Assosa Water Supply Project
CHAPTER FIVE
5.0 R I V E R D I V E RS I O N H E AD
D I V E R S I ON
W OR KS
W E AR
5.1General
A wear is an obstruction or barrier constructed across a river the obstruction is of small
height in comparison with the dam, it is used for raising water level in river to divert the
water into the intake.
In this case the Hoha diversion weir site is located about 200m u/s of the bridge over the
Hoha River on the Assossa – Kurmuk road. The bed level of this 1460m a.s.l
The river bed is massive basaltic rock foundation.
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5.2 Selection of Types of Weir
Weir can be classified according to the material of construction and certain design futures
as
 Masonry weirs with vertical drop or vertical drop weir
 Concrete weir with d/s glacis
 Rocks fill weirs with sloping aprons
They are also classified as gravity and non gravity according to design aspect.
From the above types of weir for our scheme we select massonary weir with vertical drop,
because it is suitable for any type of foundation
5.3 Selection of site for Weir
Having decides up on the location of the weir the actual site is selected with the following
consideration.
1) A narrow, straight, well defined channel with defined banks is the best.
2) Availability of material of construction
3) Accessibility of the site to the road
4) Arrangement of diversion of the river
5.4 Design of Vertical Drop Weir
The complete design of a vertical drop weir consists of the design calculation for the
following
i) Hydraulic calculation for fixing varios elevation
ii) Design of a weir wall
iii) Design of impervious apron
iv) Design of inverted filter and d/s talus
i) Hydraulic calculation
Available data
 Qd = maximum flood discharge
= 20m3/s
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Assosa Water Supply Project
 River bed level = 1460m a.s.l
 Afflux = 1 (assumed)
 Lacey’s silt factor =1(assumed)
 Discharge concentration factor = 20%
 Coefficient of discharge =0.58
 Intake level of main supply line = 1460m a.s.l
Procedures
 Length of water way
L = 4.7 Qd3/2 ,where Qd =20m3/s
= 4.7 (20)3/2
= 13.67m
 Discharge (q) per unit width of the river
q= Q/L = 20/14
= 1.43m3/s/m
 Regime scour depth is calculated from lacey’s formula
q 2 1/ 3
R  1.35( )
f
1.43 2 1 / 3
R  1.35(
)  1.71m
1
, where f= silt factor
 Regime velocity (V) and velocity head (ha)
V 
q 1.43

 0.836m / s
R 1.71
V 2 (0.836) 2
ha 

 0.0356m
2g
2 * 9.81
 Design discharge over the crest of the weir is given by
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Assosa Water Supply Project
q = 1.7Hd3/2
Head over crest weir Hd = (q/1.7)2/3 = 0.891m
Water level and Total energy levels are calculated as follows
 Elevation of T.E.L = River bed level + head over crest weir +Weir height
= 1460 +0.89 + 1.5
= 1462.39m a.s.l
 Elevation of u/s H.F.L = u/s T.E.L -Va2/2g
= 1462.39- 0.0356
= 1462.3564m a.s.l
 Elevation of d/s T.E.L = u/s T.E.L – Afflux
= 1462.354 – 1
= 1461.354m a.s.l
 Elevation of d/s H.F.L = u/s H.F.L – Afflux
= 1462.354 – 1
=1461.354m a.s.l
5.4.1 Design of under sluice portion
Head over under sluice portion = u/s TEL - river bed level
=1462.39 – 1460.0
=2.39m
The under sluice portion is so designed that it can pass 20% of the maximum flood
=0.2*20 =4m3/s
Provide one under sluice with one meter width at the bottom of diversion
q= Q sluice/L sluice
=4m3 / 1m =4m3/s/m
Scour depth (R) for the sluice section
R=1.35(q2/f)1/3 =1.35(4/1)1/3 =3.402m
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5.4.2 Design of weir wall
The design of a vertical drop weir wall is similar to the of the over flow section of gravity
dam ; Incase of weir greater the discharge the discharge the smaller is the difference of
level between the head and tail water surfaces.
Preliminary section of weir wall
 top width
 bottom width (picture)
Top width (a)
Is fixed as the largest of the following
i) a= d/√G,, where d = maximum depth of water over the crest on u/s side)
d = u/s HFL – crest level
d = 1462.354 –1461.5
d = 0.354m
G = specific gravity of wall material, 2.24kN/m3
a = 0.354 /√2.24 =0.57
ii) a =0.552(H)*(√d) = 0.552*√1.5*√0.85
iii) a =3/2*d / √G =3/2*0.854/√2.24 =0.57m
Therefore take top width 0.65m which is the largest of the three results.
Stability check for weir
Under two states
State 1) when the u/s is at crest level or at level of crest shutter and there is no flow
State 2) when the water is passing over the weir crest and the weir is submerged
Bottom width (b)
 Can be calculated by equating over turning moment to the resisting moments taken at
the outer third point
 To find moment, we shall consider all the two states mentioned above.
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State 1) Head water at the crest level and no water to the d/s side
-overturning moment with no shutter (assuming no shutter is provided)
Mo =γH3/6
-Resisting moment
M r = γ HG/6(b2+ab-a2) , G=2.24KN/m3
Equating overturning and resisting moments
γH3/ 6 =γ HG/6(b2+ab+a2)
H2=G (b2+ab-a2)
1.52 = 2.24(b2+0.65b-0.652)
b2+0.65b- 1.427 =0
Solving quadratic equation, b=0.913m
State -2) The weir is just submerged
- Overturning moment
Mo= hH2/2
Resisting moment
M r = H (G-1)/12*(b2+ab-a2)
Equating the two moments
γhH2/2 = γ H/12 (G-1)(b2+ab-a2)
hH = (G-1/6)(b2+ab-a2)
where h=d, when the tail water is at crest level ,d and h be equal for this case the value of
d(head over the crest ) is given by drowned weir formula.
q= 2/3*cd*g2*d3/2
d=(1.52*g2/cd2*g2)2/3 ,where Cd – coefficient of discharge =0.58
d= 1.52*1.432/0.582*19.62
d=0.886≈1m= h
1*1.5 = (2.24-1)/6*(b2+0.65b-0.652)
By solving quadratic equation
b≈2.465m
Therefore take bottom width,
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Assosa Water Supply Project
B=2.46m (which is larger)
Depth of piles
 Bottom level of d/s pile=d/s HGL after retrogation-2R
Where d/s HGL after ret rogation = d/s TEL –Velocity head
=1461.39 -0.0356
=1461.354m a.s.l
 Bottom level of d/s pile =1461.354 – 2*1.71
=1457.934 m a.s.l
 Depth of d/s pile
d2 = bed level –bottom level of d/s pile
=1460m -1457.934m
=2.066m
Provide 2.5m depth of d/s pile
 Bottom level of u/s pile =u/s HFL -1.5R
=1462.354 - 1.5*1.71
=1459.789m a.s.l
 Depth of u/s pile= Bed level – bottom level of u/s pile
=1460m – 1459.789
d1=0.211m
Provide 1m depth of d/s pile
5.4.3 Design of Impervious Apron
For the under seepage, the worst condition will be when the water at the u/s is up to the
crest and there is no tail water.
If the floor of the weir is designed on’ Bligh Creep Theory’
 The total creep length(l)
L = CH
L = 9*1.5 =13.5m, where C= Creep coefficient = 9 (for gravel mixed with
sand and bolder)
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Assosa Water Supply Project
 D/s impervious flow
l1=2.21C*(H/13)1/2
=2.21C*(H/13)1/2
=6.75m
 U/s impervious floor length
l2=18-6.75-(2.46+2*1+2*2.5)
=1.783m (take 2m)
 Total length of d/s floor
l3=18*C*((H/10)*(q*75))1/2
=18*9*((1.5/10)*(1.43*75))1/2
=8.66m (take 9m)
Length of filter +launching apron=9-6.75
(Length of d/s protection work)=2.25m
But, minimum length of d/s concrete block = 1.5d2 = 1.5*2.5 = 3.75m
An inverted filter is provided immediately d/s of the d/s impervious floor beyond the d/s pile
to relive the pressure so that piping does not occur. The filter is properly graded with finer
layer at the bottom and its total thickness is usually between 50cm to 70cm (irrigation
structure and water power eng ’g).
Thus provide 1m*1m*1m concrete block over 0.5m thick inverted filter and the open joint
with 10cm gap are filled with river sand.
- Minimum length of d/s launching apron
= 1.5d2= 1.5*2.5=3.75
Length of d/s protection work =3.75+3.75 =7.5
U/s protection work
Minimum length of u/s concrete blocks = d1 = 1m
Provide 1m*1m*1m thick concrete block.
Minimum length of u/s launching apron at horizontal position = 1.5 * d1= 1.5*1 = 1.5m
Thickness of horizontal launching apron T
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T
5 * d1 * t
5 *t
5 *1


 1.118  1.2m
2d 1
2
2
Hence provide launching apron of length 1.5m and thickness of 1.5mand thickness of
1.2m
Actual creep length (L)
L = 2d1+L2+B+Ld+2d2
L = 2*1+2+2.46+6.75+2*2.5
L = 18.21m
 Residual head at the toe of the weir wall or at point B
hH
H
(2d1  Lu  B)
L
1.5
h  1.5 
(2 *1  2  2.46)
18.21
h  0.968m
Thickness of impervious floor (t) using Bligh’s theory
t
4
h
4 0.968
*
 *
 1.04m
3 G  1 3 2.24  1
Therefore provide thickness of 1.04m for d/s floor from just near its junction with weir wall.
hH
 Residual head at point C,(that is 3m from toe)
Thus thickness (t ) 
4
h
4
0.81
*
 *
 0.77  0.8m
3 G  1 3 2.24  1
H
* (2d1  Lu  B  3)
L
1.5
h  1.5 
* (2 *1  2  2.46  3)
18.21
h  0.72m
Therefore, provide thickness of 0.8m for the next 3m from point c
 Residual head at point D, at a distant of 5.5m from the toe of the weir wall
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hH
H
* (2d1  Lu  B  5.5m)
L
1.5
h  1.5 
* (2 *1  2  2.41  5.5)
18.21
h  0.51m
Thus thickness (t ) 
4
h
4
0.64
*
 *
 0.548m
3 G  1 3 2.24  1
Therefore provide thickness of 0.6 for the last 1.25m of d/s impervious floor.
And also provide a nominal thickness of 0.5m below the u/s floor and below the weir wall.
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Fig 5.1 dimensions of same of weir body components
Check for thickness by Khosla’s theory referring the above figure.
I. Exit gradient
II.
Uplift pressure
Exit gradient
- Total length of the impervious floor
b = 2m+2.41m+6.75m = 11.16m
Depth of d/s pile (d2) = 2.5m
Thus exit gradient (GE) =

b 11.16

 4.464
d2
2.5
H
1
1
1.5

 0.114  1      safe!
*
*
d 2   2.5  2.79
Because safe exit gradient for course sand
with mixed gravel of weir foundation is
1
1
1  (1   2 ) 1  (1  4.464 2 )

 2.787 between 5 and 6 .(source
2
2
and water power eng ‘g, PUNMIA)

irrigation
structure
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Assosa Water Supply Project
II. Up lift pressure
a) u/s pile

b= 11.17m
b 11.16

 11.16
d1
1
1  (1   2 )
 6.1
2
1
 2 1
1  6.1  2 
 E 1  * cos 1 

  * cos 

 6.1 
   
 E 1  26.53%

d1 = 1m
 C1  100   E1  100  26.53  73.47%
 D1  100   D where D 
  1 1
1  6.1  1 
* cos 1 
  * cos 


   
 6.1 
1
 D  8.48%
 D1  100   D  100  18.48  81.52%
Correction for C1
Correction due to mutual interference
Correction = 19 *
D (d  D)
*
b'
b
Where b’= the distance between two piles
D=the depth of the pile line, the interference of which has to be determined on the
neighboring pile of depth d. D is to be measure below the level at which interference is
desired.
b= total floor length
b’=b=11.21
D= 2.5-0.5=2m and d = 1-0.5 = 0.5m
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Assosa Water Supply Project
Correction = 19 *
D ( d  D)
2
(0.5  2)
*
 19 *
*
 3.78%(ve)
b'
b
11.21 11.21
Thickness correction
    C1 
 81.52  73.47 
 * t  
  D1
 * 0.5  4.025%(ve)
d
1
m


1


Correction due to slope at C1 is nil, as C1 is neither situated at the start nor at the end of
the slope.
Therefore corrected  C1  73.47  4.025  3.78  81.275%
B. d/s pile
b=11.21m

d2= 2.5m
b 11.21

 4.484
2.5
d2
1  (1   2 )
 2.79

2
1
 2 1
1  2.79  2 
 E 2  * cos 1 
  * cos 


   
 2.79 
1

 E 2  * 73.39 *
 40.77%
180

1
  1 1
1  2.789  1 
 D 2  * cos 1 

  * cos 

 2.789 
   
1

 D 2  * 50.09 *
 27.83%
180

 C 2  0%
Correction for  E 2
Correction due to mutual interference
b=b’=11.21
D=1-0.6=0.4 &
d=2.5-0.6=1.9
Correction = 19 *
0.4 (1.9  0.4)
D ( d  D)
*
*
 19 *
 1.62%(ve)
11.21
11.21
b'
b
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Thickness correction
    D2 
 40.77  27.83 
 * t  
  E 2
 * 0.5  2.588%(ve)
d2
2.5m




Note:-the correction is negative because the pressure at E2 shall be less than at E2’
Correction due to slope
Correction due to slope is nil, as the point E 2 is neither situated at the start nor at
the end of a slope.
Hence corrected  E 2 =40.77% - 1.62%- 2.588% = 36.57%
 Percentage pressure at toe B, C,D
Percentage pressure at B
    E2 
 A  E 2   C 1
 * Ac '
b


 81.27  36.57 
 A  36.57  
 * 6.75  63.48%
11.21


Thus residual head at pt, B = 2*63.48% = 1.269m
Thickness of floor at B, t 
h
1.2965

 1.04 ………………………………..ok!
G  1 2.24  1
 Percentage pressure at pt, C
    E2 
 B  E 2   C 1
 * Bc '
b


 81.27  36.57 
 B  36.57  
 * 3.75  51.52%
11.21


Thus residual head at C = 2*51.52% = 1.03m
Thickness of floor at B, t 
h
1.03

 0.83 >0.8 ……………………….ok!
G  1 2.24  1
There fore take t =0.83m from the 2.5m of the d/s impervious floor.
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 Percentage pressure at D
    E2 
 C  E 2   C 1
 * Cc'
b


