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**Ert 318 : unit operations operations involving particulate solids**

Dr. Akmal Hadi Ma’ Radzi School of Bioprocess Eng.

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**outline Introduction to particulate solids**

Characterization of solid particles Particle size Screen analysis Tyler standard screen analysis

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Introduction Large quantities of particles handled on the industrial scale is frequently define as a whole Thus, it is necessary to know the distribution of particle sizes in the mixture and able to define a their mean size which represents the behaviour of the particulate mass as a whole It is frequently necessary to reduce the size of particles or form them into aggregates or sinters to ease the handling process Sometimes it might necessary to mix two or more solids, or sometimes need to separate a mixture into its own component or according to the sizes of the particles Important operations relating to systems of particles include storage in hoppers, flow through orifices and pipes, and metering of flows

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**Characterization of solid particles**

Individual solid particles are characterized by: (a) Shape – regular – e.g. spherical or cubical irregular – e.g. a piece of broken glass (b) Size – influence the properties such as : - the surface per unit volume - the settling rate of particle in fluid (c) composition - determines properties such as density and conductivity

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Particle shape The shape of an individual particle is expressed in terms of sphericity, which is independent of particle size - For spherical particle of diameter Dp; = 1 - For nonspherical particle; Dp = nominal diameter of particle Sp = surface area of one particle Vp = volume of one particle (1)

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**- For most of crush materials 0.6 < > 0.8**

-For particles rounded by abrasion; > 0.95 - For a cube and cylinder; = 1 Source ; Unit Operation of Chemical Engineering (McCabe,Smith & Harriot) ( Table 7.1 page 164)

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**Particle size Mixed particle sizes**

In a sample of uniform particles of diameter Dp the number of particles in a sample ,N is given by; From Eq. (1) and (2), the total surface area of particle sample, A can be computed by (2) (3) These equation applicable to mixtures of particles having various sizes and densities, the mixture is sorted into fractions, each of constant density and approximately constant size

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**Specific surface of mixture**

The specific surface, Aw ( the total surface area of a unit mass of particles) if and are constant is given by; (4) where subscripts = individual increments = mass fraction in a given increment = number of increments = average particle diameter, taken as arithmetic average of smallest and largest particle diameters in increment

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**Average particle size Volume-surface mean diameter ( ) given by**

The volume-surface mean diameter, is related to the Aw is given by By substituting Eq. (4) in Eq. (5); (5) (6) To calculate effective mean diameter, Dpm:

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**Arithmetic mean diameter ( ) Mass mean diameter ( ) **

where is the number of particles in the entire sample

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**Volume mean diameter ( )**

By dividing the total volume of the sample with the number of particles in the mixture, gives the average volume of a particle The diameter of such a particle is the volume mean diameter *For samples consisting of uniform particles, these average diameters are all the same. For mixtures containing particles of various sizes, however the several average diameters may differ widely from one another.

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**Number of particles in mixture**

For a given particle shape, the volume of any particle is proportional to its “diameter” cubed or where a is the volume shape factor. Unlike , it is different for various regular solids which are; (1) for a sphere (2)0.785 for a short cylinder (height = diameter) (3)1.0 for a cube. Assuming a is independent of size,

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screen analysis Used to measure the size of particles in the size range between in.(76 mm and 38µm) A set of standard screens is arranged serially in a stack, with the smallest mesh at the bottom and the largest at the top The sample is placed on the top screen and the stack shaken mechanically for 20 min The particles retained on each screen are removed and weighed, and the masses of the individual screen increments are converted to mass fractions or mass percentages of the total sample Any particles that pass the finest screen are caught in a pan at the bottom of the stack

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**SIEVE TRAYS Pan (the finest particles retained here)**

Sieve (different mesh size) Sieve trays were put on a stack (shaker)

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Screen Analysis Table

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Column 1 – mesh size Column 2 – width of opening of the screens Column 3 – the mass fraction of the total sample that is retained on the designated screen Xi – i is the number of the screen starting at the bottom of the stack; thus i = 1 for the pan, and screen i + 1 is the screen immediately above screen I Column 4 – average particle diameter Dpi in each increment Column 5 – cumulative fraction smaller than each value of Dpi In screen analysis, cumulative fractions are sometimes written starting at the top of the stack and express as the fraction larger than a given size

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**Tyler standard screen analysis**

Based on the opening of the 200-mesh screen, which is established at mm The area of the openings in any one screen in the series is exactly twice that of the openings in the next-smaller screen The ratio of the actual mesh dimension of any screen to that of the next-smaller screen is, (2)^1/2=1.41

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Example 1 A mixture contains three sizes of particles: 25% by volume of 25 mm size, 40% of 50 mm, and 35% of 75 mm. The sphericity is Calculate the effective mean diameter.

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Solution 1 The following data are given: x1 = 0.25, Dp1 = 25 mm; x2 = 0.40, Dp2 = 50 mm; x3 = 0.35, Dp3 = 75 mm, Фs = 0.68.

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Example 2 The screen analysis shown in Table I applies to a sample of crushed quartz. The density of the particles is 2,650 kg/m3 ( g/mm3), and the shape factors are a = 0.8 and Фs = For the material between 4-mesh and 200-mesh in particle size, calculate: Aw in square millimeters per gram and Nw in particles per gram Dv Ds Dw and Ni for the 150/200-mesh increment What fraction of the total number of particles is in the 150/200-mesh increment

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Solution 2 To find Aw and Nw,

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For the 4/6-mesh increment Dpi is the arithmetic mean of the mesh openings of the defining screens; or, from the Table, ( )/2 = mm. For this increment xi = ; hence xi/Dpi = /4.013 = and xi/Dpi3 = Corresponding quantities are calculated for the other 11 increment to give Σ xi/Dpi= and Σ xi/Dpi3= Therefore, Aw = 3282 mm2/g and Nw = particles/g.

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(b) (c) The volume-surface mean diameter is calculated from: (d) Mass mean diameter Dw is obtained from this equation. 0.4845 8.7932 1.208 0.8278

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For this, from the data in the Table, (e) The number of particles in the 150/200-mesh increment is found from this equation: This is 2074/4148 = 0.5, or 50 percent of the particles in the top 12 increments. For the materials in the pan fraction, the number of particles and specific area are enormously greater than for the coarser materials, but they cannot be accurately estimated from the data in Table.

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THANK YOU

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