Page 1 of 2
2. Design Criteria
Description
1. Rapid Mixing
- Time
- G - Value
- GT
Unit
Design Criteria
Design Flow
Max. Flow
(5000 m3/day)
(7500 m3/day)
sec
1 to 3
2.17
1.45
sec-1
500 to 700
612
1328
1100
1595
2. Flocculation Basin
- Type
- No. of Stage
No.
- Energy Input
sec-1
20 - 60
- Detention Time
minute
3. Sedimentation Basin
- Type
- Detention Time
hour
- Surface Loading
m/min
- Water Depth
m.
- Mean Velocity
m/min
- Method of Sludge Drain
- Inlet Diffuser Wall
Port Velocity
m/sec
Port Spacing
m.
Port Diameter
mm.
WATER TREATMENT PLANT DESIGN2
Page 2 of 2
Description
- Effluent Weir Loading
4. Filtration Basin
- Number of Filter Basin
- Filtration rate
- Filter Flow
Control System
- Under Drain System
- Filter Media
Type of Media
Effective Size
Uniformity Coefficient
Depth
- Back Wash Rate
- Surface Wash System
Type
Rate
Surface Jet Pressure
1.5 Type of Water Treatment Plant
- Hydraulic Design System
WATER TREATMENT PLANT DESIGN2
General DATA
Page 1 of 4
3. Coagulation Basin Design
3.1 Raw Water Pipe
∴
Design Flow (Q Design)
Velocity in Pipe
Use
3
=
5,000 m /day
= 1.8 - 2.0
m/s (Kawamura)
=
1.8
m/s
Pipe Diameter (D)
=
=
Use Pipe Diameter (D)
∴
Acture Velocity
3.2 Rapid Mixing
Type
Design Criteria
- Detention Time
- G - Value
- GT
4Q
πv
4 x5,000
3.14 x1.8 x3600 x 24
= 0.2023751
=
0.2
m.
m.
Q
=
A
= 1.8420711
m/s
= Hydraulic Type
= Static Mixer
=
1-3
= 500 - 700
= 350 - 1500
sec
sec-1
3.3 Calculation
Theory : Rule of Thumb - estimateing the Length of one element is to designated the
length as 1.5 - 2.5 times the pipe diameter the base on this criteria,the length of one element is in
the range of 3 - 5 ft (Kawamura, page no. 88)
- 2 stage = 2 element
Static Mixer Length = 1.5 xDiameter( m) x 2 (element ) − 0.5 xDiameter( m)
WATER TREATMENT PLANT DESIGN2
Coagulation(PWA)
Page 2 of 4
- 3 stage = 3 element
Static Mixer Length = 1.5 xDiameter( m.) x3 (element ) −
Diameter ( m.)
- 4 stage = 4 element
Static Mixer Length = 1.5 xDiameter( m.) x 4 (element ) − 1.5 xDiameter(m.)
- 5 stage = 5 element
Static Mixer Length = 1.5 xDiameter(m.) x5 (element ) − 2.0 xDiameter( m.)
- 6 stage = 6 element
Static Mixer Length = 1.5 xDiameter(m.) x 6 (element ) − 2.5 xDiameter( m.)
- 7 stage = 7 element
Static Mixer Length = 1.5 xDiameter(m.) x 7 (element ) − 3.0 xDiameter( m.)
- 8 stage = 8 element
Static Mixer Length = 1.5 xDiameter( m.) x8 (element ) − 3.5 xDiameter( m.)
- 9 stage = 9 element
Static Mixer Length = 1.5 xDimerter( m.) x9 (element ) − 4.0 xDiameter(m.)
=
200
mm.
=
3
element
=
0.7
m.
Q
=
A
= 1.842
m/s
= 0.380007 sec.
=
250
mm.
=
3
element
= 0.875
m.
Q
=
A
= 1.179
m/s
Detention Time (t)
= 0.7422013 sec.
Bring Acture velocity to Find Head loss from PWA Graph
Head Loss Across Static Mixer =
1
m.
ρL
= 0.819
m/s
Detention Time (t)
= 0.9160884 sec.
Bring Acture velocity to Find Head loss from PWA Graph
Head Loss Across Static Mixer = 0.25
m.
ρL
=
μ =
Theory
G =
Gxt
∴
∴
∴
∴
∴
3
o
997.1 kg/m at 25 C
0.000895 kg/m.s (N/m.s) at 25 oC
h f xgxρ
μxt
-1
= 1727.0052 sec
too hight
= 1582.0895
too hight
=
400
mm.
=
2
element
=
1
m.
Q
=
A
= 0.461
m/s
Detention Time (t)
= 2.1714688 sec.
Bring Acture velocity to Find Head loss from PWA Graph
Head Loss Across Static Mixer = 0.08
m.
Try Static Mixer Diameter
Use 2 stage
Static Mixer Length (L)
Acture Velocity
ρL
=
μ =
Theory
G =
Gxt
3
o
997.1 kg/m at 25 C
0.000895 kg/m.s (N/m.s) at 25 oC
h f xgxρ
μxt
-1
= 634.54224 sec
OK
= 1377.8887
OK
WATER TREATMENT PLANT DESIGN2
Coagulation(PWA)
Page 1 of 8
3. Coagulation Basin Design
3.1 Raw Water Pipe
∴
Design Flow (Q Design)
Velocity in Pipe
Use
3
=
5,000 m /day
= 1.8 - 2.0
m/s (Kawamura)
=
1.8
m/s
Pipe Diameter (D)
=
=
Use Pipe Diameter (D)
∴
Acture Velocity
3.2 Rapid Mixing
Type
Design Criteria
- Detention Time
- G - Value
- GT
4Q
πv
4 x5,000
3.14 x1.8 x3600 x 24
= 0.2023751
=
0.2
m.
m.
