Question

A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs are inboard channels set on a diameter of 18m. for a flow of 12900m^3/d, calculate the overflow rate, detention time,and weir loading.

Answer 1

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading  114.06 m^3/d/m

Explanation:

calculation for single clarifier

$sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d$

$surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2$

$surface area = 314.16 m^2$

volume of tank$V = A\times side\ water\ depth$

                             $=314.16\times 2 = 628.32m^3$

$Length\ of\ weir = \pi \times diameter of weir$

                       $= \pi \times 18 = 56.549 m$

overflow rate =$v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2$

Detention time$t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr$

weir loading$= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m$