratio not greater than 2:1 (eggs for example) Deff = ΦDsph = d p (3-72) • For irregular particles with one dimension shorter. This representation ensures almost the same flow resistance in both beds. HANDOUT 3.1 100 10 1 1000 100 10 APPROX. HUMAN HAIR MACRO Radio waves VISIBLE TO EYE 8 7 6 MICRO 1 Millimeter BEACH SAND PARTICLE FILTRATION 5 MICROSCOPE Infrared OPTICAL COAL DUST MILLED FLOUR TOBACCO SMOKE COLLOIDAL SILICA VIRUS YEAST CELLS BACTERIA PAINT PIGMENT POLLEN RED BLOOD CELLS TALC CLAY Visible Ultraviolet MACRO MOLECULAR ELECTRON MICROSCOPE 4 1000 1 Micron1 100 3 10 1 1 Nanometer MICROFILTRATION 200k MOLECULE 2 1 0 X-rays CARBON BLACK PYROGEN ALBUMIN PROTEIN ULTRAFILTRATION 20k 200 100 SUGARS METAL IONS ATOMS AQUEOUS SALTS 1 Angstrom ELECTROPARTICLE SIZE MAGNETIC PARTICLE RANGE LOG SCALE SPECTRUM IONIC REVERSE OSMOSIS COMMON MATERIALS SEPARATION PROCESS . WT. GRAVEL POLYMER POWDERS -LUCITE -GEON -ETC. MOLEC. 0029 0.417 0.065 0.185 0.701 0.094 0.131 0.833 0.023 0.033 0.175 0.17 0.0058 0.495 0.074 0.HANDOUT 3.295 0.36 1.0041 0.028 0.991 0.104 0.0024 0.061 0.043 0.016 0.014 0.0017 0.589 0.039 0.246 0.047 0.0015 .351 0.056 0.65 1.33 2.0097 0.0069 0.2 STANDARD MESH SIZE Tyler 4 6 8 10 12 14 16 20 24 28 32 35 42 48 60 80 100 150 200 250 325 400 US 4 6 8 12 14 16 18 20 25 30 35 40 45 50 60 80 100 140 200 230 325 400 mm 4.012 0.038 Inches 0.70 3.020 0.40 1.147 0. and 2.HANDOUT 3. xd Drag diameter Diameter of sphere that has the same resistance to motions at the same velocity as the particle.. xM Martin’s diameter Length of the line which bisects the projected image of the particle (the two halves of the image have equal areas).2. DEFINITIONS OF STATISTICAL DIAMETERS xF Feret’s diameter Distance between two tangents on opposite sides of the particle. Solid-Liquid Separation. . 1990.2) xA Sieve diameter Largest diameter sphere that can pass through the square aperture of the sieve screen. xf Free-falling diameter Diameter of sphere of same density as the particle with the same free-falling speed in the same liquid. Svarovsky.1. Butterworths. 2. xSt Stoke’s diameter Same as xf but for when Stoke’s Law applies (Re < 0. DEFINITIONS OF EQUIVALENT CIRCLE DIAMETERS xz Projected area diameter Projected area if the particle is resting in a stable position. DEFINITIONS OF EQUIVALENT AND STATISTICAL DIAMETERS. xSV Surface to Volume Ratio Diameter of sphere that has the same surface area to volume ratio as the particle. Symbol Name Definition DEFINITIONS OF EQUIVALENT SPHERE DIAMETERS xv Volume diameter Diameter of sphere with the same volume as the particle. xp Projected area diameter Projected area if the particle is randomly oriented.7 in L. xSH Shear diameter Particle width obtained with an image shearing eyepiece. xc Perimeter diameter Diameter of a sphere with the same projected perimeter as the perimeter of the projected outline of the particle. London. 2. xs Surface diameter Diameter of sphere with the same surface area as the particle. xCH Maximum chord Maximum length of a line limited by the contour of the projected diameter image of the particle.3.3 Taken from Tables 2. 3rd Ed. HANDOUT 3.8 – 200 xv Field flow fractionation 0.3 – 50 Cyclonic (wet or dry) 5 .4 LABORATORY METHODS OF PARTICLE SIZE MEASUREMENTS METHOD APPROX SIZE TYPE SIZE.5 – 30 xd Mesh obscurtion method 5 – 25 xA Laser Doppler phase shift 1 – 10.003 – 3 Equiv laser diameter Scanning infrared laser 3 – 100 Chord length Aerodynamic sizing nozzle flow 0.1 – 40 Equiv laser diameter Photon correlations spectroscopy 0.8 – 150 xz.001 – 100 xd Hydrodynamic chromatography 0.10 xSt. xf Centrifugal sedimentation 0. xCH Electron 0.001 – 5 Gravity sedimentation 2-100 xSt. xf Gravity elutriation (dry) 5 . µm Sieving (wet or dry) xA Woven wire 37-4000 Electro formed 5-120 Microscopy Optical 0. xf Flow Classification xSt. xM xSH.100 Centrifugal elutriation (dry) 2 .) 0.000 Equiv laser diameter Time of transition 150 – 1200 Equiv laser diameter Surface area to volume ratio Calculated xSV Permeametry Hindered settling Gas diffusion Gas adsorption Adsorption from solution Flow microcalorimetry TYPE OF SIZE DISTRIBUTION By mass By number By mass By mass By mass By mass By mass or by number By mass By number Depends upon detector Depends upon detector By volume By volume By number By number By number By number Mean only By number By number mean .01 – 50 xd Fraunhofer diffraction (laser) 1 – 2000 Equiv laser diameter Mie theory light scattering (laser) 0. resist. xF.01 .50 Coulter principle (elect.50 Impactors (dry) 0. Basic components of the Coulter Counter. Voltage and current are measured to quantify the resistance using Ohm’s Law: V = IR.4 to 1200 micrometers. Particles in the aperture bend the electrical current flux lines. Figure 1. As the particles pass through the aperture opening. thus causing a longer length for the current to pass and thus a higher resistance to the current (Figure 2). The resistance to the flow of electrical current through a small aperture is calibrated to the change in resistance depending upon the particle size (Figure 1). APERTURE OPENING WITHOUT PARTICLE APERTURE OPENING WITH PARTICLE Figure 2.5 ELECTRONIC PARTICLE COUNTER The electronic particle counters can measure particle sizes ranging from 0.HANDOUT 3. This method requires the particles to be placed in a stirred electrolyte solution. they bend the current flux lines around the particles. . 473 1.53 1.(3-13) and (3-14) the frequency and cumulative frequency distributions are calculated.33333/0.598 1. fdx is determined by 7/21=0.508 2.75 1. The particle sizes are added up in ∆x increments of 0.18 1.55 cm and are assigned to the average size of 1.952381 0.047619 0.55 1.575 9 2.45 1 0.5 and 1.58 1. The size ranges start with 1.45 to 1.555 2.511 2.725 1 fdx f F 0. F is determined by cumulative summing the values fdx.71 1. . Plot the frequency distribution and the cumulative frequency distribution of the average diameter of the candies.50 cm.475 1 2.35 1.7 Diameter.675 2 3.2 0 1.095238 1.952381 1 Frequency Distribution of M&Ms Frequency Distribution 10 8 6 4 2 0 1.614 1.4 1.63 1. cm Size < Avg size No.525 cm.659 2. All M&Ms of size less than 1.37 1. Mass and diameter distribution of M&M’s.047619 0.565 2.525 7 2. and so on. The results of the summation are plotted in Figure 3-4.586 2.57 1.94 1.6 1.06 1.380952 0.672 1.501 2. Grams Dia.50 are counted in the first increment.590 2. Using an average density of 1.501 2.47 1.047619 0.904762 0.588 2.8 0.18 1. cm Figure 3-4.33333.5 1.668 1.544 1. Using the formulas in Eqs.HANDOUT 3.570 2. The values for nj are determined by counting the number of M&Ms that fall in a given size increment and are assigned to the average size in the increment.36 1.857143 0.55 are in the second increment.571429 0.6 1.6 0. Table 3-1.21 1. f is 0.428571 8.952381 0.540 2.55 1. For example.7 1.01 1.550 2.05 cm.99 1.66667.75 f F 1. Plot of frequency and cumulative frequency distributions for M&M’s.6 EXAMPLE 3-1 A sample of M&M’s ™ with peanuts are weighed as listed in Table 3-1.22 1.65 1.809524 0.05 = 6.59 1. all M&Ms with size between 1.5 1.047619 0.952381 0. there are 7 M&Ms in the size increment range of 1.666667 0.542 2. the average candy diameter (assuming spherical shape) is calculated.49 1.42 1.333333 6.578 2.4 0.23 grams per cubic centimeter. 2.65 1.625 1 2.5 to 1. x h . Comparison of mean size distributions where the various means are defined by: g( x ) = ∫ g ( x )dF 1 0 g(x) = x x2 x3 log x 1/x NAME OF MEAN ARITHMETIC MEAN.7 MODE HARMONIC MEAN ARITHMETIC MEAN MEDIAN f QUADRATIC MEAN CUBIC MEAN f x Figure 3. x c GEOMETRIC MEAN.5. CUBIC MEAN. xq xg HARMONIC MEAN.HANDOUT 3. x a QUADRATIC MEAN. 00 2.10 0.14 192.08 0.04 0.36 0.10 0.59 72.45 0.88 0.42 0.92 0.06 0.01 0.10 0.01 0.10 0.20 0.20 0.14 0. number and area fractions are calculated. MM SIZE.29 0.23 0.14 0.35 0.00 0.00 Comparison of the fractional distributions of the particle size distributions.23 0.60 1.67 0.00 1.58 0.15 0.60 0.04 0.60 0.8 Sieve analysis of a sample of particles. and area fractions are calculated.31 0.70 0.05 0.20 0.38 0.00 0.10 0.14 0.11 153.40 0. TRAY.20 0.18 0.06 5660.03 38.64 27.00 0.00 58141.80 1.01 518.HANDOUT 3.71 0.85 0.24 0.40 0.19 0.03 3.19 0.30 0.00 153782.00 1384.00 Mass & Volume Frac Number Frac Area Frac Avg Particle Size.20 0.25 0. Sieve analysis of a sample of particles.06 76.82 1.50 0.03 38.10 0.15 0.06 1.03 0.69 45.26 0. 0.11 0.16 0.98 0.46 0.19 269.01 0.25 0.42 0.19 269.18 0.05 0. Note 1 Note 2 VOLUME AREA SIEVE AVG SIEVE MASS ON VOLUME V1 NUMBER NUMBER A1 TRAY AREA SIZE.90 0.06 1286. mm .50 0.00 0. number.00 0.00 67293. Mass.46 0. Mass.60 320.50 0.02 1243.11 0.25 346.79 254.01 1.44 0. MASS.00 TOTAL MASS 3.92 0.83 0.40 504.13 0.65 0.17 982.25 0.05 0.59 0.40 0. MM g FRAC MM^3 FRAC MM^3 FRAC MM^2 MM^2 FRAC pan 0 0.30 Fraction 0.21 1266.00 4889.70 0.01 21045.01 1.63 10.62 1. 000).44 (Newton’s Law range 1000<Rep<100.HANDOUT 3.1 0. and C D = 0.9 100000 Expl curve 10000 Stokes Intermediate 1000 Newton Law Cd 100 10 1 0.1 1 10 Re 100 1000 10000 100000 Figure 3.5 / Rep (Intermediate range for 3/ 5 1<Rep<1000).9. C D = 18. . The three approximate curves from left to right are C D = 24 / Rep (Stoke’s Law range for Rep<1). Drag coefficient for spheres versus Reynolds number.001 0.01 0. 6 to 0. jagged 0.10 Table 3-3 Sphericity of Some Common Materials (McCabe & Smith. jagged 0.HANDOUT 3. pg928.75 Flint sand.0 Cube 0. pg 5-54).83 Wilcox sand.87 Berl saddles 0.65 Glass. 5th ed.65 Mica flakes 0.28 Sand Average for various types 0. crushed 0. PARTICLE MATERIAL SPHERICITY Sphere 1.3 Raschig rings 0. natural (up to 3/8 inch) 0.6 Most crushed materials 0.8 . rounded 0.65 Sand.81 Short Cylinder (Length=Diameter) 0. Perry’s Handbook 6th ed.3 Coal dust. E+04 Rep Where 2 C d Rep =4 3 N GA Rep = ρd p u µ N GA = 3 dp ρ (ρ p − ρ )g µ2 D p is the equivalent diameter of a sphere with the same volume as the particles.11 1.2 0.E+03 1.E+09 1.E+06 CdRep^2 1.HANDOUT 3.E+10 1.E+07 Plot to determine drag coefficients of irregularly shaped particles at terminal velocity.0 1.E+00 1. Shape is accounted for by the sphericity.E+02 1.E+01 1.E+08 1.E+00 1.E+01 1.E+03 1.8 1.E-01 1. SPHERICITY 0.6 0. .E+04 1. The particles are randomly oriented relative to the flow direction.E+05 1.E+02 1. xv.4 0. 1 0. 2ed Butterworth.12 100 Sphericity = 1.. page 81 Where 2 ⎡ ⎤ ρg ut * = ut ⎢ ⎥ ⎢ µ (ρ s − ρ g )g ⎦ ⎥ ⎣ 1/ 3 and ⎡ ρ g (ρ s − ρ g )g ⎤ d p* = d p ⎢ ⎥ µ2 ⎣ ⎦ 1/ 3 .01 1 10 100 dp* 1000 10000 Date taken from Kunii & Levenspiel Fluidization Engineering.5 10 1 0.HANDOUT 3.23 0.9 . 0. Boston.0 0.123 0. 1991.026 Sphericity for Disks only ut* 0.043 0.. SOLIDS NOTES 4. Chase. as a first approximation. George G. Qs and Qf.3 Effective Heat and Mass Transport Properties Where available you should always use experimentally measured values for thermal conductivity and diffusivity. A thin differential element of a slurry with differential thickness ∆x . and a continuous fluid phase with conductivity kf. The surfaces of the two phases in contact with the boundaries at each side of the differential element are Af and As. ks. 4-17 . 4. This is analogous to saying that in the ideal model. At zero flow conditions. several correlations are suggested below. Q is the sum of the total heat fluxes through the separate phases. you might expect that the thermal energy will pass through the slurry with an effective thermal conductivity that is proportional to the volume fraction of the materials. The ratio of fluid area to total area equals the porosity. The University of Akron 4. as shown in Figure 4-11 that the total heat flux. ε = A f A . For first estimates and where experimental data are not available.1 THERMAL CONDUCTIVITY Suppose you have a slurry of solid particles having a thermal conductivity. Hence we conclude that the effective thermal conductivity is given by k 0 = εk f + (1 − ε )k s T0 (4-32) Surface Temperature Direction of Heat T1 Flow Area of Fluid Phase Af Fluid Phase Area of Solid Phase As Solid Phase ∆x Figure 4-11 . The temperature change across the differential element is given by ∆T = T1 − T0 .3. as given by Q = = = Q f + Qs ∆T Af k f ∆x + As k s ∆T ∆x ∆T ∆x (4-31) A(εk f + (1 − ε )k s ) It is assumed in this idealized case that the temperature profiles through the differential element are linear and that the heat flux is only in the x-direction. Chase.(4-33) may be adequate.(4-32) by setting ks equal to zero. Maxwell.4 0.8 0. 1.8 1 Ideal Maxwell Porosity. 1 0. A Treatise on Electricity and Magnetism.(4-33) and (4-34). The University of Akron If the solid phase is non-conducting then one would expect the effective thermal conductivity to be related to the porosity by ko =ε kf (4-33) as deduced from Eq. 3rd ed. ε Figure 4-12.1 0 0 0..3 0.9 0.5 0.SOLIDS NOTES 4. New York.2 0. From the plot we see that the idealized case from Eq. George G. Vol.2 0. Dover.4 to 0. Maxwell (J.6 0.C. In many engineering applications Eq. Effective conductivity versus porosity based on Eqs.6 porosity range. 4-18 .7 0.4 0. 1954) experimentally tested the analogous electrical conductivity problem and derived the correlation k o ⎛ 2ε ⎞ =⎜ ⎟ kf ⎝3−ε ⎠ (4-34) Equations (4-33) and (4-34) are plotted in Figure 4-12.6 k 0/k f 0.(4-33) follows the same trend as determined from Maxwell and over predicts by about 20% in the 0. 023R P 0.SOLIDS NOTES 4. The bulk density and viscosity were R r e 0 k0 µ k0 discussed previously. and .3. = . h.(4-35) the Reynolds number is defined as. this expression is limited to packed beds of porous inorganic materials such as alumina and silica gel with a porosity of approximately 0. the slurry mass transfer coefficients at the wall are analogous to those given by Eqs. The bulk heat capacity is given by where N u = 0 Cp = εC pf + (1 − ε )C ps (4-37) 4. The heat transfer Nusselt number Nu is replaced by the mass transfer Sherwood number Sh.3 HEAT TRANSFER COEFFICIENT For gas phase flow through a packed bed of solids.3. Unfortunately. The University of Akron 4. George G.3. For dilute slurries.3.4 MASS TRANSFER COEFFICIENT By analogy.5 0. Nu = hd p kg 0.333 r ⎛ µ0 ⎞ ⎜ ⎟ ⎜ µ0 ⎟ ⎝ w⎠ 0. and the Prandtl number Pr is replaced by the Schmidt number Sc.8 e 0. where G is the mass flux (mass per area of bed per time).14 (4-36) 0 0 Cp µ ρ 0VD hD = P . By replacing the thermal conductivity terms with diffusivity terms in Eqs.2 MASS DIFFUSIVITY Mass diffusivity in a multiphase system is analogous to thermal conductivity. In the Sieder Tate equation. Rep = Gd p µ . is estimated by the following empirical equation (McCabe and Smith. the Sieder Tate equation can be used as an approximation. the wall heat transfer coefficient.94 Rep Pr (4-35) In Eq. the bulk slurry properties are substituted for the fluid properties: N u = 0.33 = 1. These numbers are defined by Sh = kx D cDAB (4-38) 4-19 .(4-35) and (4-36) for heat transfer coefficients. ibid). 4. Chase. for turbulent flow in a tube. Other correlations are needed to take into account the wide range of porosities that are possible with slurries.(4-31) and (4-34) you have the analogous expressions relating effective diffusivity to the porosity and the diffusivities of the individual phases. Chase. kg. The University of Akron Sc = µ ρDAB (4-39) 4. 4-20 . DISPLACED FLUID FALLING PARTICLES Figure 4-13. or an impregnated catalyst. This is shown in Figure 4-13.SOLIDS NOTES 4. For example.1Rep Pr kg (4-40) This shows that the effective thermal conductivity is affected by the flow rate. silica gel.5 DISPERSIVITY (FLOW EFFECTS) Numerous correlations are available in references and textbooks. Hindered settling: as a particle falls its displaced fluid moves upward and slows the observed settling rate of neighboring particles. In a concentrated system this causes an upward fluid motion which interferes with the motion of other particles.4 Hindered Settling What happens when particles settle in concentrated solutions? As each particles falls it displaces fluid which in turn must move upward. George G. 4. due to a dispersion mechanism. the effective thermal conductivity of the bed is proportional to the gas phase thermal conductivity. Dispersion is caused by the fluid following tortuous paths and becoming intermixed in the lateral directions normal to the prevailing direction of flow. in McCabe and Smith (ibid) for a porous inorganic material such as alumina.3. given by k0 ≈ 5 + 0. C D Figure 4-14. the particles are no longer settling though the sediment may compact due to the weight of the overburden. Maude & Whitmore (Br.D interface is approximately that of “loose packing” as given by the correlation in Figure 4-5. In zone D the sediment has particles in contact with each other. Not all four zones are present in all settling processes.5 for spherical particles and has a dramatic effect on the calculated values for the hindered settling velocity. T. Volume 1. Unfortunately n is not a constant but varies as a function of the particle geometry and the Reynolds number. In zones B and C the particles are in hindered settling. 1997. page 224) notes that zone B settles in mass and the relative motion of fluid to particles is analogous to flow through a packed bed. Phy. hindered. Coe and Clevenger (Trans. 55. 477-482. 1958) modeled the hindered settling process as a power law in the concentration (volume fraction of the liquid phase) us = ut ε n (4-41) where for dilute solutions ε → 1 and us → ut . Allen (Particle Size Measurement. Inst. The concentration of the particles in zone D near the C . Zone D = sediment. A B Zone A = clear liquid zone. Min. Appl. Zone C = variable composition zone. J. Here u t is calculated as in Chapter 3 for a single particle falling through a clear fluid and ε n accounts for the hindered settling effects. The parameter n is determined experimentally. 356. Perry’s Handbook (6th ed. Met. Eng. settling rate for the group of particles as compared to the free settling terminal velocity of one particle by itself. 1916) observed that during a batch settling operation. 5th ed. The University of Akron The net effect is a slower.SOLIDS NOTES 4. Chase. George G. pg 5-68) shows that n varies from 2. Chapman & Hall.3 to 4. In zone A the particles are in low concentration and settle at their terminal velocity. hence Eq. the sedimenting fluids develop several “zones” (Figure 4-14). London. 9. Am. 4-21 . Zone B = constant composition zone. Zones of settling observed by Coe & Clevenger. (4-25) could be used here to model the motion of zone B. Let the z-direction be the direction of gravity. This approach is a preferred alternative to the Maude & Whitmore approach. George G. larger volumes of fluid are displaced causing an upward fluid velocity. ut = v s − v f . We can write this as 2 ⎞ s ⎛ πd p f ⎜ ⎟ v = − Av ⎜ 4 ⎟ ⎠ ⎝ (4-44) where A is large and hence v f is small compared to v s . In hindered settling the volume rate of flow of particles is related to the fluid phase velocity through the solid phase volume fraction (1 − ε ) and the vessel cross sectional area. then the particles have a positive velocity in the z-direction and the fluid has a negative velocity opposite to the direction of gravity. The University of Akron In the section that follows a rational approach to hindered settling is described in which the particle settles through the slurry instead of the clear fluid. At steady state the volume rate of flow of particles downward must equal the volume rate of flow of fluid upward.4. A.SOLIDS NOTES 4. Fk = C d A KE 2 ⎛π dp ⎞⎛ 1 ⎞ = CD ⎜ ρ ut2 ⎟ ⎜ 4 ⎟ ⎟⎜ ⎠ ⎝ ⎠⎝ 2 (4-42) where ut is the observed velocity of the particle relative to the stationary vessel walls. ut actually represents the velocity difference between the particle and the stationary fluid phase. Chase. The settling velocity to an external observer is different than the effective velocity difference be the two phases. 4. by (1 − ε )Av s = −εAv f or (4-45) 4-22 . (b) In high concentrations. (4-43) When settling occurs in a large vessel of cross-sectional area A the displaced fluid velocity is negligible. and (c) Velocity gradients in the fluid near the particle surfaces are increased as a result of the concentration of the particles. Terminal velocity of a single particle is correlated through the drag coefficient CD as given by the defining equation relating the kinetic force acting on the particle and the particle’s kinetic energy.1 RATIONAL ANALYSIS OF HINDERED SETTLING The primary reasons for the phenomena of hindered settling are: (a) Large particles fall at a different rate relative to a suspension of smaller particles so that the effective density and viscosity of the fluid are increased. then we must take into account the upward motion of the fluid phase as done in Eq. In the limit when only one particle is present. 236.SOLIDS NOTES 4. Geankoplis (Transport Processes and Unit Operations. Englewood Cliffs. Mech. (4-48) and the velocity observed by an external observer is This is consistent with the extreme case of dilute solutions. George G. (4-46) Since the terminal velocity is the relative velocity difference between the solid particle downward motion and the fluid phase upward motion. in the Stoke’s Law range the observed velocity is given by the modifying Stoke’s Law. 3ed. Chase. ε → 1. Davis and Hill (J. the observed velocity approaches the terminal velocity. then ut = vs − v f (1 − ε ) s = vs + v = vs ε (4-47) ε v s = ε ut .(3-61) to be v =u = s o t 2 ρp − ρ0 gd p ( 18µ 0 ). For neutrally buoyant particles in the slurry (but the falling particle is not neutrally buoyant). v s → ut . 1992) studied hindered settling with spheres falling through slurries of neutrally buoyant particles. pg 820) suggests that we replace the fluid phase density and viscosity. ε . (4-51) If all of the particles in the surrounding slurry are also setting. 513-533. The results of their work show the velocity effects are nearly independent of particle size ratio. as related through Eqs. where ρ 0 = ε ρ + (1 − ε )ρ p and (4-49) (4-50) µ 0 = µ f (ε ) where f (ε ) is a function of the fluid phase volume fraction. Prentice Hall. For more concentrated solutions the particles interfere with the drag coefficients on each other. Their work assumes Brownian motion and interparticle attractive/repulsive forces are negligible. ρ o and µ o . Fluid. The University of Akron vf =− (1 − ε ) ε vs.(4-3) to (4-9). ρ and µ in the hindered settling correlations with the slurry bulk density and bulk viscosity. Effectively we are saying that the fall of a single particle in a slurry is the same as if all of the other particles in the slurry are part of the surrounding fluid phase.(4-40) which gives 4-23 . 1993. Eq. (4-3) through (4-9). Since the velocities are different. we must relate all of the velocities to the volumetric rate of displacement. hence we are actually interested in the volumetric flow rate so we can relate it to the fluid phase mass. Let v is be the observed velocity of the ith type of particles (of size d pi and density ρi ) which occupy a solid phase volume fraction ε is (volume of all ith particles divided by total volume). Qs = Aε s v s = A∑ ε is v is (4-54) The fluid phase displacement is given by Q f = Aεv f = − Q s (4-55) We are not interested in a mass average solid phase velocity. Chase. The terminal velocity of the ith size particle is uti = v is − v f which can be manipulated as (4-56) 4-24 . The drag coefficient correlation reflects the fact that we are accelerating the fluid around the particle. George G. How can we approach this problem if the particles have a variation in size and density? If there is a variation in the size or density of the particles. ε gd p 18µ 0 (4-52) This assumes all of the particles are approximately the same size and density.SOLIDS NOTES 4. The bulk density becomes ρ 0 = ε ρ + ∑ ε is ρ i (4-53) where ∑ ε is = ε s = (1 − ε ) is the total volume fraction occupied by the solid phase. We have no additional information on the bulk viscosity so we use the same models as given in Eqs. The University of Akron v s = ε u to = 2 (ρ p − ρ 0 ) . then the different types of particles will settle at different rates. (4-58) one may use the modified forms (substitute bulk density and bulk viscosity for the fluid properties) of Stoke’s Law. we get the expression ε uti − ∑ ε js v sj v is = (ε s i + ε) j≠i (4-58) which is the expression that applies to settling Zone C.