Head and Cross Regulators

Hydraulic Structures Head and Cross Regulators Dr. Bahzad M.A. Noori 29-Dec-11

1

Head and Cross Regulators

The supplies passing down the parent canal and off take channel are controlled by cross regulator

and head regulator respectively.

Functions of Cross Regulators

1. Regulation of the canal system.

2. Raising the water level in the main canal in order to feed the off take channels.

3. To facilitate communication by building a road over the cross regulator with little extra

cost.

4. To absorb the fluctuations in the canal system.

Functions of Head Regulators

1. To regulate and control supplies entering the off take channel (distributary) from the

main (parent) canal.

2. To control silt entering into the distributary.

3. To serve for measurement of discharge.

Alignment

The best alignment of the off take channel is when it makes angle zero with the parent canal

initially and then separates out in a transition. See Fig. 13.1. In this case there is a transition

curve for both off take and parent channel to avoid silt accumulation.

Another alternative by making both channels an angle with respect to parent channel upstream.

Fig. 13.2

Cross Regulator

OFF TAKE CHANNELDistributary channel

Head Regulator

Abutment

Pier

Gate


2

In case of obligatory straight alignment of the parent channel, the usual angle of the off take

channel is 60 to 80 (in most important works needs a model study). For excessive silt entry into

the off take channel. Fig. 13.3.

Design Criteria

1. Waterway

The effective waterway of head regulator should not be less than 60% of bed width of off

taking channel and mean velocity should not exceed 2.5 m/sec.

2. Crest level

Crest level of the distributary head regulator is generally kept 0.3 m to 0.6 m higher than

crest level of cross regulator (C.R.). The crest level of C.R. is provided at bed level of

parent canal.

eH should be worked out from the formula

3/2

e eQ C B H (4.1)

where

C Coefficient of discharge

eB Effective length of crest 2t p a eB N K K H (4.2)

tB Net length of the crest

N Number of piers

pK Pier contraction coefficient

aK Abutment contraction coefficient


3

Table: Coefficients of contraction for piers and abutments.

Type of pier pK

Square nosed pier 0.02

Round nosed pier 0.01

Pointed nosed pier 0.01

Type of abutment aK

Square abutment 0.2

Round abutment 0.1

3. Coefficient of discharge (C)

The coefficient of discharge C is 1.84 for crests of width less than or

equal to 23 e

H . In case of submerged falls, C should be reduced

depending on the drowning ratio, see Fig. 6.5.

1.840 (H.R. crest)

1.705 (C.R. crest)

C

C

4. Shape of crest

The u.s face of the crest should be given a slope of 1:1. The d.s. sloping glacis should not

be steeper than 2:1.

5. Crest width should be kept equal to 23 e

H .

6. Vertical cut-offs

The cut-off should be provided at the end of u.s. and d.s. floors for safety against scour,

undermining and exit gradient. Due to Laceys scour depth

T.E.L.

He

>2/3 He

Broad Crested Weir

Round nosed pier

Pointed nose pier

Square nosed pier


4

Table 6.1 Minimum depth of u.s. and d.s. cut-offs

Canal capacity

cumec

Min. depth of u.s. cut-off

below bed level or G.L.

whichever is lower (m)

Min. depth of d.s. cut-off

below bed level or G.L.

whichever is lower (m)

Up to 3 cumec 1.0 1.0

3.1 - 30 1.2 1.2

30.1 - 150 1.5 1.5

Above 150 1.8 1.8

7. Thickness of top coat

Table 6.2 Thickness of top coat.

Canal capacity (cumec) Thickness of top coat (mm)

1.5Q 100

1.5 30Q 150

30 150Q 200

150Q 300

8. Freeboard Table: Minimum Freeboard

Canal capacity (cumec) Freeboard (m)

1.0Q 0.3

1 10Q 0.4

10 30Q 0.6

30 150Q 0.8

150Q 1.0

9. Protection works

Downstream of floor, properly designed filter loaded by concrete blocks should be

provided. The length of inverted filter is kept equal to 2D (D is the depth of d.s. cut-off

below d.s. bed). Details of minimum thickness of the filter are given in Table 6.3. The

width of gabs between the blocks shall not be more than 50 mm which should be packed

with biggest size of pebbles available. Beyond the filter, an apron of 1.5D length shall be

provided. Similar protection is also provided in the u.s. in a length equal to D. the cubic

content of material in launching apron should be equal to 32.25D m m m length.


5

Design Example

Design a cross regulator and a suitable head for a distributary which takes off at an angle

of 60 from a canal which discharges 120 cumec.

