CE404 05 Canal Falls

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Hydraulic Structures Canal Falls January 16, 2011 1 Canal Falls It is neither necessary nor economical to design every fall as meter flume. A vertical fall may also be used as a meter when working head is more than 0.33 H and the d.s F.S.L. is up to or lower than the crest level. (H is the depth of crest below T.E.L.). Hydraulic Design of Canal Falls 1. Vertical drop fall The energy is dissipated by means of impact and deflection of velocity, suddenly from the vertical to the horizontal direction. Cistern length 1/ 2 5 c L L HD Cistern depth 2/3 1 4 larger of 3 L c X H D X d X 2 3 c q d g Where: is the length of cistern, is the depression below d.s. bed, is the depth of crest below u.s. T.E.L., is drop in meters. c L L X D H 2. Glacis fall The energy is dissipated by the formation of hydraulic jump. From H L and q, Ef 2 can be found from Blench curves Fig. 3.5. To make sure that the hydraulic jump will form on the glacis, the depth of cistern is increased by 25% of Ef 2 .

Transcript of CE404 05 Canal Falls

Page 1: CE404 05 Canal Falls

Hydraulic Structures – Canal Falls January 16, 2011

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Canal Falls

It is neither necessary nor economical to design every fall as meter flume. A vertical fall may also be

used as a meter when working head is more than 0.33 H and the d.s F.S.L. is up to or lower than the

crest level. (H is the depth of crest below T.E.L.).

Hydraulic Design of Canal Falls

1. Vertical drop fall

The energy is dissipated by means of impact and deflection of velocity, suddenly from the

vertical to the horizontal direction.

Cistern length 1/2

5c LL H D

Cistern depth

2/31

4 larger of

3

L

c

X H D

Xd

X

2

3c

qd

g

Where: is the length of cistern,

is the depression below d.s. bed,

is the depth of crest below u.s. T.E.L.,

is drop in meters.

c

L

L

X

D

H

2. Glacis fall

The energy is dissipated by the formation of hydraulic jump. From HL and q, Ef2 can be

found from Blench curves Fig. 3.5.

To make sure that the hydraulic jump will form on the glacis, the depth of cistern is

increased by 25% of Ef2.

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In case the d.s. bed level is lower than cistern level, the cistern level should be provided at

bed level. Where the d.s. bed level is higher than cistern level, the cistern level is joined at a

slope 5H:1V.

Sufficient length of cistern is provided so that turbulence dies out and subcritical flows takes

place before the water leaves the floor. Length of cistern of 5Ef2 is provided for normal soils

and 6Ef2 for sandy soils.

2(5 6)cL Ef

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Design of Sharda Type Fall

1. Crest

i. Length of crest:

The length of crest is kept equal to bed width, it is also possible to extend the length

to bed width plus depth.

ii. Shape of crest

For Q < 15 cumec, the section is kept rectangular with the d.s. face absolutely

vertical. The top width 10.55tB D and minimum base width

1 2D where D1 is

the height of crest above d.s. bed level.

It may be capped with 25 cm 1:2:4 cement concrete with its both ends rounded.

For 15Q cumec , a trapezoidal section with top width 10.55tB D D with u.s.

side slope of 1H:3V and segmental top conforming to a quadrant of a circle of radius

0.3 m at d.s. edge of crest width and d.s. slope of 1H: 8V is adopted.

a. Rectangular crest fall

b. Trapezoidal fall

iii. Crest level

The following equation is used to determine the height of the crest:

1/6

3/2

t

t

DQ C L D

B

For submerged flow conditions (above 33% submergence) discharge passing over

crest is:

3/2 1/2

2

22 2

3d t L a a d t L aQ C L g H h h C L h g H h

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where,

Lt = Length of crest

Bt = Width of crest

HL = Drop in water surface

h2 = Depth of d.s. water level over top of crest

ha = Head due to velocity of approach 2 2aV g

C = 1.84 for rectangular crest

2.26 for trapezoidal crest

Crest level (free fall) d.s. F.S.L. ah D

Crest level (Submerged falls) 2d.s. F.S.L. a Lh H h

Degree of submergence 1 dh

D

2. U.S. Approaches

For discharges larger than 15 cumec, the wing walls are kept segmental with radius equal to

5 6 times D making an angle 60º at center, and carried tangentially into the berm. The

foundations of wing walls are laid on impervious floor itself.

