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Design of Canal Outlets and Their Calibration
CHAPTER
DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
8
An outlet can be defined as a device through which water is released from a distributary channel into a water course. The discharge through an outlet is usually less than 0.085 m3/s (3.0 cusecs) (IS 7986, 1976). Various types of canal outlets have been developed from time to time to obtain a suitable performance. No one type has come out to be suitable universally. In fact, it is difficult to achieve good design with respect to “flexibility” and “sensitivity” because of various indeterminate conditions both in the distributary channels and the water course, namely, discharge levels, silt charge, capacity factor, rotation of channels, and regime condition of distributary channels. Any variation in any of these factors affects proper functioning of the outlet. Even a particular type of outlet considered suitable upstream of a control structure in a canal may not be suitable in the downstream reach of the same canal. There are different types of outlets used in the canal systems that can be classified as follows.
8.1 CLASSIFICATION OF OUTLETS Outlets may be classified into the following three types: 1. Nonmodular outlets: Nonmodular outlets are the outlets whose discharge is a function of the difference in water levels in the distributary channel and the water course and variation in either affects the discharge. These outlets consist of rectangular or circular openings and pavement. The effect of downstream water level is more with a short pavement, although even with a long pavement it cannot be entirely eliminated. The common examples of this type of outlets are: (a) open sluice and (b) drowned pipe outlet. 2. Semimodular outlets: Semimodular outlets are the outlets whose discharge depends on the water level in the distributary channel and not on the water level in the water course so long as the working head is available. The working head for an outlet is the difference between the water level of the distributary channel and that of the center of the pipe or outlet. Common examples of this type of outlets are pipe outlet, venturi flume, open flume, and orifice semimodule. 3. Modular outlets: Modular outlets are the outlets whose discharge is independent of water levels in the distributary channel and the water course, within reasonable working limits, i.e., for such outlets or modules, the discharge is constant within reasonable working limits irrespective of the fluctuations in the water levels in the distributary channel and/or the water course. This type of outlets is either with or without moving parts. In the latter case these are called rigid modules. Modular outlets with moving parts are not simple to design and construct and are, thus, expensive. These are liable to derangement due to increase in friction, rusting of the moving parts, and any obstruction in the working of moving parts caused by silt and weeds being carried in the flowing water. Gibb’s module is a common example of this type of outlet or module. Planning and Evaluation of Irrigation Projects. http://dx.doi.org/10.1016/B978-0-12-811748-4.00008-X Copyright © 2017 Elsevier Inc. All rights reserved.
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CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
8.2 PERFORMANCE OF MODULE OR OUTLET Performance of the modules or outlets can be assessed by using the following criteria:
8.2.1 FLEXIBILITY Flexibility can be defined as the ratio of the rate of change of discharge of the outlet to the rate of change of discharge of the distributing channel. Numerically, it can be expressed as follows: F¼
dq=q dQ=Q
(8.1)
where q is the outlet discharge and Q is the discharge in the distributary channel. If H is the head acting on the outlet, the discharge through the outlet, q, can be expressed in its general form as: q ¼ C1 H m1
(8.2)
where C1 is the coefficient and m1 is the exponent or outlet index. For pipe outlet: q ¼ Cd Að2gÞ1=2 H 1=2 ¼ C1 H 1=2 Therefore C1 ¼ CdA(2g)1/2 and m1 ¼ 1/2. For weir type outlets: q ¼ 23Cd Að2gÞ1=2 H 1=2 ¼ 23Cd ðbHÞð2gÞ1=2 H 1=2 ¼ C1 H 3=2 Therefore C1 ¼ 23Cd Bð2gÞ1=2 and m1 ¼ 3/2. The discharge passing through the distributary channel can be expressed as: Q ¼ C2 ym2
(8.3)
BS1=2 n
In case of a wide open channel, C2 ¼ and m2 ¼ 5/3. Differentiating Eq. (8.2) with respect to H, the result is: dq ¼ C1 m1 H m1 1 dH
(8.4)
Differentiating Eq. (8.3) with respect to y, we get: dQ ¼ C2 m2 ym2 1 dy Substituting Eqs. (8.2)e(8.5) into Eq. (8.1) and solving, we obtain: dq=q m1 y dH F¼ ¼ $ $ dQ=Q m2 H dy
(8.5)
(8.6)
Since a change in the water depth of the distributary channel (dy) would result in equal change in the working head for the outlet (dH), dy ¼ dH. Considering this, Eq. (8.6) reduces to the following form: m1 y F¼ $ (8.7) m2 H
8.2 PERFORMANCE OF MODULE OR OUTLET
321
8.2.2 PROPORTIONALITY AND SETTING The outlet is said to be proportional if the rate of change of outlet discharge equals the rate of change of discharge in the distributary channel, i.e., dq dQ ¼ q Q
(8.8)
This condition leads to F ¼ 1; therefore F¼
m1 y $ ¼1 m2 H
(8.9)
or H m1 Outlet index ¼ ¼ y m2 Channel index
(8.10)
The setting can be defined as the ratio of depth of the sill level or invert level of the outlet below the full supply level (FSL) of the distributary channel to the full supply depth (FSD) of the channel. Numerically, it can be expressed as follows: Setting ¼
H Outlet index m1 ¼ ¼ y Channel index m2
(8.11)
This ratio helps in the fixing of outlet on the bank, considering the proportionality. Case I: Pipe outlet For a pipe outlet and distributary channel, the exponents are m1 ¼ 1/2 and m2 ¼ 5/3. Therefore H y
1=2 1 ¼m m2 ¼ 5=3 ¼ 0:30 or H ¼ 0.30y. It refers to a pipe outlet that can behave proportionately when the
working head for the outlet, H, is equal to 0.30 times the FSD of the distributary channel measured below the water surface. Case II: Weir-type outlet For a weir-type outlet and distributing channel, the exponents are m1 ¼ 3/2 and m2 ¼ 5/3. 3=2 1 Therefore Hy ¼ m m2 ¼ 5=3 ¼ 0:90 or H ¼ 0.90y. It refers to a weir-type outlet that can behave proportional when the working head for the outlet, H, is equal to 0.90 times the FSD of the distributary channel measured below the water surface. Hyperproportional outlet: F > 1; i.e.;
m1 H H m1 > or < y y m2 m2
or Settingð¼ H=yÞ < ðm1 =m2 Þ: Subproportional outlet: F < 1; i.e.;
m1 H H m1 < or > y y m2 m2
or Settingð¼ H=yÞ > ðm1 =m2 Þ:
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CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
8.2.3 SENSITIVITY It is defined as the ratio of the rate of change of discharge through the outlet to the rate of change of water level in the distributary channel, referred to as the normal depth of the channel. Numerically, it can be expressed as: S¼
dq=q dG=y
(8.12)
where G is the gauge of the distributary channel above the outlet and y is the normal depth or FSD. For rigid outlets or modules, the outlet discharge is fixed, irrespective of the water level in the distributary channel to the extent, therefore the sensitivity is zero. For semimodular outlets or flexible outlets where discharge through the outlet is a function of water level in the distributary channel, a gauge can be fixed and calibrated so as to indicate its reading G ¼ 0 when q ¼ 0. Sensitivity and flexibility can be related as follows: S ¼ m2 F
(8.13)
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS In this section, only nonmodular and semimodular types of outlets have been discussed as they are common in practice.
8.3.1 NONMODULAR OUTLET A pipe outlet with the exit end of the pipe submerged in water in the water course works as a nonmodular outlet. The pipes are placed horizontally and at right angles to the centerline of the distributing channel (Fig. 8.1). The discharge through the pipe outlet is computed using the following formula: pffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gHL (8.14) where q is the discharge (m3/s) of an outlet, A is the cross-sectional area of the pipe (m2), g is the acceleration due to gravity (m/s2), HL is the difference in water levels in the distributary channel and
HL = Working head = Head causing flow
FSL
FSL d
D
Distributary Channel Horizontal Pipe L Canal Bed
FIGURE 8.1 Nonmodular pipe outlet (submerged exit).
Water Course Water Course Bed
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
323
the water course (m), and Cd is the coefficient of discharge which depends on the friction factor, length, and size of the pipe outlet. The value of Cd can be computed using the following relationship: vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u d u Cd ¼ 0:05 u (8.15) t 1:5d f Lþ 400f where f ¼ the coefficient of fluid friction for pipes. It can be taken as 0.005 for clean iron pipes and 0.01 for slightly encrusted iron pipes. For earthenware pipes the value of f can be considered as 0.0075; L ¼ the length of pipe (m); and d ¼ the diameter of pipe (cm). For computational ease, an average value of Cd proposed by IS 7986 (1976) equal to 0.73 can be considered for the submerged flow condition, whereas for free flow condition as the case of semimodular outlet, its value can be considered as 0.62. Example 8.1: Design a direct outlet off-taking from the main canal having the following information: Culturable command area (CCA) of the outlet ¼ 65.44 ha Water allowance factor ¼ 0.00097 cumecs/ha FSD of the main canal ¼ 70 cm FSL of the mail canal ¼ 100.0 m FSL of the water course ¼ 99.92 m Solution: The following steps will be used to design the outlet: 1. Estimate the full supply discharge of the outlet to supply irrigation in 65.44 ha of land as: q ¼ Water allowance CCA ¼ 0:00097 65:44 ¼ 0:0635 m3 =s ¼ 63:5 lps ¼ 2:25 cfs 2. Compute the available head across the outlet as follows: HL ¼ FSLMain FSLWC ¼ 100:00 99:92 ¼ 0:08 m Since the available head for the outlet, HL, is quite less, the nonmodular outlet (i.e., submerged pipe outlet) will be provided. 3. Compute the pipe size using the following discharge formula: pffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gHL or " 4q pffiffiffiffiffiffiffiffiffiffiffi d¼ pCd 2gHL
#1=2
¼
1=2 4 0:0635 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:297x0:30 m 3:14 0:73 2 9:81 0:08
Hence, the pipe size used will be of 30 cm diameter. 4. Determine the invert level of the outlet: The bed level of the main canal is ¼ 100 0.70 ¼ 99.30 m. The pipe can be fixed at 20 cm below the FSL of the main canal. The invert level of or sill level of the outlet will be ¼ 100.0 0.20 0.30 ¼ 99.50 m. The outlet will be fixed at 99.50 99.30 ¼ 0.20 m (or 20 cm) above the bed level of the main canal.