(4-58) reduces to the solids velocity equals the terminal velocity.SOLIDS NOTES 4. has a volume fraction of essentially zero. ε1s → 0 . 4-25 . Since there is only one particle. the neutrally buoyant particles affect the bulk viscosity of the slurry. there is only one particle size. Equation (4-58) reduces to v1s = (1)u t1 − 0 = u t1 (0 + 1) (4-59) where the summation term ∑ε j≠i s j v sj is zero because no terms exist for j ≠ 1 . as expected. 4. Chase. This is the simplest case. depending upon the Reynold’s number for the particle size. The settling particle. In this case.2 CASE STUDY COMPARISONS IN HINDERED SETTLING CASE 1: One particle in free settling. and the fluid viscosity is the same as the bulk viscosity. The neutrally buoyant particles are of a concentration represented by volume fraction ε 2s ≠ 0 which is non zero.(4-50) is the most general form we can apply it to several example cases to demonstrate its utility. Newton’s Law or intermediate range. as in Case 1. Hence Eq. George G. ρ = ρ 0 . µ = µ 0 . the fluid density is the same as the bulk density. The University of Akron uti = v is + =v s i ∑ Aε + s i Qs Aε s i v is Aε (4-57) s j ⎛ ε ⎞ = v is ⎜1 + ⎟ + ε⎠ ⎝ ∑ε j≠i v sj ε or upon rearrangement.4. In estimating values for u ti in Eq. but not the bulk density. Since Eq. The fluid occupies a significantly greater volume hence ε = 1 and ε1s = 0 . CASE 2: One particle settling in a slurry of neutrally buoyant particles. In this case there are two particle sizes. As a consistency check. This case is more complex than Case 2 because now there are many particles settling. v1s = (1 − ε1s ) ut1 − ε 2s ut 2 . Eq. The observed velocity becomes v1s = ε ut 1 where the terminal velocity is a function of the bulk density and viscosity. Equation (4-58) gives the estimated velocities to be v1s = ε ut1 − ε 2s 0 = ut1 (0 + ε ) (4-60) Where the velocity of the neutrally buoyant particles is zero because the remain motionless with the fluid phase. the summation term in the numerator of Eq. The volume fractions are related by 1 = ε + ε1s + ε 2s .(4-64) to eliminate the velocity of particle 2 from the right side of Eq. the velocity of the neutrally buoyant particles to be zero.(4-58) to be s ε ut1 − ε 2s v2 v = (1 − ε 2s ) s 1 (4-61) (4-62) (4-63) (4-64) and s = v2 ε ut 2 − ε1s v1s (1 − ε1s ) where we see that the two velocities are interdependent. There are no neutrally buoyant particles present in this system. CASE 4: Two sizes of particles settling at different velocities. v 2 CASE 3: Group of particles of same size all settling at the same rate. If we use Eq. This is the most complex case that will be considered here. not just one particle. This results in 1 = ε + ε1s . The University of Akron Since all of the volume fractions must sum to 1. Since only one type of particle is present in the system. George G. This tells us that the particle settles at its modified terminal velocity rate where u t1 = u t1 (ρ 0 . denoted 1 and 2. µ 0 ). The observed velocities are determined from Eq.(4-63) then we get the velocity of particle 1 as a function of the terminal velocities of both particles.(4-50) summed over j ≠ i is identically zero.SOLIDS NOTES 4. The volume fraction.(4-58) gives s = 0. we get ε = (1 − ε 2s ) . Neither of the particles are neutrally buoyant. ε1s occupied by the particles is not zero. (4-65) 4-26 . Chase. The University of Akron With Eq.18 g/cm3) and sand (1 mm.(3-61) modified with the bulk density and viscosity u latex = 2 gd latex (ρ latex − ρ 0 ) 18µ 0 .394x10-5 m/s using Eq. Richardson.” Trans.4) the slurry bulk viscosity is estimated to be 1. Estimate the observed velocities of the latex and the sand. the terminal velocity of the latex particles is calculated to be 4. the bulk density is calculated using Eq. The limitations of the results are reiterated here for emphasis. the terminal velocity for the sand is estimated from Stoke’s Law to be 0. given by Eq.0013 using Re = ρ 0 u latex d latex µ0 .05 as the The mixture is uniformly mixed (initially) with ε latex volume fractions occupied by each type of particle.(3-62) modified with the bulk density and viscosity gives a terminal velocity of 0. The water viscosity is 1. The water volume fraction is s s given by ε = 1 − ε latex − ε sand = 0. and R. A check of Newton’s Law range. intrinsic density of 2. (4-66) The Stoke’s Law assumption is checked by the Reynold’s number which calculates to be 0. If there are any disturbances in the slurry then the eddy currents will disrupt the flow patterns. (4-67) Similarly. “Sedimentation and Fluidization Part III. See for example: J. George G. the intermediate range Eq. Engr. The Sedimentation of Uniform Fine Particles and of TwoComponent Mixtures of Solids. Inst Chem.85 . Suppose a mixture of latex spheres (39 microns.(3-60) must be used.(4-65) we can calculate the observed settling velocities for the two types of particles present in the slurry. 39. EXAMPLE 4-3.SOLIDS NOTES 4. Chase. Meikle.0 cP and using Eq.1649 m/s and Reynold’s number of 150. Sedimentation of multicomponent mixtures has been the subject of numerous papers in literature. these results only hold as long as the volume fractions are constant. also indicates the intermediate range.5 g/cm3) are settling in water at room temperature.A. Calculate the observed settling velocities of a mixture of latex spheres and sand in water. 348-356.(4.(4-2) to be 1093 kg/m3. Hence. s s = 01 . 1961. F. These results assume laminar. slow flow conditions. which is modified as 4-27 .462 m/s but the Reynold’s number is 421 which places the sand in the intermediate range. With the water density of 1000 kg/m3.409 cP. the concentrations will vary with time and position. Assuming Stoke’s law range. Also. and ε sand = 0. intrinsic density of 1. The settling process is inherently unsteady hence these results only hold at the moment in time that the concentrations are those used in the equations. 240255 0.38602 1.(4-57) the observable velocities of the latex and the sand are s s = 0. 215.128614 99. 4-28 .5046 0. The negative value for the v sand observed latex velocity indicates that the velocity is upward! The observable velocity of the water may be determined from Eq.017591 0.37079 1.218523 0.SOLIDS NOTES 4.120002 0.116492 0. George G.11649 m/s.(4-58) with some manipulation as vf =− ∑ε s s i i v = −0.116583 0. calculate the drag coefficient.239837 0.240418 0. Since the terminal velocity in Eq.238446 0.116498 0.35 271. Chase. then an iterative solution is needed.70) for which the minus sign indicates upward flow.08867 1.43677 1.116498 90. Using successive substitution a guess for the terminal velocity is used to calculate the Reynold’s number. and calculate a new guess for the velocity.64107 0.60617 1. Using Eq.16204 125.116518 0.16204 0.240429 0.116802 90.36486 1.6987 1. 1995) argues that in sedimentation the smallest particles may move upward with this fluid motion as indicated by the negative fluid and latex velocities.76932 1. This set of calculations is easily computed using a computer spreadsheet.116492 90.116583 90.36623 1.116802 0. Indust. The University of Akron u sand = 4 3 d sand g ⎛ ρ sand − ρ 0 ⎞ ⎜ ⎟ ⎟ CD ⎜ ρ0 ⎝ ⎠ (4-68) with the drag coefficient given by 3/ 5 C D = 18.(4-68) depends upon itself.5 / Rep (4-69) from Figure 3-12.11649 0.11649 Hence the terminal velocity of the sand is u sand = 0.116518 90.24038 0.17311 1. Chem. through the Reynold’s number.16889 0.23382 0.00578 m/s. guess u Re Cd calc u 0.11066 m/s and v latex = −0. 23. Flotats (Hungarian J.117532 91.120002 93.128614 0.117532 0.11649 90.00617 m / s ε (4. M. The ER fluid is made up of a non-conducting liquid phase and a conducting or semiconducting particulate phase. This is shown in Figure 4-15. 1983) and the Encyclopedia of Fluid Mechanics are two examples of good references that discuss slurry flows. Roco ed. Therefore the bulk viscosity correlations described previously in Section 4. Some slurries can be influenced by electrical and magnetic forces that change their flow behavior. Figure 4-15. Gupta. Electrorheological fluids (H.P. Michigan. Another way that slurry behavior differs from the carrier fluid is when phase separation occurs.T.. Boston. Ann Arbor.C. With no electrical field the fluid flows like a Newtonian. even when the suspensions are spherical. “Structure and Mechanisms of ElectroRheological (ER) Fluids. The mechanism by which this works is due to the particles lining up when an electrical field is applied.Y.”Chapter 12.SOLIDS NOTES 4. This is especially true for gas-liquid systems.5 Slurry Flows Cheremisinoff’s book (N. Handbook of Fluids in Motion. Butterworth. This is one example of how slurry behavior is not a simple extension of the carrier fluid’s properties. Many concentrated suspension display non-Newtonian flow behavior. Conrad. George G. 1993) are fluids that change properties under the influence of electrical fields. with the electrical field the fluid flows like a yield-stress fluid. Chase. “Rheology of 4-29 . Kao (D. Cheremisinoff and R.2 must be used with some caution. The University of Akron 4. Ann Arbor Science. Kao. Alumina particles in a silicone oil line up in fibers when an electrical field of 1000 volts per centimeter is applied. in Particulate Two-Phase Flow. Macro-mixed dispersions have regions of high concentration of particles and regions of low concentrations of particles. 1983) classifies suspension flow behavior as: Single Phase Fine Dispersions Coarse Dispersions Macro-mixed Stratified Homogeneous Pseudo-Homogeneous Heterogeneous Heterogeneous Heterogeneous A single phase by definition is homogeneous. Michigan. in Handbook of Fluids in Motion. Gupta eds. Ann Arbor Science. Fine dispersions are termed pseudo homogeneous because even though two phases are present. N. Ann Arbor. Stratified flows have regions of high concentrations of particles in a layer located at the bottom of a pipe. at a local scale the two phases are well among each other. 4-30 . Chase. Coarse dispersions are typically those with large particles (relative to the size of the slurry pipeline) and the larger particles may have a tendency to settle quickly. for example. and low concentration at the top of the pipe. George G. The University of Akron Suspensions.. because its material content does not vary with position.SOLIDS NOTES 4. and these regions are located somewhat randomly throughout the slurry. Cheremisinoff and R.” Chapter 33.P. of (a) a pile of powder. The University of Akron 4. Many factors can influence the way particles stack hence it is difficult to predict. α c. α a. particle intrinsic density. There are more appropriate methods for designing hoppers which will be discussed in detail later.1 Brief Overview of Some Bulk Properties 4. The “collective” properties are the measurable properties of groups of particles. The angle of repose is considered to be mostly a measure of the internal friction between the particles as a whole.1. and (c) powder in a rolling drum. The way that particles stack when poured into a pile is a function of the size/shape. Figure 4-1. but not between individual particles. surface forces (stickiness. It is used in a number of correlations and estimates for the behavior properties of the bulk solids. normally a simple measurement can be made to determine the angle of repose. 4-1 . One example given in Coulson and Richardson’s text relates the angle of repose to the height of the longest movable plug in a piston. Some important examples are discussed here. The angle of repose may is often incorrectly be used to estimate the angle required for the bottom of a hopper to ensure proper discharge. Chase. Some of these properties are analogous to properties that are measured on individual particles while other properties may not be defined for individual properties. BULK PROPERTIES OF POWDERS AND SLURRIES In the process industries economics usually requires us to handle and process many particles at a time rather than individually. 4. α b. and roughness of the particles. but the list is not complete. Because of this we must have a working knowledge of the collective or “bulk” properties of these materials. electrostatic). George G. α. Angle of repose.SOLIDS NOTES 4. (b) powder in a container.1 ANGLE OF REPOSE The angle of repose is a characteristic of solids which characterizes the piling or stacking nature of the particles. Figure 4-2. Baldy Mountain. George G. Figures 4-2 and 4-3 show typical piles of such materials. Chase. The top of the mountain is largely loose rock and stone that prevent plants from taking root. either volume or ground area that such a pile will occupy. Figure 4-3. The University of Akron The best use of the angle of repose is to determine the size of a pile of powder or granular materials. New Mexico. Large gravel piles at rock and stone facility near Marblehead. 4-2 . Ohio.SOLIDS NOTES 4. Philmont Boy Scout Ranch. SOLIDS NOTES 4, George G. Chase, The University of Akron 4.1.2 POROSITY (VOLUME FRACTION) Porosity is the fraction of volume space that is NOT occupied by the solid particles. If you have a two phase system of solids and liquid, you can add the respective volumes to obtain the total volume of your mixed system. VSOLIDS + VLIQUUID = VMIXTUR Figure 4-4. Addition of phase volumes yields the total volume. The porosity is defined as the volume fraction of the fluid phase: ε= V LIQUID V MIXTURE (4-1) Since porosity is defined as a fraction it must have a value between 0 and 1 inclusive. ε is the fluid phase volume fraction. (1-ε) is the solid phase volume fraction. Their sum is 1. Handout 12 (Appendix B of Foust et.al., Principles of Unit Operations, Wiley, NY, 1960) provides correlations for estimating porosity of packed beds as a function of sphericity. 4.1.3 BULK DENSITY Bulk density is the effective density of a powder or particulate solid taking into account the volume occupied by both the solid and fluid phases. The bulk density is calculated from the porosity and the intrinsic densities of the fluid and solid phases: ρ o = ε ρ + (1 − ε )ρ p . (4-2) EXAMPLE 4-1. An example of applying bulk density is determining the weight of sand in a bucket. The intrinsic density of one sand particle is about the same as that of glass, 2.6 g/cm3. If sand packs with a porosity of 0.4, how much will a twenty five liter bucket filled level to the top with dry sand weigh? The mass of sand in the bucket is given by Mass = ρ oV BUCKET Applying Eq.(4-2), neglecting mass of the air (air density is about 1/2600 that of the sand), we get ρ o = (1 − 0.4)2600kg / m 3 = 1560kg / m 3 4-3 SOLIDS NOTES 4, George G. Chase, The University of Akron Mass = (1560kg / m 3 )(25l )(m 3 / 1000l ) = 39.0kg Weight = Force = ma / g c = (39.0kg )(9.807m / s 2 ) /(1kg m / N s 2 ) = 382.5 N (or 86 lbf ) 4.2 Momentum Transport Properties There are several properties of dispersed multiphase mixtures that are important to predicting handling and transporting properties. These include bulk viscosity, coefficient of friction, Janssen’s Coefficient, and permeability. 4.2.1 BULK (SLURRY) VISCOSITY Slurries, which are mixtures of fluids and solids, display a number of interesting properties including Bingham Plastic, Power Law, Dilatent, and time dependent behaviors. Bingham Plastic (Yield Stress) flow occurs when particles in the slurry resist motion between each other and with the pipe or container wall. The shear stress must exceed a certain value (the yield value) before the fluid will flow. (see Patel, R.D., "Non-Newtonian Flow," in Handbook of Fluids in Motion, N.P. Cheremisinoff and R. Gupta eds., Chapter 6; Ann Arbor Science, 135-178, 1983). Shear thickening and/or shear thinning behavior can occur in slurries made up of fibrous or granular materials, respectively. Thickening occurs when the particles become interlocked. Thinning occurs when the particle separate to allow movement of the slurry. The particular property I want to discuss is viscosity. Assuming that other effects are negligible, can we predict how the slurry’s effective bulk viscosity changes with solids concentration (or equivalently, porosity)? Einstein (Ann. Phys., 19, 289, 1906; 34, 591, 1911) derived the theory for dilute suspensions of rigid spheres. He shows that the slurry bulk viscosity is related to the liquid viscosity by µ 0 = µ (1 + 2.5(1 − ε )) in the low concentration limit as ε → 1 . Thomas (Ind. Eng. Chem., 45, 87A, 1953) extended Einstein’s relation empirically to higher concentrations of solids (on the order of ε → 0. 95 (4-3) µ 0 = µ 1 + 2.5(1 − ε ) + 10.05(1 − ε )2 + 0.00273e16.6 (1−ε ) which offers a reasonable fit. ( ) (4-4) For highly concentrated slurries, near a critical void fraction at which the particles are in contact forming a structural bed, ε c , the bulk viscosity may be 4-4 SOLIDS NOTES 4, George G. Chase, The University of Akron approximated by (Art Etchels, personal communication, DuPont, Delaware, 1994) ⎛ (1 − ε ) ⎞ µ = µ⎜ ⎟ ⎜1 − (1 − ε ) ⎟ c ⎠ ⎝ 0 −2.5 (4-5) which says that the closer the porosity gets to the critical porosity ε → ε c the more the slurry behaves like a solid structure. Another correlation is Shook Eq. listed in homework problem E3. To estimate the values of ε c one can use the porosity of a loosely packed bed. Foust, Appendix B, gives a correlation between sphericity and porosity and Loose, Normal, and Dense packing (A.S. Foust, L.A. Wenzel, C.W. Clump, L. Maus, and L.B. Andersen, Principles of Unit Operations, Wiley, New York, 1960). The data taken from Foust’s figure is plotted in Figure 4-5 in a slightly different format. (Handout 4.1) 1 0.9 0.8 0.7 Porosity 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Sphericity 1 Loose Normal Dense Figure 4-5. Porosity as a function of sphericity and packing structure. Loose packed materials are ones that have not had much time to settle due to vibrations. A normal packed material is one that may have sat for several days or weeks. Dense packed materials are ones that have sat for a year or more and have had ample time to settle and vibrate into its most densely packed structure under force of gravity. The plot in Figure 4-5 is obtained by curve fitting data. The loose packed porosity is a function of sphericity given by ε Loose = 0.4411Φ 2 − 1.1042Φ + 1.0873 (4-6) All three curves are generated by defining a packing parameter, λ , such that 4-5 1 0.08))ε Loose for any packing condition.74 0.0 9..(4-6) should only be used if experimental data are not available. A plot of the data in Figure 4-6 gives a rough approximation between A and εc as A ≅ 10. Chase.626 0.B. New York.(4-5) is a special case of the Krieger-Dougherty Equation (I. 2nd ed.SOLIDS NOTES 4.297λ (Φ − 0.4 0. (R.5 for Normal Packing ⎪ 1 for Dense Packing ⎩ and the porosity is given by (4-7) ε = (1 − 0.8 0. Bird et.8 3. Rheol. Dougherty.39 0.al. W. Lightfoot. Transport Phenomena.7 0. 4-6 . Eq.25 9.31 0. George G.E. Wiley.629 0.3 0. (4-8) Etchel’s equation. 3.7 3.5 0.25 5.N. 137-152. (Handout 4.107x 2 R = 0. The University of Akron ⎧ 0 for Loose Packing ⎪ λ = ⎨0.65 0. 2002) list values for parameter A from several references.0 εc 0.87 5.29 0. Stewart.618 . 1959) which has the form ⎛ (1 − ε ) ⎞ − µ = µ⎜ 1 ⎜ (1 − ε ) ⎟ ⎟ c ⎠ ⎝ 0 − A (1−ε c ) (4-9) where A is a material parameter.8 5. (4-10).767 10 y = 10.6 0. Krieger and T.2 εc 0.45 0.9 Figure 4-6 Relation between parameter A and εc in Eq.03 6. and E. Bird.Soc.M.1ε c (4-10) The data points in the plot are scattered meaning that Eq.2) 12 Material A 2.28 3.J.0 9.4134 8 6 4 2 Submicron spheres 40 micron spheres Ground Gypsum TiO2 Laterite Glass Rods 30x700µm Glass Rods 100x400µm Quartz grains 53-76 µm Glass fiber aspect ratio 7 Glass fiber aspect ratio 14 Glass fiber aspect ratio 21 A 0 0 0.. Trans.732 . Powders often display fluid-like properties and the “no-slip” boundary condition is sometimes observed. Comparison of boundary conditions for no-slip. Intermediate slip condition: the velocity at the wall is proportional to the shear stress. τ ⋅ n surface = 0 τ xy surface = − µ ∂ vx =0 ∂y (4-12) (4-13) The solids movement is not zero at the wall. 1960.SOLIDS NOTES 4.2 WALL COEFFICIENT OF FRICTION In fluid flow we frequently use the “no-slip” condition at a wall surface to model the boundary condition. No slip condition: the solid particles have the same velocity as the wall (zero in this case). some slippage occurs but the solids movement is slowed as energy is dissipated. perfect-slip and intermediate slip at a wall surface.2. The particles do not stick to the wall. pg 67). The University of Akron 4. Chase.. The intermediate slip condition is one of empirical convenience. From elementary physics 4-7 . This implies that the fluid velocity at the surface of contact with the wall has the same velocity as the wall. In soil mechanics triaxial shear testers can be used to measure the wall stress-strain and slip behavior of particulate solids. et. The frictional forces acting between the particulate solids and the wall surface are complicated functions of the particle geometry. β is determined empirically. 3. For a stationary wall these boundary conditions may be summarized as listed in Table 4-1. or the wall surface may be very rough and hence is a shearing action the particles next to the wall surface get caught in ruts on the surface. Solid particles may stick to walls due to adhesive or electrostatic forces. the wall is slippery (like Teflon coating). Table 4-1. The wall may redirect the flow but it does not slow down or dissipate the kinetic energy of the fluid-like solid particles v =0 s (4-11) 2. 1. Other boundary conditions include total slip condition and an intermediate condition.al. We would like to determine a functional relationship between β and the coefficient of friction given in physics. packing arrangement. Perfect slip condition: the motion of the solid particles parallel to the wall surface are not affected by the surface. and stresses. hence the gradient in the velocity normal to the surface is zero. George G. β v z surface = τ rz surface (4-14) β is the proportionality constant and is a function of the normal stress acting on the wall surface (Bird. The final force to keep the surfaces in relative motion increases with increasing velocity. 3-59. there may be attractive forces (static charge. If you recall the drag coefficient Stokes’ Law for a sphere. but at a sufficient small scale the surfaces are rough and jagged.SOLIDS NOTES 4. Two surfaces that are in contact may appear to be smooth. Once in motion the force required to keep them in motion is less than the initial force. we deduce that in Stoke’s law range the drag force is proportional to the velocity. Eqs. etc. NY. 4-8 . Chase. 1974. The University of Akron we know that frictional forces oppose movement (Haliday & Resnick. The interlocking of the jagged edges between the surfaces causes the friction. Also. Fundamentals of Physics. Hence by analogy we should not be surprised to find the coefficient of friction to be a function of velocity (though for powdered materials it may be a weak function of velocity).) that resist motion. pg 78). Wiley. and 3-42. An initial force is required to overcome the resistances due to the interlocked jagged edges and to cause separation between the surfaces so that they may move relative to each other. 3-58. George G. George G. Effect of forces acting on a block of mass M resting on a horizontal surface. Acceleration stops when F is again balanced by the frictional force. N. (4-19) Figure 4-7. g SURFACE N= -Mg (a) Block of mass M on surface. The static and kinetic forces are functions of the normal force. Once the motion starts the force F required to maintain motion usually decreases. the kinetic friction varies with velocity (otherwise the kinetic friction could never balance the applied force as shown in Figure 4-7d). Motion. the block accelerates. and the motion becomes steady. F fs f s(max) > f k . The University of Akron g BLOCK Consider a block of mass M resting on a horizontal surface (Figure 4-7a). 