A channel of rectangular cross section (width 2m) has a longitudinal slope of 0.0012. The channel bed consists of a mixture of fine sands (D50 = 0.2 mm); Manning's parameter n =0.01. The flow rate is Q = 4 m3/s and uniform flow conditions are achieved; water temperature is 20 C.

Will bed-load motion take place (assume s = 1800 kg/m3)?

Question

A channel of rectangular cross section (width 2m) has a longitudinal slope of 0.0012. The channel bed consists of a mixture of fine sands (D50 = 0.2 mm); Manning's parameter n =0.01. The flow rate is Q = 4 m3/s and uniform flow conditions are achieved; water temperature is 20 C.

Will bed-load motion take place (assume s = 1800 kg/m3)?

BlackTom AI · Accepted Answer

We start by restating the problem: a rectangular channel with width 2 m, bed slope 0.0012, sand D50 = 0.2 mm, Manning n = 0.01, discharge Q = 4 m^3/s, and temperature 20 C. We ask whether bed-load motion will take place (s = 1800 kg/m^3).

Option A: True. To assess bed-load motion, we estimate the flow depth y from Manning’s equation and the given discharge, then compute the bed shear stress and the Shields parameter.
- Step 1: Solve for depth using Manning’s equation for a wide channel. Discharge Q = v A with A = b y = 2y, and wetted perimeter P ≈ b + 2y ≈ 2 + 2y, so hydraulic radius R = A/P = 2y/(2+2y) = y/(1+y).
- Step 2: Velocity from Manning: v = (1/n) R^(2/3) S^(1/2), where S is the slope (0.0012). Using a trial depth around y ≈ 0.95 m gives A ≈ 1.90 m^2, P ≈ 3.90 m, R ≈ 0.487, R^(2/3) ≈ 0.613, S^(1/2) ≈ 0.0346. Thus v ≈ (1/0.01) × 0.613 × 0.0346 ≈ 2.12 m/s. Then Q ≈ v A ≈ 2.12 × 1.90 ≈ 4.0 m^3/s, consistent with the given Q. So a depth on the order of about 0.95 m is reasonable.
- Step 3: Bed shear stress: τ_b = ρ g R S (a common expression for bed shear in open-channel flow). With ρ ≈ 1000 kg/m^3, g ≈ 9.81 m/s^2, R ≈ 0.487, and S = 0.0012, we get τ_b ≈ 1000 × 9.81 × 0.487 × 0.0012 ≈ 5.7 Pa.
- Step 4: Shields parameter: θ = τ_b / [ (ρ_s − ρ) g d50 ]. Here ρ_s = 1800 kg/m^3, ρ ≈ 1000 kg/m^3, d50 = 0.0002 m. The denominator is (800) × 9.81 × 0.0002 ≈ 1.57. Therefore θ ≈ 5.7 / 1.57 ≈ 3.6.
- Step 5: Interpretation. A Shields parameter on the order of a few units (≫ the typical critical Shields value around 0.03–0.05 for incipient motion) clearly indicates that bed-load motion would occur. Thus, bed-load motion taking place is consistent with the given data.

Option B: False. If one argued that bed-load motion would not occur, one would have to show that the bed shear stress is below the critical value or that the Shields parameter is below its critical value. However, using the estimated depth from Manning’s equation for the given discharge yields a substantial bed shear stress and a Shields parameter well above typical critical values, indicating incipient or ongoing bed-load movement. The alternative conclusions would require incorrect depth estimation, incorrect use of Manning’s equation, or an erroneous interpretation of Shields parameters, none of which align with the provided data.

In summary, the calculated bed shear stress and Shields parameter strongly support bed-load motion under the stated conditions; therefore, the statement 'True' is the correct interpretation based on these analyses. The alternative 'False' rests on misapplication or misinterpretation of the flow-depth or transport criteria, which is inconsistent with the quantitative checks above.

CVEN90051_2025_SM2 Module #2 Practice Quiz

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