CHAPTER 15 COMPUTATION OF WATER SURFACE PROFILES
15.1. A wide rectangular river with a bed slope of 14 × 10 carries a discharge of 3.6 3 m /s/m`. A dam is built across the river at which the afflux is found to be 1.0 m. Using the numerical integration method, estimate the water depth at a distance of 500 m upstream of the dam. Take Manning roughness coefficient, n, as 0.025. -4
Solution:
⎛ q 2 n 2 ⎞ ⎟⎟ y 0 = ⎜⎜ S ⎝ 0 ⎠ y c =
3
q2 g
=
0 .3
⎧ (3.6) 2 (0.025) 2 ⎫ =⎨ ⎬ −4 ⎩ 14 ×10 ⎭
(3.6) 2
3
9.81
0 .3
=1.69 m
=1.1 m
Therefore, M1 curve will be formed ahead of the dam. The depth of water at the dam = 1.69 + 1.0 ≈ 2.7 m. Let yi = 2.7 m and take dl1 = -500 m. Then, 0.0014 −
(3.6) 2 (0.025) 2
y`i = 1−
(2.7)10 / 3 (3.6)
2
=
0.000988 0.81878
= 0.0012
9.81× (2.7) 3
Assume y`i+1 = y`i yi+1 = 2.7-0.0012 × 500 = 2.1 m 0.0014 − y i +1 = 1−
(3.6) 2 (0.025) 2 (2.1)10 / 3 (3.6) 2
=
0.000525 0.857
= 0.000612 . Then,
9.81× (2.1) 3
⎛ 0.0012 + 0.000612 ⎞ ⎟ × 500 = 2.247 m 2 ⎝ ⎠
y i +1 = 2.7 − ⎜
Using the new value of yi+1, y`i+1 is evaluated again, 0.0014 − y i +1 = 1−
(3.6) 2 (0.025) 2 (2.247)10 / 3 (3.6)
2
=
0.000686 0.883
= 0.000776 , and
9.81× (2.247) 3
⎛ 0.0012 + 0.000923 ⎞ ⎟ × 500 = 2.175 m 2 ⎝ ⎠
y i +1 = 2.7 − ⎜
Using the new value of yi+1, y`i+1 is evaluated again,
Elementary Hydraulics Hydraulics
0.0014 −
y `i +1 = 1−
(3.6) 2 (0.025) 2 (2.175)10 / 3 (3.6) 2
=
0.000645 0.8769
= 0.000736 . Then,
9.81× (2.206) 3
⎛ 0.0012 + 0.000906 ⎞ ⎟ × 500 = 2.2174 m 2 ⎝ ⎠
y i +1 = 2.7 − ⎜
Using the new value of y is reevaluated: 0.0014 − y `i +1 = 1−
(3.6) 2 (0.025) 2 (2.174)10 / 3 (3.6) 2
=
0.0006556 0.8785
= 0.000746 . Then,
9.81× (2.216) 3
⎛ 0.0012 + 0.000746 ⎞ ⎟ × 500 = 2.214 m 2 ⎝ ⎠
y i +1 = 2.7 − ⎜
The depth of water at a distance of 500 m upstream of the dam is 2.174 m. As it can be seen, this method involves iteration procedure. The use of a computer code is therefore recommended. 2
15.2. A wide channel, having a bed slope of 0.01, carries a specific discharge of 5.0 m /s. A gate is located in the midway of the channel where the depth of water at the venacontracted section is 0.4 m. Draw the water surface profile and estimate the length of the rising curve formed after the gate using the numerical integration method. Take n = 0.025. Solution:
⎛ q 2 n 2 ⎞ ⎟⎟ y 0 = ⎜⎜ S ⎝ 0 ⎠ y c = 3
q
0.3
2
g
=
3
⎧ (5.0) 2 (0.025) 2 ⎫ =⎨ ⎬ 0.01 ⎩ ⎭ (5.0)
0.3
⎧ 0.015625 ⎫ =⎨ ⎬ ⎩ 0.01 ⎭
0.3
=1.14 m
2
9.81
=1.366 m
Therefore, the channel has a steep bed slope. The water surface profile is shown in the Figure below. S3 profile will will be formed after the gate. It is now required to calculate the length of S3 curve between y = 0.4 m and y = 1.14 m. Let yi = 0.4 m and take dl1 = 50 m.
262
Chapter 15. Computation of Water Surface Profiles
0.015625
0.01 −
(0.4)10 / 3
y i ` =
− 0.32135 = 0.00828 − 38.82
=
2
1−
(5)
0.01 −
0.015625
9.81× (0.4) 3 Assume y`i+1 = y`i. yi+1 = 0.4 + 0.00828 × 50 = 0.814 m
(0.814)10 / 3
y `i +1 = 1−
(5)
=
2
− 0.02103 = 0.005646 . Then, − 3.725
9.81× (0.814) 3
⎧ 0.00828 + 0.005646 ⎫ ⎬ × 50 = 0.612 m 2 ⎩ ⎭
yi+1 = 0.4 + ⎨ Another iteration:
0.015625
0.01 −
(0.612)
y `i +1 = 1−
(5)
10 / 3
=
2
− 0.0703 = 0.0069 . Then, − 10.18
9.81× (0.612) 3
⎧ 0.00828 + 0.0069 ⎫ ⎬ x 50 = 0.78 m 2 ⎩ ⎭
yi+1 = 0.4 + ⎨ Again,
0.015625
0.01 −
(0.78)10 / 3
y `i +1 = 1−
(5)
2
9.81 x (0.78)
=
− 0.02577 = 0.0059 . Then, − 4.37
3
⎧ 0.00828 + 0.0059 ⎫ ⎬ × 50 = 0.755 m 2 ⎩ ⎭
yi+1 = 0.4 + ⎨ Another iteration: 0.01 − y
i +1
= 1−
0.015625 (0.755)10 / 3 (5)
2
=
− 0.02987 = 0.00607 . Then, − 4.9215
9.81× (0.755) 3
⎧ 0.00828 + 0.00607 ⎫ ⎬ × 50 = 0.759 m 2 ⎩ ⎭
yi+1 = 0.4 + ⎨
Now, we consider that the correct value of yi+1 is 0.76 which is the depth of water after 50 m from the vena-contracted section. Let yi = yi+1 and take dl2 = 70 m.
263
Elementary Hydraulics
0.01 − y`i = 1−
0.015625 (0.76)10 / 3 (5)
=
2
− 0.029 = 0.00604 . Then, − 4.805
9.81× (0.76) 3 yi+1 = 0.76 + 0.00604 × 70 = 1.183 m
Another iteration: 0.01 − y`i +1 = 1−
0.015625 (1.183)10 / 3 (5)
=
2
− 0.001076 = − 0.002 . Then, − 0.539
9.81× (1.183) 3
⎧ 0.00604 − 0.002 ⎫ ⎬ × 70 = 0.901 m 2 ⎩ ⎭
yi+1 = 0.76 + ⎨ Another iteration: 0.01 − y`i +1 = 1−
0.015625 (0.901) (5)
10 / 3
=
2
− 0.01212 = − 0.00488 . Then, − 2.484
9.81× (0.901) 3
⎧ 0.00604 − 0.0048 ⎫ ⎬ × 70 =1.142 m 2 ⎩ ⎭
yi+1 = 0.76 + ⎨ Another iteration: 0.01 − y `i +1 = 1−
0.015625 (1.42)10 / 3 (5)
2
9.81× (1.42)
=
− 0.000037 = − 0.000052 . Then, − 0.711
3
⎧ 0.00604 − 0.000052 ⎫ ⎬ x 70 = 0.973 m 2 ⎩ ⎭
yi+1 = 0.76 + ⎨ Another iteration: 0.01 − y`i +1 = 1−
0.015625 (0.973)10 / 3 (5)
2
=
− 0.0071 = − 0.00402 . Then, − 1.7665
9.81 x (0.973) 3
⎧ 0.00604 − 0.00402 ⎫ ⎬ × 70 =1.112m 2 ⎩ ⎭
yi+1 = 0.76 + ⎨
264
Chapter 15. Computation of Water Surface Profiles
Another iteration: 0.01 − y`i +1 = 1−
0.015625 (1.112)10 / 3 (5) 2
=
− 0.00097 = − 0.00114 . Then, − 0.8533
9.81× (1.112) 3
⎧ 0.00604 − 0.00114 ⎫ ⎬ × 70 =1.01 m 2 ⎩ ⎭
yi+1 = 0.76 + ⎨ Again, 0.01 − y`i +1 = 1−
0.015625 (1.01)10 / 3 (5)
2
=
− 0.005115 = − 0.00347 . Then, − 1.4735
9.81× (1.01) 3
⎧ 0.00604 − 0.00347 ⎫ ⎬ × 70 =1.09 m 2 ⎩ ⎭
yi+1 = 0.76 + ⎨
Repeating the iteration, we get yi+1 = 1.06 m at a distance of 120 m from the venacontracted section. Now, we can get the exact length by extrapolation. A length of 70 m gives a water rise of 0.3 m (1.06 – 0.76). Then, 70 x 0.38 dl 2 = = 88.7 m 0.3 The total length of S3 curve = 50 + 88.7 = 138.7 m ≈ 140m. 15.3. A wide channel carrying a discharge of 32 ft /s/ft is laid at a slope of 12 × 10 . The channel has a Manning roughness coefficient of 0.018. The channel terminates into a free over fall at which the depth of flow is equal to 1.1 yc. Determine the depth of flow 400 ft upstream of the fall using the numerical integration method. 3