 81.27  36.57 
 C  36.57  
 * 21.25  41.55%
11.21


Thus residual head at D = 2*41.55% = 0.83m
Thickness of floor at B, t 
h
0.83

 0.63  0.6...................ok !
G  1 2.24  1
Therefore take t=0.63m from the last 1.25m of d/s impervious floor.
%age presser at D
ØD=Ø+
c1   E 2
b
=36.57+ (
* DD /
81.27  36.57
) *1.25 =41.55%
11.21
Therefore; Residual head at D=2*41.55%
=0.83m
Therefore; Thickness of floor at D
t=
h
0.83
=
=0.63……………..Ok
G  1 2.24  1
t=0.63m>0.6m
Therefore take t =0.63m from the 1.25 of downstream impervious floor.
5.5 Water profile at weir site
5.5.1 Water profile u/s of the weir
 Water profile u/s of the river
 Effect of back water curve on structure
 To suggested weather river training works are required or not
 To determine the minimum height of river banks u/s of weir.
 To find out whether the water surface is high enough to deliver the required
discharge to the off take canals
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Water profile d/s of the weir
Required, to
 Carry out the stability analysis of the weir.
 Design the weir structurally
 Design the d/s ring wall and protector
Hydraulic jump
To check weather the hydraulic jump is occurs or not, if the d/c between u/s and d/s
energy grade line is very high, there fore the energy must be dissipation be fore it reaches
the natural river course, other wise it causes damage to the banks and d/s of the apron.
Hydraulic jump is used to dissipate energy which is caused to increase by (He)
He= (
q 0.66667
Vo 2
)
=hd+
1 .7
2g
= 0.891m
Applying Bernoulli equation between section 1-1 and section 2-2 take reference datum
river bed level Zo=Z1
Zo+P+hd+
V 12
2g
Where Zo=Z1 and He= hd+Va2/2g
P+hd+
Va 2
V 12
=Y1+
2g
2g
P+He= Y1+
2.39= Y1+
V 12
V 12
=1.5+0.89= Y1+
2g
2g
V 12
……………………….Equation 1
2g
Applying continuity equation i.e. Q=VA
Where,
Q=q/B=V*B*Y,
 V1=q/Y1………………………………………Equation 2
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Solving equation1 and 2 we get;
19.62Y13-46.89Y12+2.045=0 …………...Equation 3
Solving equation 3 by trial and error
Y1=0.219
V1=q/Y1= 1.43 =6.53m/s
0.219
Fr1 (Fraud number)=

V1
g * y1
=
6.53
9.81 * 0.259
=4.455
Critical depth (Yc)=(q2/2g)0.66667=(1.432/2*9.81)0.66667=0.6575
Y1=0.219<Yc: 0.22; therefore there will be jump.

The conjugate(sequent) depth(i.e. Y2)
Y2= Y 1 *[ (1  8Fi 2 )  1 ] =0.1095[ (1  8 * 4..455)  1
2
Y2=1.274m
Head loss (HL);
HL=
(Y 2  Y 1)
4Y 1Y 2
=
(1.274  0.219) 3
=1.032m
4 *1.274 * 0.219
Length of jump
L=5(y2-y1) = 5(1.274-0.219) = 5.275
= 5.275 < 6.75(d/s impervious length)…….ok!
The down stream floor length (Apron length )should be greater than or equal to the length
of the jump to accommodate the hydraulic jump in it.
The Apron length determined for satisfying seepage requirement is enough to
accommodate the jump.Finally, it is calculated that measures such as using chute blocks
and end sill or extending the length of the down stream impervious floor which are used to
bound the jump in the downstream impervious floor are not necessary.
5.6 STRUCTURAL ANALYSIS OF WEIR
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Forces acting on weir body
 Hydrostatic pressure of the water
 Soil reaction at weir base
 Uplift pressure
 Friction force at the base which balance the horizontal force
 Weight of weir and water wedges
The weir should be designed stable against
- Overturning
- Sliding
- Over stress
Where, The analysis of the weir is done for static condition.
Static condition
This is the case where by Water is impounded u/s of the weir and no over flow condition is
observed.
γw =unit weight of water = 10kN/m3
pw =water pressure
M = unit weight of masonry (23KN/m3)
Forces acting on the weir
Hydrostatic force
Pw = ½*γw*H2*L =1/2*9.81 KN/m3*1.52*1 , where γw= 9.81KN/m3
=11.036KN
- Weight of weir
W 1 =M*A*L =24 KN/m3*0.65*1.5*1 , where M= unit wt of masonry
=24KN/m3
W 1=23.4KN
W 2=M*A*L=24 KN/m3*1/2*1.81*1.5*1
W 2=32.58KN
- Uplift pressure
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U=1/2* γw *H*b =1/2*9.81*1.5*2.46
U= 18.099 KN
Type of Vertical force(KN)
Horizontal force(KN)
force
Level
Moment about the
arm
toe(KNm)
from toe
(+ve)
(-ve)
Pw
(+ve)
(-ve)
11.036
(m)
(+ve)
0.5
5.518
(-ve)
W1
23.4
2.135
49.96
W2
32.58
1.206
39.29
U
18.099
∑H =55.98
∑v=18.099 ∑H=11.036
1.64
29.68
∑M=35.198 KN.m
∑M=89.25 KN.m
∑M=∑M-∑M
∑M=54.052KN.m
1) Check for overturning
∑M+/∑M- ≥1.5, 89.25/35.198 =2.53 > 1.5 ……..ok! Safe
2) Check for sliding
∑H/∑v < 0.75, 11.036/37.881=0.2913 < 0.75 ...ok! Safe
3) Check for over stresses
In order to avoid lifting up the structures heel forces and tension occurrence at the base. The
forces must pass through the middle third of the structures base.
. The forces must pass through the middle third of the structures base (i.e. (e=
B
-X<B/6))
2
X=∑M /∑V =54.052/37.881=1.426
e=
2.46
B
-x =
-1.426=0.196
2
2
e<B/6=2.46/6=0.46
0.196=e<=B/6=0.46, therefore it’s ok!
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 Thus the structure is safe for all condition.
5.7 Design of Retaining Wall
This is a structure place at stone angle normal to the direction of the fill of the water to
retain the end of the approach and, to support the loads of the structure built . In order to
prevent out flanking of water towards the main canal and safely over pass the discharge
during high flood time. It is usually constructed from masonry or gabions and provided at
both banks.
Data available
Height of retaining wall above the foundation
=H++freeboard
=1.5+0.946+0.5m (assumed)=2.9463m
Top width of retaining wall=0.4m (with vertical water face)
Assuming back slope of 1V:0.5H
Angle of repose =30o
 Ranking coefficient of active earth pressure
Ka=
1  sin  1  sin 30
=
=1/3
1  sin  1  sin 30
5.7.1Structural analysis of retaining wall
Force acting on a retaining wall
- Weight of retaining wall
- Weight of soil
- Active earth pressure
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- Water pressure
- Uplift earth pressure
1) Weight of retaining wall
W1=A1*m*l=0.5*3*24*1=36KN , where m= Unit weight of massonary
= 24KN/m3
W2= A2*m*l=0.5*3*1.5*24*1=54KN
W3=A3*m*l=0.5*2.46*24*1=29.52KN
2) Weight of soil
W4=A4*s*1=0.5*1.5*3*19*1=42.75 KN
, where s = Unit weight of soil
= 19KN/m3
W5=A5*s*1=0.5*3*19*1=28.5 KN
3) Active earth pressure
PA=0.5*Ka*s*H2=0.5*1/3*19*32=28.5KN
4) Water pressure
Pw=0.5*w*H2*1=0.5*9.81*1.52*1=11.036KN
5) Uplift earth pressure
U=0.5*H*w*B*1=0.5*3*9.81*2.46*1=36.199KN
Table 5.3 Forces and moments on retaining wall
Type
Vertical
Horizontal
Level
of
forces(KN)
forces(KN)
arm
forces +ve
-ve
+ve
-ve
Moment about toe KN.m
from toe +ve moment -ve moment
(m)
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W1
36
0.25
9
W2
54
1
54
W3
29.52
1.23
36.309
W4
42.75
1.5
64.125
W5
28.5
2.25
64.125
PA
28.5
Pw
11.06 0.5
U
∑
36.199
∑V+=190.77
28.5
5.518
1.64
∑H+=11.036 ∑H-
∑V-=36.199
∑V
1
59.36
∑M+=227.56 ∑M-=93.378
=28.5
=
154.571KN.m
∑M=∑M+-∑M-=227.56-93.378=134.182KN.m
,
∑H =17.464KN.m
a) Check the stability against over turning
M
Fso=
M
1. For no water condition


=
227.56
=7.78>1.5 ….OK!
28.5
, where ∑M-=excluding hydrostatic and
uplift force
2. Normal flow condition
Fso=∑M+/∑M-=227.56/93.378=2.44>1.5…… ok!
b) Check stability against sliding
∑H/ ∑V=17.46/154.571= 0.113<0.75…….Ok safe!
CHAPTER SIX
INTAKE AND PUMP
6.1 INTAKE
6.1.1 GENERAL
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The intake or intake works are the structures placed in a surface water source which
consists of the opening strainers or grating through which the water entered and the
conduit conveyed the water, usually by gravity , to a well or sump. From the well or sump,
the water is pumped to the mains or treatment plant. The structure may be of stone
masonry, brick masonry and R.C.C. or concrete blocks. It is to be constructed water tight
and it should be designed for all forces likely to come upon it including the pressure due
to water wave action, wind, floating derbies etc
6.1.2 SITE SELECTION
Success of an intake greatly depends up on its location while selection of sie for an
intake, the following mentioned points should be kept in mind
1. The intake site should be such that the water available is of best quality and can be
easily and economically purified.
2. The intake should be located at such a site where sufficient quality of water remains
available under all circumstances.
3. Site should be such that the intake work can provide more quality of water if required in
future for expansion of water works
4. At selected site the velocity the velocity of flow in the source should gentle, otherwise
heavy currents may endanger the safety of the intake towers and stirrup the silt from
the
bottom.
5. The site should be easily accessible without any obstruction and should be free from
the
effects of loads.
6. It should be near the treatment plant so that conveyance cost from the source to the
works can be minimized.
7. It should not located in navigation channels generally polluted and may also damage
the structure.
8. It should not be on curve in case o meandering river. If there is no alternative, then
intake should be located on the outer bank and not on the inner bank. Water
concentrated more near the outer bank although erosion problem will be
there. Inner bank always keeps on silting and this may bock the intake and put
it out of commission.
9. It should be located on the upstream side of the town. Water will not be contaminated
due to sewerage disposal.
6.1.3 River Intake Types
Types of river intake are
1. Intake wells
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2. Weir intakes
3. Pipe intakes
1. Intake wells: this is a circular masonry tower, two to six meters in diameter, provided
on
the bank. The water flows in to the intake wells through the
penstocks located at
different levels. The penstocks can be
closed or opened with valves. These penstocks
face down stream
to avoid the entry of silt. This type of intake is adopted for rivers,
which have sufficient flow through out the year.
2. Weir Intake: water is drawn from the through a channel into a sump well. For rivers
which are more or less dry in summer, a weir is built across the river to the required
height, to store excess floodwater during the rainy season. A channel is provided from
the river to the intake well from where water flows by gravity or is pumped.
3. Pipe intake: when a small quantity of water to be drawn the pipe intake ids
economical.
The required number of intake pipes is laid across the river
bed. The pipes are
supported on masonry blocks at regular
intervals. The ends are provided with strainers
and anchoreerd firmly by
masonry blocks. The pipes carry water to a jack well.
6.1.4 Design consideration of Intakes
The design of intake structure is generally specific. Rarely can a standard design be
adopted for a given site without major modification s. Some design considerations that
must be addressed in the design of each structure are listed below.
1. Intake Velocities: The velocity water entering the intake port is the single most important
design value to be selected by an engineer. High intake velocities increase headless,
entrain suspended matter , trap fish other aquatic animals .low velocities on the other
hand require the intake port to be larger and so add to the cost of the structure
2. Intake Port Location: properly designed intake structures should provide water provide
water treatment plant operators the flexibility to draw water from the stream with the
best
water quality. In order to achieve this multiple intake ports set at
various levels are
generally provided.
3. Gates: are used in intake structure to control inflow of water from the raw water source
into the water conveyance system. Gates typically in intake structures are sluice gates.
4. Coarse screen-trash rack: intake ports should be equipped with a coarse screen to
prevent large objects form entering the conveyance systems
5. Fine screen: Fine screens are used to remove smaller objects that nay pumps or other
equipment.
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6.1.5 Design of Intake structures
6.1.5.1 The inlet opening is sized such that from maximum day demand can be withdrawn
from a single level at a maximum recommended velocity of 0.15 m/sec (Water Treatment
Plant Design, AWWA)
Q=0.119m/s, V=0.15m/s
A=Q/V = 0.119/0.15 =0.79m2
Using square opening the sixe will be, b= √A =√0.79 =0.89m
Therefore, square opening of 0.89*0.89 is to be provided
6.1.5.2. Design of Coarse Screen
The coarse screen will be located at the intake slightly projected away from the intake
gate. This will prevent derbies from interfering with the operation of the gate. The coarse
screen will be constructed of flat bars attached to concrete seat projected from the intake
tower. It is typically installed vertically
The recommended spacing between bars is (50-100mm) Taking spacing of 50mm.
Provide bar with size 10mm*50mm at a center to center spacing of 50mm.
Check of head loss
Using Kirecher’s formula for head loss through bar screen
hl=  (
w 4/3 * Va 2
)
*Sin  <0.15 ,(maximum head loss)
b
2g
where hl = head loss
w = width of bars
b = clear space between the bars
Va = velocity head

