Unit Operation and PorcessesType II Settling

Type II SettlingType II settling is the settling of flocculent particles in a dilute suspension. The particle flocculate during settling; thus they increase in size and settle at a faster velocityThe example of type II settling are the primary settling of wastewater and the settling of chemically coagulated waters and wastewatersBatch settling test are usually required to evaluate the settling characteristics of a flocculent suspension. The column should be at least 130 to 205 mm. Sampling ports are provided at equal intervals in height.The suspension must be mixed thoroughly and poured rapidly into the column in order to insure that a uniform distribution of the particles occurs throughout the height of the column. Samples are removed through the ports at periodic time intervals and the suspended solids concentrations are determined.

The overflow rates, V0, are determined for the various settling times (ta, tb, and so on) where the R curves intercept the x axis. For example, for the curve Rc, the overflow rate is:Where H is the height of the column and tc is the intercept of the Rc curve and the x-axis. The fractions of solid removed. RT, for the times ta, tb, and so on are then determined. For example, for time tc, the fraction removed RT would be:Where H2 represents the height that the particles of (Rd Rc) size would settle during tc. These would intercept the sludge zone in a basin, as in Figure 9.13. By using the various fractions removed, RT, a graph of the overflow rates versus fractions removed can be constructed. Also, a graph of the fractions removed versus detention times can be made. In applying the curves to design a tank, scale-up factors of 0,65 for the overflow rate and 1,75 for detention time are used to compensate for the side-wall effects of the settling column (Eckenfelder, 1980)

Example : Primary ClarifierA primary clarifier is to be designed to treat an industrial wastewater having 320 mg/l suspended solids and a flow of 7.550 m/d. A batch settling test was performed using a 205 mm- diameter column that was 3,05 m long and had withdrawal ports every 0,61 m. The reduced data giving the percent removals are shown in Table 9.3.Determine:The design detention time and design surface loading rate if 65% of the suspended solids are to be removedThe diameter and depth of the tank

SolutionA plot of the percent removals at the various depths and times is shown in Figure 9.17. Interpolations have been made to locate the 20, 30, 40, 50, 60, and 70% removal curves.

The 20% curve intersects the x-axis at 16 min; thus the surface loading at that time is:

The detention time in hours is 16/60 or 0,27 h. The point midway between the 20 and 30% curves at 16 min is located as shown and is at a depth of 2,04 m. In a like manner, the points midway between the 30 and 40, 40 and 50, 50 and 60, and 70% curves are located, and the respective depths are 0,88; 0,61; 0,40; and 0,24 m. These values give the total fraction removed (RT) at 16 min (0,27 h) as:

RT = 20 + (2,04/3,05)(30-20) + (0,88/3,05)(40-30) + (0,61/3,05)(50-40) + (0,40/3,05)(60-50) + (0,24/3,05)(70-60)= 33,7%

Similiarly, the surface loading rates, detention times, and total fractions removed are computed for the 30, 40, 50, and 60% curves, and a summary of the reduced data is shown in Table 9.4

A plot of the fraction removed (RT) versus detention time (t) is shown in Figure 9.18. Also, a plot of the fraction removed (RT) versus surface loading rate (V0) is shown in Figure 9.19.

For 65% removal the detention time is 1,22 h; thus the design detention time is (1,22)1,75=2,14 h.For 65% removal the surface loading is 58,0 m/d-m; thus the design surface loading is (58,0)0,65=37,7 m/d-m.The required area is:

The required depth, H, is:

Latihan SoalDirencanakan sebuah bak pengendap untuk mengendapkan air limbah dengan SS 350 mg/L dan debt 7.500 m/hari. Uji laboratorium dilakukan terhadap air limbah tersebut dengan kolom pengendapan berdiameter 20 cm dan tinggi 300 cm. Pada setiap 60 cm terdapat port (sampling point). Hasil tes kolom adalah sebagai berikut:Tentukan:Waktu detensi dan surface loading agar diperoleh 65% pengendapanDiameter dan kedalaman bak

Langkah-langkah PengerjaanUbah data laboratorium menjadi % removalPlot tabel poin 1 sehingga membentuk grafik isoremovalAmbil waktu tertentu dan hitung penyisihan total (RT) dan surface loading (V0) pada waktu tersebutDengan cara yang sama (poin 3), tentukan removal total dan surface loading pada t (waktu) yang lainPlot hubungan % RT vs t dan % RT vs surface loadingTentukan waktu dan surface loading yang diperlukan untuk mendapatkan 65% pengendapanHitung luas permukaan bakHitung diameter bak, jika bak berbentuk lingkaranHitung kedalaman bak