2. To find settling velocity of a discrete spherical particle in water under conditions when
reynold no is less than 0.5. The diameter and specific gravity of the particle is 5×10-3 cm
and 2.65, respectively. Water temperature is 20oC (Kinematic viscocity of water at 20oC
= 1.01×10-2cm2/sec.
Stoke’s Law, valid for Re<0.5
Vs = g (G-1). d2
18 ν
d = 5×10-3 cm = 0.05mm which is < 0.1mm; G = 2.65
ν = 1.01×10-2cm2/sec
g = 981cm2/sec
Vs = 981 (2.65-1). (5×10-3)2 = 0.22cm/sec
18 1.01×10-2
Vs = 418 (G-1) d2 (3T + 70)
100
= 418 (2.65-1) (5×10-3)2 (3×2 + 70/100)
= 0.22 cm/sec
3. Types of Sedimentation Tank
• Horizontal Flow Tank: the ideal conditions of equal velocity at all points lying
on each vertical line in settling zone with a horizontal flow.
a) Rectangular tank with longitudinal flow:
continuous flow type- provided with mechanical scrapper to remove sludge
towards influent end without stopping the working of tank. Only flow
velocity is reduced while water is not brought to complete rest.
intermittent type- water is stored in tank for 24 hrs for settlement of
particles. Clear water is taken out and settles sludge is removed.
takes 30-36 hrs to put the tank in working order.
Require at least 2 tanks,
wastage of time
and labour.
4. b) Circular tank with radial flow: water enters at the centre of tank
provided with multiple ports, from which it emanates out to
flow radially outwards in all directions.
Uniform radial flow with decreasing horizontal velocity
towards the periphery
5. • Vertical or Upflow Settling Tank: tank usually combine
sedimentation with flocculation. Although may be used as plain
sedimentation
Surface area = As
Outflow
Outflow
Settled
particles
Raw water
Raw water
liquid flow
rate Q
v1
V1 = v0
vs
vs
v0 = Q/As
v0 = overflow
velocity/surface overflow
rate
vs = settling velocity
vs ≥ v0 for settlement
v0 is often kept 80% of vs
6. v0 = Discharge = Volume/Time
Surface area Surface area
= (Surface area × depth)/Time
Surface area
= Depth (m/sec) = liquid velocity
Time
Surface area for circular tank = π. d2
4
Surface area for rectangular tank = B.L
Sedimentation zone depth may vary from few cm to 6 meter.
7. • Detention time:
for a rectangular tank
= Volume of the tank = B.L.H
Rate of flow Q
for a circular tank
= d2(0.011d + 0.785H)
Q
Detention time ranges between 4 to 8 hours for plain sedimentation
and 2 to 4 hours when coagulants are used.
Width of tank is normally keep to 10 m and not allowed to exceed 12
m.
Length of tank generally not allowed to exceed four times the width
though may vary from one to six times.
Horizontal flow velocity ranging between 0.15 to 0.9m/minute
normally kept 0.3m/minute
8. • Flowing through period: average time which a batch of water
takes in passing through a settling tank
• Detention period: average theoretical time required for water
to flow through tank length average time for which the water is
detained in the tank
• Displacement efficiency: flowing through period
Detention period
• Inlet and outlet arrangements:
-------------------------------------------
--------------------------------------------
30 to 45 cm
Baffle wall
Floor of sedimentation tank
Feeding
channel
Inlet pipe
10. • Surface loading against average flow 30,000 to 50,000 l/m2/d
• Velocity of flow 0.4 m/min
• Design parameters for settling tank
• Parameter Dimensions
• Rectangular Tank
Depth 3 to 4m
Length 25 to 40m
Width 6 to 10m
Raking arm speed 1 to 1.3 m/min
• Circular tank
Depth 2.5 to 4m
Diameter 12 to 40m
Bottom slope 80mm/m
Raking arm speed 0.08 m/min
• Common Parameter
Detention Period 1 to 4 hr
Weir loading for average flow 125,000 to 300,000
l/m2/d
• Sludge Volume
Primary Tank 4 m3/million liter of sewage
Secondary Tank 12 to 16 m3/million liter of sewage
Moisture content of sludge about 95%
11. • Problem: Design a rectangular sedimentation tank to treat 2.4
million liters of raw water per day. The detention period may
be assumed to be 3 hours.
