Head and cross regulators
v The
process of proper distribution of water to different branches and
distributaries requires special structures called regulators.
v Regulators
in canal may be of two types:
1)
Cross regulators, and
2)
Head regulators
- Cross
regulator – provided on the parent channel to regulate the supplies of
the parent channel.
- Head
regulator – provided at the head of the off-taking channel to regulate
the supplies of the off-taking channel.
- They control the silt entry into the off-taking canal.
- They serve as a meter for measuring the discharge entering into the off-taking canal
- They
help in shutting off the supplies in the off-taking canal when not needed,
or when it is required to be closed for repairs.
Functions of Cross Regulator
- The
effective regulation of the whole canal system can be done with help of
cross-regulator.
- During
the periods of low discharges in the parent channel, the cross-regulator
raises water level of the u/s and feeds the off-take channel in rotation.
- It
helps in closing the supply to the d/s of the parent channels for the
purposes of repairs, etc.
- They
help in absorbing fluctuation in various sections of the canal system, and
in preventing the possibilities of breaches in the tail reaches.
- Incidentally,
bridges and other communication works can be combined with it.
Component Parts
v A
regulator essentially consists of piers placed across the canal at regular
intervals with grove, in which either planks or steel gates can be used to
control the supplies.
v Planks
can be used for small channels only, as the maximum height of the plank, which
can be handled manually, is about 2 meters.
v For
large channels either hand operated or mechanically operated steel gates can be
used.
v The
hand operated gates can have span of 6 to 8m while the mechanically operated
gates can be as wide as 20 meter or so.
v A
light bridge platform has to lie on the piers for operation of planks or gates.
v It
is often economical to combine a regulator with a bridge and also to flume the
channel.
Design of Head and Cross Regulator
The design components:
1. Design of crest
v The
crest of a cross regulator is generally kept equal to the upstream bed level of
the parent channel.
v For
the distributary head regulator, the crest level is kept 0.3 to 1m higher than
the crest level of the cross regulator. The crest is joined to the d/s floor
with a sloping glacis of 2:1.
2. Water Way
v The
waterway/discharge is determined by using the drowned weir formula given below.
Q=2/3
C1L {√2g} [(h + ha)3/2–ha3/2
] +C2Ld[√{2g(h+ha)}]
Where,
Q
= discharge;
L
= length of water-way;
h
= difference in water level u/s and d/s of the channel, in meter;
ha
= head due to velocity of approach;
d
= depth of d/s water level in the channel measured above the crest
C1=0.577
and C2=0.80
3. Design of d/s floor
v The
level and length of the d/s floor is determined under two flow conditions:
1. Full
supply discharge passing through both the head regulator and cross regulator,
and
2. The
discharge in parent channel being insufficient, the cross regulator gate is
partially opened and the off-taking channel is running full. Or, the head
regulator gate is fully open.
v For
both of these conditions, the discharge intensity q and the head loss HL(=h)
are known. Hence,
D/s
floor level = d/s T. E. L. – Ef2≈ d/s F. S. L. – Ef2
Where,
Ef2=energy
head at the d/s [found from the Blench curves (Plate III)]
v However,
the d/s floor level calculated from the above relation should never be provided
higher than the d/s Bed level. If found higher, the floor is provided at the
bed itself.
v Now,
Ef1 = Ef2 + HL
v Hence,
the depth D1 and D2 corresponding to Ef1 and Ef2,
respectively are found from specific energy curves(plate IV)
v Then
length of d/s floor = 5(D2 - D1)
v However,
the d/s floor should be at least 2/3rd of the total impervious
length of the floor.
4. Length of impervious floor
v
Total
length of the impervious floor should be found from consideration of
permissible exit gradient.
v
The
depth of u/s cutoff,
d1 = [1/3 u/s water depth] + 0.6m
v
The
depth of d/s cutoff,
d2 =
[1/2 d/s water depth] + 0.6m
v
Maximum
static head
Hs =
u/s F.S.L. - d/s floor level
GE =
[1/Ï€√λ][Hs/d2]
v
From
which 1/Ï€√λ is known.
v
Then,
from the exit gradient curves(Plate II), α=bd2 is known.
v
Hence,
the total length b of the impervious floor is known.