`
-4
Solution:
⎛ q 2 n 2 ⎞ ⎟ y o = ⎜⎜ 2 ⎟ ⎝ (1.49) S 0 ⎠ y c =
3
q2 g
=
3
0 .3
(32) 2 32.2
⎧ (32) 2 (0.018) 2 ⎫ = ⎨ −4 ⎬ ⎩ 2.22 ×12 ×10 ⎭
0 .3
⎧ 0.149 ⎫ = ⎨ −4 ⎬ ⎩12 ×10 ⎭
0.3
= 4.25 ft
= 3.168 ft
Therefore, the channel has a mild bed slope and M2 curve will be formed before the fall. Let yi = 1.1, yc = 3.5 ft and take dl1 = -50 ft. 0.149 0.0012 − (3.5)10 / 3 − 0.00109 y`i = = = − 0.00422 31.8 0.258 1− (3.5) 3
265
Elementary Hydraulics
Assume y`i+1 = y`i, then yi+1 = 3.5 - 0.00422 × (-50) = 3.71 ft 0.149 0.0012 − (3.71)10 / 3 − 0.000685 y`i +1 = = = − 0.00182 . Then, 31.8 0.37726 1− (3.71) 3
⎧ 0.00422 + 0.00182 ⎫ ⎬(−50) = 3.65 ft 2 ⎩ ⎭
yi +1 = 3.5 − ⎨ Another iteration:
0.0012 − y`i +1 =
1−
0.149 (3.65)10 / 3 − 0.00079 = = − 0.00228 . Then, 31.8 0.346
(3.65) 3
⎧ 0.00422 + 0.00228 ⎫ ⎬(−50) = 3.66 ft 2 ⎩ ⎭
yi +1 = 3.5 − ⎨
The last value of y is very close to the previous one. Then this is the correct value. Now, let yi = yi+1 = 3.66 ft na d take dl2 = -100 ft. yi+1 = 3.66-0.00228 (-100) = 3.89 ft 0.149 0.0012 − (3.89)10 / 3 − 0.000409 y = = = − 0.0089 . Then, 31.8 0.459 1− (3.89) 3
⎧ 0.00228 + 0.00089 ⎫ ⎬(−100) = 3.818 ft 2 ⎩ ⎭
y`i +1 = 3.66 − ⎨ Another iteration:
0.0012 − y`i +1 =
1−
0.149 (3.818)10 / 3 − 0.000513 = = − 0.001197 . Then, 31.8 0.4286
(3.818) 3
⎧ 0.00228 + 0.001197 ⎫ ⎬(−100) = 3.83 ft 2 ⎩ ⎭
yi +1 = 3.66 − ⎨
This is the correct value for y at a distance of 150 ft upstream the fall. Now, let yi = yi+1 = 3.83 ft and take dl3 = -200 ft. yi+1 = 3.83-0.001197 (-200) = 4.07 ft.
266
Chapter 15. Computation of Water Surface Profiles
0.0012 −
y`i +1 =
1−
0.149 ( 4.07) 31.8
10 / 3
=
− 0.000184 0.528
= − 0.0003483 . Then,
( 4.07) 3
⎧ 0.001197 + 0.0003483 ⎫ ⎬ (−200) = 3.98 ft 2 ⎩ ⎭
yi +1 = 3.83 − ⎨ Another iteration:
0.0012 − y`i +1 =
1−
0.149 (3.95)10 / 3 − 0.0003294 = = − 0.00068 31.8 0.484
(3.95) 3
Then,
⎧ 0.001197 + 0.00068 ⎫ ⎬ (−200) = 4.018 ft 2 ⎩ ⎭
yi +1 = 3.83 − ⎨ Another iteration:
0.0012 − y `i +1 =
1−
0.149 ( 4.018)10 / 3 − 0.00024484 = = − 0.000048 31.8 0.5098
( 4.018) 3
Then,
⎧ 0.001197 + 0.00048 ⎫ ⎬ (−200) = 4.0 ft 2 ⎩ ⎭
yi +1 = 3.83 − ⎨
Therefore, the depth of water at a distance of 350 ft upstream the fall is 4.0 ft. -4
15.4. A wide rectangular channel under a bed slope of 14 x10 carries a discharge of 3.6 2 m /s. An afflux of 1.0 m is measured at a dam located downstream of the river. Determine the distance upstream of the dam where the water depth is 2.2 m. Use the direct step method and take 1/n = 40. Solution:
⎛ q 2 n 2 ⎞ ⎟⎟ y o = ⎜⎜ S ⎝ 0 ⎠
0.3
0.3
⎧ (3.6) 2 (0.025) 2 ⎫ = ⎨ ⎬ =1.69 m −4 14 10 × ⎩ ⎭
Therefore, M1 curve will be formed ahead of the dam. The calculation for the required distance is given in Table P15.4. The depth of water at the dam = 1.69 + 1.0 ≈ 2.7 m. y c = 3 S f =
q2 g
=
3
(3.6) 2 9.81
= 1.1 m
q2n2 y 10 / 3
267
Elementary Hydraulics
Table P15.4: Calculations for the distance between y = 2.2 m and y = 2.7 m.
∆E
y
V
E
2.7
0.556
2.716
Sf × 10
-4
0.6
2.518
0.652
2.322
0.682
10.603
-186.7
4.432
9.568
-204.8
5.447
8.552
-114.6
5.043 -0.098
2.2
3.397
-4
3.82 -0.196
2.3
(So- S f ) × 10
2.955 -0.198
2.5
dl
S f × 10-4
2.224
5.58 L = 506 m
Therefore, the length of M1 curve between y = 2.2 m and y = 2.7 m is 506 m.
15.5. A wide river having a bed slope of 20 cm/km conveys a specific discharge of 4.1 2 m /s. A dam is built across the river where the afflux at the dam site is 1.55 m. The Chezy roughness coefficient is 45 (metric). Find the length of the backwater curve using the direct step method. Solution:
y0 = 3 y c = 3
q2 C 2 S 0 q2 g
=3
( 4.1) 2 ( 45) 2 × 20 × 10− 5
= 3.46 m
=1.23 m
The channel has a mild bed slope and M1 curve will be formed before the dam. The depth of water at the dam = 3.46 + 1.55 = 5.0 m. q2 S f = 2 3 C y Table P15.5. Calculations for M1 profile. y
V
E
3.5
1.17
3.57
∆E
Sf × 10
-5
1.025
4.05
0.911
4.54
0.82
5.034
16.165
3.835
12516
11.04
8.96
5468
7.875
12.125
4074
9.11 0.494
5.0
dl
-5
12.97 0.49
4.5
(S0- S f ) × 10
19.36 0.48
4.0
S f × 10-4
6.64 L = 22058 km
Therefore, the length of the M1 curve is 22.058 m
268
Chapter 15. Computation of Water Surface Profiles
15.6. A weir is constructed across a wide channel to elevate the water level. The channel 2 conveys a specific discharge of 3 m /s and has a Chezy roughness coefficient of 50 (metric). Assuming that the water depth at the weir remains constant at 4.5 m, estimate the length of the backwater curve using the direct step method for the following two cases: (a) The bed slope = 5 cm/km, and (b) The bed slope = 10 cm/km. Solution:
The critical depth does not depend on the channel slope and hence will be constant for the two cases. y c = 3 S f =
q2 g
=
3
(3) 2 9.81
= 0.972 m
q2 C 2 y 3
(a) The bed slope is 5 cm/km, y 01 = 3
q2 C 2 S 01
=
(3) 2 ×10 5
3
(50) 2 × 5
= 4.16 m
The channel has a mild bed slope and M1 profile will be formed. The calculations of the length of the backwater curve are presented in Table P15.6a. q2 S f = 2 3 C y Table 15.6a: Length of M1 profile (So = 5 cm/km). y
V
E
4.18
0.718
4.206
∆E
Sf × 10-5
0.714
4.226
0.697 0.682 0.667
4.424 4.523
0.306
4.528
15891 4.377
0.623
4.226 0.099
4.5
32353 4.694
0.099 4.4
0.106
4.859
4.325
dl
18868 4.894
0.099 4.3
(So - S f ) × 10-5
4.929 0.02
4.2
S f × 10-5
10867 4.089
0.911
3.951 L = 77,979 m
(b) The bed slope is 10 cm/km, y 02 = 3
q2 C 2 S 01
=
3
(3) 2 ×10 5 (50) 2 × 5
= 3.30 m
The channel has a mild. M1 profile will be formed. The calculations of the length of the backwater curve for this case are presented in Table P15.6b.
269
Elementary Hydraulics
Table 15.6b: Length of M1 profile (S o = 10 cm/km).