Q
=
A
= 1.8420711
m/s
= Hydraulic Type
= Static Mixer
=
1-3
= 500 - 700
= 350 - 1500
sec
sec-1
3.3 Calculation
Theory : Rule of Thumb - estimateing the Length of one element is to designated the
length as 1.5 - 2.5 times the pipe diameter the base on this criteria,the length of one element is in
the range of 3 - 5 ft (Kawamura, page no. 88)
- 2 stage = 2 element
Static Mixer Length = 1.5 xDiameter( m) x 2 (element ) − 0.5 xDiameter( m)
WATER TREATMENT PLANT DESIGN2
Coagulation(Theory)
Page 2 of 8
- 3 stage = 3 element
Static Mixer Length = 1.5 xDiameter(m.) x3 (element ) −
Diameter ( m.)
- 4 stage = 4 element
Static Mixer Length = 1.5 xDiameter( m.) x 4 (element ) − 1.5 xDiameter(m.)
- 5 stage = 5 element
Static Mixer Length = 1.5 xDiameter( m.) x5 (element ) − 2.0 xDiameter( m.)
- 6 stage = 6 element
Static Mixer Length = 1.5 xDiameter(m.) x 6 (element ) − 2.5 xDiameter( m.)
- 7 stage = 7 element
Static Mixer Length = 1.5 xDiameter(m.) x 7 (element ) − 3.0 xDiameter( m.)
- 8 stage = 8 element
Static Mixer Length = 1.5 xDiameter( m.) x8 (element ) − 3.5 xDiameter( m.)
- 9 stage = 9 element
Static Mixer Length = 1.5 xDimerter( m.) x9 (element ) − 4.0 xDiameter(m.)
WATER TREATMENT PLANT DESIGN2
Coagulation(Theory)
Page 3 of 8
=
200
mm.
=
3
element
=
0.7
m.
Q
=
A
= 1.842
m/s
= 0.380007 sec.
997.1 kg/m at 25 C
0.000895 kg/m.s (N/m.s) at 25 oC
= 547256.34
Re
Theory
f
f
= 0.048 xRe
= 0.193xRe
∴
−0.2
−0.35
∴
10 4 < Re < 10 6
if
3x10 3 < Re < 10 4
= 0.0034167
f
Theory
Darcy Formular
if
hf
=
f
L v2
D 2g
= 0.0121391
hf
Theory
h f xgxρ
G =
μxt
-1
= 667.53577 sec
Gxt
= 198.74458
WATER TREATMENT PLANT DESIGN2
Coagulation(Theory)
Page 1 of 1
WATER TREATMENT PLANT DESIGN2
Graph Static Mixer(PWA)
Page 1 of 6
4. Flocculation Basin Design
4.1 Hydraulic Mixing
Type :
Round and End Baffle Wall
Design Plant Capacity
Design Operation Flow
3
5000 m /day
24 hr.
:
:
4.2 Design Criteria
sec-1
- G - Value
- No. of Stage
- Detention Time
=
=
=
20 - 60
2-7
20 - 40
minute
4.3 Calculation
No. of Flocculation basin
=
2
Tank
Flow Rate per Tank
=
2500
m3/day
=
=
104.2
30
m3/hr
min
Give Detention Time
Theory
(Gopt ) 2.8
Where :
C is Optimum Dose Alum
td is Detention Time
Use
Give :
=
44 x10 5
Cxtd
=
=
30
30
mg/l
min
Gopt
=
20.78
sec-1
Gopt
≈
25
sec-1
No. of Stage
=
4
G1 - Value
=
60
sec-1
G2 - Value
=
35
sec-1
WATER TREATMENT PLANT DESIGN2
Flocculation
Page 2 of 6
∴
G3 - Value
=
20
sec-1
G4 - value
=
15
sec-1
Water Depth of Flocculation Basin =
Flocculation basin Volume = Qxt
=
∴
52.08
Flocculation Area
=
26.04
Baffle Area 15 % of Flocculation Basin Area
=
∴
∴
∴
∴
3.90625
2
( G is reduce 50 %)
m.
m3
m2
m2
m2
Total Area
=
29.95
Give Width of Flocculation Basin =
4.5
m.
Then Length of Flocculation Basin =
6.7
m.
Use
=
7.0
m.
Give No. of Baffle at Width =
7.0
Give No. of Baffle at Length =
4.0
Total No. of baffle
- No. of baffle at Width
= Width of Flocculation Ba sin xNo. of Baffle
=
31.50
- No. of baffle at Length
Total No. of baffle
=
=
=
Give Width of Concrete
=
0.08
m.
∴
Total Area of Baffle
=
4.76
m2
∴
Acture Flocculation Area
=
26.74
m2
∴
∴
Acture Floculation Volume
Acture Detention Time
=
=
53.48
31
m3
Length of Flocculation Ba sin xNo. of Baffle
WATER TREATMENT PLANT DESIGN2
28.00
59.50
min
Flocculation
Page 3 of 6
Give No. of Stage
=
Give No. of Baffle per Stage =
Stage 1
4
10
Acture Flocculation Volume
4
Acture Volume in stage1 =
G1 - Value
=
60
sec-1
at Q Design
=
104.2
m3/hr
= 0.028935185
No. of stage
m3/s
Theory
ΔH
Where :
∴
∴
γ
= 0.898 x10 −6
=
Head loss in each bend(slit) =
Theory
0.15
0.015
= K
v2
2g
m 2 / s at
25 o C
m.
m.
