20. A trapezoidal canal, 6m at the bottom and having side slopes 2:1 carries water to a depth of 40cm and mean velocity of 0.60 m/s. what length of a rectangular contracted weir 85cm high should be placed in the middle of the canal if the depth of water upstream is to be 1.20m? Given: Find: L =? V W.G. P = 0.85 ↓ do = 0.40 m T 6m original ws 2.4m aem- d=1.20m
20. A trapezoidal canal, 6m at the bottom and having side slopes 2:1 carries water to a depth of 40cm and mean velocity of 0.60 m/s. what length of a rectangular contracted weir 85cm high should be placed in the middle of the canal if the depth of water upstream is to be 1.20m? Given: Find: L =? V W.G. P = 0.85 ↓ do = 0.40 m T 6m original ws 2.4m aem- d=1.20m
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Need help asap. How did you come up with 7.6? Or please show the correct calculation for this. And kindly explain. Thank you. Answer from the book: 4.30 m.
![20. A trapezoidal canal, 6m at the bottom and having side slopes 2:1 carries
water to a depth of 40cm and mean velocity of 0.60 m/s. what length of a
rectangular contracted weir 85cm high should be placed in the middle of
the canal if the depth of water upstream is to be 1.20m?
Given:
Find: L =?
H
d=1.20m
P = 0.85
B
6m
= ½ (6 + 7.6) (0.4) (0.6)
= 1.63m3/s
Q = 1.84 L'H3/2
1.63 = 1.84 L' (0.35) 3/2
4
L2H
72
L = 4.28 +0.20(0.35)
L = 4.35m
original W. S
2,4m
Solution:
L' = 4.28 m
Qo = AoVo
and
48m-
contrations
do= 0.40 m
T
H = 1-20 0.85
H = 6.35](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90a775b4-ca1e-495d-9143-56cc1899f56a%2F7f77bc34-3590-440e-9f93-d88753d0b1d4%2Fbitoisr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:20. A trapezoidal canal, 6m at the bottom and having side slopes 2:1 carries
water to a depth of 40cm and mean velocity of 0.60 m/s. what length of a
rectangular contracted weir 85cm high should be placed in the middle of
the canal if the depth of water upstream is to be 1.20m?
Given:
Find: L =?
H
d=1.20m
P = 0.85
B
6m
= ½ (6 + 7.6) (0.4) (0.6)
= 1.63m3/s
Q = 1.84 L'H3/2
1.63 = 1.84 L' (0.35) 3/2
4
L2H
72
L = 4.28 +0.20(0.35)
L = 4.35m
original W. S
2,4m
Solution:
L' = 4.28 m
Qo = AoVo
and
48m-
contrations
do= 0.40 m
T
H = 1-20 0.85
H = 6.35
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