∆E
y
V
E
3.32
0.904
3.362
Sf × 10
4
0.857
3.537
0.789
3.818
0.732
4.127
0.667
19796
7.478
2.522
11142
5.892
4.108
7522
4.587
5.413
7135
5.223 0.396
4.5
0.884
6.56 0.309
4.1
9.116
-5
8.396 0.281
3.8
(So - S f ) × 10
9.837 0.175
3.5
dl
S f × 10-5
4.523
3.951 L = 45595 m
Therefore, the length of the curve decreased when the slope increased. Shorter profiles are generally encountered as the bed slope increases. 15.7. A wide rectangular channel has a bed slope of 8 cm/km and carries a discharge of 3 ` 1.5 m /s/m . The channel terminates into a sudden fall. Determine the length of the water profile, assuming a critical depth at the sudden fall. Take the Chezy coefficient as 40 (metric). Solution:
y 0 = 3
q2 C 2 S o
=
3
(1.5) 2 ×10 5 ( 40) 2 × 8
= 2.6 m and y c =
3
q2 g
=
3
(1.5) 2 9.81
= 0.612 m
The channel has mild bed slope and M2 will be formed before the free fall. It is required to evaluate the length of the M2 profile between yc = y1 = 0.612 m and yo = y5 = 2.6 m. Consider three other depths y2 = 1.0 m, y3 = 1.5 m, and y4 = 2.0 m. Table P15.7 is developed to calculate the required length. Table P15.7: Calculation of the length of M2 profile . y
V
E
0.612
2.45
0.92
∆E
Sf × 10
1.5
1.15
1.0
1.55
0.75
2.03
0.577
2.63
dl
3.635
-3.555
-63.3
0.913
-0.833
-480.2
0.296
-0.216
-2222.2
0.128
-0.048
-12500
0.176 0.60
2.6
× 10-3)
0.416 0.48
2.0
(So- S f
1.41 0.40
1.5
S f × 10-3
6.13 0.23
1.0
-5
0.08 L = -15265.7 m
The total length of M2 profile is 15.265 km. The length of the profile stretches when the flow approaches the normal depth.
270
Chapter 15. Computation of Water Surface Profiles
15.8. A triangular channel with a bed slope of 20 cm/km and a side slope of 1:1 carries a 3 discharge of 0.6 m /s. The Chezy coefficient is 50 (metric) and the channel ends with a sudden fall. Determine the normal and the critical depths. Hence, find the length of the water profile if the flow has a critical depth at the fall. Solution:
Q = AC RS o
⎛ y 0 ⎞ −5 1 / 2 ⎟ × ( 20 × 10 ) ⎝ 2.83 ⎠
y 02 x 50 ⎜
0.6 = 5/2
(yo) = 1.427, or yo = 1.153 m The critical depth is calculated from the equation of critical flow condition. Q g
= y
0.6
y c2
2 c
=
2 y c y c5 / 2
9.81
2
=
y c5 / 2 2 2/5
. Hence, yc = (0.27) = 0.59 m.
The channel has a mild bed slope and the M2 profile will be formed. Let y1 = 0.59, y2 = 0.7, y3 = 1.0, and y4 = 1.153 m. Table P15.8 is developed to estimate the length of M2. The slope of the total energy line, Sf , in the above table is calculated from the Chezy equation which, for this case, can be expressed as V 2 Sf = ( 29.7) 2 y Table P15.8: Calculations of the length of the M2 profile in the triangular section. y
A
V
E
0.59
0.348
1.724
0.742
∆E
Sf × 10
-3
0.49
1.22
0.776
1.00
0.60
1.01
1.33
0.451
1.163
-3.85
-8.83
1.40
-1.20
-195.0
0.303
-0.103
-1485.44
0.407 0.153
1.153
4.05 2.4
0.234 1.00
(So- S f ) × 10-3 dl
5.7 0.034
0.70
S f × 10-3
0.20 L = -1689.27 m
The total length of M2 profile is 1.689 km. 15.9. A wide channel having a bed slope of 20 cm/km carries a specific discharge of 2.2 2 m /s. A gate is located along the channel where the depth of flow at the vena-contracted section is 0.1 m. Determine whether a hydraulic jump will be formed after the gate or not. At what distance will the jump be formed downstream the gate? Take the Manning roughness coefficient as 0.02 (metric).
271
Elementary Hydraulics
Solution:
⎛ q 2 n 2 ⎞ ⎟⎟ y 0 = ⎜⎜ ⎝ S o ⎠ y c =
3
q2 g
=
0 .3
3
⎛ (2.2) 2 (0.02) 2 ⎞ ⎟⎟ = ⎜⎜ −5 ⎝ 20 ×10 ⎠
(2.2) 2 9.81
0 .3
= 1.98 m
= 0.79 m
The channel has a mild bed slope. A hydraulic jump will be formed if the conjugate depth to the normal depth is greater than the vena-contracted depth (0.1 m). Knowing the high stage depth of the jump (y = 1.98 m), the low stage depth is calculated as y1 =
y 2 ⎧ ⎪
8q 2 ⎫ ⎪ 1.98 ⎧⎪
⎨− 1 + 1 + 3 2 ⎪ gy 2 ⎩
8( 2.2) 2
⎬= ⎨− 1 + 1 + 2 9.81(1.98) 3 ⎪⎭ ⎪⎩
⎫⎪ ⎬ = 0.23 m ⎪⎭
M3 profile will be formed through which the depth of flow increases from 0.1 m to 0.23 m. Let y1 = 0.1, y2 = 0.14, y3 = 0.18, and y4 = 0.23 m. Table P15.9 is developed to calculate the length of M3. Table P15.9: Calculations of the length of M3 . y
V
E
0.1
22
24.8
∆E
Sf × 10-4 4.17
-12.1 0.14
15.7
12.7
12.2
7.77
9.56
dl
2.765
-2.7648
4.38
0.973
-0.9728
5.07
0.423
-0.4228
7.36
0.586 -3.11
0.23
(So- S f ) × 10-4
1.36 -4.93
0.18
S f × 10-4
4.66
0.259 L = 16.81 m
The length of M3 is 16.8 m after which the jump will be located.
15.10. A trapezoidal channel having a bed slope of 15 cm/km conveys water at a rate of 3 12.0 m /s. The channel has a bed width of 6.0 m and a side slope of 2:1. The Chezy coefficient is 40 (metric). A dam is built across the channel at which the water depth is 3.2 m. Sketch the water surface profile and estimate its length. Solution:
Q = AC RS o 2 o
12.0 = (6yo + 2 y ) 40 24.5 = (6yo + 2 y ) 2 o
(6 y o + 2 y o2 ) 6 + 2 y o 1 + 4
× (15 ×10 −5 )1 / 2
(6 y o + 2 y 02 6 + 4.47 y o
272
Chapter 15. Computation of Water Surface Profiles
Solving by trial and error, we get yo = 2.065 m. The critical depth is obtained from the critical flow condition as follows: Q g
= A
A B
. Hence,
12.0 9.81
= 3.83 = (6 y c + 2 y ) 2 c
(6 y
+ 2 y c2 ) 6 + 4 yc c
Solving by trial and error, we get yc = 0.685 m. The normal depth, yo, is greater than the critical depth, y c. Therefore, the channel has a mild bed slope and M1 profile will be formed before the dam. To calculate the length of the profile, let y1 = 3.2, y2 = 3.0, y3 = 2.6, and y4 = 2.1 m. Table P15.10 is then developed. Table P15.10. Calculations of the length of M1 profile. y
A
P
R
V
E
3.2
39.68
20.3
1.95
0.302
3.205
∆E
Sf x 10-5
36.0
19.41
1.85
0.333
3.005
29.12
17.62
1.65
0.412
2.609
21.42
15.39
1.39
0.560
3.335
11.665
-1714
5.09
9.91
-3996
10.265
4.735
-10412
6.43 -0.493
2.1
dl
3.75 -0.396
2.6
(So- S f ) -5 x 10
2.92 -0.20
3.0
S f -10-5
2.116
14.1 L = -16122 m
The total length of M1 profile is 16.122 km. The profile is presented below. The student is advised to resolve this problem with either different or more depth intervals and compare the results.
15.11. In Problem 15.10, calculate the depth of flow at a distance of 8.0 km upstream of the dam. Solution:
Referring to Table P14.8, at a distance of 1714 + 3996 = 5710 m the depth of flow is 2.6 m. This means that we need to assume a depth interval that would result in a distance of 8000 –5710 = 2290 m. One may also assume that the depth at the required point is 2.4 m and calculate the length of the profile between y = 2.6 m and y = 2.4 m. The calculations are given in Table P15.11 in which the first two steps are acquired from Table P15.10.
273
Elementary Hydraulics
Table P15.11. Calculations for the depth at a distance of 8 km from the dam. y
A
P
R
V
E
3.2
39.68
20.3
1.95
0.302
3.205
∆E
Sf x 10
36.0
19.41
1.85
0.333
3.005
29.12
17.62
1.65
0.412
2.609
25.92
16.73
1.55
0.463
dl
3.335
11.665
-1714
5.09
9.91
-3996
7.535
7.465
-2652
-5
6.43 -0.198
2.4
(So- S f )x 10
3.75 -0.396
2.6
S f x 10-5
2.92 -0.20
3.0
-5
2.411
8.64 L = -8362 m
Therefore, the depth of 2.4 m is located at a distance of 8.362 km upstream of the dam. This means that at a distance of 8.0 km, the water depth will be slightly bigger than 2.4 m. One may also interpolate to get the required depth as follows: dy 4 =
0.2 x 2290
= 0.173 m 2652 The required depth = 2.6 – 0.173 ≈ 2.43 m. 15.12. A wide channel having a bed slope of 0.045 and the Manning roughness 3 coefficient of 0.025 carries a discharge of 2.0 m /s. A dam is built across the channel at which the depth of water is 1.6 m. Draw the water surface profile and estimate the distance between the dam and the high stage depth of the formed hydraulic jump. Solution: The normal and critical depths are calculated first.
