CHAPTER 11 RESISTANCE IN OPEN CHANNELS
11.1. Water
is flowing in a trapezoidal earthen channel width 2 m and side slopes 1.5 H /1 /1 3 V . The channel is carrying a discharge of 50 m /s and is running on a slope of 0.0025 m/m. If the roughness coefficient is 0.030, what is the normal depth in the the channel? Solution: 2/3
(2 y o
(2 y o
⎡ 2 y o + 1.5 y o2 ⎤ 0.03(50) 2 + 1.5 yo ) ⎢ = ⎥ (0.0025) 0.5 ⎢⎣ 2 + 2 y o 1 + t 2 ⎥⎦ 2/3 ⎡ 2 y o + 1.5 y o2 ⎤ 2 + 1.5 y o ) ⎢ ⎥ = 30 2 3 . 605 + y o ⎦ ⎣
Therefore, yo = 3.207 m 3-m wide rectangular irrigation channel carries a discharge of 25.3 m3/s at a uniform depth of 1.2 m. Determine the slope of the channel with n = 0.02. If the discharge is increased to 40 m3/s, what is the normal depth of flow?
11.2. A
Solution:
A = b y = 3 x 1.20 = 3.6 m2, and P = b + 2y = 3 + 2 x 1.20 = 5.4 m A 3.6 R= = = 0.667 m p 5.4 2
⎛ Q x n ⎞ S = ⎜ 2 / 3 ⎟ = 0.041 ⎝ R A ⎠ A = b yo = 3yo, and P = b + 2yo = 3 + 2yo 3 y o A = R= p (3 + 2 y o )
⎛ 3 y o ⎞ ⎜ ⎟ Q= ⎜ 0.022 3 + 2 y ⎟ o ⎠ ⎝ 1
2/3
(0.041)1 / 2 (3 y o )
Hence yo = 1.69 m 11.3. A
trapezoidal earthen channel with bottom width 5 ft and 2 H on on 1 V side side slope is 3 carrying a discharge of 200 ft /s at a normal depth. If the channel is running a slope of 0.001 ft / ft and has a Manning n value of 0.025, find the normal depth.
Solution:
(5 y o
⎡ 5 y o + 2 y o2 ⎤ 2 + 2 y o ) ⎢ ⎥ ⎢⎣ 5 + 2 y o 1 + 2 2 ⎥⎦
2/3
=
0.025(200) 1.49(0.0001)1 / 2
Elementary Hydraulics
⎡ 5 y o + 2 y o2 ⎤ (5 y o + 2 y ) ⎢ ⎥ ⎣ 5 + 4.47 y o ⎦
2/3
= 335.57
2 o
yo = 7.183 ft. 11.4. An
earthen canal in good condition having a bottom width of 4 m and side slopes of 2 H on IV is designed to carry a discharge of 6 m3/s. If the slope of the canal is 0.39 m/km, what is the normal depth? Solution:
So = 0.39 m/km = 0.00039 m/m, and n = 0.03
⎡ 4 y o + 2 y o2 ⎤ 2 + 2 y o ) ⎢ ⎥ ⎢⎣ 4 + 2 y o 1 + t 2 ⎥⎦
(4 yo
⎡ 4 y o + 2 y o2 ⎤ 2 (4 y o + 2 y o ) ⎢ ⎥ ⎣ 4 + 4.47 y o ⎦
2/3
=
0.03(6) (0.00039)1 / 2
2/3
= 9.11
yo = 1.405 m 11.5. Determine
the normal depth in a trapezoidal channel with side slope of 1.5:1, bottom width of 25 ft, and channel bed slope of 0.00088, if the discharge is 1510 ft3/s and n = 0.017. Solution:
Q=
1.49 n
R 2 / 3 s 1 / 2 A =
25y + 1.5y2 x (
1.49
A 5 / 3 p 2 / 3 s 1 / 2
n 25y + 1.5y 2
25 + 3.6y
)2/3 =
0.017 x 1510 1.49 x (0.00088)1/2
= 580.76
yo = 6.2 ft. rectangular concrete lined 1.5 m wide is carrying 15 m3/s at a normal depth. If the channel is laid on a slope of 0.01 m /m, what would be the normal depth? 11.6. A
Solution:
For concrete, n = 0.015 (from Table 10.1)
1.5 y o
⎡ 1.5 y o ⎤ ⎢1.5 + 2 y ⎥ o ⎦ ⎣
2/3
=
0.015(15) (0.01)1 / 2
yo = 2.209 m
188
= 2.25
Chapter 11. Resistance in Open Channels
A trapezoidal channel is to be designed to carry a discharge of 250 ft 3/s at a normal depth of 3.5 ft. The channel is running on a slope of 0.005 ft/ft and the side slopes can be no steeper than 2.5 H on IV. The channel is earthen and will have an estimated Manning n value of 0.03. What is the minimum bottom width to affect the required normal depth? 11.7.