4-9 . Normally we model the relationship between the forces as f s ≤ µs N f k = µk N F f s(max) N= -Mg (4-16) (4-17) (c) No motion occurs until F exceeds fs(max). g (b) Force F is balance by the static friction fs. (4-15) N= -Mg If force F is held constant. Also. fk. We assume a linear dependence by letting µ k = µ k ' to obtain f k = µ k ' vN .SOLIDS NOTES 4. fk. dv/dt > 0 g F fk where µ s and µ k are the coefficients of static and kinetic friction. holding the block against the surface. the block will accelerate due to the imbalance between the forces until the kinetic frictional force increases to balance F (Figure 4-7d). N= -Mg (d) As long as force F exceeds the kinetic friction. which means that the kinetic frictional force is less than the maximum static frictional force.c). When we apply a force parallel to the surface the block does not move until the applied force exceeds the maximum static frictional force (Figure (4-7b. Chase. (4-18) At steady state we get F = µ k ' vN . SOLIDS NOTES 4, George G. Chase, The University of Akron s s s + τ zy and N = normal stress = τ yy where Recalling that F = shear stress ⋅ area = τ xy the area is the surface at y = constant, hence we get a relationship between the shear and normal stresses: (τ s xy s + τ zy ) surface 2 = µk ' v x + v z2 ( ) 1/ 2 s τ yy surface . (4.20) If the motion on the y=constant surface is only in the x direction this equation simplifies to s τ xy surface = β vx surface (4-21) s β = µk ' τ yy where surface . Now we can empirically relate the motions of the particles near a surface to their effective stresses. Internally, there are frictional forces occurring between the particles. At the bulk scale these frictions could be defined the same way except instead of a wall or boundary surface the friction is due to relative motion with other particles. In these cases the flows may be modeled similar to yield stress fluids (Bingham Plastic, Bird et. al. 1960, chapter 1). An example of this is shown in Figure 4-8 in the case of funnel flow in a hopper and the formation of rat holes. In the center of the hopper exit the powdered solids are able to freely flow downward while the material near the hopper walls is stationary because the shear stress due to gravitational force is too small to overcome the static frictional force between the solid particles. SHEAR STRESS IS TOO SMALL TO OVERCOME STATIC FRICTION AT TH SURFACE STATIONARY POWDER STATIONARY POWDER FLOW Figure 4-8. Solid particles near the hopper wall remain stationary while the material in the center flows downward. 4-10 SOLIDS NOTES 4, George G. Chase, The University of Akron 4.2.3 JANSSEN’S COEFFICIENT (AXIAL TO TANGENTIAL STRESS RATIO) Janssen’s coefficient (also called the coefficient of lateral pressure; Jaroslav, F. Mechanics of Particulate Materials, Elsevier, Amsterdam, 1982, Figure 1.3, page 23) provides a largely empirical relationship between the axial and normal solid phase stresses relative to the direction of flow. For motion in a tube the axial stress is the zz component in the z axis direction and the normal stress is the rr component in the r direction. Janssen’s coefficient is defined as K= σ rr σ zz (4-22) where σ rr and σ zz are the stress components in the r and z directions. These stresses are the total stresses acting on the multiphase material. These stresses are the sum of the fluid phase and solid phase stresses. Hence, this definition applies whether the material is totally liquid, totally solid, or a mixture in-between. This coefficient represents the ratio of the measurable stresses, as can be measured with a strain gauge to measure the force acting on a defined probe surface. For a pure liquid we know that the probe would measure an equal pressure in all directions within the liquid at a given point within the liquid at stagnant conditions. Hence for a pure liquid K=1. For a totally solid material K=0 because we know we can set a weight on top of a solid block and all of the stresses within the block are aligned vertically. Figure 4-9 gives a rough correlation showing the Janssen’s coefficient approximate values for several fluid-solid mixture materials. This correlation is crude, it only gives a rough approximation. For most granular materials we can take Janssen’s coefficient to be approximately 0.4. (Handout 4.3) 1 LIQUIDS LOW CONCENTRATION SLURRIES HIGH CONCENTRATION SLURRIES K FILTER CAKES PACKED BEDS GRANULAR MATERIALS POLYMERS 0 0 SOLIDS 1 VOLUME FRACTION LIQUID 4-11 SOLIDS NOTES 4, George G. Chase, The University of Akron Figure 4-9. Janssen’s Coefficient for various materials. For powdered materials, in air, Janssen’s Coefficient is easily interpreted as the ratio of the stresses only on the solid phase. When the fluid is liquid and the liquid is under high pressure the interpretation of the data is more difficult because the fluid pressure can also influence the probe measurement. 4.2.4 PERMEABILITY Darcy’s law relates the pressure drop to flow through a packed bed with the permeability coefficient, k . Given the packed bed shown in Figure 4-10, the permeability is defined by k ⎛ Po − PL ⎞ Q ⎜ ⎟= . µ⎝ L ⎠ A (4-23) FLOW IN PACKED BED P L AREA, A FLOW OUT Figure 4-10. Typical Packed Bed. 4-12 Material Permeability Coefficient. 1960). Transport Phenomena.) clay sandstone granite The Darcy’s law expression provides a means of estimating the flow rate for a given pressure drop of fluid. George G.. 1991. Chase.4) Table 4-2. A few correlations are available for predicting the permeability. Oxford. Coulson & Richardson (Chemical Engineeering. The data reported by Coulson & Richardson show that the permeability depends significantly upon porosity and specific surface area of contact between the fluid and solid phases. is the height (or depth) of the bed. The permeability coefficient must be determined from experiment.al. New York. The University of Akron Typical permeability values for some common materials are listed in Table 4-2. (Handout 4. flow rate divided by the cross sectional area of Q the bed A .SOLIDS NOTES 4. and 4-13 L Vo . silt. Pergamon. One of the more common correlations is Ergun’s equation (Bird et. Wiley.75 2 L ρVo (1 − ε ) Φd V ρ p o (4-24) µ where ∆P is the pressure drop (pressure at inlet minus pressure at outlet). loam peat filter aides (diatomaceous earth. is the approach velocity. k (m2) 10-12 to 10-9 10-10 to 10-9 10-16 to 10-12 10-13 to 10-11 10-14 to 10-12 10-16 to 10-13 10-16 to 10-11 10-20 to 10-18 Clean sand. sand-gravel mixture Non-woven glass fiber filter media fine sand. Table 4. Permeabilities of typical materials. ∆Pg c Φd p ε 3 150(1 − ε ) = + 1. Also.1. 4th ed. etc. page 133) has permeability values for common shapes given as B where k=B/µ. Volume 2. Fundam.(4.al. ft lb m kg m in 2 in FPS units or 1 N s2 lb f s Ergun’s equation relates the pressure drop to a quadratic expression in the superficial velocity (equivalent to Q / A ) as a function of the particle size and the bed porosity. Replace Eq.23). (4-24) should be replaced with 180 and the 1." Ind. 18(3) 199-207.80 = 2 L ρVo (1 − ε ) Φd V ρ p o (4-25) µ Lets define the Reynolds number and packed bed friction factor as Rep = ρVo d p µ (1 − ε ) (4-26) f = ∆Pg c Φd p ε 3 L ρVo2 (1 − ε ) (4-27) then the Ergun equation can be written as f = 180 + 1.75 coefficient should be replaced with 1.80 for smooth particles.SOLIDS NOTES 4. For rough particles the 1.80 ΦRep (4-28) Note that we can deduce a model for permeability from Eq.75 should be replaced with 4. One could make a comparison between Darcy’s Law and Ergun’s equation to get a relation for the permeability as a first order function of the superficial velocity.0. George G.. The University of Akron gc is the gravity conversion factor 32174 . We get 4-14 .(4-25). Chem. This gives ∆Pg c Φd p ε 3 180(1 − ε ) + 1.(4-25) with µ Vo k gc ∆P in L from Eq. where Vo = Q / A and (Po − PL ) = ∆P . 1979) studied data on a wider range of particles and concluded that the 150 coefficient in Eq. MKS units. ("Flow Through Porous Media-The Ergun Equation Revisited. Chase. MacDonald et. 1m ⎞⎛ Ns 2 ⎞⎛ 1Pa ⎞ ⎟ =⎜ 2 ⎟⎜ 10000 Pa ⎟⎜ ⎜ 60s / min ⎟ ⎟⎜ ⎜ ⎟⎜ 1N / m 2 ⎟ ⎜ 1kgm ⎟ m ⎝ ⎠ ⎠ ⎠ ⎠⎝ ⎝ ⎠⎝ ⎝ = 1.(4-23) Q L 1 µ A ∆P g c k= ⎛ 0.7 x10 −13 m 2 Substitute this value for permeability into Eq.0.46 x10 −5 m or the average particle size is about 15 microns. very slow flow of water (1 liters/minute per square meter) though a 10 cm thick packed bed of spherical Lucite particles produces a pressure drop of 10kPa. If the particles are approximately spherical the sphericity is 1.(4-30) and solving for dp gives d p = 1.001m 3 / min ⎞⎛ 1x10 −3 kg / ms ⎞⎛ 0. Chase. 4-15 . George G. and assuming normal packing the porosity is about 0.SOLIDS NOTES 4. For small Renolds number the 180 term dominates the denominator.38 (from Figure 4-5). Eq.80ΦR ] ep (1 − ε )2 (4-29) This expression is useful for estimating permeability for a powder of a particular size. Eq.(4-29) reduces to 2 k ≅ 8 x10 −4 d p (4-30) EXAMPLE 4-2 As an example. What is the approximate size of the Lucite particles? SOLUTION: Solving Darcy’s Law. The University of Akron 2 Φ 2ε 3 d p k= [180 + 1. or estimating particle size from pressure drop – flow rate data. The University of Akron Check the Reynolds number. George G.SOLIDS NOTES 4.(430) holds. 4-16 . Chase.3) 3 −6 −3 ≅ 4 x10 −4 Since Rep is much smaller than 1 then the assumption in deriving Eq. to make sure the low Reynolds number assumption holds: Rep = (1000kg / m )(0.001m / min )(min/ 60s )(15x10 m) (1x10 kg / ms )(1 − 0. SOLIDS PROCESSING HANDOUTS.5 for Normal Packing ⎪ 1 for Dense Packing ⎩ and the porosity is given by for any packing condition.9 Sphericity 1 Loose Normal Dense Figure 4-5.8 0. (4-6) (4-7) ε = (1 − 0. such that ⎧ 0 for Loose Packing ⎪ λ = ⎨0.5 0.1 1 0. The University of Akron HANDOUT 4.0873 All three curves are generated by defining a packing parameter. Chase.6 0.1042Φ + 1.1 0.9 0. George G.7 0.7 Porosity 0.1 0 0 0.4 0. The plot in Figure 4-5 is obtained by curve fitting data.4 0.2 0. λ . Loose packed materials are ones that have not had much time to settle due to vibrations.3 0.8 0.8))ε Loose (4-8) 1 .6 0. Porosity as a function of sphericity and packing structure.297λ (Φ − 0.5 0.2 0. A normal packed material is one that may have sat for several days or weeks.3 0. The loose packed porosity is a function of sphericity given by ε Loose = 0. Dense packed materials are ones that have sat for a year or more and have had ample time to settle and vibrate into its most densely packed structure under force of gravity.4411Φ 2 − 1. Chase.M. Lightfoot.39 0.2 ♦ c 0.7 3.3 0. Dougherty. 2 .5 0. 12 Material y = 10. 137-152. 2002) list values for parameter A from several references.7 0.9 Figure 4-6 Relation between parameter A and εc in Eq.626 0.E.74 0.1ε c (4-10) The data points in the plot are scattered meaning that Eq.8 0. 1959) has the form ⎛ (1 − ε ) ⎞ µ = µ⎜ 1 − ⎟ ⎜ (1 − ε ) ⎟ c ⎠ ⎝ 0 − A (1−ε c ) (4-9) where A is a material parameter.N.65 0.Soc.107x R2 = 0. Bird.. and E.1 0.0 9. Stewart. 2nd ed. (R.0 9. Trans. A plot of the data in Figure 4-6 gives a rough approximation between A and εc as A ≅ 10.732 . Rheol.767 10 8 6 4 2 Submicron spheres 40 micron spheres Ground Gypsum TiO2 Laterite Glass Rods 30x700µm Glass Rods 100x400µm Quartz grains 53-76 µm Glass fiber aspect ratio 7 Glass fiber aspect ratio 14 Glass fiber aspect ratio 21 A 0 0 0.4 0.2 Krieger-Dougherty Equation for estimating bulk viscosity of slurries (I.8 5. Wiley. Transport Phenomena. Bird et.25 5.SOLIDS PROCESSING HANDOUTS.4134 A 2. New York.618 .28 3.03 6.29 0.87 5. W.al.8 3.. Krieger and T.0 εc 0.31 0. George G.6 0. (4-10).(4-6) should only be used if experimental data are not available.629 0. The University of Akron HANDOUT 4.25 9.J.45 0. 3.B. σ rr σ zz (4-22) 1 LIQUIDS LOW CONCENTRATION SLURRIES HIGH CONCENTRATION SLURRIES K FILTER CAKES PACKED BEDS GRANULAR MATERIALS POLYMERS 0 0 SOLIDS 1 VOLUME FRACTION LIQUID Figure 4-9. this definition applies whether the material is totally liquid.4. George G. Hence. Hence for a pure liquid K=1. These stresses are the total stresses acting on the multiphase material. 3 . or a mixture in-between. For a totally solid material K=0 because we know we can set a weight on top of a solid block and all of the stresses within the block are aligned vertically. For a pure liquid we know that the probe would measure an equal pressure in all directions within the liquid at a given point within the liquid at stagnant conditions. Janssen’s Coefficient for various materials.SOLIDS PROCESSING HANDOUTS. as can be measured with a strain gauge to measure the force acting on a defined probe surface. These stresses are the sum of the fluid phase and solid phase stresses. Chase. totally solid. This correlation is crude. The University of Akron HANDOUT 4. Figure 4-9 gives a rough correlation showing the Janssen’s coefficient approximate values for several fluid-solid mixture materials. For most granular materials we can take Janssen’s coefficient to be approximately 0.3 Janssen’s coefficient is defined as K= where σ rr and σ zz are the stress components in the r and z directions. This coefficient represents the ratio of the measurable stresses. it only gives a rough approximation. sand-gravel mixture Non-woven glass fiber filter media fine sand.) clay sandstone granite Permeability Coefficient. George G.4 Table 4-2. silt.SOLIDS PROCESSING HANDOUTS. The University of Akron HANDOUT 4. Permeabilities of typical materials. Chase. loam peat filter aides (diatomaceous earth. Material Clean sand. etc. k (m2) 10-12 to 10-9 10-10 to 10-9 10-16 to 10-12 10-13 to 10-11 10-14 to 10-12 10-16 to 10-13 10-16 to 10-11 10-20 to 10-18 4 . Fixed Bed. u t < VO (c) Intermediate Flow Rate. is much smaller than the minimized fluidization velocity. Vom. In the fixed bed the particles are in direct contact with each other. and expanded beds. Fixed Bed.1 The Phenomenon of Fluidization When a fluid is pumped upward through a bed of fine solid particles at a very low flow rate the fluid percolates through the void spaces (pores) without disturbing the bed. Vom. at which the bed just begins to fluidize. and the expanded bed (c) occurs when the approach velocity is intermediate between the minimum fluidization velocity and the terminal velocity. There is a minimum fluidization velocity. mobilized. FLUIDIZATION 5. supporting each other’s weight. Mobilized Bed. The fixed bed (a) occurs when the approach velocity. The pneumatically mobilized bed (b) occurs when the approach velocity is much greater than the particle terminal velocity. Vo. or is swept out. Chase. If the upward flow rate is very large the bed mobilizes pneumatically and may be swept out of the process vessel. At an intermediate flow rate the bed expands and is in what we call an expanded state. Fixed. In the expanded bed the particles have a mean free distance between particles and the particles are supported by the drag force of the fluid.SOLIDS NOTES 5. As shown in Figure 5-1. the velocity of the fluid through the bed opposite to the direction of gravity determines whether the bed is fixed. When the (a) Slow Flow Rate. 5-1 . VOm > VO (b) High Flow Rate. VOm < VO < u t Figure 5-1. The University of Akron 5. This is a fixed bed process. The expanded bed has some of the properties of a fluid and is also called a fluidized bed. George G. ut. expanded. A bed such as this is called a homogeneous fluidized bed. Several types of instabilities are described by Kunii and Levenspeil. Large scale instabilities and heterogeneities are not observed. given by the fluid volumetric flow rate divided by the cross-sectional area of the vessel). This is typically observed when the fluid and solids have similar intrinsic densities.. Fluidized beds display a number of liquid-like properties: • • • • • Lighter objects float on top of the bed (ie. Vo (otherwise known as the empty tank velocity. 1991). When Vo < Vom then the bed remains as a fixed bed. when Vo ≥ ut . In turbulent and pneumatically mobilized beds a significant part of the bed may be carried out of the vessel. Butterworth-Heineman. At the other extreme. the bed mobilizes. 2ed. Figure 3). A spouted bed is a variation on the fluidized bed in which the flow of the fluid is localized along the center axis and the solids move downward along the vessel walls (Kunii & Levenspeil. The University of Akron approach velocity.SOLIDS NOTES 5. Boston. George G. Figure 1 page 2 (D. Chase. There are a number of ways in which to design and operate gas-fluidized systems for continuous operations. The beds have a “static” pressure head due to gravity.. given by ρ 0 gh . The solids can flow through an opening in the vessel just like a liquid. Fluidization Engineering. objects less dense than the bulk density of the bed). Levels between two similar fluidized beds equalize their static pressure heads. In many liquid-solid systems an increase in velocity above Vom results in a smooth progressive expansion of the bed. The surface stays horizontal even in tilted beds. is greater than or equal to the minimum fluidization velocity and it is less than the terminal velocity of the particles Vom ≤ Vo < u t then the bed forms a fluidized bed. Levenspeil. Figure 2). Kunii and O. For steady-state operations the particles must be recovered such as with a cyclone (Kunii & Levenspeil. When there is a large difference in the densities of the fluid and solid phases an increase in fluid velocity typically causes large bubbles or other such instabilities. 5-2 . Figure 5-2 shows a counter current column and a cross flow system. Resists rapid temperature changes. Figure 7) provide a table comparing different types of fluidized beds to the fixed bed. Counter current and cross flow methods of continuous contacting in fluidized bed designs. Beds include: • • • • • • Fixed bed Moving bed Bubbling/turbulent bed Fast fluidized bed Rotary cylinder Flat hearth The advantages of fluidized beds include: • • • • • • Liquid like behavior.SOLIDS NOTES 5. easy to control and automate. uniform temperature and concentrations. Heat and mass transfer rates are high. Chase. Rapid mixing. 5-3 . requiring smaller surfaces.2 Comparison of Contacting Methods Kunii and Levenspiel (ibid. hence responds slowly to changes in operating conditions and avoids temperature runaway with exothermic reactions. The University of Akron SOLIDS IN GAS OUT GAS OUT SOLIDS IN SOLIDS OUT GAS IN GAS IN OUT CROSS FLOW COUNTER CURRENT COLUMN Figure 5-2. Applicable for large or small scale operations. 5. Circulate solids between fluidized beds for heat exchange. George G. SOLIDS NOTES 5. 285-292. ◊ Internal flow deflectors do not improve fluidization.1). Geldart identified four regions in which the fluidization character can be distinctly defined. The bed expands considerably before bubbling occurs. 7. 5. ◊ Bubbles spit and coalesce frequently through the bed. 1973) observed the nature of particles fluidizing. Typical uses include • Reactors ◊ ◊ ◊ ◊ • • • • • • • • Cracking hydrocarbons coal gasification carbonization calcination heat exchange Drying operations Coating (example. George G. Chase. metals with polymer) Solidification/Granulation Growth of particles Adsorption/desorption Bio fluidization others 5. ◊ Gas bubbles rise more rapidly than the rest of the gas.3 Uses of Fluidization The uses for fluidized beds are limited to our imaginations. Particle comminution (breakup) is common. • 5-4 . ◊ Maximum bubble size is less than 10 cm. Pipe and vessel walls erode due to collisions by particles. Group A particles are characterized by • • Bubbling bed fluidization. Rapid mixing of solids causes non-uniform residence times for continuous flow reactors. (HANDOUT 5. The University of Akron The disadvantages of fluidized beds include: • • • • Bubbling beds of fine particles are difficult to predict and are less efficient. He categorized his observations by particle diameter versus the relative density difference between the fluid phase and the solid particles. Gross circulation of solids occurs.4 Geldart Classification of Particles Geldart (Powder Technology. 1973). and Spout from the top of the bed easily. Form channels in large beds with no fluidization. and Tend to be cohesive. Are difficult to fluidize and tend to rise as a slug of solids. Form beds whose dense phase surrounding the bubbles has low voidage. Are very large. George G. dense particles. and Have gross circulation. The University of Akron 10 B SAND-LIKE (BUBBLING BED) (EASY TO FLUIDIZE) A AERATABLE BED (EASIEST TO FLUIDIZE) C COHESIVE (DIFFICULT TO FLUIDIZE) (ρ s − ρg ) 1 3 D SPOUTABLE BED (g / cm ) 0. These beds • • • • • • • • • • • • • • Are made of coarser particles than group A particles and more dense. 7. Form bubbles as soon as the gas velocity exceeds Vom. Have bubble sizes independent of the particle size. Group B particle beds are the most common. 258-292. Geldart classification of fluidized beds.1 10 100 dp (µm ) 1000 10000 Figure 5-3. Particle properties are related to the type of fluidized beds. Group C particles Group D particles 5-5 . Chase. Form bubbles which rise slower than the rest of the gas phase. Form small bubbles at the distributor which grow in size throughout the bed. Cause slugs to form in beds when the bubble size approaches the bed diameter. (Geldart. Form bubbles which coalesce rapidly and grow large.SOLIDS NOTES 5. Powder Technology. bubbling. 5. d p * versus a dimensionless velocity u * where ⎡ ρ( ρ p − ρ ) g ⎤ dp*= dp ⎢ ⎥ 2 µ ⎢ ⎥ ⎣ ⎦ and ⎡ ⎤ ρ2 ⎢ ⎥ u* = u ⎢ µ ρp − ρ g ⎥ ⎣ ⎦ 1 3 1 3 = ( 3 4 CD Rep 2 ) 1 3 (5-1) ( ) ⎛ 4 Rep ⎞ =⎜ ⎟ ⎝ 3 CD ⎠ 1 3 . The subscripts f mean that the quantity is for a fluidized bed.SOLIDS NOTES 5. (5-2) With the data arranged this way they identify several interesting features including: • • • • Geldart’s classification. Terminal velocity. assuming that no solids are entrained and carried out of the bed. and Types of fluidization (spouted beds. At the onset of fluidization the particles are still close enough together that the pressure drop is related to the velocity by the Ergun Equation.(5-3) is the weight of the solids minus the buoyant force due to the displaced fluid. Figure 16) for classifying fluidization regimes. They plot a dimensionless particle diameter. Minimum fluidization velocity. ut . George G.5 Prediction of Minimum Fluidization A minimum velocity is needed to fluidized a bed. If the velocity is too small the bed stays fixed and operates as a packed bed. Chase. The University of Akron Kunii and Levenspeil present a more generalized diagram (ibid. If we consider a total mass balance on the solids. umf . a free body diagram tells us that the force due to pressure drop is also related to the net weight of the solids in the bed ∆P A = ρ p − ρ 1 − ε f ( )( )( A L ) gg f (5-3) c where the right side of Eq.(4-25) which relates the pressure drop to the flow rate through a packed bed. Recall the Ergun Equation is presented in dimensionless form in Eq. fast fluidized beds and pneumatic transport). Also. then the total mass of solids is constant given by M solids = ρ p 1 − ε f AL f = constant ( ) (5-4) 5-6 . and the bed height. Hence.(4-25) gives the modified Ergun Equation for fluidized beds 180(1 − ε f ) 1. vary but the rest of the terms in Eq. 1 1 2 2 (5-5) For liquids and for gases. the fluid phase density is constant. Initially if the bed is densely packed the pressure drop overshoots the fluidization pressure until the particles separate and fluidize. Substitution of Eq.SOLIDS NOTES 5. This means that at the porosities and bed heights at flow rates 1 and 2 are related by (1 − ε ) L = (1 − ε ) L . as long as the pressure drop is small. ε . L . (5-8) DENSE PACKED (FIRST TIME BED IS FLUIDIZED) P LOOSE PACKED PACKED BED OPERATION FLUIDIZED BED OPERATION Vm V Figure 5-4. George G.(5-3) is constant and thus the pressure drop in a fluidized bed is constant independent of the velocity.(5-3) into Eq.80 2 Rep f + Rep f = N GA 3 2 Φ εfΦ ε3 f (5-6) where Rep f = ρVof d p µ 3 ρ ρp − ρ g dp (5-7) and N GA = ( ) µ 2 . Chase. The University of Akron At different fluidization rates the porosity. Experimental data show this to be true. Typical pressure drop versus velocity plot for fluidized beds. the right hand side of Eq. A typical plot of the pressure drop versus the velocity is shown in Figure 5-4. 5-7 .(5-4) are constants. The University of Akron For small Rep f < 1 . George G. 5-8 . Equation (5-11) can be solved for the minimum fluidization velocity. The modified Ergun Equation. others have determined values for K 1 and K 2 . 12(3). When ε m and Φ are not known.001 < Rep m < 4000 . we can estimate the minimum fluidization conditions.7 and K2 ⎝ 2K 2 ⎠ Since the work of Wen and Yu. Using this value for ε m we can solve the Ergun Equation. 