= angle wich the bars make with the
horizontal
 = dimensional coefficient that is a function of
bars geometry and is equal to 2.42 for rectangular b
10 *10 3 4/3 (0.15)2
Hl=2.42(
) *
*Sin (90) =3.246*10-4m
3
2 * 9.81
50 *10
=3.246*10-4mthemaximum
head
loss=0.15m………………….ok!
Provide two ports bottom and top of intake tower. The bottom port should be at
least 1mabove the bed of the river to prevent rolling bodies to enter in to the ports.
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6.1.5.4 Design of Fine Screen (strainer)
Fine screens are provided on the two circular intake conduits to trap sediments that are
too small to be captured by coarse screen.
Design Criteria (Water treatment plant design, AWWA)
1. Total area of clear opening in screen 200% or more than the area of channel
protected by the screen.
2. Maximum head loss should be limited to 0.8-1.5m
3. Velocity through hole is 0.15-0.3m/s. it is recommended that the velocity be near
the
lower limit to prevent the entry of impurities.
4. Opening the strainer hole be 6-9mm.
5. Discharge through screen= 119*10-3m3/s
Assume (take) velocity through the screen 0.18m/sand height of strainer 0.45m.
Aeff = Q/V = (119*10-3m3/s) /0.18m/s
= 0.661m2
Agross = 2 * Aeff = 2 *0.661 = 1.322m2
Perimeter = Agross / height = 1.322m2/0.45m=2.94m
Diameter = perimeter / 3.14 = 0.936m
Provide cylindrical strainer of height 0.45m and diameter 0.936m
6.1.5.5 Sump Well
 In design of sump well care must be exercised
 To maintain sufficient depth of water to avoid air entry during drawdown
 To obtain uniform distribution of inflow
Design criteria
 Detention period = 30min(t)
 Velocity in suction ppipe =2m/s (1.5-3m/s)
 Ratio of bell mouth diameter of suction pipe =D/d (1.5-2)
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Assosa Water Supply Project
 Minimum width of sump well =2nD
Where n=number of units
D=diameter of suction pipe
 Submergence of pump to prevent air entry, H> 2D
 Bottom clearance of suction bell mouth= D/4 to D/2
Volume of sump well = t* Q = 30*60*119*10-3 =214.2m3
Diameter of suction pipe
Velocity in suction pipe is 2m/s
A=Q/V=0.5*0.119/2 =0.02975
A=∏d2//4
d=√4A/∏ =√4*0.02975/∏
d=0.195m
195mm diameter suction pipe is to be provided for each pump
Opening of bell mouth, D/d =1.8 D=1.8*d
D=1.8*0195 = 0.351m (351mm)
Minimum water level to prevent entry of air during drawdown is H min= 2*D =2*0.351=
0.702m
Assume depth below river bed =1.5m
Min= width of sump well =2nD =2*3*0.702 =4.212
Take width 4m
Provide free board = 0.5m
Depth from top of ground level river bed
Total depth =0.5m + 1.5m +3m =5m
Correctional area =214.2m3/5m =42.89m2 ,providing rectangular , dimension of sump is
4*11m
6.1.5.6 Design of Overflow Conduit
The pump unit installed for the first phase should have a capacity of maximum daily
demand for the year 2018. But the sump well is designed for maximum daily demand for
the year 2028. Therefore, there is an overflow discharge.
Q overflow=Incoming discharge-Pumping rate per 2018
=119-39.44
=79.56l/s
Taking velocity of 0.6m/s,
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Assosa Water Supply Project
Area of overflow conduit = Q overflow/V =79.56* 10-3m3/s/0.6m/s =0.1326m2
D=√4A/∏ =√(4*0.1326)/∏ =0.411m
450mm diameter pipe for overflow need be provided.
6.1.5.7 Design of delivery pipe
Delivery pipe designed for the maximum day demand for the year 2028 which is 119l/s.
Assuming a velocity of 0.9m/in the pipe, the required diameter of pipe will be D=
√4*0.119/0.9*∏ =0.410m , take D=450mm
So provide a 450mm cast iron main is provided.
6.2 Construction
The intake structure should be constructed from R.C.C structure to with stand the water
pressure and some impact force due to the rolling stoned and logs. The foundation depth
should be below the normal scour depth the normal scour depth. The steel sluice gates
are provided at the openings or the impacts should have a capacity to resist lateral
pressure
6.3 Pump
6.3.1 General
Pump is a mechanical device to increase the pressure energy of a liquid. In most of the
case pump is used for raising fluids from a lower level to a higher level. This is achieved
by crating a low pressure at the inlet or suction end and high pressure at the outer or
delivery end of the pump.
6.3.2 Pumping design parameter Capacity
The capacity of a pump is the volume of liquid pumped per unit of time which usually are
measured in liter per second or cubic meters per second
Head
In pump system the head refers to both pump systems having one or more pumps and
corresponding piping system. The height to which a pump can raise liquid is the pump
head and the head required to overcome the losers in a pipe system at a given flow rate is
the system head. The head against which the pump must work when water is being
pumped is called total dynamic head.
Efficiency
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Thee pump should be selected to operate near its peak efficiency point. In typical water
supply applications, pumps operate over a bond of head conditions. Therefore they can
not operate at their peak efforts all the time.
Shut off and Run out
Continued operation condition will result in damage to the pump from over heating. The
higher discharge head can dam age piped and valves from excessive pressure and higher
power requirements can overload pump driver.
6.3.3 Determination of pipe size from sump well to treatment plant (phase 1)
a) diameter of suction pipe
q = 39.44l/s
Velocity of flow in the pipe may vary from 0.8m/s to 1.8m/s
Take v = 1.8m/s
A = Q/V = (39.44*10-3m3/s)/1.8m/s
= 0.0219m2
Area=
D 2
4
, d=

4* A
=

4 * 0.0219
=0.167m=167mm
So provide d=200mm (available in the market)
 Check velocity of flow
V
Q 0.03944

 1.255m / s
A  * 0.2 2
4
V  1.255m / s  1.8m / s.........ok!
b. Diameter of delivery pipe
To determine economic diameter of pumping main we use lea formula
D  0.97to1.22 Q
Where Q is in m3/s
D-economic diameter of pipe in (m)
Assuming, D  1.00 Q  1.00 39.44 *10 3  0.199m  199mm
Provide pipe size of D=200 mm (market available)
V
 Check velocity of flow
Q 0.03944

 1.255m / s
A  * 0.2 2
4
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Assosa Water Supply Project
V  1.255m / s  1.8m / s.........ok!
6.3.3.1 Determination of total lift of the pump
The total head against which the pump should work includes discharge lift and total loss of
head due to friction, entrance, exit etc in suction and rising mains
hf 
fLV 2
2 gd
Where h f =head loss in (m)
L=length of pipe in (m)
V=velocity of flow in pipe(m/s)
g=acceleration due to gravity
f=dimensionless friction factor
Cast iron is selected for rising main. Length of pipe from pumping station to treatment
plant.
L =0.555Km=555m
f=0.02(for cast iron)
V
Q 0.03944

 1.255m / s
A  * 0.2 2
4
I) h f 
fLV 2 0.02 * 555 *1.255 2

 4.46m
2 gd
2 * 9.81 * 0.2
II) Entry loss, he 
V2
(1.255) 2

 0.040m
2 * (2 g ) 2 * (2 * 9.81)
III) Head loss due to construction such as valve pipeline (h1)
h1 
kV 2 0.5 * (1.255) 2

 0.0401m
2g
2 * 9.81
k=0.5 for circular pipe
H L  h f  he  h1  4.46  0.040  0.040  4.54m
Static head against which the pump lift is
Hs= (elevation of treatment plant –elevation of sump well) +depth of sump well below
ground level)
Hs= (1470-1460) +5=15m
Hence total dynamic head (Hf) =Hs+HL
Hf =15+4.54
=19.54m
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Therefore for the first phase two pumps one as stand by each discharging a capacity of
39.44l/s and head of 19.54m required for the second phase one additional pump with
delivery head of 19.45m is required.
6.3.4 Power required of pump (phase i )
The water hoarse power = qwh/75
= 39.44*10-3*1000*19.54/75
= 10.28Hp
The power input to the pump =
Pp = WHP/EP, where = efficiency of pump
= 10.28/0.8 = 12.85HP
Power required of pump for the second phase
I.e. from 2018 – 2028
The water horse power = QWH/75 , where Qmax = maximum flow rate at 2028 =
119l/sec
WHP = QWH/75 =119*10-3*1000*19.54/75 =31.00Hp
6.3.5 Determination of pipe size from clear water well to reservoir (phase I)
a)
Diameter of suction pipe
Q = 39.44*10-3
V = 1.8m/s…………range between 0.8 to 1.8m/s
A =Q/V = 39.44*10-3/1.8 = 0.0219m2
d=

4* A
= 0.167m =167mm
Therefore provide d =200mm (market available)
b)
Diameter of delivery pipe
To determine economic diameter of pumping main we use lea formula
D = 0.97 to 1.22 Q
, take d = 1.00 Q =0.199m =199mm
Therefore provide D = 200mm (market available)
6.3.6 Determination of total lift of the pump from clear water well to service
reservoir
(phase I)
Friction loss (hf) =flv2/2gd
Cast iron is selected for the rising of main length of pipe from clear water well to
service reservoir is 5km
L = 13888.88m=13.88Km
F = 0.02 (for cast iron)
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Assosa Water Supply Project
V= Q/A= 0.03944/3.214 *0.22/4 =1.255m/s
i) Hf = flv2/2gd = 0.02*13888.88*1.2552/2*9.81*0.2= 111.49m
ii) entry loss(he) = v22(2g)+ 0.04m
iii) loss due to valve and pipe fitting hl = kv2/2g= 0.5*1.2552/2*9.81=0.04m
HL =hf+he+hl = 111.57+0.04+0.04= 40.22m
Static head against which the pump lift
Hs = elevation of service reservoir – elevation of clear water well+4.42m
= 1720 -1470 +4.42
= 254.42m
H is to prevent air entry
Total dynamic head = 254.42 +111.57 = 365.99m
Therefore from treatment to service reservoir two pumps is needed one as stand by each
discharging a capacity of 39.44l/s and the head of 365.99m is required for the second
phase one additional pump is required to deliver the head of 365.99m
6.3.7 Power requirement of pump from clear water well to service reservoir
The water horse power =QWH/75 = 39.44*10-3*1000*365.99/75
= 192.46Hp
The power input to the pump
Pp = WHP/Ep = 192.46/0.8 240.58Hp
Where, Ep = efficiency of pump
Power required for the second phase
The water horse power =QWH/75= 119*10-3*1000*365.99/75
= 580.7Hp
6.3.7.1 Selection of Suitable type of pump Criteria for the selection of type of pump
For proper selection of a pump it is necessary to brave certain essential data on the pump
installation
The information should includes
1) Nature of liquid to be pumped
2) Capacity of pump
3) Suction condition
4) Discharge condition
5) Total head
6) Location of geometrical in doors ,outdoors ,elevation….etc
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After having the idea of essential data on the pump installation the following points may be
kept in mind while selecting any pumping necessary for the water works.
1) Reliability of service – it should be reliable and should not fail suddenly and
cause trouble
2) Capacity – it should be capable of pumping required quantity of water
3) Cost –it should be cheap in initial cost
4) Power – The which issued for running pump should be available easly at low
cost
5) Maintenance – the maintenance cost of running ump be as less as possible
6) Efficiency – pump should have high efficiency
7) Depreciation – pump should have long life and depreciation
8) Promptness – it sho0uld be prompt enough in service
Considering all the above factors, among the available type of pump we select centrifugal
pump out of the other type of pump, because centrifugal pump is used in most water
supply system due to its satisfactory and economical than the other type of pump and
fulfils all the above criteria.
Generally from the above consideration for Assosa town water supply projects to lift water
from sump well to treatment plant one working and one stand by multistage centrifugal
pump which has a capacity to lift a head of 19.54m and a discharge of 0.03944m3/s
connected parallels should be provided and also to lift a water from clear water well to
service reservoir .for the second phase two working and one stand by multistage
centrifugal pump is required which has a capacity to lift 365.99m from sump well to
treatment plant and a discharge of 0.119m3/s and which has a capacity of lift 19.54m from
clear water well to service reservoir and a discharge of 0.119m 3/s connected parallel
should be required
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CHAPTER SEVEN
7.0 WATER QUALITY AND TREATMENT
7.1. Objective of treatment
The objective treatment process is to remove all the undesirable impurities, to the extent
where they do not cause any trouble to human health and water is available to the
consumers as per health standards. Following may be the objects.
 To remove color, dissolved gasses, and murkiness of water.
 To remove objectionable taste and color.
 To kill the troublesome bacteria
 To estimate the corrosive properties of water .this treatment is essential
from
pipes and pipe fittings safely point of view.
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 To make safe for drinking and domestic purpose, and also for various
industrial purposes like brewing, dying, stream boilers etc.
7.2. Water quality analysis
The quality of raw water should be compared with the world health organization (WHO)
drinking water quality guide lines and treatment will be done for unacceptable parameters.
In order to remove suspended solids, biological and microbiological impurities, high
turbidity due to iron and floating or other non chemical pollution, color odor and taste an
appropriate and sound treatment must be carried out before supplying the water for the
community.
7.3. Treatment process
Treatment process have been given here .it is not essential that all these process will
have to be employed at all places, but it depends up on the quality of raw water .in the
case of raw water ,obtained from lakes, screening and sedimentation are not required
,because suspended and floating debris have already settled in the lake basin. But
aeration is a must because lake wastes have generally no treatment is required. Only
disinfection may be done and supplied to the consumers therefore, the character and
degree of treatment directly depends up on the nature of water or in other words on the
source.
 Screening –used to exclude floating mater, it is done just at the intake.
 Aeration – employed where elements causing taste and odor have to be removed.
 Sedimentation and coagulation – in this process suspended impurities like silt,
clay,
sand, and some bacteria are removed.
 Filtration – employed to remove very fine particles and colloidal matter which may
have escaped from sedimentation process ,micro-organisms are also removed
largely.
 Other processes – used in specific cases ;
- removing hardiness if it is beyond permissible limits.
- Removing color taste and odor if any.
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Assosa Water Supply Project
-Removing iron and manganese if present.
-maintain fluorides between 1to 1.5 p.p.m by adding if in deficiency
and removing if in excess.
7.4 Treatment plant location
Correct location of the plant plays a very important role in its success. A comparative
study of site involves a number of considerations such as
 All the processes should be located in such a sequence that water may be
flowing
from one process to the other.
 Elevations of different processes should be such that no pumping is required and
water keeps on flowing from one plant to the other.
 All plants should be located in such a way that minimum area is covered by it
.adequate place for the future extension.
 Residual colony should be located by the side of the water works. This facilitates
better working and control on the working of different units.
 A well establishment laboratory should be located at the site so that the quality of
water may cheeked, before treatment and after treatment
 Physical characteristics of the site will affect construction. Flooding, foundation
conditions, ground water level and site preparation including clearing, grading
and drainage are factors which influence directly the cost of the plant.
 In the final analysis selection of a site will depend on analyses involving
engineering and operation costs and a judgment of the factors which can not be
expressed in dollar values. As mentioned above the site should be adequate for
ultimate needs surplus land is preferable.
Lay out of treatment plant
A complete water treatment plant consists of;
 Intake works including pumping pumping plant
 Sedimentation and coagulation
 Filtration
 Disinfection
 Pure water storage reservoir
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Assosa Water Supply Project
 Pumping plant for pumping purified water from clear water storage reservoir to
the
service reservoir
 If water is hard, sometimes, softening plant is used, before the filtration plant.
7.5 Treatment plant units
The most common treatment units used in the treatment of surface water are the. They
are selected based up on the quality of the raw water to be treated.
The order of various treatment plants is
1. balancing chamber
2. mixing chamber
3. flocculation
4. sedimentation
5. filtration
6. disinfection
7. clear water well
7.6. Design of treatment plant units
7.6.1. Balancing chamber
It is located above all the components so that gravity flow to the treatment is maintained.
From the intake structure the raw water is conveyed to the balancing chamber. The
balancing chamber is recommended to make the flow current free (stabilization of
current) so that the work of the treatment plant is not disturbed highly and continuous
flow is attained.
Maximum day demand=0.055217 m3/sec
Detention time=15 min
Volume =0.055217*15*60=49.695≈50m3
Therefore provide 50m3balancing chamber in the first and second phase.
7.6.2. Mixing chamber
Flush mixer is used to provide rapid mixing so as to disperse the coagulants in the raw
water.
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Assosa Water Supply Project
A mechanical flush mixer is used because they have a reliable and continuous supply of
power and they are efficient.
Capacity of flush mixer = maximum day demand * detention time
Assuming detention period = 45sec
Capacity of flush mixer = 0.055217m3/sec*45sec=2.485m3
Therefore provide and 1.8m depth of mixing chamber.
7.6.3. Flocculation
The term flocculation is used to denote the process of floc formation. floc is a thick
insoluble gelatinous precipitate produced when coagulant is added to the water and
thoroughly mixed. The floc has the property of arresting the suspended impurities in
water during its down ward settlement to wards the bottom of the tank. In the flocculation
chamber, after sever agitation, mixer is kept slowly agitated for about 30min.during this
period, coagulants get necessary contact opportunity and develop floc after the
development of floc, the mixture is still kept slowly agitated to increase the number of
collisions or contact with water. The slow agitation of mixture after the vigorous agitation
is called flocculation.
The two types of flocculators are mechanical flocculator and hydraulic flocculator.
In Mechanical flocculator the velocity gradient is created by mechanical power input. They
also require extensive maintenance and the system is complex.
In hydraulic flocculators hydraulic energy provides the necessary velocity gradient.
However, due to these difficulties with mechanical flocculation techniques, flocculators
using hydraulic energy are appropriate for water treatment plant in developing countries.
The most widely used hydraulic flocculator is the baffled channel flocculator (Schulz and
Okun 1984) .it is widely used in developing countries and performs efficiently over a wide
range of flows.
Baffled channels can be classified in to horizontal flow and vertical flow
flocculator. Horizontal flow flocculator with around the end baffles are some times
preferred over vertical flow flocculator with over and under baffles because they are
easier to drain &clean
7.6.3.1 Design of horizontal baffled channel flocculator
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Assosa Water Supply Project
With a horizontal baffled channel flocculator, mixing is accomplished by the change in
direction of flow of water through channels this change creates head loss in the channels,
which in turn creates a velocity gradient for slow mixing.