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CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
8.3.2 SEMIMODULAR OUTLET The commonly used semimodules are: 1. 2. 3. 4.
Pipe outlet discharging freely into the water course; Venturi flume outlet or Kennedy’s gauge outlet; Open flume outlet; and Adjustable orifice semimodule (AOSM).
Here, only pipe outlet, open flume outlet, and AOSM are discussed, as these three outlets are commonly used. The suitability of the flume outlets is determined based on the ratio of canal discharge (Q) to the outlet discharge (q) and the throat width (W) as: 1. Open flume moduleðOFMÞ : Qq 20 and W 6 cm 2. Adjustable proportional moduleðAPMÞ : Qq 20 and W < 6 cm; or Qq > 20 If these conditions are not achieved, it is better to install a semimodular pipe outlet.
8.3.2.1 Pipe Outlet Discharging Freely Into the Water Course A pipe outlet works as a semimodule when discharge has free fall into the water course. This class of outlets may therefore be used as a semimodular outlet in which case the exit end of the pipe is placed higher than the water level in the water course. The working head, H0, is the difference between water level in the distributary channel and the center of pipe outlet (Fig. 8.2). The discharge is computed using the following formula: pffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gH0 (8.16) where H0 is defined in Fig. 8.2. The value of Cd can be estimated using Eq. (8.15). For general computation, the value of Cd can be considered equal to 0.62. The working head for free flowing pipe outlets, H0, should be kept as 1.5 times the diameter of the pipe; and the minimum pipe diameter should be 15 cm. For a proportional pipe outlet, the setting could be 0.30 i.e., (H/y ¼ 0.30). Example 8.2: Design a semimodular pipe outlet using the following information: CCA of the outlet ¼ 65.44 ha Water allowance factor ¼ 0.00097 cumecs/ha FSD of the main canal ¼ 70 cm FSL of the mail canal ¼ 100.0 m FSL of the water course ¼ 99.26 m FSL H0
D
d
FSL
Distributary Channel Horizontal Pipe Water Course L Canal Bed
FIGURE 8.2 Semimodular-type pipe outlets (free flow exit).
Water Course Bed
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
325
Solution: The following steps will be used to design the outlet: 1. Estimate the full supply discharge of the outlet to supply irrigation in 65.44 ha of land as: q ¼ Water allowance CCA ¼ 0:00097 65:44 ¼ 0:0635 m3 =s ¼ 63:5 lps ¼ 2:25 cfs 2. Compute the available head across the outlet as follows: HL ¼ FSLMain FSLWC ¼ 100:00 99:26 ¼ 0:74 m The available head for the outlet, HL, is sufficient to install a semimodular pipe outlet (i.e., free flow outlet). 3. Compute the pipe size considering the trail values of diameter and using the criterion that H0 1:5d. H0 should be kept as 1.5 times the diameter of the pipe of the following discharge formula: pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gH0 ¼ Cd pd 2 4 2gH0 where the value of Cd can be considered as 0.62 for the free flowing condition. For any trial value of d, the head, H0, can be computed as follows: q2 q2 q2 H0 ¼ ¼ ¼ (8.17) 2 2 2 2 CA 2gCd A Cðpd2 =4Þ where C ¼ 2gCd 2 . Trials: various trials can be considered to identify the size of pipe. The computations under different trials are presented as follows: H0 [
Trail
y (m)
q (m3/s)
d (m)
C ¼ 2gCd 2
A [ pd2/4 (m2)
(m)
(A)
(B)
(C)
(D)
(E)
(F)
1
0.70
0.0635
0.30
7.541928
2
0.70
0.0635
0.25
3
0.70
0.0635
4
0.70
0.0635
q2 CA2
Condition H0 ‡ 1:5d
Remark
(G)
(H)
(I)
0.070686
0.107004
0.450
Next trial
7.541928
0.049087
0.221884
0.375
Next trial
0.20
7.541928
0.031416
0.541709
0.300
OK
0.15
7.541928
0.017671
1.712068
0.225
Condition not acceptable
Remark [I] ¼ IF(AND(G > H,G B),“OK,” IF(AND(G > H,G > B), “Condition not acceptable,” “Next trial”)).
4. Considering these computations, the pipe size can be chosen as 0.20 m or 20 cm, and the head over the centerline of the pipe will be 0.54 m. 5. Compute the invert levels as follows: Invert/sill level of pipe ¼ FSL H0 d/2 ¼ 100.0 0.54 (0.20/2) ¼ 99.36 m. Sill level of the pipe [99.36 (100 0.70) ¼ 0.06 m ¼ 6.0 cm] is 6 cm above the bed.
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CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
6. Check the design parameters as: pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gH0 ¼ 0:62 3:41 0:22 4 2 9:81 0:54 ¼ 0:0688 m3 =s ¼ 68:82 lpsð > 63:5 lpsÞ Hence, design is OK. Example 8.3: Design a semimodular pipe outlet using the following information: CCA of the outlet ¼ 45.72 ha with an irrigation intensity of 75.83% Water allowance factor ¼ 0.00097 cumecs/ha FSD of the main canal ¼ 1.06 m FSL of the mail canal ¼ 100.76 m FSL of the water course ¼ 100.18 m Solution: The following steps will be used to design the outlet: 1. Estimate the full supply discharge of outlet to supply irrigation in 65.44 ha of land as: q ¼ Water allowance CCA ¼ 0:00097 45:72 75:83=100 ¼ 0:0336 m3 =s ¼ 33:6lps ¼ 1:19 cfs 2. Compute the pipe size considering the trail values of diameter and using the criteria that H0 1:5d. H0 should be kept as 1.5 times the diameter of pipe of the following discharge formula: pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gH0 ¼ Cd pd2 4 2gH0 where the value of Cd can be considered as 0.62 for the free flowing condition. For any trial value of d, the head H0 can be computed as follows: q2 q2 q2 ¼ ¼ H0 ¼ 2 2 2 2 CA 2gCd A C ðpd2 =4Þ where C ¼ 2gCd 2 . Trials: various trials can be considered to identify the size of pipe. The computations under different trials are presented as follows:
Trail
y (m)
q (m3/s)
d (m)
C ¼ 2gCd 2
A [ pd2/4 (m2)
q H0 ¼ CA 2 ðmÞ
Condition H0 1:5d
Remark
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
1
1.06
0.0336
0.30
7.542
0.0707
0.0300
0.450
Next trial
2
1.06
0.0336
0.25
7.542
0.0491
0.0621
0.375
Next trial
3
1.06
0.0336
0.20
7.542
0.0314
0.1517
0.300
Next trial
4
1.06
0.0336
0.15
7.542
0.0177
0.4793
0.225
OK
2
Remark [I] ¼ IF(AND(G > H,G B),“OK,”IF(AND(G > H,G > B), “Condition not acceptable,” “Next trial”)).
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
327
3. Considering the above computations, the pipe size can be chosen as 0.15 m or 15 cm, and head over the center line of the pipe will be 0.48 m. 4. Compute the invert levels as follows: Invert/sill level of pipe ¼ FSL H0 d/2 ¼ 100.76 0.48 (0.15/2) ¼ 100.205 m. Sill level of the pipe [100.205 (100.76 1.06) ¼ 0.505 m ¼ 50.5 cm] is 50.5 cm above the bed. 5. Check the design parameter as: pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gH0 ¼ 0:62 3:41 0:152 4 2 9:81 0:48 ¼ 0:0365 m3 =s ¼ 36:50 lpsð > 33.6 lpsÞ Hence, design is OK.
8.3.2.2 Open Flume Outlet The open flume outlet (Fig. 8.3) is a smooth weir with a throat constricted sufficiently to ensure the velocity above critical and long enough to ensure that the controlling section remains within the parallel throat at all discharges up to the maximum. A gradually expanding flume is provided to the outfall to obtain the maximum recovery of head (IS 12331, 1988). 22.5
y/2
57
75 Distributing Channel
R=2H R=H-y/2
Height to be kept according to H SECTION XX
120 1 in 10 39
X
57
45
Y X
Y
120 Height to be kept according to H SECTION YY
Bank Width Full Supply Level H y
0.5 TO 1
Berm
2H
75 mm thick precast RC slab 2H
12 R=2H
FIGURE 8.3 Open flume modular outlet.
15
5T O
22.5 150
Bed Level Bu
1.