Note: A minimum thickness of 0.3
to 0.5m is provided for the u/s floor from practical considerations.
5. Design of u/s and d/s Protection
v U/s
scour depth d1 is taken equal to (u/s water depth + 0.6m).
v The
d/s scour depth d2 is taken equal to ( d/s water depth + 0.6m).
v These
scour depths are below the corresponding bed levels,
v Corresponding
to these:
(a) U/s
protection:
o
The u/s protection consists of a block
protection having cubic contents = d1 cubic meters/m.
o
The cubic contents of u/s launching apron is
kept equal to 2.25d1cubic meter/meter width of regulator.
(b) D/s
protection:
o
The cubic content of d/s inverted filter is kept
equal to d2 cubic meter/meter.
o The cubic contents of d/s launching apron is kept equal to 2.25d2 cubic meter/meter width of regulator.
Design of Cross and Head Regulator Example:
Design a cross regulator and the head regulator for a
distributary channel taking off from the parent channel, for the following
data.
v Discharge
of parent channel = 100 cumecs
v Discharge
of distributary = 5 cumecs
v F.S.L.
of parent channel: (u/s)/(d/s) = 218.10m/217.90m
v Bed
width of parent channel: (u/s)/(d/s) = 42m/38m
v Depth
of water in parent channel:(u/s)/(d/s)=2.5m/2.5m
v F.S.L.
of distributary = 217.10m
v Bed
width of distributary=15 m
v Depth
of water in distributary=1.5 m
v Permissible
exit gradient = 1/5
Solution:
(a) DESIGN OF CROSS REGULATOR
Step 1. Design of crest and water-way
Crest
level = u/ s bed level of parent channel
= 218.10-2.50 = 215.60 m
v Neglecting the velocity of approach, the discharge is given by:
Q = 2/3*C1*L*[√(2g)]*h3/2
+ C2*Ld*[√(2gh)]
v Here, Q = 100 cumecs; C1 = 0.557; C2 = 0.80
v And,
h = u/s
F. S. L. - d/s F. S. L. = 218.10- 217.90 = 0.20 m
d = d/s
F. S. L. - Crest level= 217.90 - 215.60 = 2.30 m
v Substituting the values in the above
expression, we get
100 = (2/3)*0.557*L(√2*9.81)*(0.20)3/20.8L(2.30)(√2*9.81*0.2)
v From
which L = 26.2 m.
v Provide
4 bays of 7m each with a clear waterway = 28m
v Provide
3 piers of 1.5 m width each.
v Total
width of cross regulator = 28 + 4.5 = 32.5m
Step 2. Design of d/s floor
Q
= 100 cumecs
q =
100/28 = 3.58 cumecs/m
HL = h
= u/s F. S. L. - d/s F. S. L= 218.10-217.90 =0.20m
... Ef2 (Plate III) =
1.89m
v
Therefore, d/s floor level = d/s F. S. L. –
Ef2= 217.90- 1.89 = 216.01m
v This is higher than the d/s bed level.
v Hence,
adopt the d/s floor level = 215.40 (=217.9-2.5) = d/s bed level.
Ef1
= Ef2 + HL = 1.89 + 0.20 = 2.09
v From
Plate IV, D1 = 0.7m and D2 = 1.65m.
v Cistern
length (i.e. length of floor to the d/s of regulator gates) = 5(D2 – D1) = 5(1.65 - 0.7) = 4.75m.
v Total
length of impervious floor = 16m (see next step)
v Therefore,
2/3 length of impervious floor = (2/3) x 16 = 10.6m
v Hence,
provide length of d/s floor = 10.6 m
Step 3. Design of impervious floor
Depth
of u/s cutoff = d1 = (1/3) U/S water depth +0.6m = (1/3) x2.5 + 0.6 = 1.43m.
v Keep
d1 = 1.50m and width of cutoff = 0.5m
v Depth
of d/s cutoff = d2 = ½ D/s water depth +
0.6m
= ½ x 2.5 + 0.6 = 1.85m
v Keep
d2 = 2m and width of cutoff = 0.5m.