Guidelines for the design and construction of baffled channel flocculator
horizontal.
 Distance between baffles should not be less than 45cm to permit cleaning.
 Clear distance between the end of each baffle and the wall is about 1.5times the
distance between baffles, should not be less than 1.0m.
 Depth of water should not be less than 1m.
 Avoid using the asbestos-cement baffle because they corrode at the PH of alum
coagulation.
 Decay resistance timber should be used for baffle; wood construction is preferred
over metal parts.
 Detention time varies from 15 to 30 min.
 The water velocity varies from 0.1to 0.3m/s.
 Velocity gradient should vary between10 to100sec
-1
 loss of head is 15 to 60 cm
(source SCHULZ & OKUN)
The flocculator will be designed for a maximum day demand of the town and plus
treatment loss of around 5% of the maximum day demand.
The flocculator is divided in to three sections of equal volume, each section having
constant velocity gradient of 55, 40, 25sec-1 respectively.
Maximum day demand=0.055217m3/sec
Qmax = Q+5 %( Q)
= 0.055217+ (5/100)*0.055217m3/sec
=0.058m3/sec
Length of flocculator is taken as 5m
- Viscosity of water (µ) =1.14*10-3Kg/m-sec
-density of water (ρ) = 1000Kg/m3
-roughness coefficient of baffles (f) = 0.3
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Assosa Water Supply Project
-detention time for the total section = 18min
Design of the first flocculation basin section
-detention time=6min
-velocity gradient=55sec-1
-water velocity=0.3m/sec
-depth of water=1.8m
-flocculation time=18min
Total volume of flocculation
V=Q*t
V=0.058*20*60=69.6m3
Top width of flocculator,W
W=V/A
=
=
=7.73m
Width of each section = 7.73/3 = 2.6m
Number of baffles in the first flocculation section is given by:
2
 2*  *t
H * L *G 3
*
n
 
Q
  * (1.14  f ) 
 
1
Where
n = number of baffles in the basin
H = depth of water in the basin (m)
L = length of the basin (m)
G = velocity gradient (S-1)
Q = flow rate (m3/s)
t = detention time(s)
 = density of water (kg/ms)
 = dynamic viscosity (kg/ms)
f = coefficient of friction of the baffles
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Assosa Water Supply Project
 2 *1.14 *10 3 * 6 * 60 1.8 * 5 * 55  2  3
Therefore, n  
*
  74
 0.058  
 1000 * (1.14  0.3)
1
Spacing between baffles, s 
L 5

 0.067m
n 74
Since 0.067m is below the minimum range of design criteria, we take the minimum
spacing between baffles as 0.45m
So, number of baffles (n) =
L
5
=
=11.00
spacingbet weenbaffle s
0.45
Spacing between baffles and the walls is = 1.5*distance between baffles
=1.5*45cm =0.675cm
Head loss in the flocculation section
h
 * t * G 2 1.14 *10 3 * 6 * 60 * 55 2
=
1000 * 9.81
*g
= 0.126m
Design of second flocculation basin section
Detention time = 5min
Velocity gradient =40 sec-1
Number of baffles in the second flocculation section,
 2 *1.14 *10 3 * 6 * 60 1.8 * 5 * 40  2  3
*
n
  32
 0.058  
 1000 * (1.14  0.3)
1
Spacing between baffles, s,
s
L 5

 0.156m  0.45m
n 32
Since 0.156m is below the minimum range of design criteria, we take the minimum
spacing between baffles as 0.45m
So, number of baffles (n) =
L
5
=
=11.00
0.45
spacingbet weenbaffle s
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Assosa Water Supply Project
Spacing between end of baffles and the walls is = 1.5*spacing between baffles
= 1.5*0.45m
= 0.675m
Head loss h =
1.14 *10 3 * 6 * 60 * 40 2
1000 * 9.81
h = 0.0669 (say 0.07m)
Design of third flocculation basin section
Detention time = 5min.
Velocity gradient = 25sec-1
Number of baffles in the third flocculation section
 2 *1.14 *10 3 * 6 * 60 1.8 * 5 * 25  2  3
n
*

 0.058  
 1000 * (1.14  0.3)
1
n  23
Spacing between baffles, s
s
L 5

 0..217m  0.45m
n 23
Since 0.217m is below the minimum range of design criteria, we take the minimum
spacing between baffles as 0.45m
So, number of baffles (n) =
L
5
=
=11.00
0.45
spacingbet weenbaffle s
Spacing between baffles and the walls is = 1.5*spacing between baffles
= 1.4*0.45 =0.675m
Head loss =
=0.02614m
Total head loss through the flocculation sections = 0.126+0.07+0.02614
= 0.22m------ (.15-.6m) ok!
The second phase to the year 2028; Three similar horizontal flocculators as standard
above is to be constructed.
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Assosa Water Supply Project
7.6.4 SEDIMENTATION
The sedimentation process in water treatments provides for the settling and removal of
flocculated particles prior to filtration.
The removal efficiency in the sedimentation basin determines the subsequent loadings on
the filters and, accordingly has a marked influence on their capacity, the length on the
filters and quality of the filtered water.
The two major classifications for the design of sedimentation basins are:- Horizontal – flow unit
- Up flow unit
The design of such unit include:-Shape
- Detention time
-Number of basins
- Volume of sludge storage
-Dimensions
- Method of sludge removal
-Velocity
- Inlet and out let arrangements
-Direction of flow
- characteristic of the in coming flocculated water
The horizontal – flow sedimentation basin is preferable and chosen for our treatment due
to the advantage of
 It is widely used in the world because of its efficiency and inherent simplicity.
 It is with out mechanical sludge removal, so that does not require no important
equipment and labor for cleaning.
 Constriction cost is low, permitting over sizing.
 Operational and maintenance are simple.

7.6.4.1 Design of horizontal flow sedimentation
For the design purpose, the horizontal flow sedimentation tank may be divided in to four
zones.


Inlet zone
Settling zone
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Assosa Water Supply Project


Sludge zone and
Outlet zone
The sedimentation tank is designed with the assumption that the flow remains steady
and all the suspended particles are uniformly distributed on the full cross section of the
tank at right angles to the flow for the full length of the settling zone.
The inlets and outlets are designed in such a manner that they cause minimum
disturbance to the flowing water in the tank. The flow is designed to meet the following
requirements,
 The amount of water flowing out from the tank in 24 hours should at least be equal
to the daily demand.
 Velocity of flow should be so adjusted that suspended impurities of coarser nature
are removed.
The design of sedimentation tank is governed by;
-the quantity of water to be treated
-the selected detention period
-the selected surface loading rate (over flow rate)
Design of inlet zone
The velocity through the perforated baffle is about 0.2-0.3m/sec.
The head loss is estimated to be 1.7 times the velocity head.
Diameter of holes is 5cm; recommended range is 3-5cm.
-Taking a velocity of 0.25m/s
-Total area of opening =
= (0.058m3/s)/0.25m/s=0.232m2
-Area of each opening with a diameter of 50mm;
=
=0.00196m2
-Number of baffle holes required= 0.232/0.00196 = 118
-Head loss=
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Assosa Water Supply Project
Design of settling zone
As soon as a particle enters the settling zone, two forces act up on it. The flow of water
imparts horizontal force while gravitation force develops vertical component on the
particle. The particle moves under the influence of both this forces and traces a parabolic
path. Before entering the outlet zone, if a particle reaches the settling zone it is
considered removed.
The assumptions for design of settling zone are
For installation planed with new technology and good operation the surface loading rate
is taken with in a range of (30-40m/day)
[Okun, 1984].
- Surface loading rate=35m/day = 4.05*10-4m/sec
- Detention time = 2 hr (2-3hr)
-length to width ratio = 4
( >3)
Volume of tank (V) = detention period*design flow
= 2hr*0.058m3/sec*3600sec/hr =417.6m3≈418m3
Surface loading rate =
35m/day=4.05*10-4m/sec=
Therefore, W=6m
and
, where L=4*W
L=24m
Depth (d) =
Check for detention period
(2-3hr) -----ok!
T=
Check for horizontal flow
v=
(4-36m/hr)---ok!
Therefore provide a rectangular settling tank with,
Length= 24m
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Assosa Water Supply Project
Width= 6m
Depth= 2.9m
The some standard and number of unit should be provided for the phase II.
Design of sludge zone
The sludge zone is designed to accommodate all the settled particles during the process
of sedimentation. It should be located at such a position that while cleaning the sludge,
least disturbance is caused to the settling water. Normally the bottom floors of the tanks
are made to wards one side or to wards the center of the tank. Cast iron pipes, fitted with
gate valves are provided at the lowest points of the floor, from where sludge is removed
under hydrostatic pressure. To facilitate drainage of the basin, the floor should be slope
about 10% from the side walls to the center line and 5-8% from the outlet end to the inlet
end.
Provide depth of sludge zone =0.3m
Therefore total depth of sedimentation tank with a free board of 0.3m will be;
=depth of settling zone +depth of sludge zone+ free board
=2.9m+0.3m+0.3m =3.5m
Design of outlet Zone
Weir or perforated bounders are the most common structures for with drawing the effluent
water form the basin. Weir lengths should be selected to prevent high velocities of
approach and disturbance of the sludge layer.
The formula for acceptable weir length (adopted, from IRE, 1981)
L 
0.2 Q
Hvs
Where
L = Combined weir length (m)
Q = Flow rate (m3 / day)
H = Depth of tank (m)
Vs = Settling velocity (m/day)
For the phase I
L 
0.2 * 4770.75
 9.08 m  9m
2.9 * 35
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Assosa Water Supply Project
For the phase II
L 
0.2 * 10257.41
2.9 * 35
 19.53  20 m
Provide width of each weir as 1.5m.
7.6.5 FILTRATION
Filtration is a physical, chemical and biological process for separating suspended
impurities from water by passage through a porous media.
Screening of water removes larger suspended solids from water and sedimentation
following chemical coagulation removes most of the residual suspended matter. However,
there will still remain suspended matter and some fine floc particles. Flirtation of water is
done to remove them and to produce still further positive bacterial content in the water.
Filtration is also done to ensure safe, clear, and attractive water.
Filtration process is the most important part of water purification. It consists of passing
water through a thick layer of sand. During its passage through sand, the following effects
take place,
I.Fine suspended and colloidal matter which may be present in water is removed almost
completely.
II.Chemical characteristics of water get reduced.
III.The number of bacteria is considerably reduced.
These phenomena can be explained on the basis of four actions