Solution:
⎡ 3.5(b) + 2.5(3.5) 2 ⎤ 2 (3.5(b) + 2.5(3.5) ) ⎢ ⎥ ⎢⎣ b + 2(3.5) 1 + 2.5 2 ⎥⎦ ⎡ 3.5(b) + 30.625 ⎤ (3.5b + 30.625) ⎢ ⎣ b + 18.84 ⎥⎦
2/3
=
0.03(250) 1.49(0.005)1 / 2
2/3
= 71. 18
b = 4.206 ft. 11.8. A
trapezoidal channel has a bed width of 10 ft and a side slope 2:1. The channel is paved with smooth cement surface of n = 0.011. If the channel is laid on a slope of 0.001 and carries a uniform flow of depth 2 ft, determine the discharge. Solution:
A = by + ty2 = 10 x 2 + 2 (2)2 = 28 ft2 P = b + 2y R=
A p
Q = 28
=
1 + t 2 28
18.94
= 10 + 2 x 2 (1 + 2) 2 = 18.94 ft
= 1.478 ft
⎛ 1.49 ⎞ (1.478) 2 / 3 (0.0001)1 / 2 = 49.21 ft 3 /s ⎜ ⎟ ⎝ 0.011 ⎠
11.9. Water
flows uniformly in a 2-m-wide rectangular channel at a depth of 45 cm. Find the flow rate in m3/s; discus the result for the following cases. a) The channel slope 25 cm / km and n = 0.025 b) The channel slope 5 cm / km and n = 0.025 c) The channel slope 25 cm / km and n = 0.05 Solution:
By using Manning Equation Q A
1
R 2 / 3 S 1 / 2
n a) A = 0.9 m , P = 2.9 m and R = A/P = 0.3 m 1 Q = 0.9( ) (0.31) 2 / 3 (0.00025)1 / 2 = 0.261 m 3 / s 0.025 1 b) Q = 0.9( ) (0.31) 2 / 3 (0.00005)1 / 2 = 0.1166 m 3 / s 0.025 1 c) Q = 0.9( ) (0.31) 2 / 3 (0.00025)1 / 2 = 0.13 m 3 / s 0.05 Therefore, the discharge is directly proportional to the bed slope and inversely proportional to Manning Coefficient. 2
189
Elementary Hydraulics
11.10.