For falls of discharges less than 15 cumec, the approach wings may be splayed straight at an

angle of 45º.

i. U.S. protection

Brick pitching in a length equal to u.s. water depth should be laid on u.s. bed towards

the crest at a slope 10H:1V.

ii. U.S. curtain wall (cutoff)

The thickness of curtain wall equal to 1½ brick and depth equal to 1/3 u.s. water

depth + 0.6 m, be provided with min. 0.8 m also considering Lacey’s criteria (Table

6.1).

HL

d

D

P

hd

T.E.L.

T.E.L.

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3. Impervious concrete floor

i. Total floor length and its disposition:

Khosla’s method is used for large works. Bligh’s theory is used for small works.

The minimum length of floor on the d.s. side for clear falls and submergence less

than 33% is:

10.53 4.877 1.5p c LL d H

For submerged falls with submergence greater than 33%,

6.77 5.182p c LL d H

The balance of total length maybe provided under and upstream of crest.

4. Cistern

i. Length of cistern

For clear falls and submergence less than 33%,

3.8 0.415c c LL d H

For submerged falls with submergence greater than 33%,

5.2 1.067c c LL d H

ii. Depth of cistern (X)

2/31

4 larger of

3

L

c

X H D

Xd

X

5. D.S. Protection

i. Bed protection

Brick pitching about 20 cm thick resting on 10 cm ballast in a length equal 3 times

d.s. water depth.

Toe wall 1½ brick thick and of depth = ½ (d.s. water depth) with minimum 0.6 m

provided at the end of pitching.

ii. Side protection

After wing walls, the side slopes of the channel are pitched with 1 brick on edge in a

length equal to 3 times d.s. water depth. The pitching should rest on a toe wall 1½

brick thick and of depth equal to ½ d.s. water depth.

iii. Curtain wall

The thickness of curtain wall may be 1½ brick thick and depth = ½ (d.s. water depth)

with minimum 1 m, also consider Lacey’s criteria.

iv. D.S. wings

For Q > 15 cumec, d.s. wings are kept vertical for a length 5to8 LD H and may

then be gradually warped. They should be taken up to the end of pucca floor.

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Design Example 1:

Design a Sharda type fall with the data given below:

i. Full supply flow rate . .

10. .

u scumec

d s

ii. Drop 1 m

iii. Full supply depth . . 1.5

. . 1.5

u s m

d s m

iv. Bed level . . 100

. . 99

u s m

d s m

v. Bed width . . 8

. . 8

u s m

d s m

vi. Side slopes 1 H: 1V

vii. Soil Good loam

viii. Bligh’s coefficient 7

Solution

1. Length of crest

Take crest length 8tL m

2. Crest level

Since Q < 15 cumec, use rectangular crest with both sides vertical.

1/63/21.84 t tQ L D D B

Assume 0.8tB m

3/2 1/6

1/6

110 1.840 8

0.8

0.776 say 0.78

D D

D m m

Velocity of approach (with 1:1 side slopes)

100.702

8 1.5 1aV m s

Velocity head 2 20.702

0.0252 2 9.81

aVm

g

U.S. T.E.L. u.s. F.S.L. 101.5 0.025 101.525

R.L. of crest u.s. T.E.L. 101.525 0.78 100.745 say 100.75

ah m

D m

Adopt crest level = 100.75 m

3. Shape of crest

i. Top width

1 10.55 , 100.75 99 1.75

0.55 100.75 99 0.73

tB D D m

m

Adopt Bt = 0.75 m.

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Check for Q,

1/6

1.67 110 1.84 8

0.75

0.741

D

D

ii. Width of base 10.5

0.5 1.75 0.875 say 1

D

m

Its top shall be capped with 20 cm thick cement concrete.

4. Side walls

The side walls may be splayed straight at an angle of 45º from the u.s. edge of the crest and

extending by 1 m in the earthen bank from the line of F.S.L.