Mechanical string
Sedimentation
Biological metabolism
Electrolytic changes
Performance of filter unit is predicted in terms of;
-
The influent characteristics
-
The characteristics of media
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Assosa Water Supply Project
-
The mode of operation of filter
Classification of filters
Filters may be classified under two heads
1.
Pressure filters
2.
Gravity filters
I.
Slow sand filters and
II.
Rapid sand filters
I.Slow sand filters
Consists of an enclose tank, under drainage system, base material or gravel, filter
medium of sand and other ancillary fixtures for the proper functioning of the filter. They
installed in a water tight tank made from stone masonry, brick masonry or cement
concrete. The floor and sides are all coated with water proofing agent. Generally slow
sand filters;
- Requires large area for installation
- Coagulation is not required
- Construction is simple
- Cost of operation is low
- High initial cost of both land and material
- High efficient in the removal of bacteria but less efficient in the removal of color and
turbidity.
- Not flexible for meeting variation in demand.
- Skilled man power is not essential
- The filter can be constructed of local labor and material; it is suitable for small towns
and villages where land is cheaply available.
- Rate of filtration is low
II.
Rapid sand filters
Consists a rectangular water-tight tank made from stone masonry or concrete. The tank
is rendered water-tight by applying a coat of water proofing material. Generally rapid
sand filters;
- Requires small area for its installation
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Assosa Water Supply Project
- Construction is complex
- Cost of operation is high
- Low initial cost of both land and material
- Less efficient in removal of bacteria but it have high efficient in removal of color and
turbidity.
- Quit flexible for reasonable fluctuation of demand
- Method of cleaning is short and speedy.
- Skilled man power is essential
- It is suitable for big cities where land cost is high and variation in demand of water
is considerable.
- Rate of filtration is high
Based on the rate of filtration, rapid sand filter is preferable. In most cases, when
population growth is high and the town expands extensively, as for assosa town case,
rapid sand filter is recommended from economical point of view.
7.6.5.1. Design of rapid sand filter
Important purification effects in rapid sand filtration is adsorption of impurities having an
electronic charge of the filter bed material are supplemented by the electric kinetic charge
produced by the high rate of water.
The design consideration for rapid sand filter included
I. filter unit
II. Filter media
III. Under drains
IV. Back washing arrangement
V Filter control system
Filter Unit.
The size of a filter unit is determined by the required amount of water needed. It consists
of two or more units of sizes depending upon the capacity of the plant.
-
Depth of tank ranges between (2.5-3.5m)
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Assosa Water Supply Project
-
Length to breadth ratio of between (1.25-1.33)
-
Surface area of each filter ranges between (10-80m2)
-
Rate of filtration is between (4000-5000l/m2/hr)
Based on the above recommendations
-
Assume rate of filtration =5000l/m2/hr
-
Total quantity of water to be treated = 0.055217m3/hr
=0.1988*106 l/hr
-
Total area of filter bed required(A) =
= 39.76m2 ≈40m2
=
-
Number of filter units required,
(Water Supply Eng ’g Santosh K. Garg 1995)
N
Where; Q = maximum day demand in million litters per day
= 24*0.1988*106l/day
= 4.7712*106l/day
N = 1.22*
= 2.66 ≈ 3 units
At least one filter unit should be added for emergency or during repair work of any of the
other. Hence adopting four units is reasonable.
40
=13.3 m2. ( 10-80 m2) ----ok!
3
-
Area of each filter unit =
-
Assuming length to width as 1.3
A=L*W=13.3 m2
but
L=1.30*W
13.3 m2 = 1.30W 2
W=
=3.2m
L = 1.3*3.2m = 4.2m
Therefore, provide four numbers of rapid sand filters with the size of each unit 3.2*4.2.one
unit is required for stand by or during repair work of any other units.
The same standard and number of filter units will be provided in the second phase.
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Assosa Water Supply Project
Filter media
The choice of filter media is dictated by the durability required, the desired degree of
purification and the length of filter run and ease of backwash sought. Sand is the
cheapest filter medium and has been widely used. The ideal medium should have such a
size and be of such a material that it will provide a satisfactory effluent, retain a maximum
quantity of solid, and be readily cleaned with a minimum of wash water.
The sand used in rapid sand filters should be free from dirt, hard resistance, and
preferably of quartz or quartzite. Conventional rapid sand filter generally uses sand layer
of effective size varying from 0.45mm to 0.55mm having uniformity coefficient (D 60/D10)
ranging from 1.2 to 1.7. The sand depth, when used alone ranges from 600mm to 700mm
in most applications.
In standard rapid sand filters the filter medium may be underlain by 400 to 600mm gravel
which serves to support the sand, permit the filtered water to remove freely to ward the
under drain and also allows the wash water to move more or less uniformly upward to the
sand. It is placed in 5 or 6 with the finest size on top. It is also specified that it should be
hard, rounded, and durable, weight approximately 1600 Kg/m 3, be free from flat, thin or
long pieces and contain no loam, sand, clay, shells or other foreign materials.
Table 7. .A common grading and layer thickness are as follows:sand
gravel
Grading (mm)
Thickness (mm)
Grading (mm)
Thickness(mm)
40 to 60
120 to 200
2.5
70
20 to 40
80 to 120
7.5
70
10 to 20
80 to 120
15
100
5 to 10
60 to 80
30
100
2.5 to 5
60 to 80
50
160
Total depth = 400 to
600
Total=500mm
Total depth of filter media :Page 104
Assosa Water Supply Project
= (depth of and) + (depth of gravel)
= 700 mm + 500 mm
= 1200 mm
= 1.2 m
7.6.6. Under drain System
The under drain system of a filter is an important component in the design and operation.
The selection is based on the filter type and size, media characteristics, and the selected
method of back washing. The under drain contain a central manifold and lateral that are
perforated.
The selected under drain system,
-
Collect the filtered water percolating down through the sand and gravel layers
-
Durable, reliable, and cost effective
-
Ensure uniform flow distribution of the back wash
Design guide line for the design of pipe under drain system (NWS and DB 1988)
1. Ratio of length of lateral to its diameter should not exceed 60.
2. Diameter of perforations is the lateral should be from 6-12mm
3. Spacing of perforations along the lateral may vary from 75mm for the smaller holes
to 200mm for the larger.
4. Ratio of total area of perforations in the under drain system to total cross sectional
area of laterals should not exceed 0.5 for the larger holes and should decrease to
0.25 for the smaller.
5. Ratio of total area of perforation in the under drain system to the entire filter area
may be as low as 0.002.
6. Spacing of lateral may be as great as 300mm
7. Rate of washing may be varied from 0.15-0.9/min.
Design based on the guide lines
-Area of each filter = 3.2*4.2 =13.44m2
Total area of perforation =0.3% of area of filter
=0.003*13.44 = 0.04m2
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Assosa Water Supply Project
Total area of lateral= 2*area of perforation
=2*0.04 = 0.08m2
Assume diameter of lateral = 0.05m
=0.002m2
Area of lateral =
Number of lateral =
=
=40, (20 on each side of the lateral)
Area of manifold= (1.5-2)*total area of laterals
=1.5*0.08m2 = 0.12m2
Dmanifold = 0.4m
Spacing of laterals =
=
= 0.21m (< 0.3m) ----ok!
Length of lateral on each side of manifold=
Assume spacing of perforation=20cm c/c
Total area of perforation on one lateral=
=0.001m2
=
Number of perforation per lateral=
=
= 7. 5 (take 7)
Area of each perforation =
=
=1.4286cm2≈1.43cm2
Therefore diameter of perforation =
=1.349cm≈1.4cm
Check;
< 60
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Assosa Water Supply Project
< 60 ………ok!
Total number of perforation = number of perforation per lateral*number of laterals
= 7*40 = 280
Area of perforation =1.43cm2
Total area of perforation = 1.43*280 = 400cm2 = 0.04m2……..ok!
This is equal to the designed total area of perforation (0.04m2)
7.7. Back wash arrangements
A filter unit is washed when filtering medium has become so dirty. Due to this dirt, when
the water passes through the filter medium, it experiences frictional loss of resistance
known as head loss. When the head loss exceed 1.5-2.5m, the filter needs cleaning.
Washing is accomplished by reversing of water through the filter but using a much higher
rate. Back wash is provided from an elevated tank.
7.7.1. Wash water tank
The wash water tank is designed to store wash water at least for the two filters.
Assume amount of wash water as 4% of filtered water in one day
Assume rate of filtration 4m /hr
The time required for washing one filter unit is 15min
=0.25 hr is lost in one day for washing
Quantity of wash water = (4/100)*13.44*23.75*4m/hr
=50.534m3/unit
The tank should have storage capacity to store wash water for at least two units.
Therefore; capacity of wash water tank =50.534m3/unit*2unit
=101.1m3
Depth of storage tank for wash water should be between 2.3-3.5m.
Taking d=3m
For a circular tank of depth 3m, the diameter will be;
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Assosa Water Supply Project
V=A*d
D=
=
= 6.55m
There fore provide wash water tank of 6.55m diameter and 3m deep with free board of
0.3m.
7.7.2. Wash water supply main
Size of the down water pipe bringing the water to the filters for back washing be 40cm in
diameter. The time taken for washing is 10 min.
Velocity in the pipe while washing one filter unit.
=
=1.5m/sec < 3m/sec…... ok!
Therefore, provide a wash water supply main having a diameter of 40cm.
Wash water trough
Wash water troughs serve to collect and carry to the main gutter the dirty water resulting
from washing the filter.
Assume a wash water rate of 0.6m/min (0.6-0.9m/min)
Wash water discharge for one filter
=0.6*3.2*4.2
=8.064m3/min
=0.1344m3/sec
Assuming a spacing of 1.6m for wash water trough (recommended 1.5-2m), this will run
parallel to the longer dimension of the filter unit.
Number of trough =3.2/1.6 =2
Discharge per unit trough = (0.1344m3/sec)/2
=0.0672m3/sec
For the width of 0.4m the water depth of upper end is given by
h=1.73
(source Steel and GC Ghee 1979)
Q= 1.376*b*h3/2
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Assosa Water Supply Project
0.0672= 1.376*0.4*h3/2
h= 0.246m = 0.3m
providing a free board of 0.1m (0.05-0.1m)
Provide 2 troughs of 0.4m wide and 0.3m deep in each filter.
7.7.3. Pump for lifting wash water
One pump with one stand by will provide for pumping water in to the tank which has to
supply it to two filter unit in 24 hrs.
Let the total quantity of water lifted in 8hr of one shift.
= 14.135m3/hr
Total capacity of pump required=
Therefore provide one pump having 14.135m 3/hr capacities including one pump set as
stand by.
Assuming permissible velocity, v=1.3m/sec (0.6-1.55m/sec recommended)
Area of a rising pipe= Q/v = (14.135/3600)/1.3 =0.0302m 2
Diameter of rising pipe, D=
= 0.196m=200mm
Assuming the lift of water including friction losses to be 15m,
Therefore the horse power (HP) of the pump =
=
= 1718.055h.p
Assuming efficiency of the motors and pumps as 80%and 70%, the B.H.P off the pumping
sets.
B.H.P =1718.055/(0.8*0.7)
= 3067.955hp
7.8.Filter control system
It is essential to control the rate of filtration, because, a sudden increase in the rate will
make water to pass through the filter bed with out proper treatment or the dissolved gases
may get released due to sudden increase in the negative pressure. On the other hand a
sudden reduction in filtration rate will also release bubbles of gas present in the sand,
making a hole through the filter bed. Therefore, to obtain a constant rate of filtration, a
rate controlling device is fitted to the filter.
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Assosa Water Supply Project
There are two practical methods of operating filters and these differ primarily in the way
that the driving force is applied across the filter. These methods are referred to as
constant rate filtration and declining rate filtration.
In constant rate filtration, when water is taken out of service for back washing or returned
to service after back washing, the water level gradually rises or lowers in the operating
filters until sufficient head is achieved to handle the flow. Thus, the rate changes are
made slowly and smoothly without the abrupt changes associated with automatic or
manual equipment. The head loss for a particular filter is evidenced by the water level in
the filter box. When the water reaches a desired maximum level, back wash of that filter is
required.
Declining rate filtration provides significantly better filter effluent quality than constant rate
filtration. Less available head loss is needed compared with that required for constant rate
operation because the flow rate through the filter decreases toward the end of the filter
run. The head loss in the under drain and effluent piping system, therefore, decrease and
becomes available to sustain the run for a longer period than would be possible under
constant rate operation with the same available head.
The water levels in the filters should be 1.2m above the filter media at the beginning of a
filter run and raise to full available depth of 3.2m before back washing becomes
necessary.
7.9. Clear water well
Clear water well is constructed in connection with the filter, which is used to reserve
storage and to allow the plant to operate with out too frequent variation of its output rate.
The size of the clear water well mostly varies from ¼ to 1/3 of the daily capacity of the
plant.
Quantity of water to be stored =
= 7329.309/4 = 1832.33
Provide a depth of 3m, the area of the tank required
=1832.33/3 = 610m3
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Assosa Water Supply Project
Therefore, reservoir of plan area and depth
m shall be provided with addition of free
bored of 30cm
7.10. Coagulation
The single most important step in removal of very fine suspended solids from water is
proper coagulation. Once the coagulation step is performed properly, operation of the
plant is comparatively simple. Its main objective is to unit several colloidal particles
together to form bigger sized settle able flocs which may settle down easily.
The principle of coagulation may be explained from the following 2 aspects.
1- Floc formation – when coagulant is added to the water and thoroughly
mixed, it produces a thick insoluble gelatinous precipitate. This precipitate is
called floc. The floc has the property of removing fine and colloidal particles
quickly.
2- Electrical charge – the floc ions are found to posses the positive electric
charge while all the colloidal particles have negative charge, therefore flocs
attract the colloidal particles and cause their removal by settlement at the
bottom of tank.
The common chemicals generally used for coagulation are;
1.
Aluminum sulphate
2.
Sodium sulfate
3.
Ferrous sulfate
4.
Magnesium carbonate
5.
Poly electrolyte
7.10.1.DOSAGE OF COAGULANTS
The dosage of coagulants which should be added in the water depends on the following
factors;
-
Kind of coagulants
-
Turbidity of water
-
Color of water
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Assosa Water Supply Project
-
PH-value of water
-
Temperature of water
-
Mixing and flocculation time
7.11. Disinfection
A safe water supply is the prime concern of any water supply project. Safty against
disease is of utmost importance in water supply, especially the water borne, water
related, or water based diseases such as cholera, typhoid, and many others. Even though
a certain degree of disease causing pathogen removal is achived in treatment process
such as flocculation and filtration, there still exist a significant number of pathogens that
are not removed by these processes. This necessitates subsequent treatment to remove
pathogens, which are conventionally carried out by disinfection. Disinfection is generally
the final treatment process in water treatment train. Disinfection is carried out through
chemical means.
Requirements of a good disinfectant
The following are requirements of a good disinfectant:


It should be economical and easily available in bulk.

