Partcle dynamics for flow of particles in fluid
- 1. Particle Dynamics Study of particle dynamics is important in many mechanical operations - Sedimentation - Classification - Elutriation - Filtration
- 2. Assumptions 1. The particle is spherical 2. The particle is non porous, incompressible and chemically inert with fluid. 3. Density , viscosity of the fluid are constant 4. Particles is falling freely (under gravity) 5. There is no wall effect, particle is far away from wall 6. Particle is larger in size than the mean free path of fluid molecules, so slip between particle and fluid molecules.
- 3. Flow of solid particles in a liquid FR (friction) Fb(buoyancy) mg (gravity)
- 4. Motion of a Particle through a Fluid • When a particle falls in a liquid, the forces acting are: Force balance: FR (friction) (m/ρs) ρfg (buoyancy) mg (gravity) 𝒎𝒂𝒆=𝒎𝒈− (𝒎 𝝆𝒔 )𝝆 𝒇 𝒈− 𝑭 𝑹
- 5. 𝒎𝒂𝒆=𝒎𝒈− (𝒎 𝝆𝒔 )𝝆 𝒇 𝒈− 𝑭 𝑹 ae is effective acceleration FR is frictional drag with in the velocity of pa At a particular point, when total downward force is equal To the upward force. Then the particle will move with a constant velocity or zero acceleration. This velocity is known as terminal settling velocity Vt
- 6. Assuming particle to be spherical The kinematic force can be expressed in general as where, A = characteristic area of the system K = characteristic kinetic energy per unit volume f = a dimensionless parameter
- 7. Within stokes law, , for laminar regime • where, Rep = (dp Vt ρf)/ μf and is called particle Reynolds number. • Rep < 0.1 : laminar settling zone (Stokes law is applicable) • Rep > 1000 Turbulent settling zone, fd is constant and its independent of Rep • 0.1 > Rep < 1000 transition zone
- 8. The diameter of a drop of liquid fuel changes with time due to combustion according to relationship D = D0(1-t/tb). While burning, the drop falls at its terminal velocity under stokes flow regime. Derive an expression for the distance travelled by the drop. Ans:S=D0 2 (ρp-ρ)gtb/54µ
- 9. •Estimate the terminal velocity for 80- to 100- mesh particles of limestone (density= 2800 kg/m3) falling in water at 30 c.
- 10. • 2 >Nre < 500, CD = { 18.5/Nre}^0.6. • Nre>500, CD = 0.44
- 11. Problems • A mixture of coal and sand particles having sizes smaller than 0.0001 m in diameter is to be separated by screening and subsequent elutriation with water. Recommend a screen aperture such that the oversize from the screen can be separated completely into sand and coal particles. Calculate also the required water velocity. Assume that stokes law is applicable and consider reynolds number as 0.1. Density of snad = 2650 kg/m3; density of coal = 1350 kg/m3; density of water = 1000 kg/m3; viscosity of water = 0.001 kg/ms.
- 12. • A suspension of uniform particles in water at a concentration of 500 kg of solids per cubic meter of slurry is settling in a tank. Density of particles is 2500 kg/m3 and terminal velocity of a single particle is 20 cm/s. What will be the settling velocity of suspension? Consider Richardson-Zaki index is 4.6
- 13. Problem • A binary mixture of 100 µm size having densities 2 g/cm3 and 4 g/cm3 is to be classified by elutriation technique using water. Estimate the range of velocities that can do the job and recommend a suitable value
- 14. • In a mixture of quartz, specific gravity 2.65, and galena, specific gravity 7.5, the size of the particles range from 0.00052 cm to 0.0025 cm. On separation in a hydraulic classifier with water as a medium, under free settling conditions, three fractions are obtained: one consisting of only quartz, one of only galena and one mixture of quartz and galena. What are the ranges of particle sizes of the two substances in mixed portion.
- 15. Classification and Elutriation • Classification is used to designate separation of solid particles based on difference in their terminal settling velocities in fluid. • Elutriation is a process for separating particles based on their size, shape and density, using a stream of gas or liquid flowing in a direction usually opposite to the direction of sedimentation. This method is mainly used for particles smaller than 1 μm.
- 16. Hindered settling • Small particle tends to get dragged downwards by large ones. • Effective density and viscosity of concentrated suspension (ρb and μb, commonly called bulk density and bulk viscosity) are much larger than clear liquid. • Particles get closer and flocculation takes place. Each flocks now behaves as a coarser particle and settles at a high velocity. video
- 17. For a uniform suspension, the settling velocity us can be estimated from the terminal velocity for an isolated particle using Maude and Whitmore emperical equation Us = ut (e)n
- 18. The bulk density of a suspension can be computed as where, is volume fraction of liquid in the suspension. Bulk viscosity in suspension can be estimated as Substituting in Stokes law, we get = VtFs Vt the free settling velocity.