Floor Concrete
1
Top should be in level with upstream wing wall Full Supply Level Bed Level 27 15
50 According to y and bank width
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CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
The entire work can be built in brick, but the controlling section is provided with cast iron or steel bed and check plates. The throat width, W, is never kept less than 6 cm, hence it is necessary to raise the crest level of the outlet much above the bed level of the distributary canal. The formation of hydraulic jump makes the outlet independent of water level in the outlet channel (Fig. 8.4). The discharge passing through the outlet can be determined using the following formula: q ¼ KWH 3=2
(8.18)
3
where q is the discharge through the outlet (m /s), K is the constant upon the width of flume, W is the throat width (m), and H is the head over the crest. The value of K as a function of W is given as (IS 12331, 1988): 8 > < 1:60; 6 cm W 9 cm K ¼ 1:64; 9 cm < W 12 cm (8.19) > : 1:66; W > 12 cm The working head required in an open flume outlet with a 1 in 5 glacis (downward inclination) and side walls splaying (i.e., expansion) at 1 in 5 is 20% of depth of water above the crest of outlet. This type of outlet is most suitable for tail clusters and proportional distributors. Other design parameters are: • • • • • •
Length of throat (crest) ¼ 2H Setting: for weir type outlet and distributary channel, the exponents are m1 ¼ 3/2 and m2 ¼ 5/3. 3=2 1 Therefore Hy ¼ m m2 ¼ 5=3 ¼ 0:90 or H ¼ 0.90y. Minimum working head for modular outlet ¼ 0.2H Crest level of flume ¼ U/S FSL 0.9y U/S approach to throat: one curved and diverging and another straight D/S expansion: splayed to 1 in 10 to meet the bed width of the outlet channel Example 8.4: Design an open flume outlet using the following information: CCA of the outlet ¼ 65.44 ha Water allowance factor ¼ 0.00097 cumecs/ha FSD of the main canal ¼ 70 cm Discharge of the main canal at FSD ¼ 0.26 m3/s Main canal geometry ¼ most economic trapezoidal with side slope m ¼ 0.5 (H:V) Manning’s roughness, n ¼ 0.018 Canal’s longitudinal slope, S ¼ 0.00015 FB
FSL
Hs y
H
FSL Crest
C BL
C BL >H
(H+0.3 ) or 2H
FIGURE 8.4 Longitudinal flow profile of open flume modular.
Downstream Expansion
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
329
FSL of the mail canal ¼ 100.0 m FSL of the water course ¼ 99.50 m Solution: The following steps will be used to design the outlet: 1. Given parameters: Basic Parameters
Value
Unit
CCA ¼
65.44
ha
Water allowance ¼
0.00097
cumec/ha
FSL (main)
100
m
FSL (water course)
99.5
m
FSD of main canal ¼
0.7
m
Discharge in the main canal, Q ¼
0.26
cumec
m¼
0.5
(H:V)
S¼
0.00015
n¼
0.018
a ¼ 2/3
0.6667
(discharge ratio factor)
aQ ¼
0.173
cumec
Throat width, W
0.06
m
Initial flume coefficient, K ¼
1.6
(For 6 W 9)
2. Compute the required discharge in the outlet as: q ¼ Water allowance CCA ¼ 0:00097 65:44 ¼ 0:0635 m3 =s ¼ 63:5 lps ¼ 2:25 cfs 3. Check the applicability of flume by computing Q/q as: Q q
0:26 ¼ 4:094 < 20, i.e., OFM can be used. ¼ 0:0635
4. Determine the flow depth in the main canal corresponding to aQ, (i.e., 0.173 cumec) and considering the canal geometry as the most economical trapezoidal section. Design Parameter
C3
A (m2)
P (m)
R (m)
Y (m)
Bottom Width, b (m)
G
H
I
J
K
L
M
1.736
3.472
0.5
0.851
2.430
0.350
0.70
0.87
1.736
3.472
0.5
0.584
2.014
0.290
0.58
0.72
S. No.
Design Discharge (m3/s)
Bed Slope
Side Slope, m (H:V)
C0 [ b/y
C1
C2
A
B
C
D
E
F
1 (Q)
0.26
0.00015
0.5
1.236
2 (aQ)
0.173
0.00015
0.5
1.236
where C0, C1, C2, and C3 are defined in Example 7.7 (Chapter 7).
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CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
5. Compute Dy as: Dy ¼ DH ¼ ya yaQ ¼ 0:70 0:58 ¼ 0:12 m 6. Compute the head over the crest considering the ratio of q and aq and solving for H as follows: The outlet discharge, q is given by q ¼ KWH 3=2
(i)
aq ¼ KWðH DHÞ3=2
(ii)
and aq is given by:
Dividing Eq. (i) by Eq. (ii), we get 1 ¼ a
H H DH
3=2 (iii)
Solving Eq. (iii) for H, we obtained: ð1=aÞ2=3 $DH i H¼h ð1=aÞ2=3 1
(iv)
Substituting the value of a and DH in Eq. (iv), we get the value of H as: ð1=0:6667Þ2=3 0:12 i ¼ 0:507 m H¼ h ð1=0:6667Þ2=3 1
H¼
0.507 m
Crest level ¼ FSL H ¼
99.493 m
Minimum head required (¼0.2H) ¼
0.101 m
Available working head ¼ [FSL(main)FSL(WC)] ¼
0.5 m (>0.2H)
Remark: ¼IF(Available working head > minimum head required, “designed head for flume is OK”)
Designed head for flume is OK
7. Check the suitability of the computed head as follows: Since the available working head is greater than the minimum head required for the outlet, the estimated head for the outlet is sufficient to meet the full supply discharge.
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
331
8. Calculate the throat width, W, for the initial value of flume coefficient K ¼ 1.60 and finalize the value of W as: W¼
q 0:0635 ¼ ¼ 0:11 m ¼ 11:0 cm 3=2 1:60 0:5071:5 KH
Since the value of throat width, W, is more than 9.0 cm, the throat width will be recomputed for K ¼ 1.64 as: q 0:0635 W¼ ¼ ¼ 0:107 m z 11:0 cm 3=2 1:64 0:5071:5 KH 9. Check the outlet capacity with estimated design parameters, such as W, K, and H, as: qb ¼ KWH 3=2 ¼ 1:64 0:11 0:5071:5 ¼ 0:0651 m3 =s ¼ 65:1 lps Since estimated outlet discharge qb is a little higher than the required outlet discharge, the design parameters are OK. The parameters are then summarized as follows: Design Parameters
Value
Unit
(a) Throat width, W
11
cm
(b) Flume coefficient, K
1.64
(c) Head over the crest, H
50.7
cm
(d) Throat length, (L ¼ 2H)
101.4
cm
(e) Crest level of flume
99.493
m
(f) Minimum working head
10.1
cm
8.3.2.3 Adjustable Orifice Semimodules Various types of orifice semimodules have been designed so far. The one that has found popularity is called Crump’s adjustable proportionate module (APM). In this module various modifications have been made, and the latest model that is now being used in Punjab and Haryana, India, is called an AOSM. This type of an adjustable module is considered to be the best of all the modules and is mostly adopted. An adjustable orifice module consists of an orifice provided with a gradually expanding flume on the downstream side of orifice. The flow through the orifice is supercritical, resulting in the formation of a hydraulic jump in the expanding flume position. The formation of jump makes the discharge independent of water level in the water course. The principal features of an adjustable orifice module are similar to those of a flume regulator with horizontal crest and curved water approach on the upstream, and downstream wings expanding to the width of water course, b. However, unlike gates, it is provided with cast iron roof block, around which masonry is done. The opening height, y0, can be changed by suitably adjusting the roof block, which can be easily done after dismantling the masonry around it. Since roof block cannot be readjustable
332
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
Top of Bank
(A)
Roof Block
15
1.5
:1
FSL y
H 0.5 :1
Distributary Channel
1
: 1.5
Hs y0
H
Water Course Bed Level
R = 2H
Channel Bed Level
FSL Water Course
Distributary Channel
(B) W
30
65
Bed Width of Water Course
Wa 65
Wa 7.6
FIGURE 8.5 Crump’s adjustable proportional module (all dimensions in centimeters). (A) Longitudinal section and (B) plan.
without breaking the masonry around it, the opening, y0, and hence the outlet discharge, cannot be easily tempered with by the cultivators. The module is thus perfectly rigid, and at the same time adjustable in dimensions at a slight cost of redoing the masonry. A typical layout of this type of outlet is depicted in Fig. 8.5. The discharge through such an outlet can be computed using the following formula: pffiffiffiffiffiffiffiffiffiffi q ¼ Cd $ðW$y0 Þ$ 2gHs (8.20) where q ¼ the discharge through the outlet (m3/s); W ¼ the width of throat (m); y0 ¼ the height of the orifice opening (m), generally kept as 1.5e2 times of W; Hs ¼ the head measured from the upstream water level in the distributary to the lowest point of the roof block (m); and Cd ¼ the coefficient of discharge, whose value varies between 0.8 and 1.05 for throat width (W) varying between 0.06 and 0.3 m. It can be considered as 0.91 for normal throat width of 0.12 m. By adopting the value of Cd as 0.91, the formula (Eq. 8.20) for discharge through the outlet will be reduced as follows: pffiffiffiffiffiffi (8.21) q ¼ 4:03$ðW$y0 Þ$ Hs
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
333
Table 8.1 Value of k as a Function of Q Q (m3/s)
k
<0.283
1.00
0.283e1.415
1.25
1.415e5.660
1.50
>5.660
2.00
These types of adjustable modules are provided in eight different standard widths, W ¼ 0.06, 0.075, 0.10, 0.12, 0.15, 0.19, 0.24, and 0.30 m. The minimum modular head loss involved in such a module is given by the following formula: HL ¼ 0:82Hs 0:5W (8.22) Originally, when this module has a setting (i.e., H/y) of 6/10, it aims at the exact proportionality and, therefore, is called APM. The throat width, W, is fixed according to the ratio q/Q as follows: q Wa ¼ k ðBu þ y=2Þ (8.23) Q where Wa ¼ the setting forward of the d/s wing wall of the approach (m); q ¼ the discharge through the outlet (m3/s); Q ¼ the discharge of the distributary channel (m3/s); Bu ¼ the bed width of the distributary channel just upstream of the outlet (m); y ¼ the depth of water level in the distributary channel or FSD; and k ¼ the ratio between the mean velocity for the entire distributary channel and the mean velocity in the part of the distributary channel, wherein the outlet has to be installed. The values of k can be taken as a function of Q from Table 8.1. The following conditions are required for the performance of the modular: 1. the ratio Hs/y should be 0.375e0.48 for the proportionate distribution of silt; and 2. the ratio Hs/y should be 0.80 or less for modular working. Disadvantage: The waterway in this type of outlet is either deep or narrow which could get blocked easily, or is shallow and wide in which case it does not draw its fair share of silt. Example 8.5: Design an APM using the following information (Fig. 8.6): CCA of the outlet ¼ 59.81 ha Water allowance factor ¼ 0.00097 cumecs/ha FSD of the main canal ¼ 1.60 m Discharge of the main canal at FSD ¼ 3.322 m3/s Main canal geometry ¼ most economic trapezoidal with side slope m ¼ 0.5 (H:V) Manning’s roughness, n ¼ 0.018 Canal’s longitudinal slope, S ¼ 0.00025 FSL of the mail canal ¼ 97.328 m FSL of the water course ¼ 96.528 m
334
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
FSL Hs H y
FSL y0 H CBL
CBL 2H
FIGURE 8.6 Typical design parameters of the adjustable proportional module.