from water and make it safe for the consumers.
It should be able to destroy all the harmful bacteria and other organisms

Disinfectant should immediately attack bacteria when mixed with water.

odor and taste.

treated water could be easily determined.
After disinfection, the water should not become toxic and carry objectionable
Disinfectant should be such that their strength or concentration in the
Their dose should be such that there is always some residual concentration
for protection of water from contamination during storage, and conveyance,
through distribution system.
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Assosa Water Supply Project
In water supply, chlorine is the disinfectant almost universally, since it is the final safe
guard for the quality the water, the necessity of the continuous and effective application of
chlorine cannot be over emphasized. The objective of chlorination is to destroy pathogens
in water and provide some additional protection against subsequent contamination.
Chlorine dissolves more readily in cold water. In the liquid form it has a yellow color and a
specific gravity of 1.4. Reaction rate increases with rise in temperature and lowering oof
PH value. The gas is toxic, and irritating to eyes, impairs respiration, and can cause death
at high dosages.
Chlorination can be achieved in two ways: direct solution feed, where chlorine gas is fed
directly into the treated water, or else using hypochlorite salts such as calcium
hypochlorite and sodium hypochlorite. Hypochlorite tablets are used for emergency
disinfection in times of disaster, when normal water treatment is not operating.
Factors affecting efficiency of chlorine are
1.
Nature of pathogens to be destroyed
2.
Type and concentration of disinfectant
3.
Temperature of water
4.
Time of contact
5.
Physical and chemical properties of water to be treated
6.
PH (alkalinity/acidity) of water
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Assosa Water Supply Project
CHAPTER EIGHTH
DISTRIBUTION RESERVOIR
8.1Service reservoir
Distribution reservoir is also called service reservoir. Service reservoirs are used for
storing the treated water with in or near to the demand (distribution) area, distributing the
treated water to the customers. They are also designed to meet the water demand during
fire break, pumps failure, repair, etc and used to balance the hourly fluctuation of water
demand.
8.2 Types of reservoir
Distributions reservoirs are classified based on their support, shape, and material of
construction.
A) Based on support of ground
Distribution reservoirs are of two general types:
1. Surface reservoirs
These reservoirs have little or no elevation above the ground and which are
usually constructed of earthen, masonry, or a combination of both, or reinforced
cement concrete. They can be circular or rectangular in shape.
2. Elevated reservoirs
These reservoirs built entirely above the ground such as stand pipes and
elevated Tanks. They help to reduce pumping cost by giving head for gravity
flow distribution. They are usually of steel reinforced concrete or wood. Many
surface reservoirs are built on hills and thus comprise elevated storage.
B) Based on geometry
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Assosa Water Supply Project
 Circular reservoir
 Rectangular reservoirs
C) Based on material of construction
 Steel
 Reinforced concrete
 Masonry etc.
8.3 Purposes of reservoir
Distribution reservoirs serve a variety of purposes as described below:
A) With regard to water quantity
1. Storage for fluctuating demand:
Reservoirs are filling when the rate of pumping exceeds the demand rate and are
emptying when the reverse occurs. This action permits pumps to operate at
constant rates throughout any one day and thereby allows the use of pumps of
less capacity.
2. Fire storage
The immediate availability of large quantities of water within the water supply
system
is required to safe guard the community against fire breakout.
3. Emergency / break down storage
The storage of sufficient amount of water with in the system gives protection
the failure of the pump, power driving motors, or a supply conduit. This
against
quantity
of
water depends on the time taken during the repair work, and as such it is variable,
therefore, it is very difficult to estimate.
B)
With regard to pressure distribution
1. Equalizing pressure in the system.
2. Raising pressures at remote points from pumping stations.
3. Equalizing heads on pumps if the reservoirs are nearly pumping plant station.
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Assosa Water Supply Project
8.4 Location of distribution reservoir
Generally distribution reservoirs are located near the central portion of the distribution
area. It is always better to construct them on high ground of city or town at such place
where it can be constructed economically and also by considering the elevation of the
highest building to be supplied
The location of the proposed reservoir site selected should be to fully the gravity
distribution system because it reduces the cost of pumping and also to enable a good
flow to be maintained to those top most places
8.5 Accessories of reservoir
Following are the various accessories, which are commonly provided in the reservoirs
i. Inlet pipe for the entry of water.
ii. Outlet pipe for the withdrawal of water.
iii. Over flow pipe to prevent over flow of water.
iv. Float switch to stop the pump when the tank is full.
v. Float gauge to show the depth of water in the tank.
vi. Washout pipe for washing out the suspended impurities in the tank
vii. Manhole for providing entry into the tank.
viii. Access ladder to inspect the top & bottom of the tank.
ix. Ventilation for fresh air circulation in the tank.
x. Chlorinator incase when the water is directly pumped in the over head reservoir
from the tube wells
8.6 Determination of storage capacity of the reservoir
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Assosa Water Supply Project
The required reservoir capacity is determined by using
1. Analytical method
2. Mass curve
3. Sequent peak algorithm
For the Assosa water supply scheme, the analytical and the mass curve methods are
selected for their simplicity of calculation and interpretation.
The long study of 24 hourly peak factors of the Assosa town
Time
1
2
Hourly factor
0.3 0.3 0.3 0.3 0.3 0.6 1.1 2
1.8 1.6 1.5 1.4
Time
13
21
Hourly factor
1.3 1.3 1.3 1.5 1.6 1.4 1.2 0.9 0.7 0.5 0.4 0.3
14
3
15
4
16
5
17
6
18
7
19
8
20
9
10
22
11
23
12
24
When water is supplied for balancing the variable demand against a constant rate of
pumping for 24 hrs
The analysis of storage capacity can be calculated as follows.
Phase I (2008 – 2018)
Total demand of the town =3407.7m3/day
Total demand of the town in liters per day=34077
Hourly demand of the town = 141987.5 liters
Pumping hours=16
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Assosa Water Supply Project
The calculation is as shown bellow in table
Table 8.1 the analytical calculation of storage capacity for phase I
Time
Hourly
(hr)
Hourly
Factor
(1)
(2)
(3)
supply(liters)
Hourly
demand(liter
s)
Cumulative
hourly
supply(liters)
Cumulative
hourly
demand(liters)
(5)
(6)
(4)
Excess
supply
(liters) (5)
– (6) +ve
(7)
demand(lit
ers)
(6)
(5) +ve
–
(8)
1
0.3
0
42596.3
0
42596.25
42596.3
2
0.3
0
42596.3
0
85192.5
85192.5
3
0.3
0
42596.3
0
127788.75
127789
4
0.3
0
42596.3
0
170385
170385
5
0.3
212981.25
42596.3
212981.25
212981.25
6
0.6
212981.25
85192.5
298173.75
127788.7
5
454360
184583.7
5
425962.5
7
1.1
212981.25
156186
638943.75
8
2
212981.25
283975
851925
738335
113590
9
1.8
212981.25
255578
1064906.25
993912.5
70993.75
10
1.6
212981.25
227180
1277887.5
1221092.5
56795
11
1.5
212981.25
212981
1490868.75
1434073.75
56795
12
1.4
212981.25
198783
1703850
1632856.25
70993.75
13
1.4
0
198783
1703850
1831638.75
127789
14
1.3
0
184584
1703850
2016222.5
312373
15
1.3
0
184584
1703850
2200806.25
496956
16
1.5
212981.25
212981
1916831.25
2413787.5
496956
17
1.6
212981.25
227180
2129812.5
2640967.5
511155
18
1.4
212981.25
198783
2342793.75
2839750
496956
19
1.2
212981.25
170385
2555775
3010135
454360
20
0.9
212981.25
127789
2768756.25
3137923.75
369168
21
0.7
212981.25
99391.3
2981737.5
3237315
255578
22
0.5
212981.25
70993.8
3194718.75
3308308.75
113590
23
0.4
212981.25
56795
3407700
3365103.75
42596.25
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Assosa Water Supply Project
24
0.3
0
42596.3
3407700
3407700
Thus the reservoir capacity for phase I from the above table will be
 Maximum value of excess supply = 184583.75liters
 maximum value of excess demand = 511155liters
 capacity of reservoir = 184583.75+511155
=695738.75liters=695.739m3
 For fire requirement (10%) = 69.574m3
 Miscellaneous losses (3%) = 20.87217m3
 Total recommended reservoir capacity = 786.1852m3
Say=800m3 of standard reservoir capacity
Using mass curve method
mas curve diagram for phase one
cumulative supply and demand in miliun liters
4
3.5
3
2.5
2
cumulative supply
comulative demand
1.5
1
0.5
0
0
2
4
6
8
10
12
14
16
18
20
22
24
-0.5
time in (hrs)
Figure 8.1 mass curves for phase I
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Assosa Water Supply Project
Phase II (2018-2028)
 Total demand of the town =7329.309m3/day
 Total demand of the town in liters per day=7329309Hourly demand of the town =
305387.875liters
 Pumping hours=20
 Pumping rate=366465.45liters/hour
Thus the reservoir capacity for phase I from the above table will be






Maximum value of excess supply = 855086.05liters
maximum value of excess demand = 977241.2liters
capacity of reservoir = 855086.05+977241.2=1832327liters
=1832.327m3
For fire requirement (10%) = 183.233m3
Miscellaneous losses (3%) = 54.96982m3
Total recommended reservoir capacity = 2070.53m3
Say=2100m3 of standard reservoir capacity
Provide additional two reservoirs with capacity of 500m3 and 800m3 for second phase
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Assosa Water Supply Project
Table 8.2: the analytical calculation of storage capacity for phase II
Excess
Time
Hourly
(hr)
Hourly
Factor
supply(liters)
Hourly
demand(liters)
(1)
(2)
(3)
(4)
Cumulative
hourly
supply
(liters)
Cumulative
hourly
demand(liters)
(6)
(5)
supply (liters)
(5) – (6) +ve
value only
(7)
demand
(liters) – (5)
+ve
value
only
(8)
1
0.3
0
91616.363
0
91616.3625
91616.363
2
0.3
0
91616.363
0
183232.725
183232.73
3
0.3
366465.5
91616.363
366465.45
274849.0875
91616.3625
4
0.3
366465.5
91616.363
732930.9
366465.45
366465.45
5
0.3
366465.5
91616.363
1099396.35
458081.8125
641314.538
6
0.6
366465.5
183232.73
1465861.8
641314.5375
824547.263
7
1.1
366465.5
335926.66
1832327.25
977241.2
855086.05
8
2
366465.5
610775.75
2198792.7
1588016.95
610775.75
9
1.8
366465.5
549698.18
2565258.15
2137715.125
427543.025
10
1.6
366465.5
488620.6
2931723.6
2626335.725
305387.875
11
1.5
366465.5
458081.81
3298189.05
3084417.538
213771.513
12
1.4
366465.5
427543.03
3664654.5
3511960.563
152693.938
13
1.4
0
427543.03
3664654.5
3939503.588
274849.09
14
1.3
0
397004.24
3664654.5
4336507.825
671853.32
15
1.3
366465.5
397004.24
4031119.95
4733512.063
702392.11
16
1.5
366465.5
458081.81
4397585.4
5191593.875
794008.47
17
1.6
366465.5
488620.6
4764050.85
5680214.475
916163.62
18
1.4
366465.5
427543.03
5130516.3
6107757.5
977241.2
19
1.2
366465.5
366465.45
5496981.75
6474222.95
977241.2
20
0.9
366465.5
274849.09
5863447.2
6749072.038
885624.84
21
0.7
366465.5
213771.51
6229912.65
6962843.55
732930.9
22
0.5
366465.5
152693.94
6596378.1
7115537.488
519159.39
23
0.4
366465.5
122155.15
6962843.55
7237692.638
274849.09
24
0.3
366465.5
91616.363
7329309
7329309
0
0
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Assosa Water Supply Project
Using mass curve method
mass curve diagram
8
cumulative supply and demand in 10^6 liters
7
6
5
4
cumulative supply
comuulative demannd
3
2
1
0
0
2
4
6
8
10
12
14
16
18
20
22
24
-1
time in(hrs)
Figure 8.2 mass curve diagrams for phase II
8.7 Depth and shape of service reservoir
8.7.1 Depth of service reservoir
For any given quantity of water either shallow reservoir having long walls and large floor
are, or alternatively a deep reservoir constructed with high retaining wall and smaller floor
area.
Factors influencing a depth of a given storage are
 Depth at which suitable foundation conditions are encountered
 Depth at which out let main must laid
 Slope of ground, nature and types of backfill
 The slope and size of the land available
8.7.2 Shape of service reservoir
Circular reservoir is geometrically the most economical shape give, the least amount of
walling for a given volume and depth, it has the attraction of allowing of construction thin
reinforce concrete.
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Assosa Water Supply Project
8.9 Structural Design of service reservoir
8.9.1 Water Tank (reservoir) Design Consideration
The primary purpose of design is to achieve acceptable probabilities that the structure
being designed will not become unfit in any way for the use it is intended. To
accommodate the required amount of water and to ensure water tight structure, the
reservoir must be designed using reinforced cement concrete that accounts for tensile
forces as well as those due to bending.
The different types of reservoirs depending on the geometry (Circular and Rectangular),
supports (resting on the ground, under ground and elevated), and end restraints (free
sliding, hinged and fixed at top and/or base) should be compared and selected based on
their suitability and economic condition during the design of water containing reservoirs.
For small capacities rectangular tanks are usually used. And for bigger capacities circular
tanks are generally used to ensure economical and efficient system of work.
The design of tank should also consider both full and empty conditions, and the
assumptions regarding the arrangement of loading conditions so as to cause the most
critical effects. When the structure is empty it must have strength to withstand the active
pressure of any retained earth. The possible resistance of the earth, never certain to act,
is generally ignored when designing for structure full of water.
Important considerations have to be given in limiting the size of crack (mostly with no
cracks) so that leakage does not take place. The design generally governed by the
requirements of the elastic design method, but stability considerations are particularly
important. The design has to take careful account of the construction methods to be
used.
The requirements for the elastic design method are listed as follows.
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Assosa Water Supply Project
Concrete grade class as C- 30 with a minimum cement content of 330 kg/m3 and has
maximum water cement ration = 0.55, but the concert should not be more than the
maximum cement content to keep the shrinkage down.
 The maximum cracked width is limited to 0.2mm for all faces of liquid containing
or excluding structures, unless aesthetic appearance is critical, where 0.1mm is
required to avoid staining of concrete.
 Minimum cover = 40mm
 Maximum bar spacing = 300mm
 The allowable tensile stresses in concrete than control cracks in concrete grade
of C-30 shall be 1.44 N/mm2 and 2.02 N/mm2 due to direct tension and bending
respectively.
 The allowable tensile stresses in steel taken as 100 N/mm2 using deformed
bars for alternate wetting and drying exposure conditions.
 The maximum steel area in each of the two directions at right angle are 0.3% of
the concrete area (0.15% near each face) for deter med bars.
The operational processes with in the water and other industries dealing with fluids often
require circular structures to ensure their systems of work carried out efficiently and
economically. Hence circular tank is chosen for the design of Assosa town water supply
project.
The primary stresses set up with in the structure are usually a result of the ring tension
generated by the contained liquid and the main reinforcement therefore consists of bond
of circular steel hops. The ability of the cylinder to increase in diameter is resisted,
however at the base where restraints occur. If out ward movement is prevented by a
fixed joint the ring tension will be zero and vertical bending movement and sheared force
will occur.
Out of three types of base conditions i.e. free sliding, pined and fixed base, fixed base is
the most effective due to the above reasons i.e. the ring tension will be zero and only
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Assosa Water Supply Project
vertical bending moment and shear force will occur. Therefore, due to its effectiveness,
fixed base type is selected for the design of the circular reservoir
8.9.2 Design of 800m3 circular reservoir
Type: Fixed base and free ends
At the top of the will, shear force and bending moments are zero, and at the base
of the wall, slope and deflection is zero.
F
W * H
Figure 8.3 fixed base and free end reservoir
 Capacity of reservoir = 800m3
 Height of reservoir = 4m
 Assumed free board = 0.2m
 Total height of reservoir = 4m+0.2m =4.2m
 Diameter of reservoir =
 D=
4V
=
 *H
4 * 800
=16m
3.1416 * 4
 Assume top thickness of wall =200mm
 Assume bottom thickness of wall=250mm
 The average thickness of wall
 tavg=
200  250
=-225mm
2
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Assosa Water Supply Project
8.9.2.3 Design of reservoir wall
Internal force acting on the rank wall
For determining internal forces of the tank wall of fixed base and free top tank
,coefficients are available in standard tables for a given
The value of
h2
d .t
H2
=4.4444
D * t avg
Then the corresponding value of moment and hoop tension coefficients is calculated as
follows.
Table 8.3 Coefficients for vertical moments in circular reservoirs
0.1H
0.2H
0.3H
0.4H
0.5H
0.6H
0.7H
0.8H
0.9H
1H
4
0.0003
0.0015
0.0028
0.0047
0.0.0066
0.0077
0.0069
0.0023
-0.0080
-0.0268
4.444
0.000212
0.001192
0.002272
0.00390
0.00572
0.003908
0.00646
0.002852
-0.007032
-0.02477
0.0046
0.0059
0.0059
0.0028
-0.0058
-0.022
H2
D * t avg
8
5
0.0001
0.0008
0.0016
0.0029
Table 8.4 Coefficient for hoop tension
H 2 0H
D * t avg
4
0.067
0.1H
0.2H
0.3H
0.4H
0.5H
0.6H
0.7H
0.8H
0.9H
0.16
0.256
0.339
0.403
0.429
0.409
0.334
0.210
0.073
0.2511
0.3421
0.4171
0.4503
0.409
0.3624
0.2318
0.0814
0.245
0.346
0.428
0.477
0.409
0.398
0.259
0.092
4
4.444
0.0484
0.15
2
5
0.025
0.13
7
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Assosa Water Supply Project
Considering alternative exposure of wetting and drying the allowable stress based on BS:
5337 (elastic design method)
For strength design
 fs, allow= 100N/mm2 (direct tension)
 fc, allow=11N/mm2 (due to bending)
 n=15
For no crack design
 fct, allow=2.02N/mm2 (due to bending)
 fct, allow=1.444N/mm2 (due to direct tension)
8.9.2.4 Detail Reinforcement
a) Vertical section of the tank wall: -It is treated as tension member with cracked
section i.e. only reinforcement resists the maximum hoop tension. The maximum hoop
tension occurs at 0.5H with a coefficient 0.4503.
I.
horizontal reinforcement
For the analysis purpose let divided wall into two pars
Case1:
For the top half of the wall, from 0.0H to 0.5H
 From the table above the minimum coefficients is 0.4503
 Hence the hoop tension can be calculated as
 TH = max .coeff *  w * H * r where r= D/2=16/2=8m
 =0.4503*9.81*4*8=141.3582KN per meter depth of wall
Therefore area of hoop tension steel required assuming concrete section is cracked
TH
141.3582 *10 3
As 