⎛ q 2 n 2 ⎞ ⎟⎟ y 0 ⎜⎜ S ⎝ o ⎠ yc = 3
q2 g
0 .3
=3
⎧ (2.0) 2 (0.0025) 2 ⎫ =⎨ ⎬ 0.045 ⎩ ⎭ 4 9.81
0 .3
= 0.42 m
= 0.74 m
The channel has a steep bed slope. Since the water depth at the dam is 1.6 m, then a hydraulic jump will be formed. The high stage depth of the jump, y2, can be calculated as follow. y 2 =
0.42 ⎧ ⎪
8( 2) 2
⎨− 1 + 1 + 2 ⎪ 9.81(0.42) 3 ⎩
⎫⎪ ⎬ =1.2 m ⎪⎭
S1 profile will be formed after the jump, through which the water depth will increase from 1.2 m to 1.6 m as shown in the Figure. Table P15.12 is developed to calculate the length of S1.
274
Chapter 15. Computation of Water Surface Profiles
Table P15.12: Calculations of the length of S1 profile.
∆E
y
V
E
1.6
1.25
1.68
Sf x 10
-5
1.33
1.59
1.43
1.504
1.54
1.421
1.67
44.4
-2.03
73.0
44.27
-1.94
93.04
44.07
-1.88
120.58
43.79
-1.8
104.47 -0.079
1.2
58.285
81.61 -0.083
1.3
dl
64.39 -0.86
1.4
(So- S f )x 10-3
52.18 -0.09
1.5
S f x10-5
1.342
136.69 L = -7.68 m
The total length of the S1 profile (distance between the dam and the high stage depth of the hydraulic jump) is 7.68 m. This length is relatively short as the channel has a very steep bed slope. One may always note that the steeper the bed slope the shorter the length of the water surface profile. 3
15.13. A wide rectangular channel carries a discharge of 2.5 m /s has a bed slope of 0.008 and a Manning roughness coefficient of 0.025. At a certain point the bed slope changes to become 0.05. Sketch the water surface and estimate the length of the water profile. Solution: The normal depths for the two reaches and the critical depth are calculated.
⎧ (2.5) 2 (0.025) 2 ⎫ y 01 = ⎨ ⎬ 0.008 ⎩ ⎭ y c = 3
q2 g
=
3
( 2.5) 2 9.81
0 .3
= 0.81 m , and y 02
⎧⎪ (2.5) 2 (0.025) 2 ⎫⎪ =⎨ ⎬ 0.05 ⎪⎩ ⎪⎭
0.3
= 0.465m
= 0.86 m
Therefore, the channel has a steep bed slope; the second reach is steeper than the first. An S2 curve will form in the second reach as shown in the Figure below. The bed slope of the second reach (0.05), where the entire profile is located, will be considered in the calculations of the length of S2 as shown in Table P15.13.
275
Elementary Hydraulics
Table P15.13: Calculations for the length of the S2 profile. y
V
E
0.81
3.09
1.297
∆E
Sf x 10
-3
S f x 10-3
3.57
1.35
4.17
1.486
5.00
1.774
5.38
1.34
17.145
32.855
4.14
30.425
19.515
14.71
44.795
13.205
12.57
39.37 0.166
0.465
39.467
21.48 0.288
0.5
10.355 12.81
0.136 0.6
dl
4.17 0.053
0.7
-3
(So- S f ) x 10
1.94
50.22 L = 32.76 m
The length of S2 profile is about 33 m. 15.14. A trapezoidal channel with a bed width of 2.0 m and a side slope of 3:2 carries a 3 –5 discharge of 300,000 m /d. The channel has a bed slope of 12 x 10 and a Manning -3 roughness coefficient of 0.018. At a certain point the bed slope increases to 20 x 10 . Draw the water surface profile and determine the length of the water profiles in the two reaches. Solution:
Q=
300000 24 x 60 x 60
= 3.472 m 3 / s
⎧ by 0 + ty 02 1 2 / 3 1/ 2 2 1 ⎪ Q = A R S 0 = (by 0 + ty 0 ) ⎨ n n ⎪ b + 2 y 1 + t 2 ⎩ 2 ⎧⎪ by 01 + ty 01 2 3.472 = (by 01 + ty 01 ) ⎨ 0.018 ⎪ b + 2 y 1 + t 2 01 ⎩
1
⎫⎪ ⎬ ⎪⎭
⎫⎪ ⎬ ⎪⎭
2/3
S 01 / 2
2/3
(12 x 10 −5 )1 / 2
Solving by trial and error, we get y01 = 1.5 m. The normal depth in the second reach is calculated as follows:
⎧⎪ by + ty 2 2 02 02 0.442 = by 02 + ty 02 ) ⎨ ⎪⎩ b + 2 y 02 1 + t 2
⎫⎪ ⎬ ⎪⎭
2/3
Solving by trial and error, we get yo2 = 0.39 m. The critical depth can be calculated as Q g
= A
3.472 9.81
A B
= 1.1085 = (2 y c + 1.5 y ) 2 c
2 y c + 1.5 y c2 2 + 3 y c
Solving by trial and error, get yc = 0.58 m.
276
Chapter 15. Computation of Water Surface Profiles
Therefore, the first reach has a mild bed slope while the second one has a steep bed slope. Figure P15.14 presents the water surface profile along the channel length. The M2 and S2 falling profiles will be formed in the first and second reaches, respectively. To calculate the length of the M2 and S2 profiles, Tables P15.14a and P15.14b are developed. Table P15.14a. Calculations of the length of M2 profile .
∆E
y
A
P
R
V
E
0.58
1.665
4.1
.406
2.09
0.803
0.75
2.34
4.71
.497
1.48
0.862
Sf x 10
-5
5.61
.624
0.99
1.05
1.25
4.844
6.51
.744
0.717
1.28
7.41
.860
0.545
-313.5
-18.82
119.8
-107.8
-174.40
42.15
-30.15
-762.85
18.25
-6.25
-3760.0
24.7 .235
6.375
325.5
59.6 .230
1.50
dl
180 .188
3.50
(So- S f ) -5 x 10
471 .059
1.00
S f -10-5
1.515
11.8
L = -4716.07 m Table P15.14b. Calculations of the length of S2 profile.
∆E
y
A
P
R
V
E
0.58
1.665
4.1
.406
2.09
0.803
Sf x 10
1.375
3.81
.361
2.53
0.826
1.204
3.62
.333
2.88
0.872
1.084
3.41
.318
3.20
0.912
dl
639
13.61
1.69
986
10.14
4.54
1346
6.54
6.12
1164 0.040
0.39
(So- S f ) -5 x 10
807 0.046
0.45
S f -10-5
471 0.023
0.50
-5
1528
L = 12.35 m
The length of M2 profile is about 4716 m and the length of the S2 profile is 12.35 m. One may note that S type profiles are generally much shorter than M type profiles. The total length of the two profiles is 4728.35 m or about 4730 m.
277
Elementary Hydraulics
15.15. Water is issued into a wide rectangular channel from a large reservoir where the water level is 6.0 m above the channel bed. The channel has a bed slope of 0.01 and a Manning roughness coefficient of 0.021 (metric). Draw the water surface profile and estimate the length of the water profile formed after the channel inlet. Solution: For wide channels, E = 3/2 yc = 6.0 m. Hence, yc = 4.0 m
y c = 3 yo = (
q2
2
g
3
2
. Hence, q = g yc , Therefore, q = 25.06 m /s.
25.06 2 x 0.0212 0.01
) 0.3 = 2.7 m.
yo is less than y c. The channel has a steep bed slope. The water surface profile is presented here. To calculate the length of S2 profile, Table P15.15 is developed.
Table P15.15: Calculations for the length of the S2 profile. y
V
E
4.0
6.27
6.0
∆E
Sf x 10-3
6.77
6.04
7.83
6.33
9.28
6.87
5.8
4.63
5.37
54
7.86
2.14
355
5.73 0.76
2.7
3.13 3.53
0.29 3.2
(So- S f ) x 10
2.73 0.04
3.7
-3
dl
S f x 10-3
7.09
10 414.8 m
The total length of S2 profile is 414.8 m. 2
15.16. A specific discharge of 2.1 m /s is issued into a wide channel through a sluice gate where the depth of flow at the vena-contracted section is 0.2 m. The channel has a bed -4 slope of 8 x 10 and a Chezy coefficient of 40 (metric). Determine whether a hydraulic jump will be formed or not? What is the minimum length of the concrete apron required after the gate to contain the jump assuming that the length of the jump is seven times its height? Solution:
y o = 3
(2.1) 2 ( 40) 2 8 x10 −4
= 1.51 m , and y c =
The low stage depth of the jump is given as;
278
3
( 2.1) 2 9.81
= 0.766 m
Chapter 15. Computation of Water Surface Profiles
y 01 =
1.51 ⎧ ⎪
⎨− 1 + ⎪⎩
2
⎫⎪ ⎬ = 0.33 m 9.81(1.51) 3 ⎪ ⎭ 8(2.1) 2
Therefore, a hydraulic jump with a height of 1.51 - 0.33 = 1.18 will be formed and an M3 profile will be encountered before the jump. It should be noted that if yo1 is found less than the depth of flow at the vena contracted section, then the gate will be submerged and no jump will be formed. Table P15.16 is developed to calculate the length of M3. Table P15.16: Calculations for the length of M3 profile.
