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Chapter 15

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.

15.39

Step-by-Step

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/(AR^{2/3}_ H ). Does it make sense to you that the resistance of a channel to flow would increase with Manning coefficient?