Give K =
1.6
(Dr. Kawamura)
=
0.43
m/s
The required width for each slit in the stage 1 channel is calculate to be
v
Q
∴
(G ( s −1 )) 2 xγ (m 2 / s ) xV (m 3 )
g (m / s 2 ) xQ(m 3 / s )
ΔH
ΔH
∴
∴
=
=
Av
when
A = width for each slit(m) xwater depth(m)
width of each slit in stage 1 =
0.033
m.
=
33.4811 mm.
Stage 2
Acture Volume in stage1 =
Acture Flocculation Volume
4
G2 - Value
=
35
sec-1
at Q Design
=
104.2
m3/hr
= 0.028935185
WATER TREATMENT PLANT DESIGN2
No. of stage
m3/s
Flocculation
Page 4 of 6
Theory
ΔH
Where :
∴
∴
γ
=
Head loss in each bend(slit) =
Theory
0.05
0.005
= K
v2
2g
m 2 / s at
25 o C
m.
m.
Give K =
1.6
(Dr. Kawamura)
=
0.25
m/s
The required width for each slit in the stage 2 channel is calculate to be
v
Q
∴
= 0.898 x10 −6
ΔH
ΔH
∴
∴
=
(G ( s −1 )) 2 xγ (m 2 / s ) xV (m 3 )
g (m / s 2 ) xQ(m 3 / s )
=
Av
when
A = width for each slit(m) xwater depth(m)
width of each slit in stage 2 =
0.057
m.
=
57.3961 mm.
Stage 3
Acture Volume in stage1 =
Acture Flocculation Volume
4
G3 - Value
=
20
sec-1
at Q Design
=
104.2
m3/hr
= 0.028935185
No. of stage
m3/s
Theory
ΔH
Where :
∴
∴
γ
=
(G ( s −1 )) 2 xγ (m 2 / s ) xV (m 3 )
g (m / s 2 ) xQ(m 3 / s )
= 0.898 x10 −6
=
Head loss in each bend(slit) =
Theory
ΔH
ΔH
= K
WATER TREATMENT PLANT DESIGN2
0.02
0.002
m 2 / s at
25 o C
m.
m.
v2
Flocculation
Page 5 of 6
ΔH
∴
∴
Give K =
1.6
(Dr. Kawamura)
=
0.14
m/s
The required width for each slit in the stage 3 channel is calculate to be
v
Q
∴
v2
= K
2g
=
Av
when
A = width for each slit(m) xwater depth(m)
width of each slit in stage 3 =
0.100
m.
=
100.4432 mm.
Stage 4
Acture Volume in stage1 =
Acture Flocculation Volume
4
G4 - value
=
15
sec-1
at Q Design
=
104.2
m3/hr
= 0.028935185
No. of stage
m3/s
Theory
ΔH
=
γ
=
(G ( s −1 )) 2 xγ (m 2 / s ) xV (m 3 )
g (m / s 2 ) xQ(m 3 / s )
Where :
∴
∴
0.000000898
0.01
ΔH =
Head loss in each bend(slit) =
0.001
Theory
ΔH
∴
∴
v2
2g
at 25o C
m.
m.
Give K =
1.6
(Dr. Kawamura)
=
0.11
m/s
The required width for each slit in the stage 4 channel is calculate to be
v
Q
∴
= K
m2/s
=
Av
when
A = width for each slit(m) xwater depth(m)
width of each slit in stage 4 =
0.134
m.
=
133.9243 mm.
5.1.2 Upflow type (Radial-Upflow type)
Circular or square in shape
Surface Loading
= 1.3 - 1.9 m/hr
Water Depth
=
3 - 5 m.
=
1 - 3 hr.
Settling Time
Weir loading
5.1.3 Reactor Clarifiers
Flocculation Time
Settling Time
Surface Loading
Weir loading
Upflow Velocity
5.1.4 Sludge Blanket Clarifier
Flocculation Time
Settling Time
Surface Loading
Weir loading
Upflow Velocity
WATER TREATMENT PLANT DESIGN2
=
7
m3/m.hr.
= approx 20 min
= 1 - 2 hr.
=
2 - 3 m/hr
=
=
3
7.3 - 15 m /m.hr.
< 50 mm./min.
= approx 20 min
= 1 - 2 hr.
=
2 - 3 m/hr
=
=
3
7.3 - 15 m /m.hr.
< 10 mm./min.
Sedimentation-Inlet
Page 2 of 4
Slurry circulation rate
= up to 3 - 5 times the raw water inflow rate
5.2 Calculation
Type : Horizontal rectangular Tank
Design Plant Capacity
Design Operation Flow
:
:
5000
24
Design Criteria for Horizontal Rectangular Tank
5.2.1 Inlet and Outlet of the Basin
Headloss through the ports = 0.3 - 0.9
The Size of Ports in Diameter = 0.075 - 0.20
The Ports spacing approx = 0.25 - 0.5
Velocity through Diffuser wal = 0.15 - 0.60
Weir Loading rate
=
6 - 11
m3/day
hr.
mm.
m.
m.
m/s
(Prof. Munsin)
m3/hr.m (Prof. Munsin)
5.2.2 Horizontal rectangular Tank Design (Dr.Kawamura)
Minimum number of tank =
2
= 3 - 4.5
m.