v Worst
case for under-seepage will be when full supply level is maintained in the
parent channel u/s of cross regulator, but no water to its d/s side.
v Maximum
static head = Hs = u/s F. S. L. - D/S
floor level
= 218.10 -
(217.90 - 2.5) = 2.7m
v From
exit gradient equation,
GE
= [1/Ï€√λ][Hs/d2]
1/5
= 1/Ï€√λ *(2.7/2)
v Therefore,
1/Ï€√λ
= 2/(5*2.7) = 0.148
v From
exit gradient curves (Plate II)
α
= 8 when 1/Ï€√λ = 0.148
b/d2
= 8 or b =8d2=8 x 2= 16m.
v 2/3
of this length = (2/3)*16 = 10.6m which is more than 4.75 m of d/s length
suggested earlier.
v Hence,
the total length of 16m will be provided as under:
(1) D/s
floor length = 10.6 m
(2) D/s
glacis length with 2:1 slope = 2*(215.60 – 215.40) = 0.4 m
(3) Balance to be provided upstream = 5 m.
Uplift Pressure
calculations:
1) U/s cutoff:
d1 =
1.5 m; b = 16 m
1/α =
d1/b = 1.5/16 = 0.094
ΦE =
1/π {Cos -1[(λ-2)/ λ] } and ΦD = 1/π {Cos -1[(λ-1)/ λ]}
Where: λ = {1+[√(1+α2)]}/2
ΦC1
= 100- ΦE =100 - 28 = 72% and
ΦD1
= 100- ΦD = 100 - 19 = 81%
v
Assuming a minimum floor thickness of 0.5m at
the u/s, corrected pressure,
ΦC1
= 72 + [(81 – 72)/1.5] x 0.5 = 75%
(2) D/s cutoff:
d2
= 2m; b = 16m
1/α
= d2/b = 2/16 = 0.125
ΦE2 =
31% and ΦD2 = 22%
Assuming a minimum thickness of 0.6m at d/s, corrected
ΦE2 =
31 – [(31 – 22)/2] x 0.6 = 28.3 %
(3) Toe of the glacis:
Percentage
pressure = 28.3 + [(75 - 28.3)/16] x l0.6 = 59.3%
Static
pressure head = 0.593 x 2.7 = 1.6 m
Calculations for
thickness:
(1) Minimum thickness for u/s floor = 0.5 m.
(2) Thickness of floor near d/s cutoff
= (0.283
x 2.7)/(2.24 – 1) = 0.61
Hence, provide 0.70m thick floor for the last 2.1m length.
(3) Thickness of floor at the toe of the glacis
=
16/(2.24 – 1) = 1.29m
Provide
1.30m thick floor at 4m length.
(4) Thickness of floor at 4 m from toe of glacis
Percentage
pressure = 28.3+[(75-28.3)/16] x 6.6= 47.5 %
Therefore, thickness = (0.475 * 2.7)/(2.24 - 1) =1. 04 m
v Hence,
provide 1.10 m thick floor for the next 2m length of the floor.
(5) Thickness of
floor at 6m from toe of glacis
%
pressure = 28.3 + [(75 – 28.3)/16] x 4.6 = 41.8 %
Therefore,
thickness = (0.418 x 2.7)/ (2.24 – 1) ≈ 0.9 m
v Hence,
provide 0.9 m thickness for the next 2.5m.
Summary of floor thickness
1) 0.5m
thickness of u/s floor
2)
1.30m thick floor for a distance of 4m from the toe of glacis
3) 1.10
m thick floor for the next 2m length
4)
0.90 m thick floor for the next 2.5 m length.
5)
0.70 m thick floor for the last 2.1 m length.
6)
Thickness under the gates, for a distance of 1m = 1.0m.
Step 4. Design of u/s protection
v Depth
of u/s cutoff, d1 = 1/3 F.S.D. + 0.6 or
= 1/3 x 2.5 + 0.6 = 1.43m
(1) Volume of block protection = 1.43 cubic meter/m
v Keep
the thickness of protection = 1m
v Therefore,
length = 1.43 m
v Hence,
provide 2 rows of 0.8m x 0.8m x 0.6m thick
v concrete
blocks over 0.4m thick apron, in a length of 1.6m.
