Question 10.13: A rectangular channel 8 m wide, with a flow rate of 30 m^3/s...

A rectangular channel 8 m wide, with a flow rate of 30 m^3/s, encounters a 4-m-high sharpedged dam, as shown in Fig. E10.13a. Determine the water depth 2 km upstream if the channel slope is S_0 = 0.0004 and n = 0.025.

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First determine the head H produced by the dam, using sharp-crested full-width weir theory, Eq. (10.56):

Wide sharp-crested weir: C_d \approx 0.564 + 0.0846 \frac{H}{Y}       for        \frac{H}{Y} \leq 2                                     (10.56)

Q = 30  m^3/s = C_dbg^{1/2}H^{3/2} = \left(0.564 + 0.0846 \frac{H}{4  m}\right) (8  m)(9.81  m/s^2)^{1/2}H^{3/2}

Since the term 0.0846H/4 in parentheses is small, we may proceed by iteration or EES to the solution H ≈ 1.59 m. Then our initial condition at x = 0, just upstream of the dam, is y(0) = Y + H = 4 + 1.59 = 5.59 m. Compare this to the critical depth from Eq. (10.30):

y = y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{Q^2}{b^2g}\right)^{1/3}                                          (10.30)

y_c = \left(\frac{Q^2}{b^2g}\right)^{1/3} = \left[\frac{(30  m^3/s)^2}{(8  m)^2(9.81  m/s^2)}\right]^{1/3} = 1.13  m

Since y(0) is greater than y_c, the flow upstream is subcritical. Finally, for reference purposes, estimate the normal depth from the Chézy equation (10.19):

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

Q = 30  m^3/s = \frac{\alpha}{n} by R^{2/3}_h S^{1/2}_0 = \frac{1.0}{0.025} (8  m)y_n \left(\frac{8y_n}{8 + 2y_n}\right)^{2/3} (0.0004)^{1/2}

By trial and error or EES solve for y_n ≈ 3.20 m. If there are no changes in channel width or slope, the water depth far upstream of the dam will approach this value. All these reference values y(0), y_c,  and  y_n are shown in Fig. E10.13b.

Since y(0) > y_n > y_c, the solution will be an M-1 curve as computed from gradually varied theory, Eq. (10.51), for a rectangular channel with the given input data:

\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} \approx \frac{S_0 – n^2Q^2/(\alpha^2 A^2 R^{4/3}_h)}{1 – Q^2b_0/(gA^3)}             \alpha = 1.0            A = 8y              n = 0.025               R_h = \frac{8y}{8 + 2y}               b_0 = 8

Beginning with y = 5.59 m at x = 0, we integrate backward to x = -2000 m. For the Runge-Kutta method, four-figure accuracy is achieved for Δx = -100 m. The complete solution curve is shown in Fig. E10.13b. The desired solution value is

At x = -2000 m:                    y ≈ 5.00 m

Thus, even 2 km upstream, the dam has produced a “backwater” that is 1.8 m above the normal depth that would occur without a dam. For this example, a near-normal depth of, say, 10 cm greater than y_n, or y ≈ 3.3 m, would not be achieved until x = -13,400 m. Backwater curves are quite far-reaching upstream, especially in flood stages.

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