Problem 1: Primary Clarifier A batch settling test has been performed on an indu
ID: 1583830 • Letter: P
Question
Problem 1: Primary Clarifier A batch settling test has been performed on an industrial wastewater having an initial suspended solids concentration of 597 mg/L to develop criteria for the design of a primary clarifier (i.e., sedimentation basin). The test column was 125 mm in diameter and 2.44 m high, and sampling ports were located 0.61, 1.22, 1.83, and 2.44 m from the water surface in the column. The suspended solids remaining after the various sampling times are given in the Table below. The wastewater flow is 9,460 m3/day. Determine: a) The design overflow rate and detention time if 65% of the suspended solids are to be removed. Use scale-up factors of 0.65 for the overflow rate and 1.75 for the detention time. b) The diameter if a circular clarifier is used. c) The depth of the tank. C (mg/L) DEPTH10 min 20 min30 min 45 min60 min 0.61 394 352 243 182 48 1.22 m 460 406 337295 216 1.83 m 512 429 376 318 306 2.44 m 1018 1142 1208 1315 1405
Explanation / Answer
Part A:
the overflow rate = total flow per unit area.
area, A = (2*pi*d*h)/2
A = (2*3.14*0.125*2.44)/2 = 0.9577 m2
65% of the suspended solids are to be removed.
Hence,
dischare = 9460*0.65 = 6149 m3/d
overflow rate = 6149/0.9577= 6420.59 m/d
But,
detention time = volume of tank / flow rate
volume, V = (pi*d2*h)/4
V = (3.14*0.1252*2.44)/4
V = 0.0299*1.75
volume of tank, V = 0.0523 m3
flow rate = 6149 m3/d
Therefore,
detention time = 0.0523/6149 = 0.00008619 days
= 0.00008619*24*60*60
Detension Time = 7.44 sec
Part B:.
if a circular clarifier is used volume will remain constant
Hence, V = 0.0523 m3
(3.14*d2*2.44)/4 = 0.0523
d = 0.165m
Diameter of the circular clarifier, d = 165 mm
Part C:
similarly the depth, h can be obtained as
(3.14*.1252*h)/4 = 0.0523
The depth of the tank, h = 4.26 m