head and cross regulators

Design of Hydraulic Structures 3rd year/ water Resources Engineering

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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 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.


2

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

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.


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Design Criteria

1 . W a t e r w a y : 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.

The effective head (He) should be worked out form the formula:Q = C.Be.H

1.5

where

C = coefficient of discharge.

Be = effective length = Bt – 2(N*Kp + Ka)*He

Bt = Net length of crest.

N = number of piers.

Kp = piers contraction coefficient

Ka = abutment coefficient

Table: Coefficients of contraction for piers and abutments.

Type of pier Kp

Square nosed pier 0.02Round nosed pier 0.01Pointed nosed pier 0.01Type of abutment Ka

Square abutment 0.2Round abutment 0.1

Square nosed pier

Pointed nosed pier


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C = 1.840 (H.R. Crest)

C = 1.705 (C.R. Crest)

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 2/3 He.

6. Level & Length of D/S Floor

They are fixed for the conditions:

a – Full supply, when all the gates of C.R. and H.R. fully opened (generally is

governed).

b – The discharge of the parent channel is insufficient but the off take channel is

running full by partial opening of the C.R. gates or vice versa.

Downstream Floor Level:

For these conditions of flow, the discharge intensity q and HL is worked

out. The corresponding value of Ef2 is read from fig. 13.9. The d/s floor level is

then d/s T.E.L. – Ef2 .

3. Coefficient of discharge (C): The coefficient of discharge C is 1.84 for

crests of width less than or equal to eH3

2. In case of submerged falls, C should

be reduced depending on the drowning ratio, see Fig. 6.5.


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7. Length of the downstream floor.

Ef1 = Ef2 + HL

The values of D1& D2 are found from energy curve for the corresponding

values of Ef1 & Ef2. The length of the downstream floor = 5 to 6 (D2 – D1). This

length should not be less than 2/3 of the total floor length.

8. 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.

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.

Min. depth of d.s. cut-off

below bed level or G.L.

Up to 3 cumec 1.0 1.03.1 - 30 1.2 1.2

30.1 - 150 1.5 1.5Above 150 1.8 1.8

9. Total floor Length & Exit gradient

If the d is the depth of d/s cut off, and H is the maximum static head when

channel is closed and F.S.L. (Full Supply Level) is maintained in the u/s for

feeding the off take channel (i.e.) H = u/s F.S.L. – d/s floor level.

1

d

HGE ; or

H

dGE

1

Find the value of λ and find α from

2

11 2

2/12112

Therefore the total floor (b) is found from:

b = α*d

If b is large, the depth of d/s cut off should be increased and floor length

correspondingly reduced keeping safe exit gradient.


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10.Pressure Calculations & Floor Thickness

Use Khosla's method to calculation pressure at key points.

a – Static head pressure

b – Dynamic head pressure

Dynamic head at toe of glacis

H = 50 %*( D2 – D1) + at toe * HL

The minimum thickness floor concrete excluding the thickness of the top

coat shall be as:

For Q > 1.5 cumecs 600 mm

Q < 1.5 cumecs 300 mm

The floor concrete from u/s cut off to d/s cut off should be laid by top coat of:

Q incumecs

Thickness oftop coat (mm)

Up to 150 300150 – 30 20029 – 1.5 150Less 1.5 100

11. Free Board

The following free board shall be adopted:

Capacity of channel (cumecs) F.B. (m)

Less than 1.0 0.31.0 – 10 0.410 – 30 0.630 – 150 0.8Over 150 1.0

12. Protection Works

Properly designed filter loaded by concrete blocks should be provided. The

length of inverted filter is kept 2D (D is the depth of d/s cut off below d/s bed).

Details of minimum thickness of filter are given in table 6.3 (page 214). The

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


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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 length

equal to D. The cubic content of material in lunching apron should be equal to

2.25D cu. m/m length.

Design Example:

Design a cross regulator and suitable head for distributary which take off at

angle 60o from a canal with discharge of 120 m3/sec.

Discharge of distributary = 10 m3/sec

Bed width = 10 m

Water depth = 1.2 m

F.S.L. of distributary = 310.2 m

F.S.L. of parent channel u/s = 311 m

d/s = 310.85 m

Bed width of parent channel u/s = 60 m

d/s = 56 m

Water depth in parent channel u/s = 2.0 m

d/s = 2.0 m

GE = 1/5

Design:

a. Design of Cross Regulator

1. Fixation of crest level and waterway

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

= 311 – 2 = 309 m

The u/s floor shall be provided at the same level.

Degree of submergence = hd / He

hd = 311 – 310.85 = 0.15 m

He = u/s F.S.L. – crest level


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= 311 – 309 = 2 m

075.02

85.310311

He

hd

From fig. 6.5 page 252

95.0705.1*56.0

56.0

CModified

C

Cs

2/3.. ee HBCQ

Be = effective width

2/32**95.0120 eB

Be = 44.6 m ≈ 45 m

Assume 6 bays of 8 m each.

