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 A_s 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:

Parameter	Value	Source
Surface area, A_s	18,000 ft²	Existing geometry
Sediment storage volume	12,000 ft³	Below permanent pool
Drainage area	6.0 ac	Disturbed
Design storm	10-yr, 24-hr	NCDEQ standard
Design peak inflow, Q	14 cfs	TR-55, computed separately
Soil texture	silt loam	SSURGO Web Soil Survey
Influent PSD	silt-loam reference	NRCS texture triangle

Step 1 — Surface overflow rate

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

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

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, ρ_s = 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 A_s/Q = 1,285.7 (s/ft × 1):

Bin	d (mm)	v_s (ft/s)	% of yield	η_i	Trapped fraction
1 (sand)	0.500	2.30 × 10⁻¹	8%	≫ 1	100%
2 (very fine sand)	0.100	2.95 × 10⁻²	12%	38	100%
3 (coarse silt)	0.050	7.38 × 10⁻³	18%	9.5	100%
4 (medium silt)	0.020	1.18 × 10⁻³	22%	1.52	100%
5 (fine silt)	0.010	2.95 × 10⁻⁴	15%	0.379	37.9%
6 (very fine silt)	0.005	7.38 × 10⁻⁵	12%	0.0949	9.5%
7 (clay)	0.002	1.18 × 10⁻⁵	13%	0.0152	1.5%
Weighted trap efficiency			100%	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.

Option	Mechanism	Resulting η
PAM dosing at 5 ppm	Flocculates fine silt and clay into ~0.05 mm aggregates; bins 5–7 effectively captured	~95%
Increase A_s to 36,000 ft² (2× larger)	η₅ rises 38% → 76%; bins 6–7 still pass through; net rises modestly	~75%
Increase A_s 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
A_s = 18,000 → 27,000 ft² (50% larger)	Q/A_s drops; capture particle 0.0148 → 0.0121 mm; η rises to ~72%
Add forebay (12% main A_s)	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

Camp, T. R. (1946). Sedimentation and the Design of Settling Tanks. ASCE Transactions 111: 895-958.
Stokes, G. G. (1851). On the effect of the internal friction of fluids on the motion of pendulums.
NCDEQ. NC Erosion and Sediment Control Planning and Design Manual. §6.61 (Sediment Basins).
EPA. Storm Water Management for Construction Activities. EPA-832-R-92-005.

— 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.