 1413.582mm2
100
fs , allow
As= 1413.582mm2  Asmin=
ok
0.3 * b * t 0.3 *1000 * 225
=675mm2…

100
100
Spacing of 14 ring bars required
s
b * as 1000 *153.86

 108.844mm2  Smax=300mmor t=225mm
As
1413.582
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Assosa Water Supply Project
Provided: 14 ring bars at 70mm c/c in one layer placed on water face of for upper half
wall
Check thickness of wall for no cracks
Thickness of wall is determined using requirement of resistance to crack;
therefore thickness of wall is determined by limiting tensile stress in the concrete to
allowable stress value as
f ct 
TH
141.3582 *10 3

 0.5775 N / mm2  f ct ,allow  1.44 N / mm2
Ac  (n  1) * As 1000 * 225  (15  1) *1413.582
It is ok. Therefore, thickness is adequate for no concrete crack
Case 2:
For hoop reinforcement applied on bottom half of the wall, hoop tension
developed at depth of (0.6H=2.4m) from top may be used as given below
TH  max .coeiff *  w * H * r  0.409 * 9.81* 4 * 8  128.3933KN / m
Area of reinforcement required the hoop for unit strip of wall
As 
TH
128.3933 *10 3

 1283.933mm2  As min  675mm2
100
fs , allow
Check thinness of wall for no crack
f ct 
TH
128.39333 *10 3

 0.5284 N / mm2  f c ,allow  1.44 N / mm2
Ac  (n  1) * As 1000 * 225  (15  1) *1283.933
It is ok. Therefore trial thickness is adequate for crack
Spacing of ring bar using 14
S
as * b
 S max  300mm or t=225mm
As
S
153.86 *1000
 119.835mm
1283.933
Therefore, provide 14 ring bars @ 120mm c/c in one layer placed on water face for
bottom half of wall.
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Assosa Water Supply Project
b) Horizontal section of the tank wall: -The internal forces in this section are (+ve) and
(-ve) moments with maximum coefficients of 0.006908 and -0.02477 respectively
i.
Vertical reinforcement
The maximum positive and the maximum negative moment is calculated as follows
M max  max .coeiff *  w * H 3
(+ve) max.coeff=0.006908 @0.8H
(-ve) max coeff=-0.02477@1H
Therefore, (+ve) moment =0.006908*9.81*43=4.3371KN-m per meter width
(-ve) moment =-0.02477*9.81*43=-6.4040KN-m per meter width
Take absolute value of the result, the maximum moment is therefore
M max   6.404
=6.404KN-m per meter width
Check thickness of wall for flexure
fs,allow=130N/mm2
fc,allow=11N/mm2
Design constant of balanced section are:
r
f s ,allow
f c ,allow
jb  1 

kb 
130
 11.82
11
15
n

 0.539
n  r 15  11.82
kb
 0.8136
3
Rb  0.5 * f c,allow * kb * jb  0.5 *11* 0.539 * 0.8136  2.5028N / mm2
Then the effective depth of section (dreq) is given by
d req  M max
Rb *
6
 6.4040 *10
2.5025 *1000
 50.5638mm
Thickness of wall taking 14 bars and 40mm cover
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Assosa Water Supply Project
t t req  d req  cov er   / 2  50.5638mm  40mm  14mm / 2  97.5638mm  t assu  225mm it is
ok . Then t(req) will be
t req  t assu  40mm  14mm / 2  178mm
Vertical reinforcement to be placed on water face is obtained for maximum negative
moment
Then area of reinforcement
(ve) As 
M max
6.4040 *10 3

 340.1551mm2  As min  675mm2
f s ,allow * jb * d req 130 * 0.8136 *178
Hence take As=Asmin=675mm2
t=250
d=178
b=100
Figure 8.4 Transformed section or equivalent section
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Assosa Water Supply Project
 A *Y
A
Taking a unit width of reservoir and 250mm bottom thickness of the centriods of section is
x
i
i
i

1000 * 250 *125  15 * 675 *178
 127.063mm
1000 * 250  15 * 675
Moments of area of equivalent concrete sections
I ce
b * x 3 b * ( D  x)


 n * As * (d  x) 2
3
3
I ce 
1000 *127 3 1000 * (225  127) 3

 15 * 675 * (178  127) 2
3
3
=1022860124mm4=1022.860*106mm4
f ct 
M max* ( D  x) 6.404 *10 6 * (250  127)

 0.77008 N / mm2  f ct , allow  2.02 N / mm2
Ice
1022.860124 *10 6
Therefore the thickness is adequate for no crack
Spacing of 14 bars
S
as * b 153.86 *1000

 228.056mm  Smax=300mm or t=225mm
As
675
Therefore provide: 14 bars at 225mm c/c vertically in the inner face.
Vertically reinforcement on outer
(+ve)Mmax=4.337KN-m per meter width
Using t=225mm, 14 bars
D= 225-14/2-40=178mm
Hence
(+ve)As=
(ve) M max
4.337 *10 6

 230.369mm2  As min  675mm2
f s ,allow * jb * d 130 * 0.8136 *178
Therefore As=675mm2 should be taken
Spacing of 14 bars
S
as * b 153.86 *1000

 228mm  Smax=300m or t=225mm
As
4 * 675
Therefore spacing S= 225mm should be taken.
Provide 14 bars at 225mm c/c in the outer face
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Assosa Water Supply Project
8.10 Design of roof slab
Cover slab of service tank (reservoir) is treated as roof slab, which is simply supported by
the circular wall.
Consider cover slab as freely supported at edges and load uniformly
Using strength limit design state
For C-300MPa,
f cd 
For S-300MPa f yd 

0.67 f cu
fy
m


0.67 * 30
 13.4MPa
1.5
300
 260.87 MPa
1.15
Where  w  1.5
fcu=300MPa
fy=300MPa
For S-300MPa maximum design constant for single reinforcement according to ACI
max  0.437,  max  0.34,  max  0.75 b
Dead load (own weight) = t*  c , where
t=thickness of slab=200mm
=0.2*25=5 Kn/m2
Live load (LL) =0.5 KN/m2 (based on EBCS; 1995)
Therefore the design load on the slab
qd= 20 =1.3DL+1.6LL=1.3*5+1.6*0.5=7.3 KN/m2
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Assosa Water Supply Project
qd= 7.3KN/m2
D=16m
Figure 8.4 load on roof slab
Design load on the base slab
1 Design moment at the center of slab
3 * Wd * x 2 3 * 7.3 * 8.15 2
M r  M 

 90.916 KN .M
16
16
2 At the edge of the slab
M r  0, M  
wd 2 7.3 * 8.15 2
x 
 60.611KN  M
8
8
Check thickness for flexure
b 
0.0028
0.0028

 0.5828
260.87
f yd
0.0035 
0.0035 
2 *10 3
Es
Balanced mechanical reinforcement ratio
max  0.75 * b  0.75 * 0.5828  0.437
Then effective depth of section (dreq) is given by
d req 
M max
90.916 *10 6

 140.952mm
f cd *  max * b
13.4 * 0.3145 *1000
Using 14 bars and cover of 40 mm
Treq  d req  cov er   / 2
=140.952+40+14/2=187.952mm
Treq  187.952mm  t assum  225mm
And
d  t assum  40   / 2
=(200-40-14/2)mm=153m
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Assosa Water Supply Project
Reinforcement
a) At the center of slab
M r  M   90.916kn  m
M
90.916 *10 6


 0.290
f cd * b * d 2 13.4 *100 *1532
  1  (1  2 )  1  (1  2 * 0.290)  0.352
Then
As 
 * f cd * b * d
f yd

0.352 *13.4 *1000 *153
 2766.4mm2  As min  293.25mm2
260.87
Spacing using 14 bars
S
a s * b 153.938 *1000

 55.646  S max  300mmor400mm
As
2766.4mm 2
Therefore, provide 14 bars at 55mmc/c in the form of mesh at the center of slab
b) At the edge of the slab
M   M r  60.611KN  M

M
60.611 *10 6

 0.193
f cd * b * d 2 13.4 *1000 *1532
  1  (1  2 )  1  (1  2 * 0.193)  0.216
Then
Therefore
As 
 * f cd * b * d
f yd

0.216 *13.4 *1000 *153
0.5 * b * d
 1692.563mm2  As min 
 225mm2
f yk
260.87
Spacing of bars using 14 bars
S
a s * b 153.938 *1000

 90.682
As
1697.563mm 2
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Assosa Water Supply Project
Therefore provide 14 bars at 90mm c/c in the form of ring at the edge of slab just above
mesh reinforcement.
Check thickness for shear
vmax  0.5 * wd * x  0.5 * 7.3 * 8.15  29.748KN
Shear capacity of the slab, Vc
vc  0.4 * f ctd * (1  50 ) * b * d
f ctd 

0.35 * f ctk
1.5

0.35 * 30
 1.278 N / mm2
1.5
As
1697.563

 0.011
b * d 1000 *153
 /2 
0.011
 0.005
2
vc  0.4 *1.278 * (1  50 * 0.0053) *1000 *153  97.767 KN  Vmax  29.748kn
Therefore thickness is adequate for shear.
8.11 Design of circular base –slab
Section of base slab design considering simply supported at edge by ring beam and
monolithic with the wall of the tank supported uniform load due to own weight, roof slab,
and weight of water when the tank is full of water i.e
qd =wt. of base slab + wt. of water +wt. of roof slab
Considering exposure condition for base slab continuous liquid contact thus the allowable
stress used for design are
fs,allow=130N/mm2, fc,allow=11N/mm2, and n=15
Design constant of balanced section for flexural member
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Assosa Water Supply Project
r
f s ,allow
f c ,allow
jb  1 

