Canal Aqueduct

Design of Syphon Aqueduct

Sri Sunflower College of Engineering & technology 12

Chapter-3

Selection of type of aqueduct

The above data of hydraulic particulars, a type-3 aqueduct is designed. It is only purely from an economical aspect that we go in for the type-3 aqueduct. For major drains it will be uneconomical to go in for a type-3 aqueduct as the extra cost of barrel with a large number of vents will be more than the cost of other works necessary for a type-3 aqueduct.

So, whenever an aqueduct are to be actually constructed, comparative costs are to be worked out for a type-2 and a type-3, and whichever is economical is to be chosen and adopted.

In the case of a type-3 aqueduct, the canal will be flumed and taken through a masonry or reinforced concrete trough supported on piers and abutments. The maximum velocity through the trough is generally taken as twice the normal velocity or 1.5m/s whichever is less.



Chapter - 4

Design of canal trough

Discharge, Q = 35 m3/s

Average velocity = 0.83 m3/s

Design velocity = 2% average velocity = 2*0.83=1.66 m/s

But the maximum design velocity = 1.5 m/s

Adopt design velocity = 1.5 m/s

Q= A x v

35= Ax1.5

A=23.3 m2

Depth of flow, y= F.S.D= F.S.L – B.L= 42.00-40.00=2m

Bottom level = Ultimate bed level of canal = 39.75

Top level = Ultimate F.S.L + 0.5 = 42.50 + 0.50 = 43.00



Chapter - 5

Design of drainage water way

Let us provide 3 vents of 2.5 m wide length of water way, l=3X2.50=7.5 m

Sill level of canal trough = 39.75 m (given)

Thickness of bottom slab = 250 mm (assume)

Thickness of wearing coat = 80 mm (assume)

Bottom level of the canal trough = sill level- thickness of wearing coat-

Thickness of bottom slab

= 39.75-0.08-0.025=39.64 m

Bottom level of slab = 39.75-0.25 = 39.50

Average bed level of the drain = 38.00

Since the M.F.L of the drain = ultimate bed level of the canal

Let us adopt depressed bed level of the drain = 37.00

Depth of water = y1 = Bottom level of the canal trough slab- depressed B.L

Of drain=39.45-37=2.45 m

Let the design velocity in the drain, V1= 3.25 m/s (assume)

Discharge in the drain, Q1 = 60 m3/s

Q1=A1 X V1,

60= A1 X 3.25

A1=18.46 m2

A1=L X y1

18.46 = L X 2.45

L = 7.54 m

Length of barrel = B+ 2X (thickness of side wall)

= 12 + 2X (0.3) =12.60 m

Sri Sunflower College of Engineering & technology

Check for loss of head in the canal due to fluming of canal water way

In a type-1 or type-2 aqueduct, the canal water way is not reduced and is taken over the drain as it is. Hence, there is no loss of head

However, in case of a typeresulting in an increase in velocity through the trough. Unless there is a difference in water levels before entry and after exit, attainment of increased velocity in the trough is not possible.

In aqueduct of short lengths, by limiting the velocity to twicthe loss of head may be very small or almost negligible and hence it is generally ignored. The assumption is that the upstream water surface will in course of time assume a flatter slope to the extent required to drive the flow

However, in large and longer aqueducts it is not so. In order to economies in cost of the canal trough, we may be forced to increase the velocity through the trough. In addition, the length of the trough is an additional factor. These two factors combine to indicate a significant loss of head, which will have to be provided for, while formulating the canal hydraulic particulars. Structures constructed ignoring this aspect will not function properly.

In the present case, to illustrate this aspect, the loss of head in the canal is computed.

Consider section A-A

Canal bed level = +40.00

Full supply level = +42.00

Average velocity, V1 = 0.83 m/s

Velocity head = V

Design of Syphon


Chapter - 6

Check for loss of head in the canal due to fluming of canal water way through the trough

2 aqueduct, the canal water way is not reduced and is taken over the drain as it is. Hence, there is no loss of head in the canal.

