Head and Cross Regulators

Hydraulic Structures Structures – Head and Cross Regulators

December 12, 2010

Head and Cross Regulators The supplies passing down the parent canal and off take channel are controlled by cross regulator and head regulator respectively. A utment Pier Gate

Cross Regulator

Head Regulator

OFF TAKE CHANNEL Distributary channel

Functions of Cross Regulators 1. 2. 3. 4.

Regulation of the canal system. Raising the water level in the main canal in order to feed the off take channels. To facilitate communication by building a road over the cross regulator with little extra cost. 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

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Hydraulic Structures – Head and Cross Regulators

December 12, 2010

In case of obligatory straight alignment of the parent channel, the usual angle of the off take channel is 60º to 80º (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. H e should be worked out from the formula Q  C B e H e3/2

(4.1)

where C 

Coefficient of discharge

B e 

Effective length of crest  B t  2  N K p  K a  H e

B t 

Net length of the crest Number of piers

N  K p 

K a 

Pier contraction coefficient Abutment contraction coefficient 2

(4.2)


December 12, 2010

Table: Coefficients of contraction for piers and abutments.

Type of pier

Square nosed pier Round nosed pier Pointed nosed pier

K p

Type of abutment

0.02 0.01 0.01 K a

Square abutment Round abutment

0.2 0.1

Square nosed pier

Round nosed pier

3. Coefficient of discharge (C ) The coefficient of discharge C is 1.84 for crests of width less than or

equal to 23 H e . In case of submerged falls, C should be reduced

Pointed nose pier

depending on the drowning ratio, see Fig. 6.5. C  1.840

(H.R. crest)

C  1.705

(C.R. crest) . . . He

>2/3 He Broad Crested Weir 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 H e . 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 Lacey’s scour depth

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December 12, 2010

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

Min. depth of u.s. cut-off below bed level or G.L. whichever is lower ( m) 1.0 1.2 1.5 1.8

Canal capacity cumec

Up to 3 cumec 3.1 - 30 30.1 - 150 Above 150

Min. depth of d.s. cut-off below bed level or G.L. whichever is lower ( m) 1.0 1.2 1.5 1.8

7. Thickness of top coat

Table 6.2 Thickness of top coat.

Canal capacity ( cumec) Q  1.5

Thickness of top coat ( mm) 100

1.5  Q  30 30  Q  150 Q  150

150 200 300

8. Freeboard Table: Minimum Freeboard

Canal capacity ( cumec) Q  1.0

Freeboard ( m) 0.3

1  Q  10 10  Q  30 30  Q  150 Q  150

0.4 0.6 0.8 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 2 D ( 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 1.5 D 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 2.25 D m 3  m m length.

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December 12, 2010

Design Example Design a cross regulator and a suitable head for a distributary which takes off at an angle of 60º from a canal which discharges 120 cumec. Discharge of distributary Bed width of distributary Water depth of distributary Full supply level of distributary Full supply level of parent channel (F.S.L.)

= 10 cumec = 10 m = 1.2 m = El. 310.2 m u.s. 311.00 m



Bed width of parent channel



Depth of water in parent channel



Safe exit gradient (G.E.)



d.s. u.s. d.s. u.s. d.s.



 

310.85 m 60 m 56 m 2m 2m

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  309 m hd  T.E.L.u.s.  T.E.L.d.s.

 311  310.85  0.15m H e  F.S.L.u.s.  crest level

=311  309  2 m Degree of submergence hd 0.15   0.075 H e 2 From Fig. 6.5 C s  0.56, where C  1.705 C C s  1.705  0.56  0.95 Q  CB e H 3/2 3/2

120  0.95  B e  2 

Be  44.66 m say 45 m

Assume 6 bays of 8 m each = 48 m 5


December 12, 2010

B t  B e  2(N K p  K a )H e

 45  2  5  0.01  0.2   2  46 m  48 m O.K. Provide 5 piers with rounded nose of width 1.6 m each. Total waterway  6  8  5 1.6  56 m O.K. 2. Level and length of downstream floor Q  120 cumec q

Q B e



120 48

 2.5 m 3 / sec / m

H L  T.E.L.u.s.  T.E.L.d.s.  F.S.L.u.s.  F.S.L.d.s.