Water Depth
= 0.3 - 1.7
m/min
Mean Flow Velocity
= 0.15 - 0.91 m/min (Prof. Munsin)
= 0.02 - 0.06 m/min (Prof. Munsin)
Surface Loading
= 1.4 - 3.4
m/hr
Detention Time
= 1.5 - 4
hr.
= 2-4
hr.
Length/Width Ratio (L/W) = Minimum of 4:1
Water Depth/Length Ratio = Minimum of 1:15
Sludge Collector Speed
m/min
For the Collection path = 0.3 - 0.9
= 1.5 - 3.0
m/min
For the Return
WATER TREATMENT PLANT DESIGN2
where :
W = Launder Depth (m.)
3
Q = Total flow rate of discharge (m /sec)
B = inside width of the Launder (m.)
W =
Use W =
0.19
0.3
m.
m.
Use V-noych weir 90 degree
Theory Discharge of water over V-notch weir
Q =
where :
8
θ 5
C d 2 g tan H 2
15
2
3
Q = Overflow Discharge (m /s)
Cd = Discharge Coefficient = 0.584
θ = V - notch angle
H = Heigh of flow (m.)
90 degree
0.05 0.10
Give 1 V-notch weir have length =
∴
Total V-notch weir =
0.15
67
m.
Flow rate per V-notch weir =
1.56
m3/hr
H
∴
H
5
2
= 0.000194
= 0.033
WATER TREATMENT PLANT DESIGN2
m.
Sedimentation Outlet
Page 1 of 4
6. Fiter Tank Design
Design Criteria (Dr. Kawamura)
6.1 Number of Fliter
Theory
= minimum
2
N = 1.2Q 0.5
Where :
N
=
Q =
6.2 Size of Filter
6.2.1 Ordinary gravity filters
- Width of Filter cell
- Length to width ratio
- Area of Filter cell
- Depth of the filter
6.2.2 Self-backwash filters
- Depth of the filter
- Length to width ratio
- Area of Filter cell
- Depth of the filter
6.3 Filter Bed
Type of Medium and Depth
L/d e
Total number of filters
Maximum plant flow rate in (mgd)
= 3-6
= 2 : 1 to 4 : 1
m.
= 25 - 100
= 3.2 - 6
m2
m.
= 3-6
= 2 : 1 to 4 : 1
m.
= 25 - 80
= 5.5 - 7.5
m2
m.
>
1000
for ordinary monosand and media bed
6.4 Filtration Rate
Filter rate
WATER TREATMENT PLANT DESIGN2
= 15 - 20
m3/hr/m2
Filtration
Page 2 of 4
6.5 Headloss across the filter
2.7 - 4.5
- Total Headloss across each filter (for ordinary gravity filter) =
- Net Headloss available for filtration (for ordinary gravity filter) = 1.8 - 3.6
6.6 Filter washing
- Ordinary rapid sand bed
- Ordinary dual media bed
6.8 Filter Media
6.8.1 Medium Sand for rapid sand filter
m/hr.
- Filter rate
= 7.0 - 7.5
- Effective Size
= 0.45 - 0.65 mm.
- U.C.
= 1.4 - 1.7
m.
- Depth
= 0.6 - 0.75
- S.G.
=
2.63
6.8.2 Multimedia filter
10 - 30
m/hr.
- High rate filtration
Sand
- Effective Size
= 0.45 - 0.65 mm.
- U.C.
= 1.4 - 1.7
- Depth
=
0.3
m.
Anthacite Coal
- Effective Size
= 0.90 - 1.4 mm.
- S.G.
= 1.5 - 1.6
- Depth
=
0.45
m.
WATER TREATMENT PLANT DESIGN2
Filtration
Page 3 of 4
6.9 Underdrain
6.9.1 Normal backwash filters
- Pipe lateral
- Headloss at ordinary backwash rate = 0.9 - 1.5 m.
6 - 10 mm.
- Ordinary size (diameter) =
- Lateral spacing
=
12
inch
- Orifices are spaced 3 - 4 in. apart and 45o down-angle from the horizontal
on both sides of the lateral
- Maximum lateral length of 20 ft
- Precast concrete laterals
- Headloss at ordinary backwash rate
8 - 10 mm.
- Orific size (diameter)
=
- 12 in lateral spacing
- 3 in. orifice spacing on eather side of the lateral
- Maximum lateral length of 16 ft.
6.9.2 Self - backwash type of filter
- Headloss at design backwash rate : 0.15 - 0.3 m.
Gravel Support Bed
Layer Number
1
2
3
4
5
Size
20 - 40 mm.
12 - 20 mm.
6 - 12 mm.
3 - 6 mm.
1.7 - 3 mm.
WATER TREATMENT PLANT DESIGN2
Depth of Size
100 - 150
75 mm.
75 mm.
75 mm.
75 mm.