However, in case of a type-3 aqueduct, the canal water-way is flumed or reduced, n an increase in velocity through the trough. Unless there is a difference in water levels

before entry and after exit, attainment of increased velocity in the trough is not possible.

In aqueduct of short lengths, by limiting the velocity to twice the normal canal velocity, the loss of head may be very small or almost negligible and hence it is generally ignored. The assumption is that the upstream water surface will in course of time assume a flatter slope to the

through the trough with that bit of extra velocity.

However, in large and longer aqueducts it is not so. In order to economies in cost of the canal trough, we may be forced to increase the velocity through the trough. In addition, the length of

is an additional factor. These two factors combine to indicate a significant loss of head, which will have to be provided for, while formulating the canal hydraulic particulars. Structures constructed ignoring this aspect will not function properly.

present case, to illustrate this aspect, the loss of head in the canal is computed.

Canal bed level = +40.00

Full supply level = +42.00

= 0.83 m/s

V12/2g = 0.832/2x9.81 = 0.037 m


16

Check for loss of head in the canal due to fluming of canal water way

2 aqueduct, the canal water way is not reduced and is taken over the drain as

way is flumed or reduced, n an increase in velocity through the trough. Unless there is a difference in water levels

before entry and after exit, attainment of increased velocity in the trough is not possible.

e the normal canal velocity, the loss of head may be very small or almost negligible and hence it is generally ignored. The assumption is that the upstream water surface will in course of time assume a flatter slope to the

through the trough with that bit of extra velocity.

However, in large and longer aqueducts it is not so. In order to economies in cost of the canal trough, we may be forced to increase the velocity through the trough. In addition, the length of

is an additional factor. These two factors combine to indicate a significant loss of head, which will have to be provided for, while formulating the canal hydraulic particulars.

present case, to illustrate this aspect, the loss of head in the canal is computed.



Total energy line at A-A = F.S.L + (V12/2g) = 42.00+0.037 = 42.037

Consider section B-B

Canal width = 20 m

Depth of water, y =F.S.L- Bed level = 42.00 – 40.00 = 2.00

At B-B the canal is rectangle in cross section, area of flow a= 20x2 = 40 m2

Discharge, Q = 35 m3/s

Velocity of flow, V2 = Q/a = 35/40 = 0.87 m/s

Head loss from A-A to B-B due to change in the velocity V1 to V2

Head loss = (V22-V1

2)/2g

= (0.8752-0.832)/2x9.81

= 0.004

On the U/S end the transition is abrupt and not smooth. So the entire eddy

Loss is taken in to consideration.

The T.E.L at B-B = T.E.L at A-A – eddy loss = 42.035 – 0.004 = 42.031

Section at the entrance of trough C-C

Width of the canal = 12 m (assume)

Depth of canal, y = 2 m

A3 = 12x2 = 24 m2

Velocity, V3 = 3 =3524 = 1.46 m/s

There is a gradual change in c/s from B-B to C-C. There is a loss of head from

B-B to C-C due to change in velocity.

Head loss = 0.25 x ( V32- V2

2)/2g = 0.25 x (1.462-0.8752)/2 x 9.81

= 0.018

T.E.L at C-C with reference to section B-B = 42.035 – 0.018 = 42.017



Consider section D-D

From C-C to D-D, there is a uniform velocity. The loss of head in the trough is only friction loss which manifests itself as surface fall to sustain the velocity.

