Following solved example will clearly illustrate the method of designing the water treatment plant:
Example:
A water works has to supply the water to a city having population of 5.3 lakhs the average per capita demand of the town is 145 litres/day.
If the maximum demand is 1.5 times the average demand, design the complete water treatment plants from the following data:
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(a) Maximum turbidity in water in monsoons
= 1500 ppm
(b) Turbidity in water in remaining period
= 150 ppm
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(c) Raw water has hardness of
= 100 mg/litre as CaCo3
(d) Limiting hardness in supply water
= 50 mg/litre
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(e) Ion exchange capacity of the resin
= 11 kg of hardness/cu. m
(f) Salt required for regeneration of the resin
= 40 kg/cu.m. of resin
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(g) Softening plant works for 2 shifts of 8 hours per day.
(h) There are no objectionable tastes and odours in the water.
Solution:
Proposed treatment units are:
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(i) Sedimentation with coagulants.
(ii) Rapid gravity filters.
(iii) Water softening plant,
(iv) Chlorinators.
Average quantity of water required by the city:
= 5.3 × 105 × 145 litres/day
= 76.85 × 106 litres/day
= 76.85 × 103 cu.m./day
= 76.85 × 103/24 × 60 × 60 = 0.89 cu. m. sec.
Maximum demand of the city = 1.5 × 76.85 × 103 cu.m/day
= 1.333 cu.m/day
Design of Sedimentation Aided with Coagulants:
(a) Chemical Feeding:
Using alum as coagulant. The optimum quantity of dose will be determined by actual jar test in the laboratory. Let the optimum dose be 5 to 8 mg/litre.
The average quantity of alum required
= (76.85 × 103 × 103 × 5)/106 to (76.85 × 103 × 103 × 8)/106
= 385 to 615 kg/day.
The maximum requirement in summer will be 1.5 time more
= 577 to 615 kg/day
This quantity of alum shall be first mixed with the water to form a solution of 5% strength and then will be added through solution feed device.
The maximum capacity of the solution feed device:
= 923 × 20 litre/day
= 923 × 10/60 × 24 litres/min.
= 12.82 litres/min.
Minimum dose which will be feeded during average demand
= 12.82/1.5 = litres/min. = 8.55 litres/min.
Quantity of solution to be feeded in one shift of 8 hours
= 12.82 × 60 × 8 litres = 6153.6 litres.
Providing two solution tanks, the capacity of each tank
= 3077 litres
= 3.077 cu.m. (Say 3.1 cu.m.)
Keeping depth of the solution in the tank as 1.0 m and 15 cm as free board, the side of the square tank
= √3.1
= 1.75 m (1.8 m say)
... Size of each tank solution = 1.8 × 1.8 × 1.15m.
Both the tanks shall be utilized in preparation of the solution alternately.
(b) Design of Approach Channel:
Two clarifiers have been proposed with two separate mixing flocculation tanks. The flow of raw water will be divided into two channels.
The maximum flow in each channel
= 1.333/2 cu.m./sec. = 0.667 cu.m./sec.
Providing a velocity of 60 cm/sec. the cross-sectional area of the approach channel
= 0.667/0.6 sq.m. = 1.11 sq.m.
Provide 1.2 × 1.0 m channel with 92.5 cm water depth, having free board of 7.5 cm at the maximum flow.
(c) Design of Mixing Tanks:
Mechanical flash mixers will be used for mixing the coagulant solution with the water. Assuming the detention period of one minute.
Capacity of each flash mixer =0.667 × 60 cu.m. = 40.02 cu.m.
Providing depth of 3.0 m, the side of the square plan mixing tank
= √40.02/3 m = 3.65 m
... Provide 3.70 × 3.70 × 1.0 w size of the mixing tank with each flocculator.
(d) Design of Flocculating Tanks:
2 Numbers flocculating tanks will be provided, one with each mixing tank. Slow moving paddles will be provided with variable speed electric motors. Assuming the flocculating time of 30 minutes, the capacity of each flocculating tank
= 0.667 × 60 × 30 cum = 1200.6 cu.m.
