The design of some of the treatment units have also been illustrated by means of suitable solved example.
Example:
For a city having population of 2.75 lacs, design a suitable sewage treatment plant, with the following data:
(i) Sewage flow = 200 litres/capita/day (average).
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(ii) Hourly variation
Max. Flow/Average flow = 1.40
and Minimum flow/Average flow = 0.65
(iii) Suspended solids in the sewage = 560 ppm
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out of which 100 ppm is settle able grit.
(iv) B. O. D. of the average = 300 ppm.
(v) Climatic condition of the town-hot and dry. Average atmospheric temperature.
= 27°C
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(vi) Required quality of effluent after treatment
B.O.D. 30ppm
suspended solids 50 ppm
(vii) Characteristics of sewage-purely domestic.
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Solution:
Full sewage treatment is to be given for obtaining the desired quality of the effluent.
Following sewage treatment units are proposed to be provided:
(a) Screens
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(b) Grit chamber
(c) Primary settling tank
(d) Trickling filters
(e) Secondary settling tanks.
The sludge produced in these units will be digested, dried on open sand beds and used as fertilizer.
Average daily sewage flow = 200 × 2.75 × 106 litres/day
= 55 x 106 litres/day
= 55 x 106 litres/day
= 0.637 cu.m./sec.
The average sewage flow is quite large and therefore mechanical units can be provided in all the sewage treatment processes.
Design of Screens:
Mechanically screened bar screen will be provided, with 2.5 cm clear opening between the bar. Assuming the velocity of flow at the time of maximum hours flow as 80 cm/sec.
Net affective area of the screens =1.4 x 0.637 / 0.80
= 1.115sq.m.
Providing depth of flow as 1.20 m, the required effective width
= 1.115/1.2 = 0.929 m (1.9 m say)
Providing 5 cm x 1.0 cm. flat C.I. bars overall width required
= 1.41 m.
The screen will be fixed inclined at an angle of 30° to the vertical and with a drop of 10 cm will be given to the floor of the screen chamber on just stream side of it.
Design of Grit Chamber:
Two grit chambers will be provided after the screen. Assuming detention period of 1 minute at the average flow.
Capacity of the grit-chamber = 0.637 x 60 cu.m.
= 38.22 cu.m.
Assuming surface loading of 1250 m3/m2/day
Surface area = Average daily flow/ Surface loading
55 x 103/1250
= 44 m3/m2 at average flow
Assuming horizontal velocity of flow
= 0.25 m/sec.
Length of grit chamber = 0.25 m/sec. = 15 m.
Cross-section area of each grit chamber
38.22 /2x 15 =12.7 sq.m.
Provide Nos. grit chamber of size 15 x 1.2×0.1 m size each. The inlet and outlet channel weirs should be in addition to the above size.
Design of Primary Settling Tanks:
Assuming detention period of 2 hours at average flow
Total capacity = 2 x 60 x 60 x 0.637 cu.m.= 4586.4 cu.m.
Providing 2 Nos. Circular settling tanks of 3.3 m depth at the sides. Bottom floor will have slope of 1 in 15 towards centre. The volume of the conical hopper bottom will act as sludge storage.
It is given that grit chamber will remove 100 ppm of suspended solids. The primary settling tanks will remove 65% of the remaining which is equal to
= 65/100(500-100) = 292.5ppm
The primary settling tank will also remove 30% of B.O.D
= 300 x 30/100= 90 ppm
... The effluent coming out from the primary settling tank will contain suspended solid
= 450 – 292.5 =157.5 ppm.
= 300-90 = 210 ppm
Quantity of suspended solids removed
292.5 x 55000 x 103 /106
= 16087.5 kgm/day
Assuming that solid content is only 5%, the quantity of sludge
= 16087.5 x 100/5 = 3,21,750 kgm/day
Assuming the density of the sludge
= 1000 kgm/cu.m.
Volume of the sludge produced = 321750/1000 =321.75 kgm/day
Design of Trickling Filters:
B. O. D. load in the primary effluent = (210 ppm)
= 210 x 55000 x 103/106 = 1.1550 kgm/day
Providing high rate trickling filters with recirculation arrangements using 5 Nos. trickling filters of filter bed 40.0 m diameter and 1.2 m deep.
Total volume of filter media = 75.36 cu.m.
B.O.D. load on filter r = 11550 /6280
= 1.839 kgm/m2/day (safe)
Surface loading = 550x3/6280 = 8757.962 litres/m2/day (safe)
Providing Recirculation ratio as 1.5 Hydraulic recirculation factors
= I + 1.5= 2.5
... Average hydraulic load on filters
55000 x 2.5/1.62 = 84876.5 cu.m/day (safe)
Design of Secondary settling tanks:
Quantity of effluent to be handled (including recirculation effluent)
= 55000 x 2.5 = 137500 m3/day
Assuming detention period of 2 hours
Total capacity of tank =4/24 x 137500 = 114583
Providing 4 Nos. radial flow tanks with 3.0 m depth.
Capacity of each tank = 11458 over 4 = 2864.5 m3
The bottom floor will be made at a slope of 1 n 12 towers the centre and the capacity in the hopper will be kept reserved for collection of sludge.
Surface loading
To decrease the weir loading radian trough 8 Nos. 3.5 m. long each are provided to collect the effluent in addition to the circumferential weir.
The total weir length 4 x 4 r (35.0) + (4 x 3.5 x 2 x 8) = 439.82 + 224.00 = 663.82 m
... Weir loading 137500/663.82 = 207.132 m3/m/day (safe)
Primary effluent contains 210 ppm of B.O.D. and assuming that secondary effluent is to contain 50 ppm B. O. D.
:. Quantity of sludge removed in the secondary settling tank
Assuming a solid content of 5% in the sludge, total quantity of sludge.
Further assuming that density of sludge is 100 kgm/cm3
Design of Digester Units:
Quantity of sludge produced
= (Quantity of sludge produced in P.S. T. + Quantity of sludge produced in P.S. T.)
= 321.75 + 176.0 = 497.75 cu.m.
As after digestion the sludge volume, reduces to about one-fourth the original volume, there are average volume of the sludge after digestion will be equal
Now assuming that in the average atmospheric temperature the time required for the digestion of the sludge being 35 days the required volume of the digester
= 5/8 × 35 × 497.75 = 10888 m3
Assuming that digested sludge will be removed once a week for drying on the sand beds, the storage space required
7 × ¼ × 497.75 = 871m3
Providing 20% additional space for 5 cu.m. and grit accumulation the total volume of the digesters required.
= 1.2(10888) + 871= 13937 cu.m.
In other words, the volume in terms of m2/capita
Provide sludge digester tanks of 16.00 m diameter and 17/0 m high (overall), with conical hopper at the top and bottom.
... No. of digester tanks 13937/3416.3 = 4 Nos.
The digested sludge shall be dried on open sand beds, because the climate is both hot and dry as given. The area will be = population x 0.0094 musp3
= 275000 x 0.0094 m2
= 2585 m2
Providing 6 Nos. of sand beds. Area of each sand bed
2285/6 = 430.8 m2
... Provide sand beds of 14.5 x 30 each. Ans.