Q (5-8)/ Water flows uniformly in an earth Trapezoidal cross section openchannel. Data available are:Q=10 (m³/s), B=5 (m). n=0.0229 {s/m(13)), S, = 0.0015 and sideslope (H:V = 1:1).(mm) 15.1-TCalculate values of normal water depth ye and normal flow velocity vein the channelANSWER/(m) 0.5-oy olavig (2) pa ni li commen?despooncarni tuzen ad (m) ola(m) ILI=oy, smol stod f(m) 8E2-8Z=1A=5 ye+y₂²P=5+2.828 ycR = A/P = (5y + y.3)/(5+2.828 y.)Q=Vc. A10=(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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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

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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