Critical Settling Velocity & settling Velocity (Overflow rate) · Example: Design the plain...

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



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)



Area: = W* L = W* 4W = 4W2

Thus Width = 10m;

length = 40m

Assume detention time = 3hrs

H = t*V0 = (3*3.6 × 10−4m

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



Example: Repeat the previous example using circular tank.

Note: the maximum tank diameter is 45 m

Solution:

𝑉𝑠 = 3.6 × 10−4

𝑚

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 × 10−4m

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



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



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 m3/m/d within the

range (150:300) .OK



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



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



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



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



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

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Critical Settling Velocity & settling Velocity (Overflow rate) · Example: Design the plain...

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