Solution: 1. Given parameters: Basic Parameters
Value
Unit
CCA ¼
59.81
ha
Water allowance ¼
0.00097
cumecs/ha
FSL (main)
97.328
m
FSL (water course)
96.950
m
FSD of main canal ¼
1.60
m
Discharge in the main canal, Q ¼
3.322
cumecs
m¼
0.5
(H:V)
S¼
0.00025
n¼
0.018
a ¼ 2/3
0.6667
(Ratio factor)
aQ ¼
2.215
cumecs
Throat width, W
0.06
m
Flume coefficient
4.03
2. Compute required discharge in the outlet as: q ¼ Water allowance CCA ¼ 0:00097 59:81 ¼ 0:058 m3 =s ¼ 58:0 lps
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
335
3. Check the applicability of flume by computing Q/q as: Q 3:322 q ¼ 0:058 ¼ 57:27 > 20, i.e., APM can be used. It also shows that the APM outlets are more applicable in head reaches. 4. Determine the flow depth in the main canal corresponding to aQ (i.e., 2.215 cumec) and considering the canal geometry as the most economical trapezoidal section. Design Parameters Side Slope,
Design
Bottom
S.
Discharge
Bed
m (H:
C0 [ b/
No.
(m3/s)
Slope
V)
y
C1
C2
A
B
C
D
E
F
1
3.322
0.00025
0.5
1.236
2.215
0.00025
0.5
1.236
A
P
R
Y
Width,
C3
(m2)
(m)
(m)
(m)
b (m)
G
H
I
J
K
L
M
1.736
3.472
0.5
4.444
5.555
0.800
1.60
1.98
1.736
3.472
0.5
3.258
4.757
0.685
1.37
1.69
(Q) 2 (aQ)
where C0, C1, C2 and C3 are defined in Example 7.7 (Chapter 7).
5. Compute Dy, which is equal to DHs as: Dy ¼ DHs ¼ ya yaQ ¼ 1:60 1:37 ¼ 0:23 m 6. Compute the head over the crest considering the ratio of q and aq and solving for H as follows: The outlet discharge, q, is given by pffiffiffiffiffiffi q ¼ 4:03$ðW$y0 Þ$ Hs (i) and aq is given by
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi aq ¼ 4:03$ðW$y0 Þ$ Hs DHs
Dividing Eq. (i) by Eq. (ii), 1 ¼ a
Hs Hs DHs
(ii)
1=2 (iii)
Solving Eq. (iii) for H, we obtain: ð1=aÞ2 $DHs i Hs ¼ h ð1=aÞ2 1 Substituting the value of a and DH in Eq. (iv), we get the value of H as: ð1=0:6667Þ2 0:23 i ¼ 0:414 m Hs ¼ h ð1=0:6667Þ2 1 The computed value of Hs is less than 0.8y, and is therefore OK.
(iv)
336
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
7. Compute y0 considering W ¼ 6.0 cm as follows: q 0:058 pffiffiffiffiffiffiffiffiffiffiffi ¼ 0:373 m pffiffiffiffiffiffi ¼ y0 ¼ 4:03 W Hs 4:03 0:06 0:414 8. Compute H considering W ¼ 6.0 cm, y0 ¼ 0.373 m, and Hs ¼ 0.414 m as follows: H ¼ y0 þ Hs ¼ 0:373 þ 0:414 ¼ 0:787 m Crest level of the flume ¼ FSL H ¼ 97.328 0.787 ¼ 96.541 m Hs ¼
0.414 m
y0 ¼
0.373 m
H ¼ y0 þ Hs
0.787 m
Crest level ¼ FSL H ¼
96.541 m
Minimum modular head (¼0.75Hs) ¼
0.311 m
Available working head ¼ [FSL(main) FSL(WC)] ¼
0.378 m (>0.75Hs)
Remark: ¼IF(available working head > minimum head required, “designed head for flume is OK”)
Designed head for flume is OK
Since the available working head is greater than the minimum head required for the outlet, the estimated head for the outlet is sufficient to meet the full supply discharge. 9. Check the outlet capacity with estimated designed parameters, such as W, k, and H as: qb ¼ 4:03Wy0 Hs 1=2 ¼ 4:03 0:06 0:373 0:4140:5 ¼ 0:05803 m3 =s ¼ 58:03 lps Since the estimated outlet discharge qb is a little higher than the required outlet discharge, the designed parameters are OK. The parameters are then summarized as follows: Designed Parameters (a) Throat width, W
Value
Unit
6
cm
(b) Head over the crest, H ¼
78.7
cm
(c) Crest level ¼
96.541
m
(d) y0 ¼
37.3
cm
(e) Hs ¼
41.4
cm
(f) Minimum working head ¼
31.1
cm
8.5 CONCLUDING REMARKS
337
8.4 CALIBRATION OF OUTLET Calibration of the outlet is nothing but the development of a relationship between the opening of the outlet and actual (or measured) discharge passing through it for a particular gauge or water level in the parent canal (distributary channel), and its comparison with design discharge as per the standard design formula (Fig. 8.7). The measurement of discharge through the outlet can be done through cut-throat flumes, weirs, or notches, depending on the situation. Measurements can also be carried out through digital current meters, which give the velocity followed by discharge after multiplying with the wetted area. For measuring the water level in parent canal or distributary channel as well as in the water course in the case of nonmodular outlets, staff gauge will be used. A format used for the calibration of the outlet is provided in Table 8.2. In general, the outlets work satisfactorily, unless there is a damage in pipes.
8.5 CONCLUDING REMARKS In the present chapter, the procedural computation of designing the canal outlets has been explained. The chapter includes the design of three main types of outlets: nonmodular, semimodular, and modular outlets with explanatory worked examples. For detailed knowledge in the subject, readers can refer to other published literature by the United States Bureau of Reclamation and other national and international standards. Other than the described outlets, integrated controlled outlets are also available at present that can give discharge passing through them and are generally used in canal automation.
Measured discharge (m 3/s)
0.060
0.050 0.040 0.030
0.020 0.010 0.000 0.000
FIGURE 8.7 Calibration chart for the outlet.
0.010
0.020 0.030 0.040 Rated discharge (m3/s)
0.050
Table 8.2 Format Used for Outlet Calibration (a) Name of Minor/ Distributary/Main canal:
LMC
(b) RD:
220 Ch
(c) Type of outlet:
SemiModular (2)
Note: Non-Modular ¼1, Semi-Modular¼2, Modular ¼3
(d) Outlet section:
Pipe
Note: Rectangular or Pipe
(e) Size of outlet:
15 cm
(f) Length of pipe:
3m
(g) Invert level of pipe:
0m
Sample
Water Level in Distributing Channel (m)
Water Level in Distributing Channel Above Pipe Invert Level (m)
Water Level in Water Course (m)
Height of Opening of Outlet for Rectangular Outlet (m)
Percent Opening in Case of Circular Outlet
Working/ Operating Head for Outlet (m)
Rated Discharge (m3/s)
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
1
1.25
1.25
0.15
50
2
1
1.00
0.15
3
1.00
1.00
4
0.9
0.90
Measurement of Discharge Through Outlet (Cut-Throat Flume)
Measured Discharge Through Outlet (m3/s)
Flume Size
ha (cm)
Q (lps)
(viii)
(ix)
(x)
(xi)
(xii)
1.175
0.026
C-1
14.0
19.94
0.020
75
0.925
0.035
C-1
18.0
31.69
0.032
0.15
100
0.925
0.047
C-1
23.0
49.79
0.050
0.15
50
0.825
0.022
C-1
15.0
22.65
0.023
FURTHER READING
339
REFERENCES IS 12331, 1988. General Requirements for Canal Outlets [WRD 13: Canals and Cross Drainage Works]. Bureau of Indian Standards, Manak Bhawan, New Delhi. IS 7986, 1976. Code of Practice for Canal Outlets [WRD 13: Canals and Cross Drainage Works]. Bureau of Indian Standards, Manak Bhawan, New Delhi.
FURTHER READING CWC, 1984. Manual on Irrigation and Power Channel. Central Water Commission, Ministry of Water Resources, New Delhi, India. Central Board of Irrigation and Power (Publication No. 171), Malcha Marg, New Delhi. Garg, S.K., 2005. Hydrology and Water Resources Engineering. Khanna Publication, Delhi. USBR, 1967. Design Standards No. 3: Canal and Related Structures. United States Department of Interior Bureau of Reclamation, Colorado, USA.
DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
8
An outlet can be defined as a device through which water is released from a distributary channel into a water course. The discharge through an outlet is usually less than 0.085 m3/s (3.0 cusecs) (IS 7986, 1976). Various types of canal outlets have been developed from time to time to obtain a suitable performance. No one type has come out to be suitable universally. In fact, it is difficult to achieve good design with respect to “flexibility” and “sensitivity” because of various indeterminate conditions both in the distributary channels and the water course, namely, discharge levels, silt charge, capacity factor, rotation of channels, and regime condition of distributary channels. Any variation in any of these factors affects proper functioning of the outlet. Even a particular type of outlet considered suitable upstream of a control structure in a canal may not be suitable in the downstream reach of the same canal. There are different types of outlets used in the canal systems that can be classified as follows.
8.1 CLASSIFICATION OF OUTLETS Outlets may be classified into the following three types: 1. Nonmodular outlets: Nonmodular outlets are the outlets whose discharge is a function of the difference in water levels in the distributary channel and the water course and variation in either affects the discharge. These outlets consist of rectangular or circular openings and pavement. The effect of downstream water level is more with a short pavement, although even with a long pavement it cannot be entirely eliminated. The common examples of this type of outlets are: (a) open sluice and (b) drowned pipe outlet. 2. Semimodular outlets: Semimodular outlets are the outlets whose discharge depends on the water level in the distributary channel and not on the water level in the water course so long as the working head is available. The working head for an outlet is the difference between the water level of the distributary channel and that of the center of the pipe or outlet. Common examples of this type of outlets are pipe outlet, venturi flume, open flume, and orifice semimodule. 3. Modular outlets: Modular outlets are the outlets whose discharge is independent of water levels in the distributary channel and the water course, within reasonable working limits, i.e., for such outlets or modules, the discharge is constant within reasonable working limits irrespective of the fluctuations in the water levels in the distributary channel and/or the water course. This type of outlets is either with or without moving parts. In the latter case these are called rigid modules. Modular outlets with moving parts are not simple to design and construct and are, thus, expensive. These are liable to derangement due to increase in friction, rusting of the moving parts, and any obstruction in the working of moving parts caused by silt and weeds being carried in the flowing water. Gibb’s module is a common example of this type of outlet or module. Planning and Evaluation of Irrigation Projects. http://dx.doi.org/10.1016/B978-0-12-811748-4.00008-X Copyright © 2017 Elsevier Inc. All rights reserved.
319
320
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
8.2 PERFORMANCE OF MODULE OR OUTLET Performance of the modules or outlets can be assessed by using the following criteria:
8.2.1 FLEXIBILITY Flexibility can be defined as the ratio of the rate of change of discharge of the outlet to the rate of change of discharge of the distributing channel. Numerically, it can be expressed as follows: F¼
dq=q dQ=Q
(8.1)
where q is the outlet discharge and Q is the discharge in the distributary channel. If H is the head acting on the outlet, the discharge through the outlet, q, can be expressed in its general form as: q ¼ C1 H m1
(8.2)
where C1 is the coefficient and m1 is the exponent or outlet index. For pipe outlet: q ¼ Cd Að2gÞ1=2 H 1=2 ¼ C1 H 1=2 Therefore C1 ¼ CdA(2g)1/2 and m1 ¼ 1/2. For weir type outlets: q ¼ 23Cd Að2gÞ1=2 H 1=2 ¼ 23Cd ðbHÞð2gÞ1=2 H 1=2 ¼ C1 H 3=2 Therefore C1 ¼ 23Cd Bð2gÞ1=2 and m1 ¼ 3/2. The discharge passing through the distributary channel can be expressed as: Q ¼ C2 ym2
(8.3)
BS1=2 n
In case of a wide open channel, C2 ¼ and m2 ¼ 5/3. Differentiating Eq. (8.2) with respect to H, the result is: dq ¼ C1 m1 H m1 1 dH
(8.4)
Differentiating Eq. (8.3) with respect to y, we get: dQ ¼ C2 m2 ym2 1 dy Substituting Eqs. (8.2)e(8.5) into Eq. (8.1) and solving, we obtain: dq=q m1 y dH F¼ ¼ $ $ dQ=Q m2 H dy
(8.5)
(8.6)
Since a change in the water depth of the distributary channel (dy) would result in equal change in the working head for the outlet (dH), dy ¼ dH. Considering this, Eq. (8.6) reduces to the following form: m1 y F¼ $ (8.7) m2 H
8.2 PERFORMANCE OF MODULE OR OUTLET
321
8.2.2 PROPORTIONALITY AND SETTING The outlet is said to be proportional if the rate of change of outlet discharge equals the rate of change of discharge in the distributary channel, i.e., dq dQ ¼ q Q
(8.8)
This condition leads to F ¼ 1; therefore F¼
m1 y $ ¼1 m2 H
(8.9)
or H m1 Outlet index ¼ ¼ y m2 Channel index
(8.10)
The setting can be defined as the ratio of depth of the sill level or invert level of the outlet below the full supply level (FSL) of the distributary channel to the full supply depth (FSD) of the channel. Numerically, it can be expressed as follows: Setting ¼
H Outlet index m1 ¼ ¼ y Channel index m2
(8.11)
This ratio helps in the fixing of outlet on the bank, considering the proportionality. Case I: Pipe outlet For a pipe outlet and distributary channel, the exponents are m1 ¼ 1/2 and m2 ¼ 5/3. Therefore H y
1=2 1 ¼m m2 ¼ 5=3 ¼ 0:30 or H ¼ 0.30y. It refers to a pipe outlet that can behave proportionately when the
working head for the outlet, H, is equal to 0.30 times the FSD of the distributary channel measured below the water surface. Case II: Weir-type outlet For a weir-type outlet and distributing channel, the exponents are m1 ¼ 3/2 and m2 ¼ 5/3. 3=2 1 Therefore Hy ¼ m m2 ¼ 5=3 ¼ 0:90 or H ¼ 0.90y. It refers to a weir-type outlet that can behave proportional when the working head for the outlet, H, is equal to 0.90 times the FSD of the distributary channel measured below the water surface. Hyperproportional outlet: F > 1; i.e.;
m1 H H m1 > or < y y m2 m2
or Settingð¼ H=yÞ < ðm1 =m2 Þ: Subproportional outlet: F < 1; i.e.;
m1 H H m1 < or > y y m2 m2
or Settingð¼ H=yÞ > ðm1 =m2 Þ:
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CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
8.2.3 SENSITIVITY It is defined as the ratio of the rate of change of discharge through the outlet to the rate of change of water level in the distributary channel, referred to as the normal depth of the channel. Numerically, it can be expressed as: S¼
dq=q dG=y
(8.12)
where G is the gauge of the distributary channel above the outlet and y is the normal depth or FSD. For rigid outlets or modules, the outlet discharge is fixed, irrespective of the water level in the distributary channel to the extent, therefore the sensitivity is zero. For semimodular outlets or flexible outlets where discharge through the outlet is a function of water level in the distributary channel, a gauge can be fixed and calibrated so as to indicate its reading G ¼ 0 when q ¼ 0. Sensitivity and flexibility can be related as follows: S ¼ m2 F
(8.13)
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS In this section, only nonmodular and semimodular types of outlets have been discussed as they are common in practice.
8.3.1 NONMODULAR OUTLET A pipe outlet with the exit end of the pipe submerged in water in the water course works as a nonmodular outlet. The pipes are placed horizontally and at right angles to the centerline of the distributing channel (Fig. 8.1). The discharge through the pipe outlet is computed using the following formula: pffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gHL (8.14) where q is the discharge (m3/s) of an outlet, A is the cross-sectional area of the pipe (m2), g is the acceleration due to gravity (m/s2), HL is the difference in water levels in the distributary channel and
HL = Working head = Head causing flow
FSL
FSL d
D
Distributary Channel Horizontal Pipe L Canal Bed
FIGURE 8.1 Nonmodular pipe outlet (submerged exit).
Water Course Water Course Bed
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
323
the water course (m), and Cd is the coefficient of discharge which depends on the friction factor, length, and size of the pipe outlet. The value of Cd can be computed using the following relationship: vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u d u Cd ¼ 0:05 u (8.15) t 1:5d f Lþ 400f where f ¼ the coefficient of fluid friction for pipes. It can be taken as 0.005 for clean iron pipes and 0.01 for slightly encrusted iron pipes. For earthenware pipes the value of f can be considered as 0.0075; L ¼ the length of pipe (m); and d ¼ the diameter of pipe (cm). For computational ease, an average value of Cd proposed by IS 7986 (1976) equal to 0.73 can be considered for the submerged flow condition, whereas for free flow condition as the case of semimodular outlet, its value can be considered as 0.62. Example 8.1: Design a direct outlet off-taking from the main canal having the following information: Culturable command area (CCA) of the outlet ¼ 65.44 ha Water allowance factor ¼ 0.00097 cumecs/ha FSD of the main canal ¼ 70 cm FSL of the mail canal ¼ 100.0 m FSL of the water course ¼ 99.92 m Solution: The following steps will be used to design the outlet: 1. Estimate the full supply discharge of the outlet to supply irrigation in 65.44 ha of land as: q ¼ Water allowance CCA ¼ 0:00097 65:44 ¼ 0:0635 m3 =s ¼ 63:5 lps ¼ 2:25 cfs 2. Compute the available head across the outlet as follows: HL ¼ FSLMain FSLWC ¼ 100:00 99:92 ¼ 0:08 m Since the available head for the outlet, HL, is quite less, the nonmodular outlet (i.e., submerged pipe outlet) will be provided. 3. Compute the pipe size using the following discharge formula: pffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gHL or " 4q pffiffiffiffiffiffiffiffiffiffiffi d¼ pCd 2gHL
#1=2
¼
1=2 4 0:0635 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:297x0:30 m 3:14 0:73 2 9:81 0:08
Hence, the pipe size used will be of 30 cm diameter. 4. Determine the invert level of the outlet: The bed level of the main canal is ¼ 100 0.70 ¼ 99.30 m. The pipe can be fixed at 20 cm below the FSL of the main canal. The invert level of or sill level of the outlet will be ¼ 100.0 0.20 0.30 ¼ 99.50 m. The outlet will be fixed at 99.50 99.30 ¼ 0.20 m (or 20 cm) above the bed level of the main canal.