∆E
y
V
E
0.20
10.5
5.82
Sf x 10
-3
8.4
3.85
7.0
2.80
6.36
-433.7
4.54
216.25
-215.45
4.87
131.65
130.85
3.13
152.5 -0.41
0.33
434.5
-3
280 -1.05
0.30
(So- S f ) x 10
589 -1.97
0.25
dl
S f x 10-3
2.39
110.8 L = 12.54 m
The length of the M3 profile is 12.54 m. The minimum length of the concrete apron = 12.54 + 7 x 1.18 = 20.8 m ≈ 21.0 m. -3
15.17. A wide rectangular channel having a bed slope of 4.85 x 10 carries a discharge of 3 2.5 m /s. The Chezy roughness coefficient is 45 (metric). A dam is built across the channel at which the depth of water is measured as 1.8 m. Draw the water surface profile and estimate the length of the backwater curve. Solution: The normal and the critical depths are calculated first.
y 0 = 3
y c = 3
q2 C 2 S 0 q2 g
=
=3
3
( 2.5) 2 ( 45) 2 4.85 x 10 −3 ( 2 . 5) 2 9.81
= 0.86 m
= 0 .86 m
The channel has a critical bed slope and C1 profile will be formed before the dam. The water surface profile is presented below. The calculations for the length of C1 are given in Table P15.17.
279
Elementary Hydraulics
Table P15.17: Calculations for the length of C1 profile. y
V
E
1.8
1.39
1.9
∆E
Sf x 10
1.67
1.64
2.08
1.42
2.91
72.4
412.6
-63.0
134.9
350.1
-62.8
331.5
153.5
-84.7
-5
178 -0.13
0.86
(So- S f ) x 10
91.8 -0.22
1.2
dl
S f x 10-5
53 -0.26
1.5
-5
1.29
485 L = -210.5 m
The total length of the C1 profile (backwater curve) is about 211 m. 15.18. A rectangular channel with a bed width of 6.0 ft and a water depth of 4.0 ft 3 conveys a discharge of 30 ft /s. Determine the bed slope assuming the Manning roughness coefficient n = 0.025. At a certain point along the channel, the bed slope is reduced such that the normal depth is increased to 6.0 ft. Draw the water surface profile and estimate the distance through which the depth of flow varies. Solution:
Q = A
1.49 2 / 3 1 / 2 R S 01 n 2/3
1.49 ⎛ 24 ⎞ 1/ 2 30 = 24 x ⎜ ⎟ S 01 0.025 ⎝ 14 ⎠ -5 Therefore, S01 = 21.4 x 10 . The critical depth can be calculated as y c = 3
q2 g
=
3
(5) 2 32.2
= 0.92 ft
Since the critical depth is less than the normal depth, the bed slope is mild. M1 curve will be formed in the first reach as shown in the figure below. Table P15.18 provides the calculations for the length of M1 curve. The slope of the total energy line, Sf , in that Table is given as V 2 S f = 3552.16 R 4 / 3
280
Chapter 15. Computation of Water Surface Profiles
Table P15.18: Calculations for the length of M1 curve. y
A
P
R
V
E
6.0
36
18
2.0
0.833
6.01
∆E
Sf x 10
-5
33
17
1.94
0.909
5.513
30
16
1.875
1.00
5.016
27
15
1.80
1.11
4.52
24
14
1.71
1.25
11.82
-4204.7
10.895
9.605
-5174.4
14.01
6.49
-7642.5
18.67
1.83
15.84 -0.496
4.0
8.68
12.18 -0.496
4.5
dl
9.61 -0.497
5.0
(So- S f ) -5 x 10
7.75 -0.497
5.5
S f x 10-5
4.024
21.5
27103.8 L = -44125.4 ft
Therefore, the length of the gradually varied flow reach is 44125 ft. 15.19. A trapezoidal channel has a bed width of 4.0 ft, a side slope of 2:1 and a bed slope -4 3 of 12 x 10 . The channel carries a discharge of 90 ft /s and its Manning roughness coefficient, n, is 0.021. A dam is built across the channel where the heading up (afflux) is 2.22 ft. Determine the depth of flow 1500 ft upstream of the dam. What is the total length of the profile? Solution:
Q = (by 0 + ty 0 ) 2
1.49 ⎛ ⎜
90 = (4 y 0 + 2 y 02 )
⎞ ⎟ 2 ⎟ 1 + t ⎠
by 0 + ty 02
n ⎜ b + 2 y 0 ⎝
2 1.49 ⎛ 4 y 0 + 2 y 0 ⎞
2/3 1/ 2
S 01
2/3
⎜ ⎟ 0.021 ⎜⎝ 4 + 2 y o 5 ⎠⎟
(12 x 10 −4 )1 / 2
Solving by trial and error, we get yo = 2.78 ft. The critical depth is obtained from the critical flow condition:
⎛ 4 y c + 2 yc2 ⎞ ⎟⎟ . Hence, 36.617 = ( 4 y c + 2 y )⎜⎜ = A B 4 4 y + g c ⎠ ⎝
Q
90 32.2
A
1/ 2
2 c
= 15.86 = (4 y c + 2 y ) 2 c
4 y c + 2 y c2 4 + 4 y c
Solving by trial and error, we get yc = 1.85 ft. Therefore, the normal depth is bigger than the critical depth and the channel has a mild bed slope. M1 curve will be formed before the dam. The depth of flow at the dam is: y1 = 2.78 + 2.22 = 5.0 ft. To calculate the depth at a distance of 2500 ft upstream of the dam, Table P15.19 is developed.
281
Elementary Hydraulics
Table P15.19: Calculations for the water surface profile. y
A
P
R
V
E
5.0
70.0
26.36
2.66
1.29
5.026
∆E
Sf x 10
-6
58.5
24.12 2.43
1.54
4.537
48.0
21.89 2.19
1.88
4.055
38.5
19.65 1.96
2.34
3.586
30.0
17.42 1.72
3.00
3.141
26.58
16.43
1.62
3.39
-451.5
194.9
100.5
-479.6
344.6
85.5
-548.5
655.6
54.4
-818.0
1033.8
166
-1090.3
867.6 -0.181
2.78
108.3
443.5 -0.445
3.0
116.95
245.6 -0.469
3.5
dl
144.2 -0.482
4.0
(So- S f ) -6 x 10
89.7 -0.489
4.5
S f x 10-6
2.96
1200 L = -3387.9 ft
The first three steps give a distance of 1479.6 ft measured upstream of the dam. Therefore, the depth at a distance of 1500 ft from the dam is about 3.5 ft. The total length of the M1 profile is 3388 ft. 15.20. A 12.0 m rectangular channel, having a bed slope of 50 cm/km and a Manning roughness coefficient of 0.015, conveys water at a normal depth of 1.0 m. A dam is built across the channel at which the water depth is 2.0 m. Calculate the depth of flow 2.0 km upstream of the dam using the step by step method. Solution:
1
2/3
(50 x 10 − )
2 / s
. Hence, y c =
Q=
x 12.0(0.857) 0.015
q=
Q 16.14 b
=
12.0
=1.345 m
5 1/ 2
=16.14 m 3 / s 3
q2 g
=
3
(1.345) 2 9.81
= 0.57 m
Therefore, the channel has a mild bed slope and M1 curve will be formed before the dam. To calculate the depth of flow at a distance of 2.0 km upstream the dam, Table P15.20 is developed. Table P15.20: Calculations for the depth of flow 2.0 km upstream a dam using the step by step method. y
A
P
R
V
E
2.0
24.0
16.0
1.5
0.673
2.023
∆E
Sf x 10
19.2
15.2
1.26
0.841
1.636
15.6
14.6
1.06
1.035
1.355
14.4
14.4
1.00
1.12
dl
0.88
4.12
-939.25
1.68
3.31
-848.50
2.52
2.48
-366.35
2.20 -0.091
1.2
(So- S f ) -4 x 10
1.17 -0.281
1.3
S f x 10-4
0.59 -0.387
1.6
-4
1.264
2.82 L = -2154.1 m
282
Chapter 15. Computation of Water Surface Profiles
From the last depth interval in the above table we deduce that dy of 0.1 m gives a length of 366.35 m. Assuming a linear relationship between the values of dy and dl, then the required value of dy for the last interval can be calculated as: dy =
0.1 x (200 − 939.25 − 848.50)
= 0.058 m ≅ 0.06 m 366.35 Therefore, the depth of flow at a distance of 2.0 km upstream the dam = 1.3 – 0.06 = 1.24 m. 15.21. A trapezoidal channel of a bed width of 7.0 m and a side slope of 3:2 is laid on a 3 slope of 0.001 and carries a discharge of 30.0 m /s. The channel terminates to a free fall at which the depth is reduced to the critical depth. If the Manning roughness coefficient is 0.02, compute and plot the flow profile upstream from the over fall to a section where the water depth is 0.9 yo. Use the step by step method. Solution:
Q = 30.0 = (7 y 0 + 1.5 y 02 ) 18.973 =
2 1 ⎧ 7 y 0 + 1.5 y 0 ⎫
⎨ ⎬ n ⎩ 7.0 + 3.6056 y 0 ⎭
2/3
(0.001)1 / 2
(7.0 y 0 + 1.5 y 02 ) 5 / 3
(7 + 3.6056 y 0 ) 2 / 3 Solving by trial and error, we get yo = 1.71 m. Then 0.9 yo = 1.539 m. The critical depth is calculated from the equation of the critical flow: Q 2 B (30) 2 (7.0 + 3 y) = 1 . Hence, =1 2 gA3 9.81(7.0 y c + 1.5 yc ) 3 Solving by trial and error, we get yc = 1.132 m. The channel has a mild bed slope and an M2 falling profile will be formed before the sudden fall. Table P15.21 presents the calculations for the length of the M2 profile using the step by step method. Table P15.21: Calculations for the length of M2 profile using the step be step method. y
A
P
R
V
E
1.13
9.845
11.07
0.889
3.05
1.604
1.15
10.03
11.15
0.900
2.99
1.606
∆E
Sf x --4 10
11.33
0.932
2.84
1.611
1.25
11.09
11.51
0.964
2.71
1.624
11.69
0.995
2.58
1.640
1.40
12.74
12.05
1.057
2.35
1.681
12.55
1.142
2.09
38
-28
-1.78
32.5
-22.5
-5.78
28
-18
-8.89
24
-14
-29.29
17.5
-7.5
-113.89
20 0.082
14.33
-0.63
26 0.041
1.54
-32
30 0.016
11.64
42
35 0.013
1.30
dl
41 0.005
10.56
(So- S f ) x -4 10
43 0.002
1.20
S f x 10-4
1.763
15
L = -160.26 m
283
Elementary Hydraulics
The total length of the profile is about 160 m. A better accuracy can be achieved if smaller dy intervals are taken as the flow approaches the normal depth rather than as the flow approaches the sudden fall (or the critical depth). 15.22. The flow in a very wide channel tends to a free fall at which the depth is reduced 3 to 0.67 times the normal depth. The channel conveys a specific discharge of 4 m /s and has a bed slope of 10 cm/km. Estimate the distance upstream of the fall where the water depth is 0.9 the normal depth. Take Manning coefficient, n, as 0.025 and use the step by step method. Solution: The normal and critical depths are calculated first as follows:
⎛ q 2 n 2 ⎞ ⎟⎟ y 0 = ⎜⎜ S ⎝ 0 ⎠
0.3
⎛ (4) 2 (0.025) 2 ⎞ ⎟⎟ = ⎜⎜ −5 10 ⎝ ⎠
0.3
= 3.981 m and
y c
=3
q2 g
=
3
(4) 2 9.81
= 1.177 m
The channel has a mild bed slope and an M2 profile will be formed. The depth at the fall is 0.67 times the normal depth = 2.667 m. The calculations for the profile length between y = 2.667 m and y = 0.9 yo = 3.582 m are given in Table P15.22. Table P15.22: Calculations for the length of M2 profile.
