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Design a BNR process with anaerobic, anoxic, and aerobic reactor.. Anoxic will receive effluent from anaerobic and recycle from aerobic.. The sludge pump and piping for RAS shall be des

Activated Sludge Process Design

Information Checklist

1 Select the type of biological treatment process.

2 Conduct a material mass balance and determine expected range of flows (minimum, average, and peak) and

loadings (COD, TSS, nutrients, etc.).

3 Determine biological kinetic coefficients (lab studies).

4 Develop a preliminary site plan, piping layout, and

location of collection boxes, return sludge pumps, etc.

5 Obtain design criteria.

6 Obtain effluent quality criteria (BOD5, TSS, TN and TP).

7 Develop data on settling characteristics of the biological

solids.

8 Obtain list of equipment manufacturers and provide

equipment selection guide.

Biological Reactor

Design Criteria

1 Design a BNR process with anaerobic, anoxic, and aerobic

reactor.

2 Anaerobic will receive influent and returned sludge from

clarifier Anoxic will receive effluent from anaerobic and recycle from aerobic.

3 The effluent will have BOD5/TSS/TN/NO3--N/NH4+-N/TP of

10/10/10/8/1/1 mg/L or better, respectively

4 Provide four independent process trains in parallel.

5 Anaerobic and anoxic must be deep, square and >1.5 hr.

6 Provide equipment for measuring raw wastewater flow, return

activated sludge, waste sludge, and air supply.

7 Blowers shall be capable of delivering max air requirements

considering the largest single unit out of service.

8 Aeration equipment shall provide complete mixing of MLSS

and shall be capable of maintaining a min of 2.0 mg/L DO.

Design Criteria - continued

9 Diffusers and piping shall be capable of delivering 150% of

the average air requirements.

10 The sludge pump and piping for RAS shall be designed to

provide capacity up to 150% of average design flow.

11 Internal recycle between aeration and anoxic basins will have

a capacity 200% of the average design flow.

12 All sidestreams from the sludge-handling facilities (thickeners,

digesters, and dewatering units) shall be returned to the

aeration basin

13 The sludge wasting shall be achieved from the common

collection box containing the effluent MLSS from aeration

basins.

14 The basin hydraulics shall be checked at peak design flow plus

the design return sludge flow when only three basins are in

operation.

Design Criteria - continued

12 The biological kinetic coefficients and operational parameters

shall be determined from laboratory studies.

Sludge age, c = 12 days

Schematic Flow and Piping Arrangement

Design Strategy

1 Conduct a material mass balance

2 Calculate  of anaerobic zone

3 Calculate c, biomass increase, and  of anoxic zone

4 Calculate c, biomass increase, and  of aerobic

zone

5 Calculate the dimensions of anaerobic, anoxic, and

aerobic reactors

6 Design the required mixing equipment

7 Design aeration system

8 Select influent and effluent structure, develop

hydraulic profile

Design CalculationsFirst iteration

1 Calculate influent flow (stream 1)

Flow = 0.44 m 3 /sec × 86,400 sec/d = 38,016 m 3 /d

BOD 5 = 38,016 m 3 /d×250 g/m 3 ×kg/1,000 g = 9,504 kg/d

TSS = 38,016 m 3 /d×260 g/m 3 ×kg/1,000 g = 9,884 kg/d

Org-N= 646 kg/d , NH 4+ -N= 722 kg/d ,

NO3--N= 0 kg/d , TN= 1,369 kg/d , TP= 228 kg/d

2 Calculate primary sludge characteristics (stream 3)

BOD5 ( 34% removal ) = 9,504 kg/day × 0.34 = 3,231 kg/d

Design Calculations - continued

3 Calculate primary treated effluent (stream 4)

4 The characteristics of streams 2 and 5 are the same as those for

streams 1 and 4, respectively

5 Effluent standards (stream 6) were given as design criteria

6 Calculate WAS (stream 7)

Eff BOD5 = Eff sol BOD5 + BOD5 of eff biological solids

Design Calculations - continuedIncrease in TVSS due to overall BOD 5 removal = Y/(1+kd·c) (S0 - S1) Q

Design Calculations - continued

Compute Org-N, NH 4 + -N, NO 3 - -N, TN in WAS

8 Q recycle into anoxic zone

Total NO3- -N lost by denitrification = TN influent (stream 4) - TN effluent - TN

WAS

552 kg/d = (22,729 m³/d + Qrecycle m³/d )× 8 g NO3--N /m³ × 10 -3 kg/g

Q recycle = 46,271 m³/d

Q recycle /Q= 46,271  37,882 = 1.22 1.25

9 This design considers a stripping of P from WAS (stream 8) In order to

reduce the P buildup in the BNR system, PO43- ions are released under

anaerobic condition, thus alum precipitation of PO 4 -P as AlPO 4 and

denitrification of NO3--N in WAS are achieved A summary of how this process affect the whole process performance is shown below:

.

