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## Q. 15.2

Water ﬂows through a horizontal rectangular channel as shown in Figure 15.15A before encountering a depression 1 ft deep. If the ﬂow depth and speed upstream are 4 ft and 1 ft/s, respectively, ﬁnd the water depth and ﬂow speed over the depression. ## Verified Solution

We can solve this problem by using equation 15.7:

$\frac{1}{2g}\left[\frac{V_1y_1}{y(x)} \right]^2+y(x)+h(x) = \frac{V^2_1}{2g} +y_1 +h(x_1)$

where point 1 is upstream and x is the location of the arbitrary second point. At a point anywhere along the depression where the bed height is z = h2, the water depth is y2. Upstream the bed height is z = h1 and the water depth is y1. Thus we have the following relationship between variables at these two points:

$\frac{1}{2g}\left[\frac{V_1y_1}{y_2} \right]^2+y_2+h_2 = \frac{V^2_1}{2g} +y_1 +h_1$

Multiplying this equation by $y^2 _2$ and rearranging, we have

$y^3_2 +y^2_2\left[h_2-\frac{V^2_1}{2g} -y_1-h_1\right]+\frac{1}{2g}(V_1y_1)^2=0$

It is convenient to choose the coordinate origin and datum elevation at the level of the upstream bed as shown in Figure 15.15A. Then taking point 2 in the depression, we have h1 = 0 ft , h2 = −1 ft, and h2 − h1 = −1 ft. The remaining known values are V1 = 1 ft/s , y1 = 4 ft. Inserting these values into the preceding equation, we obtain

$y^3_2 +y^2_2\left[-1\ \mathrm{ft}-\frac{(1\ \mathrm{ft}/s)^2}{2(32.2\ \mathrm{ft}/s^2)} -4\ \mathrm{ft}-0\ \mathrm{ft}\right]+\frac{1}{2(32.2\ \mathrm{ft}/s^2)}[(1\ \mathrm{ft}/s)(4\ \mathrm{ft})]^2=0$

as the equation to be solved to determine the value of y2. This equation simpliﬁes to

$y^3 _2 −(5.015\ \mathrm{ft})y^2 _2 +0.2484\ \mathrm{ft}^3 =0$

Solving this cubic equation yields three possible values of the water depth: − 0.217 ft, +0.228 ft, and +5.00 ft. The negative value is not physically possible so can be ignored. But how do we decide whether the water depth is +0.228 ft, or +5.00 ft? The answer can be found by recalling that the behaviors of the depth slope and the free surface slope depend on the value of the local Froude number. We can use Eq. 15.8 to calculate the local Froude number upstream of the depression:

$Fr_1=\frac{V(x)}{\sqrt{gy(x)} }$             (15.8)

$Fr_1=\frac{V_1}{\sqrt{gy_1} }=\frac{1\ \mathrm{ft}/s}{\sqrt{\left(32.2\ \mathrm{ft}/s^2\right)(4\ \mathrm{ft}) } }=0.88$

Since the ﬂow is subcritical the water depth must increase (see Eq. 15.9a or Figure 15.14C). We conclude that the correct answer in this case is a water depth over the depression of 5.00 ft, as illustrated in Figure 15.15B. The ﬂow speed over the depression can be determined by using Eq. 15.4: V(x) = V1y1/y(x). Inserting the data, we have

$\frac{dy}{dx}=\frac{−dh/dx }{[1−Fr^2]}$                                (15.9a)

$V_2=\frac{(1\ \mathrm{ft}/s)(4\ \mathrm{ft} )}{5.00\ \mathrm{ft}}=0.8\ \mathrm{ft}/s$

Before concluding this example, notice carefully that if you choose to use a different datum from which to measure bed height, say one such that the upstream bed height is z = h0, then you would write h1 = h0, h2 = h0 − 1 ft, and get h2 − h1 = (h0 − 1 ft ) − h0 = −1 ft. This leads to the same equation for y2, and thus the same answer. This is to be expected, since the choice of datum should not affect the answer.