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QA check silt loam design review 6 acres

Trap-efficiency check for an existing sediment basin: Camp's equation, back-solved

A common scenario: the basin geometry comes from a prior project with similar drainage area. The reviewer wants the trap-efficiency calc before submittal. Given As and Q, back-solve Camp's equation for the capture particle, then run the 7-bin Stokes/Camp to get the weighted trap efficiency.

Given: 18,000 ft² basin, 14 cfs peak (10-yr), silt-loam soil. Back-solved capture particle: 0.015 mm at 100% (with 1.2 short-circuit factor). 7-bin trap efficiency: 67.0% — below NCDEQ's 80%. The basin captures the design particle fine; the deficit is the fine-silt and clay tails of the PSD. Add a forebay or PAM and the system clears 80%.

What you have

The basin geometry and design hydrology are fixed (someone else did the prior work). Your job is to verify it before submitting:

ParameterValueSource
Surface area, As18,000 ft²Existing geometry
Sediment storage volume12,000 ft³Below permanent pool
Drainage area6.0 acDisturbed
Design storm10-yr, 24-hrNCDEQ standard
Design peak inflow, Q14 cfsTR-55, computed separately
Soil texturesilt loamSSURGO Web Soil Survey
Influent PSDsilt-loam referenceNRCS texture triangle

Step 1 — Surface overflow rate

The fundamental Camp equation: a particle is captured if its settling velocity vs exceeds the surface overflow rate Q/As:

$\frac{Q}{A_s} = \frac{14 \text{ cfs}}{18{,}000 \text{ ft}^2} = 7.78 \times 10^{-4} \text{ ft/s}$

Convert to SI for the Stokes solve:

$\frac{Q}{A_s} = 7.78 \times 10^{-4} \div 3.281 = 2.37 \times 10^{-4} \text{ m/s}$

Step 2 — Back-solve Stokes for the capture particle

Stokes' law for spherical sediment in 20°C water, ρs = 2650 kg/m³, ρ = 1000, μ = 0.001 Pa·s:

$v_s = \frac{g(\rho_s - \rho)d^2}{18 \mu}$

Solve for d:

$d^2 = \frac{18 \mu v_s}{g(\rho_s - \rho)} = \frac{18 \cdot 0.001 \cdot 2.37 \times 10^{-4}}{9.81 \cdot 1650} = \frac{4.27 \times 10^{-6}}{16{,}187}$
$d = \sqrt{2.64 \times 10^{-10}} = 1.62 \times 10^{-5} \text{ m} = 0.0162 \text{ mm}$

Apply Camp's 1.2 short-circuiting factor (real basins don't behave like ideal plug flow):

$d_{design} = 0.0162 \div \sqrt{1.2} = 0.0148 \text{ mm}$

The basin captures 100% of all particles ≥ 0.0148 mm. That's finer than NCDEQ's 0.020 mm standard criterion — a good sign for design particle, but not the whole story.

Step 3 — 7-bin trap efficiency on a silt-loam PSD

The single design particle isn't the answer; you need the weighted efficiency across the full grain-size distribution. With As/Q = 1,285.7 (s/ft × 1):

Bind (mm)vs (ft/s)% of yieldηiTrapped fraction
1 (sand)0.5002.30 × 10⁻¹8%≫ 1100%
2 (very fine sand)0.1002.95 × 10⁻²12%38100%
3 (coarse silt)0.0507.38 × 10⁻³18%9.5100%
4 (medium silt)0.0201.18 × 10⁻³22%1.52100%
5 (fine silt)0.0102.95 × 10⁻⁴15%0.37937.9%
6 (very fine silt)0.0057.38 × 10⁻⁵12%0.09499.5%
7 (clay)0.0021.18 × 10⁻⁵13%0.01521.5%
Weighted trap efficiency100%67.0%

Sum: 60% (bins 1-4) + 5.69% (bin 5) + 1.14% (bin 6) + 0.20% (bin 7) = 67.0%.

The basin captures the design particle but not the regulatory target. The 13% mass in the <0.002 mm clay fraction passes through nearly untouched (1.5% capture). This is structural — gravity basins can't fix it.

Step 4 — Closing the gap to 80%

To reach 80% trap efficiency on this silt-loam influent, you have to capture more of the fine-silt fraction. The basin already captures 100% of bins 1–4; the deficit lives entirely in bins 5–7.

OptionMechanismResulting η
PAM dosing at 5 ppmFlocculates fine silt and clay into ~0.05 mm aggregates; bins 5–7 effectively captured~95%
Increase As to 36,000 ft² (2× larger)η5 rises 38% → 76%; bins 6–7 still pass through; net rises modestly~75%
Increase As to 72,000 ft² (4× larger)Sized for the 0.010 mm bin; bin 5 now 100% captured, bin 6 partial~83%
Add a 12% forebay (cleanout zoning)Concentrates coarse for easy maintenance — does not meaningfully raise η; both cells see the same Q so coarse the forebay catches was going to be caught anyway. See the worked counter-example.~67% (no change)

Recommendation: PAM dosing is the cheapest and most reliable option for this basin. If the project doesn't allow chemical addition, the only path to 80% is a 4× larger main basin — which often forces a re-layout of the site.

What changes if you tweak the inputs

If you change…The result moves…
Q = 14 → 22 cfs (25-yr storm)Capture particle moves coarser to 0.0186 mm; trap efficiency drops to 60.5%
Soil silt loam → sandy loam (more sand in PSD)Higher % in bins 1-3; trap efficiency rises to ~78% with same basin
As = 18,000 → 27,000 ft² (50% larger)Q/As drops; capture particle 0.0148 → 0.0121 mm; η rises to ~72%
Add forebay (12% main As)System η rises ~16 percentage points from forebay capture of bins 1–3 alone

Run the back-solve in HydroComplete

Drop in any A_s and Q — get capture particle, full 7-bin trap efficiency, and the forebay/PAM what-if in one screen.

Open in app 14-day free trial

Sources and further reading

— Michael Flynn, PE
When the basin geometry is fixed, the design problem becomes "what does this actually catch." Camp's equation runs both directions; the second direction is the one the reviewer asks about.

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