(2) Cubic content of launching apron
= 2.25d1 = 2.25 X 1.43 =
3.21 m3/m.
v Hence
provide 1m thick and 3.5m long launching apron.
Step 5. Design of d/s protection
v Depth
of d/s cutoff, d2 = ½ F.S.D. + 0.6
= ½ x 2.5 + 0.6
= 1.85m
(1) Volume of inverted filter = 1.85m3/m
v Keep
thickness of concrete blocks = 0.6 m kept over 0.6m thick graded filler, making
a total thickness of 1.2 m.
Length
= 1.85 / 1.2 = 1.54m
v Hence,
provide 2 rows of 0.8m x 0.8m x 0.6m C.C block over 0.6m graded filter for a
length of 1.6m.
(2) Launching apron
volume
= 2.25 x d2 = 2.25 x 1.85 = 4.15m3/m
v Hence,
provide 1m thick launching apron for a length of 4.5m.
v Provide
a 0.4m thick and 1.5m deep toe wall between the filter and the launching apron.
(b) DESIGN OF DISTRIBUTARY HEAD REGULATOR
Step 1. Design of crest and water-way
v Crest
level =
u/s bed level + 0.5
= (218.10 - 2.5) + 0.5 =
216.10m
v Neglecting
the velocity of approach, the discharge is given by:
Q
= 2/3*C1*√(2g)*h3/2+C2*L*d*√(2gh)
Where,
Q = 15cumecs, C1 = 0.577 and C2 = 0.80
h = u/s F.S.L. – d/s F.S.L. = 218.10 – 217.80 = 1.0m
v Crest
level=u/s bed level + 0.5= (218.10 -
2.5) + 0.5 = 216.10m
d = d/s F.S.L. – Crest level = 217.10 –
216.10 = 1.0m
v Substituting
these values in the discharge formula,
15=2/3*0.577*L*[√(2*9.81)]*(1.02/3)+
0.8*L*√(2*9.81*1.0)
v From
which L = 2.87m
v This
is too less in comparison to 15m width of the off taking channel.
v Hence,
provide 2 bays of 3.5m each with 1m thick pier in between.
v Therefore,
over all width of regulator = 8m.
v The
crest will have u/s glacis of 1:1 slope and d/s glacis of 2:1 slope.
v Keep the width of crest = 1.0m
Step 2. Design of d/s floor
Q
= 15 cumecs
q = 15/7.5 = 2
cumecs/m
HL = u/s F.S.L.
– d/s F.S.L. = 218.10 -217.10 = 1.0m
v Hence,
for q = 2 and HL = 1.0, we get from Plate III, Ef2 = 1.58m
v Therefore,
R.L
of d/s floor = d/s F.S.L. – Ef2 = 217.10 – 1.58 = 215.52m
v Keep
the R.L. of d/s floor = 215.50m
v Again,
v Ef1
= Ef2 + HL = 1.58 + 1 = 2.58m
v For
Ef1 = 2.58 and Ef2 = 1.58m,
D1 = 0.42m & D2 = 2.55m
v Therefore,
Length of cistern = 5(D2 – D1) = 5(2.55 – 0.42) = 10.5m
Total Length of impervious floor = 14m (from step 3)
v 2/3
of impervious floor = (2/3) *14m = 9.3m
v Hence,
provide length of cistern (i.e. floor to the d/s of glacis) = 10.5 m.
Step 3. Design of impervious floor
v Depth
of u/s cutoff = 1/3 u/s water depth =
0.6m
d1 = 1/3 x 2.5 + 0.6 ≈
1.5m
Hence,
provide 0.5 m wide and 1.5m deep cutoff.
v Depth
of d/s cutoff = ½ d/s water depth + 0.6 m
d2 = ½ x 1.5 + 0.6 = 1.35m
v Now,
in this case, the seepage head
= u/s F. S. L. - d/s floor
level
Therefore,
Hs = 218.10- 215.50 = 2.60m.