This loss of head is calculated using manning’s formula, using the value of 0.014 for n

Length of R.C trough = (3x 2.5) + (2x1) + (2x0.5) = 10.5 m

Sectional area in the trough A4 = 12x2 = 24 m2

Velocity developing = 3524 = 1.46 m/s

Wetted perimeter = p = (2x2) + 12 = 16 m

Hydraulic mean depth = R = 4

= 2416 = 1.5 m

Manning’s formula, V = 1 (R)2/3 (S)1/2

1.46 = 1

0.014 (1.5)2/3 (S)1/2

S = 1

4110 Head loss = S x length =

14110 x 10.5 = 0.003 m

T.E.L at D-D = T.E.L at C-C – Head loss = 42.017 – 0.003 = 42.014 m

Consider section E-E

Neglecting the frictional loss in the exit transition the eddy loss in the transition is calculated as follows

Q = 35 m3/s

A5 = (B +n y) y

Here B = 20 m, y = 2 m, n = 12

A5 = (20 + 12 2) 2 = 42 m2

V5 = 5 = 3542 = 0.83 m/s

There is a gradual change in the section from D-D to E-E. There is a change in the velocity.



Head loss = 0.25 x (V42- V5

2)/2g = 0.25 x (1.462- 0.832)/2 x 9.81= 0.018 m

T.E.L at E-E = T.E.L at D-D – Head loss = 42.014 – 0.018 = 41.99 m

Velocity head = V52/2g = 0.832 / 2x 9.81 = 0.0351 m/s

T.E.L at E-E = H.F.L at TEE + Velocity head

41.996 = HFL at TEE + 0.0351

HFL at TEE = 41.961 m

F.S.L at A-A = 42.00 m

Total loss of head = loss of head from A-A to E-E

= TEL at A-A – TEL at E-E

= 42.037 – 41.961 = 0.039 m

Total head loss from A to E = 0.004 + 0.0175 + 0.003 + 0.018 = 0.0425

The total head loss is very small. Hence it can be neglected.



Chapter - 7

Fixing the M.F.L of the drainage

The MFL of the drain in rear of the siphon barrel is 39.75. So, the barrel flows full under maximum flow conditions. The necessary afflux required to push through 60 m3/s with a velocity of 3.25 m/s is calculated by unwins inverted syphon formula,

Afflux, d = (1+f1 +f2 ) 2

2 Here v= velocity in the drain = 3.25 m/s

g= 9.81 m/s2

L = length of drain = 12.60 m

C/S area = 3 x 2.5 x 2.45 = 18.375 m2

Wetted perimeter = p = 3 x (2 x (2.5+2.45)) = 29.7 m

R= hydraulic radius = = 18.375

29.7 = 0.62 m

f1= 0.0505

f2 = a (1+ 0.3 ) = 0.003 (1+ 0.3 0.1

0.62 ) = 0.003145

d = (1+0.505+ 0.00315 12.600.62 ) (

3.25 3.252 9.81 ) = 0.85 m

M.F.L on U/S = D/S M.F.L + afflux = 39.75 + 0.85 = 40.60 m



Afflux at the drop of bed

Bed level of drainage = 38.00

At crossing with canal, bed level of drainage = 37.00

Drop in the bed level = 38 – 37 = 1

M.F.L of the drain = 39.75 m

Bottom level of the canal trough = 39.45 m

Afflux required = 39.75 – 39.45 = 0.3 m

Here drop is 1 m then it is treating it as a drowned weir (submerged weir)

For submerged weir, q = 3.54 y2 d10.5 + 1.77 d1

3/2

y2 = depth of water = 2.60 m

q = 2 = 6011 = 5.45 m2/s

B2 = width of drainage b/w wing walls

5.45 = 3.45 (2.60) d10.5 + 1.77 d1

3/2

5.45 = 8.97 d10.5 + 1.77 d1 d1

0.5

5.45 = d10.5 (8.97+1.77 d1)

5.452 = d1 (8.97+1.77 d1)2

29.70 = d1 (80.46+3.14 d12+31.75 d1)