(v 2 2 g is negligible)

 311  310.85  0.15m From Blench Curves, Fig. 3.5 Ef 2  1.435 m

d.s. floor level  F.S.L.d.s.  Ef 2

 310.85  1.435m  309.415 m Actual cistern level  310.85  2  308.85 m lower than 309.415 m O.K. Cistern length Lc  5  D 2  D1  firm soil

 6( D 2  D1 ) weak soil Ef 1  Ef 2  H L  1.435  0.15  1.585 m

From energy of flow curves (Fig. 2.7) Ef 1  1.575 m gives D1  0.534 m Ef 2  1.435 m gives

D 2  1.3 m

Lc  5 1.3  0.534   3.83 m

3. Vertical cutoffs Q  120cumec

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  307.5 m 4. Total floor length and exit gradient H 1 G E  d   H  F.S.L.u.s.  G.L.d.s.

 311  308.85  2.15m Trial and error d (m) 1.5 1.6

λ

α

L   d (m)

5.20 4.57

9.15 8.08

14 12.93

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1 5



2.15

1

1.6  

December 12, 2010

,  =4.57 1/2

2    2  1  1  

1/2

2   2  457  1  1   Total floor length L   d

 8.08

 8.081.6  12.93 m say 13m d.s. cutoff is at El.  308.85  1.6  307.25 m d.s. floor length d.s. glacis length with (2H:1V) slope u.s. floor length Total floor length

= 6.5 m = 2(309-308.85) = 0.3 m = 6.2 m = 13 m

5. Pressure calculation a. Upstream cutoff d  1.5 m ,b  13 m

1



d



1.5

 0.115  b 13  D1  100  20  80%  C 1  100  28  72%

Correction of  C for floor thickness 

t

 D   C  d 0.6   80  72   3.2% 1.5

 C corrected  72  3.2  75.2% of H b. Downstream cutoff d  1.6 m , b  13 m 1



d



1.6

 b 13  E  31%

 0.123

 D  22%

Correction of  E for floor thickness  

0.6 1.6

 E corrected  31 3.38  27.62% of H

7

 31  22   3.38%


December 12, 2010

6. Floor thickness a. d.s. floor 1. at 2 m from d.s. end

% pressure  27.62 

75.2 27.62 12

 1.5

 33.6%of H Head  H  % pressure =2.15 

33.6 100

 0.722 m of water

Minimum concrete thickness t min 

Head



 c   w

0.722 2.25  1

 0.58 m

Provide 0.6 m thick concrete floor for 1.5 m. 2. at 4.5 m from d.s. end % pressure  27.62 

75.2  27.62 12

 4.0

 43.48%of H Head  H  % pressure =2.15 

t min

43.48

 0.934 m of water 100 Head 0.934    0.75 m  c   w 2.25  1

Provide 0.8 m thick concrete floor for 2.5 m. 3. at 6.5 m from d.s. end 75.2  27.62 % pressure  27.62   6.0 12  54.41%of H t min 

2.15  0.5141

 0.9 m 2.25  1 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 m  0.8 m  0.6 m C.C. blocks over 0.6 m thick inverted filter. ii. Launching apron 3 2 Volume per meter length  2.25 D m m  m

If thickness is 1.0 m, then the required length  Provide 3.5 m long launching apron. 8

2.25 D 1.0



2.25 1.5 1

 3.37 m


December 12, 2010

8. Downstream protection i. Block protection Length of protection  2 D  2 1.6  3.2 m

Provide 4 rows of 0.8 m  0.8 m  0.6 m blocks over 0.6 m inverted filter. ii. Launching apron

Volume per meter length  2.25 D m 3 m  m 2 If thickness is 1.0 m, then Required length 