∆E
y
V
E
2.667
1.5
2.782
--4
Sf x 10
1.379
3.00
1.25
3.28
1.117
3.646
3.34
-2.34
-931.6
2.475
-1.45
-1931
1.745
-0.745
-4912.7
2.07 0.366
3.582
-4
2.88 0.280
3.20
(So- S f ) x 10
3.8 0.218
2.90
dl
S f x 10-4
1.42 L = -7775.3 m
The length of the M2 profile between the two given depths is about 7775 m. 15.23. A broad-crested weir is located in a 40 m wide rectangular channel, as shown in 3 -3 Figure 14.P9. The channel conveys a discharge of 500 m /s under a bed slope of 1.8x10 . The Manning roughness coefficient, n, is 0.032. The sill height of the weir is 2.0 m. The weir discharge is given as: 3/2
Q = 1.705 CdBh Where B is the weir width, Cd is the discharge coefficient = 0.86, and h is the water head above the weir. Determine the length of the backwater curve using the step-by-step method. Solution: The normal and the critical depths are evaluated first.
1
q = y 0 R n
2/3
⎛ 40 y 0 ⎞ ⎜ ⎟ S . Hence, 12.5 = y 0 0.032 ⎜⎝ 40 + 2 y 0 ⎠⎟ 1
1/ 2 0
284
2/3
(1.8 x 10 −3 )1 / 2
Chapter 15. Computation of Water Surface Profiles
⎛ 40 y 0 ⎞ ⎟⎟ 9.428 = y 0 ⎜⎜ 40 2 y + 0 ⎠ ⎝
2/3
Solving by trial and error, get yo = 4.14 m y c =
3
q2 g
=
3
(12.5) 2 9.81
= 2.516 m
Therefore, the channel has a mild bed slope and an M1 curve will be formed. The water head over the weir is deduced from: 3/2
3/2
q = 1.705 Cd h . Then, 12.5 = 1.705 x 0.86 h 2/3 h = (8.5248) = 4.17 m The depth of flow at the weir = 2.0 + 4.17 = 6.17 m
M1 profile will therefore be formed before the weir. The calculations for the profile length are tabulated in Table P15.23. Table P15.23: Calculations for the backwater curve. y
A
P
R
V
E
6.17
246.8
52.32
4.717
2.03
6.38
∆E
--4
Sf x 10
220
51.0
4.314
2.27
5.76
200
50.0
4.00
2.50
5.32
180
49.0
3.673
2.78
4.89
166
48.3
3.437
3.01
11.58
-535.4
8.80
9.2
-478.3
12.02
5.98
-719.1
15.93
2.07
-1352.7
13.96 -0.28
4.15
6.42
10.08 -0.43
4.50
dl
7.51 -0.44
5.00
(So- S f ) -4 x 10
5.33 -0.62
5.50
S f x 10-4
4.61
17.89
L = -3085.5 m
The total length of the backwater curve is 3085.5 m. 15.24. Considering the data given in Problem 15.3, estimate the length of M2 profile using the step-by-step method. Solution:
It is required to calculate the length of the M2 profile between y = 4.25 ft and y = 3.5 ft. The slope of the total energy line is estimated from; V 2 n 2 S f = 2.22 y 4 / 3 The calculations are presented in Table P15.24. The total length of M2 curve is 2190 ft.
285
Elementary Hydraulics
Table P15.24: Calculations for the length of M2 profile.
∆E
y
V
E
3.5
9.14
4.8
-3
Sf x 10
8.42
4.9
8.0
5.0
7.62
5.1
7.53
-122.3
1.608
-4.08
-245.1
1.3605
-1.605
-623.0
1.225
-0.25
-1200
1.25 0.03
4.25
-8.175
1.471 0.1
4.2
2.0175 1.745
0.1 4.0
(So- S f ) x 10
2.29 0.1
3.8
-4
dl
S f x 10-3
5.13
1.2 L = -2190.4 ft
-5 15.25. A trapezoidal channel with a bed slope of 18x10 and a side slope of 2:1 carries a 3 discharge of 600 ft /s. The channel has a best hydraulic section and terminates into a free over fall. Determine the water depth 2750ft upstream of the fall using the step-by-step method. Take the Manning roughness coefficient as 0.025 and assume a critical depth at the fall. b + 2ty Solution: For the best hydraulic section, = y 1 + t 2 2
b + 4y = 4.472y, or b = 0.472y. Also R = y/2 1 Q = A R 2 / 3 S 01 / 2 n 1.49 ⎛ y 0 ⎞ 600 = (by 0 + ty ) ⎜ ⎟ 0.025 ⎝ 2 ⎠
2/3
2 0
⎛ y ⎞ 750.36 = 2.472 y ⎜ 0 ⎟ ⎝ 2 ⎠
(18 x 10 −5 )1 / 2
2/3
2 0
3/8
yo = (481.85) = 10.14 ft, and b = 4.79 ft The critical depth is calculated from: ( 4.79 y c + 2 y ) 2 c
(4.79 y
Q g
= A
A B
+ 2 y c2 ) =105.736 4.79 + 4 y c c
Solving by trial and error, we get yc = 4.57 m. Then the channel has a mild bed slope and an M2 profile will be formed as the flow approaches the free over fall. To determine the water depth at a distance of 2750 ft upstream of the fall Table P15.25 is developed. The depth of water at a distance of 2750 ft is around 8.2 ft.
