Lecture-6 Motion of a Particle Through Fluid (One dimensional Flow) 1
Equation of Motion of a spherical Particle (one dimensional Flow) On Board 2
Terminal Velocity Particle reaches a maximum velocity which is attainable under given circumstances and is called terminal velocity du dt = g (ρ p ρ f ρ p ) C Du 2 o ρ f A p 2m (1) Under gravitational settling, g is constant, drag increases with velocity, equation (1) shows that acceleration decreases with time and approaches zero. 3
Put du/dt = 0 in equation(1): 0 = g ( ρ p ρ f ρ p ) C Du 2 o ρ f A p 2m g ( ρ p ρ f ρ p 2gm ( ρ p ρ f ρ p ) ) = C Du 2 o ρ f A p 2m 1 C D ρ f A p = u o 2 u t = For case of centrifugal forces: 2gm ( ρ p ρ f ρ p ) 1 C D ρ f A p (2) u t = ω 2rm ρ p ρ f 1 (3) ρ p C D ρ f A p 4
Evaluation of Terminal velocity(influence of Drag Coefficient) Determination of terminal velocity requires the numerical values of Drag Coefficient, C D. Drag coefficient can be determined from the graph (Fig. 7.6) as a function of Reynold s number under some restricted conditions: Particle must be a solid sphere. Particle is far from other particles and from wall surface. Moving with terminal velocity. Problems Due to assumption Written above Underestimate: The drag coefficients for accelerating particles are appreciably greater than those reported hence terminal velocity differs considerably than predicted. Particle shape: Variation in particle shape and change in orientation during free motion of particles through fluid consumes energy and increases the effective drag on the particle. As a result terminal velocity (especially disks and platelike particle) is less than would be predicted from curves of fixed orientation. 5
The conditions during settling of particles also affect the value drag coefficient. The settling of spherical particle depends can be divided into two based on the environment the particle is facing. Free settling: When particle is at sufficient distance from the boundaries of container and from other particles so that its fall is not affected by them, the process is called free settling (Drag is low) Hindered settling: If motion is impede by other particles and wall of the container then which will happen, when the particles near each other even though, they may not actually be colliding, the process is called hindered settling (Drag is higher). 6
Being smaller in size poses another problem i.e. Brownian Movement (this is random motion imparted to particle by collisions between the particle and the molecule of surrounding fluid) It is appreciable at a particle size of about 2-3 μm and pre-dominates over the gravity with particle size 0.1 μm. Brownian movement tends to suppress the effect of force of gravity, so settling does not occur. But the application of centrifugal force reduces the relative effect of Brownian movement. 7
Quantitative Analysis of motion of spherical particles: For sphere of diameter D P and m = π/6 D p3 ρ p A p = π/4 D p 2 So the settling velocity under the action of gravity: (putting into equation-2), It becomes: u t = 4g ( ρ p ρ f ) Dp (4) ρ f 3C D As general case the terminal velocity can be determined by trial and error after getting Re,p (Particle Reynolds's Number) to get an initial estimate of C D (given as follows) u t = 4g ( ρ p ρ f ) Dp ρ f 3C D Guess Re,p Read C D from Graph Calculate U t No Compare calculated Re,p with Guess Recalculate Re,p Terminate Yes 8
The previous algorithm can be bypassed for limiting cases of Reynolds number (either very low Reynolds number or very high Reynolds number. Low Reynolds number(less than 1): Drag Coefficient varies inversely with Re,p and the pertinent equations are given below C D = 24 / Re,p (5) F D = (3πμu t D P )/g c (6) and u t = g(ρ p ρ f )D p 2 18μ (7) (Stokes law) Above equation (stoke s Law) can be applied for the Re,p < 1. At Re,p = 1, C D = 26.5 instead of 24 in the above equation. Stokes law gives about 5% error. The stokes law can be modified to predict the velocity of a small sphere in a centrifugal field by substituting rω 2 for g 9
Higher Reynolds Number (1000 < Re,p < 200,000 ) For 1000 < Re,p < 200,000 the drag coefficient is constant then: C D = 0.44 (8) F D = (0.055 ρπd p2 u 2 t )/ g c (9) U t = 1.75 g ( ρ p ρ f ρ f ) D p (10) Newtonian Law applicable for large particle falling in gas or low viscosity fluids. The terminal velocity, ut, varies with D p 2 in stokes region whereas in the Newton s region it varies with D p 0.5 10
Criterion for settling regimes To identify the range in which motion of the particle lies, the velocity term is eliminated from Reynolds number by substituting u t from equation 7, For Stock s Region Re,p = D pu t ρ µ = D p µ (ρ fgd 2 p (ρ p ρ f ) 18µ ) (11) = ρ fgd p 3 (ρ p ρ f ) 18 µ 2 Then the equation (12) will become K = D p ( ρ fg(ρ p ρ f ) µ 2 ) 1/3 (12) Re p = 1 18 K3 Setting Re,p = 1 gives K = 2.6, for know size of particle K can be calculated from equation (12) If K is less than 2.6 then Stoke s law applicable. 11
Substituting for U t from equation (10) into equation (11) gives the following result setting Re,p = 1000 Re, p = 1.75 K 1.5 K = 68.9 and Re,p = 200,000 gives K = 2360 So; 68.9 < K < 2360 Newton s law would be applicable When the K is greater than 2360, the drag coefficient may change abruptly with small changes in the fluid velocity. Above this range and between 2.6<K<68.9 the terminal velocity can be calculated by trial and error procedure indicated on the slide # 8 12
Algorithm (Criteria of Settling) After having all properties Find values of K Stoke s Region K < 2.6 Newton s Region 2.6 < K < 68.9 If the K > 2360 or 2.6<K<68.9 In both cases assume Re, p Calculate u t (from equation 4) Check Re,p YES NO 13
Hindered Settling
In Hindered Settling, the velocity of the particle is affected by the presence of nearby particles, so normal drag correlations do not apply. Particles in settling displaced liquid, which flows upward and makes the particle velocity relative to fluid greater than the absolute settling velocity. For Uniform suspensions the settling velocity can be determined by following empirical correlation developed by Maude & Whitmore n = 4.6 n = 2.5 Stoke s Region Newton s Region u s u t = Ɛ n Here Ɛ = void fraction For Small particles For Large particles u s u t = 0.62 (Ɛ = 0.9) u s u t = 0.77 (Ɛ = 0.9) u s u t = 0.095 (Ɛ = 0.6) u s u t = 0.28 (Ɛ = 0.6) 15
For Fine Suspension u t should be calculated by using density and viscosity of fine suspension. Ɛ can be taken as volume fraction of suspension instead of void fraction. e.g; suspension of very fine sand in water are used in seperating coal from heavy mineral, and the density of suspension is adjusted to a value slightly greater than that of coal to make the coal particles rise to the surface, while the mineral articles sink to the bottom. In that case suspension effective viscosity can be calculated as: µ s µ = 1 + 0.5(1 Ɛ) Ɛ 4 The above equation applies only when ε < 0.6 and is most accurate when ε > 0.9