324
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
8.3.2 SEMIMODULAR OUTLET The commonly used semimodules are: 1. 2. 3. 4.
Pipe outlet discharging freely into the water course; Venturi flume outlet or Kennedy’s gauge outlet; Open flume outlet; and Adjustable orifice semimodule (AOSM).
Here, only pipe outlet, open flume outlet, and AOSM are discussed, as these three outlets are commonly used. The suitability of the flume outlets is determined based on the ratio of canal discharge (Q) to the outlet discharge (q) and the throat width (W) as: 1. Open flume moduleðOFMÞ : Qq 20 and W 6 cm 2. Adjustable proportional moduleðAPMÞ : Qq 20 and W < 6 cm; or Qq > 20 If these conditions are not achieved, it is better to install a semimodular pipe outlet.
8.3.2.1 Pipe Outlet Discharging Freely Into the Water Course A pipe outlet works as a semimodule when discharge has free fall into the water course. This class of outlets may therefore be used as a semimodular outlet in which case the exit end of the pipe is placed higher than the water level in the water course. The working head, H0, is the difference between water level in the distributary channel and the center of pipe outlet (Fig. 8.2). The discharge is computed using the following formula: pffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gH0 (8.16) where H0 is defined in Fig. 8.2. The value of Cd can be estimated using Eq. (8.15). For general computation, the value of Cd can be considered equal to 0.62. The working head for free flowing pipe outlets, H0, should be kept as 1.5 times the diameter of the pipe; and the minimum pipe diameter should be 15 cm. For a proportional pipe outlet, the setting could be 0.30 i.e., (H/y ¼ 0.30). Example 8.2: Design a semimodular pipe outlet using the following information: CCA of the outlet ¼ 65.44 ha Water allowance factor ¼ 0.00097 cumecs/ha FSD of the main canal ¼ 70 cm FSL of the mail canal ¼ 100.0 m FSL of the water course ¼ 99.26 m FSL H0
D
d
FSL
Distributary Channel Horizontal Pipe Water Course L Canal Bed
FIGURE 8.2 Semimodular-type pipe outlets (free flow exit).
Water Course Bed
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
325
Solution: The following steps will be used to design the outlet: 1. Estimate the full supply discharge of the outlet to supply irrigation in 65.44 ha of land as: q ¼ Water allowance CCA ¼ 0:00097 65:44 ¼ 0:0635 m3 =s ¼ 63:5 lps ¼ 2:25 cfs 2. Compute the available head across the outlet as follows: HL ¼ FSLMain FSLWC ¼ 100:00 99:26 ¼ 0:74 m The available head for the outlet, HL, is sufficient to install a semimodular pipe outlet (i.e., free flow outlet). 3. Compute the pipe size considering the trail values of diameter and using the criterion that H0 1:5d. H0 should be kept as 1.5 times the diameter of the pipe of the following discharge formula: pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gH0 ¼ Cd pd 2 4 2gH0 where the value of Cd can be considered as 0.62 for the free flowing condition. For any trial value of d, the head, H0, can be computed as follows: q2 q2 q2 H0 ¼ ¼ ¼ (8.17) 2 2 2 2 CA 2gCd A Cðpd2 =4Þ where C ¼ 2gCd 2 . Trials: various trials can be considered to identify the size of pipe. The computations under different trials are presented as follows: H0 [
Trail
y (m)
q (m3/s)
d (m)
C ¼ 2gCd 2
A [ pd2/4 (m2)
(m)
(A)
(B)
(C)
(D)
(E)
(F)
1
0.70
0.0635
0.30
7.541928
2
0.70
0.0635
0.25
3
0.70
0.0635
4
0.70
0.0635
q2 CA2
Condition H0 ‡ 1:5d
Remark
(G)
(H)
(I)
0.070686
0.107004
0.450
Next trial
7.541928
0.049087
0.221884
0.375
Next trial
0.20
7.541928
0.031416
0.541709
0.300
OK
0.15
7.541928
0.017671
1.712068
0.225
Condition not acceptable
Remark [I] ¼ IF(AND(G > H,G B),“OK,” IF(AND(G > H,G > B), “Condition not acceptable,” “Next trial”)).
4. Considering these computations, the pipe size can be chosen as 0.20 m or 20 cm, and the head over the centerline of the pipe will be 0.54 m. 5. Compute the invert levels as follows: Invert/sill level of pipe ¼ FSL H0 d/2 ¼ 100.0 0.54 (0.20/2) ¼ 99.36 m. Sill level of the pipe [99.36 (100 0.70) ¼ 0.06 m ¼ 6.0 cm] is 6 cm above the bed.
326
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
6. Check the design parameters as: pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gH0 ¼ 0:62 3:41 0:22 4 2 9:81 0:54 ¼ 0:0688 m3 =s ¼ 68:82 lpsð > 63:5 lpsÞ Hence, design is OK. Example 8.3: Design a semimodular pipe outlet using the following information: CCA of the outlet ¼ 45.72 ha with an irrigation intensity of 75.83% Water allowance factor ¼ 0.00097 cumecs/ha FSD of the main canal ¼ 1.06 m FSL of the mail canal ¼ 100.76 m FSL of the water course ¼ 100.18 m Solution: The following steps will be used to design the outlet: 1. Estimate the full supply discharge of outlet to supply irrigation in 65.44 ha of land as: q ¼ Water allowance CCA ¼ 0:00097 45:72 75:83=100 ¼ 0:0336 m3 =s ¼ 33:6lps ¼ 1:19 cfs 2. Compute the pipe size considering the trail values of diameter and using the criteria that H0 1:5d. H0 should be kept as 1.5 times the diameter of pipe of the following discharge formula: pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gH0 ¼ Cd pd2 4 2gH0 where the value of Cd can be considered as 0.62 for the free flowing condition. For any trial value of d, the head H0 can be computed as follows: q2 q2 q2 ¼ ¼ H0 ¼ 2 2 2 2 CA 2gCd A C ðpd2 =4Þ where C ¼ 2gCd 2 . Trials: various trials can be considered to identify the size of pipe. The computations under different trials are presented as follows:
Trail
y (m)
q (m3/s)
d (m)
C ¼ 2gCd 2
A [ pd2/4 (m2)
q H0 ¼ CA 2 ðmÞ
Condition H0 1:5d
Remark
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
1
1.06
0.0336
0.30
7.542
0.0707
0.0300
0.450
Next trial
2
1.06
0.0336
0.25
7.542
0.0491
0.0621
0.375
Next trial
3
1.06
0.0336
0.20
7.542
0.0314
0.1517
0.300
Next trial
4
1.06
0.0336
0.15
7.542
0.0177
0.4793
0.225
OK
2
Remark [I] ¼ IF(AND(G > H,G B),“OK,”IF(AND(G > H,G > B), “Condition not acceptable,” “Next trial”)).
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
327
3. Considering the above computations, the pipe size can be chosen as 0.15 m or 15 cm, and head over the center line of the pipe will be 0.48 m. 4. Compute the invert levels as follows: Invert/sill level of pipe ¼ FSL H0 d/2 ¼ 100.76 0.48 (0.15/2) ¼ 100.205 m. Sill level of the pipe [100.205 (100.76 1.06) ¼ 0.505 m ¼ 50.5 cm] is 50.5 cm above the bed. 5. Check the design parameter as: pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ¼ Cd A 2gH0 ¼ 0:62 3:41 0:152 4 2 9:81 0:48 ¼ 0:0365 m3 =s ¼ 36:50 lpsð > 33.6 lpsÞ Hence, design is OK.
8.3.2.2 Open Flume Outlet The open flume outlet (Fig. 8.3) is a smooth weir with a throat constricted sufficiently to ensure the velocity above critical and long enough to ensure that the controlling section remains within the parallel throat at all discharges up to the maximum. A gradually expanding flume is provided to the outfall to obtain the maximum recovery of head (IS 12331, 1988). 22.5
y/2
57
75 Distributing Channel
R=2H R=H-y/2
Height to be kept according to H SECTION XX
120 1 in 10 39
X
57
45
Y X
Y
120 Height to be kept according to H SECTION YY
Bank Width Full Supply Level H y
0.5 TO 1
Berm
2H
75 mm thick precast RC slab 2H
12 R=2H
FIGURE 8.3 Open flume modular outlet.
15
5T O
22.5 150
Bed Level Bu
1.
