Part iX Sanitary engineering Sedimentation
Page 1 اعداد د. محمد يونس و د. غسان الدليمي
Type of sedimentation:
1- Plain sedimentation: no chemical is used
2- Chemical sedimentation: chemical is used to enhance the efficiency
Critical Settling Velocity & settling Velocity (Overflow rate)
- Settling velocity Vs or as called overflow rate
Particles move horizontally with the fluid (all particles have the same horizontal
velocity)
Particles move vertically with terminal settling velocity(different for particles
with different size, shape and density)
All particle with Vs,o>Vcwill be completely settled.
All particle with Vs,o<Vcwill be removed in the ratio Vp/Vc
Part iX Sanitary engineering Sedimentation
Page 2 اعداد د. محمد يونس و د. غسان الدليمي
Example 1:
Find the terminal settling velocity of a spherical discrete particle (sand) with diameter
0,02 mm and specific gravity of 2.65settling through water at 200C?
If Vs = 3.6 × 10−4 𝑚/𝑠 Then using a safety factor of 1.4 (to account for inlet and
outlet looses) determine the required grit chamber area (rectangular tank) to remove
the sand particle if the flow rate is 0.1 m3/s
𝜌𝑤𝑎𝑡𝑒𝑟 = 998.2 kg/m3
µ𝑤𝑎𝑡𝑒𝑟 = 1.002 × 10−3 𝑁𝑠/𝑚2
Solution:
Vs = 9.81
18×1.002 ×10−3 × (2650 − 998.2) × (0.02 × 10−3)2 = 3.6 × 10−4 𝑚
𝑠
Re = 𝜌 𝐷 𝑣𝑠
µ = =
1000×2.2×10−3 ×0.02×10
1.002×10−3
−3
= 0.04 thus the flow is laminar
Since Q = Vs * A
Then the area = 0.1
3.6×10−4 = 277.8 m
2
Using the safety factor the required area = 1.4* 277.8= 389 m2
Thus tank dimension (taking width to length ratio = 1:4)
Part iX Sanitary engineering Sedimentation
Page 3 اعداد د. محمد يونس و د. غسان الدليمي
Area: = W* L = W* 4W = 4W2
Thus Width = 10m;
length = 40m
Assume detention time = 3hrs
H = t*V0 = (3*3.6 × 10−4 m
s ) × 3600= 3.8m ~ 4m............ water level in the tank
•Vh= Q/(L*W) = 0.1
40×10 = 2.5 × 10−4 m
s= 0.9 m/hr
•Take weir loading rate = 250 m3/m.d:
L weir = Q/Wload = 0.1× 24 × 60 × 60/250 = 34.5m , Use suspended troughs inside
the tank
Part iX Sanitary engineering Sedimentation
Page 4 اعداد د. محمد يونس و د. غسان الدليمي
Example: Repeat the previous example using circular tank.
Note: the maximum tank diameter is 45 m
Solution:
𝑉𝑠 = 3.6 × 10−4𝑚
𝑠= 31.1𝑚/𝑑𝑎𝑦
*Area = 389m2
since the tank is circular :
•As = πD2/4= 389 m
2 D = 22.25 m ≈22.5 m < 45 m
Assume detention time = 3hrs
H = t*V0 = (3*3.6 × 10−4 m
s ) × 3600= 3.8 m ~ 4m............ water level in the tank
Check horizontal velocity at the beginning and end of settling zones:
•Vh = Q/(π Din H) = 0.1/(π× 3.8 × 3.8) = 0.132 m/ min(End of inlet zone)
•Vh = Q/(π DoutH) = 0.1/(π× 26.05 × 3.8) = 0.02 m/min (beginning of outlet zone)
•Take weir loading rate = 250 m3/m.d:
L weir = Q/Wload = 0.1× 24 × 60 × 60/250 = 34.5m , Use suspended troughs inside
the tank.
Max Dtotal = 45m ...ok since the design Dtotal = 30.1m
Available Length = πD total = π(2H + D) =π *(2*3.8+22.25) = 93.8m> 45
AvailableWload =0.1/93.8= 92.1 m3/m.d< 250 m3/m day ......O.K
Available length = Circumference
of Circleمحيط الدائرة
Part iX Sanitary engineering Sedimentation
Page 5 اعداد د. محمد يونس و د. غسان الدليمي
Typical sedimentation tanks design dimensions :
A- RectangularTank
- depth: 3-5 m
- length: 15-90 m
- width: 3-24 m
B- Circular Tank
- depth: 3-5 m
- diameter: 4-60 m
C- Retention time 2-4 hr
D- Tank bottom slope 1-2%in rectangular tank; 2-4% in circular tank
E- The weir (الهدارات) length should be 1/7 of total tank length
F- Surface Load Rate (S.L.R) = Q/ (total surface area) = 𝑄
𝑛.𝐴 = over flow rate
= 20-40m3/m
2/d
Where n = number of tanks
G- Hydraulic load Rate (H.L.Rate) on outlet weir = weir loading rate = over load =
H.L =𝑄/(𝑛. 𝐿𝑤) = 150 − 300 m3/𝑚/𝑑
H- Horizontal velocity ≤ 0.3 𝑚/𝑚𝑖𝑛
Part iX Sanitary engineering Sedimentation
Page 6 اعداد د. محمد يونس و د. غسان الدليمي
Example:
a circular clarifier (sedimentation tank) will be used to treat 3000m3hourly, assume
the surface area of the tank is 2400 m2
estimate the tank dimensions?
