Example 5.4 Determine the terminal settling velocity of a worn sand particle having a measured sieve diameter of 0.6 mm and specific gravity of 2.65. Assume the settling is type 1 and the temperature of the water is 22°C.
d = 0.6( 10 ) m : 9.2(10-4) k for worn sands, ¡3 = 0.86
V22 = 9.2 (10-4) kg/m-s = 9.2 (10-4) N-s/m2, d = 1.24£0 333d„,
Therefore, d = 1.24 (0.860333 )(0.6)(10-3) = 0.71 (10-3) m v = ¡4 981.[2650-997] [0.71 (10^ = 01)24 ^ . ^ Cd(997) Je»
CD = 24 + -L + 0.34 (for 1 < Re < 104) Re JRe
Re = ^ = dv(997) = i,083,695.70dv = 769.42v V 9.2( 10-4)
Solve by successive iterations:
0.90 0.131 100
0.88 0.132 101
A raw water that comes from a river is usually turbid. In some water treatment plants, a presedimentation basin is constructed to remove some of the turbidities. These turbidity particles are composed not of a single but of a multitude of particles settling in a column of water. Since the formulas derived above apply only to a single particle, a new technique must be developed.
Consider the presedimentation basin as a prototype. In order to design this prototype properly, its performance is often simulated by a model. In environmental engineering, the model used is a settling column. Figure 5.8 shows a schematic of columns and the result of an analysis of a settling test.
At time equals zero, let a particle of diameter do be at the water surface of the column in a. After time to, let the particle be at the sampling port. Any particle that arrives at the sampling port at to will be considered removed. In the prototype, this removal corresponds to the particle being deposited at the bottom of the tank. to is the detention time. The corresponding settling velocity of the particle is vo = ZJto, where Zo is the depth. This Zo corresponds to the depth of the settling zone of the prototype tank. Particles with velocities equal to vo are removed, so particles of velocities equal or greater than vo will all be removed. If xo is the fraction of all
particles having velocities less than vo, 1 - xo is the fraction of all particles having velocities equal to or greater than vo. Therefore, the fraction of particles that are removed with certainty is 1 - xo.
During the interval of time to, some of the particles comprising xo will be closer to the sampling port. Thus, some of them will be removed. Let dx be a differential in xo. Assume that the average velocity of the particles in this differential is vp. A particle is being removed because it travels toward the bottom and, the faster it travels, the more effectively it will be removed. Thus, removal is directly proportional to settling velocity. Removal is proportional to velocity, so the removal in dx is therefore (vp/vo)dx and the total removal R comprising all of the particles with velocities equal to or greater than vo and all particles with velocities less than vo is then rx0 Vp r = 1- xo + f -idx (5.17)
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