Use rounded piers & square abutment; kp = 0.01, ka = 0.2, net length

Bt = Be + 2(N*Kp + Ka)*He

= 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

The total water way = 6.0 * 8 + 1.6 * 5

= 48 + 8 = 56 m

2. Level and length of downstream floor

Q = 120 cumecs

q = discharge intensity between piers.

sec/5.248

120 2mB

Qq

Head loss HL = u/s F.S.L – d/s F.S.L.

= 311 – 310.85 = 0.15 m

From fig 2.5 (Blanch curve) for q = 2.5 m2/sec and HL = 0.15 m we get:

Ef2 = 1.435 m

D/s floor Level = d/s F.S.L. – Ef2

= 310.85 – 1.435 = 309.415 m


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Adopt cistern level = 310.85 – 2 = 308.85 m is lower than 309.415 m.

design d/s floor level on El. = 308.85 m

Ef1 = Ef2 + HL

Ef1 = 0.15 + 1.435 = 1.585 m

From energy curve we get: D2 = 1.3 m

D1 = 0.534 m

Cistern length = 5*(D2 – D1)

= 5*(1.3 – 0.534)

= 3.83 m

3. Vertical Cut offs

Referring ton table 6.1, the min. depth of u/s & d/s cut off for this range of

discharge = 1.5 m.

(u/s cut off is at El. = 309 – 1.5 = 307.5 m, and d/s cut off is at El. 308.85 –

1.6 = 307.25 m)

4. Total floor length and exit gradient

u/s F.S.L = 311 m, d/s floor = 308.85 m

max. static head (H) = F.S.L. u/s – floor level d/s

= 311 – 308.85

= 2.15 m

d = depth of d/s sheet pile (cut off) = 1.6 m

1

d

HGE

1

6.1

15.2

5

1 ; λ = 4.57

2

11 2

22112

2/12112

α = 8.08

Total floor length required b = α*d = 8.08 * 1.6 = 12.98 m.


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= 12.8 from book

d/s floor length = 6.5 m

d/s glacis length with 2:1 slope

= 2(309 – 308.85) = 0.3 m

u/s floor length including crest = 6.2 m

Total = 13 m

5. Pressure Calculation

a. Upstream Cut off

d = 1.5 m

b = 13 m

115.013

5.11

b

d

D1 = 100 – 20 = 80% of H

C1 = 100 – 28 = 72 % of H

Correction of C for thickness

Assume t = 0.6 m

%2.3)7280(5.1

6.0 CDt

d

tC

Correction due to interference of d/s cut off may be neglected

C corrected = 72 + 3.2 = 75.2 % of H

b. Downstream Cut offd = 1.6 m

b = 13 m

123.013

6.11

b

d

E = 31 % of H

D = 22 % of H

Correction of E for thickness

Assume t = 0.6 m

%38.3)2231(25.2

6.0tC


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Correction due to interference of u/s cut off may be neglected

E corrected = 31 – 3.38 = 27.62 % of H

6. Floor thickness

a. Downstream floor

I. At 2 m from d/s end

% pressure = 6.33)62.272.75(12

5.162.27 % of H

Unbalanced head = H * % pressure= 2.15 * 0.336 = 0.722

Floor thickness required = 58.025.1

722.0 m of concrete

Provide 0.6 m thick floor

II. At 4.5 m from d/s end

% pressure = 48.43)62.272.75(12

462.27 % of H

Floor thickness = 75.025.1

4348.0*15.2 m of concrete

Provide floor thickness = 0.8 m of concrete

III. At 6.5 m from d/s end (toe of glacis).

% pressure = 41.51)62.272.75(12

662.27

Floor thickness = 88.025.1



b. Upstream floor

0.6 m thick shall be provided for the u/s floor. It should thickened to 1 m

under crest

7. Protection

I. Upstream Protection

a. Block Protection

Block protection in u/s is provided in length (L) = D

L = 1.5 m


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For Q = 120 cumecs

see table 6.3

Provided 2 rows of 0.8 * 0.8 * 0.6 c.c. block

Over 400 (or 600) mm thick apron in length = 1.6 m

b. Lunching apron

Vol. / m length = 2.25*D m3/m

= 2.25*1.5 = 3.37 m3/m

If thickness = 1 m

length required = 3.37 m

Provide launching apron in 3.5 m length

II. Downstream Protection

I. Inverted Filter (D = 1.6 m)

Length of filter (L) = 2*D

= 2*1.6 = 3.2 m

From table 6.3, provide 4 rows of 0.8 * 0.8 * 0.6 c.c. block over 600 mm

thick graded

II. Lunching apron

Vol. / m = 2.25*D m3/m

= 2.25*1.6 = 3.6 m3/m

If the thickness of apron be 1 m, the length required 3.6 m. provide 4 m

length. Provide 0.4 m thick and 1.2 m deep toe wall between inverted filter and

launching apron.


