of 60

Wastewater Treatment

Published on March 2017 | Categories: Documents | Downloads: 5 | Comments: 0
94 views

Comments

Content


  

  

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

Baffle Channel Baffle Channel
2-7
4

Baffle Channel
4

20 - 40

stage 1 = 60
stage 2 = 35
stage 3 = 20
stage 4 = 15
31

110
64
37
27
21

Rectangular
1.5 - 4
0.02 - 0.06
3.0 - 4.5
< 1.5
Manual

Rectangular
2.5
0.019
3
0.129
Manual

Rectangular
1.67
0.029
3
0.193
Manual

0.15 - 0.60
0.40 - 0.70
100 max

0.165
0.5
75

0.247
0.5
75
Design Criteria

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

Unit

Design Criteria

Design Flow

Max. Flow

(5000 m3/day)

(7500 m3/day)

m3/m./hr.

12 max

8.68

13

No.
m/hr

min. 2
5 to 7

3
5.5

3
8.33

Constant Rate
Influent Level Control
Pipe Lateral
Sand
0.55 - 0.65
1.40 - 1.70
700

mm.
mm.
m./min.

m./min.
m.

WATER TREATMENT PLANT DESIGN2

0.60 - 0.70

0.7

0.12 - 0.16
15 - 20

Fixed Nozzle
0.15
15

Design Criteria

Page 1 of 1
1. ขอมูลพื้นฐาน
1.1 คุณภาพน้ํา
- Turbidity
- Alkalinity
- pH
- Temperature
- Fe
- Mn
- Hardness

=
=
=
=
=
=
=

NTU
mg/l as CaCO3
o

C
mg/l
mg/l
mg/l as CaCO3

1.2 แหลงน้ํา

= Surface Water

1.3 Jar Test
- Alum Dose
- Alkalinity
- pH

=
=
=

- Temperature =

mg/l
mg/l as CaCO3
o

C

1.4 Design Flow
Design Plant Capacity =
Design Operation Flow =

5000
24

m3/d
hr

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.)

WATER TREATMENT PLANT DESIGN2

Coagulation(PWA)

Page 3 of 4




Try Static Mixer Diameter
Use 3 stage
Static Mixer Length (L)
Acture Velocity



Detention Time (t)




Try Static Mixer Diameter
Use 3 stage
Static Mixer Length (L)
Acture Velocity





=
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

=

μ =

Theory
G =

Gxt




too less

Try Static Mixer Diameter
Use 2 stage
Static Mixer Length (L)
Acture Velocity

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
= 2848.0866 sec

too hight

= 2113.8535

too hight

=
=
=
=

WATER TREATMENT PLANT DESIGN2

300
2
0.75

mm.
element
m.

Q
A

Coagulation(PWA)

Page 4 of 4
A





= 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.




Try Static Mixer Diameter
Use 3 stage
Static Mixer Length (L)
Acture Velocity



Detention Time (t)



Check Renolds Number(R e) =

D p ρ L vs

μ

ρL

=

μ =

3

o

997.1 kg/m at 25 C
0.000895 kg/m.s (N/m.s) at 25 oC

= 410442.26

Re

Theory
f
f

= 0.048 xRe
= 0.193xRe



−0.2

−0.35



10 4 < Re < 10 6

if

3x10 3 < Re < 10 4

= 0.003619

f

Theory
Darcy Formular

if

hf

=

f

L v2
D 2g

= 0.0021906

hf

Theory

h f xgxρ

G =

μxt
-1
= 251.00527 sec

Gxt

too less

= 95.383772

WATER TREATMENT PLANT DESIGN2

Coagulation(Theory)

Page 4 of 8
=
250
mm.
=
3
element
= 0.875
m.
Q
=
A
= 1.179
m/s
= 0.7422013 sec.




