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7-7-AttachGrowth_F09.doc
Biological Treatment
Types of Secondary Treatment Systems
- Suspended Growth Systems (Reactors) e.g. Activated Sludge processes - Conventional - Completely mixed
- Attached Growth Systems (Reactors)
e.g. Trickling Filters, Rotating Biological Contactor (RBC) Submerged Rotating Biological Contactor (SBC)
Attached Growth System
Principles - Schematic diagram showing principles of the biological process in a trickling filter.
Nutrient Organic carbon Medium Surface
Air O 2 O 2
: : Wastewater
End products CO 2
Anaerobic Aerobic 0.1 - 0.2 mm Biofilm
Biological process on medium surface in a Trickling Filter
7-7-AttachGrowth_F09.doc
Biofilm images (x 1000): dead (red) and live (green) bacterial cells grown on an MFC electrode. Left: heavy growth with clustered live cells on the anode membrane. Right: sparse attachment to a control membrane (an anode-material membrane unattached to the electrical circuit).
Examples
- Trickling filters,
- Rotating Biological Contactor (RBC)
- Submerged Rotating Biological Contactor (SBC)
7-7-AttachGrowth_F09.doc
Characteristics:
- Facultative system (Aerobic to anaerobic)
- Biofilm
- bacteria, protozoans, fungi, rotifers, algae, sludge worms, filter-fly larvae.
- Thin aerobic film 0.1-0.2 mm
- Biological variation with depth of the filter
- Algae uptake at the upper surface
- Nitrification near the bottom 5. Sloughing – breaking off of biofilm
Advantages and Disadvantages of TF
Advantages
- Low energy input
- Accepts qualitative and quantitative shock loads
- Accepts toxic load (to some extent)
- Good sludge settling at secondary clarifier
Disadvantages
- Poor performance in winter
- Fair performance in summer
- Land requirement
National Research Council (NRC) formula (SI Unit) (3rd^ DC 379; 4th^ DC 486)
For a single-stage filter or the first stage of a two-stage filter,
E
Q C
V F
in
(^1) 0 5
.
where E 1 = BOD 5 removal efficiency for the first stage filter at 20°C, including recirculation and sedimentation, fraction Q = wastewater flow rate, m^3 /s Cin = influent BOD 5 , mg/L V = volume of filter media, m^3 F = recirculation factor
F
R
R
2
where R = recirculation ratio = Qr/Q Qr = recirculation flow rate, m^3 /s Q = wastewater flow rate, m^3 /s
For the second stage filter,
E
E
QC
VF
e
2
1
0 5
.
where E 2 = BOD 5 removal efficiency for the second stage filter at 20°C, fraction E 1 = BOD 5 removed in first stage, fraction Ce = effluent BOD 5 from first stage, mg/L
Effect of Temperature on the efficiency
E T = E 20 θ ( T −20)
where ET = BOD removal efficiency at temperature T (°C) E 20 = BOD removal efficiency at 20°C θ = 1.
2 1
E QC^ e E VF
+^ =
−^ ^
2 2 1
E E QC^ e E VF
+ ^ =
− ^
2 2 1
E QC e E E VF
=^ −
( )
2 1 2 2
QC e (^) E E VF E
^ −
=^ −
( )
2 1 2 2
e 4.
E
QC VF E
E
( )
2 1 2 2
V QC^ e E F E E
^
Example 5-5 (3rd^ DC 380); Example 6-12 (4th^ DC 487)
Using the NRC equations, determine the BOD 5 of the effluent from a single-stage, lower-rate trickling filter that has a filter volume of 1,443 m^3 , a hydraulic loading of 1, m^3 /d, and a recirculation factor of 2.78. The influent BOD 5 is 150 mg/L.
(Solution) To use the NRC equation, the hydraulic loading must first be converted to the correct units.
Q
m day
day m
,^3 0 022^3
, sec
sec
E
QC
VF
in
(^1) 0 5 0 5
..
where Q = 0.022 m^3 /s Cin = 150 mg/L V = 1,443 m^3 F = 2.
The concentration of BOD 5 in the effluent is
Ce = (1 - 0.8943)(150 mg/L) = 15.8 mg/L
Check:
( )( )
( )
V
QC
F
E
E
= in m −
1 1
2 2
3
Example
Given Design Criteria: Design BOD load = 15 lb/1000 ft^3 - day Hydraulic loading = 2 - 4 Mgal/acre - day Filter Depth = 5 - 7 ft Over flow rate for primary sedimentation tank= 600 gpd/ft^2 Over flow rate for final clarifier = 800 gpd/ft^2
Given Data: Q = 1.5 MGD, BODave = 180 mg/L, Qr = 0 MGD, T = 17°C BOD removal efficiency in the primary sedimentation tank = 35%
Design the single stage TF and compute the overall WWTP efficiency for BOD removal.
Primary sedimentation Secondary sedimentation
Q = 1.5 MGD BOD = 180 mg/L
Trickling filter
(Solution)
- BOD loading to the primary sedimentation tank.
(180 mg/L) (1.5 MGD) (8.34) = 2,252 lb/d
- BOD loading to TF = (1 - 0.35)(2,252 lb/d) = 1,464 lb/d
1,464 lb/d
- Volume of TF media required, V = ------------------------- 15 lb /1000 ft^3 - d = 97,600 ft^3
- Surface area of TF with the TF depth h = 5 ft
A
5 ft
A = V / h = 97,000 ft^3 / 5 ft = 19,400 ft^2 A = 19,400 ft^2 (1 acre / 43,560 ft^2 ) = 0.45 acre
1.5 MGD
- Hydraulic loading = Q/A = ----------------- = 3.4 Mgal/ acre-d 0.45 acre
4'. Surface area of TF with the TF depth h = 6 ft
A = V / h = 97,000 ft^3 / 6 ft = 16,200 ft^2 A = 16,200 ft^2 (1 acre / 43,560 ft^2 ) = 0.37 acre
1.5 MGD 5'. Hydraulic loading = Q/A = ---------------- = 4.1 Mgal/ acre-d 0.37 acre
- Depth of TF - Let’s build two (2) TF units with the diameter of 100 ft. V = π r^2 h = 3.14 (50 ft)^2 h = 97,600 ft^3 / 2 = 48, 800 ft^3 Solve for the depth h, h = 6.2 ft
- BOD removal efficiency
E
W
VF
(^1) 0 5 0 5
^
^
Note: Design BOD load, W/ V = 15 lb/1000 ft^3 – day
( ) ( )
F
R
R
- Effect of temperature on the efficiency
E T = E 20 θ ( T^ −^ 20)^ = ( 82 2%)(1035.. )(^17 −20)=74%
- Overall plant efficiency, ET, including primary and secondary treatment
at 20°C
ET = 100 - 100 [ (1 - Prim. Sed. eff) ( 1 - TF eff.) ] = 100 - 100 [ (1 - 0.35) (1 - 0.82) ] = 88.3 %