- 19. Problem • A glass sphere (RD = 2.60) 2.0 mm in diameter is observed to have a fall velocity of 1.25 cm/s in oil of density 917 kg/m3. Estimate the coefficient of dynamics viscosity of the oil. 0.293 Pa.s
- 20. Effect of particle shape • The sphericity is considered. • Empirical correlation: Pettyjohn and Christiansen; for Rep<0.05, • For turbulent settling zone at Rep>2000,
- 21. Problem • Compute the maximum velocity at which particles of silica 0.04 mm in diameter (sp. Gr. 2.65) will fall through water that fills a 50 mm ID glass cylinder, if • The slurry is so dilute that free settling prevails • The mass ratio of water to silica is 2.0 • 1. Assume the particles to be essentially spherical. • 2. Assume particle to be cube of side length 2 mm.
- 22. Problem • A suspension of uniform particles in water at a concentration of 500 kg of solids per cubic meter of slurry is settling in a tank. Density of prticles is 2500 kg/m3 and terminal velocityof a single particle is 20 cm/s. What will be the settling velocity of suspension. Assume the value of exponent, n = 4.6
- 23. Classification • The term classification is used for separation of solid particles based on difference in their terminal settling velocities in a fluid. • Sizing: particle of same density are separated according to their size. • Sorting: material of same size are separated according to density.
- 24. • Particles (say, A and B) having the same terminal velocity in a fluid are called “equally falling” particles. • Ratio of sizes when they are equally falling is called the settling ratio. • Derivation see book - student task. • da/db in laminar and turbulent settling zone.
- 25. Classification equipment • Gravity settling classifier • Cone classifiers • Bowl classifiers • Centrifugal classifiers
- 26. Cyclone separator • Uses centrifugal force without having any moving or rotating part. • The dust laden gas is admitted tangentially into the cyclone separator. • Gas-solid separation • Particle size range > 5-10 μm • Separation efficiency : 98% Embedded video
- 27. Hydro cyclone • Liquid – solid particle suspension • Liquid – liquid separation
- 28. • For effective separation • Upward force on the particles should be as large as possible. • The particles get along the wall, the gas comes out clean. • Upward force on the particle (centrifugal force) is the tangential velocity of the gas. m = mass of the particle. r = radius of rotation of the particle.
- 29. • Inward drag on the particle (estimated by Stokes Law) Where, Dp = diameter of the particle. = Viscosity of gas. Vr = radial component of the velocity of gas
- 30. • At equilibrium = Assuming particle to be spherical Terminal velocity of particle, laminar free settling by Stokes law Since, particle density is much greater than air density, ≈
- 31. On working further: see book D0=Diameter of top outlet Z= depth of the separator D = diameter of the cyclone G = mass flow rate of gas Ai = Area of cross section of gas inlet Vt = minimum terminal velocity
- 32. Estimate the cut diameter and overall collection efficiency of cyclone given the particle size distribution of dust from cement kin. Particle size distribution and other data is given below Dp (µm) 1 5 10 20 30 40 50 60 >60 Wt.Per cent 3 20 15 20 16 10 6 3 7 Gas viscosity = 0.02 Cp; Specific gravity of particle = 3; inlet gas velocity = 48 ft/s; Effective number of turns Cyclone diameter = 8 ft; inlet width = 2 ft Overall collection Efficiency : Sigma (Wi * Ei) Wi = weight fraction
- 33. • If gravitational free settling velocity of the particle is equal or more than (Vt)min. Separation will take place. • Therefore,
- 34. Problem 1 • A cyclone separator 1.0 m in diameter and 3.0 m in depth has been used to handle 144,000 kg/day of dust-laden gas. The gas inlet is 0.3 m square and gas outlet in 0.4 in diameter. If the size distribution of the particles (sp. Gr. 1.2) in the feed gas is as given below, estimate what percent of particles will be separated from the gas stream: Particle size (microns) Mass fraction - 11.7 + 8.3 0.098 - 8.3 + 5.9 0.234 - 5.9 + 4.2 0.277 - 4.2 + 2.9 0.149 - 2.9 + 2.1 0.101 - 2.1 0.141 Viscosity of gas = 0.03 centipoise and its specific gravity = 0.001
- 35. • Ai = (0.3x0.3) = 0.09m2 • Do = 0.4 m • ρg = 1.0 kg/m3 • µg = 3 x 10-5 kg/m.s • Z = 3.0 m • D = 1.0 m • ρs = 1200 kg/m3 • G=144000/(24 x 3600) = 1.667 kg/s • (Dp)min = 4.3 microns