3.14 d13 + 31.75 d1

2 + 80.46 d1 – 29.70 = 0

d1= 0.326

M.F.L over the drop = U/S M.F.L + afflux = 40.60 + 0.32 = 40.92 m



Chapter - 8

Design of side walls of canal trough

Bottom level of the side wall = bottom level of the canal trough

Top level of the side wall = U/S F.S.L + 0.5 = 42.50 + 0.5 = 43.00

Depth of water h = U/S F.S.L – Sill level = 42.50 – 39.75 = 2.75 m

Let top thickness = 2 m

It is designed as a cantilever wall

� of water = 10 KN/m3

Water pressure = p = � x h = 10 x 2.75 = 27.5 KN/m2

Let us consider 1 m length of side wall

Total pressure, p= Area of pressure diagram x length of wall

= 12 x 27.5 x 2.75 x 1 = 37.81 KN

Centre of pressure,y = ℎ3 =

2.753 = 0.917 m

Bending moment, m = p x y = 37.81 x 0.917 = 35KN-m

Mu = 1.5 x 35 = 52.17 KN-m

Adopting M-20 grade concrete & Fe-415 grade steel



= 0.48

D = 300 mm, b = 1000 mm, d1 = 40 mm, d = D - d1

d = 300 – 40 = 260 mm

Mu, limit = 0.36

bd2 fck (1 – 0.42

)

= 0.36 x 0.46 x 1000 x 2602 (1 – 0.42 x 0.48)x20

= 186.6 KN-m

Mu< Mu, limit (O.K)

Design of steel reinforcement

(a) Main steel :1) Minimum area of steel,

Ast, min = 0.12% of gross area

= 0.12100 x 1000 x 300

= 360 mm2

2) Maximum area of steel ,Ast, max = 4% of gross area

= 4

100 x 1000 x 300

= 12,000 mm2

3) Mu1 = 0.87 fyAst d (1 –

)

52.17 x 106 = 0.87x415x260x Ast1x

(1-- 415

1000 260 20)

555.75 = Ast1- Ast12 (7.98 x 10-6)

Ast1 = 582.86 mm2 , 11947.25 mm2

Ast1< Ast1, max (O.K) Ast1 > Ast1, min (O.K)

For Ast1 = 11947.25 mm2

=

0.87 0.36 =

0.87 415 11947.250.36 20 1000 260 = 2.304

= 0.48

>

(Not O.K)



For Ast1 = 583 mm2

=

0.87 415 5830.36 20 1000 260 = 0.112

<

(O.K)

Ast1 = 583 mm2

Let us adopt 12 mm dia bars, �1 = 12 mm

Area of one bar, A � = 3.14

4 x 122 = 113 mm2

Number of bars, n1 = Ast1A � =

583113 = 5.16

Spacing, S1 = 1000

n1 = 193.8 mm

Adopt 12 mm � @ 180 mm C/C

Check for shear design

Nominal shear force, τv = = 37.81 1031000 260 = 0.145 N/mm2

Percentage of steel provided = 100

=100 583

1000 260 = 0.224%

As per IS 456 – 2000

For, M-20, 100

= 0.224% then τc = 0.81 N/mm2

τv<τc (O.K)

As per IS – 456 – 2000, For M-20 grade concrete

τc, max = 2.8 N/mm2

τv<τc, max (O.K)

Hence provide minimum shear reinforcement,

as per IS-456-2000, Sv is the least of

1) = 0.4

0.87 Let us adopt 4 legd stirrups of 8 mm dia

Asv = 4 x 3.14

4 x 82 = 200 mm2



2001000 =

0.40.87 415

Sv = 180 mm2) 0.75 X d = 0.75 x 260 = 195 mm3) 300 mm

Sv = 180 mm Adopt 4 legd vertical stirrups @ 8 mm dia @ 150 mm C/C

Distribution steel

Providing steel on both faces

1) Area of steel , Ast2 =

2 = 360

2 = 180 mm2

2) Let us adopt dia of bar, �2 = 8 mm

Area of one bar, A �2 = 3.14

4 x 82 = 50 mm2

Number of bars, n2 =Ast2A � =

18050 = 3.6

Spacing, s2 = 1000

n2 = 1000

3.6 = 277.7 mm

Adopt 8mm dia @ 250 mm C/C



Chapter - 9

Design of bottom slab of canal trough

It is designed as a continuous slab.

