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Background...

Abbreviations...

Q (m³ / h)	Flow - rate
Q_o (m³ / h)	Influent flow - rate
Q_e (m³ / h)	Effluent flow - rate
Q_r (m³ / h)	Recycled sludge flow - rate
Q_w (m³ / h)	Wasted sludge flow - rate
BOD (mg / L)	Biochemical oxygen demand
BOD_o (mg / L)	Influent biochemical oxygen demand
V_a (m³)	Aeration tank volume
SS (mg / L)	Suspended solids (SS)
SS_e (mg / L)	Effluent SS
SS_{r , w} (mg / L)	Recycled and wasted sludge SS
A (m³ / h)	Air flow - rate
MLSS (mg / L)	Mixed liquor suspended solids
t (h)	Hydraulic retention time
OL (kg BOD / m³ . day)	Organic loading
F / M (kg BOD / kg MLSS . day)	Food to microorganism ratio
r	Recycle ratio
SA (day)	Sludge age
ASR (m³ / kg BOD)	Air supply rate
E (%)	BOD removal efficiency

Equations...

(1) Normally ;

Q_w << Q_o and Q_e = Q_o = Q

(2) Aeration time or hydralic retention time = Volume / Flow - rate

t = V_a / Q

Typical aeration times : Conventional activated sludge process = 4 - 8 hrs ; Contact - Stabilization = 0.5 - 1.0 hrs (contact tank) ; Extended aeration = 24 hrs (aeration times are used to determine volume of aeration tanks)

(3) BOD or organic loading = Mass of BOD applied per day / Tank volume

OL = ( Q_o )( BOD_o ) / V_a

Typical OL values : Conventional activated sludge process = 0.3 - 0.6 kg BOD / m³ . day ; Contact - Stabilization = 1.0 - 1.2 kg BOD / m³ . day ; Extended aeration = 0.16 - 0.4 kg BOD / m³ . day (this is an alternative method for determining volume)

(4) Food to microorganism ratio = Mass of BOD applied per day / Mass of suspended solids in aeration tank

F / M = ( Q_o )( BOD_o ) / ( V_a ) ( MLSS )

Typical F / M values : Conventional activated sludge process = 0.2 - 0.4 kg BOD / kg MLSS . day ; Contact - Stabilization = 0.2 - 0.6 kg BOD / kg MLSS . day ; Extended aeration = 0.05 - 0.15 kg BOD / kg MLSS . day (used to determine the needed concentration of MLSS)

(5) Recycle ratio = Recycle flow - rate / Influent flow - rate

r = Q_r / Q_o

Typical recycle ratio values : Conventional activated sludge process = 0.25 - 0.50 ; Contact - Stabilization = 0.25 - 1.0 ; Extended aeration = 0.75 - 1.5 (used to control concentration of MLSS)

(6) Mean cell residence time or sludge age = Mass of suspended solids in aeration tank / Mass of solids leaving the system

SA = [ ( V_a )( MLSS ) ] / [ ( Q_e ) ( SS_e ) + ( Q_w ) ( SS_w ) ]

Typical sludge age values : Conventional activated sludge process = 4 - 15 day ; Extended aeration = 20 - 30 day (an estimate of the length of time that average suspended solids or biomass stays in the process, used to select volume and MLSS and solids wasting rate)

(7) Air supply rate = Volume of air per unit time / Influent flow

ASR = A / Q_o

or

ASR = A / [ ( Q_o ) ( BOD_o ) ]

(8) BOD removal efficiency ;

E ( % ) = [ ( BOD_o - BOD_e ) / BOD_o ] 100

(BOD removal efficiency is a function of MLSS, sludge age, or F/M ; there is usually an optimum range of MLSS, sludge age, and F/M)

Basic Calculations...

Treatment Efficiency - Example 1...

Given : (a) BOD₅ of domestic wastewater entering primary clarifier = 200 mg / L, (b) BOD₅ removal efficiency of activated sludge system (based on total BOD₅ in the effluent from the secondary clarifier) = 90%
Required : (a) Effluent BOD₅ and (b) Overall BOD₅ removal efficiency (primary clarifier + activated sludge system)
Solution : BOD₅ removal efficiency of primary clarifier ranges from 30% - 40%. Assume BOD₅ removal efficiency of 35%.
Primary effluent BOD₅ = (200 mg / L) (1 - 0.35) = 130 mg / L
Compute BOD₅ in final clarifier effluent
(a) Final effluent BOD₅ = (130 mg / L) (1 - 0.90) = 13 mg / L
Note : Computation of primary effluent BOD₅ concentration did not consider the primary clarifier sludge withdrawal flow. In addition, this approach does not consider any BOD associated with recirculation flows.
(b) Overall efficiency = (200 mg / L - 13 mg / L) / (200 mg / L) = 93.5 %

Activated Sludge Mass Balance - Example 2...