Design Calculati ons - continued

Org.-N kg/d

NH4+ -N kg/d

NO3- -N kg/d

TN kg/d

TP kg/d

Increase in flow rate is caused by addition of alum, increase in TSS is

caused by precipitation of AlPO4 and Al(OH)3, biodegradable fraction =

0.54, organic fraction = 0.66 This example only considers the first iteration

10 Calculate combined sludge (stream 3 + stream 8)

Q = 134 m3 /d + 648 m 3 /d = 782 m 3 /d

BOD 5 = 3,231 kg/d + 1,515 kg/day = 4,746 kg/d

TSS = 6,227 kg/d + 2,921 kg/d = 9,148 kg/d

11 PS and WAS are mixed and blended Dilution water is needed to maintain

hydraulic loading in the thickener (9.8 m³/m²-d) Final effluent is used as

dilution water It is assumed a solid loading of 46.9 kg/m²-d

Thickener area= 9,148 kg/d  46.9 kg/m²-d = 195 m²

Flow to the thickener= 9.8 m³/m²d × 195 m²= 1911 m³/d

Flow of dilution water= 1,911 – 782 = 1,129 m³/d

12 Characteristics of blended sludge (stream 9)

BOD 5 and TSS in dilution water = 379 kg/d×1,129 m³/d37,882 m³/d = 11 kg/ d

#5 #6 #5

10 Calculate combined sludge (stream 3 + stream 8)

Q = 134 m3 /d + 648 m 3 /d = 782 m 3 /d

BOD 5 = 3,231 kg/d + 1,515 kg/d = 4,746 kg/d

TSS = 6,227 kg/d + 2,921 kg/d = 9,148 kg/d

11 PS and WAS are mixed and blended Dilution water is needed to maintain

hydraulic loading in the thickener (9.8 m³/m²-d) Final effluent is used as

dilution water It is assumed a solid loading of 46.9 kg/m²∙d

Thickener area= 9,148 kg/d  46.9 kg/m²∙d = 195 m²

Flow to the thickener= 9.8 m³/m²∙d × 195 m²= 1911 m³/d

Flow of dilution water= 1,911 m3 /d – 782 m 3 /d = 1,129 m³/d

12 Characteristics of blended sludge (stream 9)

BOD 5 and TSS in dilution water = 10 mg/L × 1,129 m³/d = 11 kg/d

#5 #6 #5

Schematic Flow and Piping Arrangement

13 Calculate thickened sludge (stream 10)

TSS = 0.85(efficiency)× 9,159 kg/day = 7,785 kg/day

Q = 7,785 kg/d × 1,000 g/kg  0.06 g/g  1.03  1 g/mL 106

= 126 m3/d

Org.-N = (194(stream 3) + 235(stream 8)) × 0.85 = 366 kg/d

NH 4 + -N = (2.6(stream 8) + 0.6(stream 8) + 1.1(dilution water) kg/d) × 126 m³/d  1,911 m³/d = 0.3 kg/d

15 Characteristics of anaerobically digested sludge (stream 12 and

supernatant)

BOD 5 and TS in the supernatant = 3,000 and 4,000 mg/L, respectively; TS

in digested sludge = 5%; TVS reduced = 52%; BOD5 stabilization = 60%,

Org.-N into to NH 4+-N = 15%, Org.-N into to soluble Org.-N = 10%, gas

Design Calculations - continued

Others components in supernatant

BOD 5 = 3000g/m³ × 34 m³/d ×10 -3 kg/g = 102 kg/d

Org.-N = 366(stream 10)× (1- 0.1- 0.15)×(136 kg/d4,870 kg/d) + 366 kg/d × 0.1× (34 m³/d126 m³/d)= 18 kg/d

16.Characteristics of dewatered sludge

The sludge dewatering facility consists of filter presses It is assumed that

the sludge cake has 25% solids; dewatering facility captures 95% solids;

specific gravity of sludge cake is 1.06; organic polymers added to the sludge for conditioning are 0.5% of sludge solids; and 80% added chemicals are

incorporated into the sludge cake Total volume of belt wash water and

chemical dilution water = 35 L/kg TSS No BOD5 is added by chemical

conditioning.

