M2Q1

The flow conditions in a natural river are such that the shear stress is
𝜏
0
= 0.085 Pa. The sediment size is 30 mm, the relative density is s = 2.6, and the water density
𝜌
= 1000 kg/m3. Using the formulae of Nielsen (1992; see below), what can we conclude about the bed load sediment transport rate?

Question

M2Q1

The flow conditions in a natural river are such that the shear stress is 
𝜏
0
 = 0.085 Pa. The sediment size is 30 mm, the relative density is s = 2.6, and the water density 
𝜌
 = 1000 kg/m3. Using the formulae of Nielsen (1992; see below), what can we conclude about the bed load sediment transport rate?

BlackTom AI · Accepted Answer

Restating the problem in my own words: we have flow conditions with shear stress tau0 = 0.085 Pa in a natural river. The sediment size is 30 mm, relative density s = 2.6, water density rho = 1000 kg/m^3. Using Nielsen (1992) formulas, we must determine the bed load transport rate q_s.

Option A: Transport rate is about 0.0023 m^2/s. This would imply a finite, positive bed load rate predicted by the equation. To accept this, the right-hand side of Nielsen’s expression must yield a positive value large enough to give q_s / sqrt((s-1) g d_s^3) ≈ some positive number. However, whether this specific numerical value is obtained depends on plugging in the given tau0, rho, s, and d_s. If the right-hand side evaluates to a negative value or a zero, this option would be incorrect.

Option B: Nothing, because the equation cannot be applied as the sediment size is out of range. This would be argued if 30 mm were outside Nielsen’s applicable size range. But Nielsen’s framework is typically applied to a broad range of gravel sizes, and there is no explicit indication in the problem that 30 mm is outside its applicability range. Without a stated range violation, this explanation is not justified.

Option C: Rate is zero because the equation returns a negative value. This hinges on evaluating the Nielsen expression on the right-hand side: if the bracketed term [12 tau0 /( rho (s-1) g d_s) - 0.05] becomes negative for the given tau0, rho, s, and d_s, then the entire q_s expression would be set to zero (since a negative transport rate is not physical). The structure of the equation often implies that when the forcing parameter is below threshold, bed load transport is zero. If the computed bracket is negative, this would indeed yield a zero transport rate.

Option D: Bed load is very small but not negligible. This would be the case if the right-hand side is positive but small, giving a small positive q_s. Without performing the actual numeric evaluation, this cannot be confirmed. If the math yields a positive but tiny value, this option would be plausible; otherwise it would be incorrect.

Now, examining the structure of Nielsen’s formula: the left-hand side q_s divided by sqrt((s-1) g d_s^3) equals a function of tau0, rho, s, and d_s on the right. The right-hand side includes a term (12 tau0)/(rho (s-1) g d_s) minus 0.05, multiplied by sqrt(tau0 /(rho (s-1) g d_s)). Depending on the given tau0, s, d_s, and rho, the first factor may be less than 0.05, making the bracket negative. If so, the right-hand side would be negative, and since a transport rate cannot be negative, the physically consistent result is q_s = 0. Hence Option C becomes the correct interpretation under that condition.

In summary, the most consistent interpretation with the given formulation and typical threshold behavior of bed load transport is that the equation yields a negative value for the driving term, leading to zero transport rate. The other options either assert a concrete positive rate without confirmation, rely on the idea of the formula being inapplicable without evidence, or claim a small but nonzero rate without numeric support. The logic hinges on the sign of the bracketed expression in Nielsen’s right-hand side and the convention that transport cannot be negative.

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

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