Example Solved Problems [Activated Sludge Process and Extended Aeration Systems]

 

PROBLEMS

1. An average operating data for conventional activated sludge treatment plant is as follows:

Determine:

(i) Aeration period (hours).

(ii) F/M kg BOD per d/kg MLSS.

(iii) Percentage efficiency of BOD removal.

(iv) Sludge age (days).

Solution:

(i) Aeration period (0) in hours:

(ii) F/M ratio:

(iii) Percentage efficiency of BOD removal:

(iv) Sludge age in days (θ_c):

2. 

 For the above data, determine (i) Aeration Period, (ii) F/M ratio, (iii) BOD removal efficiency, (iv) Sludge Age.,

Solution:

(i) Aeration Period

(ii) F/M ratio

(iii) Percentage efficiency of BOD removal

(iv) Sludge age

3. Design a conventional activated sludge plant to treat settled domestic sewage with diffused air aeration system for the following data.

i) Population - 1,20,000

ii) Per capita sewage contribution - 160 lpcd

iii) Settled sewage BOD_{5  -} 200 mg/l 

iv) Effluent BOD₅ required - 15 mg/l.

Solution:

Step 2: Volume of Aeration Tank.

Step 3: Check for HRT (Hydraulic Retention Time).

Step 4: Check for Volumetric Loading.

Step 5: Return Sludge Ratio.

Step 6: Tank Dimensions.

Assume depth = 3 m, Width = 4.5 m.

Length of aeration tank = Volume / Depth x Width 6400 / 3x4.5 = 474 m.

Provide continuous channel with six baffles, 7 sections of length = 68 m.

Total length, L= 7 x 68=476 m.

Thickness of each baffle = 0.25 m.

Total Width = (7 x 4.5) + (6 x 0.25) = 33 m.

Provide free board = 0.5 m.

Overall inner dimensions of tank = 68 m x 33 m x 3.5 m.

Step 7: Air requirement and arrangement of diffuser plates.

Air needed = 100 m3/day per kg BOD, removed.

100 (200-15)×19.2 /24×60

= 250 m³/min.

Choose standard diffuser plates of size 0.3 m × 0.3 m x 25 mm passing 1.2 m³ of air/

min/m2 with 0.3 mm pores.

No. of plates required = 250 / 0.3×0.3×1.2 = 2315 nos.

Provide 2320 plates.

Plate concentration of 30% extra provided in first half of tank to prevent frequent

No. of plates in first zone = 1.3 x 2320/2 = 1508.

Centre to centre distance of rows of plates = 1.2 m

No. of rows in first half length (234 m) = 234/1.2 = 195

Provide 8 plates in a row of 1.2 m spacing

Plates provided in first 234 m length = 195 rows x 8 in each row = 1560.

Balance plates to be provided = 2320 - 1560 = 760.

These plates are provided in balance length of (476-234) 242 m.

Spacing of plates: = 242x8 / 760 = 2.55 m (2.5 m).

 

Step 8: Check for minimum air availability:

In second half, 760 diffuser plates give air = 760 x 0.3 × 0.3 x 1.2

= 82.08 m³/min.

In a length of 242 m, air available per metre length of channel = 82.08 /242  = 0.34 m³/min/m

length > 0.25 m3/min/m length.

Hence satisfactory.

 

4. Design a conventional activated sludge plant to treat domestic sewage with diffused air aeration system, given the following data.

Population - 35,000.

Average sewage flow - 180 lpcd.

BOD of sewage - 220 mg/l.

BOD removed in primary treatment -30%.

Overall BOD reduction desired -  85%.

 

Solution:

(a) Design of Aeration Tank

(i) Daily Sewage Flow

Q=180 lpcd x 35000 = 6300 x 10³ l/d

= 6300 m³/d.

 

(ii) BOD of Sewage = 220 mg/l

BOD of sewage entering aeration tank.

Yo=70% of 220 mg/l = 154 mg/l.

(30% BOD removed in primary settling).

BOD of effluent, Y_E = 15/100 x 220=33 mg/l.

(Overall BOD removal - 85%).

BOD removal efficiency in ASP = 154 x 100 = 79%.

Assumptions

F/M = 0.35

MLSS (X) = 2000 mg/L / for ƞ between 85 - 95%.

