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Question 12.12: The rectangular channel in Fig. 12–27a is made of unfinished......

The rectangular channel in Fig. 12–27a is made of unfinished concrete and has a slope of S_0 = 0.035. When it reaches the check dam, the flow backs up as shown in Fig. 12–27b. At a specific location before the dam, the water has a depth of 1.25 m and the flow is Q = 0.75 m³/s. Classify the surface profile for the flow. Take n = 0.014.

fig 12-27b
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Fluid Description.   The flow is steady and nonuniform. The water is considered incompressible.
Analysis.   To classify the surface profile, we must determine the critical depth y_c, the normal flow depth y_n, and the critical slope S_c. Using Eq. 12–7, with q = Q/b = (0.75 m³/s)/(2 m) = 0.375 m²/s,

y_c  =  (\frac{q^2}{g})^{1/3}  =  [\frac{(0.375  m^2/s)^2}{9.81  m/s^2}]^{1/3}  =  0.2429  m

Since y = 1.25 m > y_c = 0.2429 m, the flow is subcritical. The depth y_n that will produce normal or uniform flow for Q = 0.75 m³/s is determined from the Manning equation. Since
P = (2y_n + 2 m)
then Eq. 12–20 becomes

Q  =  \frac{kA^{5/3} S_0 ^{1/2}}{n P^{2/3}};                         0.75  m^3 /s  =  \frac{(1)[(2  m)y_n]^{5/3} (0.035)^{1/2}}{0.014(2y_n  +  2  m)^{2/3}}

 

                                  \frac{y_n^{5/3}}{(2y_n  +  2  m)^{2/3}}  =  0.017678

Solving by trial and error, or using a numerical
procedure, we get
y_n = 0.1227 m
The critical slope is now determined from Eq. 12–22,

S_c  =  \frac{n^2gA_c}{k^2b_{top}R_{hc}^{4/3}}  =  \frac{(0.014)^2 (9.81  m/s^2)(2  m)(0.2429  m)}{(1)^2 (2  m)[\frac{2  m(0.2429  m)}{2(0.2429  m)  +  2  m}]^{4/3}}

=  0.004118

Since y = 1.25 m > y_c > y_n and S_0 = 0.035 >  S_c, then according to Table 12–2 the water surface in Fig. 12–27 is an S1 profile.

TABLE 12–2 Surface Profile Classification

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