130
 11.82
11
kb 
n
15

 0.539
n  r 15  11.82
kb
 0.8136
3
Rb  0.5 * f c,allow * kb * jb  0.5 *11* 0.539 * 0.8136  2.5028N / mm2
Shear at the base of wall = direct tension developed in the base slab
H2
Thus, the coefficients of
 4.444 from table is 0.12999
D *t
Therefore direct tension developed in the base of slab is calculated as follow
TH  0.1299 *  W * H 2  0.1299 * 9.81* 4 2  20.3885KN
Per
meter
width of the slab
Section of base slab monolithic with the wall of the tank is designed for combined action
of direct tension caused by water pressure on the wall. Maximum moment of slab caused
by uniform load on the base slab and restraints moments of the wall caused by water
pressure
Consider thickness of wall t= 400mm
Effective depth of slab considering 16 bar and 40mm cover
 D=t- cover-  / 2 =400-40-8=352mm
 Wt. of base slab =0.352m*25KN/m3=8.8KM/m2
 Wt. of roof slab=
 Wt. of water
=7.3KN/m2
=4m*9.81KN/m3=39.24KN/m2
 Total load =qd=8.8+7.3+39.34=55.34KN/m2
For simply supported circular slab (with radius =4m) the maximum radial and
circumferential moments are obtained by
Mr  M 
3 * q d * a 2 3 * 55.34 * 4 2

 166.020 KN  m at the center of the slab
16
16
Msupport= (-ve) Mbase,wall=6.4040KN-m …. Restraint moment at the base of
wall
Reinforcement of base of slab
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Assosa Water Supply Project
Minimum area of reinforcement for base of slab of unit width (BS: 5337)
Asmim=
0.3 * b * t 0.3 *1000 * 400

 1200mm2
100
100
At the center of the slab design for combined action of TR= 20.3855KN and
M=166.02KN-m
Therefore the required reinforcement is determined by
(+ve)As=
Spacing of 20 bar
S
T
166.02 *10 6
20.3885 *10 3
M
 R 

 4616.1014mm2
130
f s * j * d f s 130 * 0.8136 * 352
b * as 1000 * 314.16

 68.057mm  Smax=300mm
As
4616.1014
Provided: 20 bars at 65 mm c/c in the form of mesh placed at the bottom of slab
At the support:
Design for combined action of T=20.3885KN and M=-6.4040KN-m Therefore required
reinforcement is determined by
(-ve)As=
M
6.404 *10 6
20.3885 *10 3


 328.845mm2  Asmin =1200mm2
f s * j * d 130 * 0.8136 * 352
130
There fore take As=1200mm2
Therefore extra bars are required in addition to vertical bars of wall.
Check trail thickness of slab for no crack due negative moment & direct tension
 ( A * Y )  400 *1000 * 200  15 *1200 * 352  206.5455mm
400 *1000  15 *1200
A
Location of neutral axis of equivalent un-crack concrete section of wall
Y
Moments of inertial and area of equivalent un-cracked concrete section of wall
b* y
b * ( D  y) 3


 n * Ast * (d  y) 2
3
3
3
I ce
1000 * 206.54453
(400  206.5445) 3
I ce 
 1000 *
 15 *1200 * (352  206.5445) 2
3
3
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Assosa Water Supply Project
I ce  5731296970mm4
Then tensile stress due to moment and direct tension
f ctb
M * ( D  Y ) 6.404 *10 6 * (400  206.5445)


 0.2162 N / mm 2
5731296970
I ce
f ct 
TR
20.3885 *10 3

 0.05097
Ace
400 *1000
Then check tensile stress interaction equation for no concrete crack
f ctb
f ctb,allow

f ct
f ct ,allow

0.2162 0.05097

 0.14245  1….. Ok
2.02
1.44
Therefore trial thickness of wall is adequate for no concrete crack!
The design of 500m3 is done in a similar fashion of design of 800m3. For the calculation of
bill of quantities we considered proportional amount as their volume.
CHAPTER NINE
Distribution system
9.1 General
After the water has reached to the service reservoir, it becomes necessary to distribute it
to a number of houses, industries and public places by means of network of distribution
system. The distribution systems, the distribution systems consist of pipes of various
sizes, valves, meters, distribution reservoirs, pumps, hydrants etc. the pipe lines carry
water to each and every street and road. Valves control the flow of water through the
pipes. Meters are provided to measure the quantity of water consumed by the individual
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Assosa Water Supply Project
as well as by the town. Hydrants are provided to connect the water to the fire fighting
equipment during fine accident. Service connection is done to connect the individual
building with the water line passing through the streets. Pumps are provided to pump the
water to the elevated service reservoir or directly in the water mains to obtain the required
pressure in the pipe lines. The layout of the roads and the elevation of the town also
considered to select the layout of the distribution systems respectively the following are
the requirement of a good distribution system
 It should convey the reacted water to the consumers with the same degree of
purity.
 The water should reach to every consumer with the required pressure head.
 Sufficient quantity of reacted water should reach for the domestic and industrial
use.
 The distribution system should be economical and easy to maintain and operate.
 It should be able to transport sufficient quantity of water during emergency shuch
as
fire fighting.
 During repair work, it should not cause obstruction to the traffic.
 It should be safe against any future pollution. The pipe lines as much as possible
should not be laind below the sewer lines.
 The quality of the pipe should be good and it should not burst.
 It should be water-tightt and the water losses dur to leakage should be bare
minimum as much as possible.
 For efficient distribution it is required that water should each to every consumers
with required rate of flow. There fore, some pressure in pipe lines is necessary
which should force the water to reach at every place.
9.2 Methods of distribution
For efficient distribution it is required that water should reach to every consumer with
required rate of flow. Therefore, some pressure in pipelines is necessary to force the
water to reach at every node. This can be done by one of the following methods, as local
conditions or other considerations may dictate.
I gravity distribution system
This is possible when the source of the supply is at some elevation above the city so that
sufficient pressure can be maintained in the mains for domestic and fire service. This is
the most reliable method if the conduit leading from source to city is adequate in size and
well safe guarded against accidental breaks.
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Assosa Water Supply Project
II pumping system with out storage
In this system water is directly pumped in mains. Since the pumps have to work at
different rates in a day, the maintenance const increases. It is preferred to have number
of pumps and only the required numbers may work at various times to meet the varying
demand, in place of proving pump of variable speed. High lift pumps are required and
their operations are continuously watched.
If the power fails, the whole supply of town will be stopped. Therefore, it is better to have
a diesel pumps in addition to the electric pumps as standby. During fires, the water dan
be pumped in the required quantity by the stand by units also.
III combined gravity and pumping system (dual system)
In this system the pump is connected to the mains as well as to an elevated reservoir. At
the beginning when the demand is small the water is stored in the elevated reservoir,
when the demand increases the flow in the distribution system comes from both the
pumping station as well as elevated reservoir. In this system water comes from two
sources from reservoir and for pumping station, which is called dual system this system is
more reliable and economical because it requires uniform rate of pumping but meets low
as well as maximum demand, the water stored in elevated reservoir meets the
requirements of demand during breakdown of pumps and for fire fighting.
Following are the main advantages of this system:i) The balance reserve in the storage reservoir will be utilized during fire. In case of
fire demand is more and if required the water supply of few localities may be
closed.
ii) This system is the overall best system. It is economical efficient and reliable.
iii) This system has the advantages that during power failure, the balance water
stored in the reservoir will be supplied to the town.
iv) The pumps have to work at constant speed with out any variation in their speed.
In Assosa town the water is distributed to the community by the gravity system
because the service reservoir is located at high elevation of the town. So the
water reaches to every consumer in sufficient quantity.
9.3 layout of distribution system
Depending upon their layout direction of supply, they are classified as follows:i. Dead end or three system:- it is suitable for irregular developed towns or cities. In this
system one man starts from service reservoir along the main road. Sub mains are
connected to the main in both the direction along other roads, which meet the roads
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Assosa Water Supply Project
carrying sub-mains, branches and minor distributors are laid and are connected to sub
mains. From these branches service connections are made to individual houses. the
advantage of these system are cheap in initial cost and easy determination of pipe
diameter, value size etc. but the main disadvantage of f this system is formation of dead
ends and if pipe breaks down or is closed for repair, the whole locality beyond the point
grows without water. Since the water is reaching at each point from one side only it can
not meet the fire demand.
ii. Grid-iror system: - this system is most convenient for towns having rectangular layout of
roads. Actually, this system is an improvement over dead end system. All the dead ends
ae interconnected with each other and water circulated freely through out the system. In
this system mainline is laid along the main road. Sub-mains are taken in both directions
along other minor roads and streets. From these sub-mains branches are taken out and
are interconnected to each other and water circulates freely through out the system. This
system removes all the disadvantages of dead end system.
Following are main advantages of this system
i.As the water is supplied from both the sides to every point, very small area will be
affected during repair.
ii.Since the water reaches every point form more than one route, the friction losses and
the sizes of the pipes are reduced.
iii.All the dad ends are completely eliminated, therefore the water remains in continuous
flow and there is no stagnation and chance of pollution is reduced to minimum.
iv.In case of fire, more quantity of water can be diverted towards the affected area, by
closing the valves of near by localities.
As this system has many advantages it also has some disadvantages:i.More number of valves and longer length of pipe is required in this system, there by
increasing the overall cost.
ii.If one section is to be repaired more number of valves are required to be closed.
iii.The design is difficult and costlier.
Considering the above advantage and layout of roads of Assosa town this method is
adopted.
III circular or ring system:- this system is adopted only in well planned locality of cities. In
this system each locality is divided into square and the water main are laid around all the
four sides of the square. All the sub-mains and branches ae taken off from the boundary
mains and are are inter connected. This system is the best of the other system but it
requires many valves and more pipe length. The ring system is most suitable for towns
and cities having well planned road.
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Assosa Water Supply Project
iv.Radial system:- actually this is the reverse of ring system and water flows towards outer
periphery from one point. The entire district is divided in to various zones and one
reservoir is placed for each zone which is placed at the center of the zone. The water
lines are laid radically from it.
9.4 design procedure of the distribution system
First of all the layouts of the distribution pipes are prepared and the position of valves, fire
hydrants etc, are also marked on it. The reduced levels of the town at various points are
also marked on it. The reduced levels of the town at various point are also marked on the
plan. Now the total population to be served at each node is also marked on the plan. After
this the minimum water pressure required at the tail end and near the highest building of
the city are also determined and noted on the plan. After completing the above work, the
main work is to determine the sizes of distribution popes, which would be capable to carry
the required quantity of water at the desired pressure.
9.5 selections of pipe materials
The types of pipe used for distributing water under pressure includes: ductile iron, cast
iron, asbestos, cement concrete, steel and PVC. Small diameter pipes for house
connection are usually plastic. For use in transmission and distribution system pipe
materials must have the following characteristics.
 Adequate tensile and bending strength to with stand external loads that result from
trench back fill and earth movement caused by freezing, thawing or unstable soil
condition
 Ability to resist impact loads encountered in transportation, handling and installation.
 Smooth, non corrosive interior surface for minimum resistance to water flow.
 Pipe materials that can be provided with tight joints and easy to tap for making
connections.
Among different types of pipes, existing on the market PVC pipes are selected for
the
distribution system. This is because PVC pipes are light in weight, cheap,
easy to join
and install durable, good electric insulators and free from corrosion.
Additionally, DCI
pipes are extended from borehole to service reservoir due to
its strength and resistance to corrosion.
9.6 design of pipelines
Till date no direct methods are available for the design of distribution pipes. While doing
the design first of all the diameters of pipes are assumed, the terminal pressure heads
which could be made available at the end of each pipe section after allowing for the loss
of pressure head in the pipe section when full peak flow discharge is flowing are then
determined. The determination of the friction loss in each pipe section is done. The total
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discharge flowing through main pipe is to be determined in advance. While design of
pipes the following points should be kept in mind: The main lines should be designed to carry and times the average demand of the
city.
 The water demand at various pints in the city should be noted.
 The velocity in pipelines should be less than 2m/sec.
 In the distribution system the minimum size of distribution pipe is as 50mm and
service pipe of 20mm is used for giving house connection.
9.7 pressure in distribution
When the water enters in the distribution main the water head continuously is lost due to
friction in pipes, at entrance of reduces, due to valves, bends, meters etc till it reaches
consumers tap. The net available head at the consumers tap is the head at the entrance
of the water minus all the losses in the way. The effective head available at the service
connection to the building is very important because, the height up to which the water can
rise in the building will depend on this available head only. The greater the geed the more
will be the height up to which it will rise.
If adequate head is not available at the connection to the building, the water will not reach
the upper storey, to overcome this difficulty the required effective head is maintained lithe
street pipe lines.
The water should reach for each and every consumer. Therefore it should reach on the
upper most storeys. The pressure which is required to be maintained in the distribution
system depends up on the following factors: The height of the highest building up to which water should reach without boosting.
 The distance of the locality form the distribution reservoir
 Supply is to be metered or not, higher pressure will be required to compensate for
the high loss of head in meters.
 How much pressure will be required for fie hydrants.
 The funds available for the project work.
9.8 nodal demand computation
The average consumption in the odes can be determined in two ways
1. From the geographical map, by dividing the system in to a number of areas
assumed to be supplied from the corresponding node.
2. By calculating the average number of consumers per meter of pipe in each loop.
Even dispersion of house connections through the system is assumed in both
cases. Location of the source connections to the system will depend on the route
of secondary mains, more over;
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


They should be concentrated towards the areas of higher demand
The pipe route should be as short as possible.
The pipe should be easily accessible; passing underneath the building
should be avoided.
9.9 computer analysis of distribution system
The distribution network is simulated using Epanet 2.0. The simulation was carried out for
extended period analysis by taking in to consideration the hourly demand variation pattern
on maximum and average day. The analysis began by feeding assumed diameter in the
computer; and the pressure, velocity and headless are checked for peak and average
flow. As the result of Epanet 2.0 analysis is shown in the appendix.
.
9.10 appurtenances
The different device required for controlling the flow of water for preventing leakage and
other similar purposes in water distribution network is called appurtenances. In this
network the following appurtenance are used:1. Valves: - in distribution network it is used to control flow of water, regulates
pressure, releases and admits air, prevents flow of water in opposite direction and
so on. Valves such as check valves, pressure valves, gate valves, air relief valves
and drain valves are to be provided where they are needed.
2. Pipe fittings:- the various pipe fittings such as beds, crosses, tees, elbows, caps,
ripples, plugs, flanges are to be provided during the laying of the distribution.
3. Manholes: - for inspection and maintenance purpose manholes are to be provided
at every junction of pipes and at places where valves are to be installed and when t
here is a change in direction of pipes.
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