1966) noted that K 1 and K 2 stay nearly constant over a wide range of particles and for 0.8 and K 2 = 3 3 2 ε mΦ ε mΦ Wen and Yu (AIChE J. Φ . we can neglect the Rep f 2 term and get the Blake-Kozeny expression 180(1 − ε f 2 ε3 fΦ ) Rep f = N GA for Rep f < 1 (5-9) or Vof = 2 (ρ p − ρ )g ε 3f Φ 2 dp 180µ (1 − ε ) f (5-10) which relates the fluidization velocity to the void volume fraction of the expanded bed. ⎟ = 33. provided K 1 and K 2 are known. thus giving a prediction of Vo mf with a 34% standard deviation.0408 . can be estimated from Figure 4-1 for loose packing and known sphericity. To estimate the onset of fluidization. Eq. 610-612. Vo m . such as with very small particles.(5-6) is rewritten as K 2 Rep m + K 1 Rep m = N GA 2 (5-11) where K 1 = 180(1 − ε m ) 1. Table 5-1 summarizes values for fine and coarse particles. Chase.(5-6) for the minimum fluidization velocity. we can still estimate the minimum fluidization velocity.SOLIDS NOTES 5. Eq. gives Rep m ⎛ K1 ⎞ ⎛ K1 ⎞ 1 = ⎜ ⎟ ⎟ + K N GA − ⎜ ⎜ 2K ⎟ ⎜ 2K ⎟ 2 ⎠ 2 ⎝ 2⎠ ⎝ 2 (5-12) ⎛ K1 ⎞ 1 for which Wen and Yu determined ⎜ = 0. The minimum fluidization porosity. ε m . Eq. Chem. To obtain a conservative estimate. Vo m should be estimated for the largest particle. 610-612. Now lets consider what happens when there is a large size distribution of particles in a fluidized bed. as fluid flows upward and the flow is increased. Vo m . 12(3).6 Wide Size Distributions of Particles The previous discussion applies predominately to beds of narrow size distribution of particles. This may differ significantly from the size distribution in the fresh feed due to elutriation of fines. 5.0408 0..(5-12). or agglomeration of particles. The boiling point is not fixed. The University of Akron Table 5-1. These conditions are shown in Figure 5-6. However. Estimating Vo m for a wide size range of particles is analogous to measuring the boiling point of a liquid mixture. (HANDOUT 5.1984. Chitester et. You must also check the terminal velocity of the smallest particles to make sure that you do not entrain fines and carry them out of the top of the bed. Values for constants in Wen and Yu’s correlation.2) 5-9 .. This partial fluidization will occur at a smaller velocity than the average Vo m . If a bed of particles has a bimodal distribution. George G. In such a bed the minimum fluidization velocity. it has two size ranges as for example given in Figure 5-5. must be determined for the particular size distribution in actually in the bed. Sci. 39.7 1 K2 SOURCE Fine Coarse 0.0494 Wen and Yu.SOLIDS NOTES 5.al.253. attrition. 1966. the fine particles in the voids between the larger particles will fluidize before the larger particles. but varies with the composition. PARTICLES ⎛ K1 ⎞ ⎜ ⎟ ⎝ 2K 2 ⎠ 33. to fluidize the whole bed. Chase. Eng.7 28. One can estimate Vo m by using the average particle size (a permeability average is most appropriate). Several fluidization conditions can exist for fluidized beds with bimodal size distributions. AIChE J. The University of Akron Bimodal Distribution Number Particle Size Figure 5-5. Bimodal distribution of particle sizes showing two peaks (modes) in the number of particles of each size.SOLIDS NOTES 5. Chase. 5-10 . George G. George G. The average particle size may gradually vary throughout the depth of the bed. The University of Akron dp small dp small dp large dp large ε FLOW dp avg (a) Complete segregation of particles into a region of predominately small particles and a region of predominately large particles. Fluidized beds with bimodal size distribution. Figure 5-6.SOLIDS NOTES 5. 5-11 . dp small dp mixed dp small dp mixed dp large dp large ε FLOW dp avg (b) Partial segregation into two regions with different particle sizes separated by a layer of mixed particle sizes. dp avg ε FLOW (c) No segregation of particles. dp mixed dp mixed. The segregation may also be characterized by an abrupt change in bed porosity. Chase. Particle properties are related to the type of fluidized beds. 1 . (Geldart. 7. 258-292.SOLIDS PROCESSING HANDOUTS. Geldart classification of fluidized beds.1 10 B SAND-LIKE (BUBBLING BED) (EASY TO FLUIDIZE) A AERATABLE BED (EASIEST TO FLUIDIZE) C COHESIVE (DIFFICULT TO FLUIDIZE) (ρ s − ρg ) 1 3 D SPOUTABLE BED (g / cm ) 0.1973). Powder Technology. George G. The University of Akron HANDOUT 5.1 10 100 dp (µm ) 1000 10000 Figure 5-3. Chase. Fluidized beds with bimodal size distribution. George G. The University of Akron HANDOUT 5. dp avg ε FLOW (c) No segregation of particles. 2 . Chase.SOLIDS PROCESSING HANDOUTS. dp mixed dp mixed. The average particle size may gradually vary throughout the depth of the bed.2 dp small dp small dp large dp large ε FLOW dp avg (a) Complete segregation of particles into a region of predominately small particles and a region of predominately large particles. dp small dp mixed dp small dp mixed dp large dp large ε FLOW dp avg (b) Partial segregation into two regions with different particle sizes separated by a layer of mixed particle sizes. Figure 5-6. The segregation may also be characterized by an abrupt change in bed porosity. Solids Notes 6. This occurs because small particles that are expelled from the top of the bed have high velocities and they require greater distance to slow down and turn around to return to the bed. and M is an empirical constant. Fine particles are present in fluidized beds from several sources: Feed streams Mechanical attrition or breakage of larger particles Temperature stress cracking Size reduction due to chemical reactions. Chase. Leva (Chem. fines can often be recovered such as with cyclones or hydrocyclones. He found that (1) When the column height above the bed is small. However. 6-1 . Co is the initial concentration. ELUTRIATION OF PARTICLES FROM FLUIDIZED BEDS Elutriation is the process in which fine particles are carried out of a fluidized bed due to the fluid flow rate passing through the bed. the elutriation rate is high. When fines elutriation is a significant problem and modifications to the bed design cannot aid in reducing the problem. 47. fine particles are elutriated out of a bed when the superficial velocity through the bed exceeds the terminal velocity of the fines in the bed. (2) The elutriation process causes a decrease in particle concentration. elutriation can also occur at slower velocities. Prog.. etc. Engr. George G. shrinkage. 1951) measured the rate of elutriation (total mass per time) from a bed of particles with a bimodal size distribution. The concentrations may be empirically modeled by an Arrhenius type expression as C = C0e − Mt (6-1) where C = concentration at time. Typically. The University of Akron 6. t. 39. But if the height exceeds a certain minimum size then the rate is a constant minimum value (Figure 6-1). The University of Akron Free Space Above the Bed Fluidized Bed Elutriation Rate Air flow Free Space Height Above the Bed Figure 6-1. 6-2 . The higher the velocity. Gs [kg/m2/s] The bulk density of the solids in the exiting gas stream is called the holdup.1 Definition of Terms Lets define some terms that we can use to describe the elutriation process. Free Board Height. The greater the boiling rate. the more volatile material leaves the mixture at the lower boiling temperature. to cool and slow down the elutriated particles and return them to the mixture. Hf. elutriation may be used to remove dusts or very fine particles from coarser particles. By analogy with boiling of liquid mixtures. George G. Not all elutriation bad. the greater the rate at which the low boilers will leave the bed. Chase. Sometimes elutriation can be helpful. The free space height above the bed serves as a condenser. ρ0 [kg/m3] We note that the entrainment is related to the holdup and the superficial velocity by (6-2) G s = ρ 0Vo Hf Lean Phase Dense Phase Air flow rate Q = AVo Figure 6-2. Gs 6. Fluidized bed with flux rate Gs. finer particles have a lower boiling temperature than larger particles.Solids Notes 6. is defined to be the measure of the free space above the boundary between the dense phase and the lean phase (Figure 6-3). A fluidized bed behaves similar to a mixture of liquids with different volatilities. Elutriation Rate (total mass/time) vs the free space height above the bed. hence the greater free space height. The flux of solids carried out of the top of the column is called entrainment. The boiling temperature is analogous to the fluidization velocity. the greater capacity that is needed of the condenser. In the liquid-liquid mixture. For example. TDH depends upon the superficial velocity and the particle properties. Air flow rate ρ0 Q = AVo Figure 6-3. The Transport Disengagement Height (TDH) is the height above the dense-phase/leanphase boundary above which entrainment and bulk density do not change appreciably. George G. If Hf < TDH then coarse particles will be carried out of the column. When Hf > TDH then the holdup and entrainment rates are close to their minimums. The fine particles. Both fines and coarse particle are entrained in the Transport Disengagement Height (TDH) region. The TDH is the height at which the kinetic energies of particles due to the collisions in the bed has been expended against gravity potential. 6-3 . A fluidized bed usually has two regions or phases: dense bubbling phase and lean dispersed phase (Figure 6-2). Above the TDH only fines are entrained. Usually Hf = TDH is the most economical design height for the fluidized bed. whose terminal velocities are less than the superficial velocity. Only fines are entrained in this region Hf TDH Both fines and coarse particles are entrained in this region Dense Phase Bulk density. we need to know the rate of entrainment and the size distribution of the entrained particles in relation to the size of the particles in the bed. Chase. continue to be entrained out of the column. The University of Akron For design.Solids Notes 6. and the coarse particles whose terminal velocities are greater than the superficial velocity are able to fall back down to the bed. (HANDOUT 6. Kunii and Levenspiel give a good review of this literature. We can model the data on the chart in the form of Log (TDH / d t ) = mLog (d t ) + Log (b ) where dt is the vessel diameter. see Fluidization notes 5) there are several methods discussed in literature. In these notes you are only given a brief introduction.64Vo (6-1) (6-2) (6-3) Finally. The University of Akron 6.115Vo − 0. Chase. METHOD 1. 472. For beds of fine particles (Geldart A classification. Dimensionless TDH/dt vs vessel diameter. For catalyst pellets in 20 to 150 micron size range. we take the above equations and try to compress the chart by plotting the dimensionless TDH/dt versus the Froude Number. 4. Taking data points from the chart and plotting them we can curve fit to find the parameters m and b: m = −0.1) 6-4 . Zenz and Weil (AICHE J. Fr = Vo2 / d t / g to get the plot shown in Figure 6-5. George G. 1958) proposed a correlation between dimensionless TDH and the vessel diameter. on a log-log plot. the relation between the dimensionless TDH and the vessel diameter are nearly linear (see Kunii and Levenspiel Figure 5 page 173) as indicated in Figure 6-4.Solids Notes 6.2 Estimation of TDH for Geldart A Particles. Vo TDH/dt dt Figure 6-4.587 b = 4. dt. then we can relate TDH Vo2 = 23. (Can.Solids Notes 6.. 1973) independently determined.4 0. Fournol et. dp=58 microns.(6-5) is likely to be material specific.8 1 1. the TDH for this material to be given by Vo2 = 0.2 0. Chem. we get Vo2 = constant = 0.8972 Figure 6-5. Dimensionless TDH/dt versus Froude Number.171x 2 R = 0. J.6 0.4 Vo2/dt/g y = 23. for a fluidized bed of fine catalyst particles.171 (6-4) dt dt g or. 401. upon rearrangement.001 TDH g (6-6) 6-5 .al. Chase. George G.2 1. The University of Akron 30 25 20 TDH/dt 15 10 5 0 0 0. Engr. 51.0432 TDH g (6-5) The constant in Eq. If we take the linear fit in Figure 6-5 to represent the material behavior. Chemical Engineering Thermodynamics. Chase.E. Balzhiser. the total entrainment rate is given by * G s = ∫ G si P ( d p ) dd p (integration over all particles entrained) (6-10) To apply this procedure. and J. the flux rates are analogous to the vapor pressures.Solids Notes 6.. eds. Prentice Hall.R. (AICHE J. M. Eliassen.al. xi.Y. 1972) equates the partial pressure of component i in the vapor phase to the mole fraction in the liquid phase times the pure fluid vapor pressure. 1980) we assume that the flux rate of solid size dpi is proportional to its mass fraction. Following the work by Zenz et. 1958. P(dp). * Gsi = xi Gsi (6-7) * where G si is the flux rate from an imaginary bed of all particles of size dpi. 3. and Fluidization III. Divide the size distribution into narrow intervals and find which intervals have terminal velocities greater than the superficial velocity (these are the particles that are entrained. 4. Raoult’s Law for an ideal fluid mixture (R. 472. * 2. The procedure to determine the flux rate from a bed with known particle size distribution is as follows: 1. The University of Akron 6. This approach extends the analogy between fluidized beds and boiling of a liquid mixture (discussed in the introduction to Section 6). New Jersey. Englewood Cliffs. G si 6-6 . George G. a correlation such as shown in Figure 6-6 is required to find * .D.3 Entrainment Rate from Tall Vessels There are several methods for estimating entrainment rates.. N. Grace & Matsen. Pi = xi Pi * (6-8) hence. Find G si for each size range. Plenum. The total entrainment is given by * Gs = ∑ xi Gsi (6-9) In terms of a continuous size distribution. because Hf > TDH). Samuels. George G. For Geldart class B. 58 (38).1 G * s 1 10 1 100 ρ g u o Figure 6-6. 65. Boston. 1991). Fluidization Engineering. 2ed. Chem. Boston. 1969) Lewis et.01 0. 2. Butterworth. Prog. Japan. or D particles (the larger particles) other predictive models are available. Chase. 1962) Wen and Chen (AICHE J. 84.Solids Notes 6. Butterworth. Data taken from Kunii & Levenspeil. 6-7 . Eng. From this plot the value of G*s may be determined for Geldart A particles and for fines removed from larger particle beds. 28. al.. 117. 1982) Kunii and Levenspiel (Fluidization Engineering. C. 2ed. Recommended references: Kunii and Levenspiel (J. 1991 (figure 6 page 175). Symposium Series. The University of Akron 10000 10000 1000 Fines removed from larger particle beds 1000 (u o − ut )2 gd p 100 100 2 10 5 u o gd p 10 Geldart A beds with most particles entrainable 10 1 0. Eng. (Chem. 171x 2 R = 0.6 0. 2ed.2 1. Dimensionless TDH/dt versus Froude Number. George G. Boston. 1 .1 G * s 1 10 1 100 ρ g u o Figure 6-6.4 0.SOLIDS PROCESSING HANDOUTS.8972 Figure 6-5.8 1 1. Fluidization Engineering.2 0. Butterworth. Chase. Data taken from Kunii & Levenspeil. From this plot the value of G*s may be determined for Geldart A particles and for fines removed from larger particle beds. The University of Akron HANDOUT 6.4 Vo2/dt/g y = 23.1 30 25 20 TDH/dt 15 10 5 0 0 0. 10000 10000 1000 Fines removed from larger particle beds 1000 (u o − ut ) gd p 100 2 100 2 10 5 u o gd p 10 Geldart A beds with most particles entrainable 10 1 0. 1991 (figure 6 page 175).01 0. 81.M. 22-28. These stages are: PRETREATMENT SOLIDS CONCENTRATION SOLIDS SEPARATIONS POST TREATMENT Table 7-1 further breaks these topic down into sub categories (HANDOUT 7. England.” Chem. 7-1 . SOLID/LIQUID SEPARATIONS 7. Eng. Purchase. Tiller. H. R.C. 1981. Eng.R. “Tackle Solid-Liquid Separation Problems..M. 3. Because of its complexity an number of approaches have been developed. As part of the introduction I summarize these approaches for you. George G. 117-119. 1991. Solid/Liquid Separation Technology.1 INTRODUCTION There are large varieties of separation equipment available to industry for Solid/Liquid separations.” Chem.. Romans. Talcott. 87(7).SOLIDS NOTES 7. M. Prog. Smith. 2. There are four stages to S/L Separations. Croydon. Not all of these stages are present in all processes. D.S. 1974. Uplands Press. Ernst. A frequent question is: “Which is the right choice for my application”? A number of authors have published selection guides to help the practicing engineers. The University of Akron 7. Chase. REFERENCES: 1. F. “Bench Scale Design of SLS Systems.B. G.1). The University of Akron Table 7-1. remove moisture. CLARIFICATION granular bed precoat drum FILTRATION CENTRIFUGATION vacuum sedimenting centrifuge gravity filtering centrifuge pressure cyclone expression To remove solubles. PRETREATMENT To increase particle size.SOLIDS NOTES 7. Chase. or prepare material for downstream processes. Four Stages of Solid Liquid Separations. George G. THICKENING gravity sedimentation cross flow filtration cyclones periodic pressure filters To separate the solids from the liquid. reduce viscosity. to form cakes of dry solids or to produce particulate free liquid. reduce cake porosity. PHYSICAL washing drying repulping deliquoring CLARIFICATION gravity sedimentation SOLIDS CONCENTRATION SOLIDS SEPARATION POST TREATMENT 7-2 . CHEMICAL coagulation flocculation PHYSICAL crystallization aging freezing/heating filteraid admix To reduce the volume of material to process. 1 2/99 = 2. Chase. George G. 91% of the liquid was removed and the mixture yet to be treated was only 10% of the volume of the original feed. In each of these steps how much of the original slurry is processed and how much liquid is removed? Basis: 100 unit volumes of slurry feed Table 7-2. You run this slurry through a thickener that concentrates it to 10% solids. Pressurized equipment such as pressure filters are more expensive to operate than thickeners. (HANDOUT 7.1) Table 7-3. suppose you have slurry feed that is 1% solids. Pretreatment techniques. Effect of concentrating a slurry on the volume. Step Solids Volume Volume Volume % Solids Liquid In Mixture Feed 1 1 99 Thickener Filter Deliquoring 10 25 50 1 1 1 9 3 1 Volume Liquid Removed 0 90 6 2 % Liquid Removed from Feed 0 90/99 = 90. (HANDOUT 7. there is incentive to optimize the process by setting up the equipment in series. The University of Akron 7. for example. Pretreatment aids in reducing costs by prethickening the slurry.0 Volume of Mixture (After Treatment) 100 10 4 2 Hence. Relative costs of filters and thickeners as related to size. There are a number of pretreatment techniques that can be used and that have this effect. Cost s Filter Thickener Size Figure 7-1. The centrifuge is then used to deliquor the cake with a resulting cake of 50% solids. you can see that even though the thickener only concentrated the solids from 1 to 10%.SOLIDS NOTES 7. followed by a filter centrifuge that filters it to 25% solids.2) For example. as in the previous example. These pretreatment techniques are summarized in Table 7-3.9 6/99 = 6. 7-3 . Hence.2 Pretreatment The cost of S/L separations is directly related to the volume of material that must be processed. Rate of filtration increased. Micro-flocs agglomerate into larger flocs. Size of crystals increase. cake moisture) Reduce viscosity Prevents bubbles from forming Destabilize colloidal suspensions. faster filtration rate. improve drainage and decrease moisture content. Solid particles Solids concentration Solid-liquid interaction Heat treatment/pressure cooking Freeze/thaw Ultrasonics Ionized radiation Wetting agents 7-4 . More porous cake. Chase. The University of Akron Act Upon Liquid Treatment Technique Heating Dilution with solvent Degassing by chemical additive Coagulation by chemical additives Flocculation by shear forces Ageing Increase concentration with a thickener Classify to eliminate files Filter Aid body feed Effect Reduce viscosity (reduce resistance to flow. but thicker cake Physical conditioning of sludge to induce coagulation and flocculation. Reduce interfacial surface tension.SOLIDS NOTES 7. allow particles to agglomerate. George G. reduced load on filter Filtration rate increased and reduced moister content. The generalizations in Table 7-4 may be of assistance in making your choice. Low Cake Moisture Content Difficult to Continuously Discharge Cake Equipment is Expensive.SOLIDS NOTES 7. Gravity Simple Low Operating Cost Very Bulky Still High Volume of Liquid After Separation Vacuum is Easy to Produce Effective up to ∆P of about 0. Electrokinetic Vacuum Pressure Centrifugal Other 7-5 . Comparison of Forces Used in SLS. George G.3 WHICH FORCE TO USE There are four primary driving forces that are used to separate particles form liquids. Low Operating Costs -Efficiently Falls for Particles <10µm Centrifuges Have Longer Residence Time: -More Efficient. Output is Low Moisture Content of Cake May Be High Volatile Liquids Difficult to Handle Greater Ouput Per Unit Area Smaller Equipment. Chase.8 atm Improved Rates Over Gravity Equipment is Simple. The University of Akron 7. But Bulky and Expensive Compared to Pressure Filters.4) Table 7-4. The choice of which method to use in any particular application must take into account many diverse factors. (HANDOUT 7. Even for Fine Particles -High Throughputs -Low Residual Moisture Sonic. High Operating Costs Maximum Separating Forces (High-Gravity) Construction is simple for Cyclones: -Compact. Surface charges and electrostatic forces can keep particles separated.SOLIDS NOTES 7. Typical zeta potential values are shown in Figures 7-3 and 7-4. Several instruments are available for measuring zeta potential such as the Zeta Meter and the Anton Paar EKA. and as cakes are formed. This stress on the solids matrix can cause the matrix to compact. (HANDOUT 7.4 RANGE OF APPLIED FORCES AND COMPACTIBILITY As the slurries are thickened.5) Electrical Field Surrounding the particles Low Concentration Slurry. George G. Filtration: Equipment Selection Modeling and Process Simulation. Zeta potential is influenced by material properties of the particles as well as the pH and salt concentration of the solution. Particle contact as a function of concentration. very small particles can also have large repulsive forces due to surface charges and electrostatic effects. However. to one in which particles are in contact. (HANDOUT 7. (HANDOUT 7. Especially in filter cakes. Wakeman and E.6) Fibrous materials are also known to compact. requiring compressive forces to push particles together. Some data reported in literature is shown in Figure 7-5. Compaction occurs when particles rearrange their relative positions to each other to pack more closely together.S. The University of Akron 7. Elsevier. At the isokinetic point the zeta potential is zero and the particles can pack more closely together with the least amount of force (R. The smaller the pores (typically with smaller particles) the greater the force required. Tarleton. One measure of the surface charge is Zeta Potential.7) 7-6 . Their compaction may be a combination of surface charges and bending of the fibers. the mixture transitions from a condition in which there is a relatively large mean-free distance between particles. the drag force of the fluid on the particles can cause significant stresses to develop on the skeletal matrix of the solids.J. The degree of compactability has a profound effect on the way the solid matrix reacts to changes in the applied pressure. 1999). Oxford. Chase. Little Particle-Particle Contact High Concentration Mixture (Low Porosity) With Significant Particle-Particle Contact Figure 7-2. A certain amount of force is required to squeeze the fluid phase out of the pore spaces between the particles. 14 M Illite . Patel.0 -10.0 -20.0 2 3 4 5 6 7 pH 8 9 10 11 12 Zeta Potential [mV] Kaolinite .0. 2001). Halloysite and Illite Clayes in 0 M and 0.0 -50. The University of Akron Zeta Potential versus pH 10 5 0 -5 0 -10 -15 -20 -25 -30 -35 5 10 15 loose-unglued sand run 1 loose-unglued sand run 2 loose-unglued sand run 3 loose-unglued sand run 4 loose-unglued sand run 5 loose-unglued sand run 6 crushed-glued sand run 7 crushed-glued sand run 8 crushed-glued sand run 9 crushed-glued sand run 10 ZP (mV) pH Figure 7-3. The University of Akron. Measured zeta potentials for sand particles with and without glue (the glue is used to hold the sand particles together in a sand cartridge). Characterization of Consolidated Sand Cartridge Filter. Dissertation. A.0 0. 