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Background...

Abbreviations...

Q (m3 / h) Flow - rate

Qo (m3 / h) Influen t flow - rate

Qe (m3 / h) Effluent flow - rate

Qr (m3 / h) Recycled sludge flow - rate

Qw (m3 / h) Wasted sludge flow - rate

BOD (mg / L) Bi ochemical oxygen demand

BODo (mg / L) Influent biochemical oxygen demand

Va (m3) Aeration tank volume

SS (mg / L) Suspended solids (SS)

SSe (mg / L) Effluent SS

SSr , w (mg / L) Recycled and wasted sludge SSA (m3 / h) Air flow - rate

MLS S (mg / L) Mixed l iquor suspended soli ds

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

Qw

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t = Va / Q

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

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

OL = ( Qo )( BODo ) / Va

Typical OL values : Conventional activated sludge process = 0.3 - 0.6 kg BOD / m 3 . day Contact - Stabilization = 1.0 - 1.2 kg BOD / m 3 . 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 = ( Qo )( BODo ) / ( Va ) ( 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 = Qr / Qo

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 tocontrol 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 = [ ( Va )( MLSS ) ] / [ ( Qe ) ( SSe ) + ( Qw ) ( SSw ) ]

Typical sludge age values : Conventional activated sludge process = 4 - 15 day Extended aeration = 20 - 30 day (an estimate of the length of time that averagesuspended 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 / Qo

or

ASR = A / [ ( Qo ) ( BODo ) ]

(8) BOD removal efficiency

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E ( % ) = [ ( BODo - BODe ) / BODo ] 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) BOD5of domestic wastewater entering primary clarifier = 200 mg / L, (b) BOD5 removal efficiency of activated sludge system (based on total BOD5 in theeffluent 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 BOD5concentration 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) Qw = 0.064 mgd, (d) S0 = 200 mg / L, (e) X =

2,800 mg / L, (f) Se = 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 tobe 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) Influentsuspended 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

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(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 ft 3 = (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)] / [(SSe) (Qe) + (SSw) (Qw)]

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 = QR / 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) Calculatedimensions 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 / dayAssume 35% of the B OD5is 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.ft 3)] = 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 / ft 2

Total surface area = (3x106 gal / day) / (800 gpd / ft2) = 3,750 ft2

Surface area per clarifier = 3,750 ft 2 / 2 = 1,875 ft2

[(n) (D2)] / 4 = 1,875 ft2

D = 48.9 ftUse a side water depth of 11 ft

Detention time = [(1,875 ft

2

) (11 ft)] / [(3x10

6

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 gpdRequired :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 ft 3