Provide 4 channels laid in parallel to one another, capacity of each channel
= 300.15 cu. m.
Providing 3.0 m depth of water, and 30 m length of the channel, its width
300.15/3 × 30 = 3.335 m
... Provide 4 channels of size 3.5 × 3.0 × 3.0 m with water depth of 2.86 and 14 cm as free board in continuation on each flash mixer.
(e) Design of Settling Tanks:
Providing 2 Nos. circular radial flow sedimentation tanks with a surface loading of 2000 litres/hr/m2 of plan area.
Surface area of each tank = 0.667 × 60 × 60 × 103/20000 sq.m. = 120.6 sq. m.
... Diameter of the tank = √1200.6/π/4
= 39.1 m = 39 m say
... The weir loading in the settling tank at average flow.
Providing a detention period of 2.5 hours at the time of maximum flow, the capacity of the tanks required
= 1.333/2 × 60 × 2.5 cu.m = 5998.5/4 × 392 = cu. m.
= 5.02 (say 5.0 m) which is reasonable
... Provide 2 Nos. circular settling tanks of 39 m diameter and 5.3 m depth with 30 m as free board. The bottom of the tank will be given a slope of 1 in 12 for pushing the sludge towards the centre with the help of scrapers from where it will be removed by the sludge outlet under hydrostatic pressure of the water filled in the tank.
The water will be collected from the settling tank over its circumferential channel along the weir provided all around the water from this channel shall be taken to the rapid gravity filters in covered channels of 2.0 × 1.0 m.
(f) Design of Rapid Gravity Filter:
Quantity of water to be treated
= 1.5 × 76.85 × 106 litre/day
Assuming the rate of filtration of rapid gravity filter as 4500 litres/m2/hour, and also assuming that 30 minutes shall be utilized daily in the back washing of the filter, the total filter area required
Providing 16 units of rapid gravity filter with 2 Nos. as stand-bye units.
Surface area of each unit = 1090/14 = 77.862 sq. m.
... Provide each unit of size =10 × 8m
Wash water tank. Overhead tank will be provided for the back washing of the filters, with its bottom at about 20 m above the bottom of the strainers. Also assuming the quantity of wash water as 2.5% of the total water filtered.
Quantity of wash water = 2.5/100 × 10 × 8 × 4500 × 23.5 litres/unit
= 2.1 × 105 litres/unit
= 211.5 cu.m/unit
The wash water tank should have storage capacity to store wash water for two units at least.
... Capacity of the wash water tank = 423 cu.m.
Assuming the depth of water as 3.0 m in the tank, and providing circular tank, the diameter of the tank
... Provide wash water tank of 13.5 m diameter and 3.3 m deep with 30 cm as free board.
Pumps for lifting wash water. As the pumps have to supply the water to the water tank, which has to supply it to 14 No. of units in 24 hours. Let the total quantity of water be lifted in 8 hours of one shaft.
Total capacity of the pumps required
... Provide 4 pumping units each of 125 cu.m./hour capacity, including one pumping set as stand-bye of 33% about.
H.P. of pumping sets (Motors). Let the diameter of the rising mains be 30 cm.
Now assuming the total lift of the water including friction losses be 25 m.
Assuming efficiency of the motors and pumps as 90% and 70%, the B.H.P. of the pumping sets
Wash water supply main. Let the size of the down water pipe bringing the water to the filters for back washing be 30 cm in diameter. The rate of flow of wash water while washing one filter nit (time taken in washing being 30 minutes).
= 1.663 m/sec < 3 m/sec.
The loss of head due to friction in this down take pipe
i.e., 1 m head loss in every 14.75 length of pipe. Total loss of head due to friction in 20 m length of pipe
= 20/14.75 = 1.356 m
Wash water droughts design. The main wash water trough will be laid in the middle of the filter near water surface. Side small trough at about 1.80 m c/c shall be connected to it from both the sides, which will collect the wash water and will bring it to the main trough.
As the size of the filter is 10 × 8 m, therefore each cross trough will collect the water from the surface area of 4 × 18= 12 sq.m.
