The flow rate in Example Problem 6.10 decreases at a rate of dQ/dt = –0.5 cfs/s at the reservoir. Make a table that shows the depth in the channel at both its beginning and at a position 2000 ft downstream therefrom as a function of time.

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In order to solve the depth at the origin, with the flow rate known here as a function of time, it is necessary to solve the equation F = A − Q/(w + [latex]V_{o} − w_{o}[/latex]) = 0 by trial using the definition of w as related to the depth Y for each entry in the table. At another position x, the sum of [latex]t_{1}[/latex] + Δt = t (the time for which this line in the table is to apply) must be solved. When known values are substituted for [latex]t_{1}[/latex] and Δt, this produces the equation: F = ([latex]Q_{o}[/latex] − Q)/|dQ/dt| + x/(w + [latex]V_{o} − w_{o}[/latex] + c) − t = 0, in which Q = VA, and V = V(0, t) = w(0, t) + [latex]V_{o} − w_{o}[/latex], and the area A is obtained from the equation A = [b + mY(0, t)]Y(0, t). The computer programs whose listings are given below carry out this solution. The program is given twice; first in FORTRAN and then in PASCAL. The first portion of the programs read in the problem variables, then the stage function w is integrated over the range of depths from 0.01 to [latex]Y_{o}[/latex]. Thereafter, the two equations above are solved, and the time, the flow rate, the depth, and the velocity associated with each of these times are printed. The solutions are given below.

Question 6.12: The flow rate in Example Problem 6.10 decreases at a rate of...

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