L5- Stabilisation Ponds

Question 1

Stabilization pond

Answer 1

Artificial large shallow basin to treat WW by physiochemical and biological processes

Question 2

Advantages and disadvantages of stabilisation ponds and construction issues

Answer 2

Adv: Simple, low cost for equipment & O&M, high removal of pathogens, high algae production

Dis: algae inhibit discharge, large SA required, quality of effluent low, bad odour (anaerobic)

Issues: enbankment, isolation, winds and waves

Question 3

Overview of Types of ponds

Answer 3

Anaerobic: all water under anaerobic conditions

Aerobic: aerobic conditions maintained for whole volume- oxygen supplied via photosynthesis

Facultative: surface zone aerobic; lower anoxic zone anaerobic; oxygen supplied by photosynthesis

Aerated: Surface aerators supply oxygen and maintain water or part of it in continuous mixing

Question 4

Pond specifications

Answer 4

Anaerobic: depth 5-10m, production of gas, odour control issues, effluent can’t be discharged, 20-50d residence time

Aerobic: shallow basin (0.3-0.6m)- light penetrate to bottom, exploit photosynthetic power

Facultative: depth 1.5-2.5m, (facultative zone- mixture of aerobic and anaerobic zones)

Aerated: depth 2-6m, treat high organic loads, high efficiency, low growth of algae

Question 5

Facultative pond zones and bacteria types

Answer 5

Upper aerobic: aerobic bacteria use oxygen from algae and surface aeration
Lower anoxic: anaerobic and aerobic bacteria degrade organics using several oxygen sources
Sludge zone: sludge degradation by anaerobic bacteria

Question 6

Mechanical aeration system

Answer 6

Mixing zone: maintenance of solids in suspension (inner circle)

Oxygenation zone: diffusion of oxygen in liquid but not mixing (outer circle area)

Considerations: homogeneous distribution, aerators at inlet, no aeration at outlet

Question 7

Aerobic degradation of organics equation/products

Answer 7

organics+O2+m/o -> new m/o + CO2 + H2O + PO4^3- + NH4+

Question 8

Anaerobic degradation of organics equation/products

Answer 8

Organics + m/o -> CH4 + H2S + NH4 + CO2
NH4 -> NO2 -> NO3 -> N2 increase
H2S + m/p -> sulphur smells

Question 9

Operational factors for rate of biological reactions

Answer 9

Sunlight -> O2 and CO2
O2 -> rate
CO2 -> pH -> rate

Question 10

Removal of pollutants

Answer 10

BOD removal: dissolved (aerobic degradation by bacteria) or particulate (sedimentation, anaerobic degradation)

Suspended solids: mainly in algae, can’t be controlled

Nitrogen: nitrification and denitrification and algae consumption

Phosphorus: low consumption from algae; Al2(SO4)3 added

Question 11

Temperature effect on BOD removal coefficient & define variables in equation

Answer 11

During summer effect of temperature increased as kinetic rates of biochemical reactions increased

KT- BOD removal coefficient at T (1/d)
K20- BOD removal coefficient at 20deg C
T- water temperature
pheta- temperature coefficient

Question 12

Complete mix reactor type and formula

Answer 12

CSTR (complete mix = 100%)

S = S0/(1+Kt)^n for qual cells in series

Question 13

No mix reactor type and formula

Answer 13

Plug flow (PFR)

S = S0*exp(-Kt)

Question 14

Axial dispersed flow formula

Answer 14

S = S0[(4ae^(1/2d)]/[((1+a)^2)exp(1/2d) - ((1-a)^2)*exp(-a/2d)]

a = sqrt(1+(4Ktd)

Question 15

Surface area formula

Answer 15

A = V/D = Qt/D

Question 16

Organic load formula

Answer 16

Lorg = QS0/A = S0D/tow

Question 17

Volumetric load formula

Answer 17

Lv = BODmass/vol x time = Q*S0/V = S0/tow

Question 18

Dispersion factor equation & define variables

Answer 18

d = 1/(L/B)

L = length of pond
B = breadth of pond

Question 19

Dispersion factor for different reactors

Answer 19

d = (D/mu*L)
u = liquid velocity, L = distance between in/out, D= dispersion

d->0 for PFR (no mixing)
d->infinity for CSTR (complete mixing)

Facultative ponds -> d = 0.3-1

Question 20

Design steps for stabilisation ponds

Answer 20

Given values: S0, k, d, S, Q

Calculate tow
Calculate A = Q*tow/d

Question 21

Oxygen requirements, OR - define variables

Answer 21

OR- oxygen requirements (kg O2/d)
a- coefficient (0.8-1.2 kg O2/kg BOD)
Q- volumetric flow (m3/d)
S0- total influent BOD concentration (g/m3)
S- soluble effluent BOD concentration (g/m3)
1000- unit conversion

Question 22

Power requirements, P - define variables

Answer 22

P- power requirements (kW)
OEfield- field operating conditions (0.55-0.65 of OE)

where OE is oxygenation efficiency of aerator at 1.2-2 kg O2/kWh at 20deg C

Question 23

Removal of pathogens - define variables

Answer 23

N, N0 = concentration of pathogens at effluent and influent (number/100 mL)
KT - microorganisms removal coefficient (1/d)
tow - treatment time (d)

Question 24

Pathogens inactivation factors

Answer 24

Solar light
High temp and pH
Algae are toxic to m/o
Deficiency of nutrients
Other m/o that consume pathogens

Question 25

(For Lorg) What is 1 ha converted into m2?

Answer 25

1 ha = 10,000 m2

Question 26

Reason for series stabilisation pond systems

Answer 26

Optimal efficiency

Question 27

Reason for parallel stabilisation pond systems

Answer 27

Uniform distribution of flow and solids

Question 28

Facultative ponds in a row- typical set up

Answer 28

1: initial anaerobic for solids sedimentation
2-4: secondary facultative ponds for BOD removal
5- maturation/disinfection ponds for pathogen removal