Floor Concrete
1
Top should be in level with upstream wing wall Full Supply Level Bed Level 27 15
50 According to y and bank width
328
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
The entire work can be built in brick, but the controlling section is provided with cast iron or steel bed and check plates. The throat width, W, is never kept less than 6 cm, hence it is necessary to raise the crest level of the outlet much above the bed level of the distributary canal. The formation of hydraulic jump makes the outlet independent of water level in the outlet channel (Fig. 8.4). The discharge passing through the outlet can be determined using the following formula: q ¼ KWH 3=2
(8.18)
3
where q is the discharge through the outlet (m /s), K is the constant upon the width of flume, W is the throat width (m), and H is the head over the crest. The value of K as a function of W is given as (IS 12331, 1988): 8 > < 1:60; 6 cm W 9 cm K ¼ 1:64; 9 cm < W 12 cm (8.19) > : 1:66; W > 12 cm The working head required in an open flume outlet with a 1 in 5 glacis (downward inclination) and side walls splaying (i.e., expansion) at 1 in 5 is 20% of depth of water above the crest of outlet. This type of outlet is most suitable for tail clusters and proportional distributors. Other design parameters are: • • • • • •
Length of throat (crest) ¼ 2H Setting: for weir type outlet and distributary channel, the exponents are m1 ¼ 3/2 and m2 ¼ 5/3. 3=2 1 Therefore Hy ¼ m m2 ¼ 5=3 ¼ 0:90 or H ¼ 0.90y. Minimum working head for modular outlet ¼ 0.2H Crest level of flume ¼ U/S FSL 0.9y U/S approach to throat: one curved and diverging and another straight D/S expansion: splayed to 1 in 10 to meet the bed width of the outlet channel Example 8.4: Design an open flume outlet using the following information: CCA of the outlet ¼ 65.44 ha Water allowance factor ¼ 0.00097 cumecs/ha FSD of the main canal ¼ 70 cm Discharge of the main canal at FSD ¼ 0.26 m3/s Main canal geometry ¼ most economic trapezoidal with side slope m ¼ 0.5 (H:V) Manning’s roughness, n ¼ 0.018 Canal’s longitudinal slope, S ¼ 0.00015 FB
FSL
Hs y
H
FSL Crest
C BL
C BL >H
(H+0.3 ) or 2H
FIGURE 8.4 Longitudinal flow profile of open flume modular.
Downstream Expansion
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
329
FSL of the mail canal ¼ 100.0 m FSL of the water course ¼ 99.50 m Solution: The following steps will be used to design the outlet: 1. Given parameters: Basic Parameters
Value
Unit
CCA ¼
65.44
ha
Water allowance ¼
0.00097
cumec/ha
FSL (main)
100
m
FSL (water course)
99.5
m
FSD of main canal ¼
0.7
m
Discharge in the main canal, Q ¼
0.26
cumec
m¼
0.5
(H:V)
S¼
0.00015
n¼
0.018
a ¼ 2/3
0.6667
(discharge ratio factor)
aQ ¼
0.173
cumec
Throat width, W
0.06
m
Initial flume coefficient, K ¼
1.6
(For 6 W 9)
2. Compute the required discharge in the outlet as: q ¼ Water allowance CCA ¼ 0:00097 65:44 ¼ 0:0635 m3 =s ¼ 63:5 lps ¼ 2:25 cfs 3. Check the applicability of flume by computing Q/q as: Q q
0:26 ¼ 4:094 < 20, i.e., OFM can be used. ¼ 0:0635
4. Determine the flow depth in the main canal corresponding to aQ, (i.e., 0.173 cumec) and considering the canal geometry as the most economical trapezoidal section. Design Parameter
C3
A (m2)
P (m)
R (m)
Y (m)
Bottom Width, b (m)
G
H
I
J
K
L
M
1.736
3.472
0.5
0.851
2.430
0.350
0.70
0.87
1.736
3.472
0.5
0.584
2.014
0.290
0.58
0.72
S. No.
Design Discharge (m3/s)
Bed Slope
Side Slope, m (H:V)
C0 [ b/y
C1
C2
A
B
C
D
E
F
1 (Q)
0.26
0.00015
0.5
1.236
2 (aQ)
0.173
0.00015
0.5
1.236
where C0, C1, C2, and C3 are defined in Example 7.7 (Chapter 7).
330
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
5. Compute Dy as: Dy ¼ DH ¼ ya yaQ ¼ 0:70 0:58 ¼ 0:12 m 6. Compute the head over the crest considering the ratio of q and aq and solving for H as follows: The outlet discharge, q is given by q ¼ KWH 3=2
(i)
aq ¼ KWðH DHÞ3=2
(ii)
and aq is given by:
Dividing Eq. (i) by Eq. (ii), we get 1 ¼ a
H H DH
3=2 (iii)
Solving Eq. (iii) for H, we obtained: ð1=aÞ2=3 $DH i H¼h ð1=aÞ2=3 1
(iv)
Substituting the value of a and DH in Eq. (iv), we get the value of H as: ð1=0:6667Þ2=3 0:12 i ¼ 0:507 m H¼ h ð1=0:6667Þ2=3 1
H¼
0.507 m
Crest level ¼ FSL H ¼
99.493 m
Minimum head required (¼0.2H) ¼
0.101 m
Available working head ¼ [FSL(main)FSL(WC)] ¼
0.5 m (>0.2H)
Remark: ¼IF(Available working head > minimum head required, “designed head for flume is OK”)
Designed head for flume is OK
7. Check the suitability of the computed head as follows: Since the available working head is greater than the minimum head required for the outlet, the estimated head for the outlet is sufficient to meet the full supply discharge.
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
331
8. Calculate the throat width, W, for the initial value of flume coefficient K ¼ 1.60 and finalize the value of W as: W¼
q 0:0635 ¼ ¼ 0:11 m ¼ 11:0 cm 3=2 1:60 0:5071:5 KH
Since the value of throat width, W, is more than 9.0 cm, the throat width will be recomputed for K ¼ 1.64 as: q 0:0635 W¼ ¼ ¼ 0:107 m z 11:0 cm 3=2 1:64 0:5071:5 KH 9. Check the outlet capacity with estimated design parameters, such as W, K, and H, as: qb ¼ KWH 3=2 ¼ 1:64 0:11 0:5071:5 ¼ 0:0651 m3 =s ¼ 65:1 lps Since estimated outlet discharge qb is a little higher than the required outlet discharge, the design parameters are OK. The parameters are then summarized as follows: Design Parameters
Value
Unit
(a) Throat width, W
11
cm
(b) Flume coefficient, K
1.64
(c) Head over the crest, H
50.7
cm
(d) Throat length, (L ¼ 2H)
101.4
cm
(e) Crest level of flume
99.493
m
(f) Minimum working head
10.1
cm
8.3.2.3 Adjustable Orifice Semimodules Various types of orifice semimodules have been designed so far. The one that has found popularity is called Crump’s adjustable proportionate module (APM). In this module various modifications have been made, and the latest model that is now being used in Punjab and Haryana, India, is called an AOSM. This type of an adjustable module is considered to be the best of all the modules and is mostly adopted. An adjustable orifice module consists of an orifice provided with a gradually expanding flume on the downstream side of orifice. The flow through the orifice is supercritical, resulting in the formation of a hydraulic jump in the expanding flume position. The formation of jump makes the discharge independent of water level in the water course. The principal features of an adjustable orifice module are similar to those of a flume regulator with horizontal crest and curved water approach on the upstream, and downstream wings expanding to the width of water course, b. However, unlike gates, it is provided with cast iron roof block, around which masonry is done. The opening height, y0, can be changed by suitably adjusting the roof block, which can be easily done after dismantling the masonry around it. Since roof block cannot be readjustable
332
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
Top of Bank
(A)
Roof Block
15
1.5
:1
FSL y
H 0.5 :1
Distributary Channel
1
: 1.5
Hs y0
H
Water Course Bed Level
R = 2H
Channel Bed Level
FSL Water Course
Distributary Channel
(B) W
30
65
Bed Width of Water Course
Wa 65
Wa 7.6
FIGURE 8.5 Crump’s adjustable proportional module (all dimensions in centimeters). (A) Longitudinal section and (B) plan.
without breaking the masonry around it, the opening, y0, and hence the outlet discharge, cannot be easily tempered with by the cultivators. The module is thus perfectly rigid, and at the same time adjustable in dimensions at a slight cost of redoing the masonry. A typical layout of this type of outlet is depicted in Fig. 8.5. The discharge through such an outlet can be computed using the following formula: pffiffiffiffiffiffiffiffiffiffi q ¼ Cd $ðW$y0 Þ$ 2gHs (8.20) where q ¼ the discharge through the outlet (m3/s); W ¼ the width of throat (m); y0 ¼ the height of the orifice opening (m), generally kept as 1.5e2 times of W; Hs ¼ the head measured from the upstream water level in the distributary to the lowest point of the roof block (m); and Cd ¼ the coefficient of discharge, whose value varies between 0.8 and 1.05 for throat width (W) varying between 0.06 and 0.3 m. It can be considered as 0.91 for normal throat width of 0.12 m. By adopting the value of Cd as 0.91, the formula (Eq. 8.20) for discharge through the outlet will be reduced as follows: pffiffiffiffiffiffi (8.21) q ¼ 4:03$ðW$y0 Þ$ Hs
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
333
Table 8.1 Value of k as a Function of Q Q (m3/s)
k
<0.283
1.00
0.283e1.415
1.25
1.415e5.660
1.50
>5.660
2.00
These types of adjustable modules are provided in eight different standard widths, W ¼ 0.06, 0.075, 0.10, 0.12, 0.15, 0.19, 0.24, and 0.30 m. The minimum modular head loss involved in such a module is given by the following formula: HL ¼ 0:82Hs 0:5W (8.22) Originally, when this module has a setting (i.e., H/y) of 6/10, it aims at the exact proportionality and, therefore, is called APM. The throat width, W, is fixed according to the ratio q/Q as follows: q Wa ¼ k ðBu þ y=2Þ (8.23) Q where Wa ¼ the setting forward of the d/s wing wall of the approach (m); q ¼ the discharge through the outlet (m3/s); Q ¼ the discharge of the distributary channel (m3/s); Bu ¼ the bed width of the distributary channel just upstream of the outlet (m); y ¼ the depth of water level in the distributary channel or FSD; and k ¼ the ratio between the mean velocity for the entire distributary channel and the mean velocity in the part of the distributary channel, wherein the outlet has to be installed. The values of k can be taken as a function of Q from Table 8.1. The following conditions are required for the performance of the modular: 1. the ratio Hs/y should be 0.375e0.48 for the proportionate distribution of silt; and 2. the ratio Hs/y should be 0.80 or less for modular working. Disadvantage: The waterway in this type of outlet is either deep or narrow which could get blocked easily, or is shallow and wide in which case it does not draw its fair share of silt. Example 8.5: Design an APM using the following information (Fig. 8.6): CCA of the outlet ¼ 59.81 ha Water allowance factor ¼ 0.00097 cumecs/ha FSD of the main canal ¼ 1.60 m Discharge of the main canal at FSD ¼ 3.322 m3/s Main canal geometry ¼ most economic trapezoidal with side slope m ¼ 0.5 (H:V) Manning’s roughness, n ¼ 0.018 Canal’s longitudinal slope, S ¼ 0.00025 FSL of the mail canal ¼ 97.328 m FSL of the water course ¼ 96.528 m
334
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
FSL Hs H y
FSL y0 H CBL
CBL 2H
FIGURE 8.6 Typical design parameters of the adjustable proportional module.