A particularly filled pipe 1.0-m in diameter laid on a uniform slope 1:2000, calculate the maximum flow that can run through this pipe (n = 0.016). Using Chezy equation,
Solution:
Q = A C R S o
=
C
A 3 P
S o
For a given channel cross section, C and So will be constant. The maximum velocity is satisfied when the value of (A3/P) is maximum. Therefore, differentiating with respect to θ, and knowing that r 2 A= (θ-sinθ), and P = r θ, then 2 d d θ
Then,
⎛ A 3 ⎞ ⎜ ⎟= 0 ⎜ P ⎟ ⎝ ⎠ dA ⎞ ⎛ 3 dP − P 3 A 2 ⎜ A ⎟ d θ ⎟ 0 ⎜ d θ = ⎜ ⎟ P 2 ⎜ ⎟ ⎝ ⎠
Or, A r – 3r θ r 2
r 2
2
θ
(1- cosθ) = 0, which can be written as, r 2
(θ-sin θ) r - 3r θ (1- cosθ) = 0, hence 2 2 θ - sinθ - 3 θ + 3 θ cosθ = 0, or 3 θ cosθ - sinθ − 2θ = 0 Solving the above equation we get θ = 308o (5.37 rad). r 2 So, A = (θ-sinθ) = 0.7705 m2 2 r ⎛ sin θ ⎞ R = , ⎜1 − ⎟ = 0.2866 m , Then: 2 ⎝ θ ⎠ 1 Q = A R 2 / 3 S 1 / 2 = 0.4681 m3/s n 11.11. A
rectangular channel 15 ft wide flow at a normal depth of 6 ft. The discharge is 530 ft /s and the Manning n value is 0.025. What is the rate of energy dissipation feet per feet channel length in this channel? 3
Solution:
A = 15 x 6 = 90 ft2, P = 15 + 2 x 6 = 27 ft2 , R = 3.33 ft 530 V= = 5.89 ft / s 90 1.49 2 / 3 1 / 2 V= , or R S o n For uniform flow So = Sf
190
Chapter 11. Resistance in Open Channels
2
2 ⎛ (0.025 x 5.89) ⎞ nV ⎞ ⎛ ⎜⎜ ⎟ = 0.00196 ft/ft Sf = ⎜ = ⎟ 2/3 2/3 ⎟ 1 . 49 1 . 49 x ( 3 . 33 ) R ⎝ ⎠ ⎝ ⎠
11.12.
A rectangular canal with a bed slope 8 cm/km ad a bed width of 100 m. If at a depth of 6 m, the canal carries a discharge of 860 m3/s, find the Manning’s roughness, n, Chezy coefficient, C, and the coefficient of friction, f . Also find the average shear stress on bed. Solution:
A = b x y = 100 x 6 = 600 m2, P = b + 2y = 100 + 2 x 6 = 112 m A 600 R= = = 5.36 m P 112 Q = AC RS Hence, C = 69.22 1
Q=A
n
1
860 = C=
R 2 / 3 S 1 / 2
n
(600)5/3 (112)2/3 (0.00008)1/2
8 g f
Hence, n = 0.019
Hence, f = 0.016
τo = γRS = 9810 x 5.36 x 8 x 10-5 = 4.2 N/m2 11.13. Show
that for a circular culvert of diameter D the velocity of flow will be the maximum when the depth of flow y at the center is 0.81 D. Use the Chezy formula. Using Chezy equation,
Solution:
Q = C R S o
= C
A P
S o
For a given channel cross section, C and So will be constant. The maximum velocity is satisfied when the value of (A/P) is maximum. Therefore, differentiating with respect to θ, d d θ
Then,
⎧ r 2 ⎫ ⎪⎪ (θ − sin θ ) ⎪⎪ d ⎧ r (θ − sin θ ) ⎫ 2 ⎨ ⎬ = 0 , or ⎨ ⎬=0 2 r d θ θ θ ⎩ ⎭ ⎪ ⎪ ⎪⎩ ⎪⎭
2r (θ − sin θ )
− 2 θ (1 − cos θ )
4θ 2
=0
Which can be written as,
θ - sinθ − θ + θ cos θ = 0. Hence, θ cos θ = sinθ, or θ = tan θ Solving the above equation we get θ = 257o (4.485 rad). Hence,
191
Elementary Hydraulics
(360 − 257)
α = y −
2
= 51.5 o , and then
D
2
D
= cos α , Hence, y = 0.81 D
2 11.14. A
sewer of diameter D = 0.6 m has a bed slope of 1:200. What is the possible maximum velocity of flow in this pipe? What is the discharge ay this velocity? Take C = 55. Solution:
For maximum velocity, θ = 257o (4.485 rad) , as shown in Problem 11.13. r 2 A= (θ − sin θ ) = 0.246 m 2 2 r ⎛ sin θ ⎞ R = ⎜1 − ⎟ = 0.183 m , Then: 2 ⎝ θ ⎠ V = C RS = 1.662 m/s and Q = AV = 0.409 m3/s
11.15. It
is desired to design a trapezoidal channel with a bottom width of 10 ft and 2 H on 1 V side slope. Sieve analysis revealed a grain size distribution, which result in an allowable bed shear stress of 0.5 lb/ft2 and a Manning n value of 0.03. If the channel is to be designed to run at normal depth of 5 ft, what will be the resulting discharge? Solution:
1 + (2) 2 = 32.36 ft, and R = A/P = 3.09 ft.