1999).SOLIDS NOTES 7.0 -60. Zeta potential measurements for Kaolinite. Examination of Deep Bed Filtration.0 M Illite .14 M NaCl solution.0.0. George G. The University of Akron.0 M Kaolinite .0 -30.14 M Halloysite .14 M Figure 7-4. 7-7 .0 M Halloysite . Thesis. Chase.0 -40. Stephan. (H. 10. (E. The Supercel (processed diatomaceous earth) is slightly compactable with a fragile structure which disintegrates under sufficient pressure or shear force. A diagram of the ranges of pressures encountered in SLS and corresponding porosities are shown in the Figure 7-6. The University of Akron 1 0. kPa Figure 7-5. 367-377. Compressibility of packed beds of selected materials (H. ultimate strength. The fine silica and talc are highly compactable. 49.5 0.9 0. (HANDOUT 7. Eng. George G. then we typically see changes in porosity in the range of 0. Chem. At his high pressure range the compaction occurs due to particle deformation and crushing (where the material’s intrinsic coefficient of elasticity.9 to 0. The kaolin is moderately compactable.8. 1953). Its open structure results from the unusual shape of diatoms which are the basic raw material. Prog. “Resistance and Compressibility of Filter Cakes (Part I and II). In sedimentation and thickening. Chase. in high-pressure expression. If a material is highly compactable.6 Supercel Fine Silica 0..SOLIDS NOTES 7.P.8 Grade E Carbonyl Iron Porosity 0. At the other extreme. the forces occur in the low pressure range. and the flocculated latex is super-compactable. and loss of bound water are important).4 0.7 Kaolin Latex Talc 0.7) 7-8 . The carbonyl iron typifies particles which form incompressible beds and have sufficient internal strength to resist crushing. such as a floc.3 1 10 100 1000 10000 Applied Compressive Stress. Particles flow into unoccupied voids and flocs are squeezed. 303-318. the effective pressure may range up to hundreds of atmospheres. Grace. space. Li. and costs. and cake dryness. Because of this difficulty a master selection chart is yet to be developed. The charts also give a ranking or score of 0 (worst) to 9 (best) on four different criteria: crystal breakage (ie. Tiller and W. 4th ed.01 0. 7. ATM Figure 7-6. University of Houston. Croydon. washability of the formed cake. This is due to the multiple operating factors and that often the “right” choice is a combination of 2 or more items operating in series or in parallel.4 CENTRIFUGE EXPRESSION 0. Table 7-5 gives a selection chart for selecting between different types of filters. The University of Akron 1 THICKENER 0. (F.1 1 10 100 1000 Compressive Stress.2 PARTICLE DEFORMATION 0 0. which is superior in all respects to other equipment. Compressive Stress relation to separation operation. On the other hand. George G. Uplands Press. 7. Theory and Practice of Solid/Liquid Separation.8.5 Selecting the Right Equipment It is rare if ever true that there is an absolute “right” choice for SLS equipment. (HANDOUTS 7. 2002.. It is in this respect that the available charts are of the most value. 7-9 .6 VACUUM FILTER PRESSURE FILTER 0.SOLIDS NOTES 7.8 GRAVITY FILTER Porosity 0. These charts are based on typical operating ranges in solid mass fractions and particle sizes. it is possible to separate out the “wrong” choices. least amount of breakage of crystal particles).B. Solid/Liquid Separation Technology.9) Other similar tables are available in literature (D. clarity of the filtrate. Table 7-6 gives a similar chart for selecting between different types of centrifuges. Purchas. Chase. 1981) that considers criteria such as power. 1 – 80 5 – 500 1 –500 0. Solid/Liquid Separation Technology. TYPICAL OPERATING RANGE PERFORMANCE SCORE 0 TO 9 (9 IS BEST) % (mass) OF PARTICLE SIZE SOLIDS IN OF SOLIDS.001 – 1 0. Purchas.B.001 – 0.0.2 2 – 90 5 – 90 0.001 – 0. Filtr.5 – 80 1 – 100 0. Filter Selection Chart. Sep 7(1) 76-79.1 0.001 – 0. Uplands Press.5 – 100 1 – 500 5 – 100.1 – 80 5-6 5 5 5 6-7 5-6 4-5 2-3 5-7 - 6 8 7 7 7 7 7 2 9 - 7-8 8-9 7-8 7-9 7-8 8-9 7 7-8 8 6 7 6 7-9 8 8 8 8 8 8 8 8 - 7-10 . (data from D.5 .SOLIDS NOTES 7. Chase.1 0.01 – 1 0.005 – 0.001 – 50 0.001 – 0.1 0.5 TO 500 0. The University of Akron Table 7-5. Croydon. Davies.05 2 – 90 0.000 20 – 100. George G.5 – 100 0.000 0. THE FEED MICRONS Filter CAKE DRYNESS WASH PERFORMANCE FILTRATE CLARITY CRYSTAL BREAKAGE LEAF PLATE CANDLE/PRECOAT CARTRIDGE FILTER PRESS SHEET FILTER STRAINER VACUUM DRUM VACUUM DRUM WITH PRECOAT VACUUM DISC BAND. E. 1970)..1 0. PAN. 1981.200 0.001 . TABLE WATER SCREEN DEEP BED 0.001 – 0.1 0. Filtr.1 – 100 0.01 – 1 0.000 2 – 20.1 – 100 0.000 100 – 20.000 1 – 20.SOLIDS NOTES 7. Centrifuge TYPICAL OPERATING RANGE PERFORMANCE SCORE 0 TO 9 (9 IS BEST) CAKE DRYNESS WASH PERFORMANCE FILTRATE CLARITY CRYSTAL BREAKAGE % (mass) OF SOLIDS IN THE FEED PARTICLE SIZE OF SOLIDS..000 2 – 20. Purchas. Chase.B.000 100 – 30.1 – 100 9 9 9 7-9 9 7 6 4 3 3 3 - 5 6 5 5 6 4 5 4 4 5 5 4 5 4 3 6 7 5 - 5 3 - 4 4 6-7 6-7 6-7 6-7 7-11 . ).000 100 – 20.05 – 5 50 – 80. Sep 7(1) 76-79. The University of Akron Table 7-6.02 – 2 0.003 –0. MICRONS PUSHER PEELER/SCRAPER WORM SCREEN OSCILLATING SCREEN BASKET CONICAL SCREEN SCREEN BOWL DECANTER TUBULAR BOWL DISC (MANUAL) DISC (SELF CLEANING) NOZZLE BOWL 8 – 90 5 – 80 3 – 90 3 – 90 2 –80 5 – 90 2 – 70 2 – 70 0. E. 1970).1 – 100 0. Davies. 1981.08 0. Solid/Liquid Separation Technology.000 0. Centrifuge Selection Chart (data from D.000 90 – 20. Croydon. Uplands Press. George G. Talcott. filter aid admix A paper by Ernst et. (HANDOUT 7. Ernst. continuous feed pusher centrifuge 0. 87(7). disc.1 to 10 cm/sec Rapid Filtering Gravity pans.M. Cake Formation Rate (rate at Separation type Equipment which cake height grows in a gravity filter). 7-12 . H. granular beds. pan filters.1 to 10 cm/min Medium Filtering Vacuum drum. Progr. G. screens. Eng. precoat drums. 1977. George G. Selection based on cake formation rate (F.SOLIDS NOTES 7. peeler centrifuge 0. Romans. (M. Tiller. 22-28.R. 65-76).1 to 10 cm/hr Slow Filtering Pressure filters. disc and tubular centrifuges. Chem. 0. The University of Akron Another way to select equipment type is on the rate of cake formation as suggested by F. horizontal belt.10) Table 7-7.S. R. Tiller in Table 7-7.al. CEP. horizontal belt or top-feed drum filter. 1991). Chase. This ranking allows us to narrow down options so that only a small number of types of equipment need to be evaluated. Smith. Oct. Tackle Solid-Liquid Separation Problems.C. The selection method eliminates unsuitable equipment and it ranks suitable equipment on how well they can perform the separation. sedimenting centrifuges Negligible cake Clarification Cartridges. CLARIFICATION gravity sedimentation SOLIDS SEPARATION freezing/heating filteraid admix SOLIDS CONCENTRATION THICKENING gravity sedimentation cyclones cross flow filtration periodic pressure filters To separate the solids from the liquid. cake moisture) Dilution with solvent Reduce viscosity Degassing by chemical additive Coagulation by chemical additives Flocculation by shear forces Ageing Increase concentration with a thickener Classify to eliminate files Filter Aid body feed Prevents bubbles from forming Destabilize colloidal suspensions. allow particles to agglomerate. faster filtration rate. Chase. FILTRATION CENTRIFUGATION vacuum sedimenting centrifuge gravity filtering centrifuge pressure cyclone expression To remove solubles. reduced load on filter Filtration rate increased and reduced moister content. or prepare material for downstream processes. to form cakes of dry solids or to produce particulate free liquid. remove moisture. but thicker cake Physical conditioning of sludge to induce coagulation and flocculation. Size of crystals increase. Reduce interfacial surface tension. CHEMICAL PHYSICAL coagulation crystallization flocculation aging To reduce the volume of material to process.1 Table 7-1.SOLIDS PROCESSING HANDOUTS. More porous cake. Rate of filtration increased. improve drainage and decrease moisture content. Treatment Technique Effect Heating Reduce viscosity (reduce resistance to flow. Micro-flocs agglomerate into larger flocs. PHYSICAL washing repulping drying deliquoring Table 7-3. Four Stages of Solid Liquid Separations PRETREATMENT To increase particle size. George G. reduce cake porosity. The University of Akron HANDOUT 7. CLARIFICATION granular bed precoat drum POST TREATMENT Act Upon Liquid Solid particles Solids concentration Solid-liquid interaction Heat treatment/pressure cooking Freeze/thaw Ultrasonics Ionized radiation Wetting agents 1 . Pretreatment techniques. reduce viscosity. Effect of concentrating a slurry on the volume. you can see that even though the thickener only concentrated the solids from 1 to 10%. Filter Costs Thickener Size Figure 7-1. Step Solids Volume Volume Volume % Liquid Volume Solids Liquid Liquid Removed % In Remove from Feed Mixture d Feed 1 1 99 0 0 Thickener Filter Deliquorin g 10 25 50 1 1 1 9 3 1 90 6 2 90/99 = 90. 2 . George G. The University of Akron HANDOUT 7.2 Table 7-2. Relative costs of filters and thickeners as related to size.0 Volume of Mixture (After Treatment) 100 10 4 2 Hence. 91% of the liquid was removed and the mixture yet to be treated was only 10% of the volume of the original feed.SOLIDS PROCESSING HANDOUTS. Chase.9 6/99 = 6.1 2/99 = 2. Low Cake Moisture Content Difficult to Continuously Discharge Cake Equipment is Expensive. The University of Akron HANDOUT 7. Gravity Simple Low Operating Cost Very Bulky Still High Volume of Liquid After Separation Vacuum is Easy to Produce Effective up to ∆P of about 0. Low Operating Costs -Efficiently Falls for Particles <10µm Centrifuges Have Longer Residence Time: -More Efficient. Chase. Even for Fine Particles -High Throughputs -Low Residual Moisture Sonic.8 atm Improved Rates Over Gravity Equipment is Simple. But Bulky and Expensive Compared to Pressure Filters.SOLIDS PROCESSING HANDOUTS. Comparison of Forces Used in SLS. George G.4 Table 7-3. High Operating Costs Maximum Separating Forces (High-Gravity) Construction is simple for Cyclones: -Compact. Electrokinetic Vacuum Pressure Centrifugal Other 3 . Output is Low Moisture Content of Cake May Be High Volatile Liquids Difficult to Handle Greater Ouput Per Unit Area Smaller Equipment. Little Particle-Particle Contact High Concentration Mixture (Low Porosity) With Significant Particle-Particle Contact Figure 7-2. requiring compressive forces to push particles together. Particle contact as a function of concentration. The University of Akron HANDOUT 7. George G. 4 .SOLIDS PROCESSING HANDOUTS. Surface charges and electrostatic forces can keep particles separated.5 Electrical Field Surrounding the particles Low Concentration Slurry. Chase. The University of Akron HANDOUT 7. A.0 M Illite .0 M Halloysite .6 Zeta Potential versus pH 10 5 0 -5 0 -10 -15 -20 -25 -30 -35 5 10 15 loose-unglued sand run 1 loose-unglued sand run 2 loose-unglued sand run 3 loose-unglued sand run 4 loose-unglued sand run 5 loose-unglued sand run 6 crushed-glued sand run 7 crushed-glued sand run 8 crushed-glued sand run 9 crushed-glued sand run 10 ZP (mV) pH Figure 7-3. 10. 5 . Halloysite and Illite Clayes in 0 M and 0.14 M NaCl solution. 2001).0 0.14 M Zeta Potential [mV] 8 9 10 11 12 Halloysite . Thesis.0. (E. Chase.0.0 2 3 4 5 6 7 pH Kaolinite .0 M Kaolinite . Measured zeta potentials for sand particles with and without glue (the glue is used to hold the sand particles together in a sand cartridge).0 -50.0. (H. Stephan.14 M Figure 7-4.0 -30.0 -10. Zeta potential measurements for Kaolinite.0 -20.SOLIDS PROCESSING HANDOUTS. Dissertation. The University of Akron. Characterization of Consolidated Sand Cartridge Filter. Patel.14 M Illite .0 -60.0 -40. George G. The University of Akron. 1999). Examination of Deep Bed Filtration. 1953).8 0. 6 .SOLIDS PROCESSING HANDOUTS.5 0. “Resistance and Compressibility of Filter Cakes (Part I and II). Li. 4th ed.1 1 10 100 1000 Compressive Stress. 2002. ATM Figure 7-6. Theory and Practice of Solid/Liquid Separation. University of Houston.P. Chem.. Compressive Stress relation to separation operation. (F. 367-377.9 0.4 0.7 1 0. kPa Figure 7-3. Tiller and W.01 0. 49. George G.2 PARTICLE DEFORMATION 0 0.. Eng. Chase.6 VACUUM FILTER PRESSURE FILTER 0.7 0. The University of Akron HANDOUT 7.6 0. Compressibility of packed beds of selected materials (H. Prog.3 1 10 100 1000 10000 Grade E Carbonyl Iron Kaolin Latex Talc Supercel Fine Silica Porosity Applied Compressive Stress. 1 THICKENER 0.8 GRAVITY FILTER Porosity 0.4 CENTRIFUGE EXPRESSION 0. Grace. 303-318. 1 0.1 – 80 5 – 500 1 –500 0. THE FEED MICRONS CAKE DRYNESS WASH PERFORMANCE FILTRATE CLARITY CRYSTAL BREAKAGE LEAF PLATE CANDLE/PRECOAT CARTRIDGE FILTER PRESS SHEET FILTER STRAINER VACUUM DRUM VACUUM DRUM WITH PRECOAT VACUUM DISC BAND. George G. TYPICAL OPERATING RANGE PERFORMANCE Filter SCORE 0 TO 9 (9 IS BEST) % (mass) OF PARTICLE SIZE SOLIDS IN OF SOLIDS.5 . Filter Selection Chart. Uplands Press.5 TO 500 0.0.001 – 0.B. Chase. Croydon.8 Table 7-5.1 0. Purchas.1 0.01 – 1 0.001 – 0.05 2 – 90 0.001 – 0.1 0.005 – 0.5 – 80 1 – 100 0. Solid/Liquid Separation Technology. 1981).2 2 – 90 5 – 90 0.000 0.001 .1 – 80 5-6 5 5 5 6-7 5-6 4-5 2-3 5-7 - 6 8 7 7 7 7 7 2 9 - 7-8 8-9 7-8 7-9 7-8 8-9 7 7-8 8 6 7 6 7-9 8 8 8 8 8 8 8 8 - 7 . (data from D.5 – 100 0.001 – 1 0.1 0. The University of Akron HANDOUT 7.SOLIDS PROCESSING HANDOUTS.5 – 100 1 – 500 5 – 100. PAN.200 0.000 20 – 100. TABLE WATER SCREEN DEEP BED 0.001 – 0.001 – 50 0.001 – 0. Chase.000 1 – 20.003 –0.000 90 – 20. Uplands Press.1 – 100 0.1 – 100 9 9 9 7-9 9 7 6 4 3 3 3 - 5 6 5 5 6 4 5 4 4 5 5 4 5 4 3 6 7 5 - 5 3 - 4 4 6-7 6-7 6-7 6-7 8 .05 – 5 50 – 80.000 0. 1981).9 Table 7-6.01 – 1 0. The University of Akron HANDOUT 7.000 2 – 20.1 – 100 0. Centrifuge Selection Chart (data from D.000 100 – 30. Solid/Liquid Separation Technology. George G.1 – 100 0. MICRONS PUSHER PEELER/SCRAPER WORM SCREEN OSCILLATING SCREEN BASKET CONICAL SCREEN SCREEN BOWL DECANTER TUBULAR BOWL DISC (MANUAL) DISC (SELF CLEANING) NOZZLE BOWL 8 – 90 5 – 80 3 – 90 3 – 90 2 –80 5 – 90 2 – 70 2 – 70 0.SOLIDS PROCESSING HANDOUTS. Croydon.000 100 – 20. Centrifuge TYPICAL OPERATING RANGE PERFORMANCE SCORE 0 TO 9 (9 IS BEST) CAKE DRYNESS WASH PERFORMANCE FILTRATE CLARITY CRYSTAL BREAKAGE % (mass) OF SOLIDS IN THE FEED PARTICLE SIZE OF SOLIDS. Purchas.000 100 – 20.B.000 2 – 20.08 0.02 – 2 0. Tiller. disc. horizontal belt or top-feed drum filter. sedimenting centrifuges Negligible cake Clarification Cartridges. Oct. continuous feed pusher centrifuge 0. horizontal belt. Selection based on cake formation rate (F. precoat drums. 1977. Progr. The University of Akron HANDOUT 7. disc and tubular centrifuges. 0. Eng.1 to 10 cm/min Medium Filtering Vacuum drum. screens. Cake Formation Rate (rate at Separation type Equipment which cake height grows in a gravity filter). pan filters. Chase. filter aid admix 9 .SOLIDS PROCESSING HANDOUTS.10 Table 7-7. 65-76).1 to 10 cm/hr Slow Filtering Pressure filters. granular beds.1 to 10 cm/sec Rapid Filtering Gravity pans. Chem. peeler centrifuge 0. George G. " Sep.SOLIDS NOTES 8. The viscosity of the liquid affects the viscosity of the slurry (and its associated pressure losses for pumping this slurry) and it affects the pressure loss for the liquid passing through the cake filter medium.. As a liquid passes through a filter its absolute pressure decreases.1 0. 8. There are several correlations available for estimating the slurry viscosity as given in Solids Notes 4: Bulk Properties of Powders and Slurries. PRETREATMENT OF S/L MIXTURES The separation of solids from liquids can often be enhanced by pretreating the slurry. A few of these techniques are expanded upon here.S. The greatest advantage in changing temperature is obtained with highly viscous materials.001 TEMPERATURE CHANGE Figure 8-1.01 0. Another pretreatment of the liquid is degassing. There are a number of factors that influence the separation and which can be improved by the pretreatment. Chase and M.1 Treatment of the Liquid The most common treatment of the liquid is to change its viscosity. In practice there are two techniques available for reducing the viscosity -raise the temperature -dilute with a less viscous liquid. Pressure drop increases as much as 300% have been reported. 26(1). 1000 100 VISCOSITY IN POISE 10 100 C 1 0. etc. we can see that a change in the liquid viscosity will result in a proportional change in the slurry viscosity.G. Typically. Sci. but it also has its drawbacks. the effect of temperature on a liquid viscosity is shown in the generalized curve in Figure 8-1. 117-126. A summary of pretreatment techniques are listed in Table 7-3. Willis. A large reduction in viscosity can be obtained from a small increase in the temperature. If dissolved gases are in the liquid then these gases may form bubbles within the cake (G. Chase. George G. Most of these correlations are of the form µ o = µ f (ε ) (8-1) From this. "Flow Resistance in Filter Cakes Due to Air. 1991). Technol. These bubbles reduce the area available for the liquid flux through the cake and hence cause the pressure drop to increase. The University of Akron 8. Viscosity – temperature relationship for liquids. Dilution with a less viscous liquid is a useful technique. 8-1 . For example. the viscosity of water can be reduced by almost 50% by heating it from 20° to 55° C. It increases the slurry volume and the filtrate may require additional processing such as liquid-liquid extraction. SOLIDS NOTES 8, George G. Chase, The University of Akron 8.2 Treatment of the Solid Particles 8.2.1 Coagulation and flocculation Terminology: Coagulation Describes the phenomena in which very fine particles of colloidal size adhere directly to each other as a consequence of Brownian motion (once repulsing electrical surface forces are sufficiently depressed by addition of ions). Formation of open aggregates formed by coagulations through bridging action of polymers between separate particles. Particle smaller than 1 µm. Dispersion of colloids (aerosol- dispersion of colloidal droplets in air) Particles <0.2 µm. Although colloids are larger than molecules, they are too small to be seen under a microscope Dispersions of larger particles Flocculation- ColloidSolutionSupercolloids- Suspensions- The separation of the very small particles in solutions present more problems than do larger particles. Hence techniques have been developed to agglomerate these particles to improve the separations. Solutions can be classified into either lyophilic (hydrophilic) or lyophobic (hydrophobic) colloids. LyophobicExamples – clays, hydrated oxides can be formed chemically or with mechanical mixingsensitive to addition of electrolytes and can be coagulated. Examples – proteins, humic acidsless sensitive to electrolytes. Very high concentrations of electrolyte salts required for precipitation. Lyophilic- The basic requirement for the formation of larger particles is for the small particles to come together. All particles carry a residual charge. Attracting forces such as Van der Waal’s and London forces are opposed by the repulsion between like charges. If the charge on the particles can be reduced, then close approach is possible. REPULSION BETWEEN PARTICLES + + - + - ATTRACTION BETWEEN PARTICLES Figure 8-2. Attraction and repulsion forces between particles. 8-2 SOLIDS NOTES 8, George G. Chase, The University of Akron NET REPULSION FORCE ELECTROSTATIC – PROPORTIONAL TO r -2 NET ATTRACTION LONDON-VDW – PROPORTIONAL TO r -6 DISTANCE BETWEEN PARTICLES Format 8-3. Relative attractive and Repulsive forces acting on small particles. 8.2.2 The Colloidal Model Most particles carry a residual charge on their solid surface. Usually it is negative (minerals and clay) but in some cases it is positive (sewage sludge). There are 3 postulated mechanisms which cause this charge: (1) Crystal Lattice Defects – Thus an excess of anions (negative) or cations (positive) exist at the surface, (2) Sparingly Soluble Ionic Crystals – when dispersed in water they exist in equilibrium with a concentration of product ions. The charge potential of the solid ψ0 is determined from the Nernst equilibrium condition c⎞ ⎛ RT ⎞ ⎛ ⎟ Ψ0 = ⎜ (8-2) ⎟ ln⎜ ⎜ ⎝ νF ⎠ ⎝ c0 ⎟ ⎠ where ν is the valence, F is the Faradays constant, and c0 is the zero charge concentration. (3) Adsorption of Ions From Solution – such as via hydrogen bonding when large organic molecules are absorbed. Such agents are useful in changing positively charged particles into negatively charged particles. The Double Layer The model for a net negatively charged particle is shown in Figure 8-4. The negative surface is surrounded by a layer of attached positive ions. This in turn is surrounded by a layer of loosely attracted negative ions (hence “double” layer). The charge extends out into the bulk and becomes diffuse and random (a Boltzmann distribution is assumed). A shear layer on surface exists between the attached positive layer and the loosely attracted negative layer. The Nernst potential is the potential between the bulk fluid at a long distance from the particle surface and the potential of the solid surface. The zeta, ζ, potential is the potential difference between the bulk fluid and the potential of the shear plane, which is the plane between the attached positive layer and the loosely attracted negative layer. The ζ potential is directly measurable whereas the Nernst potential is not easily measured. 8-3 SOLIDS NOTES 8, George G. Chase, The University of Akron + + + SHEAR SURFACE + Neg. Particle ATTACHED POSITIVE LAYER + + - - LOOSE NEGATIVE LAYER + ZETA POTENTIAL DISTANCE 1/ 2 POTENTIAL CURVE There are a number of mathematical models that describe the double layer model. They allow for calculating and estimating the potential and charge density on the surface of the particle, and the thickness of the double layer. The Gouy-Chapman model is dψ ⎛ 32πnkT ⎞ = ⎜− ⎟ ε ⎠ dx ⎝ where k NERNST POTENTIAL Figure 8-4. Double Layer Model. ⎛ νeψ ⎞ sinh ⎜ ⎟ ⎝ kT ⎠ (8-3) ν ε ψ n e N I = Boltzamnn constant = valence = bulk concentration of the ν ion = dielectric constant of the liquid bulk phase = double layer potential at distance x = electronic charge = Avogadro’s number = Ionic strength 8-4 George G.4 A practical result of this is that the smaller the double layer thickness the closer the particles may come together and consequently are easier to coagulate.(8-3) gives Ψ = Ψ0 exp(− χx ) where χ is the Debeye-Huckel function (8-4) χ= 8πe 2 N 2 I 1000εRT (8-5) The potential approaches zero (becomes small enough that it is essentially zero) at x ≈ 3/χ. The University of Akron Integration of Eq. and in presence of electrolyte. For small colloid particles of low charge. For water d≅ 2. Chase. µm 900 31 15 0.SOLIDS NOTES 8. is approximately 1/χ. Table 8-1. d. The addition of electrolyte (as long as it does not absorb on the surface) will reduce the double layer thickness. 8-5 . Some example values are listed in Table 8-1. we see that the thickness is inversely proportional to the square root of the ionic strength.3x10 −9 cm I (8-6) Hence. Examples of Double layer Thickness Medium Distilled water 10-4M NaCl 10-4M MgSO4 Sea Water d=1/χ. the thickness of the double layer. Chase. Hence. The University of Akron 8. electrical repulsion forces are absent. One obvious limitation to orthokinetic coagulation is if the velocity gradient becomes too large it can actually break up the flocs. The relative velocities of particles bring the particles together so that they can coagulate.SOLIDS NOTES 8. Paddles (stirred tanks) Baffles (in pipes. differential settling Various handbooks are available for sizing and designing such equipment. static mixers) Particles (packed or fluidized beds. Perikinetic Coagulation – coagulation due to Brownian motion Brownian motion alone has too low a collision rate (theory not presented here).2. or tanks) Pipes (wall affects.3 The Rate of Aggregation The rate of aggregation (i.. then aggregation depends upon the motion of the particles to collide. Orthokinetic Coagulation – coagulation due to stirring and agitation Orthokinetic coagulation occurs when velocity gradients exist within the system (Figure 8-5). If. it will not produce flocs of suitable size (1 mm) within a reasonable time. George G. 8-6 . with pretreatment of electrolyte. Velocity Profile Particles moving with the fluid Velocity Figure 8-5. Mixing and agitating devices can be divided into 4 groups. channels. bubble swarms. coagulation) depends upon the rate of collisions between particles and the probability that collided particles will stick together.e. Paddle Example Calculate the mean velocity gradient for a 4blade paddle in a 100 m3 tank of water at 20°C. Figure 8-7.5m × 3m ) 2(100m 3 )(10 −3 kg / m / s ) Which is adequate for the design.419 m/s Then 3 3m 0.2 (assumed) µ = 1. = 36 / s 8-7 . ⎛ min ⎞ V p = rn = r (2π )(rpm )⎜ ⎟ ⎝ 60 sec ⎠ The Mean Velocity Gradient is given by G= = Power Volume x Viscosity of Mixture C D ρ (V p − V ) A p 3 4rpm 1m (8-8) 2Vµ The rule of thumb for effective coagulation is 20 s-1 < G < 75 s-1. Paddle Wheel Stirrer.5 m G= 1. ρ = 1 × 103 kg/m3 CD = 1.8) 〈Vp〉 Vp = Mean paddle speed r Figure 8-6. Example Paddle Wheel Stirrer. George G.SOLIDS NOTES 8.0 × 10-3 kg/ms Vp = (1 m)(2π)(4 rpm)(8 min/60 s) = 0.2(10 2 kg / m 2 )(0. Chase.33m / s ) (4 × 0. The University of Akron Paddle Coagulation For the paddle wheel design the power is given by Power = Force × Distance/Time = Force × Velocity = C D ρ (V p − V ) Ap 2 n TOTAL AREA Ap = ∑ A 2 Drag Force × (V p − V ) Relative velocity between paddle and fluid (8-7) A A where CD= Drag Coefficient Ap = Projected paddle area of all paddles (Vp-V) ≈ (0. the Ergun Equation. Re ≤ 1 . Chase.