Program for uniform depth in rectangular channel

The uniform (normal) depth $y_n$ of a rectangular channel is found from Manning's equation. It is non-linear in $y$, so Newton-Raphson is used.

$$f(y)=AR^{2/3}-\frac{nQ}{\sqrt{S_0}}=0,\qquad A=by,\quad P=b+2y,\quad R=\frac AP$$ $$\frac{df}{dy}=R^{2/3}\left[\frac53b-\frac43R\right],\qquad y_{new}=y-\frac{f(y)}{f'(y)}$$

Algorithm

1. Start.
2. Read $b$, $n$, $Q$, $S_0$.
3. Compute the target $K_0=nQ/\sqrt{S_0}$ and take the initial guess $y=1$ m.
4. Repeat steps 5-8 up to 50 times:
5. Compute $A=by$, $P=b+2y$, $R=A/P$.
6. Compute $f=AR^{2/3}-K_0$ and $f'=R^{2/3}[(5/3)b-(4/3)R]$.
7. $y_{new}=y-f/f'$; if $y_{new}\le0$ take $y_{new}=y/2$.
8. If $|y_{new}-y|<10^{-6}$ stop; otherwise set $y=y_{new}$.
9. Print the uniform depth $y_n$.
10. Stop.

Program (C)

#include <stdio.h>
#include <math.h>

/* Newton-Raphson for the uniform (normal) depth of a rectangular channel */
int main(void)
{
    double b, n, Q, S, y, A, P, R, f, dfdy, ynew, target;
    int i;

    printf("Enter b (m), n, Q (m3/s), S0: ");
    scanf("%lf %lf %lf %lf", &b, &n, &Q, &S);

    target = n * Q / sqrt(S);        /* required value of A*R^(2/3) */
    y = 1.0;                         /* initial guess */

    for (i = 1; i <= 50; i++) {
        A = b * y;
        P = b + 2.0 * y;
        R = A / P;
        f = A * pow(R, 2.0 / 3.0) - target;
        /* d(A R^(2/3))/dy = R^(2/3) * ( (5/3) b - (2/3) R * 2 ) */
        dfdy = pow(R, 2.0 / 3.0) * ((5.0 / 3.0) * b - (4.0 / 3.0) * R);
        ynew = y - f / dfdy;
        if (ynew <= 0.0)             /* keep depth positive */
            ynew = 0.5 * y;
        if (fabs(ynew - y) < 1e-6) {
            y = ynew;
            break;
        }
        y = ynew;
    }
    printf("Uniform depth yn = %.4f m (after %d iterations)\n", y, i);
    return 0;
}

Sample run

Input: 5 0.015 15 0.0004 (b = 5 m, n = 0.015, Q = 15 m³/s, S0 = 0.0004)

Output: Uniform depth yn = 2.0708 m (after 4 iterations)

Check: $A=10.354$ m², $P=9.142$ m, $R=1.1325$ m, $AR^{2/3}=11.25$, $nQ/\sqrt{S_0}=11.25$ ✓.