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b. Design of Distributary head Regulator

1. Fixation of crest level and waterway

The crest level of distributary head shall be provided 0.5 m higher than u/s

floor level

u/s level = bed level of parent channel = 309 m

Crest level = 309 + 0.5

= 309.5 m

He = 311 – 309.5

= 1.5 m

hd = u/s F.S.L. – d/s F.S.L. (distributary)

= 311 – 310.2 = 0.8 m

533.05.1

8.0

He

hd

5.698.0 figfromC

Cs

C = 1.84 for sharp crested weir,

Coefficient of discharge

8.198.0*84.1 Cs

mBe 3)5.1(*8.1

105.1 , This is very small

Provide 60% of distributary width

= 0.6 * 10 = 6 m

Use 2 bays of 3 m each separated 1 m thick piers.

The overall water way = 3 * 2 + 1 = 7 m

2. Level and length of downstream floor

Q = 10 cumecs

L = 6 m (net water way)

sec/67.16

10 2mB

Qq

HL = 311 – 310.2 = 0.8 m


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From Blanch curve for q = 1.67 m2/sec and HL = 0.8 m we get:

Ef2 = 1.37 m.

Floor level required = d.s. F.S.L. – Ef2

= 310.2 – 1.37 = 308.83 m

Provide d/s floor level at El. 308.8 m

Ef1 = Ef2 + HL

Ef1 = 1.37 + 0.8 = 2.17 m

From energy curve we get: D2 = 1.32 m

D1 = 0.32 m

Length of cistern = 5*(D2 – D1)

= 5*(1.32 – 0.32)

= 5 m

Provide cistern length = 6 m

3. Vertical Cut offs

a. U/S cut off

Provide u/s cut off depth = 1.5 m.

The bottom R.L. of cut off = 309 – 1.5 = 307.5 m

b. D/S cut off

The min. d/s cut off depth (see table 6.1) = 1.2 m below d.s. floor.

Provide d/s cut off depth = 1.6 m from safe exit gradient consideration.

4. Total floor length and exit gradient

Max static head (H) = F.S.L. u/s – floor level d/s

= 311 – 308.8

= 2.2 m

Depth of d/s cut off (d) = 1.6


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G.E. = 1/5

1

d

HGE

1

6.1

2.2

5

1 ; λ = 4.79

2/12 )1)1*2(( = 8.52

Total floor length required (b) = α*d = 1.6 * 8.52 = 13.63 m

13bال نأخذ لالستمرار مع نتائج المث =

d/s floor length = 6 m

D/s glacis length 2:1 slope = 2(309.5 – 308.8) = 1.4 m

Crest width = 1m

U/s glacis length with 1:1 slope = 1(309.5 – 309) = 0.5 m

U/s floor = 4.1 m

Total length = 13 m

5. Pressure Calculation

a – Upstream cut off

d = 1.5 m

b = 13 m

115.013

5.11

b

d

D1 = 80 % of H

C1 = 72 % of H

Correction due to thickness

Assume floor thickness t = 0.6 m

%2.3)7280(5.1

6.0tC

C1 corrected = 72 + 3.2 = 75.2 % of H

b. Downstream Cut off

d = 1.6 m

b = 13 m


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13.013

6.11

b

d

E = 32 % of H

D = 22 % of H

Correction due to thicknessAssume t = 0.6 m

%75.3)2232(6.1

6.0tC

E corrected = 32 – 3.75 = 28.25 % of H

6. Floor Thickness

a. D/S floor thickness

Max H = 311 – 308.8

= 2.2 m

I. at 2 m from d/s end

% pressure = 12.34)25.282.75(12

5.125.28 % of H

Min floor thickness = 6.024.1


Provide floor thickness = 0.6 m.

II. at 4 m from d/s end

% pressure = 5.41)25.282.75(12

5.325.28 % of H

Min floor thickness = 73.024.1



III. at toe of glacis or 6 m from d/s end

% pressure = 77.49)25.282.75(12

5.525.28 % of H

Unbalanced head = 0.4977 * 2.2 = 1.095 m

Unbalanced head due to dynamic condition =50 %*( D2 – D1) + at toe * HL

= 8.0*100

77.49)32.032.1(

100

50 = 0.898 m

Design as static head = 1.095 m


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Min fool thickness = 88.024.1

095.1 m of concrete

Provide 1 m thickness under toe.

b. U/S floor thicknessMin. thickness of 0.6 m shall be provided in the upstream which should be

thickened under the crest.

7. Protection

a . Upstream Protection

Same as provided in u/s of cross regulator

b. Lunching apron

I. Inverted filter (Block protection)

Length of filter = 2D

= 2*1.7 = 3.4 m

Provide 5 rows of 0.6 * 0.6 * 0.4 m c.c. blocks over 0.4 m thick graded

filter.

II. Launching apron

Vol. / unit length = 2.25*D m3/m

= 2.25*1.7 = 3.83 m3/m or (m2)

Assume the thickness of apron = 0.8 m

Length required = m5.48.0

6.3

Provide 5 m length of lunching apron.

Masonry wall = 0.4 * 1 deep

Shall be provide between filter and lunching apron


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head and cross regulators

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