WASTE STABILIZATION POND WASTE STABILIZATION POND (WSP)(WSP)

WASTE STABILIZATION POND WASTE STABILIZATION POND (WSP)(WSP)

Advantages:Advantages:

SimplicitySimplicity simple to constructsimple to construct simple to operate and maintainsimple to operate and maintain only unskilled labour is neededonly unskilled labour is needed

Low CostLow Cost cheaper than other wastewater treatment cheaper than other wastewater treatment

processesprocesses no need for expensive equipmentno need for expensive equipment

High EfficiencyHigh Efficiency BOD removals > 90%BOD removals > 90% Total nitrogen removals is 70-90%Total nitrogen removals is 70-90% Total phosphorus removal is 30-45%Total phosphorus removal is 30-45% Efficient in removing pathogensEfficient in removing pathogens

Types of WSPTypes of WSP

Anaerobic pondAnaerobic pond Facultative pondFacultative pond Maturation pondMaturation pond

Anaerobic PondAnaerobic Pond

2-5 m deep2-5 m deep Receive high organic loading (usually > Receive high organic loading (usually >

100 g BOD/m100 g BOD/m33 d) – contain no dissolved d) – contain no dissolved oxygen and no algaeoxygen and no algae

Primary function is BOD removalPrimary function is BOD removal Retention times are short (e.g. 1 day)Retention times are short (e.g. 1 day)

Facultative PondFacultative Pond

1-2 m deep1-2 m deep Two types:Two types: 1. Primary Facultative Pond 1. Primary Facultative Pond – – receive raw wastewater. receive raw wastewater. 2. Secondary Facultative Pond 2. Secondary Facultative Pond – – receive settled wastewater receive settled wastewater (e.g. effluent from anaerobic pond)(e.g. effluent from anaerobic pond) The primary function is the removal of The primary function is the removal of

BODBOD

3 zone exist:3 zone exist:

A surface zone where aerobic bacteria and A surface zone where aerobic bacteria and algae exist in a symbiotic relationship. The algae exist in a symbiotic relationship. The algae provide the bacteria with oxygen and the algae provide the bacteria with oxygen and the bacteria provide the algae with carbon dioxide.bacteria provide the algae with carbon dioxide.

An anaerobic bottom zone in which An anaerobic bottom zone in which accumulated solids are decomposed by accumulated solids are decomposed by anaerobic bacteria.anaerobic bacteria.

An intermediate zone that is partly aerobic and An intermediate zone that is partly aerobic and partly anaerobic in which the decomposition of partly anaerobic in which the decomposition of organic wastes is carried out by facultative organic wastes is carried out by facultative bacteria.bacteria.

Maturation PondMaturation Pond

1-1.5 m deep1-1.5 m deep Receive the effluent from a facultative Receive the effluent from a facultative

pondpond Primary function is the removal of Primary function is the removal of

pathogenspathogens

BOD RemovalBOD Removal

in in anaerobic pondsanaerobic ponds BOD removal is achieved by BOD removal is achieved by sedimentation of settleable solidssedimentation of settleable solids

in in secondary facultative pondssecondary facultative ponds that receive that receive settled water (anaerobic pond effluent), the settled water (anaerobic pond effluent), the remaining non-settleable BOD is oxidized by remaining non-settleable BOD is oxidized by heterotrophic bacteriaheterotrophic bacteria

in in primary facultative pondsprimary facultative ponds (receive raw (receive raw wastewater), the above functions of anaerobic wastewater), the above functions of anaerobic and secondary facultative ponds are combinedand secondary facultative ponds are combined

in in maturation pondsmaturation ponds only a small amount of only a small amount of BOD removal occursBOD removal occurs

Pathogen RemovalPathogen RemovalBacteriaBacteria Faecal bacteria are mainly removed in Faecal bacteria are mainly removed in

facultative and especially maturation pondsfacultative and especially maturation ponds The principal mechanism for faecal bacteria The principal mechanism for faecal bacteria

removal are:removal are:1- Time and temperature1- Time and temperature

- Faecal bacteria die-off in ponds increase - Faecal bacteria die-off in ponds increase with both time and temperaturewith both time and temperature

2- High pH2- High pH - Faecal bacteria (except Vibrio Cholerae) die- Faecal bacteria (except Vibrio Cholerae) die very quickly (within minutes) at pH>9very quickly (within minutes) at pH>9

