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Critical stress for Re > 400 and critical diameter

Short answer

For Re* > 400, τ* ≈ 0.06, so τc = 0.06(γs−γ)d. Equating τc to actual bed shear γRS0 yields d_cr = γRS0 / [0.06(γs−γ)]. For quartz sediment (specific gravity 2.65), this simplifies to d_cr ≈ 10 R S0.

Derive an expression for critical stress for Re* > 400 (fully developed turbulent flow). Hence, finally express critical diameter of sediment in terms of geometric properties of channel (i.e. prove dcr=10RS0d_{cr} = 10 R S_0, where R is hydraulic radius and S0S_0 is bed slope).

Critical shear stress for Re∗>400Re_* > 400

From the Shields diagram, the dimensionless critical shear stress is

τ∗=τc(γs−γ) d\tau_* = \frac{\tau_c}{(\gamma_s - \gamma)\,d}

For Re∗=u∗dν>400Re_* = \dfrac{u_*d}{\nu} > 400 (fully rough, fully developed turbulent flow) the viscous sublayer is negligible and τ∗\tau_* becomes independent of Re∗Re_* and equal to a constant:

τ∗≈0.056≈0.06\tau_* \approx 0.056 \approx 0.06

Therefore the critical shear stress is

τc=0.06 (γs−γ) d\boxed{\tau_c = 0.06\,(\gamma_s - \gamma)\,d}

Critical diameter

The average boundary shear stress in uniform flow is

τ0=γRS0\tau_0 = \gamma R S_0

At the limit of stability τ0=τc\tau_0 = \tau_c for the critical diameter dcrd_{cr}:

γRS0=0.06 (γs−γ) dcr\gamma R S_0 = 0.06\,(\gamma_s - \gamma)\,d_{cr} dcr=γRS00.06 (γs−γ)=RS00.06 (s−1)d_{cr} = \frac{\gamma R S_0}{0.06\,(\gamma_s - \gamma)} = \frac{R S_0}{0.06\,(s - 1)}

where s=γs/γs = \gamma_s/\gamma is the specific gravity of the sediment. For quartz sand and gravel, s=2.65s = 2.65, so s−1=1.65s - 1 = 1.65:

dcr=RS00.06×1.65=10.1 RS0d_{cr} = \frac{R S_0}{0.06 \times 1.65} = 10.1\,R S_0 dcr≈10 R S0\boxed{d_{cr} \approx 10\,R\,S_0}

with dcrd_{cr}, RR in the same units (for example metres). Grains larger than dcrd_{cr} remain at rest; smaller grains move.

✓ Answer checked by Claude Opus 5.5 (high effort)

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