(4-25) in Chapter 4 simplifies to the Carman-Kozeny equation as ∆P 180µV0 (1 − ε ) 2 = 2 3 L dp ε gc and G becomes (8-13) G= Q∆Pg c power = fluid volume x viscosity εALµ 180(1 − ε ) 2 V02 (1 − ε )V0 G= = 6 5 2 4 d pε d pε 2 The same criterion for pipe flow applies here 104 < G t < 105 where the residence time is given by ( ) (8-14) t= pore volume εL = volumetric flow rate V0 (8-15) 8-8 . the pressure loss is given by ∆P ρ where = hf hf = 4 f L V2 D 2g c (8-9) (8-10) f = Friction factor L = Pipe length D = Diameter gc = gravity constant The mean velocity gradient is given by G= Q∆Pg c power = volume x vis cos ity LAµ (8-11) Substituting in Eqs.SOLIDS NOTES 8. Eq. For spherical particles and low Reynolds number. Packed Bed Coagulation Steep velocity gradients are possible in packed beds.(8-9) and (8-10) gives G= 2 fV 3 Dµ (8-12) In this geometry the criterion for G is 104 < G t < 105 for coagulation where t=L/V is the fluid residence time. The University of Akron Pipe Coagulators In straight horizontal pipes. George G. 00139m / s Gt = (432 s )(140 / s ) = 6.com/nutrition/pdfs/zeta. Cf. The particles migrate in the electrical field due to the strength of the field. Electro-phoretic movement of small particles due to a voltage potential See website http://www. One way is to measure the electro-phoretic effect.4) ⎝ 3600 s / hr ⎠ 0 .6 m t= = 432 s 0.5 mm ε = 0.pdf for more information on Zeta Potential. The basic feature is to observe and measure the velocity of particles through a microscope. Lists of Cf for electrolytes and solutions exist in literature. The observed velocity. Electrical field applied to a solution with particles. 8. A large number of salts such as Al2(SO4)3 and Na2P2O7 are available. There is a critical concentration. Normally we want ς to be zero ( so that there is minimal repulsing force due to charge). The usual dosage for a useful coagulant in water is 100 to 400 mg/liter. George G.SOLIDS NOTES 8. Several techniques are possible.0005m(0. Figure 8-8.4) ⎛ 5m / hr ⎞ ⎟ = 140 / s 2 ⎜ 0.3 Zeta Potential A practical way to characterize the double layer is to measure its zeta potential. Commercial equipment are available (such as the Zeta Meter). which is the concentration required of the salt to cause coagulation.6 m Vo = 5 m/hr G=6 5 (1 − 0. this is appropriate because large flocs could clog the sand bed. Chase.4 L = 0.oxywave. VE. 8-9 . However. The University of Akron Packed Bed Example: Calculate Gt and G for a fixed sand bed filter where dp = 0.04 x10 4 G is a relatively high value so large flocs will not form. Gt is in the design range criteria. can be related to the zeta potential by VE = εEς 4πµ (8-16) ε = dielectric constant of fluid E = external field ς = zeta potential Changes in zeta potential ς can be observed when electrolyte is added as shown in Figure 7-4. Na+ O O Poly quaternary ester R1 Notes: RNH2 HC-C=O R-C-H Amino Carbonyl Aldehydic O C=C-H Vinylic CH2=CHCOOH Acrylic Acid (Reacts with base and GST sodium acrylate) C-O-C2H4-N+ ClCH2=CH R2 R3 POLYMERS Non-Ionic Polyacrylamides (CH2-CH) CONH2 Polyethyleneoxide (O-CH2-CH2) x x Anionic Acrylamide Co-Polymer (CH2-CH) COOH x (CH2-CH) CONH2 y Polyacrylics (CH2-CH) COOH x Cationic Polyamines (CH2-CH2NH-CH2CH2NH) x 8-10 .SOLIDS NOTES 8. Chase.4 Flocculation by Polyelectrolytes The term flocculation is being used here to refer to the effects that polyelectrolytes (polymers) have on causing small particles to form large aggregates. George G. Chemical Nature of Synthetic Polyelectrolytes Types of synthetic polymer flocculants fall into three categoreis: non-ionic (neutral) anionic (negative) cationic (positive) MONOMERS Acrylamide Sodium Acrylate CH2=CH-CONH2 CH2=CH-C-O. The University of Akron 8. One of the greatest advances in recent years in solid-liquid separation has been the development of polymers with remarkable properties to flocculate solutions when added only in trace quantities. The University of Akron Acrylamide Co-Polymers (CH2-CH) CONH2 x (CH2-CH) CH2 N+ ClCH3 CH3 CH3 y (CH2-CH) CONH2 x (CH2-CH) CO O. Positively charged polymers wrapping around negatively charged particles to form flocs.SOLIDS NOTES 8. CRC Press. “Polyelectrolytes for Water and Waste Water Treatment”. Chase. Molecular weights are classified as High Medium Low Very Low 20x106 10x106 5x106 <1x106 More on chemistry of polyelectrolytes is given by W. 8-11 . Mode of Action and Application The mechanism of flocculation is charge neutralization and bridging. and (2) Charge density distribution within the polymer molecule. FL.Na+ y The anionic/cationic character can be altered by co-polymerization of the various monomers. Boca Raton. Any one product is characterized by (1) Average molecular weight.L. George G. + + + + + Increasing Ionic Strength + + + + + + + + + + + + Figure 8-9.K. 1981. Schwuyer. The University of Akron The best choice of polyelectrolyte is best maybe after laboratory trials of samples of the liquid to be clarified. the polymer with the highest molecular weight will give the fastest sedimentation rate. for a given charge density. Chase. recycle settled solids into the stream 8-12 .SOLIDS NOTES 8. Some practical points on efficient polyelectrolyte usage: Add the polyelectrolyte to the main stream in a very dilute (<0. add recycled or other dilution water to the system At low solids concentrations.1%) solution Add as near as possible to the point where flocculation is required Add at points of local turbulence Add in stages at different points Add across whole stream to be treated Avoid turbulence after floc formation At high solids concentrations. Usually. George G. If a mass of particles is disturbed (by stirring. 9. it is easier for smaller particles to move into the void spaces between larger particles during the disturbance. 9-1 . density. vibrating) such that the individual particles move relative to each other. For coarse particles. Chase. Segregation is important in handling of powders because in many operations you want to avoid concentrations of one particle type or another. Differences in size. PERCOLATION OF FINES (also called sifting). Segregation processes are the opposite of mixing processes. groups of fine particles can travel farther than single large particles. or other material properties occurs when concentrations of one type of particle in a hopper or pile is much greater than the average for the entire mass. in a storage hopper there is a tendency for small particles to migrate downward in the direction of gravity. which in the latter the particles are blended into relatively uniform consistency. Segregation upon pouring a powder into a pile occurs frequently due to the percolation (as well as trajectory) mechanisms (Figure 9-1). George G. density. A very small size difference is sometimes sufficient to cause a measurable segregation. If a powder has a size distribution. SEGREGATION Separation articles by size. shape and resilience all contribute to segregation. Hence. • HIGHER CONCENTRATION IN SMALLER PARTICLES HIGHER CONCENTRATION IN LARGER PARTICLES Figure 9-1.1 Mechanisms We need to understand the mechanisms that cause segregation before we can design systems to minimize the effect. The University of Akron 9. a rearrangement in the packing occurs. Four primary mechanisms cause segregation: • TRAJECTORY SEGREGATION. When a powder cloud is in flight. the larger particles tend to travel the farthest. particle size difference is the most important parameter. Segregation of particles when poured into a pile due to percolation. The process of separating the particles into different relative concentrations is referred to as segregation is the opposite.SOLIDS NOTES 9. By far. shaking. Usually segregation is not a problem for particles smaller than about 30 microns for particle densities in the 2000 to 3000 kg/m3 range. The air velocity may exceed the terminal velocity of the smaller particles. However. Upon vibrating the beaker the ball bearing rises to the top of the sand. This occurs even when the larger particle is much denser than the fine particles. This effect can be demonstrated by submerging a large steel ball bearing in sand in a beaker. • 9. o Avoid vibrations. George G.SOLIDS NOTES 9. This can cause the fines to suspend while the coarse particles settle out. stirring (unless done to promote mixing). if the surface forces cause the particles to agglomerate into larger particles. Fibers and flakes do not segregate as easily as spherical shapes. The University of Akron • RISE OF COARSE PARTICLES UPON VIBRATION. • • Use continuous mixing. or is poured into the top of a hopper. a large volume of air is displaced. For more dense particles the size must be smaller than 30 microns. ELUTRIATION SEGREGATION (also called fluidization). Use materials with large aspect ratios. 9-2 . o Use mass flow hoppers.2 Reduction in Segregation Methods to reduce segregation are aimed at reducing the effects that cause segregation. Avoid situations that promote segregation o Avoid pouring where the powder forms a sloping surface. When a powder is discharged from a hopper. Rough surfaces also reduce the segregation effect. • • Make sizes of components as close as possible to the same size. o Avoid pouring into the core of a hopper. For very small particles the surface forces act to minimize the segregation effect on individual particles. shaking. Chase. • • A small amount of liquid may be added to a powder to increase its attractive forces between particles and minimize fines migration. then segregation may become a problem again. Reduce the absolute size of the particles. The fines eventually settle on top of the suspension and form a layer on top of the powder that is concentrated in the fines. Change the particle size mixture at the supply site to a more uniform size. Is the material likely to segregate? Yes. George G. 9-3 . The University of Akron 9.3 Example A transport truck is loaded with a well mixed powder with a wide size distribution. Chase. due to vibrations from the transport. What options do you have to minimize the segregation at the delivery site? Additives to reduce segregation (liquid? Fibers?) Do the mixing at the delivery site. The truck delivers the product to a facility 20 miles from the its loading point. What material will exit first? Last? Fines are more likely to exit first and coarse to exit last.SOLIDS NOTES 9. Cheremisinoff and R. 3. shown in Figure 10-3. J. October 1961. though not necessarily all with the same velocity. Wiley. The University of Akron 10. The characteristics and differences between the flows are depicted in Figure 101. New York. chapter 31. His primary works are published in "Gravity Flow of Bulk Solids". hoppers must be designed such that they are easy to unload. 1990. More importantly. His work identified the criteria that affect material flow in storage vessels.M. mass flow and funnel flow. Midland. if it flows at all.. M. 10-1 . Figure 10-2 shows some of the more common designs found for mass flow hoppers. “Flow of Solids in Bunkers. These issues and others discussed here are important to consider when designing storage hoppers. Bridgwater and A. George G. N. This was all changed by the research of Andrew W. The primary difference between mass and funnel flow is that in mass flow all of the material in the bin is in motion. Ann Arbor. including the equations and measurement of the necessary material properties. a variety of designs are possible for funnel flow hoppers. the way the hopper is designed affects how much of the stored material can discharge and whether there mixing of solid sizes or dead space that reduces the effective holding capacity of the hopper. 2.” in Handbook of Fluids in Motion.1 Flow Modes There are two primary and distinct types of flow of solids in hoppers. UK. Holdich. Hoppers come in a variety of shapes and designs. November 1964. * Other references that give a good summary on this topic include: 1. In funnel flow only a core of material in the center above the hopper outlet is in motion while material next to the walls is stationary (stagnant). Butterworth. 2002. Gupta eds. at least as far back as man has harvested and stored crops. 807-846. Chase. R. Fundamentals of Particle Technology. Also. The way the hopper is designed affects the rate of flow of the powder out of the hopper. Rhodes. Jenike developed the theory and methods to apply the theory. 1983. Prior to the 1960s storage bins were designed largely by guessing. Loughborough. Also. Jenike in the 1960s. There is also a special case that is a combination of these two flows called expanded flow. University of Utah Engineering Experiment Station.SOLIDS NOTES 10.P. 10. not just conical. These flows get their names from the way in which solids move in the hoppers. Principles of Powder Technology. Bulletin 108. and Bulletin 123. Hoppers must be designed such that they are easy to load.* Hoppers are used in industry for protection and storage of powdered materials. Scott. HOPPER DESIGN People have stored powdered materials for thousands of years. 75 TO 1 x D STAGNANT REGION FUNNEL FLOW MASS FLOW (A) MASS FLOW (B) FUNNEL FLOW (C) EXPANDED FLOW Figure 10-1. The University of Akron D ACTIVE FLOW CHANNEL MINIMUM LEVEL FOR MASS FLOW IN HOPPER IS 0. In funnel flow (B) the material moves in a central core with stagnant material near the walls.SOLIDS NOTES 10. Expanded flow (C) is a combination of mass flow in the hopper exit and funnel flow in the bin above the hopper (normally used in retrofit situations). George G. 10-2 . Common designs for mass flow hoppers. Chase. β Da (A) CONICAL HOPPER (B) SQUARE OPENING (C) CHISEL βp L Dp (E) WEDGE (F) PYRAMID Figure 10-2. In mass flow (A) all material moves in the bin including near the walls. The problems that we would like to solve or avoid are • RATHOLING/PIPING. The University of Akron Dp (A) PYRAMID. The resulting effect is a sluggish flow of solids as the air penetrates in a short distance freeing a layer of material and the process starts over with the air penetrating into the freshly exposed surface of material (Figure 10-4c). either the material does not discharge adequately from the opening in the hopper or the material segregates during the flow. but strong enough that the material discharge rate slows down while air tries to penetrate into the packed material to loosen up some of the material. FLOW IS TOO SLOW. Ratholing or piping occurs when the core of the hopper discharges (as in funnel flow) but the stagnant sides are stable enough to remain in place without flowing. The material does not exit from the hopper fast enough to feed follow on processes. NO FLOW DUE TO ARCHING OR DOMING. Common designs for funnel flow hoppers. FLUSHING. George G. The material is cohesive enough that the particles form arch bridges or domes that hold overburden material in place and stop the flow completely (Figure 10-4b).SOLIDS NOTES 10. Dead spaces in the bin can prevent a bin from complete discharge of the material.2 Hopper Design Problems Hopper design problems are normally one of two types. Flushing occurs when the material is not cohesive enough to form a stable dome. 10. SQUARE OPENING Dc (B) CONICAL (C) CYLINDRICAL FLAT-BOTTOMED SLOT OPENING D c (D) CYLINDRICAL FLAT-BOTTOMED CIRCULAR OPENIN Figure 10-3. INCOMPLETE EMPTYING. • • • • 10-3 . Chase. leaving a hole down through the center of the solids stored in the bin (See Figure 10-4a). PARTICULATE SOLIDS BIN WALLS SOLIDS OUT AIR IN (A) RATHOLING OR PIPING (B) BRIDGING/DOMING (C) FLUSHING Figure 10-4. For many materials. Mass flow is not necessary in all cases. 10-4 . if allowed to sit in a hopper over a long period of time the particles tend to rearrange themselves so that they become more tightly packed together.SOLIDS NOTES 10. The required cone angle from the vertical axis for mass flow to occur ranges from 40° to 0°. Moisture in the air can react with or dissolve some solid materials such as cement and salt.J. A good description of this effect is given by Griffith (E. The consolidated materials are more difficult to flow and tend to bridge or rat hole. NY. • TIME CONSOLIDATION. George G. Different size and density particles tend to segregate due to vibrations and a percolation action of the smaller particles moving through the void space between the larger particles.3 Predicting Mass Flow Many of the problems associated with bin and hopper design can be avoided by designing the hopper to operate in mass flow mode. Another important effect is called caking. This effect is referred to as Dense Packing by Foust in the bed porosity in Figure 4-5. In some situations a mass flow hopper design is not practical due to the head room required. VCH Publishers. Griffith. CAKING. When the air humidity changes the dissolved solids re-solidify and can cause particles to grow together. Cake Formation in Particulate Systems. In most applications if you have a choice you want mass flow. Common problems in bin/hopper design. 1991). • 10. The University of Akron • SEGREGATION. Table 10-1 summarizes the key advantages and disadvantages of both mass flow and funnel flow hoppers. Chase. But in the extreme cases or in cases in which mass flow is not really necessary then you may opt for the shorter funnel flow hopper design. Caking refers to the physiochemical bonding between particles what occurs due to changes in humidity. Some of these have been mentioned above.SOLIDS NOTES 10. Interfacial forces. Chase. Usually. George G. Binding mechanisms include: 1. The University of Akron Table 10-1.3. 2. Adhesion and Cohesion 3.1 BINDING MECHANISMS There are a number of mechanisms that cause solid materials to bind together and thus make flow difficult if not impossible. Caking) • • • • • • Mineral Bridges Chemical reactions Partial melting Binder hardening Crystallization of dissolved substances There are a number of effects that are lumped together and are termed adhesion and cohesion. Solids Bridge (ie. very small particles display adhesion properties. These include mechanically deformable particles that can plastically deform and bind to each other or with bin walls. 10. Advantages and Disadvantages of Mass and Funnel Flow Hoppers MASS FLOW ADVANTAGES • • • • • • • • Flow is more consistent Reduced radial segregation Stresses on walls are more predictable Effective use of full bin capacity First-in = First-out More wear of wall surfaces Higher stresses on the walls More head room required FUNNEL FLOW • Low head room required DISADVANTAGES • • • • • • • Rat holing Segregation First-in = Last-out Time consolidation effects can be severe Poor distribution of stresses on walls may cause silo collapse Flooding Reduction of effective storage capacity. 10-5 . but if the particles are in close enough contact the attractive forces are stronger. One of the more common test apparatus is the Jenike Shear Tester.they become entangled. liquid and gas) and interfacial tensions. Chase. • 5.2 TESTING REQUIREMENTS To design storage hoppers. the following material properties are needed: • • • • Internal friction coefficient Wall friction coefficient Permeability Compressibility Other factors that should be considered include temperature and moisture content along with phase diagrams if caking may be a problem. The movement of the sample holder causes shear between a powder sample and a sample of the hopper wall as in Figure 10-5(a) to determine the wall friction coefficient. These effects are due the contact surfaces between three phases (solid. George G. The University of Akron • Interfacial forces include liquid bridges and capillary forces between particles. There are also short range repulsive forces. Or.SOLIDS NOTES 10. 10-6 . However. the angle of repose alone is not sufficient to account for all of the mechanisms affecting hopper performance. Attractive forces. The Jenike Shear Tester has similarity to the Triaxial shear tester mentioned in Chapter 4.3. and its popularity among engineers is not due to its usefulness but due to the ease with which it can be measured. Interlocking forces are due to the geometric entanglement that occurs with fibrous materials. The angle of repose is only useful in determining the contour of a pile. 4. the movement causes a shear internally in the powder sample as in Figure 10-5(b) to determine the internal coefficient of friction. 10. Interlocking forces. The powder sample is placed in a sample holder. analogous to what happens when you store coat hangers in a box . as well as longer range electrostatic and magnetic forces. • Many earlier bin designs were based upon the angle of repose (see Chapter 4). Attractive forces include intermolecular forces such as van der Waal’s force. the coefficient of friction is expressed as the “angle” of wall friction given by φ as φ = arctan( µ ) . The friction tests are simple application of physics to determine the friction coefficients as discussed in Chapter 4 where the shearing force F is related to the normal force N by the coefficient of friction µ in the equation F = µN (10-1) As commonly practiced. Chase. (10-2) Other shear test devices are available commercially. which operates by placing a sample between two circular disks and rotating the disks about their center axis relative to one another. The powder sample is sheared within itself.S. The powder sample is in contact with a wall sample. The University of Akron BRACKET APPLIED WEIGHT (NORMAL FORCE) COVER SHEAR FORCE RING SHEAR FORCE SAMPLE WALL MATERIAL POWDER SAMPLE (a) Wall Friction Test.Z.SOLIDS NOTES 10. BRACKET APPLIED WEIGHT (NORMAL FORCE) COVER SHEAR FORCE RING SHEAR FORCE POWDER SAMPLE (b) Internal Friction Test. such as the rotating disk test called Peschl Tester. developed by I.A. Figure 10-5. Internal and wall friction tests with the Janike Shear Tester. Shearing occurs between the powder and wall samples. This has an advantage over the Jenike Shear tester because the 10-7 . George G. Peschl of The Netherlands. giving the particles opportunity to compact. Chase. The University of Akron shearing can occur for longer periods of time. acting on the cross sectional area. The measured results have not been fully tested for use in bin design and sealing powders can be a problem. Differential force balance on a cylindrical storage bin. There are also shear stresses.3 STRESSES IN HOPPERS AND SILOS Consider the equilibrium of forces acting on a differential element.SOLIDS NOTES 10. This device is good for elastic materials and for pastes. dz . the annular shear cell has infinite travel. Pv . The cover is rotated relative to the rest of the assembly. The sample is placed in the annular space between two cylinders. causing shearing of the sample. In the stationary situation the surrounding fluid (air) pressure acts uniformly on all solid particles throughout the silo. However. there are compressive normal stresses.3. We list the various components contributing to the force in the z-direction: Overburden normal stress acting downward on the surface at z Normal stress acting upward on the surface at z + ∆z Shear stress acting on the silo walls acting upward Gravity force acting downward on the differential element Pv A z z D Pv − ( Pv + dPv ) A z z + dz Pv + dPv τR − τ R π D dz ρ o A dz g Figure 10-6. A. George G. As with the rotating disk tester. due to the overburden of material above the volume element. of the solid phase acting on the silo walls. Another test that has been around for many years is the annular shear cell. 10. However friction between the moving parts of the sample holder can cause doubt about the accuracy of the measured results and there are some doubts about the assumed zero stress at the center of the rotating disks. 