The design of water softening plants will be clear from the following solved example:
Design a water softening plant for a water works having the following data:
(i) Hardness in the water = 450 mg/litre as CaCO3,
(ii) Quantity of water to be treated
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= 25000 litres/hour
(iii) Allowable hardness after treatment in the work
= 50 mg/litre as CaCO3
(iv) Ion exchange capacity of the resin to be used in the plant
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= 10 kg of hardness/cu m
(v) Salt required for regeneration of the resin
= 45 kg/cu.m of the resin
The water softening plant works for 2 shifts of 8 hours per day. Assume any suitable data where necessary.
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Solution:
Quantity of water to be softened per shift of 8 hour
= 25000 × 8
= 2 × 105 litres
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Now as the hardness of the water is to be removed from 450 mg/litre to 50 mg/litre i.e., net hardness to be removed is 400 mg/litre.
= 400/450 × 100 = 88.9%
But as the water softening plant removes 100% hardness, therefore from the above requirement it is clear that the hardness of 88.9% water is removed and then it is to be mixed in the remaining 11.1% raw water, to get the treated water hardness of 50 mg/litre.
... Quantity of raw water whose 100% hardness is to be removed per shift
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= 88.9/100 × 2 × 103 litres = 1.778 × 105 litres = 177.8 cu.m
Quantity of hardness removed = (178.8 × 450 × 1000)/(1000 × 1000) = 80 kg.
... Quantity of cation exchange resin required
80/10 = 8 cu.m.
Provide 6 units of water softening plant, each of 1.6 cu.m. cation exchange resin. The surface area of each unit shall be 1.0 sq.m. and depth of the resin 1.6 m. Thus five units will treat the required quantity of water; one unit will work as stand by during repair of other units. Ans.
Regeneration:
After softening the water for 7 hours, the regeneration of the unit shall be done in the remaining 1 hour of the shift by passing 10% solution of the common salt (NaCl) through it. The water of the balancing tank shall be used for this regeneration. The stand by unit will work to compensate for it.
Quantity of salt required for regeneration
= 45 kg/cu.m of zeolite resin
= 45 × 8 kg = 360 kg
Using 10% brine solution, the volume of brine solution
= (360 × 100)/10 = 3.6 cu.m.
Two numbers of balancing tanks of 1.8 cu.m. capacity each can be provided for the regeneration work.
Keeping the depth of water as 1.0 w and square section, the side of the tank
B = √18 = 1.3416 m = 1.35 say
Providing free board of 15 cm, the overall dimensions of the each tank will be 1.35 × 1.35 × 1.15 m. Ans.
Check for contact period. Rate of flow of water over the soften bed.
(Volume of water treated per shift)/(7 hours working of softener) = 25400 litre/hour
Rate of filtration = (Rate of volume of water be passed)/(Surface area of units)
= 25400/(5 × 1.0) litres/m2/hour
= 25400/(5 × 1.0 × 60) litres/m2/hour
= 84.67 litres/m2/hour
= 84.67 × 10-3 cu.m/m2/min.
Now since the depth of each bed is 1.6 m, the average time of travel through the bed on the contact-period.
= 1.6/(84.67 x 10-3) minutes
= (18.89 minutes)/sufficient (O.K.)