Question 15.13: A culvert made of unfinished concrete and measuring 2 m × 2 m...

The culvert is sketched in Figure 15.38. From Table 15.1 the Manning coefficient for unfinished concrete is n = 0.014. Since we know that the slope of the channel is described in terms of an elevation change over its length, we can use Eq. 15.34 to calculate the bed slope as S_B = (h₁ − h₂)/L = 0.5/100 = 0.005.

S_B = tan θ = \frac{h_1-h_2}{L}              (15.34)

The volume flowrate is given by Eq. 15.53b, Q = (C₀/n)AR^{2/3} _H S^{1/2} _B, with C₀ = 1, since we are working in SI units. Next we use Figure 15.6 to write the formulas A(y) = wy and R_H(y) = wy/(w + 2y). Inserting these into the formula for Q and rearranging yields

(wy)\left(\frac{wy}{w+2y} \right)^{2/3} =\frac{nQ}{S^{1/2}_B}                                    (A)

The desired depth satisfies this equation. Inserting Q = 4.7 m³/s, n = 0.014, S_B = 0.005, and w = 2 m, we obtain

2y\left(\frac{2y}{2+2y} \right)^{2/3} =0.931

as the equation to be solved. Using a symbolic math code, we find y = 0.8 m.

We see that a flow of 4.7 m³/s in this culvert occurs at a depth of 0.8 m. The flow area in that case is A = wy = 2 m (0.8 m ) = 1.6 m², and the velocity is V = Q/A = 4 .7 m³/s/1.6 m² = 2.94 m/s. The Froude number is Fr = V/\sqrt{gy} = 2.94 m/s/\sqrt{(9.81\ m/s^2)(0.8\ m)} = 1.05. Thus the flow is supercritical.

Note that to iterate this problem by hand we can recognize that the answer must be a depth of 2 m or less. Several iterations are shown:

for y = 0.6:

2y\left(\frac{2y}{2+2y} \right)^{2/3} =0.624

for y = 0.9:

2y\left(\frac{2y}{2+2y} \right)^{2/3} =1.33

for y = 0.7:

2y\left(\frac{2y}{2+2y} \right)^{2/3} =0.77

for y = 0.8:

2y\left(\frac{2y}{2+2y} \right)^{2/3} =0.932

TABLE 15.1 Values of the Manning Roughness Coefficient, n