 

(iii) Find Volume of Aeration Tank.

(within the limits of 4 to 8 hr, hence O.K.)

(iv) Check for Aeration Period (or) HRT (θ).

(within permissible range of 0.3 to 0.7 kg/m3, hence O.K.)

(v) Check for volumetric loading.

(vi) Check for SRT (θ_c).

(vii) Check for Return Sludge Ratio.

 (SVI ranging between 50-150 ml/gm say 100 ml/gm)

(Within the range 25 to 50%). Hence O.K.

However 33% return sludge provided at SVI = 125.

 

(viii) Tank Dimensions.

Adopt aeration tank of depth = 3 m and width = 4.5 m.

Length of aeration tank = Volume/ BxD =  1386 / 4.5x3 = 105 m.

Provide continuous channel, with 3 aeration chambers, each of 35 m length.

Total width = 4.5 x 3+2 x 0.25 = 14 m.

Total depth = 3 m + 0.5 m freeboard = 3.5 m.

Overall dimensions of aeration tank = 35 m x 14 m X

(ix) Rate of air supply required.

Assume air requirement = 100 m3 of air per kg of BOD removed

= (154-33)×6300×100 / 1000 mg/lxm³ xm³/kg

Blower capacity = 121 x6300×100 /1000 = 76230 m³/d

= 76230 /24×60 m³/min.

= 53 m³/min.

Assume standard diffuser plates of 0.3 m x 0.3 m x 25 mm size releasing 1.2 m³ of air/ min/m² with 0.3 mm pores are provided.

Total number of plates required = 53/ 1.2m³/min/m²×0.3×0.3

= 491 500 (say).

More plates in initial length of tank.

1^st one-third length - 45% of plates.

2^nd one-third length - 30% of plates.

3^rd one-third length - 25% of plates.

Plates in 1st chamber - 45% x 500 = 225

Plates in 2nd chamber - 30% × 500 = 150

Plates in 3rd chamber - 25% x 500 = 125

Assume spacing of plates as 1.2 m.

In channel of 35 m length,

No. of spacings = 35/1.2  = 29 Nos.

No. of plates = 29 - 1 = 28 plates.

225 plates in 35 m length = 225/28 =8 plates in each row.

❖ 28 rows @ 1.2 m C/C distance, 8 plates in each row, total 225 plates in 35 m length in first chamber.

❖ In second chamber - 150 plates, 8 plates in each row.

Spacing = 35mai (150/8) = 1.8 m C/C.

In 3^rd chamber, 125 plates.

Spacing = 35m/(125/8) = 2.25 m C/C.

 

(b) Design of Secondary Sedimentation Tank.

Assume SLR (surface loading rate) of 20 m³/d/m² at average flow of 6300 m3/d.

 (i) Surface area required = 

Adopting solids loading [range 100 to 150 kg/d/m²] of 125 kg/d/m² for MLSS of 2000 mg/l.

Solids loading rate = Total solids applied/ Surface area of tank kg SS/m²/d 

(ii) Surface area required = 6300 m3/dx2000 mg/l / 1000 

Higher surface area of 315 m² is adopted.

Assume Circular Sedimentation Tank.

(c) Design of Sludge Drying Beds.

Quantity of excess wasted sludge ?

For 10 kg/m³ suspended solids concentration in secondary sludge.

Excess secondary sludge volume

 (Taken to sludge drying beds)

Activated-Sludge Treatment Systems.

(i) Conventional Process.

(ii) Tapered Aeration Process.

(iii) Step Aeration Process.

(iv) Contact Stabilisation Process.

(v) Completely Mixed Process.

(vi) Modified Aeration Process.

(vii) Extended Aeration Process.

(i) Conventional ASP:

• Plug Flow - It is a system where waste water moves down along aeration tank, unmixed with rest of tank contents.

• O₂ (oxygen) demand at the head of aeration tank is high and reduces down the prim stir length.

• Air is provided uniformly along the basin length with porous diffusers. This leads to either oxygen deficiency in the initial zone (or) wasteful application of air in the subsequent zones.

• BOD removal is 85-95%.

• The plug flow is achieved by a long and narrow aeration tank with length equal to 5 to 50 times the width.

• Limitation – Aeration tank volume is large. It is difficult to meet the high oxygen demand at the head of aeration tank.