10-8 . A cover is placed on top of the sample to hold the powder in the annular space and to apply the desired normal force. τ R . in a straight sided cylindrical silo (Figure 10-6). where A = π D2 4 .A. George G.SOLIDS NOTES 10. This latter expression is known as the Janssen Equation. Janssen. This gives the balance of forces as A( Pv ) − A( Pv + dPv ) − τ R π Ddz + ρ o Adz g = 0 which reduces to − A(dPv ) − τ R π Ddz + ρ o Adz g = 0 (10-4) (10-3) From physics we relate the shear stress at the wall to the lateral normal stress acting in the radial direction at the wall. ⎜ 1 − exp⎜ − ⎝ D ⎠⎠ 4 µ Kg c ⎝ (10-9) where the g c is the gravity constant conversion factor to convert the result from units of mass to units of force. 39. Pw . Zeitschrift.(10-5) into (10-4) gives − A(dPv ) − µ Pw π Ddz + ρ o Adz g = 0 which has both Pw and Pv terms. where Pw = KPv . τ R = µ Pw Substitution of Eq. (10-7) into (10-6) and rearranging. dPv = − D ⎝ v 4µ K ⎠ Equation (10-8) is integrated with the boundary condition that Pv = 0 at z = 0 . (10-7) Substituting Eq. with the coefficient of friction. to obtain ρ o gD ⎛ ⎛ 4 µ Kz ⎞ ⎞ Pv = ⎟⎟ . Versuche über Getreidedruck in Silozellen. Verein Deutcher Ingenieure. Chase. µ . (10-5) (10-6) Janssen solved this equation (H. 1045-1049) by assuming that the vertical normal stress is proportional to the lateral normal stress (section 4. we get (10-8) 4µ K ⎛ ρ o gD ⎞ ⎜P − ⎟ dz . 10-9 . When we plot the pressure in the silo as a function of depth from the free surface of the granular material at the top we get a plot as shown in Figure 10-7.3). The University of Akron At steady state (no accelerations.2. August 1985. or neglecting inertial terms) the sum of the forces must equal zero. This is one reason why commercial silos are designed tall and narrow rather than short and squat. The University of Akron BULK SOLIDS HYDROSTATIC z PRESSURE Figure 10-7. Utah.θ ) .W. Utah Engineering Experiment Station. 10) as indicated in Figure 10-8. (10- θ r HOPPER OPENING σ Figure 10-8. 123. Andrew Jenike (A. in the hopper is a function of position (r . Note that in Figure 10-7 the asymptotic pressure for large depth is only a function of the silo diameter and not on depth. University of Utah. The stress. σ . Chase. 1964) postulated that the magnitude of stress in the converging section is proportional to the distance from the hopper apex (as well as a dependence on the angle). 10-10 . At the bottom of the silo is the converging hopper section. Storage and Flow of Solids. Bulletin No. Vertical normal stress profile in a silo. Jenike. Salt Lake City. George G.SOLIDS NOTES 10. The stress is written as σ = σ (r .θ ) . the results of those calculations shown in Figure 10-9 give us insight as to the conditions at the bottom of the silo at the hopper discharge. George G. Hopper stress field including the stresses in the converging hopper discharge section.SOLIDS NOTES 10. The University of Akron The rigorous calculations applying the radial stress field assumption are beyond the scope of this discussion. BULK SOLIDS STRESS STORAGE SILO VERTICAL POSITION HYDROSTATIC STRESS HOPPER STRESS Figure 10-9. Chase. However. Figure 10-9 shows that there is essentially no stress at the hopper outlet. 10-11 . This is good because it allows dischargers such as screws and rotary valves to turn easily. (10-9).4 psi) The result in (b) is a factor of about 13 times greater than the normal wall stress calculated in (a). APPLICATION OF JANSSEN’S EQUATION A large welded steel silo 4 meters in diameter and 20 meters high is to be built.364)(0.1 psi) (b) If the silo was filled with water instead of granular solids. Chase.(10-9) we get the vertical stress at the bottom of the silo: Pv = (560 kg/m3 )(9. is assumed to be 0. George G.267 N/m2 (2.4) ⎞ ⎞ ⎜ 1 − exp⎜ = ⎟⎟ 2 ⎜ ⎟ 4(0. Eq. The plastic pellets have the following characteristics ρ o = 560 kg / m 3 φ = 20 o SOLUTION: (a) The Janssen Equation.364)(0. Estimate the pressure on the wall at the bottom of the silo if the silo is filled with (a) plastic pellets.4. Substituting these quantities into Eq.668 N/m2 (or 5. is for silos of circular cross section. The coefficient of wall friction is obtained by inverting Eq.4(20 m)(0.364 K.2 psi) To estimate the normal stress on the wall we apply Janssen’s assumption ρ o gD ⎛ ⎛ 4 zµ K ⎞ ⎞ ⎜ 1 − exp⎜ − ⎟⎟ ⎜ 4µ Kg c ⎝ D ⎠⎟ ⎝ ⎠ Pw = KPv = 0. This is due to the wall friction exerting a vertical upward force on the granular solids. The University of Akron EXAMPLE 10-1.668 N/m2 ) = 14. but it is not often measured. and (b) water. The Janssen coefficient can vary with material as indicated in Figure 4-4.807 m/s 2 )(4 m) ⎛ ⎛ . Diameter and height are given in the problem statement.(10-2) as µ = tan( 20o ) = 0.4 (35. 10-12 . The silo has a central discharge on a flat bottom. the Janssen Coefficient.4)(1kg m/ N s ) ⎝ 4m ⎠⎠ ⎝ = 35.SOLIDS NOTES 10.140 N / m 2 (28. the pressure at depth H is given by P = ρ gH gc 1000 kg / m 3 (9.807m / s2 )(20 m) = 1 kg m / N s2 = 196. One way of analyzing stresses in a solid is through Mohr Circles.3. George G.SOLIDS NOTES 10. (c) Mohr Circle. Mohr circles relate shear stress to the normal stress. The properties that are used in the design of a mass flow hopper are the effective angle of internal friction. Chase.4 HOPPER ANGLE AND OUTLET SIZE FOR MASS FL0W HOPPERS To size and design a hopper we determine the design necessary for mass flow operation based upon the material properties. Given normal and shear stresses on a solid block (Figure 10-10). 10. 10-13 . the material flow function. In a mass flow hopper during discharge the stress distribution is such that a stable arch or funnel flow do not occur and therefore the flow will not stop.4.1 The Material Flow Function Whether a hopper operates in mass flow or funnel flow depends on the flow properties of the powder material and how it interacts with the hopper walls. (b) Triangle shows maximum and minimum normal forces with no shear. it is possible to find an angle ( θ ) of a surface within the block such that the normal stress on the new surface is a maximum (or minimum) and the shear forces are zero. and the angle of wall friction between the powder material and the wall material.3. (a) Angle θ is a surface with maximum/minimum normal force with zero shear. This analysis can be used in the design of a new hopper or to check the suitability of an existing hopper for use with a particular material. Stresses on a block. Norma l σ xx σ yx σ xy σ yy SHEAR STRESS AXIS Shea r Ma x (b ) Mi n σ yy MAX NORMAL σ xy σ yx (a) θ MOHR CIRCLE NORMAL AXIS σ xx (c) MIN NORMAL Figure 10-10. The University of Akron 10. The University of Akron Similarly. Several different maximum loads are applied to generate at least three different JYL curves.SOLIDS NOTES 10. the JYL is not a straight line and does not pass through the origin. Below the JYL the normal stress is too large for the powder to flow at the give shear stress. 10-14 . The maximum and minimums are plotted on the stress plot (Figure 10-10 (c)). we can find an angle at which the shear stress is a maximum or minimum. Usually this occurs 90° from the maximum normal stress. George G. For powder flow in a hopper. we want to know the shear stress needed to initiate flow (overcome the coefficient of static friction). The JYL represents a surface that divides between operating conditions. A material’s flowability depends upon the shear strength and how the shear strength changes with compacting stresses. Chase. Jeniky Yield Locus (JYL) curve plotted from experimental data of Shear Stress versus Normal Stress for a given starting Critical Point. In experiments on a shear tester you must pre-stress the sample by applying a maximum load (critical point on the Jeniky Yield Locus (JYL) curve) and then reducing the applied stress during the experiments (Figure 10-11). • If the material is cohesive. When extrapolated down to the zero shear stress the JYL crosses perpendicular to the Normal Stress axis. • Shear Stress JYL Critical Point (End Point) Normal Stress Figure 10-11. Above the JYL the shear stress is sufficient to cause powder movement. fc . We are interested in two in particular. Mohr Circle passing through origin and tangent to the JYL. This is used to determine the value of f c . 10-15 . The diameter of this Mohr Circle is the Unconfined Yield Stress (UYS). 1. The Mohr Circle passing through the origin and tangent to the JYL represents the stress needed to initiate flow in an arch (ie. Chase. to cause the arch to collapse) (Figure 10-12). JYL Shear Stress Normal Stress fc Figure 10-12. the Unconfined Yield Stress (UYS). The University of Akron Several Mohr Circles may be drawn that are tangent to the JYL.SOLIDS NOTES 10. George G. George G. The University of Akron 2. Stresses below the MFF and the arch is stable. The f c versus σ 1 data are plotted to obtain the Material Flow Function (MFF) (Figure 10-14). Each JYL gives an f c and σ 1 . For points above the MFF the material is flowing and the frictional analysis in deriving the MFF does not apply. 10-16 . The Mohr Circle passing through the critical point (end point) represents the state of the material at the compacting stress (Figure 10-13). Critical Point (End Point) JYL Shear Stress Mohr circle Intersects Critical Point and is tangent to the JYL Normal Stress σ1 Figure 10-13. Mohr Circle passing through the critical point and tangent to the JYL. This represents the conditions for failure without volume change.SOLIDS NOTES 10. This is used to determine the value of σ 1 . We assume it represents the stress conditions of the powder as it flows downward towards the outlet. Chase. The MFF curve represents the stress needed to make an arch collapse as a function of the compacting stress under which it was formed. Chase. Critical Points Multiple JYL curves Shear Stress δ Normal Stress Figure 10-15.SOLIDS NOTES 10.2 The Effective Angle of Internal Friction The effective angle of internal friction. Non-cohesive materials have small f c values. 10-17 . The Material Flow Function.3. More cohesive materials have larger f c values. George G. The University of Akron More cohesive fc MFF Less cohesive Non-cohesive σ1 Figure 10-14. δ .4. δ is the angle of the slope of the line through the origin that is tangent to the Mohr Circles at the critical point. The angle δ is the angle of the tangent line to the Mohr Circles at the critical points that passes through the origin. 10. as shown in Figure 10-15. is determined from the JYL plot. 3. 1964) showed that for an element at any position inside of a mass flow hopper. Shear Stress δw Normal Stress Figure 10-16. δ w . ff = σ compacting stress = 1 applied shear stress AS (10-11) Jenike published charts from which ff is determined. Storage and Flow of Solids. and (2) the shear stresses in the material tend to make it flow.W. The University of Akron 10. Charts for symmetrical slot outlet hoppers and for conical hoppers are shown in Figures 10-17 and 10-18. This property is determined from experiments run with the shear tester as shown in Figure 10-5(a) where the measured shear force is plotted versus to the normal load (Figure 10-16).4. Plot to determine the wall angle. George G. between the powder and the wall of the hopper. University of Utah. Bulletin No. Often the data are linear. If they are nonlinear then the smallest angle is used. Salt Lake City..3 The Angle of Wall Friction The last property needed is the wall friction. Utah. Utah Engineering Experiment Station.4.SOLIDS NOTES 10. Chase. Jenike (A.4 Determining the Minimum Hopper Outlet Size The forces acting on the powdered material stored in a hopper tend to (1) compact the powder (i. 10-18 . reduce its bulk density). φ w .3. 10. Jenike. 123.e. the ratio of the compacting stress to the shear stress has a constant value that he called the flow factor: flow factor. Chase. Flow Factor. George G. ff 10-19 .1 δ=40 1 60 70 0 0 10 20 30 40 50 Semi-included angle.6 δ=30 1.7 Wall Friction.19 . φ w = 22 o and δ = 50 o gives θ = 30.9 Wall Friction Curves 40 1.3 δ=50 δ=60 δ=50 δ=60 1.8 δ=40 δ=35 1. For example (dashed arrows). Design chart for symmetrical slot outlet hoppers.2 δ=30 1. degrees Figure 10-17. degrees 30 1. The University of Akron 50 δ=60 δ=50 δ=30 2 Flow Factor Curves 1.SOLIDS NOTES 10.5 o and ff = 1.4 δ=45 10 1.5 20 δ=40 1. Design chart for conical outlet hoppers. For example.6 20 1. George G. ff 10-20 .SOLIDS NOTES 10.8 δ=30 δ=35 Wall Friction. φ w = 22 o and δ = 50 o gives θ = 20. The University of Akron 40 δ=60 δ=50 Wall Friction Curves 2 1.1 0 0 5 10 15 20 25 30 35 40 45 1 Semi-included angle.5 o and ff = 1.29 .5 1.9 Flow Factor Curves δ=40 30 δ=30 1. Flow Factor.4 δ=40 1. Chase.3 10 δ=50 δ=60 δ=30 δ=40 δ=50 δ=60 1. degrees Figure 10-18.2 1. degrees 1.7 1. Conical Hopper with outlet size D and semi included angle θ .SOLIDS NOTES 10. Therefore the hopper should operate where the 1/ff curve is above the MFF curve. the opening diameter.4. ie. Correlations relating the outlet size to the CAS are provided. The University of Akron The intersection of the MFF curve in Figure 10-14 with the line through the origin having a slope of 1/ff is the Critical Applied Stress (CAS) (Figure 10-19). George G. Recall that the condition of flow (no arching) occurs for points on or above the MFF curve. Chase. D . the outlet opening size must ensure that the applied stress exceeds the CAS. Intersection of the 1/ff and MFF curves defines the Critical Applied Stress (CAS). For conical hoppers. above or to the right of the CAS. Figure 10-20. AS MFF σ1 Figure 10-19. Hence. is given by D = H (θ ) CAS ρ g / gc o (10-12) (10-13) D θ Semi included angle H (θ ) = 2 + θ 60 Where θ is in degrees. 1/ff curve CAS Unconfined Yield Stress. Typical values for H are about 2. f c or Applied Stress. 10-21 . from the charts in Figures 10-17 or 10-18. The stress in the outlet of the hopper may drop below the CAS if the opening is too small. Figure 10-20. ∆y (kPa) 1.0 1.6 1.SOLIDS NOTES 10. EXAMPLE 10-2. These are the minimum dimensions for the outlets to ensure mass flow. Chase.97 0. Determine the wall slope and opening size to ensure mass flow in a conical hopper for this material.73 10-22 . δ (Figure 10-15) Data taken from tangent line to Mohr Circle at critical point σ1 (kPa) 2.85 0. In practical design the angle θ is reduced by 3o as a margin of safety. Shear Stress Internal Friction JYL Curve number (Figures 10-12 and 10-13) 1 2 3 4 Wall Friction Measurements (Figure 10-16) 1 2 Effective Angle of Internal Friction. Larger openings may be used for greater throughput and still maintain mass flow. Symmetrical slot outlet hopper of opening size W x L. In the next section some correlations are given for estimating the flow throughput of the solids through the opening.4 2. George G.03 Run. EXAMPLE HOPPER DESIGN The Shear Stress – Yield Stress JYL plots for a certain material yield the data in Table 10-2.0 fc (kPa) 0.78 Shear Force (kPa) 0.3 Normal Force (kPa) 2.689 1.91 0.0 3. ∆x (kPa) 1. Assume the bulk density is 1300 kg/m3. Table 10-2. The University of Akron For symmetrical slot outlet hoppers the opening size is determined from W = H (θ ) CAS ρ o g / gc (10-14) L H (θ ) = 1 + θ 180 (10-15) (10-16) W L > 3W Figure 10-21.0 Rise. Experimental Shear Stress data on example powder. Tan(δ ) = ∆y 1.0 or. The angle is easily found from the rise/run as before Tan(δ w ) = ∆y 1.6 0. the angle is δ w = arctan(0.73 or.2 0 0 1 2 3 4 From which we see the data are linear.SOLIDS NOTES 10.2 1 0.3433 ∆ x 3 . George G.578) = 30 o To get the wall friction angle the given data points are plotted as in Figure 10-16. the tangent of the angle is equal to rise/run. the angle is δ = arctan(0. 1.8 0.0 = = 0.4 0. The University of Akron SOLUTION: To get the effective angle of internal friction.76923 ∆x 1.84 .03 = = 0.3433) = 19 o From Figure 10-18 we get the semi included angle θ = 28 o and the flow factor ff = 1. 10-23 . Chase. 161m Hence.2 1 CAS 0. for a mass flow conical hopper the minimum diameter of the opening is 0.83 kPa.8 0. and the ff value is used to plot the 1/ff curve as in Figure 10-19. Chase.4 0. From Eq. George G. the CAS = 0.2 0 0 1 σ1 2 3 Hence. As a margin of safety.4 1.47 From Eq.161 m. the semi included angle is reduced by 3 degrees and the hopper design angle is 24 degrees.47 ⋅ (0. fc 0.807m / s 2 ) N s2 = 0. The University of Akron The f c vs σ 1 data are plotted to obtain the Material Flow Function (MFF).83kPa)(1000 Pa / kPa)(1 N / m 2 / Pa) 1kg m 1300kg / m 3 (9.SOLIDS NOTES 10.6 Stress. 10-24 . 10-13 H = 2. to obtain the CAS value 1.10-12 D = 2. A few of the equations are provided here. Table 10-3. The Beverloo Equation is & = 0. H. Chem Eng Sci. ASME.A. Beverloo.4. derived from fundamental principles (Trans. J.5 m where dp = particle diameter (m) (10-18) 10-25 . 224-230 (1966) is & = ρoA m Bg 2(1 + m)Tan(θ ) (10-17) where θ = semi included angle of the hopper & = discharge rate (kg/sec) m ρ o = bulk density (kg/m3) g = gravity acceleration (9.1 COARSE PARTICLES For coarse particles (particles > 500 microns in diameter) there are two equations commonly used. 69-80.(10-18) Parameter Conical hopper Symmetric slot hopper B A m D.58ρ o g 0. 115. Engrs. AIME. 10.SOLIDS NOTES 10.5 ( D − kd p ) 2.A. Leniger. 1961) tested a variety of seeds and derived an empirical equation. Trans. van de Velde. Chase. (1965). The University of Akron 10. one for mass flow and one for funnel flow MASS FLOW – JOHANSON EQUATION The Johanson equation.807 m/s2) Depending on whether a conical or symmetric slot opening hopper the remaining parameters in the equation are given in Table 10-3. Eq. diameter of outlet W WL π 4 D2 1 0 FUNNEL FLOW – BEVERLOO EQUATION A theoretical expression for funnel flow discharge is not available. The Flow of Granular Solids Through Orifices. 232. Beverloo (W. 260-269. George G. Parameters in the Johanson Equation. Min.4 Rate of Discharge from Hoppers There are a number of methods for calculating discharge rates from silos or hoppers. For non-circular outlets the hydraulic diameter is used D= 4(cross sectional area ) (outlet perimeter) The remaining parameters are defined as in Eq.SOLIDS NOTES 10. 1972) ρ 3 µ 3 Vo 4Vo2 sin θ + 15 5 B ρ pd p3 & = ρ o AVo m 1 2 4 3 =g (10-19) where Vo = average velocity of solids discharging (m/s) A.4. µ = air density and viscosity ρ p = particle density ρ o = bulk density of the powder bed The remaining parameters are defined with Equations (1017) and (10-18).4 if not discharge rate data are available. 10. D = outlet diameter (m).2 FINE PARTICLES Fine particles (dp < 500 microns) tend to flow slower by a factor of 100 to 1000 than that predicted by the Johanson equation. Carleton gives an expression for predicting the velocity of the solids as (Powder Tech. The reason for this is the effect of air drag on the motion of the particles is much greater for fine particles. For fine particles the pore diameters in the powder bed are small and there is a significant amount of air drag that resists the powder motion.. typically 1. Particle beds need to dilate (increase distance between particles) before the powder can flow. The University of Akron k = constant. The term kd p accounts for the wall effect where the particles do not fully flow at the perimeter of the outlet. George G.9 with k = 1. 6. 91-96.(10-17). Chase.3 < k < 2. B = given in Table 10-3 ρ . 10-26 . This means air must penetrate into the bed through the bottom surface of the hopper as the powder moves through the constriction formed by the conical walls. HANDOUT 10-1 σ xx σ yx σ xy σ yy σ yy σ xy σ yx (a) θ σ xx Normal Shear (b) Max Min MAX NORMAL SHEAR STRESS AXIS MOHR CIRCLE NORMAL AXIS (c) MIN NORMAL . HANDOUT 10-2 JYL Shear Stress Critical Point (End Point) Normal Stress JYL Shear Stress Normal Stress fc JYL Shear Stress Critical Point (End Point) Normal Stress σ1 . HANDOUT 10-3 Multiple JYL curves Critical Points Shear Stress δ Normal Stress More cohesive fc MFF Less cohesive Non-cohesive σ1 . degrees 30 1.5 20 δ=40 1.8 δ=40 δ=35 1. Design chart for symmetrical slot outlet hoppers.6 δ=30 1.2 δ=30 1.HANDOUT 10-4 50 δ=60 δ=50 δ=30 2 Flow Factor Curves 1.3 δ=50 δ=60 δ=50 δ=60 1. degrees Figure 10-17. For example (dashed arrows).1 δ=40 1 60 70 0 0 10 20 30 40 50 Semi-included angle.5 o and ff = 1.19 . and δ = 50 o gives θ = 30.9 Wall Friction Curves 40 1.7 Wall Friction. Flow Factor.4 δ=45 10 1. ff φw = 22o . Design chart for conical outlet hoppers.29 .9 Flow Factor Curves δ=40 30 δ=30 1.4 δ=40 1.5 o and ff = 1.8 δ=30 δ=35 Wall Friction.1 0 0 5 10 15 20 25 30 35 40 45 1 Semi-included angle. degrees Figure 10-18. For example. degrees 1.3 10 δ=50 δ=60 δ=30 δ=40 δ=50 δ=60 1.5 1. φw = 22o and δ = 50 o θ = 20.7 1. ff gives .2 1.6 20 1.HANDOUT 10-5 40 δ=60 δ=50 Wall Friction Curves 2 1. Flow Factor. Rotating Shear Test Wall material Get φw SHEAR STRESS φw NORMAL STRESS 3.HANDOUT 10-6 Steps on using the shear stress data to design a hopper. Plot 1/ff on mff curve (fc vs σ1) to get CAS CAS Get CAS fc Slope 1/ff MFF 5. Fit φw and δ to hopper correlation δ Get θ and ff. 1. σ1 . Use CAS and θ in correlations to select opening size. (θ is the theoretical angle of the hopper. Rotating Shear Test Plot SHEAR STRESS vs NORMAL STRESS JYL SHEAR STRESS Internal Friction Get fc vs σ1 Get δ δ fc NORMAL STRESS σ1 2. in final design subtract 3 degrees for margin of safety) φw δ ff θ 4. Symmetrical slot outlet hopper of opening size W x L. from the charts in Figures 10-17 or 10-18. For symmetrical slot outlet hoppers the opening size is determined from W = H (θ ) H (θ ) = 1 + L > 3W CAS ρ g / gc o (10-14) θ L (10-15) (10-16) Figure 10-21. 60 D Figure 10-20. Conical Hopper with outlet size D and semi included angle θ . Typical values for H are about 2.4. is given by D = H (θ ) CAS ρ g / gc o (10-12) H (θ ) = 2 + θ θ (10-13) Semi included angle Where θ is in degrees. the opening diameter. D .HANDOUT 10-7 For conical hoppers. Figure 10-20. 180 W . µ = air density and viscosity ρp ρ o = particle density = bulk density of the powder bed .4.3 < k < 2. D= 10.2 FINE PARTICLES (dp < 500 microns) CARLETON EQUATION 4(cross sectional area ) (outlet perimeter) ρ 3 µ 3Vo 4Vo2 sin θ + 15 5 B ρ pd p3 & = ρ o AVo m where 1 2 4 3 =g (10-19) Vo = average velocity of solids discharging (m/s) A.807 m/s2) Table 10-3. typically 1.1 COARSE PARTICLES (particles > 500 microns in diameter) MASS FLOW – JOHANSON EQUATION & = ρoA m where Bg 2(1 + m)Tan(θ ) (10-17) θ = semi included angle of the hopper & = discharge rate (kg/sec) m ρ o = bulk density (kg/m3) g = gravity acceleration (9.4 if not discharge rate data are available.(10-18) Parameter Conical hopper Symmetric slot hopper B A D.9 with k = 1.5 m where (10-18) dp = particle diameter (m) k = constant.58 ρ o g 0. diameter of outlet W WL π 4 D2 1 m 0 FUNNEL FLOW – BEVERLOO EQUATION & = 0. Parameters in the Johanson Equation. Eq.HANDOUT 10-8 10.4. B = given in Table 10-3 ρ .5 ( D − kd p ) 2. Chapter 3. Separation of particles by size is referred to as classification. Most methods of separation are not 100% effective. or metal particles from polymers). We classify to remove unwanted parts of a size distribution. London. FREQUENCY SIZE TAILINGS OVERSIZE Figure 11-1. The process of classification has been used for many years. or some other desired property).. There is usually a range of particle sizes that are separated with varying degrees of efficiency. 