Let us consider 1m wide slab

Thickness of wearing coat = 8 cm

Weight of wearing coat = 8

100 x 24 = 1.92 KN/m2

Thickness of slab = 25 cm

Weight of slab = 25

100 x 25 = 6.25 KN/m2

Depth of water = 2.75 m

Weight of water = 2.75 x 10 = 27.5 KN/m2

Total load on the slab = 1.92 + 6.25 + 27.5

= 35 KN/m2

Effective span (l) = clear water way + effective thickness of slab

= 2.5 + 0.26 = 2.76 m

Maximum B.M = M3 = 2

10 = 35 2.76 2.76

10 = 26.67 KN-m

Mu3 = 1.5 x M1 = 40 KN-m

Mu, limit = 0.36

bd2 fck (1 – 0.42

)

= 0.36x0.48x (1 – 0.42 x 0.48) x1000x260 x 20

= 186.6 KN-m

Mu3<Mu,limit (O.K)



Design reinforcement

a) Main steel 1) Minimum area of steel , Ast, min = 0.12% of gross area

= 0.12100 x 1000 x 300

= 360 mm2

2) Maximum area of steel ,Ast, max = 4% of gross area

= 4

100 x 1000 x 300

= 12,000 mm2

3) Mu3 = 0.87 fyAst d (1 –

)

40 x 106 = 0.87 x 415 x Ast3 x 260 x (1—3 415

1000 260 20)

426.10 = Ast3- Ast32 (7.98 x 10-5)

Ast3 = 441.66 mm2

= 12089 mm2

Ast3 = 442 mm2

Ast3<Ast, max (O.K) Ast3 >Ast,min (O.K)Let us adopt 12mm dia bars, �3 = 12 mm

Area of one bar, A �3 = 3.14

4 x 122 = 113 mm2

Spacing, s3 = 1000

n3 = 10003.91 = 255.6 mm

Adopt 12 mm dia bars @ 250 mm c/c

b) Distribution steelsProviding steel on both faces

Area of steel, Ast4 =

2 = 360

2 = 180 mm2

Adopt 8mm dia @ 250 mm c/c.

Check for shear design

Shear force, V = 2 = 35 2.76

2 = 48.3 KN

Nominal shear force, τv = = 48.3 10 10 10

10 10 10 260 = 0.185 N/mm2

As per IS-456-2000, for M-20 grade , τc, max =2.8 N/mm2

τv<τc, max (O.K)

Percentage of steel, 100

= 100 442

10 10 10 260 = 0.17%



As per IS-456-2000, for M-20 grade, 100

= 0.17% then

τc = 0.3 N/mm2

τv<τc (O.K)

Minimum shear reinforcement is provided.

Sv is least of

1) = 0.4

0.87 20010 10 10 =

0.40.87 415

Sv = 180 mm2) 0.75 d = 195 mm3) 300 mm

Sv = 180 mmAdopt 4 legd vertical stirrups @ 8 mm dia @ 180 mm c/c



Chapter -10

Design of tail channel

Tail channel will be always straight, its length will be 50 to 60m on either side of roads

Top level = M.F.L of drain

= 39.75m

Bottom level = bed level of the drain at crossing = 37.00m

Depth of flow, Y3 = 39.75-37.00 = 2.75m

Let us assume velocity of flow, V = 1.5m/sec

Q = A×V

60 =1.5A

A = 40m2

Slope = 12H: 1V (assume)

A = (B+ny) y

40 = (B+0.5×2.75)2.75

B = 13.17m

Adopt B = 14m

Slope of channel,

Manning’s formula, V = 1

(R)2/3 (S)1/2

Here, V = 1.5m/sec, N = 0.015 (assume)

A = (B+ny)y

= (14+0.5×2.75)

A = 42.3m2

Wetted perimeter, P = B+2 ( 2 + ) .y

= 14+2( ( . )2 + )×2.75


= 20.15m

R = = = 2.1m

V = (R)2/3 (S)1/2

1.5 = (2.1)2/3 (S)1/2

S =

Design of Syphon



32

Canal Aqueduct

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