5. D.S. expansion

Side walls shall be straight and parallel up to the end of floor and shall be kept vertical.

6. U.S. protection

Brick pitching in a length equal to u.s. water depth = 1.5 m should be laid on the u.s. with a

slope of 1:10 downwards and 3 pipes of 15 m diameter at the bed should be provided for

drainage during maintenance (cleaning).

7. Cistern elements

i. Depth of cistern

3

cdX , and

2 2

331.25

0.5429.81

c

qd m

g

0.5420.181

3X m

or

2/3

2/3

1

4

11 0.75 0.21

4

LX H D

m

Cistern depth X= 0.21 m say 0.25 m

ii. Length of cistern

3.8 0.415

3.8 0.542 0.415 1 3.47

c c LL d H

m

or

1/2 1/2

5 5 1 0.75 4.3c LL H D m

Provide 4.5 m long cistern at R.L. 98.75 m

8. Length of impervious floor

Bligh’s coefficient = 7

Maximum static head 100.75 99 1.75m

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Total floor length required 7 1.75 12.25m

Minimum d.s. floor length required

10.53 4.877 1.5

10.53 0.542 4.877 1.5 1.0

9.08 say 9.0

p c LL d H

m

Provide d.s. floor 9 m long and the balance 3.25 m under and u.s. of the crest.

9. Floor thickness

Minimum floor thickness of 0.3 m should be provided at the upstream region.

Maximum uplift head at the toe of crest

1.75

12.25 3.25 1.2912.25

m

Floor thickness required1.29

1.031.25

m

Provide 1.05 m thick concrete overlaid with 0.2 m thick brick pitching.

Max. uplift head at 2.25 m d.s. from the toe of crest

1.75

12.25 5.5 0.9612.25

Floor thickness required0.96

0.771.25

m

Provide 0.8 m thick concrete overlaid with 0.2 m brick pitching.

Floor thickness required at 4.5 m d.s. the toe of crest

1.75 12.25 7.75

0.5112.25 1.25

m

Provide 0.55 m thick concrete overlaid with 0.2 m brick pitching.

Floor thickness at 6.75 d.s. from the toe of crest

1.75 12.25 100.26

12.25 1.25m

Provide 0.3 m thick concrete overlaid with 0.2 m brick pitching.

10. Curtain walls

a. D.S. curtain wall

The curtain walls at d.s. end of floor should be 1½ brick thick and of depth 2 0.6d m

to a minimum of 1 m.

Depth of curtain wall at d.s. end floor1.5

0.6 1.352

m

Provide 0.4 1.4m m deep curtain wall.

b. U.S. curtain wall

Depth of curtain wall at u.s. end flooru.s. water depth

0.6 0.5 0.6 1.13

m

Provide 0.4 1.1m m deep curtain wall.

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11. D.S. Protection

a. Bed protection

Length of bed protection33 3 1.5 4.5D m

Provide 4.5 m long dry brick pitching resting on 10 cm ballast which should be protected

by a toe wall 0.4 m wide and 0.8 m deep (half d.s. water depth).

b. Side protection

For length similar to that of bed, provide dry brick pitching 0.2 m thick on sides resting

on a wall of 0.4 m and 0.8 m deep.

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Design Example 2:

Design an unflumed non-meter straight glacis fall with the following data:

ix. Full supply flow rate . .

30. .

u scumec

d s

x. Drop 1 m

xi. Full supply depth . . 1.4

. . 1.4

u s m

d s m

xii. Full supply level . . 102

. . 101

u s m

d s m

xiii. Bed level . . 100.6

. . 99.6

u s m

d s m

xiv. Bed width . . 23

. . 23

u s m

d s m

xv. Side slopes 1 H: 1V

xvi. Safe exit gradient 1/6

Solution

1. Length of crest

Take crest length equal to the width of channel 23tL m

2. Crest level

3/21.84 tQ L D

3/230 1.84 23

0.8

D

D m

30

0.8823 1.4 1.4

aV m s

Velocity head 2 20.88

0.0395 say 0.042 2 9.81

aVm m

g

U.S. T.E.L. u.s. F.S.L. 102 0.04 102.04

Crest level u.s. T.E.L. 102.04 0.8 101.24 say 101.2

ah m

D m

Provide crest at level = 101.2 m

3. Width of crest

2 20.8 0.53

3 3tB D m

Provide crest width Bt = 0.6 m.