3- High light intensity3- High light intensity - Light of wavelength 425 – 700 nm can- Light of wavelength 425 – 700 nm can damage faecal bacteriadamage faecal bacteria

Design of WSPDesign of WSP

1. Anaerobic Pond1. Anaerobic Pond

Volumetric BOD loading (g/mVolumetric BOD loading (g/m33d)d)

------- (1)------- (1)

WhereWhere

Li = influent BOD, mg/L (=g/mLi = influent BOD, mg/L (=g/m33))

Q = flow, mQ = flow, m33/d/d

= anaerobic pond volume, m= anaerobic pond volume, m33

aiV QL /

a

should lie between 100 and 400 g/mshould lie between 100 and 400 g/m33d d

to maintain anaerobic conditionsto maintain anaerobic conditions

to avoid odour release to avoid odour release

The mean hydraulic retention time (HRT), The mean hydraulic retention time (HRT), ttaa (day) is determined from: (day) is determined from:

--------- (2)--------- (2)

V

Qt aa /

Design values of permissible volumetric loading on and Design values of permissible volumetric loading on and percentage BOD removal in anaerobic ponds at various percentage BOD removal in anaerobic ponds at various

temperaturestemperatures

Temperature (oC)

Volumetric loading (g/m3d)

BOD removal(%)

<10 100 40

10 – 20 20T – 100 2T + 20

>20 300 60*

2. Facultative Ponds2. Facultative Ponds Surface BOD loading (Surface BOD loading (ss, kg.ha d), kg.ha d)

------ (3)------ (3)

where where

Af = facultative pond area, mAf = facultative pond area, m22

s = 10LiQ/Af

The permissible BOD loading, The permissible BOD loading, ss max max

ss max = 350 (1.107 – 0.002T) max = 350 (1.107 – 0.002T)T-25T-25 ----- (4) ----- (4)

Once a suitable value of Once a suitable value of ss has been has been selected, the pond area is calculated selected, the pond area is calculated from equation (3) and its retention time from equation (3) and its retention time (t(tff, day) from:, day) from:

ttff = A = AffD/QD/Q ---------- (5) ---------- (5)

WhereWhere DD = pond depth, m= pond depth, m QQ = wastewater flow, m= wastewater flow, m33/day/day

3. Maturation Ponds3. Maturation Ponds

(a) Faecal Coliform Removal(a) Faecal Coliform Removal

The resulting equation for a single pond is:The resulting equation for a single pond is:

Ne = Ni / (1 + kNe = Ni / (1 + kTTt)t) -------- (6) -------- (6)

WhereWhere

Ne =Ne = number of FC per 100 mL of number of FC per 100 mL of effluenteffluent

Ni =Ni = number of FC per 100 mL of number of FC per 100 mL of influentinfluent

kkT T = = first order rate constant for FC first order rate constant for FC

removal, per dayremoval, per day

t =t = retention time, dayretention time, day

for a series of anaerobic, facultative and for a series of anaerobic, facultative and maturation ponds, equation (6) becomes:maturation ponds, equation (6) becomes:

--- (7)--- (7)

Ne and Ni now refer to the numbers of FC Ne and Ni now refer to the numbers of FC per 100 mL of the final effluent and raw per 100 mL of the final effluent and raw wastewater wastewater

nmTfTaT

ie tKtKtK

NN

)1(11

The value of kThe value of kTT is highly temperature is highly temperature

dependent.dependent.

kkTT = 2.6 (1.19) = 2.6 (1.19)T-20T-20 ---------- (8) ---------- (8)

Check the BOD effluent concentrationCheck the BOD effluent concentration, le, le

--------- (9)--------- (9)

WhereWherele =le = BOD effluent concentration, mg/L BOD effluent concentration, mg/L lili = = BOD influent concentration, mg/L BOD influent concentration, mg/L

KK11 = = first order rate constant for BOD first order rate constant for BOD

removal, per dayremoval, per dayt =t = retention time, day retention time, day

11

tK

ll ie

The value of The value of KK11 is highly temperature is highly temperature

dependentdependent

------ (10)------ (10)

where where

KK11@ 20@ 20ooC = 0.3 per day and C = 0.3 per day and

= 1.05= 1.05

20)20(1)(1

TT KK

For n ponds in series, BOD effluent can be For n ponds in series, BOD effluent can be calculated as follows calculated as follows

------ (11)------ (11) nmf

ie

tKtK

ll

11 11

ExampleExample

Design a waste stabilization pond to Design a waste stabilization pond to treat 10,000 mtreat 10,000 m33/day of a wastewater /day of a wastewater which has a BOD of 350 mg/L and which has a BOD of 350 mg/L and 1x101x1088 FC per 100 mL. The effluent FC per 100 mL. The effluent should contain no more than 1000 should contain no more than 1000 FC per 100 mL and 20 mg/L BOD. FC per 100 mL and 20 mg/L BOD. The design temperature is 18The design temperature is 18ooC.C.