Discharge of distributary = 10 cumec

Bed width of distributary = 10 m

Water depth of distributary = 1.2 m

Full supply level of distributary = El. 310.2 m

Full supply level of parent channel (F.S.L.) u.s. 311.00

d.s. 310.85

m

m

Bed width of parent channel u.s. 60

d.s. 56

m

m

Depth of water in parent channel u.s. 2

d.s. 2

m

m

Safe exit gradient (G.E.) 1

5

Solution

A. Design of Cross Regulator

1. Fixation of crest level and waterway of cross regulator

Crest level of C.R. F.S.L. of parent channel - water depth

311- 2 309m

u.s. d.s.T.E.L. T.E.L.

311 310.85 0.15

dh

m

u.s.F.S.L. crest level

=311 309 2

eH

m

Degree of submergence

0.150.075

2

d

e

h

H

From Fig. 6.5

0.56, where 1.705

1.705 0.56 0.95

s

s

CC

C

C

3/2

3/2120 0.95 2

e

e

Q CB H

B

44.66 say 45eB m m

Assume 6 bays of 8 m each = 48 m

2( )

45 2 5 0.01 0.2 2

46 48 O.K.

t e p a eB B N K K H

m m


6

Provide 5 piers with rounded nose of width 1.6 m each.

Total waterway 6 8 5 1.6 56 O.K.m

2. Level and length of downstream floor

3

120

1202.5 / /

48e

Q cumec

Qq m sec m

B

2

u.s. d.s. u.s. d.s.T.E.L. T.E.L. F.S.L. F.S.L. ( 2 is negligible)

311 310.85 0.15

LH v g

m

From Blench Curves, Fig. 3.5

2 1.435Ef m

d.s. floor level d.s. 2F.S.L.

310.85 1.435 309.415

Ef

m m

Actual cistern level 310.85 2 308.85m lower than 309.415 m O.K.

Cistern length 2 1

2 1

5 firm soil

6( ) weak soil

cL D D

D D

1 2 1.435 0.15 1.585LEf Ef H m

From energy of flow curves (Fig. 2.7)

1 1

2 2

1.575 gives 0.534

1.435 gives 1.3

5 1.3 0.534 3.83c

Ef m D m

Ef m D m

L m

3. Vertical cutoffs

120Q cumec

Referring to table 6.1, the minimum depth of u.s. and d.s. cutoff = 1.5 m.

u.s. cutoff is at El. 309 1.5 307.5m 4. Total floor length and exit gradient

1E

HG

d

u.s. d.s.F.S.L. G.L.

311 308.85 2.15

H

m

Trial and error

d (m) L d (m)

1.5 5.20 9.15 14

1.6 4.57 8.08 12.93

1 2.15 1, =4.57

5 1.6


7

1/22

1/22

2 1 1

2 457 1 1 8.08

Total floor length

8.08 1.6 12.93 say 13

L d

m m

d.s. cutoff is at El. 308.85 1.6 307.25m

d.s. floor length = 6.5 m

d.s. glacis length with (2H:1V) slope = 2(309-308.85) = 0.3 m

u.s. floor length = 6.2 m

Total floor length = 13 m

5. Pressure calculation

a. Upstream cutoff

1

1

1.5 , 13

1 1.50.115

13

100 20 80%

100 28 72%

D

C

d m b m

d

b

Correction of C for floor thickness

0.6

80 72 3.2%1.5

D C

t

d

corrected 72 3.2 75.2% of C H

b. Downstream cutoff

1.6 , 13

1 1.60.123

13

31%

22%

E

D

d m b m

d

b

Correction of E for floor thickness 0.6

31 22 3.38%1.6

corrected 31 3.38 27.62% of E H


8

6. Floor thickness

a. d.s. floor

1. at 2 m from d.s. end

75.2 27.62%pressure 27.62 1.5

12

33.6%of H

Head % pressure

33.6=2.15 0.722 of water

100

H

m

Minimum concrete thickness min

Head 0.7220.58

2.25 1c wt m

Provide 0.6 m thick concrete floor for 1.5 m.

2. at 4.5 m from d.s. end

75.2 27.62%pressure 27.62 4.0

12

43.48%of H

Head % pressure

43.48=2.15 0.934 of water

100

H

m

min

Head 0.9340.75

2.25 1c wt m


3. at 6.5 m from d.s. end

75.2 27.62%pressure 27.62 6.0

12

54.41%of H

min

2.15 0.51410.9

2.25 1t m

Provide 1.0 m thick concrete floor for the rest of the glacis.

7. Upstream protection

i. Block protection (inverted filter)

Length of protection = depth of cutoff (D) =1.5 m

Provide 2 rows of 0.8 0.8 0.6m m m C.C. blocks over 0.6 m thick

inverted filter.

ii. Launching apron

Volume per meter length 3 22.25D m m m

If thickness is 1.0 m, then the required length

2.25 2.25 1.53.37

1.0 1

Dm

Provide 3.5 m long launching apron.

