An aeration basin for an activated sludge treatment plant has the following char

ID: 1921110 • Letter: A

Question

An aeration basin for an activated sludge treatment plant has the following characteristics:

Volume = 10000 gal MLSS = 1500 mg/L

Q=25000 gal/day

A secondary clarifier is used to separate the solids prior to recycling the sludge to the aeration basin.

Qr = 5000 gal/day Qw= 500 gal/day

Influent SS concentration = 100 mg/L Effluent SS concentration = 30 mg/L

Use a mass balance approach to determine Xr. Then determine the mean cell residence time. Include all flows in your mass balance.

Assume completely mixed and steady state conditions.

1) Xr

Qr /Q = Xt/Xr - Xt

Xt = 1500 mg/l

Qr = 5000 gal/day = 19 m3/day

Q = 25000 gal/day = 95 m3/day

19/95 = 1500/Xr - 1500

0.2 = 1500/Xr - 1500

0.2Xr -300 = 1500

0.2Xr = 1500-300

Xr = 1200/0.2

Xr = 6000 mg/l

2) Mean cell resistance time (MCRT)

MCRT = V . Xt/ Qw.Xr + (Q-Qw)XE

Qw = 500 gal/day = 1.9 m3/l

V = 10000 gal = 38m3

XE = 30 mg/l

MCRT = 38 x 1500/ 1.9 x 6000 +(95-1.9)30

= 57000/11400 +2793

= 4.01

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