M2Q9 PART #1 (Numerical Answer)

The discharge in a river is 32 m3/s. The river width is 10 m, and the uniform depth is 2.9m (Manning's coefficient n = 0.031 and slope = 0.0007). The depth-averaged suspended sediment concentration Cs = 0.76% and the d50 at the river bed is 0.6 mm. PART #1 (numerical calculation): What is the suspended sediment transport rate? (Assume the relative density s = 2.6, water density
𝜌
= 1000 kg/m3, viscosity 10-6 m2/s). (Use the modified shield diagram below if necessary). (Answer in m2/s).

Question

M2Q9 PART #1 (Numerical Answer)

The discharge in a river is 32 m3/s. The river width is 10 m, and the uniform depth is 2.9m (Manning's coefficient n = 0.031 and slope = 0.0007). The depth-averaged suspended sediment concentration Cs = 0.76% and the d50 at the river bed is 0.6 mm. PART #1 (numerical calculation): What is the suspended sediment transport rate? (Assume the relative density s = 2.6, water density 
𝜌
 = 1000 kg/m3, viscosity 10-6 m2/s). (Use the modified shield diagram below if necessary). (Answer in m2/s).

BlackTom AI · Accepted Answer

Here is a step-by-step analysis of the given problem, with all key ideas laid out in a clear sequence.

1) Restate the data we have from the question:
- River discharge Q = 32 m3/s
- River width b = 10 m
- Depth-averaged suspended sediment concentration Cs = 0.76% = 0.0076 (dimensionless, volumetric concentration)
- We are asked for the suspended sediment transport rate in units of m2/s (i.e., volume of solids transported per unit width per unit time).

2) Understand what the requested quantity represents:
- The suspended sediment transport rate per unit width, often denoted S′, can be obtained by dividing the volumetric suspension flux by the channel width: S′ = (volume of sediment transported per second) / width.
- The total volumetric sediment transport rate (volume per unit time of solid material) is Cs multiplied by the total discharge Q, since Cs is the fraction of the flow that is sediment by volume.

3) Compute the total volumetric sediment transport rate:
- Sediment volume flux = Cs × Q = 0.0076 × 32 m3/s = 0.2432 m3/s.

4) Convert to transport per unit width:
- To express this per unit width, divide by the river width b: S′ = (0.2432 m3/s) / (10 m) = 0.02432 m2/s.

5) Check units and plausibility:
- The resulting unit is m2/s, which matches the asked unit for the problem.
- The calculation uses straightforward volumetric concentration scaling and a per-width normalization; no other factors (density, viscosity, d50, or n) are needed for this particular expression, since Cs already accounts for the suspended fraction and Q is the total flow rate.

6) Note on available information:
- The problem statement mentions a modified shield diagram and several other geophysical details, but for the specific numerical quantity asked here (suspended sediment transport rate per unit width), the direct approach Cs × Q divided by b suffices. The other parameters (density, slope, n, d50) influence sediment transport mechanisms and thresholds but are not required for this simple volume-per-width calculation.

7) Final numerical result (as computed):
- S′ = 0.02432 m2/s (approximately 0.0243 m2/s when rounded to four significant figures).

If you needed the total sediment mass flux rather than per-unit-width flux, you would multiply by the sediment density ρs (where ρs ≈ s × ρ = 2.6 × 1000 kg/m3 = 2600 kg/m3) to obtain kg/s. However, for the requested units of m2/s, the per-width volumetric flux calculation above is the appropriate one.

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

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