Solution: 1. Given parameters: Basic Parameters
Value
Unit
CCA ¼
59.81
ha
Water allowance ¼
0.00097
cumecs/ha
FSL (main)
97.328
m
FSL (water course)
96.950
m
FSD of main canal ¼
1.60
m
Discharge in the main canal, Q ¼
3.322
cumecs
m¼
0.5
(H:V)
S¼
0.00025
n¼
0.018
a ¼ 2/3
0.6667
(Ratio factor)
aQ ¼
2.215
cumecs
Throat width, W
0.06
m
Flume coefficient
4.03
2. Compute required discharge in the outlet as: q ¼ Water allowance CCA ¼ 0:00097 59:81 ¼ 0:058 m3 =s ¼ 58:0 lps
8.3 DESIGN OF OUTLETS: DISCHARGE THROUGH OUTLETS
335
3. Check the applicability of flume by computing Q/q as: Q 3:322 q ¼ 0:058 ¼ 57:27 > 20, i.e., APM can be used. It also shows that the APM outlets are more applicable in head reaches. 4. Determine the flow depth in the main canal corresponding to aQ (i.e., 2.215 cumec) and considering the canal geometry as the most economical trapezoidal section. Design Parameters Side Slope,
Design
Bottom
S.
Discharge
Bed
m (H:
C0 [ b/
No.
(m3/s)
Slope
V)
y
C1
C2
A
B
C
D
E
F
1
3.322
0.00025
0.5
1.236
2.215
0.00025
0.5
1.236
A
P
R
Y
Width,
C3
(m2)
(m)
(m)
(m)
b (m)
G
H
I
J
K
L
M
1.736
3.472
0.5
4.444
5.555
0.800
1.60
1.98
1.736
3.472
0.5
3.258
4.757
0.685
1.37
1.69
(Q) 2 (aQ)
where C0, C1, C2 and C3 are defined in Example 7.7 (Chapter 7).
5. Compute Dy, which is equal to DHs as: Dy ¼ DHs ¼ ya yaQ ¼ 1:60 1:37 ¼ 0:23 m 6. Compute the head over the crest considering the ratio of q and aq and solving for H as follows: The outlet discharge, q, is given by pffiffiffiffiffiffi q ¼ 4:03$ðW$y0 Þ$ Hs (i) and aq is given by
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi aq ¼ 4:03$ðW$y0 Þ$ Hs DHs
Dividing Eq. (i) by Eq. (ii), 1 ¼ a
Hs Hs DHs
(ii)
1=2 (iii)
Solving Eq. (iii) for H, we obtain: ð1=aÞ2 $DHs i Hs ¼ h ð1=aÞ2 1 Substituting the value of a and DH in Eq. (iv), we get the value of H as: ð1=0:6667Þ2 0:23 i ¼ 0:414 m Hs ¼ h ð1=0:6667Þ2 1 The computed value of Hs is less than 0.8y, and is therefore OK.
(iv)
336
CHAPTER 8 DESIGN OF CANAL OUTLETS AND THEIR CALIBRATION
7. Compute y0 considering W ¼ 6.0 cm as follows: q 0:058 pffiffiffiffiffiffiffiffiffiffiffi ¼ 0:373 m pffiffiffiffiffiffi ¼ y0 ¼ 4:03 W Hs 4:03 0:06 0:414 8. Compute H considering W ¼ 6.0 cm, y0 ¼ 0.373 m, and Hs ¼ 0.414 m as follows: H ¼ y0 þ Hs ¼ 0:373 þ 0:414 ¼ 0:787 m Crest level of the flume ¼ FSL H ¼ 97.328 0.787 ¼ 96.541 m Hs ¼
0.414 m
y0 ¼
0.373 m
H ¼ y0 þ Hs
0.787 m
Crest level ¼ FSL H ¼
96.541 m
Minimum modular head (¼0.75Hs) ¼
0.311 m
Available working head ¼ [FSL(main) FSL(WC)] ¼
0.378 m (>0.75Hs)
Remark: ¼IF(available working head > minimum head required, “designed head for flume is OK”)
Designed head for flume is OK
Since the available working head is greater than the minimum head required for the outlet, the estimated head for the outlet is sufficient to meet the full supply discharge. 9. Check the outlet capacity with estimated designed parameters, such as W, k, and H as: qb ¼ 4:03Wy0 Hs 1=2 ¼ 4:03 0:06 0:373 0:4140:5 ¼ 0:05803 m3 =s ¼ 58:03 lps Since the estimated outlet discharge qb is a little higher than the required outlet discharge, the designed parameters are OK. The parameters are then summarized as follows: Designed Parameters (a) Throat width, W
Value
Unit
6
cm
(b) Head over the crest, H ¼
78.7
cm
(c) Crest level ¼
96.541
m
(d) y0 ¼
37.3
cm
(e) Hs ¼
41.4
cm
(f) Minimum working head ¼
31.1
cm
8.5 CONCLUDING REMARKS
337
8.4 CALIBRATION OF OUTLET Calibration of the outlet is nothing but the development of a relationship between the opening of the outlet and actual (or measured) discharge passing through it for a particular gauge or water level in the parent canal (distributary channel), and its comparison with design discharge as per the standard design formula (Fig. 8.7). The measurement of discharge through the outlet can be done through cut-throat flumes, weirs, or notches, depending on the situation. Measurements can also be carried out through digital current meters, which give the velocity followed by discharge after multiplying with the wetted area. For measuring the water level in parent canal or distributary channel as well as in the water course in the case of nonmodular outlets, staff gauge will be used. A format used for the calibration of the outlet is provided in Table 8.2. In general, the outlets work satisfactorily, unless there is a damage in pipes.
8.5 CONCLUDING REMARKS In the present chapter, the procedural computation of designing the canal outlets has been explained. The chapter includes the design of three main types of outlets: nonmodular, semimodular, and modular outlets with explanatory worked examples. For detailed knowledge in the subject, readers can refer to other published literature by the United States Bureau of Reclamation and other national and international standards. Other than the described outlets, integrated controlled outlets are also available at present that can give discharge passing through them and are generally used in canal automation.
Measured discharge (m 3/s)
0.060
0.050 0.040 0.030
0.020 0.010 0.000 0.000
FIGURE 8.7 Calibration chart for the outlet.
0.010
0.020 0.030 0.040 Rated discharge (m3/s)
0.050
Table 8.2 Format Used for Outlet Calibration (a) Name of Minor/ Distributary/Main canal:
LMC
(b) RD:
220 Ch
(c) Type of outlet:
SemiModular (2)
Note: Non-Modular ¼1, Semi-Modular¼2, Modular ¼3
(d) Outlet section:
Pipe
Note: Rectangular or Pipe
(e) Size of outlet:
15 cm
(f) Length of pipe:
3m
(g) Invert level of pipe:
0m
Sample
Water Level in Distributing Channel (m)
Water Level in Distributing Channel Above Pipe Invert Level (m)
Water Level in Water Course (m)
Height of Opening of Outlet for Rectangular Outlet (m)
Percent Opening in Case of Circular Outlet
Working/ Operating Head for Outlet (m)
Rated Discharge (m3/s)
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
1
1.25
1.25
0.15
50
2
1
1.00
0.15
3
1.00
1.00
4
0.9
0.90
Measurement of Discharge Through Outlet (Cut-Throat Flume)
Measured Discharge Through Outlet (m3/s)
Flume Size
ha (cm)
Q (lps)
(viii)
(ix)
(x)
(xi)
(xii)
1.175
0.026
C-1
14.0
19.94
0.020
75
0.925
0.035
C-1
18.0
31.69
0.032
0.15
100
0.925
0.047
C-1
23.0
49.79
0.050
0.15
50
0.825
0.022
C-1
15.0
22.65
0.023
FURTHER READING
339
REFERENCES IS 12331, 1988. General Requirements for Canal Outlets [WRD 13: Canals and Cross Drainage Works]. Bureau of Indian Standards, Manak Bhawan, New Delhi. IS 7986, 1976. Code of Practice for Canal Outlets [WRD 13: Canals and Cross Drainage Works]. Bureau of Indian Standards, Manak Bhawan, New Delhi.
FURTHER READING CWC, 1984. Manual on Irrigation and Power Channel. Central Water Commission, Ministry of Water Resources, New Delhi, India. Central Board of Irrigation and Power (Publication No. 171), Malcha Marg, New Delhi. Garg, S.K., 2005. Hydrology and Water Resources Engineering. Khanna Publication, Delhi. USBR, 1967. Design Standards No. 3: Canal and Related Structures. United States Department of Interior Bureau of Reclamation, Colorado, USA.