8. Downstream protection

i. Block protection

Length of protection 2 2 1.6 3.2D m


9

Provide 4 rows of 0.8 0.8 0.6m m m blocks over 0.6 m inverted filter.

ii. Launching apron

Volume per meter length 3 22.25D m m m

If thickness is 1.0 m, then

Required length2.25 2.25 1.6

3.61.0 1

Dm


Provide 0.4 m thick and 1.2 deep toe wall between filter and launching apron.

B. Design of distributary head regulator

1. Fixation of crest and waterway

The crest level should be provided 0.5 m higher than upstream floor level.

Crest level 309 0.5 309.5m

311 309.5 1.5eH m

311 310.2 0.8dh m

0.80.533

1.5

d

e

h

H

0.98sC

C (From Fig. 6.5)

1.84 0.98 1.80sC

The effective width of waterway is found by

1.5

1.510 1.8 1.5

3

e e

e

e

Q C B H

B

B m

Provide 60% of distributary width60

10 6100

m

Provide 2 bays of 3 m each separated by 1 m thick pier.

Therefore, the overall waterway 2 3 1 7m

2. Level and length of d.s. floor

10 , waterway 6

101.67

6

Q cumec m

q m sec

Head loss 311 310.2 0.8LH m

Using Blench curves (Fig. 3.5)

2 2

1 2 1

1.37 gives 1.32

1.37 gives 0.32L

Ef m D m

Ef Ef H m D m

Cistern length

2 15

5 1.32 0.32 5.0

cL D D

m

Provide cistern length = 6m


10

Floor level required d.s. 2T.E.L.

310.2 1.37 308.83

Ef

m

Provide d.s. floor level at El. 308.8 (Ground Level)

3. Vertical cutoffs

a. u.s. cutoff

Provide u.s. cutoff depth 1.5 m

The bottom of cutoff 309 1.5 307.5m

b. The minimum d.s. cutoff depth = 1.2 m

4. Total floor length and exit gradient

Maximum static head u.s. d.s.F.S.L. F.L.

311 308.8 2.2m

Exit gradient

1E

HG

d

1/2

22 1 1

1 2.2 1

5 1.2

With trial and error

d (m) L d (m)

1.2 8.50 15.99 19.00

1.4 6.26 11.47 16.05

1.6 4.75 8.50 13.62

1.7 4.24 7.42 12.60

Get 13 , 1.7L m d m

d.s. floor length (cistern length) = 6.0 m

d.s glacis length 2H:1V slope 2 309.5 308.8 = 1.4 m

Crest width 2 23 3 1.5eH = 1.0 m

u.s. glacis length 1H:1V slope 1 309.5 309.0 = 0.5 m

u.s. floor length = 4.1 m

Total floor length = 13 m

5. Pressure calculations

a. Upstream cutoff

1

1

1.5 , 13

1 1.50.115

13

100 20 80%

100 28 72%

D

C

d m b m

d

b


11

Assuming the u.s. floor thickness = 0.6 m

Correction of C for floor thickness 0.6

80 72 3.2%1.5

corrected 72 3.2 75.2% of C H

b. Downstream cutoff

1.7 , 13

1 1.70.13

13

32%

22%

E

D

d m b m

d

b

Correction of E for floor thickness 0.6

32 22 3.5%1.7

corrected 32 3.5 28.5% of E H

6. Floor thickness

a. d.s. floor

1. at 2 m from d.s. end

75.2 28.5%pressure 28.5 1.5

12

34.33%of H

Minimum concrete floor thickness min

0.3433 2.20.6

2.25 1t m

Provide 0.6 m thick concrete floor for 1.5 m length.

2. At 4 m from d.s. end


3. At the toe of glacis (beginning of the hydraulic jump)

5.5

%pressure 28.5 75.2 28.512

49.9%of H

Unbalanced head 0.499 2.2 1.095m

Unbalanced head due to dynamic condition

2 150%

1 49.91.32 0.32 0.8

2 100

0.898

toe LD D H

m

min

1.0950.88

2.25 1t m

Provide floor thickness 1.0 m.

b. d.s Floor thickness


12

Same as provided in u.s. floor for the cross regulator (minimum thickness of 0.6

m in the u.s. which should be thickened under the crest).

7. u.s Protection

Same as provided in the u.s. of cross regulator.

8. d.s. protection

i. Block protection (inverted filter)

Length of filter 2

2 1.7 3.4

D

m

Provide 6 rows of 0.6 0.6 0.4m m m C.C. blocks over 0.4 m thick graded

filter.

ii. Launching apron

Volume per meter length 32.25 2.25 1.7 3.825D m m

Assume thickness of launching apron = 0.8 m, then

Required length3.8

4.750.8

m


Masonry toe wall 0.4 m thick and 1.2 deep shall be provided between filter

and launching apron.


13


14

Head and Cross Regulators

Documents

Transcript of Head and Cross Regulators