A = 10(5) + 2(5)2 = 100 ft2, P = 10 + 2(5)
τo = 0.5 = γ R(So) = γ(3.09) (So) = 62.4 (3.09) (So), hence So = 0.00259 ft/ft Q = 100
1.49 0.03
(3.09) 2 / 3 (0.00259)1 / 2
= 536.6 ft 3 / s
11.16. It
is desired to design a trapezoidal channel with a bottom width of 15 ft and 2 H on 1 V side slopes. Sieve analysis has revealed a grain size distribution that results in maximum allowable bed shear stress of 0.75 lb/ft2. The channel must be designed to carry a discharge of 1000 ft3/s at a normal depth of 7 ft. a) At what slope should the channel be laid? b) What is the Manning n value for this channel? Solution:
A = 15(7) + 2(7)2 = 203 ft2, P = 15 + 2(7)
1 + (2) 2 = 46.3 ft, and R = A/P = 4.38 ft.
192
Chapter 11. Resistance in Open Channels
V = Q/A =
1000
= 4.926 ft/s 203 τo = γ RSf = 0.75 = 62.4(4.38) (So), hence So = 0.00274 ft/ft 2
⎛ nV ⎞ ⎛ n(4.926) ⎞ ⎟ ⎜⎜ ⎟ = = So = ⎜⎜ 0 . 00274 2/3 ⎟ 2/3 ⎟ ⎝ 1.49( R ) ⎠ ⎝ 1.49(4.38) ⎠
2
n = 0.0424 20 ft wide rectangular channel carries a discharge of 400 ft3/s at a normal depth of 10 ft. If the roughness coefficient is 0.03, what shear stress in lb/ft 2 is imparted to the channel boundary by this flow? 11.17. A
Solution:
A = 200 ft2 , V = 400/200 = 2 ft/s, P = 20 + 2(10) = 40 ft. 200 R= = 5 ft , τo = γRS 40 2
2
⎛ nV ⎞ ⎛ 0.03(2) ⎞ ⎟ ⎜⎜ ⎟ = 0.0001897 ft/ft = So = ⎜⎜ 2/3 ⎟ 2/3 ⎟ 1 . 49 1 . 49 ( 5 ) R ⎝ ⎠ ⎝ ⎠ 3 τo = 62.4 lbs/ft (5) (0.0001897) = 0.0591 lbs/ft2 11.18. Design
a channel (assuming a rigid boundaries) to irrigate 54,430 hectares at a rate of 60 m /ha/day. The soil allows side slope of 1:1, n = 0.025, and the bed slope is 10 cm/km. Find the channel dimensions for each of the following cases: a) The maximum allowable velocity being 0.7 m/s b) The maximum allowable boundary shear stress = 0.05 lb/ft2 c) V = 0.36 y0.64 3
Solution:
a)
For maximum allowable velocity being 0.7 m/s Q 37.8 = = 54.0 m 2 A= V 0.7 A = by + ty2 = 54 m2 (1) 1 V = 0.7 = R 2 / 3 s 1 / 2 So, R = 2.315 m and then P = 23.33 m n b + 2 y 1 + 12 = b + 2.828 y = 23.33 m (2) From (1) and (2) get b = 14.8 m and y = 3.0 m
b)
For maximum allowable boundary shear stress τo = 0.05 Ib/ft2 0.05 x 453.6 τ = = 0.024 gm / cm 2 = 2.4 N/m2 2 (30.48)
193
Elementary Hydraulics
τ = γRs
So, 2.4 = 9810 x R x 10-4
∴ R = 240 cm = 2.4 m Q=
1 n
R 2 / 3 s 1 / 2 A =
1 0.25
(2.4) 2 / 3 (10 − 4 ) 0.5 A
So, A = 52.72 m2
Then, P = 21.96 A = by + ty2 = 52.72 m2 b + 2 y 2
(1)
= 21.96
(2)
by the quadratic formula, y = 3.32 m, b = 12.56 m
c)
For v = 0.36y0.64 and Q = 37.8 m3/s 1 v = R 2 / 3 s 1 / 2 = 0.4 R 2 / 3 n 2/3
So,
⎛ by + y 2 ⎞ 0.64 ⎟⎟ v = 0.36y = 0.4 ⎜⎜ 2 y 2 ⎝ b + ⎠
Q=Axv
0.36y0.64 x (by + y2) = 37.8
(1)
(2)
From (1) and (2) get b = 14.30 and y = 3.0 m 11.19.