1 Ladislav Svarovsky.SOLIDS NOTES 11. The University of Akron 11. Solid-Liquid Separation. the sharp separation will be more expensive. Usually for the same production rate. GRADE EFFICIENCY Classification is the art of separating solid particles in a mixture of solids and fluid into fractions according to particle size or density by methods other than screening. A good discussion on Grade Efficiency is given by Svarovsky. George G. 11. In Figure 11-1 indicates the tailings and oversize particles that may be removed from a material for a particular purpose. as could be deduced from the 2nd Law of Thermodynamics. Size distribution indicating undesired tailings and oversize particle ranges.1 Why Do We Classify Particles? We classify particles to remove contaminants and unlike particles (wheat from chaff. 11-1 . 1990. 3rd ed. The variation in the efficiency is referred to as the Grade Efficiency Curve.1 The grade efficiency is a way of characterizing how well particles are separated according to size (density. Chase. Butterworths. Some methods make a sloppy separation while others make a sharp classification. 95 g/cc) and plastic ware (polystyrene. the milk jug plastic will float while the plastic ware will sink. As the belt rotates around the magnetized cylindrical roller the ferrous materials cling longer to the belt than non-ferrous materials fall off and thus allow separation. but rate of production may be low. Example of a magnetized separator. Air classifier for separating paper and low density materials from solid waste streams. The large drag force to gravitational force ratio of the air on the low density materials causes materials such as paper to separate from plastics and metals.05 g/cc) is dumped into a tank of water. These methods take advantage of different drag forces acting on different size particles. 11-2 . Air Air Air Plastic and metals input Metals and Plastics Exit Stream Figure 11-2. The University of Akron There are many processes that may be used to classify particles: Screening Sieving Air or water classification Physical separations • • • • • Magnetic Gravity Electrostatic Radiation Color These methods are effective. Chase. This provides a simple means for separation. Magnetized cylinder Plastic and metals input Rotating belt Paper Exit Stream Solid Trash Feed Stream Ferrous metals output Plastic and non-ferrous metals output Figure 11-3. George G. density of 0.SOLIDS NOTES 11. Some additional examples of separators are given in Figures 11-2 through 11-4. density of 1. These methods rely on physical properties other than differences in fluid drag to classify. An example of gravity separation if a mixture of plastic milk jugs (high density poly ethylene. 153-164.al.(11-2). By analogy with Eq. see the paper by Guistino et. but may not be pure. The total mass balance gives M = Mc + M f .” Separations Technology.2 Measuring Efficiency Separation efficiency is directly related to processing costs. (11-1) ET . Willis. 1995. the grade efficiency of separation of size x is defined as Gx = M cx Mx (11-3) 2 JM Giustino. Lasers scan the belt and identify the locations of plastic pieces of specific colors. 11-3 . Air jets corresponding to the locations of the identified plastic pieces apply puffs of air to blow the plastic pieces into bins. M fx . the masses of size x in each stream are noted by M x . 10. G. 5 (3). The method of separation considered here is called the Grade Efficiency (For an alternative method based on thermodynamic entropy. Chopped plastic pieces are spread in a monolayer on a perforated moving belt. x. The challenge is to determine the best way to define and measure the efficiency of the separation.S. Chase. The University of Akron Top View Air Holes Moving Belt Monolayer of Chopped plastic Moving Belt Laser scan to ID colors by location Air jets use puffs of air to blow plastic squares into separate bins Side View Air Jets Figure 11-4. M M (11-2) ET = We assume that there is no agglomeration or communition in the separator. George G. The outlet streams of the separation may be concentrated in the desired or undesired products. Chase.2). and M.G.SOLIDS NOTES 11. For a particle size. The total separation efficiency. M cx . Separation by color. “Thermodynamic Separation Efficiency and Sedimentation Criteria for Multiphase Processes: A Comparison of Rigorous and Approximate Models. is defined as Mf Mc = 1− . Most separations are not 100% efficient. A black box powder separation process is shown in Figure 11-5. The University of Akron Coarse Stream Total mass Mc of powder Feed Stream Total mass M of powder Black box Separation Process Fines Stream Total mass Mf of powder Figure 11-5. Chase. George G. the grade efficiency is related to the size distribution functions by G ( x) = (11-6) A typical plot of the grade efficiency versus the particle size is shown in Figure 11-5.SOLIDS NOTES 11. x. Black box (hypothetical) separation process to separation coarse particles from fine particles. Properties of Particulate Solids) that the mass and mass components of each stream are related by M x = M (fraction of size x) = M f x dx = M dFx similarly. (11-4) M cx = M c f cx dx = M c dFcx M c f cx . on the curve. The area under the curve plotted in Figure 11-6 represents the coarse cut (the stream with the larger particles) and the area above the plotted curve represents the fines cut. We know from the definitions of frequency distributions (see notes Section 3. Mf x (11-5) Hence. At a point. Typical S-shaped grade efficiency curve. Svrovsky1 gives more detail on defining and measuring these curves. G(x) represents the fraction of particles of size x that are separated out of the feed stream and contained in the fines product stream. 1 Fines Cut G ( x) Coarse Cut 0 x Figure 11-6. 11-4 . For an unsteady process such as filtration. consider the water filter pumps used by backpackers on a hiking trail. As an example of how you might apply the grade efficiency curve. Filter B appears to be only about 50% effective at removing 1 micron particles. Comparison of grade efficiency curves for two filters. as shown in Figure 11-7. In a typical cake filtration. microns Figure 11-8. the savy backpacker would choose filter A because it is better than 99.99% effective at removing 1 micron particles. G ( x) Increasing Time G ( x) Increasing Time x Cake Filtration x Depth Filtration Figure 11-7. One of the objectives of the filter is to remove harmful bacteria from water. George G. the cake itself improves the separation and the grade efficiency shifts with increasing efficiencies for smaller particles. 11-5 . Comparison of typical grade efficiency curves for Cake Filtration and Depth Filtration. Chase. the curve changes with time. The University of Akron For a continuous steady process the curve in Figure 11-6 is steady. though it is 99+% efficient at removing 10 micron particles. initially the filter may perform very well. Filter A is approximately 99+% efficient at removing 1 micron particles. as the filter cake depth increases. Bacteria are typically greater than 1 micron in size. Gradually the capture sites in a typical depth filter are occupied (though other mechanisms such as straining may occur) and the fine particles start to bleed through. If you have the choice of two filters with grade efficiency curves shown in Figure 11-8. In Depth filtration. A G ( x) B 1 10 x. A and B. Filter B is only about 50% efficient at removing 1 micron particles. Hence in Depth Filtration the grade efficiency curve shifts toward larger particles as the filter becomes less efficient at capturing small particles.SOLIDS NOTES 11. (11-7). Consider an arbitrary process in Figure 11-9a.SOLIDS NOTES 11. One and Two arbitrary processes in series with corresponding Grade Efficiencies. Chase. Figure 11-9. For example. This can be generalized to conclude that the more separations steps in your process the smaller the yield of your product. Mx G(x) Mcx Mfx a. then Mc2x < Mc1x. Arbitrary process that divides the feed stream M into fines and coarse streams. George G. The University of Akron One of the reasons for inventing the Grade Efficiency is that it makes some calculations regarding the separation of particles by size easier. 11-6 . in Figure 11-9b the streams exiting process 2 have amounts of size x given by M c 2 x = G2 x G1x M x M f 2 x = (1 − G 2 x )G1x M x (11-9) (11-10) Suppose Mc2x is your product in stream C2 in Figure 11-9b. The grade efficiency of this process is given by G. Two arbitrary process in series with grade efficiencies G1 and G2 . Because G < 1. 2 G2x C2 Mc2x Mx 1 G1x C1 Mf2x Mf1x b. where the amount of material of size x entering the process is given by Mx and the amount of material of size x leaving in the coarse and fine streams are given by M cx = G x M x M cx = (1 − G x )M x (11-7) (11-8) You can use the grade efficiency to determine the amounts of material of size x in various streams of processes that are cascaded in series or parallel. The amount of product in stream C1 is Mc1x = G1xMx as given by Eq. 3 Cut Size and Sharpness of Cut To compare efficiencies between steady state processes. Chase. the sharpness must be greater than or equal to unity. Real separation processes have a sharpness of cut greater than unity. These concepts are shown in Figure 11-10. denoted x 50 .SOLIDS NOTES 11. and because the grade efficiency is a monotonically increasing curve. including 1. The sharpness of cut is defined as I 80 / 20 = x80 x 20 (11-11) By this definition. 11-7 . cut size refers to the 50% cut size. 3. x50 x20 x50 x80 x x Figure 11-10. dx dG ( x ) dx 2. I x = Sum of triangular areas in Figure 11-10(b). In an idealized case in which there is a perfect separation. I 20 / 80 = x 20 which is the inverse of I 80 / 20 . (b) The idealized sharp cutoff grade efficiency curve is a vertical line Other definitions of sharpness of cut are also used. where all particles less than the 50% cut size exit in the fines stream and all particles greater than the cut size exit in the coarse stream. we define cut size and sharpness of cut. George G. Normally. Sharpness of cut is defined as a ratio of particle sizes specified at two efficiencies. typically at 20% and 80%. This is the particle size for which 50% of the particles exit the separation process in the coarse product stream and 50% exit in the fines product stream. (a)Typical grade efficiency curve with the particle sizes indicated for which the separation is 20. the sharpness of cut equals unity. x values such as 90/10 could also be defined. The University of Akron 11. Variance in the slope. 4. Slope of the grade efficiency curve. Idealized sharp cutoff with x80 =1 x20 G ( x) 80 50 20 (a) G ( x) (b) 50 The sharper the cut the smaller the triangular areas between the real and idealized grade efficiency curves. and 80 percent efficient. x80 dG . 50. George G. M fx . This approach is not very realistic because of the difficulty in obtaining monodispersed materials (especially in the very small particle size range) and because of the time and effort required. 10 lbm/hr ± ? 10. and you need at least two of the following. Fines product rate.SOLIDS NOTES 11. As indicated in Figure 11-11. Repeat for other size x (until you have enough points to construct your curve or you loose patience). It is best to have data from the smaller of the two product streams because errors in sampling are smaller than when you subtract two larger streams to get the smaller stream. Due to the error. Feed particle size distribution. Chase. it is better to measure the smaller stream directly instead of calculating it from the two larger streams. Coarse particle size distribution.990 lbm/hr ± 50 lbm/hr Figure 11-11. Fine particle size distribution. Example in which the size of the smaller product stream is smaller than the error in measurements of the other two steams. 11-8 . M cx .4 Construction of the Grade Efficiency Curve In the ideal case you would feed into your separator a material with a monodispersed particle size distribution (of size x). Calculated G ( x ) . the flow rate of the smaller stream may be less than the error of measurement of the larger streams.000 lbm/hr ± 50 lbm/hr 9. The University of Akron 11. You would • • • Measure M x . Coarse product rate. In real applications you need to measure two of the following • • • • • • Feed rate. 09 0.7 0.8 0.SOLIDS NOTES 11.075 0.4 0.2 0. microns Plot the grade efficiency curve and calculate I20/80. 11-9 .45 0.9 0.5 Example 1 A sample of the feed.10 1 Mx ∆Fcx ~0 0.20 0. The University of Akron 11. Chase.5 0.235 1 M fx The stream rates are: • • • Feed rate = 100 lbm/hr Coarse Product Rate = 60 lbm/hr Fines Product Rate = 40 lbm/hr G(x) 1 0.01 1 M cx ∆F fx ~0 0.45 0.325 0.5 362.365 0. George G. coarse.40 0.1 0 0 100 200 300 400 500 600 700 800 900 100 0 Size.3 0.30 0.6 0.5 256 ~0 0. and fines streams for a separation of a material "Hexamethyl chicken wire" is screened with the following results: Screen Size (microns) Average Particle Size Feed Size mass Fraction retained on screen Mass Rate of particles size x in Feed Stream Coarse Size mass Fraction retained on screen Mass Rate of particles size x in Coarse Stream Fines Size mass Fraction retained on screen Mass Rate of particles size x in Fines Stream G(x ) = M cx Mx ∆Fx 850 600 425 300 212 Total 725 512. SOLUTION: Recall that M x = M ∆Fx M cx = M c ∆Fcx M fx = M f ∆F fx Using these equations and a basis of 1 hour. George G. x50. The University of Akron SOLUTION A sample of the feed.075 0.45 0.8 0. Chase.528 .30 0. and 625 microns respectively.5 0.4 0.7 0.6 60 ∆F fx ~0 0.4 40 1 0.4 0.3 Plot the grade efficiency curve and calculate I20/80.01 1 M cx 0 27 27 5.1 0 0 100 200 300 400 500 600 700 800 900 100 0 x x x 20 50 80 Size. microns I 20 / 80 = 330 = 0.6 9. mass retained on the 850 micron screen is zero. In the limit Gx approaches 1 at this size (Gx = 1 at the size for which all particles exit the separator in the Coarse stream).9 0. From the curve the x 20 . and fines streams for a separation of a material "Hexamethyl chicken wire" is screened with the following results: Screen Size (microns) Average Particle Size Feed Size mass Fraction retained on screen Mass Rate of particles size x in Feed Stream Coarse Size mass Fraction retained on screen Mass Rate of particles size x in Coarse Stream Fines Size mass Fraction retained on screen Mass Rate of particles size x in Fines Stream G(x ) = M cx Mx ∆Fx 850 600 425 300 212 Total 725 512.27 0.68 0.9 0.10 1 Mx 0 30 40 20 10 100 ∆Fcx ~0 0.5 256 ~0 0.2 0. and x80 values. x50 .6 G(x) 0.40 0.09 0. The sharpness of cut is calculated to be 0. 625 BLANK FORM FOR CALCULATING GRADE EFFICIENCY 11-10 . the table is filled in.235 1 M fx 0 3 13 14.365 0. Plot Gx vs the Average particle size to determine the x20.5 362. coarse.45 0.06 The stream rates are: • • • Feed rate = 100 lbm/hr Coarse Product Rate = 60 lbm/hr Fines Product Rate = 40 lbm/hr Note.20 0. The calculated points for the Grade Efficiency are plotted on the graph and a curve is fitted to the points.SOLIDS NOTES 11. 1 0. 450.325 0. and x80 values are estimated to be 330. 7 0.9 0. George G. The University of Akron Screen Size (microns) Average Particle Size Feed Size mass Fraction retained on screen Mass Rate of particles size x in Feed Stream Coarse Size mass Fraction retained on screen Mass Rate of particles size x in Coarse Stream Fines Size mass Fraction retained on screen Mass Rate of particles size x in Fines Stream G(x ) = M cx Mx ∆Fx Mx ∆Fcx M cx ∆F fx M fx The stream rates are: • • • Feed rate = Coarse Product Rate = Fines Product Rate = 1 0.3 0.8 0.SOLIDS NOTES 11.5 0. microns Plot the grade efficiency curve and calculate I20/80.2 0. 11-11 .4 0.1 0 0 100 200 300 400 500 600 700 800 900 100 0 Size. Chase.6 G(x) 0. The University of Akron 11. 2 11-12 . from Eqs. but is implied). Chase. Hence. G1x = G2x). George G. Derive a formula in terms of Grade efficiencies for determining the fractional amount of particles of size x in the coarse stream exiting separator 2 in the compound process shown in Figure 11-12 where the grade efficiencies of both processes are the same (ie.5 Example 2 The grade efficiency represents the fractional amount of particles by mass of size x in the feed stream that exits the separator in the coarse stream. 1 G1 M feed Mc1 coarse 2 G2 Mf1 fines SOLUTION Mc2 coarse Mf2 fines The fractional amount of size x in the coarse stream exiting separator 2 can by determined by (the subscript ‘x’ is dropped from this notation.(11-7) and (11-8) M c = GM Hence and M f = (1 − G )M M f 1 = (1 − G1 )M M f 2 = (1 − G 2 )M f 1 M c1 = G1 M For process 2 and (1) M c 2 = G2 M f 1 where M f1 and (2) is the feed stream to process 2. For an arbitrary process.SOLIDS NOTES 11. by combining equations (1) and (2) we get M f 2 = (1 − G 2 )(1 − G1 )M Since G1 = G2 this expression simplifies to M f 2 = (1 − G ) M 2 or M f2 M = (1 − G ) .. These methods rely on physical properties other than differences in fluid drag to classify.HANDOUT 11-1 FREQUENCY SIZE TAILINGS OVERSIZE Figure 11-1. 11-1 . Paper Exit Stream Air Air Air Solid Trash Feed Stream Metals and Plastics Exit Stream Figure 11-2. Air classifier for separating paper and low density materials from solid waste streams. There are many processes that may be used to classify particles: Screening Sieving Air or water classification Physical separations • • • • • Magnetic Gravity Electrostatic Radiation Color These methods are effective. These methods take advantage of different drag forces acting on different size particles. The large drag force to gravitational force ratio of the air on the low density materials causes materials such as paper to separate from plastics and metals. Size distribution indicating undesired tailings and oversize particle ranges. but rate of production may be low. Example of a magnetized separator. Lasers scan the belt and identify the locations of plastic pieces of specific colors. Top View Air Holes Moving Belt Monolayer of Chopped plastic Moving Belt Laser scan to ID colors by location Air jets use puffs of air to blow plastic squares into separate bins Side View Air Jets Figure 11-4. Separation by color.HANDOUT 11-2 Rotating belt Magnetized cylinder Plastic and metals input Ferrous metals output Plastic and non-ferrous metals output Figure 11-3. Chopped plastic pieces are spread in a monolayer on a perforated moving belt. 11-2 . As the belt rotates around the magnetized cylindrical roller the ferrous materials cling longer to the belt than non-ferrous materials fall off and thus allow separation. Air jets corresponding to the locations of the identified plastic pieces apply puffs of air to blow the plastic pieces into bins. M M (11-2) Assume that there is no agglomeration or comminution in the separator. x 11-3 . Black box (hypothetical) separation process to separation coarse particles from fine particles. M = Mc + M f . (11-1) ET . For a particle size. At a point. G(x) represents the fraction of particles of size x that are separated out of the feed stream and contained in the fines product stream.HANDOUT 11-3 Coarse Stream Total mass Mc of powder Feed Stream Total mass M of powder Black box Separation Process Fines Stream Total mass Mf of powder Figure 11-5. x. M cx . on the curve. The total mass balance gives The total separation efficiency. The grade efficiency of separation of size x is defined as Gx = M cx Mx (11-3) M x = M (fraction of size x) = M f x dx = M dFx For the Coarse Steam Hence (11-4) M cx = M c f cx dx = M c dFcx G ( x) = M c f cx . x. M fx . Mf x (11-5) (11-6) 1 G ( x) Fines Cut Coarse Cut 0 Figure 11-6. is defined as ET = Mf Mc = 1− . Typical S-shaped grade efficiency curve. the masses of size x in each stream are noted by M x . 2 0.075 0.9 0.7 0.1 0 0 100 200 300 400 500 600 700 800 900 100 0 Size.5 362.20 0.325 0.6 0.5 256 ~0 0.45 0. coarse.4 0. microns Plot the grade efficiency curve and calculate I20/80.09 0.01 1 M cx ∆F fx ~0 0.10 1 Mx ∆Fcx ~0 0.45 0.HANDOUT 11-4 A sample of the feed.235 1 M fx The stream rates are: • • • Feed rate = 100 lbm/hr Coarse Product Rate = 60 lbm/hr Fines Product Rate = 40 lbm/hr G(x) 1 0.5 0.30 0. and fines streams for a separation of a material "Hexamethyl chicken wire" is screened with the following results: Screen Size (microns) Average Particle Size Feed Size mass Fraction retained on screen Mass Rate of particles size x in Feed Stream Coarse Size mass Fraction retained on screen Mass Rate of particles size x in Coarse Stream Fines Size mass Fraction retained on screen Mass Rate of particles size x in Fines Stream G(x ) = M cx Mx ∆Fx 850 600 425 300 212 Total 725 512.40 0. 11-4 .365 0.3 0.8 0. etc. sieves. By their nature. belt alignment sensors. heat sensors. Dust is mixed or suspended in air 4. pipes. Probably the most important prevention of secondary explosions is good housekeeping. Do not let powder to accumulate on floors. Most buildings will collapse if the interior pressure exceeds about 10 psi (70 kPa). Dust explosions can occur within storage buildings. The University of Akron 14. The dust concentration exceeds a minimum needed for the explosive reaction 5. Most often the burning of organic materials does not cause an explosive force. With the right conditions dust explosions can exceed 100 psi. that might cause sparks). Reinforced concrete structures typically can withstand 2 – 8 psi of static pressure. An acceptable minimum concentration of dust is 1. walkways. oats.) at a rate in proportion to the surface area. etc. but other powdered materials can cause explosions as well. soybeans. Dry dust 7.SOLIDS NOTES 14. • Magnets can remove iron and other metals that when conveyed in the powder can cause sparks. Fuel source (the dust itself) 3. the more explosive is its nature. conveying systems. Removal of these powders removes the fuel source for the secondary explosions. Catastrophic explosions (destruction of a large building) usually occur due to secondary explosions. H2O. When a dust explosion occurs seven conditions are nearly always present: 1. • Heat sensors can alert operators to trouble spots so that heat sources can be eliminated before an explosion occurs. when the conditions are right. rice. etc. grain elevators and hoppers must handle dusts in concentrations greater than the minimum required. a powdered material gives off product gases (CO2. Motion sensors. 14-1 . George G. How are these devices used to prevent dust explosions? • Motion sensors can be used to detect when machinery moves in ways that they are not designed (vibrations. When burned. etc. friction. The dryer and finer the dust. It is common knowledge that organic materials can burn. When the powdered material is confined within a grain elevator or storage building the rapid release of product gases can create a significant gas pressure within the building. higher concentrations are considered dangerous. All of the elements are in a confined space Removal of one or more of these conditions can prevent a dust explosion (elimination of three or more is preferred). Chase. What makes the powdered materials different that when burned they can cause an explosion? Powdered materials have very high surface areas per unit mass.13 g/cm3.) are well known for dust explosions. Ignition source 6. etc. The secondary explosions occur in other locations in building when powders are suddenly suspended into the air because of the vibrations and shocks caused by the primary explosion (the first explosion in a grain elevator). Proper designs of these devices include pressure relief panels that can blow away to reduce the pressure and prevent the explosion from damaging the whole building. Grain elevators (storing corn. wheat. • Belts must be properly aligned otherwise the rubbing of belts can cause friction and the friction can cause heat that could be an ignition source. hoppers. DUST EXPLOSIONS Historically dust explosions have occurred in processing plants where dry organic powdered are handled. Oxygen (air) 2. and magnets are common safety devices used to prevent dust explosions. P.A. United Kingdom. Bartknecht. 2. Englewood Cliffs. 3. Rushden. D. Chemical Process Safety: Fundamentals with Applications. Prentice Hall. 5. Dust Explosions in the Process Industries. McGraw-Hill. 1980. George G. Dust Explosions. N. 1982. J. Industrial Explosion Prevention and Protection. 1990. 4.T. NY. Eckhoff.SOLIDS NOTES 14. and Luvar. Elsevier.Y.F. Dust Explosions.. N.. Field. Springer-Verlag. Butterworth.. Chase..K. NJ. F. Crowl. 1991. Bodurtha. 14-2 .. W. R. The University of Akron References 1. 1989.Y.