Date post: | 18-Apr-2015 |
Category: |
Documents |
Upload: | ritesh-kumar |
View: | 40 times |
Download: | 5 times |
CEU- 401Environmental Engineering - II
Contents:
1. Activated Sludge Process - Design
2. Trickling filters – Details & Design
3. Oxidation Pond - Design
Modification in the ASP
1.Tapered aeration process
� Higher air-supply at the inlet
� 45%, 30% & 25% @ each 1/3 lengths
2. Step aeration process
� Sewage introduced along the length
� Require uniform O2 throughout
Q
Qr
3. Contact Stabilization process
� shorter aeration in the recycle line
� 0.5 to 1.5 hours
Modification in the ASP
Aeration tank
SST
Stab. TankQw
Qr
4. Complete Mix process
� Hourly variation high
� Biological instability problems occur
� Uniform supply & uniform with drawl
Modification in the ASP
Qr
Q
SST
Modification in the ASP
5. Extended Aeration Process
� PST is avoided
� Aeration ~ 12 to 24 hours
� BOD removal ~ 95 to 98%
� High MLSS concentration
� O2 required so high
� Sludge in endogenous respiration
Size & Volume of Aeration Tank
tc
w R
V X
Q Xθ = ( )w R y o E e tQ X Q Y Y k X Vα= − −
( )te t y o E
c
V Xk X V Q Y Yα
θ+ = −
1( ) ( )e t y o E
c
k X V Q Y Yαθ
+ = −
( )
(1 )y o E c
te c
Q Y YV X
k
α θθ
−=
+Gives, V for any assumed
values of θc & Xt
o
t
Q YF
M V X=
O2 Requirements of the Aeration Tank
O2 Reqd. for,
� Oxidation of influent organic matter
� Endogenous respiration of microbes
02
( )( / ) 1.42E
w R
Q Y YO gm d Q X
f
−= −
5 0.68u
BODf
BOD= ≈
O2 Reqd. for nitrification = 4.56 kg O2/kg NH3-N
O2 Requirements of the Aeration Tank
Aerators provided � to transfer O2
Under, T = 20°C ; P = 760 mm Hg
O2 transfer capacity (N) under field condition,
20( ) (1.024)
9.17
Ts S LN D D
Nα−−=
Ds – DO saturation value for sewage
DL – Operation DO level in aeration tank
α- correction factor for O2 transfer for sewage (0.8- 0.85)
Ns – O2 @ std. condition � 1.2 - 2.4 kg O2/kWh
Design of ASP
1.Compute daily sewage flow, YO and YE based
on desired BOD removal
2. Find the F/M ratio and MLSS conc. from
the std. table
3. From F/M, calculate V of the tank
4. Check the HRT (t = V/Q) & verify the value
Design of ASP
5. Check the SRT (θc) to be maintained
6. Check the volumetric loading = Q YO/V
7. Check return sludge ratio (QR/Q) – 0.25 to
0.50
8. Calculate tank dimensions,
d � 3-4.5 m; W � 5 to 10 m &
L � 30 to 100 m
Design of ASP
9. Calculate the rate of air supply
� 100 m3 of O2/kg of BOD removed
10. Calculate no. of diffusers reqd.
� each length, 1.2 m3 O2/min/m2
11. Design the SST by assuming the surface
loading rate; SST = 20 m3/d/m2
12. Finally design the sludge drying bed loading
Trickling Filters (TFs)
� Conventional TFs
� High-rate TFs (improved form)
o Also called as percolating or sprinkling filters
o Components:
� Coarser filtering media
� Spray nozzle or rotary distributor
�Under drain system
�Microbial layer (attached layer �slime layer)
Trickling Filters (TFs)
o Innermost layer in anaerobic
o Deficit for OM & O2
o Sloughing � creates turbidity
o Sloughing is a function of
�Organic loading – rate of metabolism
�Hydraulic loading – creates shearing velocity
o TFs constructed above GL
o Rotation of arms – 2 to 0.5 RPM
o Diameter – 30 to 60 m
Trickling Filters (TFs)
� DT – 1 to 3 min
� Filter media – 2 to 3 m
� Honey combed well
� Under drains
� Flow velocity – 0.9 m/sec
� High-rate TF � Recirculation of sewage through the filters
Trickling Filter – Merits & Demerits
o High quality effluent
o Loading high
o 75% BOD removal & 85% SS removal
o Good at warm weather & Self cleansing also
o Head-loss high
o High construction cost
o Cant treat raw sewage
o More operational troubles : (1) Odour, (2) Ponding
troubles and (3) Fly nuisance
Design of Trickling Filter
o Design of tank diameter and depth
o Design of rotary distributor
o Design of under-drainage system
Design based on,
� Hydraulic-loading rate
� 22 to 44 ML /ha/day
� 110 to 330 ML/ha/day
� Organic –loading rate
� 900-2200 kg BOD5/ha-m/day
� 6000-18000 kg BOD5/ha-m/day
Performance of Conventional TFs
o In conventional TF – highly nitrified effluent with
stabilized sludge
o BOD reduction � 80 to 90%
o Efficiency,
o u � organic loading rate in kg/ha-m/day
o Higher loading provides lesser efficiency in the
filters
100(%)
1 0.0044 uη =
+
Recirculation of Treated Sewage
o Essential in high-rate filters
o Recirculation � portion of treated or partially