Water is flowing in rectangular channel with a bed slope of 0.005 ft/ft. The channel width is 10 ft and the discharge is 300 ft3/s. Now, suppose that the depth at a given place in the channel is observed at a spot 1000 ft downstream is observed to be 4.5 ft. Estimate the value of Manning coefficient n of this channel. Solution:
A1 = 6 (10) = 60 ft 2 P1 = 12 + 10 = 22 ft 60 R 1 = = 2.73 ft. 22 300 V1 = = 5 ft/s 60 (5) 2 E1 = 6 + = 6.39 ft. 2 g ∆E = 6.39 – 5.19 1.2 ft. 1.2 Sf = = 0.0012 ft/ft 1000
A2 = 4.5 (10) = 45 ft2 P2 = 9 + 10 = 19 ft 45 R 2 = = 2.37 ft. 19 300 V2 = = 6.67 ft/s 45 (6.67) 2 E2 = 4.5 + = 5.19 ft. 2 g
194
Chapter 11. Resistance in Open Channels
2.73 + 2.37 = 2.55 ft/s 2 5 + 6.67 Vavg = = 5.835 ft/s 2 1.49 2 / 3 1 / 2 1.49 V= R S = 5.835 = (2.555) 2 / 3 (0.0012)1 / 2 n n 0.09646 5.835 = . Hence n = 0.0165 n R avg =
trapezoidal channel is designed to carry 25 m3/s on a slope of 0.0015 m/m. The channel is unlined, and in order to prevent erosion, the maximum allowable velocity is 1.5 m/s. The side slope must be no steeper than 2 H on 1 V and the Manning n value is 0.03. In order to meet these requirements, what flow depth and bottom width should be used? 11.20. A
Solutions:
Q = 25 m3/s, V = 1.5 m/s , and A = Q/V = V=
1
R 2 / 3 S 1 / 2 =1.5 =
1
25 1.5
= 16.67 m2
( R 2 / 3 ) (0.0015)1 / 2
0.03 n 2/3 1.5 = 1.29 R , then R = 1.25 m
A 16.67 ⇒ P =13.33 m R = 1.25 m = = P P A − ty 2 16.67 − ty 2 2 A = by + ty ⇒ b = = y y
1 + t 2
P = b + 2y 16.67 − 2 y 2
=
b + 2 y 5
= b + 4.47 y , therefore
+ 4.47 y = 13.33 , or 16.67-2y2 + 4.47 y2 = 13.33 y
y 16.67 + 2.47 y2 = 13.33 y, or 2.47 y2 – 13.33 y + 16.67 = 0
− b ± b 2 − 4ac 13.33 ± (13.33) 2 − 4(2.47) (16.67) y= = 2a
y=
13.33 ± 12.99 4.94
2(2.47)
=
13.33 ± 3.6 4.94
y = 3.43 m or 1.97 m 16.67 − 2(1.97) 2
= 4.52 m 1.97 If y = 3.43 m, then b is negative and hence not possible. If we let y = 1.97 m, then b =
195
Elementary Hydraulics
11.21. Show
that for the most coefficient trapezoidal channel, the side slope must be equal to 1/ 3 . Solution: 2
A = by + ty
hence, b =
A − ty 2 y
P = b + 2y 1 + t 2
⎞ ∂ ⎛ A ⎞ ∂ P ∂ ⎛ A − ty 2 = 0 = ⎜⎜ + 2 y 1 + t 2 ⎟⎟ = ⎜⎜ − ty + 2 y 1 + t 2 ⎟⎟ ∂t ∂t ⎝ y ⎠ ⎠ ∂t ⎝ y =
1
− y + 2 y ( 1 + t 2 ) −1 / 2 (2t ) 2
= − y + t=
2ty 2
1 + t
= 0. Then
2ty 2
1 + t
= y
1 3
11.22. Show
that the most efficient trapezoidal channel section will have a hydraulic radius equal to one-half the depth. Solution:
Eq. 11.20:
b = 2y ( 1 + t 2
− t ) A by + ty 2 = R= P b + 2 y 1 + t 2 2 y 2 ( 1 + t 2 - t) y + ty 2 R= 2 y ( 1 + t 2 - t) + 2y 1 + t 2 y 2 (2 1 + t 2 - 2t + t) y = = 2 y( 1 + t 2 - t + 1 + t 2 ) 2
A rectangular flume is to be built to carry a discharge of 120 ft 3/s on a slope of 0.001 ft/ft. If the flume is constructed of corrugated metal, what should be its dimensions to minimize the cost of the material? 11.23.