 

SolutionSolution

(a) Anaerobic Ponds(a) Anaerobic Ponds

From Table the design loading is given by:From Table the design loading is given by:

= 20T–100 = (20 x 18)-100 = 260 g/m= 20T–100 = (20 x 18)-100 = 260 g/m33dd

The pond volume is given by equation (1) as:The pond volume is given by equation (1) as:

= L= LiiQ/ Q/

= 350 x 10,000/260 = 13,462 m= 350 x 10,000/260 = 13,462 m33

v

a v

 

The retention time is given by equation (2) The retention time is given by equation (2) as:as:

= 13,462 /10,000 = 1.35 day= 13,462 /10,000 = 1.35 day

The BOD removal is given in Table as:The BOD removal is given in Table as:

R = 2T + 20 = (2 x 18) + 20 = 56 percentR = 2T + 20 = (2 x 18) + 20 = 56 percent

Qt aa /

 

(b) Facultative Ponds(b) Facultative Ponds

The design loading is given by equation (4) The design loading is given by equation (4) as:as:

ss max = 350 (1.107 – 0.002T) max = 350 (1.107 – 0.002T)T-25T-25

= 350 (1.107 – 0.002T)= 350 (1.107 – 0.002T)T-25T-25

= 350[1.107 – (0.002 x 18)]= 350[1.107 – (0.002 x 18)]18-2518-25

= 216 kg/ha d= 216 kg/ha d

 

Thus the area is given by equation (3) as:Thus the area is given by equation (3) as:

AAff = 10 x 0.44 x 350 x 10,000/216 = 10 x 0.44 x 350 x 10,000/216

= 71,300 m= 71,300 m22

s = 10LiQ / Af

 

The retention time is given by equation (5) as:The retention time is given by equation (5) as:

ttff = A = AffD/QD/Q

Taking a depth of 1.5 m, this becomes:Taking a depth of 1.5 m, this becomes:

ttff = 71,300 x 1.5/10,000 = 71,300 x 1.5/10,000

= 10.7 day= 10.7 day

 

(c ) Maturation Ponds(c ) Maturation Ponds

Faecal Coliform RemovalFaecal Coliform Removal

For 18For 18ooC the value of kC the value of kTT is given by equation is given by equation

(8) as:(8) as:

kkTT = 2.6 (1.19) = 2.6 (1.19)T-20T-20 = 2.6(1.19) = 2.6(1.19) -2-2 = 1.84 day = 1.84 day-1-1

 

The value of NThe value of Nee is given by equation (7): is given by equation (7):

Taking tTaking tmm = 7 days, this becomes: = 7 days, this becomes:

For n = 1, NFor n = 1, Nee = 99957 > 1000 FC/100 mL = 99957 > 1000 FC/100 mL

For n = 2, NFor n = 2, Nee = 7201 > 1000 FC/100 mL = 7201 > 1000 FC/100 mL

For n = 3, NFor n = 3, Nee = 518 < 1000 FC/100 mL , OK = 518 < 1000 FC/100 mL , OK

neN )784.11(7.1084.1135.184.11

108

nmTfTaT

ie tKtKtK

NN

)1(11

 

For a depth of 1.5 m, the area of the For a depth of 1.5 m, the area of the maturation pond is maturation pond is

AAmm = Q t = Q tmm /D /D

= 10,000 x 7/1.5= 10,000 x 7/1.5

= 46,667 m= 46,667 m22

 

BOD RemovalBOD Removal

Anaerobic Pond: Anaerobic Pond: llee = 0.44 x 350 mg/L = 154 mg/L = 0.44 x 350 mg/L = 154 mg/L

Facultative and Maturation Pond:Facultative and Maturation Pond:

dayxK /272.005.13.0 2018)18(1

= 1.6 mg/L

nmf

ie

tKtK

ll

11 11

37272.017.10272.01

154

el