Given : Flow schematic of a completely-mixed activated sludge system. (a) Q = 4.0 mgd, (b) X_R = 8,000 mg / L, (c) Q_w = 0.064 mgd, (d) S₀ = 200 mg / L, (e) X = 2,800 mg / L, (f) S_e = 7 mg / L, (g) V = 0.861 mil.gal.
Required : The recirculation ratio, R.
Solution : Note that for tis problem, there is more information than is needed to solve the problem. To solve, perform a mass balance at the junction of flows upstream of the aeration tank. Mass of VSS in = Mass of VSS out The VSS in the return sludge is 8,000 mg / L. The VSS in the influent stream may be considered to be small (i.e., equal to 0).
(Q) (0) + (R) (Q) (8,000) = [Q + (R) (Q)] (2,800)
(R) (Q) (8,000) = [Q + (R) (Q)] (2,800)
(R) (8,000) = 2,800 + (2,800) (R)
(5,200) (R) = 2,800
R = 0.54

Design of the Activated Sludge Process - Example 3...

Given : (a) Influent wastewater flow = 7.7 mgd, (b) Volume of aeration tanks = 300,000 ft³ = 2.24 mil.gal, (c) Influent total solids (TS) = 600 mg / L, (d) Influent suspended solids (SS) = 120 mg / L, (e) Influent BOD = 173 mg / L, (f) Effluent TS = 500 mg / L, (g) Effluent SS = 22 mg / L, (h) Effluent BOD = 20 mg / L, (i) MLSS = 2,500 mg/L, (j) Recirculated sludge flow = 2.7 mgd, (k) Waste sludge flow = 54,000 gpd, (l) SS in waste sludge = 9,800 mg/L.
Required : (a) Aeration period, (b) Volumetric BOD loading, (c) F / M ratio, (d) TS removal efficiency, (e) SS removal efficiency, (f) BOD removal efficiency, (g) Sludge age, (h) Recirculation ratio.
Solution : (a) Aeration period = Volume / Flow without recirculation = 2.24 mil.gal / 7.7 mil.gal / day = 0.29 day = 7.0 hr
(b) Volumetric BOD load = Quantity of influent BOD / Aeration volume = [(173 mg / L) (8.34 lb / mil.gal / mg / L) (7.7 mil.gal / day)] / 300,000 ft³ = (37 lb / day) / (1,000 ft³)
(c) F to M = Quantity of influent BOD / Mass of MLSS in Aeration Basin
F to M = [(173 mg / L) (8.34 lb / mil.gal / mg / L) (7.7 mil.gal / day)] / [(2,500 mg / L) (8.34 lb / mil.gal / mg /L) (2.24 mil.gal)] = 0.24 lb BOD / day.lb MLSS
(d) TS removal efficiency = (600 - 500) / 600 = 16.7 %
(e) SS removal efficiency = (120 - 22) / 120 = 81.7 %
(f) BOD removal efficiency = (173 - 20) / 173 = 88.4 %
(g) Sludge age = [(MLSS) (Volume)] / [(SS_e) (Q_e) + (SS_w) (Q_w)]
Sludge age = [(2,500 mg / L) (2.24 mil.gal) (8.34 lb / mil.gal / mg / L)] / [(22 mg / L) (7.7 mil.gal / day) (8.34 lb / mil.gal / mg / L) + (9,800 mg / L) (0.054 mil.gal / day) (8.34 lb / mil.gal / mg / L)] = 8.0 day
(h) R = Q_R / Q = 2.7 mgd / 7.7 mgd = 0.35

Design of the Activated Sludge Process - Example 4...

Given : Town of 25,000, (b) Per capita wastewater flow = 120 gpcd, (c) Per capita BOD contribution = 0.20 lb / cap . day
Required :(a) Volume for step-aeration process with maximum volumetic BOD loading of 40 lb BOD / 1,000 ft³ / day a minimum aeration period of 6.0 hr, (b) Calculate dimensions of two circular final clarifiers.
Solution : (a) Determine influent flow and BOD load to primary clarifier ;
Q = (25,000 cap) (120 gpcd) = 3 mgd
Influent BOD load = (0.20 lb / cap.day) (25,000 cap) = 5,000 lb / day
Assume 35% of the BOD₅ is removed in the primary clarifiers. Compute BOD load to the aeration basin.
BOD load aeration basin = (0.65) (5,000 lb / day) = 3,250 lb / day
Compute the volume based on the volumetric loading = (3250 lb / day) / [(40 lb) / (1,000 day.ft³)] = 81,250 ft³
Compute the volume based on the aeration period = [(3x10⁶ gal / day) (6 hr) (1 day / 24 hr)] / (7.481 gal / ft³) = 100,254 ft³
Use a volume equal to 100,300 ft³
(b) Typically the average overflow rate for activated sludge plants with a flow greater than 1 mgd is 800 gpd / ft²
Total surface area = (3x10⁶ gal / day) / (800 gpd / ft²) = 3,750 ft²
Surface area per clarifier = 3,750 ft² / 2 = 1,875 ft²
[(n) (D²)] / 4 = 1,875 ft²
D = 48.9 ft
Use a side water depth of 11 ft
Detention time = [(1,875 ft²) (11 ft)] / [(3x10⁶ gal / day / 2) (1 ft ³ / 7.481 gal)] = 0.10 day = 2.5 hr

Design of the Activated Sludge Process - Example 5...

Given : Q = 100,000 gpd
Required : Volume of an extended aeration process
Solution : The minimum aeration period for an extended aeration system is 24 hr.
Volume = (100,000 gal / day) (24 hr) (1 / 24 hr / day) = 13,368 ft³