Solution:
A = by = 2y2 P = b + 2y = 2y + 2y = 4y R = A/P = 2y2/4y = 0.5 y 1.49 2 / 3 1 / 2 1.49 Q=A (0.5 y) 2 / 3 (0.001)1 / 2 R S o = 2 y 2 0.025 n 8/3 120 = 2.3745 y 50.5369 = y8/3 y = 4.35 ft; b = 2y = 8.7 ft.
196
Chapter 11. Resistance in Open Channels
11.24.
Find the best hydraulic section for a rectangular section with cross-sectional area of 12 m2. Solution:
For the best hydraulic section dP/dy = 0 A A = by, and P = b + 2y = + 2 y y dp A =− 2 +2=0 dy y A 12 So, y2 = = =6 2 2 ∴y = 2.45 m and b = 4.9 m
One may also check the result through the following table. A = by P = b + 2y
6 x 2 = 12 10
3 x 4 = 12 11
1 x 12 = 12 25
12 x 1 = 12 14
4.9 x 2.45 = 12 9.8
11.25.
A concrete lined canal having one side vertical and other side is sloping at 2:1 carries a discharge of 10 m3/s with velocity of 0.715 m/s. Determine the dimensions of the canal and the bed slope for the minimum cost of construction if the cost of excavation is 2.0 $/m3 and the cost of lining is 6.0 $/m2(n = 0.014). Solution:
A =
Q
= 14.0 m 2
(1)
V y(b+y) = 14.0 m2 1 v = R 2 / 3 s 1 / 2 n
⎛ (b + y ) y ⎞ ⎜ ⎟ 0.715 = 0.014 ⎜⎝ b + 3.24 y ⎠⎟ 1
2/3
s1 / 2
(2)
Cost = A x 125 + P x 750 per meter length of the canal = y(b+y) x 125 + 750 x (b+3.24y) 14 But from (1) b = − y y 14 14 Cost = y ( − y + y ) x125 + 750 x ( − y + 3.24 y ) y y d (cos t ) = zero For minimum cost dy Hence, y = 2.5 m and b = 3.1 m Therefore, A = 14m2, P = 11.2 m and R = 1.25 m And then from equation (2) we can get S = 7.44 cm/km
197
Elementary Hydraulics
trapezoidal channel is designed to carry a discharge of 1800 ft3/s at a normal depth on a slope of 0.0005 ft/ft. The side slopes of the channel must be 1.5 H /1 V and the Manning n value is 0.035. If the channel is to be a regular earthen canal in good condition, what should be its dimensions for maximum hydraulic efficiency? 11.26. A
Solution:
For the best hydraulic section b = 2y ( 1 + t 2
− t ) = 2y ( 1 + (1.5) 2 − 1.5 ) = 0.605 y
P = b + 2y 1 + t 2 = 0.605 y + 2y
1 + (1.5) 2
= 4.21 y
A = by + ty2 = 0.605y2 + 1.5 y2 = 2.105 y2 y R= 2 ⎛ 1.49 ⎞ R 2 / 3 S 1 / 2 Q=A ⎜ ⎟ o ⎝ n ⎠
⎛ 1.49 ⎞ ⎛ y ⎞ 1800 = 2.105 y ⎜ ⎟⎜ ⎟ ⎝ 0.035 ⎠ ⎝ 2 ⎠
2/3
2
(0.0005)1 / 2
1425.95 = y8/3 Therefore, y = 15.23 ft and b = 9.2 ft. 11.27. Due
to the topography of the ground surface, a 2-km-long circular water tunnel is proposed to replace 20 km-long escape channel of trapezoidal section, side slope 1:1. The tunnel section is best discharging section (θ = 308o) and the trapezoidal section is best hydraulic section (R = y/2) and both have the same roughness and bed slope. Determine whether the construction of the tunnel instead of the channel will reduce or increase the total cost of the project. Solution:
