Jq V

Note that this equation does not state that the velocity vp must be terminal. It only states that the fractional removal R is directly proportional to the settling velocity vp. For discrete settling, this velocity is the terminal settling velocity. For flocculent settling (to be discussed later), this velocity would be the average settling velocity of all particles at any particular instant of time.

To evaluate the integral of Equation (5.17) by numerical integration, set f dx = 1Y vpAx (5.18)

This equation requires the plot of vp versus x. If the original concentration in the column is [co] and, after a time of settling t, the remaining concentration measured at the sampling port is [c], the fraction of particles remaining in the water column adjacent to the port is x = w (519)

Corresponding to this fraction remaining, the average distance traversed by the particles is Zp/2, where Zp is the depth to the sample port at time interval t from the initial location of the particles. The volume corresponding to Zp contains all the particles that settle down toward the sampling port during the time interval t. Therefore, vp is vp = -Z2---t-- p (5.20)

The values x may now be plotted against the values vp. From the plot, the numerical integration may be carried out graphically as shown in c of Figure 5.8.

Example 5.5 A certain municipality in Thailand plans to use the water from the Chao Praya River as a raw water for a contemplated water treatment plant. The river is very turbid, so presedimentation is necessary. The result of a column test is as follows:

c(mg/L) 299 190 179 169 157 110 79 28

What is the percentage removal of particles if the hydraulic loading rate is 25 m3/m2 . d? The column is 4-m deep.

Solution:

qo = hydraulic loading rate = 25 m3/m2 • d = 25 m/d = 0.0174 m/min rx«vp Cx°vp 1 x-,

R = 1- xo +1 — dx I — dx = — > vp Ax o Jn v„ Jn v„ v.^-1 p

■>0 v vo = 0.0174 m/min t (min) c (mg/L) x = [c]/[c„] vp = Zp/2t = 4/2t (m/min) —

0 60 80 100 130 200 240 420

299 190 179 169 157 110 79 28

Find the x corresponding to vo = 0.0174 m/min x -0.57 0.0174-0.02

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