Primary clarifiers are used to remove particles from wastewater through sedimentation. They typically remove 50-70% of suspended solids and 25-40% of biochemical oxygen demand (BOD). Primary clarifiers come in rectangular and circular designs and are typically sized to have detention times of 1.5-2.5 hours and overflow rates of 1-2.5 m3/m2-day. BOD and TSS removal efficiencies can be estimated based on detention time using empirical equations. Settled solids are unlikely to be resuspended if horizontal flow velocities remain below the calculated scour velocity.

1. Primary Clarifiers

2. • Primary clarifier are usually designed to remove particles with
settling rates of 0.3 – 0.7 mm/sec.
• Two common design of primary clarifier, rectangular and
circular.
• Efficiently designed and operated primary sedimentation tank
should remove from 50 -70 % of suspended solids and from
25 – 40% of BOD.

3. Typical dimensions of primary clarifier
Unit Range Typical
Rectangular Tank
Depth m 3-4.9 4.3
Length m 15-90 24-40
Width m 3-24 4.9-9.8
Sludge collector speed m/min 0.6-1.2 0.9
Circular Tank
Depth m 3-4.9 4.3
Diameter m 3-60 12-45
Sludge collector speed rpm 0.02 -0.05 0.03

4. BOD and TSS removal :
• Typical performance data for the removal of BOD and TSS in
primary sedimentation tank as a function of detention time
and concentration are presented in the figure below.
• These curves are derived from observations of the
performance of actual sedimentation tanks.
• The relationship in the figure can be modeled as follows:
R=
𝑡
𝑎+𝑏 𝑡
R : expected removal efficiency
t: nominal detention time
a , b : experimental constants

6. • Typical values of the experimental constants are as follows:
Surface loading rates:
• Sedimentation basins are normaly designed on the basis of
surface overflow rate (commonly termed “overflow rate)
expressed as m3/m2.d. Typically, 30 -50 m3/m2.d is suitable for
average flow.
• It should be emphasized that overflow rate must be set low
enough to ensure satisfactory performance at peak rates.
a b
BOD 0.018 0.02
TSS 0.0075 0.014

7. Typical design information
Rectangular Circular
Detention time (h) 1.5 -2.5 1.5-2.5
Overflow rate (m/h) 1-2.5 1-2.5
Weir load (m3/m.d) 125-500 125-500

9. Ex/ The average flow rate at a small municipal wastewater
treatment plant is 20 000 m3/d. the highest observed peak daily
flow rate is 50 000 m3/day.
a. Design rectangular primary clarifier with a channel width of 6
m, use a minimum of two clarifiers.
b. Calculate the scour velocity to determine if settled material
will become resuspended.
c. Estimate the BOD and TSS removal at average and peak flow.
Use an overflow rate of 40 m3/m2.d at average flow and a side
water depth of 4 m.

10. Solution.
a. Design
A = Q/SOR
A= 20 000/ 40 = 500 m2
L = A/W
L= 500/ (2X6)= 41.7 m =42 m.
Retention time = volume of the tank / Q
Retention time = 2 X (42X6X4) X 24/ 20 000 = 2.4 h.
Overflow rate = Q / A
Overflow rate = 20 000/ (2 X (6X42)) = 39.7 m3/m2.d

11. Retention time = volume of the tank / Q
Retention time = 2 X (42X6X4) X 24/ 50 000 = 0.97 h.
Overflow rate = Q / A
Overflow rate = 50 000/ (2 X (6X42)) = 99 m3/m2.d
b. Calculate the scour velocity:
𝑉 𝑠𝑐𝑜𝑢𝑟 =
8 𝛽 𝑔 𝑆 − 1 𝑑
𝑓
𝑉 𝑠𝑐𝑜𝑢𝑟 =
8 ×0.05 ×9.81 1.25−1 100 ×10−6
0.025
= 0.063 m/sec

12. Calculate the peak flow horizontal velocity to compare to vscour
V = Q / A where A: cross-sectional area through which the flow
is passes.
V= 50 000/2(6X4) = 1041 m/d = 0.012 m/sec.
The horizontal velocity value, even at peak flow, is substantially
less than the scour velocity. Therefore, settled matter should not
be resuspended.

13. c. estimate the BOD and TSS removal
At average flow:
𝐵𝑂𝐷 𝑟𝑒𝑚𝑜𝑣𝑎𝑙 =
𝑡
𝑎+𝑏 𝑡
=
2.42
0.018+0.02 × 2.42
= 36%
𝑇𝑆𝑆 𝑟𝑒𝑚𝑜𝑣𝑎𝑙 =
𝑡
𝑎+𝑏 𝑡
=
2.42
0.0075+0.014 × 2.42
= 58%
At peak flow:
𝐵𝑂𝐷 𝑟𝑒𝑚𝑜𝑣𝑎𝑙 =
𝑡
𝑎+𝑏 𝑡
=
0.97
0.018+0.02 × 0.97
= 26%
𝑇𝑆𝑆 𝑟𝑒𝑚𝑜𝑣𝑎𝑙 =
𝑡
𝑎+𝑏 𝑡
=
0.97
0.0075+0.014 × 0.97
= 46%

14. Ex/ a wastewater containing 250 mg/l suspended solids and 200
mg/l BOD passes through a clarifier. Removal efficiencies of BOD
and suspended solids are 37 and 70% respectively at average
flow ( 10 000 m3/d) , and 27 and 48% respectively at peak flow (
30 000 m3/d) .
Estimate:
a. Effluent BOD and TSS concentration at average and peak flow.
b.The maximum and minimum rate of sludge production
(assume sludge is 95% water).

15. Solution:
At average flow:
Effluent BOD = 200 X (1- 0.37) = 126 mg/L.
Effluent TSS = 250 X (1- 0. 7) = 75 mg/L.
Solids accumulated=0.7 X 250 X 10 000 X (103/ 106)=1750kg/day
Sludge production= 1750/0.05 =3500 kg/day.
At peak flow:
Effluent BOD = 200 X (1- 0.27) = 146 mg/L.
Effluent TSS = 250 X (1- 0. 48) = 130 mg/L.
Solids accumulated=0.48 X 250 X 30 000 X (103/ 106)=3600 kg/day
Sludge production= 3600/0.05 =72 000 kg/day.