For trapezoidal channel of best cross section: The top width is equal to twice the side length Therefore, B = 2 2 y , and Bottom width, b = 2 2 y -2y A=
B + b
y
= 1.828 y 2
2 Discharge of trapezoidal section QT, QT = A x C RS = 1.828 y2 x C
y
2
S
= 1.293 y2.5 x C S For circular channel r 2 A = (θ − sin θ ) = 3.082 r 2 2 r ⎛ sin θ ⎞ R = ⎜1 − ⎟ = 0.574 r 2 ⎝ θ ⎠
198
(1)
Chapter 11. Resistance in Open Channels
Q = AxV QC = 3.082 r 2.5 x C S QT
=
1.293 y 2.5 x C S
2.225 r 2.5 x C S Then: r = 0.7896 y QC
= 1 (same discharge)
Total volume For trapezoidal channel: (Vol)h = AT x LT = 1.828y2 x 20 x 1000 = 36560y2 m3 For circular channel: (Vol)C = AC x LC = p(0.7896y)2 x 2 x 1000 = 3917.4y2 m3 Cost of excavation of the trapezoidal channel = 36560y2 x 4 = 146240y2 Cost of excavation of the circular tunnel = 3917.4y2 x 20 = 78347y2 Reduction in the cost = 146240y2 – 78347y2 = 67893 y2 trapezoidal channel is to design to carry a discharge of 75 m3/s at maximum hydraulic efficiency. The side slope of the channel is 2 H /1 V and the Manning n value is 0.030. a) If the maximum allowable velocity in the channel is 1.75 m/s, what should be the dimensions of the channel? b) What should be the slope of the channel? 11.28. A
Solution:
A = Q/V = a)
b A A y2 y
= = = = =
75 1.75
= 42.857 m 2
2y ( 1 + t 2 − t ) = 0.472 y by + ty2 = 0.472 y2 + 2y2 = 2.472 y2 42.857 = 2.472 y2 17.336 4.16 m and b = 0.472 y = 1.96 m
1 + t 2 =1.96 + 2(4.16) 1 + 2 2 m = 20.56 m 42.85 R = A/P = = 2.08 m 20.56 1 V = 1.75 = (2.08) 2 / 3 S o1 / 2 0.03 1.75 = 54.315 S 01 / 2 P
=
b + 2y
So
=
0.00104 m/m
A trapezoidal channel is designed to convey a discharge of 30 m3/s and runs on a slope of 10 cm /km. Determine the dimensions of the most efficient section. Take Chezy’s C as 40 (metric). 11.29.
Solution:
For the most efficient trapezoidal section, b = 1.155 y, t = 1/(3)0.5, and R = y/2. Then,
199
Elementary Hydraulics
Q = ACR 1/2 S 1o/ 2 30 = (by + ty2) 40 (
y
2
)1/2 (0.0001)1/2
y 75 = (1.155 y2 + 0.577 y2) ( )1 / 2 2 5/2 106.67 = 1.732 y y = (61.24)2/5 = 5.186 m ; b = 1.155 y = 6.0 m 11.30. A
rough timber flume (n = 0.016) with a cross section of a most efficient triangular section conveys water at a depth of 1.5 ft under uniform conditions. The channel has a bed slope of 0.001. Calculate the discharge. Solution:
A = y2 , P = 2y 1 + t 2 Q=A
= 2 2 y , R =
y
2 2
1.49 2 / 3 1 / 2 1 y 2 / 3 ) (0.0001)1 / 2 R S o = y 2( ) ( n n 2 2
1.5 ⎞ ⎛ 1.49 ⎞ ⎛ = (1.5) ⎜ ⎟ ⎟⎜ ⎝ 0.016 ⎠ ⎝ 2 2 ⎠ 2
2/3
(0.001)1 / 2 = 5.34 ft3/s
200