4. D.S. glacis

Glacis slope of 2 in 1 joined tangentially to the cistern floor with a radius equal to D = 0.8 m

shall be provided.

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5. Approach

a. U.S. glacis: Glacis slope may be ½ H: 1 V joined tangentially to the crest with a radius

equal to D/2 = 0.4 m.

b. Sides: The side walls may be splayed at an angle of 45º from the end of u.s. glacis and

extended by 1 m into the earthen bank from the line of F.S.L.

6. U.S. cutoff

Depth of cutoff 1 1.40.6 0.6 1.067 say 1.1

3 3

dm m

But minimum depth in Table 6.1 is 1.2 m. Therefore, provide 0.5 1.2m m deep cutoff wall.

7. Cistern

a. Depth of cistern

2

2

301.305

23

1.0

1.22 Fig. 2.7

L

q m s

H m

Ef m

21.25 1.525Ef m

R.L. of cistern

2d.s. T.E.L. 1.25

101 0.04 1.525

99.515

Ef

m

Or R.L. of cistern 2d.s. bed level 0.25

99.6 0.25 1.22 99.295

Ef

m

Provide cistern level at R.L. 99.2 m

b. Length of cistern

5 1.22 6.1cL m

Provide cistern length = 6.4 m

The cistern shall be joined to the d.s. bed in a slope of 1:5 in a length of 2 m.

c. D.S. cutoff wall

Depth of cutoff wall below bed 3 1.40.6 0.6 1.3

2 2

dm

Provide 0.5 1.5m m deep cutoff wall and the cutoff wall shall project above bed level by

3 1.40.14 say 0.15

10 10

dm m to act as a deflector.

d. Bed protection: Nil

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e. Side protection

Length of d.s. side protection33 3 1.4 4.2d m

Provide 0.2 m thick dry brick pitching in 4.5 m length beyond impervious floor. The

pitching shall rest on 0.4 0.8m m toe wall including 0.3 m thick foundation concrete.

f. Friction blocks: No friction blocks are required.

8. Total floor length and exit gradient

Exit gradient 1 6EG

Max. static head 101.2 99.6 1.6H m

1/2

2

1

1 1.6 1

6 1.5

4.15

2 1 1

7.23

E

HG

d

Total floor length required

7.23 1.5 10.85

b d

m

Provide total floor length = 14.2 m

9. Pressure calculations

Assume u.s. floor thickness = 0.3 m and d.s. floor thickness near d.s. cutoff = 0.6

a. U.S. cutoff

1

14.2 , 1.2

1 1.20.08

14.2

100 100 18 82%D D

b m d m

d

b

1

26

100 26 74%

0.3 correction due to floor thickness 82 74 2%

1.2

E

C

C

b. D.S. cutoff

Max (H) Cutoff depth (m) Length (m)

1.6 1.3 13.10

1.6 1.4 11.90

1.6 1.5 10.85

2.0 1.5 17.90

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1.5

14.2

d m

b m

1 1

1 1.50.106

14.2

30%, 20%E D

d

b

1

1

0.6 correction for depth 30 20 4%

1.5

corrected 30 4 26%

E

E

c. At the toe of glacis

% Pressure 76 26

26 8.413.2

57.82%

Floor thickness0.5782 1.6 0.4

1.061.25

m

Provide 1.1 m thick concrete floor at the toe of glacis in a length of 2 m.

At 2m from the toe of glacis

% Pressure 76 26

26 6.413.2

50.24%

Floor thickness0.5024 1.6 0.4

0.9631.25

m

Provide 1.0 m thick concrete floor at the toe of glacis in a length of 2.5 m.

At 4.4 m from the toe of glacis,

% Pressure = 41.15%

Provide 0.9 m thickness.

At 6.9 m, % pressure = 31.68%

Provide 0.8 m thickness until the end of the floor.

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