Solution :
Water depth =
Assume tank diameter = 30m thus number of tank =
No of tank = n× 𝜋 × 302
4= 2400
Thus n = 3.4 take 3 tanks
Modify the required diameter :
3× 𝜋 × 𝑑
2
4= 2400 d = 32m
S.L.R (over flow) = 3000×24
3×𝜋 × 322
4
= 29.8 m3/m
2/d within the range (20:40)
Hydraulic Load Rate (H.L.R)= Q/Lw = 3000×24
3×𝜋 ×32 = 238 m
3/m/d within the
range (150:300) .OK
Part iX Sanitary engineering Sedimentation
Page 7 اعداد د. محمد يونس و د. غسان الدليمي
Sludge :
- It is the settleable solids separated from liquids during processing
When suspended solids (S.S) precipitate at the tank bottom its concentration is usually
2-5%.
This solid should be removed to keep the tank storage capacity using sludge cleaning
process.
This process is repeated 3-4 times daily, the solid density 1.05 : 2.1 ton/m3
Part iX Sanitary engineering Sedimentation
Page 8 اعداد د. محمد يونس و د. غسان الدليمي
Design of sludge Hopper
Sludge = water + [2-5% solids]
Assume the SS inlet the sedimentation tank = X mg/l (if not given)
Removal Ration (R.R) in the sedimentation tank and = 60-85%
Thus the amount of sludge that removed daily (Y) = 𝑄 (
𝑚3
𝑑)×𝑋 (𝑔/𝑚3
106 (𝑔/𝑡) × 𝑅. 𝑅 = 𝑡𝑜𝑛/𝑑
But sludge contain SS concentration (C) = 2-5% only t
hus the amount of Solid sludge Z (t/day) = Y/ C%
Assume the sludge specific gravity ( γ) = 1.05 – 2.1
Thus the sludge volume (m3/day) =
𝑍
γ
Volume of Sludge hopper :
For Rectangular tank:
Hopper volume =(𝑎+𝑏
2× h) × l
For Circular tank :
Volume = Aav× h
Aav = (A1 + A2)/2
=
𝜋
4 (𝐷1
2+𝐷22)
2
Usually D1 = 2:3m; D2 = 1:2m ; h = 1:2m
Part iX Sanitary engineering Sedimentation
Page 9 اعداد د. محمد يونس و د. غسان الدليمي
Design of Sludge pipe:
It is used to drain the sludge out of the sedimentation tank
Withdraw time 5:15min , velocity 1:1.5 m/s
Pipe diameter minimum = 150mm
Example:
Design the plain rectangular sedimentation tank for a water purification of an hourly
output 5000m3 , then get the amount if the sludge in the S.S if the raw wastewater has
SS concentration 80ppm and the sludge specific gravity is 1.1?
Solution:
Assume S.L.R = 30 m3/m
2/day
Assume retention time = 4hr (2:4hr)
Thus the minimum total volume = 4 * 5000 = 20,000 m3 = n.w.l.d
Thus total tank area = 5000 * 24 / 30 = 4000 m2
=No. of tank * length* width of tank
4000m2 = n.l.w
Then the water depth = 20,000/ 4,000 = 5m.
Assume number of tank = 8 then each tank area = 500 m2
then the tank width = =√500
4 = 11.2 make the width = 10m
then the length = 50m ,
then each tank volume = 10*50*5 = 2,500 m3
actual total volume = 2,500 * 8 = 20,000 m3
Check:
Actual Retention time = 20,000/5000 = 4hr .ok
Actual SLR = (5000/4000)*24 = 30 m3/m
2/d . ok
H.L = 5000 ×24
8 ×50 = 300 m3
/m/d .ok
Part iX Sanitary engineering Sedimentation
Page 10 اعداد د. محمد يونس و د. غسان الدليمي
Sludge Amount:
S.S = 80 mg/l ; assume R.R = 80%
Then the amount of sludge per day = Y 9ton/day)
= 5000 × 24 (m3/d)
×80 ×0.8
106 = 7.68 ton/day
Assume the S.S concentration in the sludge = 3%
Thus the total amount of sludge (water + SS) = 7.68/0.03 = 256 ton/day
Thus the total sludge volume = 256/1.1 = 232 m3/day
Sludge Stabilization:
Refer to power point lecture that submitted at LandMark Amman – 24/nov.
2016
Part iX Sanitary engineering Sedimentation
Page 11 اعداد د. محمد يونس و د. غسان الدليمي
Part iX Sanitary engineering Sedimentation
Page 12 اعداد د. محمد يونس و د. غسان الدليمي
Flocculent Sedimentation ( type 2 and 3):
- The design procedure for sedimentation tanks of type 2 and 3 are
the same as type 1.
- The difference is mainly in the overflow rate (vs)???
- The following table gives the design criteria of these two types.
sizes change shape change specific gravity change