286
Chapter 15. Computation of Water Surface Profiles
Table P15.25: Computations for the water depth at a distance of 2750 ft upstream of the fall. y
A
P
R
V
S f x -3 10
4.5
63.66
25.22
2.52
9.42
7.28
4.7
66.69
25.81
2.58
9.0
6.43
5.0
73.95
27.15
2.72
8.11
4.87
S f x E
3.18
5.95
131.5
36.09
3.64
4.56
166.3
40.57
4.10
3.61
2.13
173.7
41.46
4.19
3.45
-3
-3
0.06
-5.47
-10.96
-13.96
0.53
-3.32
-159.64
-173.6
0.77
-1.41
-546.1
-719.7
0.88
-0.619
-1421.6
-2141.3
0.18
-0.347
-518.7
-2660
6.55
1.04
7.32
0.558
8.20 0.527
8.2
-6.67
6.02
0.799 8.0
0.02 5.96
1.59 7.0
L
5.94
3.50 31.62
Dl
-3
5.65
100.7
(So- S f ) -3 x 10
10
6.85
6.0
∆E
0.496
8.38
15.26. A trapezoidal channel of 2.5 m bed width and 1:1 side slope has a bed slope of 0.0045. Under the normal conditions the depth of flow is equal to the critical depth, which is 0.9 m. A gate is located in the midway of the channel length. The depth of water just upstream of the gate is 1.4 m. Determine the length of the backwater curve using the step-by-step method. Take the Chezy roughness coefficient, C, as 45 (metric). Solution: Since the depth of flow under the normal conditions is equal to the critical depth, then the channel has a critical bed slope. C1 rising curve will be formed before the gate. A 2 Q = AC RS 0 , A = 3.06 m , R = = 0.60647 m P
Q = 3.06 x 45 0.60647 x 0.0045 = 7.19 m 3 / s The length of the curve will be calculated between y = 0.92 m and y = 1.4 m. The calculations are presented in Table P15.26. Table P15.26: Calculations for the length of C1 profile. y
A
P
R
V
E
1.4
5.46
6.46
.845
1.32
1.489
1.2
4.44
5.89
0.754
1.62
1.334
∆E
Sf x 10
5.33
.657
2.05
1.214
3.146
5.10
.617
2.285
dl
1.369
3.131
-49.5
2.439
2.061
-58.2
3.669
0.831
-33.7
3.159 -0.028
0.92
(So- S f ) -3 x 10
1.719 -0.120
3.5
S f x 10-3
1.018 -0.155
1.0
-3
1.186
4.179
L = -141.4 m
287
Elementary Hydraulics
15.27. A trapezoidal channel with a side slope of 1:1 and a bed width of 3.0 m is comprised of two reaches with bed slopes of 0.0004 and 0.03, respectively. The channel 3 has a Chezy roughness coefficient of 55 (metric) and coveys a discharge of 12.0m /s. Determine the length of the gradually varied flow reach in this channel. Solution: The normal depths in the two reaches of the channel are determined from the discharge equation:
⎛ 3 y 0 + yo2 ⎞ ⎟⎟ x ( S 0 )1 / 2 Q = AC RS 0 . Hence, 12 = (3 y 0 + y ) x 55 x ⎜⎜ ⎝ b + 2.828 y 0 ⎠ 2 0
Substituting for So by 0.0004 and 0.03, and solving by trial and error, we get yo1 = 2.02 m and yo2 = 0.55m, respectively. The critical depth is calculated as Q g
3 y c + y c2
3.831 = (3 y c + y ) 2 c
3 + 2 yc
Solving by trial and error, get yc = 1.02 m The first reach has a mild bed slope while the second one has a steep bed slope. The gradually varied flow reach comprises M2 and S2 profiles. The lengths of the two profiles are calculated as given in Tables P15.27a and P15.27 b, respectively. Table P14.27a: Calculations for the length of the M2 profile, S o = 0.0004. y
A
P
R
V
E
1.02
4.1
5.885
0.697
2.927
1.457
∆E
Sf x 10-3
5.04
6.394
0.788
2.381
1.489
1.4
6.16
6.96
0.885
1.948
1.593
7.525
0.978
1.63
1.735
1.8
8.64
8.09
1.068
1.389
1.898
8.656
1.155
1.20
-11.35
1.898
-1.498
-69.43
1.157
-1.17
-121.37
0.747
-0.347
-469.74
0.504
-0.104
-1682.69
0.597 0.175
10.0
-2.82
0.898 0.163
2.0
3.22
1.417 0.142
7.36
dl
2.378 0.104
1.6
(So- S f )x 10-3
4.063 0.032
1.2
S f x 10-3
2.073
0.41
L = -2354.58 m Table P15.27b: Calculations for the length of the S2 profile, S o = 0.03. y
A
P
R
V
E
1.02
4.1
5.885
0.697
2.927
1.457
∆E
Sf x 10-3
3.04
5.26
0.578
3.947
1.594
0.60
2.16
4.697
0.460
5.556
2.173
4.584
0.435
6.018
6.487
23.513
583
15.545
14.455
40.06
24.85
5.15
45.24
22.18 0.233
1.994
dl
8.91 0.579
0.56
(So- S f ) x 10-3
4.063 0.137
0.80
S f x 10-3
2.406
27.52
L = 91.13 m
The total length of the gradually varied flow reach is L = 2354.58 + 91.13 = 2445.71 m
288
Chapter 15. Computation of Water Surface Profiles
15.28. A triangular channel with a side slope of 1:1 and a bed slope of 0.02 carries a 3 discharge of 100 ft /s. A gate is located at the end of the channel at which the water depth is 7.0 ft. determine the length of the rising curve before the gate. Take Manning roughness coefficient, n, as 0.02 and use the step-by-step method. Solution: The critical depth is calculated as follows:
100 32.2
= y
y c2
2 c
2 y c
2/5
. Hence, yc = (24.922) = 3.62 ft
2 1.49 2 / 3 1 / 2 1.49 ⎛ y 0 ⎞ 2 ⎜ ⎟ Q = A R S o . Then, 100 = y 0 x n 0.02 ⎜⎝ 2.828 y 0 ⎠⎟
1/ 2
(0.02)1 / 2
yo = (18.98)3/10 = 2.418 ft Q2 Q2 A1 h1 + = A2 h 2 + gA1 gA2 (2.418) 3 3 y 23
32.2 x ( 2.418) 2
=
y 23 3
+
(100) 2 32.2 x y 22
310.559
= 4.712 + 53.117 = 57.829 y 22 Solving by trial and error, we get y2 = 5.175 ft. A hydraulic jump with a low stage depth of 2.418 ft and a high stage depth of 5.175 ft will be formed. An S1 rising profile will then be formed after the jump as shown the Figure. The calculations are presented in Table P15.28. 3
+
+
(100) 2
Table P15.28: Calculations for the length of S1 profile. y
A
P
R
V
E
5.175
26.78
14.64
1.83
3.73
5.391
∆E
Sf x 10
30.25
15.56
1.945
3.31
5.67
36.00
16.97
2.12
2.78
6.12
42.25
18.38
2.30
2.37
6.587
49.00
19.8
2.475
2.04
9.665
19.033
14.66
6.62
19.34
23.27
5.22
19.48
23.97
2.785
19.72
24.24
3.33 0.478
7.0
dl
5.11 0.467
6.5
(So- S f ) -3 x 10
8.13 0.450
6.0
S f x 10-4
11.2 0.279
5.5
-4
7.065
2.24
L = 86.64 ft.
The total length of S1 curve is about 87 ft. One may also proceed in Table P15.28 staring from y = 7.0 ft to y = 5.175 ft. The same answer should be obtained but with a negative sign.
289
Elementary Hydraulics
15.29 A wide channel terminates into a sudden fall. The channel conveys a discharge of 3 ` 4.2 m /s/m and has a bed slope of 16 cm/km. Assume that the depth of flow at the fall is equal to the critical depth and Manning roughness coefficient is 0.022, find the length of the water surface profile ahead of the fall. Solution:
⎧ (4.2) 2 (0.022) 2 ⎫ y 0 = ⎨ ⎬ −5 16 x 10 ⎩ ⎭ y c =
3
q2 g
=
3
( 4.2) 2 9.81
0 .3
= 3.3 m
=1.22 m
The channel has a mild bed slope and an M2 curve will be formed before the fall. The calculations for the profile length are presented in Table P15.29. Table 15.29: Calculations of the length of M2. y
V
E
1.22
3.44
1.823
∆E
Sf x 10-4
2.1
2.225
1.68
2.644
1.4
3.1
1.29
3.335
-26.635
-163
6.248
-4.648
-901
3.11
-1.51
-3020
1.936
-0.336
-6994
2.193 0.435
3.25
26.235
4.026 0.456
3.0
dl
8.47 0.419
2.5
(So- S f )x 10-4
44 0.402
2.0
S f x 10-4
1.679 L = 11078 m
Therefore, the length of M2 calculated between y = 1.22 m and y = 3.25 m is 11.078 km. 15.30. A trapezoidal channel with a bed width of 7.0 m and a side slope of 3:2 has a bed 3 slope of 0.001. The channel conveys a discharge of 30 m /s and has the Manning roughness coefficient of 0.02. A dam is constructed to elevate the water level. The water depth at the dam is 2.5 m. Find the distance upstream the dam where the water depth is 2.0 m . Solution: 2 1 ⎧ ⎪ 7 y 0 + 1.5 y 0 2 30 = (7 y 0 + 1.5 y 0 ) ⎨ 0.02 ⎪ 7 + 2 y 1 + (1.5) 2 0 ⎩
⎫⎪ ⎬ ⎪⎭
2/3
(0.001)
1/ 2
Solving by trial error we get yo = 1.713 m. Also, (7 y c + 1.5 y c2 ) 2 Q A 2 30 = . Hence, = 9.578 = 7 + 3 y c 9.81 g B Solving by trial and error, we get yc = 1.132 m. The channel has a mild bed slope and M1 curve will be formed before the dam. The calculations are given in Table P15.30.
290
Chapter 15. Computation of Water Surface Profiles
Table P15.30: Calculations for the length of M1 between y = 2.5 m and y = 2.0 m. y
2.5
A
B
26.875
14.5
P
R
16.01
1.68
V
1.116
∆E
E
2.563
Sf x -4 10
24.035
13.9
15.29
1.57
1.248
2.38
21.315
13.3
14.57
1.46
1.407
2.2
20.0
13.0
14.21 1.41
1.5
x 10
2.954
7.046
-259.7
4.098
5.902
-305.0
5.236
4.762
-178.5
4.781 -0.085
2.0
dl
-4
3.414 -0.18
2.1
10
-4
2.494 -0.183
2.3
S f x (So- S f )
2.115
5.692 L = 743.2 m
Therefore, the required length is 743.2 m -4
15.31. A wide channel with a bed slope of 15x10 conveys a specific discharge of 25 2 ft /s. A gate is located at the channel inlet where the depth of water at the vena contracted section is 1 ft. Determine the length of the concrete apron to be constructed after the gate to contain the jump. Take the Manning roughness coefficient as 0.018 and assume the length of the jump to be 7 times its height. Solution:
q = y 0
1.49 2 / 3 1 / 2 R S 0 n
⎛ q 2 n 2 ⎞ ⎟⎟ y 0 = ⎜⎜ 2 ( 1 . 49 ) S 0 ⎠ ⎝
0 .3
= 3.43 ft , and y c =
3
( 25) 2 32.2
= 2.687 ft
Then a hydraulic jump must be formed after the gate if the conjugate depth to the normal depth is greater than 1 ft. 3.43 ⎧ ⎪
⎫⎪ y = = 2.06 m ⎨− 1 + 1 + 3 ⎬ 2 ⎪ 32 . 2 ( 3 . 43 ) ⎪ ⎩ ⎭ 8( 25) 2
M3 curve will be formed between y = 1.0 m and y = 2.06 m. The calculations are presented in Table P15.31, noting that 0.091216 q2n2 S f = = 10 / 3 2.22 y y 10 / 3 Length of the jump = 7(3.43-2.06) = 9.59 ft The required length of the apron = 209.8 + 9.59 = 219.39 ft For safety considerations, the apron should have a length of not less than 250 ft
291
Elementary Hydraulics
Table P15.31. Calculations of M3 curve.