treated sewage
o Single-stage or two-stage recirculation process
o Recirculation given (1) continuous dosing, (2)
equalizing, (3) longer contact and (4) influent
remains fresh all the time
o Higher influent flow � wash off the filter
o Loss of nitrates
Efficiency of High-rate TFs
o Depends on � (1) volume of recirculated flow
and (2) organic loading
o Recirculation ration, = R/I
R – Vol. of sewage recirculated ;
I – Vol. of raw sewage
o Recirculation factor,
o Efficiency (%),
2
1
[1 0 .1 ]
R
IFR
I
+=
+
100
1 0.0044Y
V F
η =+
Y – Total organic loading in kg/dayV – Filter volume in ha-m
1
1 1
100
0.00441
1Y
V F
η
η
=+
−
2nd stage Efficiency
Types of High-rate TFs
1. Biofilters
� Shallow filters with 1.2 to 1.5 m deep
� Recirculation of portion of filter eff. to the PST
2. Accelo-Filters
� 1.8 to 2.4 m deep
� Direct recirculation of unsettled filter eff. to the
distributor feed
3. Aero-Filters
� recirculation only during low sewage flow
condition (depth – 1.8 m)
� Recirculation of SST eff. to the distributor feed
Steps Involved in the Design of TFs
1.Calculate total BOD to be treated per day
2.Assume organic loading rate & find volume
3.Assume, depth and find the surface area
4.Calculate, the diameter of the TF
5.Check for hydraulic loading
6.Design the rotary distributors � design for
peak flow (2.25 times average flow)
Steps Involved in the Design of TFs
7. Assume velocity of flow @ peak flow - 2 m/sec
8. Then find the surface Area
9. Design the arms � rotary spray type
(a) Length of arm = (D/2) – width of central column
(b) Assume no. of arms
(c) Calculate area of arm by assuming, v = 1.2 m/sec
(d) find, area of arm A = (Q/V)
Steps Involved in the Design of TFs
10. Design the orifice opening in the arm (no. of
orifices in the arm)
(a) Discharge in each orifice,
(b) Total no of orifices =
11. Design of under-drainage system
Area of channel =
Steps Involved in the Design of TFs
12. Assume, width of under-drain �0.25 m
� Assume, rectangular channel, then
calculate depth
13. Find, slope of the channel (S)
14. Finally, design the laterals
Oxidation Ponds
o Open flow earthen basin
o Longer detention period (few days to several days)
o Stabilization ponds � By aerobic bacteria
o O2 demand met by � algae and microbes
o Algal-photosynthesis or algal-symbiosis
o Algae � produce O2 during photosynthesis
o End products � CO2, NH3 and phosphates
o Very small depth � below 0.5 m
Oxidation Ponds
o Practically pure aerobic ponds are difficult to
operate
o Facultative ponds are used � 1 to 1.5 m
o Oxidation pond � partially treated sewage is
introduced as influent
o Sewage Lagoon � received raw sewage
o Oxidation pond treatment
� reduced BOD/OM
� algae discharged along effluent
Oxidation Ponds
o Oxidation pond
effluent not
disposed in the
U/S end of the
dams/rivers
o Oxidation Pond
effluent used for
land irrigation
Oxidation Pond - Design Criteria
o Surface area worked out assuming suitable
organic loading rate
o Hot tropical countries � 300 to 150 kg/ha/day
o Colder countries � 90 to 60 kg/ha/day
o L = 2 W & depth ~ 1 to 1.5 m
o Detention time – 20 to 30 days
o Free board � 1 m
o BOD5 loading rate depends on latitude
Latitud(o N)
BOD5loading
8 325
12 300
16 275
20 250
36 150
Oxidation Pond - Design Criteria
o Detention time (days)
o
o L � BOD of the effluent entering the pond
o Y � BOD removed
o BOD removal - 90% & Coliform removal - 99%
o Sludge accumulated in ponds – 2 to 5 cm/year
o Sludge removal required once in 6 years
o Min liquid depth to be maintained – 0.3 m
Advantages of Oxidation Pond
o Suitable for hot dry countries (> 200 days)
o Suitable where large land available at low price
o Very cheap in installation
o Maintenance cost very less
o No skilled supervision required
o Flexible in operation
o Nuisance due to mosquito breading
o Odour problem – Far from city
Design Problems1. Design a conventional ASP to treat domestic sewage
with diffused air aeration system, given the following data:Population – 35,000; Ave. sewage flow – 180 lpcd; BOD of sewage – 220 mg/L; BOD removed in PST –30%; Overall BOD reduction desired – 85%
2. Design suitable dimensions of a circular trickling filter units for treating 5 ML of sewage per day. The BOD of sewage is 150 mg/L. Also design suitable dimensions for its rotary distribution system, as well as under-drainage system.