Try Static Mixer Diameter
Use 3 stage
Static Mixer Length (L)
Acture Velocity



Detention Time (t)



Check Renolds Number(R e) =

D p ρ L vs

μ

ρL

=

μ =

3

o

997.1 kg/m at 25 C
0.000895 kg/m.s (N/m.s) at 25 oC

= 328353.8

Re

Theory
f
f

= 0.048 xRe
= 0.193xRe



−0.2

−0.35



10 4 < Re < 10 6

if

3x10 3 < Re < 10 4

= 0.0037842

f

Theory
Darcy Formular

if

hf

=

f

L v2
D 2g

= 0.0009382

hf

Theory

h f xgxρ

G =

μxt
-1
= 117.54084 sec

Gxt

too less

= 87.238961

WATER TREATMENT PLANT DESIGN2

Coagulation(Theory)

Page 5 of 8
=
300
mm.
=
2
element
= 0.75
m.
Q
=
A
= 0.819
m/s
= 0.9160884 sec.




Try Static Mixer Diameter
Use 2 stage
Static Mixer Length (L)
Acture Velocity



Detention Time (t)



Check Renolds Number(R e) =

D p ρ L vs

μ

ρL

=

μ =

3

o

997.1 kg/m at 25 C
0.000895 kg/m.s (N/m.s) at 25 oC

= 273628.17

Re

Theory
f
f

= 0.048 xRe
= 0.193xRe



−0.2

−0.35



10 4 < Re < 10 6

if

3x10 3 < Re < 10 4

= 0.0039247

f

Theory
Darcy Formular

if

hf

=

f

L v2
D 2g

= 0.0003352

hf

Theory

h f xgxρ

G =

μxt
-1
= 63.237189 sec

Gxt

too less

= 57.930857

WATER TREATMENT PLANT DESIGN2

Coagulation(Theory)

Page 6 of 8
=
400
mm.
=
3
element
=
1.4
m.
Q
=
A
= 0.461
m/s
= 3.0400564 sec.




Try Static Mixer Diameter
Use 3 stage
Static Mixer Length (L)
Acture Velocity



Detention Time (t)



Check Renolds Number(R e) =

D p ρ L vs

μ

ρL

=

μ =

3

o

997.1 kg/m at 25 C
0.000895 kg/m.s (N/m.s) at 25 oC

= 205221.13

Re

Theory
f
f

= 0.048 xRe
= 0.193xRe



−0.2

−0.35



10 4 < Re < 10 6

if

3x10 3 < Re < 10 4

= 0.0041572

f

Theory
Darcy Formular

if

hf

=

f

L v2
D 2g

= 0.0001573

hf

Theory

h f xgxρ

G =

μxt
-1
= 23.778303 sec

Gxt

too less

= 72.287381

WATER TREATMENT PLANT DESIGN2

Coagulation(Theory)

Page 7 of 8
=
200
mm.
=
6
element
=
1.3
m.
Q
=
A
= 1.842
m/s
= 0.7057274 sec.




Try Static Mixer Diameter
Use 6 stage
Static Mixer Length (L)
Acture Velocity



Detention Time (t)



Check Renolds Number(R e) =

D p ρ L vs

μ

ρL

=

μ =

3

o

997.1 kg/m at 25 C
0.000895 kg/m.s (N/m.s) at 25 oC

= 410442.26

Re

Theory
f
f

= 0.048 xRe
= 0.193xRe



−0.2

−0.35



10 4 < Re < 10 6

if

3x10 3 < Re < 10 4

= 0.003619

f

Theory
Darcy Formular

if

hf

=

f

L v2
D 2g

= 0.0040683

hf

Theory

h f xgxρ

G =

μxt
-1
= 251.00527 sec



Gxt

too less

= 177.14129

WATER TREATMENT PLANT DESIGN2

Coagulation(Theory)

Page 8 of 8
=
150
mm.
=
6
element
= 0.975
m.
Q
=
A
= 3.275
m/s
= 0.2977287 sec.




Try Static Mixer Diameter
Use 6 stage
Static Mixer Length (L)
Acture Velocity



Detention Time (t)



Check Renolds Number(R e) =

D p ρ L vs

μ

ρL

=

μ =

3

o

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.

WATER TREATMENT PLANT DESIGN2

Flocculation

Page 6 of 6
CHECK

=

G average

G1 + G2 + G3 + G4
4

=
G average xt

Design Criteria

=

60,069

1x10 4

< Gxt < 1x10 5

1x10 4

< Gxt < 15 x10 4

WATER TREATMENT PLANT DESIGN2

sec-1

32.5

OK.

