QUALITY OF RAW & TREATED DOMESTIC SEWAGE The general characteristic of sewage is assumed as shown in the table below. Anticipated general characteristics before treatment: Oil and Grease: upto 50mg/lt PH
:
6 – 9.0
BOD5
:
250-350mg/lt
COD
:
600-800mg/lt
SS
:
400 mg/lt
E coli
:
10MPN
The anticipated final water quality: PH
:
6.5 – 9.0
BOD5
:
< 10 mg/lt
COD
:
< 50 mg/lt
TSS
:
< 20 mg /lt
NH4-N
:
N-Total
:
< 5 mg/lt < 10 mg/lt
Fecal Coliform (MPN/100ML): < 100 3.2 TREATMENT METHODOLOGY ADOPTED The methodology adopted for treatment of sewage is based on Sequencing Batch Reactor (SBR) followed by disinfection.
3.4 SBR TECHNOLOGY The Sequencing Batch Reactor (SBR) process is a sequential suspended growth (activated sludge) process in which all major steps occur in the same tank in sequential order. There are two major classifications of SBRs: the intermittent flow (IF) or “true batch reactor”, which employees all the steps and the continuous flow (CF) system, which does not follow these steps. Both have been used successfully worldwide installations. SBRs can be designed and operated to enhance removal of nitrogen, phosphorus, and ammonia, in addition to removing TSS and BOD5 .The intermittent flow SBR accepts influent only at specified intervals and, in general, follows the five-step sequence. There are usually two IF units in parallel. Because this system is closed to influent flow during the treatment cycle, two units may be operated in parallel, with one unit open for intake while the other runs through the remainder of the cycles. In the continuous inflow SBR, influent flows continuously during all phases of the treatment cycle. To reduce short circuiting, a partition is normally added to the tank to separate the turbulent aeration zone from the quiescent area.
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3.5 DESIGN DETAILS OF SEWAGE TREATMENT PLANT: Total quantity of raw effluent
= 180 m3/day = 11.86 m3/hr
Standard conditions: BOD5 @ 270 C
= 250 mg / L
Expected BOD5 reduction: 1. Avg. Influent BOD5
= 200 mg/L
2. BOD5 reduction in SBR
= 95 %
Therefore BOD5 at the outlet of secondary treatment
=
15 mg/L
3. BOD5 reduction with tertiary treatment like Coagulation, filtration and disinfection
ii) SBR design calculations a) Reactor volume Assumptions for SBR kinetics * F/M ratio
= 0.15
* MLSS
= 4000 mg/l
* Net sludge yield
= 0.76 kg MLSS/kg BOD5
* Min. solids retention time = 8 days * Reactor volume decanted each day = 60% * Dissolved O2 liquor concentration
= 2 mg/L
* Oxygen co- efficients kg of O2/kg of BOD5
= 1.28
kg of O2/kg of NH3N
= 4.6
* Oxygen transfer factors ά (typical for coarse bubble diffusers) ß (typical for domestic wastewater)
= 0.85 = 0.95
* Typical O2 transfer rate for coarse bubble diffusers = 1.25 O2/KWH 4
kg
* No. of cycles/day
=4
(a) Reactor Volume * BOD5 removed (kg/day) = [(BOD influent - BOD effluent) (mg/L)] x flow (L/day)] BOD5 = 34.2 kg/day * Required aerobic mass
= BOD5 removed F/M Ratio = 228 kg MLSS
* Reactor volume (low water volume) = MLSS mass (kg) / MLSS Concentration = 57.0 m3 * Since the decant volume represents 60% of the total volume * Total reactor volume
= 13.2 m3
Provided 75 m3 x 2nos SBR tanks * Size of the unit
= 16.0sq.mX4.8m Liquid depth
b) Decant volume * Total decant volume
= Total reactor volume - reactor Volume (Low water level)
Total decant volume
= 19.8 m3
Required reactor volume = 82.0 m3 Provided 65.0 m3 * Size of the unit
= 14.0sq.mtX4.5m Liquid depth
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Aeration * Nitrogenous O2 demand (kg of O2/day) = NH3 - N oxidized (kg/day) x kg O2/kg of NH3N * Carbonaceous O2 demand (kg of O2/day) = Synthesis N x kg O2/BOD5 * NH3 - N oxidized (kg/day) = TKN removed (kg/day) - synthesis N (kg/day) * TKN removed = 6.65 kg/day * Synthesis N
=
5% waste activated sludge of total daily sludge
production * Sludge production (kg/day) = net sludge yield (kg MLSS/kg BOD5) x BOD5 removed (kg/day) Therefore sludge production
SAOR (kg O2/hr) = [AOR x Cs x teta (T-20)] / [ae x (beta x Csw - C0) x blower usage (hr/day) where SAOR = standard actual O2 requirement (kg O2/day) teta = temperature correction factor = 1.024
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Cs = O2 saturation concentration at standard temperature and pressure = 9.02 mg/L Csw = concentration correction for elevation 1000 ft. =
9.02 - 0.0003 x elevation = 8.72 mg/L
(NOTE: 0.0003 may be used as a rule of thumb describing a 0.0003 mg/L rise/drop in DO saturation concentration per every foot of elevation increase /decrease.) C0 = 2 mg/L ae
=
0.85
Beta
=
0.95
T
=
30 0C
(67 0F)
Blower usage= 16 hrs/day (Based on 4 cycles per day (6 hr/cycle), 1.0 hr fill time, 3.5 hr react time, 0.75 hr settle time, 0.5 hr decant time, and 0.25 hr idle time) SAOR
=
5.94 kg of O2/hr
* Sludge flow rate (L/day) = sludge mass flow (kg/day) / sludge density (kg/L) * Typical sludge density