Design Calculations - continued

Flow rate of belt wash and chem dilution water = 4,734 kg/d × 35 L/kg ×

10 -3 m³/L = 166 m³/d

Others constituents in belt wash and chem dilution water

BOD 5 = (379(plant effluent)kg/d  37,882 m³/d)× 166 m³/d = 1.7 kg/d

Others constituents in the sludge cake

Design Calculations - continued

Final results

The previous computational procedure must be repeated for several iteration to obtain a stable value of influent quality

(< 1% of fluctuation.)

The effluent quantity discharged from the plant will be slightly

less than the influent flow due to evaporation loss, loss of water in production of digester gases, and moisture contained in the sludge cake

Next slide shows the characteristic of streams in final iteration of material mass balance analysis

Stream Flow

m³/d

BOD5kg/d

TSS kg/d

Org.-N kg/d

NH4+ -N kg/d

NO3- -N kg/d

TN kg/d

TP kg/d

Stream Flow

m³/d

BOD5kg/d

TSS kg/d

Org.-N kg/d

NH4+ -N kg/d

NO3- -N kg/d

TN kg/d

TP kg/d

Design Calculations - continued

Dimensions of Anaerobic Zone

1 A minimum  (HRT) of 1.5 h for anaerobic zone is generally used for

typical municipal wastewater

Dimensions of Anoxic Zone

rate (max,DN ) × temp correction (F T ) × DO correction (F DO)

Design Calculations - continued

min. c,DN  1  (’max,DN – kd,DN)

Typical values:

 kd,DN = 0.04 d-1 endogenous decay coefficient

min. c,DN = 6.9 d

Design  c,DN = safety factor (1.5) × min.c,DN = 10.4 d

a conservative value of 12 d is used for the anoxic zone

Y obs,DN = YDN  (1 + kd,DN + design c,DN ) = 0.405 g VSS/g NO3--N

Typical values:

Design Calculations - continued

p x,DN= Yobs × (TNinfluent - Org.-Ncells - NO3--Neffluent – NH4+-Neffluent)

Org.-N cells = 0.122(12.2% N in cells )g Org.-N/g VSS × Yobs × (BOD5,influent

– BOD5,sol effluent) = 0.122 × 0.405 × (200 – 3.7)

Design Calculations - continued

Dimensions of Aerobic Zone (based on both BOD5 stabilization and

nitrification)

1 Calculate design  c,N for nitrification

Design Calculations - continued

2 Calculate minimum  c,N for nitrification

min. c,N  1  (’max,N – kd,N)

Typical values:

 kd,N= 0.05 d-1 endogenous decay coefficient

min. c,N = 3.8 d

3 Determine design  c,N for nitrification

Design  c,N= safety factor for process consideration (1.5) × safety

factor for kinetic consideration (2.0) × min.c,N

Design  c,N= 11.4 d

A conservative value of 12 d is used for the aerobic zone

Design Calculations - continued

4 Calculate the NH 4 + -N concentration in the effluent

CH3OH + 1.5O2 → CO2 + 2H2O: 1 mol MeOH = 1.5 mol BODL

NO3-+1.08CH3OH + H+→0.065C5H7O2N+0.47N2+0.76CO2+2.44H2OBODL consumption = 1.08 mol × 32 g/mol MeOH/14 g NO3--N × 1.5 mol BODL/mol MeOH = 3.7 g BODL/g NO3--N

BOD5,denitrif.= (0.68 g BOD5/g BODL) × (3.7 g BODL/g NO3--N )

× (TN - Org.-N - NO --N – NH +-N )

Design Calculations - continued

6 Calculate BOD 5 consumption for deoxygenatin of DO

BOD5,DO.= (0.68 g BOD5/g BODL) × (1.3 g BODL/DO) × (Rr+reclycle × DOmax,N - (1 + Rr+reclycle) × DOmax,DN)

Rr+reclycle= Qr/Q + Qrecycle/Q, we will see (slide 35) that Rr+reclycle= 0.6 + 1.7= 2.3, although the first approach was = 0.6 + 1.25 = 1.85 (see slides 12 & 13)

DOmax,N = 3.0 mg/L

DOmax,DN = 0.1 mg/L

7 Calculate sludge growth caused by BOD 5 removal, p x, BOD5

Typical values:

 kd,BOD5= 0.06 d-1

Design Calculations - continued

px,BOD 5= Yobs,BOD5×(BOD5influent - BOD5,P-release - BOD5,denitrif - BOD5,DO - BOD5,influent)

px,BOD 5= 0.349 × (200 – 0 – 44.3 – 5.8 – 3.7) = 51.0 mg VSS/L

8 Calculate sludge growth caused by nitrification, px, N

Yobs,N = YN  (1 + kd,N + design c,aerobic ) = 0.125 g VSS/g BOD5

Typical values:

