Q (5-8)/ Water flows uniformly in an earth Trapezoidal cross section open channel. Data available are: Q=10 (m³/s), B =5 (m). n=0.0229 {s/m(13)}, S, = 0.0015 and side slope (H:V = 1:1). Calculate values of normal water depth ye and normal flow velocity ve in the channel (m) 8E.2-8 ANSWER/ Z=1 A = 5 ye+y² P=5+2.828 ye R = A/P = (5y + y2)/(5+2.828 y.) Q=Vc. A 10=(1/0.0229). R(23), S. (0.5) (5 ye+ y²) 5.917 (5 ye+ye2)(1.67)/(5 +2.828 y)(1.67) Solving this Equation by Trial and error, gives, y = 1.1 (m) ng (0) på ni li camben?. qamar a tuzos cd (m) (m) 121=oy, sm) stof (m) 8E2=8
Q (5-8)/ Water flows uniformly in an earth Trapezoidal cross section open channel. Data available are: Q=10 (m³/s), B =5 (m). n=0.0229 {s/m(13)}, S, = 0.0015 and side slope (H:V = 1:1). Calculate values of normal water depth ye and normal flow velocity ve in the channel (m) 8E.2-8 ANSWER/ Z=1 A = 5 ye+y² P=5+2.828 ye R = A/P = (5y + y2)/(5+2.828 y.) Q=Vc. A 10=(1/0.0229). R(23), S. (0.5) (5 ye+ y²) 5.917 (5 ye+ye2)(1.67)/(5 +2.828 y)(1.67) Solving this Equation by Trial and error, gives, y = 1.1 (m) ng (0) på ni li camben?. qamar a tuzos cd (m) (m) 121=oy, sm) stof (m) 8E2=8
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
![Q (5-8)/ Water flows uniformly in an earth Trapezoidal cross section open
channel. Data available are:
Q=10 (m³/s), B=5 (m). n=0.0229 {s/m(13)), S, = 0.0015 and side
slope (H:V = 1:1).
(mm) 15.1-T
Calculate values of normal water depth ye and normal flow velocity ve
in the channel
ANSWER/
(m) 0.5-oy ola
vig (2) pa ni li commen?
despooncarni tuzen ad (m) ola
(m) ILI=oy, smol stod f
(m) 8E2-8
Z=1
A=5 ye+y₂²
P=5+2.828 yc
R = A/P = (5y + y.3)/(5+2.828 y.)
Q=Vc. A
10=(1/0.0229). R(23), S. (0.5) (5 y + y.²)
5.917 (5 ye+y)(1.67)/(5 +2.828 y.)(1.67)
Solving this Equation by Trial and error, gives,
Ye = 1.1 (m)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e68f5b5-6f98-4fed-9046-35296724b3f1%2F74c36dc0-8fb3-4053-88f9-e3533b15e89b%2Fg3zcl2u_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q (5-8)/ Water flows uniformly in an earth Trapezoidal cross section open
channel. Data available are:
Q=10 (m³/s), B=5 (m). n=0.0229 {s/m(13)), S, = 0.0015 and side
slope (H:V = 1:1).
(mm) 15.1-T
Calculate values of normal water depth ye and normal flow velocity ve
in the channel
ANSWER/
(m) 0.5-oy ola
vig (2) pa ni li commen?
despooncarni tuzen ad (m) ola
(m) ILI=oy, smol stod f
(m) 8E2-8
Z=1
A=5 ye+y₂²
P=5+2.828 yc
R = A/P = (5y + y.3)/(5+2.828 y.)
Q=Vc. A
10=(1/0.0229). R(23), S. (0.5) (5 y + y.²)
5.917 (5 ye+y)(1.67)/(5 +2.828 y.)(1.67)
Solving this Equation by Trial and error, gives,
Ye = 1.1 (m)
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