Wastewater treatment: Activated sludge

Activated SludgeSecondary settling Tank BS The aeration tank contains mixed liquor suspended solids MLSS, a combination of influent wastewater and return recycled activated sludge from

Biological Treatment

Activated Sludge

Activated Sludge

Secondary settling Tank BS

The aeration tank contains mixed liquor suspended solids (MLSS), a

combination of influent wastewater and return (recycled) activated sludge from the settling tank

The MLSS includes:

• Bacteria (both aerobes and facultative)

• Slime: usually in flocs composed of extracellular polymeric substances, live bacterial cells, cell debris (dead, lysed)

• Free cells

• Protozoa

• Nondegradable organic matter

• Inert suspended solids

Activated sludge floc with slime

• MLSS rapidly (20-45 min)

absorbs organic matter in

wastewater influent

• Bacteria then solubilize and

oxidize organic matter.

• The food (organic matter) to

microorganisms ratio (F/M)

dictates character of

bacteria and flocs

Activated sludge process is defined by:

1 Aeration period = detention time= 3-30 hrs

Design Considerations

3 F/M= (Q*BODin)/(V*MLSS)

2 BOD loading = (BOD entering per day)/ (V=volume of aeration tank)

w w

e e

sludge

Q SS

Q SS

V MLSS t

*

*

*

+

=

SSe= suspended solids in the effluent

SSw= suspended solids in waste sludge

Qe = effluent flow rate

Qw= sludge flow rate

4 Sludge age = mean cell residence time

Aeration Tank SST

Qw, SSw

Qe, SSe

Conventional aeration

High-purity oxygen Step feed to aeration

Extended aeration

Types of Activated Sludge Systems

Process BOD load

(lb BOD/d per 1000 ft 3 )

MLSS mg/l lb BOD/d F/M

per lb MLSS

tsludge days

Aeration period hr

R % % BOD

removal

Conventional 20-40 1000-3000 0.2-0.5 5-15 4-7.5 20-40 80-90

Step aeration 40-60 1500-3500 0.2-0.5 5-15 4-7 30-50 80-90

Extended

aeration 10-20 2000-8000 0.05-0.2 >20 20-30 50-100 85-95

High-purity

oxygen >120 4000-8000 0.6-1.5 3-10 1-3 30-50 80-90

Loading and operation procedure for different aeration processes

Consider the following activated sludge system Calculate:

1 The aeration period

2 BOD loading

3 F/M ratio

4 SS and BOD removal efficiency

5 tsludge

6 Return activated sludge rate

7 What is the required QR if the MLSS is to be 2200 mg/l

MLSS=2600 mg/l

Qw=39,000 gpd

QR=0.85mgd

SS=212mg/l

BOD=185 mg/l

Q=2.14 mgd

SSw=8600 mg/l

BOD=15 mg/l

SSe=15 mg/l

Aeration:

4 units each 40x40ft 2

15.5 ft deep

Settling tank

V= 4*40*40*15.5* (7.481/106)= 0.74 million gallon

t=V/Q=(0.74/2.99)*(24)=5.9 hr

Solution

BOD loading= [2.14*185*8.34]/[4*40*40*15.5/1000]= 33.3 lb BOD/d/1000 ft3

Note that the BOD loading from re-circulated sludge is

negligible

MLSS lb

per day BOD

lb l

mg gal

mil

l mg mgd

M

34 8 / 2600

74 0

34 8 / 185

14 2

×

×

×

×

=

SS removal= (212-15)/212= 0.93= 93%

BOD removal= (185-15)/185= 0.92= 92%

days l

mg mgd

l mg mgd

l mg gal

mil

) 34 8 / 8600 039

0 ( ) 34 8 / 15 14

2 (

34 8 / 2600 74

0

=

×

× +

×

×

×

×

=

Return sludge ratio (R)= 0.85/2.14=0.40 =40%

To find QR for a MLSS =2200, we apply mass balance on suspended solids

at the entrance of the aeration tank

2.14*212 + QR*8600 = (2.14+ QR)*2200

Solve for QR= 0.66 mgd