∆E
y
V
E
1.0
25
10.71
-4
Sf x 10
20
7.46
16.67
5.82
14.28
4.92
12.14
49.4
334.5
-319.5
49.4
188.5
-173.5
51.9
111.5
-96.5
59.1
141 -0.57
2.06
-657.5
236 -0.9
1.75
672.5 433
-2.18 1.5
(So- S f ) x 10
912 -3.25
1.25
-4
dl
S f x 10-4
4.35
82 L=209.8ft
15.32. A wide rectangular channel has a critical bed slope carries a specific discharge of 2 40 ft /s. Determine the bed slope knowing that the Manning roughness coefficient is 0.022. A gate is placed in midway of the channel. Draw the water surface profile and determine the length of the gradually varied flow reach after the gate. The depth of water at the vena-contracted section after the gate is 1.5 ft. Solution: The channel has a critical bed slope, i.e., yo = yc
y c = y 0 =
( 40) 2
= 3.676 ft 32.2 1.49 40 = (3.676) (3.676) 2 / 3 ( S 0 )1 / 2 0.022 3
2
-3
So = (0.06745) = 4.55 x 10
C3 curve will be formed after the gate. The water surface profile is shown below. The calculations for the length of C3 are presented in Table P15.32.
The length of C3 is 351 ft.
292
Chapter 15. Computation of Water Surface Profiles
Table P15.32: Calculations for the length of C3 profile.
∆E
y
V
E
1.5
26.67
12.54
Sf x 10
-4
20.0
8.21
16.0
6.47
13.33
5.76
10.96
74.8
255.25
-209.75
82.9
127.05
-81.55
87.1
68.1
-22.6
106.2
89.6 -0.24
3.65
-579.0
164.5 -0.71
3.0
624.5 346
-1.74 2.5
(So- S f )x 10
903 -4.33
2.0
-4
dl
S f x 10-4
5.52
46.6 L = 351 ft -4
15.33. A barrage is constructed across a wide river with a bed slope of 16x10 . The specific discharge of the river is 150 ft2/s. The afflux at the barrage is 5.0 ft. Find the distance upstream of the dam where the depth of flow is 14.0 ft. Take the Manning roughness coefficient as 0.026. 2
Solution: q = 150 ft /s, yc =
For wide channels, yn = (
3
150 2 g
qxn
= 8.87 ft. 150 x 0.026
)3 / 5 = ( 1/ 2
1/ 2
) 3 / 5 = 12.29 ft.
1.49 x (0.0016) 1.49 x S o So for an afflux of 5 ft, y 1 = 12.29 + 5 = 17.29 ft. M1 profile will be encountered before the barrage. So we must compute the profile between depths of 17.29 ft and 14.0 ft as given in Table 15.33. Table 15:33. Calculations of the length of M1 profile between y = 14.0 ft and 17.29 ft y (ft)
V(ft/s)
E (ft)
17.29
8.67
16.46
∆E
(ft)
Sf
9.28
17.36
10.0
16.55
10.71
15.78
So - S f
0.000588
0.00101
1089
0.000743
0.000857
945
0.000931
0.000669
1151
0.000823 0.77
14.0
S f
0.000664 0.81
15.0
___
0.000512 1.1
16.0
dl (ft)
___
0.00104 Total 3185 ft
15.34. A sharp crested weir is constructed across a wide rectangular channel that carries a 2 specific discharge of 6.5 m /s as shown in Figure P15.34. The channel has a bed slope of 20 cm/km and a Manning roughness coefficient of 0.026. The weir has a sill height of 3.0 m and the discharge per unit width over the weir is given as 2 q = C d 2g h 3/2 3
293
Elementary Hydraulics
Where Cd = 0.7 is the discharge coefficient and h is the water head over the weir. Determine the normal and critical depths. Estimate the distance upstream of the weir where the depth of flow is 4.75 m. 2
Solution: q = 6.5 m /s, yc =
yn = ( q=
2
3
6.5 2 g
6.5 x 0.026 1/ 2
(0.0002)
= 1.62 m.
) 3 / 5 = 4.43 m. 3/2
C d 2g h 3/2 = 0.67 x 0.7 x 4.43 h ; h = 2.14 m
3 Then, y1 = 3 + 2.14 = 5.14 m. So, we must compute the water surface profile between the depths of 5.14 m and 4.75 m. The calculations are presented in Tab le 15.34. Table 15.34. Calculations for the length of water surface profile y (m)
V (m/s)
E (m)
5.14
1.26
5.22
∆E
(m)
Sf
1.33
4.99
1.37
So - S f
0.000132
0.000068
3382
0.000151
0.000049
3061
0.000144 0.15
4.75
___
S f
0.000121 0.23
4.90
dl (m)
___
4.84
0.000159 Total L= 6443 m
-4 15.35. A dam is built across a wide rectangular river with a bed slope of 10 . The river 2 conveys a specific discharge of 2.833 m /s. If the depth of flow just upstream of the dam is 3.5 m, estimate the length of the backwater curve. How far upstream of the dam will the backwater curve cause a velocity reduction of 18% as compared to the velocity under normal flow conditions? Take Chezy’s roughness coefficient as 70 (metric).
2
Solution: q = 2.833 m /s, yc =
3
2.8332 g
= 0.935 m. 3/2
For wide channels, using the Chezy equation, q = C y So 3/2
1/2
1/2
2.833 = 70 x y x (0.001) , whence yn = 1.18 m, Vn = q/yn = 2.4 m/s. Then Vn – 0.18 Vn = 1.97 m/s. Then, y at that point would be 2.833/1.97 = 1.44 m. So, we must compute the M1 profile from an initial depth of 3.5 m, to the intermediate depth of 1.44 m, and then on to normal depth of 1.18 m. Use the Chezy equation for Sf : V2 Sf = 2 for wide channels. C xy The total length out to a depth of 1.4 m (approximately the 18% point) is 2365 m, while the total length of the profile to 1.2 m (approximately normal depth) is 29 33 m.
294
Chapter 15. Computation of Water Surface Profiles
Table 15.35: Calculations of the length of M1 profile. y (m)
V (m/s)
E (m)
3.5
0.81
3.53
∆E
(m)
Sf
0.88
3.24
0.98
2.95
1.09
2.66
1.23
2.38
1.42
2.10
1.67
1.84
2.02
1.61
2.36
308
0.0000804
0.000920
315
0.000114
0.000886
316
0.000170
0.000830
337
0.000270
0.000730
356
0.000465
0.000535
430
0.000771
0.000229
568
0.000595 0.13
1.2
0.000941
0.000335 0.23
1.4
0.0000585
0.000206 0.26
1.7
303
0.000134 0.28
2.0
0.000956
0.0000932 0.28
2.3
0.0000438
0.0000676 0.29
2.6
dl (m)
So - S f
0.0000494 0.29
2.9
___
S f
0.0000382 0.29
3.2
___
1.48
0.000947 Total L = 2933 m
15.36. A very wide river, with a bed slope of 10 cm/km and a Manning roughness 2 coefficient of 0.032, conveys a specific discharge of 10 m /s. A dam is constructed across the river so that the water depth behind it becomes 14.0 m with a water surface elevation of 55.0 measured from an arbitrary datum. Two villages A and B are to be built 20 km and 45 km upstream of the dam, respectively. Find the respective land elevation (relative to the datum) of the two villages so that they are not to be drowned. Use the standard step method. Solution: At the dam, y = 14 m, WS = 55 m, so the channel invert is 41.0 m. Then, with a channel slope of 0.0001 m/m, the channel invert 20 km upstream is 43 m and the invert 45 km upstream is 45.5 m. 2
At the dam, y = 14 m, q = 10 m /s, V = q/y = 0.714 m/s, E = y+ ET = 55.03 m. Sf = (
0.032 x 0.714 (14)
2/3
(0.714) 2 2g
= 14.03 or
) 2 = 0.0000155 m/m, for wide channels where R = y.
At Section 2, 20 km upstream, assume y = 13 m, WS = 56 m, V = 10/13 = 0.769 m/s, (0.769) 2 0.032 x 0.769 2 ) = 0.0000198 m/m. E = 13+ = 13.03 m, ET = 56.03 m, Sf = ( 2g (13) 2 / 3 ___
S f =
Sf1 + S f2
= 0.0000176 m/m. Then hlobs = 56.03 – 55.03 = 1 m, and hcal = 0.0000176 2 x 20,0000 = 0.352 m. Since the two estimates do not match, estimate another y at Section 2.
295