( Kawamura)
(Qasim )

Flocculation

Page 1 of 2

WATER TREATMENT PLANT DESIGN2

Moody Diagram

Page 2 of 2

WATER TREATMENT PLANT DESIGN2

Moody Diagram

Page 1 of 4
5. Sedimentation Basin Design
5.1 Design Criteria (Dr.Kawamura)
5.1.1 Rectangular Basin (Horizontal Flow)
= 0.83 - 2.5 m/hr
Surface Loading
Water Depth
=
3 - 5 m.
= 1.5 - 3.0 hr.
Detention Time
= > 1/5
Width/Length
Weir loading

=

< 11

m3/m.hr.

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

Sedimentation-Inlet

Page 3 of 4



Design Plant Capacity
Number of Sedimentation

=
=

5000
2

m3/day
Tank

Flow rate per Basin

=

2500

m3/day

5.3 Inlet Zone
Inlet Diffusion Wall
Darcy - Weisbach Formular
=

hf

f

L v2
D 2g

Relationship with G
G

Where :

2

fv 3 ρ
2 gD μ

=

G = Mean velocity gradient ( s −1 )
f

=

Friction Factor

v

=

velocity passthroug h orifice( m / s )

D =

Diameter of Orifice (m)

ρ

=

Mass Density of Water ( kg / m 3

μ

=

absolute vis cos ity ( kg / m.s )

Give G
=
Orifice made from concrete : ε =
Give Diameter of Orifice
ε/D

=
=
find f Give from Reynolds number
Give Reynolds number(R)
=
From Moody Diagram
then f
=

WATER TREATMENT PLANT DESIGN2

=
=

2
997.1 kg/m at 25 oC
0.000895 kg/m.s (N/m.s) at 25 oC

-1
10 1/sec ( Design Criteria G = 10 - 75 s )
1.22 mm.
m2
100 mm. ∴ Area = 0.007854

0.0122
17,000
0.042
Sedimentation-Inlet

Page 4 of 4




v

=

Check Renold Number ( R )
R

=

Total Area of Pores

=

Theory

Headloss

=



Give C for Orifice
Headloss

=
=

WATER TREATMENT PLANT DESIGN2

0.161 m/sec
(PWA . Criteria 0.15 - 0.20 m/sec)
vD/ν

or
17,957 O.K.

1
2g

D pδ L vs

μ

Q/v

m2

0.35888
46

m2

⎛v⎞
x⎜ ⎟
⎝C ⎠

γ

=

μ
δ

pores

2

0.65
0.0031

m.

Sedimentation-Inlet

Page 1 of 1
5.4 Horizontal Rectangular Basin









Design Plant Capacity =
Number of Sedimentation =

5000
2

m3/day
Tank

Flow rate per Basin

=

2500

m3/day

Give Water Depth
Give Surface Loading

=
3
m.
= 1.9
m/hr.
= 0.031667 m/min (Design Criteria : 0.02 - 0.06 m/min)

The Required total surface Area =

Give Tank Width
Tank Length
Give Detention Time
Tank Volume

Q(m 3 / min)
Surface Loading (m / min)

=

54.82

m2

=
=
=
=

4.5
12.18
3

m.
m.
hr.

m2

(Because Flocculation basin Width : 4.5 m)

Qxt

m3
= 312.50
Acture Tank Length
= 23.15
m.
Use Acture Tank length ≈
24
m.
Width : Length
=
1 : 5.14
Acture Surface Loading = 1.00
= 0.0167

WATER TREATMENT PLANT DESIGN2

(Design Criteria > Minimum 1:5)

OK.

m/hr.
m/min. (PWA. Design Criteria : 0.02 - 0.06 m/min)

Sedimentation

Page 1 of 2
5.5 Outlet Zone
Give Weir Loading
Weir Length
Give Outlet Zone Width

=
=
=

12
8.68
2.5

m3/hr.m.
m.
m.

Theory
Launder Size (d)
= Q 0.4
m.
Use 2 Launder per Basin
0 .4
⎛ 2500 ⎞


= ⎝ 24 x3600 x 2 ⎠


d
=
Use d =

Check Weir Length (L)
Theory

L =

0.18
0.25

m.
m.