Filtration
Page 4 of 4
6.9.3 Basic Hydraulic
1. Influent channel
2. Influent valve
3. Effluent Channel
4. Effluent Valve
5. Backwash main
6. Backwash valve
7. Surface wash line
8. Wash-waste main
9. Wash-waste valve
10. Inlet to filter underdrain lateral
Page 1 of 4
6.10 Filtration Design
Filtration type
Backwash by
: Single Filter Media
: Elevation Tank and Surface wash
3
= 5000 m /day
- Q design
Theory
(Dr.Kawamura,210 page)
= 1.2Q 0.5
N
Where :
N = Total Number of filters
Q = Maximum plant flow rate in mgd
∴
N
Give Hydraulic Loading
=
=
1.37922 Use
3
Tanks
m3/hr/m2
7
∴ Surface Area of Filter Tank = 29.76
m2
∴ Area per Tank
∴ Use Tank area
= 9.92
= 4.45
m2
x
2.23 (Length to width ratio 2 : 1 to 4 : 1)
≈
x
2.5
Use Acture Tank Area
5
m2
∴ Acture tank Area
= 12.5
∴ Flow per Tank
3
= 1667 m /day
1. Inlet Pipe Design
- Give velocity
=
Q =
D =
0.6
m2
m/s (JWWA)
Av
4Q(m 3 / s )
πv
0.20238
∴ Use inlet pipe diameter (D) = 0.2
D =
WATER TREATMENT PLANT DESIGN2
m.
m.
Filtration Design
Page 2 of 4
Acture Velocity
=
∴ Acture Velocity
Q(m 3 / s)
A(m 2 )
= 0.61434 m/s
Headloss
- Give Pipe length (L)
- Friction Loss
Theory
Hazen - William Equation
=
2.5
m.
Q(m 3 / s ) = 0.278CD( m )
∴
S
∴
= SxL
hL
⎛ 3.597Q (m 3 / s ) ⎞
⎟
= ⎜
⎜ CD 2.63 ⎟
(m)
⎠
⎝
hL
- Miner Loss
1 - Inlet
1 - Outlet
1 - Gate Valve
Total
1
0.54
⎛ 3.597Q(m 3 / s) ⎞
⎜
⎟
⎜ CD 2.63 ⎟
(m)
⎝
⎠
1.85
1.85
xL
=
120
0.00644
=
=
=
=
K - Value
0.5 (velocity head)
1 (velocity head)
0.2
1.7
WATER TREATMENT PLANT DESIGN2
=
hL
L
hL
New Pipe use C =
∴
S 0.54
⎛ Q(m 3 / s) ⎞
⎟
= ⎜
⎜ 0.278CD 2.63 ⎟
( m)
⎝
⎠
From Slope of Energy grade Line (S) =
∴
2.63
m
Filtration Design
Page 3 of 4
∴
∴
Miner Loss
=
Miner Loss
= 0.0327
Total Headloss
2. Filter Media
Sand
- Effective Size
- Uniformity Coefficient
- Sand Fliter Depth (L)
- L/de
Gravel Support Bed
Layer
Upper
Lower
Underdrain design
- Type : Pipe lateral
- Velocity in lateral pipe
- Lateral spacing
use
- Orifice diameter
use
Kv 2
2g
m
= Friction Loss + Miner Loss
= 0.03914 m
=
=
=
=
1
2
3
4
5
0.45 - 0.65 mm. ≈
0.55
1.40 - 1.70
0.65
m
1182 more than 1000
Size (mm.)
1.7 - 3.0
3-6
6-12
12 - 20
20 - 40
mm.
OK.
Depth of Layer (mm)
150
75
75
75
75
= 1.37 m/s
= 0.08 - 0.20 m (Mahidol University)
= 0.2
m
= 6.38 - 12.7 mm. (Mahidol University)
= 7
mm
WATER TREATMENT PLANT DESIGN2
Filtration Design
Page 4 of 4
- Orifice area/crossection area of filter tank
= 0.0015 - 0.005
- Orifice area/pipe area
= 0.25 - 0.5
- Number of lateral
= 25 (give lateral spacing 0.2 m. Length Tank)
- Flow rate
= Velocity rate in lateral pipe ( m / s ) xSurface Area ( m )
2
2
∴ Surface area of Pipe lateral = 0.00056 m
- Flow per Lateral
- Pipe Lateral Diameter
=
=
3
m3/hr
= 0.02679
∴ Use Pipe Lateral Diameter(D ≈ 27
- Total Orifice area/ filter area =
0.35 %
∴
∴
∴
Total Orofice Area
- Give Number of Orifice
Nx
πD(m) 2
4
N
m
4Q (m 3 / s)
πv(m / s)
m
mm
(Design Criteria : 0.2 - 1.5 %)
2
= 0.04375 m
= N
= Total Orifice Area(m 2 )
=
1137
- Number of Orifice / Lateral = 45.5 pores
∴ - Orifice Spacing
= 0.05498 m
Use Orifice Spacing = 0.05
m
WATER TREATMENT PLANT DESIGN2
Filtration Design
Page 1 of 4
3. Clear Water Pipe in Filter tank
3.1 Clear water pipe in Filter tank (Lateral pipe)
- Velocity
= 1
m/s
- Flow per tank
6. Collecter Backwash Pipe
- Flow per tank
- Velocity
- Pipe Diameter
∴
D =
Use pipe diameter
4Q (m 3 / s )
πv
= 0.091
= 100
7. Surface wash Pipe
- Surface wash rate = 0.15
Q = Av
- Flow rate
Use Flow rate
m
mm.
m/min (Design Criteria 0.12 - 0.16 m/min)
= 112.5
m3/hr
= 115
m3/hr
- Surface jet pressure = 15 - 20 m.
(headloss)
Use
= 15
m
WATER TREATMENT PLANT DESIGN2
(Design criteria)
Clear Water Pipe in Filter Tank
Page 3 of 4
Theory
h =
∴
Velocity
1 ⎛v⎞
x⎜ ⎟
2g ⎝ c ⎠
= 11.151
-Give Orifice Diameter = 5
- Orifice Area
= πD 2
2
m/s
mm.
