Question 10.9: Let us extend the data of Example 10.5 to compute a portion ...

In Example 10.5 we computed y_c = 4.27 ft. Since our initial depth y = 3 ft is less than y_c, we know the flow is supercritical. Let us also compute the normal depth for the given slope S_0 by setting q = 50 ft^3/(s · ft) in the Chézy formula (10.19) with R_h = y_n:

Q = V_0A \approx \frac{\alpha}{n} AR_h^{2/3}S^{1/2}_0                                       (10.19)

Solve for:                              y_n ≈ 4.14 ft

Thus both y(0) = 3 ft and y(L) = 4 ft are less than y_n, which is less than y_c, so we must be on an S-3 curve, as in Fig. 10.14a. For a wide channel, Eq. (10.51) reduces to

\frac{dy}{dx} = \frac{S_0 – n^2Q^2/(\alpha^2 A^2R^{4/3}_h)}{1 – Q^2b_0/(gA^3)}                                    (10.51)

\frac{dy}{dx} = \frac{S_0 – n^2q^2/(\alpha^2 y^{10/3})}{1 – q^2/(gy^3)} \approx \frac{0.0048 – (0.022)^2(50)^2/(2.208y^{10/3})}{1 – (50)^2/(32.2y^3)}                          with y(0) = 3 ft

The initial slope is y′(0) ≈ 0.00494, and a step size Δx = 5 ft would cause a change Δy ≈ (0.00494)(5 ft) ≈ 0.025 ft, less than 1 percent. We therefore integrate numerically with Δx = 5 ft to determine when the depth y = 4 ft is achieved. Tabulate some values:

The water depth, still supercritical, reaches y = 4 ft at

x ≈ 230 ft downstream

We verify from Fig. 10.14a that water depth does increase downstream on an S-3 curve. The solution curve y(x) is shown as the bold line in Fig. E10.9b.

For little extra effort we can investigate the entire family of S-3 solution curves for this problem. Figure E10.9b also shows what happens if the initial depth is varied from 0.5 to 3.5 ft in increments of 0.5 ft. All S-3 solutions smoothly rise and asymptotically approach the uniform flow condition y = y_n = 4.14 ft.