0.2Q
Hv s

where :
L = Combined weir length (m)
3
=
Q = Flow rate (m /day)
H = Depth of Tank (m)
=
vs = Settling velocity (m/day)

Give vs =
L =




Weir Length (L) =
Use weir length =
Weir length/Basin =

0.04812 m/min

=

2500
3

m3/day
m

69.2928 m/day

0.2 x 2500(m 3 / day )
3(m) x69.3(m / day )

2.41
2.5
10

WATER TREATMENT PLANT DESIGN2

m.
m.
m.
Sedimentation Outlet

Page 2 of 2
Check Launder Depth
Theory
W

=

⎛ Q (m 3 / s) ⎞
⎜⎜
⎟⎟
⎝ 1 .4 B ( m ) ⎠

2

3

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

= 0.6 - 0.74
= 0.74 - 0.9

6.7 Surface Wash Rate
- Flow rate
- Pressure

: Fix nozzle type
= 0.12 - 0.16 m/min
KPa
= 55 - 83

m.
m.

m/min
m/min

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

WATER TREATMENT PLANT DESIGN2

Ordinary Filter
(m/s)
0.6
0.91
1.52
1.52
3.05
2.4
2.4
2.4
2.4
1.37

Self-backwash filter
(m/s)
0.6
1.52
0.6
0.6
0.91
1.52
2.4
2.4
2.4
1.37

Filtration

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

= 1667

m3/day

= 69.4

m3/hr

(JWWA)

- Pipe Diameter (D)
Q =

D =



Av

4Q (m 3 / s )
πv

= 0.157
- Use pipe diameter = 200
Lateral Pipe length = 2.5

m
mm.
m.

3.2 Clear water pipe in Filter tank (Maniflow)
- Pipe Diameter (D)
Q =

D =



Use

4. Backwash Pipe
- Backwash rate

4Q (m 3 / s )
πv

= 0.2715
≈ 300

m
mm.

= 0.7

m/min (Design Criteria 0.6 - 0.7 m/min)

Q =

- Flow rate
- Velocity
- Pipe Diameter

Av

Av

= 525
= 2
D =

m3/hr
m/s

(JWWA)

4Q (m 3 / s )
πv

WATER TREATMENT PLANT DESIGN2

Clear Water Pipe in Filter Tank

Page 2 of 4


Use pipe diameter

= 0.3048
= 300

m
mm.

= 525
= 2

m3/hr
m/s

5. Lateral Backwash pipe
- Flow rate
- Velocity
- Pipe Diameter

D =



Use pipe diameter

(JWWA)

4Q (m 3 / s )
πv

= 0.3048
= 300

m
mm.

= 69.4
= 3

m3/hr
m/s (JWWA criteria 2.5 - 6.0 m/s)

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

V
ε
ω

ชั้นทรายสูง
H/L
H

=
=
=
=
=
=
=

0.9629
1.94
0.4
0.8
0.65
1.00
0.65

mm2/s
mm/s

m.
m.

- Headloss Through Gravel (Filtration)
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)
=
=
=
=

ν

V
ε
ω

Layer
1
2
3
4
5

From H/L
H

0.9629
1.94
0.4
0.8

Size (mm)
1.7 - 3
3-6
6 - 12
12 - 20
20 - 40

=
=

WATER TREATMENT PLANT DESIGN2

mm2/s
mm/s

di Depth (mm)
2.26
150
4.24
75
8.49
75
15.49
75
28.28
75
450
0.0076
0.0034

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.
Head loss and Hydraulic Profile

Page 3 of 5


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

=
hL

20

m.

⎛ 3.597Q (m 3 / s ) ⎞

= ⎜
⎜ CD 2.63 ⎟
(m)



New Pipe C =
hL

WATER TREATMENT PLANT DESIGN2

=

1.85

xL

120
0.089

m.

Head loss and Hydraulic Profile

Sponsor Documents

Or use your account on DocShare.tips

Hide

Forgot your password?

Or register your new account on DocShare.tips

Hide

Lost your password? Please enter your email address. You will receive a link to create a new password.

Back to log-in

Close