Water is flowing in a trapezoidal channel at a rate of Q = 20 m/s. The critical depth y for such a channel must satisfy the equation 0 = 1 - (Q² B)/(s A.3) where g = 9.81 m/s², Ac = the cross-sectional area (m2), and B = the width of the channel at the surface (m). For this case, the width and the TAYSN %3D cross-sectional area can be related to depth y by 3 = 3 + y and A. = 3y + y². Solve for the critical depth using bisection. For calculation use initial guesses of x = 0.5 and x, = 2.5, and iterate until the approximate error falls below 3% or the number of iterations exceeds 4. 0278 0278 278
Water is flowing in a trapezoidal channel at a rate of Q = 20 m/s. The critical depth y for such a channel must satisfy the equation 0 = 1 - (Q² B)/(s A.3) where g = 9.81 m/s², Ac = the cross-sectional area (m2), and B = the width of the channel at the surface (m). For this case, the width and the TAYSN %3D cross-sectional area can be related to depth y by 3 = 3 + y and A. = 3y + y². Solve for the critical depth using bisection. For calculation use initial guesses of x = 0.5 and x, = 2.5, and iterate until the approximate error falls below 3% or the number of iterations exceeds 4. 0278 0278 278
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Water is flowing in a trapezoidal channel at a rate of Q = 20 m3/s. The critical depth y for such a channel must satisfy the equation 0 = 1 - (Q2 B)/(g
A.3) where g = 9.81 m/s2, A, = the cross-sectional area (m2), and B = the width of the channel at the surface (m). For this cas 278.
width and the
STAYBIN
2.5, and iterate until the
cross-sectional area can be related to depth y by
initial guesses of x| = 0.5 and :
B = 3 + y and A. = 3y + y?. Solve for the critical depth using bisection. For calculation
approximate error falls below 3% or the number of iterations exceeds 4.
TAYSN
5278
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