 

(ii) Tapered Aeration Process:

•  This system overcomes the limitation in the conventional ASP.

• The quantity of air supplied matches with the demand exerted by the micro organisms (BOD). vinotinu bobivonqal A

• Air is supplied at a high rate at the head of the aeration tank to meet the high oxygen demand and then gradually reducd at rest of the length to match the reduced oxygen demand.

•  It is achieved by varying the spacing of aerators.

• Main advantage is the optimal supply of air.

• One more advantage is the avoidance of over aeration.

(iii) Step Aeration Process:

• The settled sewage is introduced along the length of the aeration tank in several steps while the return sludge is introduced at the head of tank.

• The influent discharge Q is introduced in four steps Q₁, Q₂, Q₃ and Q4 while the return sludge Q, is introduced at the head of the tank.

• Because of step addition of sewage, the oxygen demand in the system is uniform over the length of the aeration tank.

• There is reduction in the aeration tank volume without lowering the BOD removal efficiency.

(iv) Contact Stabilisation Process (Bisorption):

• This process provides reaeration of the return activated sludge from the final clarifier (sedimentation tank).

• This process utilises the absorptive properties of activated sludge.

• This method uses smaller aeration tank.

• The incoming sewage is aerated in the contact aeration tank with return sludge for a short period of 30-90 minutes. This is known as absorptive phase during which the sludge absorbs the organic matter in sewage.

• The mixed liquor is then settled in secondary clarifier.

• The contact stabilisation process is quite effective in the removal of colloidal and suspended organic matter.

• This method is well suited for the treatment of fresh domestic sewage.

(v) Completely Mixed Process:

• The completely mixed process disperses the incoming waste and the return sludge uniformly throughout the basin, so as to obtain a completely mixed flow regime.

• Completely mixing is achieved by distributing the sewage and the return sludge uniformly along one side of the aeration tank and withdrawing the aerated sewage uniformly on the opposite side.

Advantages:

• There is less variation in the organic loading resulting in uniform O, demand and uniform effluent quality. (i)

• There is dilution of the waste water into entire basin volume resulting in reduced effect of shock loadings.

• There is maximum efficiency at all time due to complete mixing.

(vi) Modified Aeration Process:

• Similar to the conventional process (or) tapered aeration process.

• However, this process is distinguished by:

(i) Short retention period (1.5 to 3 h)

(ii) High volumetric loading 

(iii) High F/M ratio

(iv) Low percentage of sludge return

(v) Low concentration of MLSS

(vi) High organic loading.

(vii) Lesser air requirements

• Modified aeration develops dispersed biological growth which does not flocculate and quickly settle.

• BOD removal is 60 to 75%.

•  Used where intermediate quality effluent is desired as that used for sewage farming.

 

(vii) Extended Aeration Process:

•  It is a completely mixed process with long hydraulic retention time, high sludge age, high MLSS concentration and low F/M ratio.

• BOD removal efficiency is high.

• Due to long detention in the aeration tank, the solids gets well stabilised.

• Excess sludge does not require separate digestion and can be directly dried on sand beds.

•  Excess sludge production is also minimum.

• Air required is high in this process.

 

Operation Difficulties in an ASP

Advantages of ASP:

(i)  Clear sparkling eflluent.

(ii) No odours during process.

(iii) No fly nuisance or odour nuisance.

(iv) Highly efficient - 90% removal of suspended solids, BOD, bacteria.

(v) Degree of stabilisation is controlled.

(vi) Low cost of installation.

(vii) Small area is required.

(viii) Excess sludge has high fertilizer value.

(ix) Head loss is low.

(x) Greater flexibility of operation (effluent quality can be controlled).

Disadvantages of ASP:

(i) Sensitive to o variations in quality of sewage particularly industrial sewage (sewage

bulking)

(ii) High cost of operation, greater power consumption,

(iii) Lot of machinery to be handled,

(iv) Skilled attention required,

(v) Uncertain results,

(vi) Large sludge produced which is difficult to dewater, digest and dispose off

(vii) Variations in sewage flow may produce adverse effects and inferior quality effluent.

(viii) Quantity of sludge recirculated has to be adjusted every time - cumbersome  operation.

(ix)  Operational problems like bulking of sludge is common.