In the case of the
cross regulator, Hs was equal to 2.70m and d2 was equal to 2m.
v Hence,
in this case also, we should provide d2 = 2.0m (as against calculated value of
1.35) otherwise the floor length will come out to be excessive (i.e. much more
than 16m length of the previous case).
v Hence,
provide downstream cutoff of 0.5 wide and 2m deep.
v Now,
GE = [1/Ï€√λ][Hs/d2]
1/5 = [1/Ï€√λ][2.6/2.0]
1/Ï€√λ =
2/(5*2.6) = 0.154
v From
exit gradient curves, value of α = 7 for 1/Ï€√λ = 0.154
v Therefore,
b
= αd2 = 7 x 2 = 14m.
v The
length of 14m will be provided as under:
1. Length
below the toe of glacis = 10.5 m
2. Length
of d/s glacis at 2:1slope,
=2*(216.10-2l5.60)
= 1.2m
- Width
of crest = 1m
- Length
of u/s glacis at 1:1 slope
=
216.10 – 215.60 = 0.50
1. U/s
floor: balance =14 - (10.5 + l.2 + 1 + 0.5)= 0.8m.
v Hence,
provide a minimum length of 2.8m to the u/s, taking a total impervious length =
16m. This increase of 2m length will accommodate the piers and u/s wings.
Pressure calculations
(i) U/s cutoff
d2 =
1.5m; b = 16m
1/α = d1/b
= 1.5 /16 = 0.094
v Hence,
ΦC1 = 100 - 28 = 72% and ΦD1 = 100 - 19
= 81%
v Assuming
a minimum floor thickness of 0.5m at u/s , corrected
ΦC1 = 72 + [(81-72)/1.5]*0.5 =75%
(ii) D/s cutoff
d2
= 1.5m; b = 16m
1/α = d2/b = 2
/16 = 0.125
v Hence, ΦE2 = 31%
and ΦD2 = 22%
v Assuming
a minimum thickness of 0.6m at d/s, corrected
ΦE2 = 31 – [(31 – 22)/2] = 28.3%
(iii) Toe of glacis
v Pressure = 28.3 + [(75-28.3)/16]*10.5 = 59.2%
Calculation of
thickness of floor
= (0.592x2.6)/(2.24-1) = 1.24m
v Provide
a thickness of 1.30m for a distance of 2m from the toe of the glacis.
iv.
Thickness of floor at 2m from the toe of glacis:
%
pressure = 28.3 + [(75-28.3)/16] X 9.5 = 53%
Therefore,
thickness = (0.53 X 2.6) /(2.24 – 1) = 1.11m
v.
Thickness of floor at 6m from the toe of the
glacis:
%
pressure = 28.3 + [(75 - 28.3) /16] X 4.5 = 41.4%
Therefore,
thickness = (0.414 X 2.6)/ (2.24-1) = 0.84m
v Provide a thickness of 0.90m for the next
2.5m.
vi.
Thickness of floor at 8.5 m from the toe of the
glacis:
%
pressure = 28.3 + [(75 – 28.3)/16] X 2 = 34.1%
Therefore,
thickness = [(0.341 X 2.6)/(2.24-1) = 0.71m.
v Hence,
provide 0.7m thick floor for its 2m length
Step 4. Design of u/s protection
v U/s
scour depth = 1/3 x 2.5 + 0.6 = 1.5m
v Hence,
provide the same protection as in the case of cross regulator.
Step 5. Design of d/s protection
v D/s
scour depth = ½ d/s water depth + 0.6m
d2 = ½ x 1.5 + 0.6 =
1.35m
(i) Inverted filter
Volume
= d2 = 1.35 m3/m
v Provide
0.5m thick cement concrete blocks over 0.5m thick graded filter.
v Length
required = 1.35/1 = 1.35 m.
v However,
provide two rows of 0.8m x 0.8m X 0.5m thick concrete blocks over 0.5 thick
graded filters.
(ii) Launching apron
v Volume
= 2.25 d2 = 2.25 x 1.35 = 3.04m3/m.
v Provide
1m thick launching apron for a length of 3.5m.
v Provide a masonry toe wall 0.4m wide and 1.20m deep between the filter and the launching apron.
1 Comments
Amazing solved problem using plate curves.
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