2.25 D



2.25 1.6

 3.6 m 1.0 1 Provide 4.0 m long launching apron. 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  309.5 m H e  311  309.5  1.5 m hd  311  310.2  0.8 m hd



H e C s C

0.8 1.5

 0.533

 0.98 (From Fig. 6.5)

C s  1.84 0.98  1.80

The effective width of waterway is found by Q  C B e H e

1.5 1.5

10  1.8  B e 1.5 B e  3 m

Provide 60% of distributary width 

60

10  6 m 100 Provide 2 bays of 3 m each separated by 1 m thick pier. Therefore, the overall waterway 2  3  1  7 m 2. Level and length of d.s. floor Q  10 cumec , waterway  6 m q

10 6

 1.67 m sec

Head loss H L  311  310.2  0.8 m Using Blench curves (Fig. 3.5)

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December 12, 2010

Ef 2  1.37 m

gives D 2  1.32 m

Ef 1  Ef 2  H L  1.37 m

gives D1  0.32 m

Cistern length Lc  5  D 2  D1 

 5 1.32  0.32   5.0 m Provide cistern length = 6 m Floor level required  T.E.L.d.s.  Ef 2

 310.2  1.37  308.83m 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  307.5 m b. The minimum d.s. cutoff depth = 1.2 m 4. Total floor length and exit gradient Maximum static head  F.S.L.u.s.  F.L.d.s.

 311  308.8  2.2 m Exit gradient G E 

H

1

d   1/2

   2  1  1 2



1 5





2.2

1

1.2  

With trial and error d (m)

λ

α

L   d (m)

1.2 1.4 1.6 1.7

8.50 6.26 4.75 4.24

15.99 11.47 8.50 7.42

19.00 16.05 13.62 12.60

Get L  13 m ,d  1.7 m

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December 12, 2010

= 6.0 m

d.s. floor length (cistern length) d.s glacis length 2H:1V slope

 2  309.5  308.8

= 1.4 m

Crest length

 23 H e  23 1.5

= 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 d  1.5 m ,b  13 m

1



d



1.5

 0.115  b 13  D1  100  20  80%  C 1  100  28  72%

Assuming the u.s. floor thickness = 0.6 m Correction of  C for floor thickness 

0.6 1.5

 80  72  3.2%

 C corrected  72  3.2  75.2% of H b. Downstream cutoff d  1.7 m , b  13 m 1



d



1.7

 b 13  E  32%

 0.13

 D  22%

Correction of  E for floor thickness  

0.6 1.7

 32  22   3.5%

 E corrected  32  3.5  28.5% of H 6. Floor thickness a. d.s. floor 1. at 2 m from d.s. end

% pressure  28.5 

75.2  28.5

12  34.33%of H

 1.5

Minimum concrete floor thickness t min 

0.3433 2.2

2.25  1 Provide 0.6 m thick concrete floor for 1.5 m length. 11

 0.6 m


December 12, 2010

2. At 4 m from d.s. end Provide 0.8 m thick concrete floor for 2.5 m. 3. At the toe of glacis (beginning of the hydraulic jump) 5.5 %pressure  28.5   75.2  28.5 12  49.9%of H Unbalanced head  0.499  2.2  1.095 m Unbalanced head due to dynamic condition  50%  D 2  D1    toe H L

1

 1.32  0.32   2  0.898 m t min 

49.9 100

 0.8

1.095

 0.88 m 2.25  1 Provide floor thickness 1.0 m. b. d.s Floor thickness 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 D  2 1.7  3.4 m Provide 6 rows of 0.6 m  0.6 m  0.4 m C.C. blocks over 0.4 m thick graded filter. ii.Launching apron

Volume per meter length  2.25 D  2.25 1.7  3.825 m 3 m Assume thickness of launching apron = 0.8 m, then 3.8 Required length   4.75 m 0.8 Provide 5.0 m long launching apron. Masonry toe wall 0.4 m thick and 1.2 deep shall be provided between filter and launching apron.

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December 12, 2010

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December 12, 2010

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Head and Cross Regulators

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