 YN= 0.2 g VSS/g BOD5

 kd,N= 0.05 d-1

px,N = Yobs,N × (TNinfluent - Org.-Ncells - NH4+-Neffluent)

Org.-Ncells = 0.122(12.2% N in cells ) g Org.-N/g VSS × px,BOD5

px,N = 3.5 mg VSS/L

Design Calculations - continued

8 Calculate the ratio of heterotrophs and nitrifiers in the MLVSS

Increase in TVSS caused by growth of total cell mass in the mixed culture

9 Determine design  BOD5 and design  N

design  BOD5= (24 × design c,aerobic × px,BOD5)  (fhet. × X)

design  N= (24 × design c,aerobic × px,N)  (fauto. × X)

fhet.:fraction of heterotrophic microorg = 0.94

fauto.:fraction of autotrophic microorg = 0.06

X: MLVSS = 3,000 mg/L

Design Calculations - continued

Design Calculations - continued

differ just a little bit from the table shown in slide 23 (stream 7) The new

values are shown below:

Design Calculations - continued

Estimation of the Q r and Q recycle

Design Calculations - continued

Oxygen requirement

1 Compute theoretical oxygen requirement (ThOR)

ThOR = Carbonaceous BODL + Nitrogenous BODL

Carbonaceous BOD L= BODL,influent – BODL,sol.effluent – BODL,cells

(BOD5,denitrif./0.68) mg BODL/L + 1.3 g BODL/g DO × [(Rr+recyle

Operating DO in aerobic zone

Max DO in anoxic zone

Design Calculations - continued

NitrogenousBOD L = 4.57 g BODL/g N × (TNinfluent - Org.-Ncells

Design Calculations - continued

2 Compute standard oxygen requirement under field conditions

where

Total ThOR = theoretical oxygen required, kg/day;

C = min DO maintained in the aeration basin, mg/L;

 = salinity surface tension factor (0.9 for wastewater) (DO saturation

wastewater/DO saturation clean water);

 = oxygen transfer correction factor for wastewater (0.8~0.9) (oxygen

transfer wastewater/oxygen transfer clean water);

α 1.024

] C/C βFF

[(C

ThOR

Total kg/day

sw s

'

Design Calculations - continued

Habs = absolute atm Pressure at elevation sea level (at 500 m, Habs= 9.7

H

H2

1F

kg 18,707

0.75 1.024

2.0)/9.15]

1.17 0.9

[(8.5

kg/day

11,718 SOR

2

20 24

Design Calculations - continued

3 Compute the volume of air required

Air weight = 1.201 kg/m3; O2 = 23.2% in air by weight

Theoretical air required under field condition

18,707 kg/day  ( 0.232 × 1.201 kg/m3) = 839,234m3/day air

Design Calculations - continuedInfluent, baffle walls, and effluent structure of the anaerobic basin

1 Select the arrangement of influent structure: a rectangular (1 m wide and 3 m deep) channel constructed along the side of the first chamber The influent

and return sludge enter the channel The channel has 16 square (20 cm x 20 cm) submerged ports (4 on the bottom and 12 on the front wall) that

distribute the influent along the width of the basin The hydraulic of the

system is established at a peak design flow condition (peak design flow +

return sludge)

Ave design flow to BNR= 42,000 m³/d= 0.486 m³/s

Peak design flow to BNR= 1.321 m³/s

Return flow= 0.486 × 0.6= 0.292 m³/s

Recycle flow= 0.486 × 1.7= 0.826 m³/s

Ave design flow to each (4) train= (0.486 + 0.292)/4= 0.195 m³/s

Peak design flow to each (4) train= (1.321 + 0.292)/4= 0.403 m³/s

Ave discharge to each port= 0.195/16= 0.012 m³/s

Peak discharge to each port= 0.403/16= 0.025 m³/s

Design Calculations - continued

2 Compute the headloss

For influent channel

Q ΔZZ

Design Calculations - continued

2 Calculate the head over the effluent weir at the third chamber.

Cd = 0.6, L: length of weir = 5.5 m, Qave = 0.195 m³/s, Qpeak = 0.403

32

Q  C d L H

Design Calculations - continued

Influent, baffle walls, and effluent structure of the anoxic basin

1 The influent structure of the anoxic chamber consists of an influent

channel along the side of anaerobic chamber It receives weir overflow

from the anaerobic chamber and recycle flow The intermediate walls and effluent structure are similar to that of anaerobic chambers

Ave design flow to each (4) train = (0.486+ 0.292 + 1.7 × 0.486)/4

= 0.402 m³/sPeak design flow to each (4) train = (1.321+ 0.292 + 1.7 × 0.486)/4

= 0.610 m³/s

2 Compute the headloss

2

d A 2gC

QΔZZ