4
=
1.9635E-05
- Flow per orifice
= 0.0002
- Number of Orofice = Qtotal
m2
m3/s
Qorifice
= 146 holes
∴
∴
∴
∴
∴
∴
- Use 2 Pipe lateral
Number of Orifice per lateral =
Orifice spacing
= 0.0658
73.0
m
- Try orifice Diameter =
mm.
Orifice area
=
6
2.82743E-05
3
- Flow per orifice
= 0.0003 m /s
Number of Orifice per lateral = 101.3198
- Use 2 Pipe lateral
Number of Orifice per lateral = 50.65992
Orifice spacing
= 0.1
m
holes
(Tank Length = 5 m)
(minus length from Wall tank = 0.1 m.
2 side = 0.2 m)
m2
holes
holes
-Surface wash pipe Diameter
Main Pipe
WATER TREATMENT PLANT DESIGN2
Clear Water Pipe in Filter Tank
Page 4 of 4
Velocity
Pipe Diameter
= 2.4
D =
∴
= 2
= 2.4
D =
∴
(Dr.Kawamura)
4Q (m 3 / s )
πv
Use Pipe Diameter = 0.1302
Use Pipe Diameter = 150
Lateral
Velocity
Pipe Diameter
m/s
m
mm.
pipe
m/s
(Dr.Kawamura)
4Q (m 3 / s )
πv
Use Pipe Diameter = 0.092
Use Pipe Diameter = 100
WATER TREATMENT PLANT DESIGN2
m.
mm.
Clear Water Pipe in Filter Tank
Page 1 of 1
WATER TREATMENT PLANT DESIGN2
Trough height
Page 1 of 2
8. Water Through Design
Theory
⎛ Q( m 3 / s ) ⎞
⎟⎟
Minimumtrough height = ⎜⎜
⎝ 1.4B(m) ⎠
Minimum trough height
∴
= W +
∴
B
W +
2
3
+
=
Equation 1
free board
B
+ free board
2
=
Equation 1
2
Equation 2
Equation 2
⎛ Q (m 3 / s) ⎞
⎜⎜
⎟⎟
⎝ 1 .4 B ( m ) ⎠
2
3
Where :
B =
water depth inside of the trough from base line (m)
inside width of the trough (m)
Q =
total flow rate of discharge per trough (m3/s)
W
=
Design Criteria backwash water and air = 0.25 - 0.7 m3/m2.min
backwash rate =
∴
Use
∴
∴
0.7
m3/m2.min
Q(m 3 / s )
Surface area (m 2 )
m3/s
=
0.14583
2 trough per Filter tank
Q
use
w
3
total flow rate of discharge per trough =
0.072917 m /s
Give Free board = 0.051 m. (free board should be a minimum of 50 mm.)
Give inside width of the trough =
0.4
m.
Minimum trough height(P) = 0.307897
m.
=
30.78971
cm.
≈
31
cm.
From Sand Layer Depth (L) = 650
mm.
WATER TREATMENT PLANT DESIGN2
=
0.65
m.
Water Trough Design
Page 2 of 2
Theory
+
0 . 75 L
P<
0.7954 <
Use
Ho
Theory
1.5 H o
Use
S
<
0.88
=
1.31497
Ho
Ho
<
=
WATER TREATMENT PLANT DESIGN2
< L
+
P
0.957897
m
< S < 2H o
S
1.53
<
1.753294
m
Water Trough Design
Page 1 of 1
Hydraulic Design
1. Head loss (Run)
1.1 Head loss Sand
1.2 Head loss Gravel
1.3 Head loss Underdrain
1.4 Head loss at outlet piping
2. Head loss (Backwash)
2.1 Head loss Sand
2.2 Head loss Gravel
2.3 Head loss Underdrain
2.4 Head loss piping from Elevation Tank
3. Head loss from Surface wash
- Calculation later from Layout and find out Hydraulic grade line and Surface & Backwash pipe
WATER TREATMENT PLANT DESIGN2
Hydraulic Design
Page 1 of 2
9. Chlorination Design
Design Criteria (Dr. Kawamura)
Dosage
:
Number of chlorine feeder :
Residual Chlorine
:
Contact time
:
pH
:
:
Chlorine solution tank
Chlorine stock
:
Safety features
:
1 - 5 mg/l (2.5 mg/l average)
Minimum of two : one stanby is required
Over 0.5 mg/l (Higher Level)
Over 30 min (longer)
6-7
Enough to produce a 1 day supply
Minimum of 15 days storage
Eye wash, shower, gas masks
Design
Use liquid chlorine concentration 1 % prepare from stock liquid chlorine 50 % feed
to main pipe before Elevation tank. Keep Contact time = 30 min (minimum)
1. Chlorine Feeder
Q - Design
= 5000
m3/d
m3/hr
= 208.3
Assume Chlorine demand of water =
1
mg/l
For residual chlorine about 0.5 - 1 mg/l
Use chlorine dosage 1.5 - 2 mg/l
Required chlorine
=
208.3 x(1.5 to 2 mg/l)
=
312.5 to
416.7 g/hr
Meaning of liquid chlorine 1 % is chlorine =
10 g/l (1 L of water = 1000 g)
Chorine feeder rate
= 31.25
to
41.67 L/hr
∴ Use Chlorine feeder rate = 35
to
40 L/hr
2. Dilution stock liquid chlorine solution 50 % to 1 % liquid chlorine solution
WATER TREATMENT PLANT DESIGN2
Chlorine Design
Page 2 of 2
Assume stock liquid chlorine solution 50 % one plastic equal =
N1 xV1 = N 2 xV2
Theory
Where :
N = Chlorine concentration (%)
V = Volume of Liquid (liters)
Give
50 %
N1 =
20
liters
V1 =
1
%
N2 =
?
