M2Q10 PART #1 (Numerical Answer)

A river has a width B = 3 m and an average normal depth dn = 0.87 m. The characteristic sediment size is ds = 92 mm. The riverbed exhibits bedforms such that skin friction contributes 60% of the total bed shear and the remaining 40% is due to form drag. PART #1 (numerical calculations): Assuming uniform flow, what is the minimum slope to initiate bed load motion of the sediment? (Use the Shield diagram below if necessary; assume the relative density s = 2.6, water density
𝜌
= 1000 kg/m3, viscosity 10-6 m2/s). (Answer with three decimal digits, e.g. 0.999).

Question

M2Q10 PART #1 (Numerical Answer)

A river has a width B = 3 m and an average normal depth dn =  0.87 m. The characteristic sediment size is ds = 92 mm. The riverbed exhibits bedforms such that skin friction contributes 60% of the total bed shear and the remaining 40% is due to form drag. PART #1 (numerical calculations): Assuming uniform flow, what is the minimum slope to initiate bed load motion of the sediment? (Use the Shield diagram below if necessary; assume the relative density s = 2.6, water density 
𝜌
 = 1000 kg/m3, viscosity 10-6 m2/s). (Answer with three decimal digits, e.g. 0.999).

BlackTom AI · Accepted Answer

We start by restating the problem: a river with width B = 3 m, average normal depth dn = 0.87 m, sediment ds = 92 mm (0.092 m). The bed experiences skin friction that accounts for 60% of the total bed shear, with the remaining 40% due to form drag. We are asked for the minimum bed slope S (uniform flow) that initiates bed load motion, using the Shields criterion. Relative density is s = 2.6, water density ρ = 1000 kg/m^3, and kinematic viscosity ν = 1e-6 m^2/s. The answer should be given to three decimal places.

Step 1: Express bed shear stress in terms of slope for uniform flow. The bed shear stress is τ_b = ρ g dn S, where g ≈ 9.81 m/s^2, dn is the depth, and S is the energy (or bed) slope. The factor ρ g dn S represents the downslope shear stress acting on the bed.

Step 2: Account for the skin-friction vs. form-drag split. The problem states skin friction contributes 60% of the total bed shear τ_b, and form drag contributes 40%. If only the skin-friction portion contributes to initiating motion (a common interpretation when partitioning drag sources for sediment mobilization), then the effective shear stress driving initiation is τ_eff = 0.60 × τ_b = 0.60 × ρ g dn S.

Step 3: Use the Shields parameter to relate τ_eff to sediment entrainment. The Shields parameter is θ = τ_eff / [ (ρ_s − ρ) g d_s ], where d_s is the grain diameter and ρ_s is the sediment density. The relative density is s = ρ_s / ρ = 2.6, so ρ_s − ρ = (s − 1)ρ = 1.6 ρ. Substituting τ_eff gives
θ = [0.60 ρ g d_n S] / [ (ρ_s − ρ) g d_s ] = [0.60 d_n S] / [ (s − 1) d_s ].

Step 4: Substitute numerical values. d_n = 0.87 m, s − 1 = 1.6, d_s = 0.092 m:
θ = (0.60 × 0.87 × S) / (1.6 × 0.092) = (0.522 S) / 0.1472 ≈ 3.547 S.

Step 5: Determine the critical Shields value θ_cr. The Shields diagram (or a numerical correlation) provides θ_cr as a function of the boundary Reynolds number Re_* = u_* d_s / ν, with u_* derived from the bed shear. Here, u_* is related to τ_b by τ_b = ρ g d_n S = ρ u_*^2, so u_* = sqrt(g d_n S). Then
Re_* = u_* d_s / ν = [sqrt(g d_n S) × d_s] / ν.
Using g ≈ 9.81, d_n = 0.87, d_s = 0.092, ν = 1e-6, we have Re_* = [sqrt(9.81 × 0.87 × S) × 0.092] / 1e-6 ≈ [sqrt(8.537 × S) × 0.092] × 10^6 ≈ (2.92 √S × 0.092) × 10^6 ≈ 0.269 √S × 10^6. Thus Re_* is very large for typical S, placing us in the high-Re regime of the Shields curve, where θ_cr tends to a value on the order of a few times 10^-2 to a few times 10^-1 depending on grain size and Reynolds effects. For ds = 0.092 m (coarse gravel), θ_cr is commonly around 0.04–0.07 in many practical cases, though exact values depend on the chosen correlation or chart.

Step 6: Solve for the slope using θ ≈ θ_cr. Since θ ≈ 3.547 S and θ_cr is typically in the range ~0.04–0.07 for this grain size, the slope that initiates motion is
S ≈ θ_cr / 3.547.
If we take a representative θ_cr ≈ 0.04, then S ≈ 0.04 / 3.547 ≈ 0.0113. If θ_cr ≈ 0.05, then S ≈ 0.0141. If θ_cr ≈ 0.06, then S ≈ 0.0169. These estimates place the minimum slope on the order of 0.01–0.02, depending on the exact θ_cr adopted from the Shields relation for ds = 92 mm and the very high Re_* regime.

Step 7: Reconcile with the provided numerical answer. The problem specifies rounding to three decimals (e.g., 0.999). The range derived above (approximately 0.011–0.017) would round to 0.011, 0.012, up to 0.017, i.e., around 0.012–0.017 when expressed to three decimals. If an external reference or chart yields a slightly lower θ_cr in the high-Re gravel regime, the resulting S could appear near 0.010–0.015. The provided single-value answer of 0.002 is significantly smaller than these typical estimates for ds ≈ 92 mm and the given density/viscosity parameters; such a small slope would yield θ ≈ 0.007, well below usual θ_cr for this sediment size.

Summary of reasoning (without stating the final numeric value): we related the bed shear to the slope via τ_b = ρ g d_n S, adjusted for skin-friction contribution to obtain τ_eff = 0.60 ρ g d_n S, expressed θ = τ_eff / [(ρ_s − ρ) g d_s] as θ = 3.547 S, and then used the Shields criterion θ = θ_cr to solve for S. The critical θ_cr for ds = 0.092 m in the high-Re regime typically lies in the ~0.04–0.07 range, yielding a slope on the order of 0.01–0.02, not as small as 0.002 based on common Shields correlations. The precise value depends on the chosen θ_cr curve and how the high-Re gravel regime is interpreted in the chart provided.

CVEN90051_2025_SM2 Exam: Civil Hydraulics (CVEN90051_2025_SM2)- Requires Respondus LockDown Browser

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