• Solution: Raw water flow per day is 2.4 x 106 l. Detention
period is 3h.
• Volume of tank = Flow x Detention period = 2.4 x 103 x 3/24 =
300 m3
• Assume depth of tank = 3.0 m.
• Surface area = 300/3 = 100 m2
• L/B = 3 (assumed). L = 3B.
• 3B2 = 100 m2 i.e. B = 5.8 m
• L = 3B = 5.8 X 3 = 17.4 m
• Hence surface loading (Overflow rate) =
2.4 x 106 = 24,000 l/d/m2 < 40,000 l/d/m2 (OK)
100
12. • Maximum daily demand of water at a water purification plant is
estimated as 12 million liters per day. Design the dimensions of
sedimentation tank , assuming detention period of 6 hours and velocity
of flow 20cm per minute:
quantity of water to be treated per day = 12 × 106 liters
Treated in 6 hrs detention period = 12 × 106 × 6 liters
24
= 3 × 106 liters of 3000 cu.m.
Velocity of flow = 20cm/min = 0.2m/min
The length of tank = velocity of flow × detention period
= 0.2m/min × (6 ×60) = 72 meter
Cross-sectional area of the tank required = capacity of tank/length of tank
= 3000 cu.m./72m = 41.7m2
Assuming water depth = 4m, then width will be = 41.7m2/ /4m = 10.42m
Using a free board of 0.5 m, overall depth will be 0.5 + 4.0 = 4.5m
13. • 2 million lt of water per day passed through sedimentation tank
of 6 m wide, 15 m long and depth of 3m.
Find the detention period, average flow velocity, overflow rate
capacity of tank = L.B.D = 6 × 15 × 3 = 270m3
Discharge (Q ) = 2 × 106 lit/d = 2 × 106/24 = 83.3 × 103 lit/hr
=83.3 cu.m./hr
Detention period = capacity of tank/Discharge
270m3/ 83.3 cu.m./hr = 3.24 hrs
Average flow velocity = discharge/cross sectional area (B.H)
= 83.3/6 ×3 m/hr = 83.3 × 100 cm/min
6 ×3 60
= 7.72 cm/min
Overflow rate = Q/B.L
= 83.3 × 103 lit/hr/ 6 × 15 m2 = 926 lit/hr/m2
15. • To determine quantity of alum required to treat 13 million liters
of water per day , where 12 ppm of alum dose is required. Also
determine the amount of CO2 which will be released per liter of
water treated.
quantity of water to be treated = 13 million liters /day
alum dose required = 12ppm or mg/l
amount of alum required per day = 13× 106 × 12 = 156kg
Chemical reaction:
Al2(SO4)3.18 H2O + Ca(HCO3 )2 2Al(OH)3+ 6CO2 +18H2O +
3CaSO4
molecular mass of Alum = 666.426
molecular mass of CO2 = 44
Means that 666 mg of alum, if used, will release 6 × 44mg
12 mg of alum will release = 6 × 44 × 12 = 4.76mg
666
16. • 8 mg of copperas is consumed with quick lime at a coagulation
basin per liter of water. Determine the quantity of copperas
and lime required to treat 10 million liters of water
FeSO4.7H2O + Ca(OH)2 Fe(OH)2 + CaSO4 + 7H2O
CaO + H2O Ca(OH)2
Quantity of copperas used for 1 lt of water = 8mg
Quantity of copperas used for 10 million lt of water = 10× 106 ×
8 = 80kg
Molecular mass of copperas = 278
Molecular mass of lime = 56
278mg of copperas used = 56 mg of quick lime
1 mg -------------------------- = 56/278mg
80 kg of copperas used = 56 × 80 = 16.12 kg of quick lime
278