liters
V2 =
50 % x 20 = 1% x V2
∴
20
liters
=
1000
liters
Use mixing tank volume =
Fill stock Liquid Chlorine =
1000
20
liters made from plastic
liters in mixing tank and fill water until
limited
1000
liters
V2
3. Period of Mixing
Maximum chlorine feeder rate
Required Liquid Chlorine
So period of mixing
=
40 Liters/hr
= 960
= 960 Liters/day
= Every Day
4. Liquid Clorine 50 % Stock
Use storage time
=
30
days ( 1 month)
Required Liquid Chlorine =
1
%
= 28800 Liters per month
Required Liquid Chlorine =
50
%
= 576 Liters per month
∴ Stock Liquid Chlorine
= 28.8 Plastic Tank per Month
∴ For Order per Month say =
29 Plastic Tank
WATER TREATMENT PLANT DESIGN2
Chlorine Design
Page 1 of 2
10. Surface Wash and Backwash System
- Use water from Elevation tank
- Water Level in Elevation tank = 22 - 25 m.
- From Site Pant : Pipe length = 35 m.
Backwash System
Use Pressure for Backwash
- Head loss due to water flowing through a sand bed fluidized
Theory
hL
= (1 − e )(S g − 1)
L
Where :
hL
=
e =
L
=
Sg
=
head loss through the media bed during backwash. (m)
porosity of the clean stratified bed at rest. (not fluidize =
depth of the stratified bed at rest. (m)
=
0.65
=
2.65
specific gravity of the media.
0.4
m
Calculation
hL
=
0.6
m
- Head loss through the supporting gravel bed fluidzed
2 2
2
2
2
2
จาก H/L = (150νV/g)[(1-ε) /ε ](1/ω) Σ(xi/di ) + (1.75V /g)(1-ε/ε )(1/ω) Σ(xi/di)
mm2/s
ν
=
0.9629
mm/s
V(Backwash rate)
=
11.67
ε
ω
=
=
WATER TREATMENT PLANT DESIGN2
0.4
0.8
surface wash and backwash
Page 2 of 2
Layer
1
2
3
4
5
From H/L
H
Size (mm)
1.7 - 3
3-6
6 - 12
12 - 20
20 - 40
di
2.26
4.24
8.49
15.49
28.28
=
=
0.0995
0.0448
- Backwash trough height
Give Fluidized Bed Expand
=
∴ Media Depth(Fluidized) =
=
∴ Expanded sand depth =
=
∴
Backwash trough height =
=
∴
Depth (mm)
150
75
75
75
75
450
25
0.65
0.8125
0.8125
0.1625
xi
0.3333
0.1667
0.1667
0.1667
0.1667
1.0000
xi/di
0.1476
0.0393
0.0196
0.0108
0.0059
0.2232
xi/di2
0.0654
0.0093
0.0023
0.0007
0.0002
0.0778
m.
%
+
m
m
0.1625
Sand Fliter Depth
Expanded Sand Depth + trough height
+ 6 in
(Munsin Tuntuvate)
0.6227971
Head loss for Backwash system =
Headloss in sand bed fluidized + gravel bed fluidized + sand depth + backwash trough
=
1.9611
WATER TREATMENT PLANT DESIGN2
m
surface wash and backwash
Page 1 of 3
11. Headloss in Piping System
Give backwash rate
=
0.7
m/min (Design criteria = 0.6 - 0.7 m/min)
m2
= 12.5
= A(m 2 ) xv(m / hr )
Acture tank Area
∴ Backwash flow rate
3
= 525 m /hr
= 300
mm
= 0.3
m
= 2.0642 m/s
Give Main Pipe Diameter
∴ Acture velocity
11.1 Friction Loss (hL) at Main Pipe
Theory Hazen-William Equation
Q(m 3 / s ) = 0.278CD( m )
∴
Give
= SxL
hL
⎛ 3.597Q (m 3 / s ) ⎞
⎟
= ⎜
⎜ CD 2.63 ⎟
(m)
⎠
⎝
1.85
xL(m)
m
m
hL
=
0.5288
m
too Low
hL
= 250
= 0.25
= 2.9724
= 1.2839
mm
m
m/s
m
OK
Try Main Pipe Diameter
∴
∴
S 0.54
hL
New Pipe
C = 120
= 0.3
Pipe Diameter
Pipe Length
= 35
∴
2.63
Velocity
WATER TREATMENT PLANT DESIGN2
Backwash in Piping System
Page 2 of 3
11.2 Velocity Miner HeadLoss
11.2.1.accessory
1 - Inlet
=
3 - 90oBend
Total
K
1
= 2.25
= 3.25
Theory
v2
2g
Headloss
=
Headloss
= 1.464
K
11.2.3. Velocity headloss in Main pipe
1. Pipe Diameter = 0.3
2. Velocity
= 2.0642
3. K
= 2.2
Headloss
= 0.4778
m
m
m/s
m
11.3 Headloss at lateral
Theory
Flow per Lateral = 21
Hazen-William Equation
hL
∴
Pipe Diameter
Headloss
m3/hr
⎛ 3.597Q (m 3 / s ) ⎞
⎟
= ⎜
⎜ CD 2.63 ⎟
(m)
⎠
⎝
= 0.0268
= 12.472
1.85
xL(m)
m
m
11.4 Headloss at Orifice
Flow per Orifice = 0.061
- Orifice diameter = 7
Velocity
= 0.4411
WATER TREATMENT PLANT DESIGN2
m3/hr
mm
m/s
Backwash in Piping System
Page 3 of 3
Theory
Headloss
=
1
2g
⎛v⎞
x⎜ ⎟
⎝C ⎠
2
∴
Give C for Orifice = 0.65
Headloss
= 0.0235
m
∴
Total Headloss
m
= 16.409
- Sand Depth
=
- Gravel Depth
=
- Sand Expansion
=
- Trough Height
=
∴ Trough Height From Base Line =
∴ Total Dynamic Head
=
- Water Level in Elevation tank =
∴ Different Head Loss
=
11.5 Use Orifice Plate
Theory
Headloss
=
1 ⎛v⎞
x⎜ ⎟
2g ⎝ C ⎠
Theory
Choose =
22
m.
