Numerical

1. Numerical Problems
WATER TREATMENT PROCESSES

2. Problem
 Determine the quantity of copperas and Lime required /year to treat
5 million liters of water on daily basis if 10 mg of copperas is
consumed with lime. The reaction is
FeSO4 . 7H2O + Ca (OH)2 Fe (OH)2 + CaSO4 + 7H2O

3. continued
Given Data:
Water to be treated = 5 million liters
Copperas required = 10 mg/L
Required Data:
Quantity of Copperas and Lime on daily basis

4. continued
Solution:
Thereactionis
FeSO4.7H2O+Ca(OH)2 Fe(OH)2+CaSO4 +7H2O
And CaO+H2O Ca(OH)2
TotalamountofCopperasneeded=5*106
*10=50000000mg=50*365=18250kg/Year

5. continued
From reaction one mg of Copperas = one mg of Ca (HCO3) = one mg of
CaO
Molecular weight of Copperas = 56 + 32 + 4*16 +7 (2*1 +16) = 278
Molecular weight of Ca O = 40 +16 = 56
Now 278 mg of copperas needed 56 mg of CaO
10 mg of copperas needed (56/278) * 10 = 2.01mg / Liter
Total amount of CaO = (2.01mg / Liter)* 5*10 6 * 365= 3668.25 Kg/ year

6. Problem
 A sedimentation tank is used for a surface overflow rate of 16.0
m/day. Design a tank to treat a flow of 625 m3/hr. having detention
time of 120 minutes. Provide a storage sludge capacity of 20 % of
effective volume. Under ideal condition what size of particles will
be removed 100% if specific gravity is 2.65 µ = 1* 10 -3 kg/sec-m
and density of water is 1000 kg/m3

7. continued
Given Data:
Vo = 16.0 m/ day
Q = 625 m3/hr
t = 120 minutes
Sludge volume = 20% of effective volume
µ = 1* 10-3 kg/m. sec
Density of water is 1000 kg/ m3
Required Data:
Design of Sedimentation Tank
Diameter of Particles to be 100% removed

8. continued
Area of Sedimentation Tank = Q /V0 = 625 m/hr * (24 hr/day) / (16.0
m/day) = 937.5 m2
Volume = Q * t = (625 m3/hr) * (120 minutes) *(hr/ 60minutes) = 1250
m3
Let W: L = 1: 5 then Area = 5w*w =937.5 or w = 13.69 m; L = 68.47 m
Depth = Volume /A = 1250/937.5 = 1.33 m
Dimensions are L = 68.47 m; W = 13.69 m; and Depth =1.33 m

9. continued
Sludge volume = 20% of effective volume = 1250 * 20/100 = 250 m3
Total Volume = Effective volume + sludge Volume =1250 + 250 = 1500 m3
To calculate Vs use Stokes's law Vs = gd2 (Ss-1) pw / 18 µ = 9.81 d2 (2.65-1)
1000 /(18*1*10-3) = 899250 d2
At 100% removal V0 - Vs => 899250 d2 = 16 m/day = 1.852 *10 -4 or d =
1.435*10 -5 m
= 0.01435 mm

10. Problem # 02
 A slow sand filtration unit has to treat a flow of 0.12 m3/ second.
Calculate the number of units if each unit having dimensions of 17 m
width and 25 m length. The filtration rate is 4.0 m3 / m2- day. During
cleaning operation if two units are not in working condition. What
will be the filtration rate? Is this rate of filtration within the range of
slow sand filter? Also calculate the quantity of sand to be removed
from 3 units by scraping 300 mm of sand during cleaning operation.

11. Solution
Given Data:
Q = 012 m3/sec
Dimensions = (17 * 25 ) m
Rate of filtration = 4.0m3 / m2 –day
Sand removal from 3 units 300 mm depth
Required Data:
No of units and volume of sand removal if one unit is not working then
check the filtration rate.

12. continued
Q = (0.12 m3/sec) * (3600 secs/hr) * (24hr/ day) = 10368.0 m3 / day
Total area = Q/ V0 = 10368 / 4 = 2592 m2
Area of one unit = 17 * 25 = 425 m2
Number of units = Total area / area of one unit = 2592 /425 = 6.10 = 6
units Answer
The number of units is six out of which two units are out of order. So
five units are in working conditions working units area = 425*4 =
1700.0 m2

13. continued
Filtration rate = Q/A = (10368 m3 /day) / (1700.0 m2) =6.099m3 / m2 –
day
The slow sand filter limit is 2.9 to 6.7 m3 / m2 –day.
Therefore it is in the range of slow sand filter
Sand removal = 300 mm *425 * 3 *(m/100 mm) = 382.4 m3

14. Problem # 05
Design a Rapid Sand Filtration unit on Peak flow rate for a population
of 100,000 to be served by ADD as 200 liters/ (capita-day) water
supply. Assume rate of filtration = 3 * 105 m3/ (ha-day). Amount of
wash water = 5% of filtered water /day. Filter dimensions of each unit
= 17.5 m *10 m. The filter needs back washing once in 24 hours.

15. continued
Given Data:
Population = 100000
ADD = 200 LPCD
V0 = 3 * 105 m3/ (ha-day)
Back wash water = 5%
Dimensions = (17.5 * 5) m
Required Data:
Design of system and back wash water

16. continued
Population = 100,000;
ADD = 200 LPCD
Peak flow = 2.70 * ADD
Q peak = P * q * 2.70 =100000 * (200 L /C-day)* 2.70 * (m3/1000 Liters)
=54000 m3/day
V0 = 3 * 105m3 / (ha-day) * (ha/104 m2) = 30 m3 /m2 –day
Total area =Q / V0 = (54000 m3/day )/ (30 m3 /m2 –day) = 1800 m2

17. continued
Area of one unit = 17.5 * 10 = 175.0
No of units = Total area / area of one unit) = 1800 / 175 = 10.28 = 10
units having dimensions of 18 * 10
Back washing = 5% of filtered water (5/100) *54000 = 2700 m3