Rate of wash water = 211.5/30 × 10 × 8 cu.m./m2/min.
= 0.0881 cu.m/m2/min
... Water collected by each side trough
= 0.881 × 7.2 = 0.634 cu.m/min.
Using 25 cm × 25 cm side troughs for collecting the wash water, its carrying capacity
Q = 327.5 by3
= 327.5 (0.25) (0.20)3
= 0.655 cu.m/min
Which is slightly more than 0.634 cu.m/sec required capacity. Hence provide 25 × 25 cm trough with 20 cm water depth will be sufficient to collect water from the sides.
The capacity of the main trough will be
= 211.5/30 = cu.m./mm. = 7.05 cu.m/min.
Providing 35 cm x 45 cm main trough with 40 cm water depth.
Its carrying capacity Q = 327.5 (35) (40)3
= 7.336 cu.m/min
which is more than 7.05 cu.m/min hence the section adopted is safe.
Sewer for carrying wash water
Rate of wash water = 211.5/30 cu.m/min. = 0.1175 cu.m/sec.
when one unit is washed at a time.
One R.C.C. pipe of 20 cm dia laid at a slope of 1 in 350 will be provided to carry the wash water.
Strainer design:
The minimum openings in the strainers should be (1/3)% of the bed area.
... Strainer openings = 10 × 8/3 × 100 = 0.267 sq.m.
Assuming that each strainer opening is 2.5 mm, number of opening in the strainer
Spacing of the strainers both way
Provide 13 cm c/c both ways strainer openings.
Design of water softening plant
Quantity of maximum soft water required per 8 hour shift
It is assumed that the softening plant will remove 100% hardness of water. But as 50 mg/litre hardness is required in the treated water, therefore, after removing 100% hardness, some quantity of untreated water will be mixed in it to attain 50 mg/l hardness. As the water contains 100 mg/litre hardness, therefore, the hardness of the 50% water will be removed and mixed in the remaining hard water.
... Quantity of water to be softened per shift
= 38425/2 cu. m = 19212.5 cu. m
... Hardness removed = 19212.5 × 1000 × 100/1000 × 1000 = 1921.25/11
Provide 30 beds of 2.2 × 2.2 m having 1.45 depth. Out of 30 Nos., 5 Nos. will work as stand-bye and will work during the repairing, refilling and regeneration of the 2 units (working).
After 7 hours of work the softening plant units will be regenerated by passing 5% brine solution in them.
The quantity of salt required = 50 kg/cu.m of resin
... Quantity required per shift = 40 × 25 × 2.2 × 2.2 × 1.45 kg= 7018 kg/shift
... Quality of brine solution = 7018 × 100/5 kg.= 140.36 cu.m.
For brine tanks of size 4.5 × 4.5 × 1.5 m shall be provided with 10 cm free board.
Check. Average flow rate over resin beds
= 19212.5/7 = 2744.62 cu. m/hour
Average velocity of flow through each unit
= 2744.62/25 × 60 × 2.2 × 2.2 = 0.378 m/mm.
The time taken to pass through the resin bed
= 1.45/0.378 = 3.84 minutes approx. (reasonable).
Design of Chlorination Plant:
The disinfection of the water will be done by post-chlorination using a vacuum type of chlorinator through which the liquid chlorine will be fed into the filtered water. The dose of chlorine to be added will vary from 0.5 to 1.0 ppm depending on the quality of water and the chlorine demand tests.
... Quantity of the chlorine required
... Liquid chlorinator having capacity of feed chlorine at rate of 6.4 to 10.0 kg/hour will be installed.
Design of Clear Water Reservoir:
Underground clear water reservoir having capacity of about 8 hours will be provided.
Quantity of water to be stored = 76 85 × 103/3 cu.m. = 25617 cu.m.
Providing a depth of 5.0 m, the area of the tank required
= 76.85 × 103/3cu.m. = 25617 cu.m.
The plan will mainly depend on the area available at the site.
Therefore, reservoir of plan are 5125 sq.m. and depth 5.0 m shall be provided with addition to free board of 50 cm.