2
Give C for Orifice = 0.65
Velocity (v) = 5.58
∴
m/s
Pipe Diameter
D =
∴
0.65
m
0.45
m
0.16
m
0.31
m
1.57
m
17.98
m
22 - 25 m.
4.02
m.
4Q(m 3 / s )
πv
Pipe Diameter = 0.182
= 182
WATER TREATMENT PLANT DESIGN2
m.
mm.
Backwash in Piping System
Page 1 of 5
12. Headloss and Hydraulic Profile
Qdesign
=
Qmax
= 1.5Qdesign
=
5000
7500
m3/d
m3/d
- Headloss at static Mixer
at Q design
=
0.08
m.
- Headloss at Flocculation
for 1st Stage
at Q design
=
0.15
m.
for 2nd Stage
at Q design
=
0.05
m.
for 3rd Stage
at Q design
=
0.02
m.
for 4th Stage
at Q design
=
0.01
m.
- Headloss at Diffuser wall (Inlet Zone Sedimentation)
at Q design
= 0.0031 m.
- Headloss Over V-notch weinr (Outlet Zone Sedimentation)
at Q design
= 0.0328 m.
- Headloss Inlet Pipe (Filtration)
WATER TREATMENT PLANT DESIGN2
Head loss and Hydraulic Profile
Page 2 of 5
at Q design
=
0.0391 m.
- Head loss pass through clean filter media (Filtration)
(5νV/g)[(1-ε)2/ε3](6/ω)2Σ(xi/di2)
จาก H/L
=
ν
Total Head loss across the filter Media at Qdesign
=
Head loss pass through clean filter media (Filtration) +
Headloss Through Gravel (Filtration)
0.652 m.
=
- Head loss for underdrain
1. Head loss at orifice
Total Number of filters
=
3
Tanks
=
3
5000 m /day
∴ Flow rate per Tank
1 Tank have Number of lateral
=
=
3
1667 m /day
25
∴ Flow rate per lateral
1 lateral have Number of orifice
=
=
3
66.67 m /day
45.47 pores
=
3
1.466 m /day
=
0.061
- Q design
∴
Flow rate per orifice
m3/hr
Theory
Q =
∴
Velocity pass through orifice
=
∴
Theory
Q =
Av
Av = C d A 2 ghL
⎛ v
⎜⎜
⎝ Cd
∴
0.441
m/s
Head loss at orifice
Q = C d A 2 ghL
∴
Av
hL
⎞
⎟⎟
⎠
=
(1)
(2)
(3)
2
=
2 ghL
1 ⎛ v
⎜
2 g ⎜⎝ C d
⎞
⎟⎟
⎠
2
WATER TREATMENT PLANT DESIGN2
(4)
Head loss and Hydraulic Profile
Page 4 of 5
2 g ⎜⎝ C d ⎟⎠
L
Cd Orifice
∴
Head loss at orifice
=
0.61
=
0.027
=
3
66.67 m /day
=
2.778
m.
2. Head loss at Lateral
Flow rate per lateral
m3/hr
Theory
Q =
∴
Velocity pass through lateral
Minor loss
K outlet
∴
Minor loss
Theory
hL
=
=
K
=
=
1
0.096
1.37
hL
Total Head loss at Qdesign
m/s
2
v
2g
⎛ 3.597Q (m 3 / s ) ⎞
⎟
= ⎜
⎜ CD 2.63 ⎟
(m)
⎠
⎝
New Pipe C =
∴
Av
m.
1.85
xL
120
=
0.296
m.
=
=
+ Minor loss
0.391 m.
hL
- Head loss at delivery Pipe (Lateral Clear water Pipe in Filtration Tank)
Theory
Q =
WATER TREATMENT PLANT DESIGN2
Av
Head loss and Hydraulic Profile
Page 5 of 5
∴
3
- Flow per tank
= 69.44 m /hr
- Use pipe diameter
=
0.2
m.
Velocity pass through lateral clear water pipe
= 0.614 m/s
- Miner Loss
K - Value
1 - Inlet
=
0.5 (velocity head)
1 - Outlet
=
1 (velocity head)
1 - Gate Valve =
0.2
90o bend
Tee
Total
Theory
Miner Loss
=
=
=
0.9
1.8
4.4
=
Kv 2
2g
=
0.085
m.
- Head loss at delivery Pipe (Maniflow Clear water Pipe in Filtration Tank)
Give Pipe Length
Theory