Question 4.1: Compute the normal depth in a trapezoidal channel having a b......

We use the above procedures one by one to determine y_{n}.

Design curves

By substituting the values of n, Q, and S_{o} into the right-hand side of Eq. 4-32, we obtain
\quad\quad\quad\quad AR^{2/3}=\frac{nQ}{C_{o}S_{o}^{1/2}}  (4-32)
\quad\quad\quad\quad \frac{nQ}{C_{o}S_{o}^{1/2}}=\frac{0.013\times 30}{1.\times (0.001)^{1/2}}\\\quad\quad\quad\quad\quad\quad\quad= 12.33
Hence, it follows from Eq. 4-32 that
\quad\quad\quad\quad AR^{2/3}=12.33
Now
\quad\quad\quad\quad \frac{AR^{2/3}}{B_{o}^{8/3}}=\frac{12.33}{(10)^{8/3}}\\\quad\quad\quad\quad\quad\quad = 0.026
For s = 2 and AR^{2/3}/B_{o}^{8/3} = 0.026, we read from Fig. 4-5 that y_{n}/B_{o} = 0.11. Hence,
\quad\quad y_{n} = 1.1 m.

Trial-and-Error Procedure

We earlier computed AR^{2/3} = 12.33 for the design curve procedure. By using the data for the channel, we obtain the following expressions for A and R:
\quad\quad\quad\quad A=\frac{1}{2}y_{n}(10+10+2sy_{n})\\\quad\quad\quad\quad\quad =y_{n}(10+2y_{n})
\quad\quad\quad\quad P=B+2\sqrt{s^{2}+1y_{n}}\\\quad\quad\quad\quad\quad=10+4.47 y_{n}
\quad\quad\quad\quad R=\frac{y_{n}(10+2y_{n})}{10+4.47y_{n}}
Now, substituting these expressions for A and R into AR^{2/3} = 12.33 and simplifying the resulting equation, we obtain
\quad\quad\quad\quad \left[y_{n}(20+2y_{n})\right]^{5/3}-12.33(10+4.47y_{n})^{2/3} = 0
A trial-and-error solution of this equation yields
\quad\quad\quad\quad\quad\quad\quad\quad y_{n} = 1.09 m

Numerical Method

To compute the normal depth in a trapezoidal channel, computer programs are developed using the Newton and the bisection methods. An initial estimate for the flow depth in the Newton method is used as 0.5 m; and YL = 0.5 m and YR = 10 m are used as initial estimates for the bisection method. The normal depth computed by both of these methods is 1.09 m.