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## Q. 15.14

A trapezoidal ﬂood control channel has a relatively clean bottom but weedy sides, as shown in Figure 15.39A. Find an expression for the equivalent Manning coefficient nE for this channel.

## Verified Solution

We can think of this channel ﬁrst as a whole and suppose that it has an equivalent Manning coefficient nE, a ﬂow area AE, and a hydraulic radius RHE. From the geometry in Figure 15.39A and the formulas in Figure 15.6 for a trapezoidal channel at normal depth yN, we have

AE = yN(w + yN cot θ)          and        $R_{HE}=\frac{y_N(w+y_N \cot θ) }{w+2y_N/\sin θ}$

By using Eq. 15.53b, $Q = (C_0/n)AR^{2/3}_ H S^{1/2} _B$, we can write the volume ﬂowrate for the whole channel as

$Q_E= \frac{C_0}{n_E} A_ER^{2/3}_{HE} S^{1/2} _B$                                         (A)

The velocity in the channel is VE = QE/AE.

Next we can think of the channel as if it were three separate channels, as indicated in Figure 15.39B, with the same velocity VE in each of them. The ﬂowrates in the three channels are then seen to be QI = VE AI, QII = VE AII, and QIII = VE AIII. The total volume ﬂowrate of the whole channel must then be given by the sum of the ﬂows occurring in the three subsections. We can write this as

QE = QI + QII + QIII                                      (B)

Finally we will assume that the volume ﬂowrate in each of the subsections may be calculated separately as a uniform open channel ﬂow in each subsection, with the ﬂow corresponding to the characteristics of the particular subsection.

Two of these channels are triangular and can be seen to have a single wetted side of length l = yN/sin θ and a free surface width s = yN/tan θ. Thus they have a ﬂow area AI = AIII = $y^2_ N$/(2 tan θ) and a wetted perimeter of PI = PIII = l = yN/ sin θ. The hydraulic radius of each of the triangular channels is RHI = RHIII = $\frac{1}{2}$yN cos θ. By using Eq. 15.53b, and writing the Manning coefficient for the weedy sides as nS, we can give the following volume ﬂowrate in each of these channels:

$Q_I=Q_{III}= \frac{C_0}{n_S} A_IR^{2/3}_{HI} S^{1/2} _B$                                (C)

The central channel can be seen to have a ﬂow area AII = wyN, wetted perimeter PII = w, and thus a hydraulic radius RHII = yN. Again from Eq. 15.53b, and writing the Manning coefficient for the clean bottom as nB, we ﬁnd the volume ﬂowrate in the central section:

$Q_{II}= \frac{C_0}{n_B} A_{II}R^{2/3}_{HII} S^{1/2} _B$                                            (D)

Inserting (A), (C), and (D) into (B) and simplifying, we obtain

$\frac{1}{n_E}\left( A_{A}R^{2/3}_{HE}\right)=\frac{21}{n_S}\left( A_{I}R^{2/3}_{HI}\right)+\frac{1}{n_B}\left( A_{II}R^{2/3}_{HII}\right)$                                  (E)

This equation allows us to calculate the equivalent Manning coefficient for the entire channel if we know the parameters describing the three subsections. After ﬁnding nE, we can use (A) to calculate the ﬂowrate in the whole channel. It is interesting to note that (E) is similar to the formula for the equivalent electrical resistance of three resistors in parallel. We see that in uniform open channel ﬂow, Eq. 15.53b, $Q = (C_0/n)AR^{2/3}_ H S^{1/2} _B$, can be thought of as equivalent to Ohm’s law I = V/R, with the volume ﬂowrate, Q, equivalent to the current I, the square root of the bed slope $S^{1/2} _B$, acting as the driving voltage V, and the resistance R equivalent to the combination n/(A$R^{2/3}_ H$ ). Does it make sense to you that the resistance of a channel to ﬂow would increase with Manning coefficient?