5 acres silt loam 10-yr storm North Carolina

Sediment basin sizing for a 5-acre residential subdivision in Mecklenburg County, NC

A clearing-and-grading permit for a 5-acre residential subdivision off Steele Creek Rd. Single basin, skimmer outlet. Walks through RUSLE annual yield, Camp's surface-area criterion sized to the 0.020 mm silt fraction, and 7-bin Stokes/Camp trap efficiency. Real numbers, every step shown.

Result: 22,500 ft² minimum surface area to capture the 0.020 mm silt fraction at the 10-yr peak (22.2 cfs); RUSLE annual yield 268 tons across the disturbed area; computed 7-bin trap efficiency = 65.5% — below the typical NCDEQ 80% target, so the basin needs a forebay, PAM dosing, or sizing to a finer design particle to comply. Step-by-step formulas, every assumption, every number below.

Site inputs

The inputs an engineer in NC actually fills out, with where each comes from:

ParameterValueSource
Disturbed area, A5.0 acProject boundary
Average slope4.0%LiDAR / contour map
Slope length, λ200 ftField measurement
Soil texturesilt loamUSDA SSURGO Web Soil Survey, site coordinates
R factor (Charlotte, NC)180USDA isoerodent map
K factor (silt loam)0.48USDA-NRCS K table
C factor (active construction)0.90RUSLE C table, bare/grading
P factor (no support practice)1.00Conservative, pre-stabilization
Design storm (basin sizing)10-yr, 24-hrNCDEQ sediment basin standard
10-yr, 24-hr depth (Charlotte)5.4 inNOAA Atlas 14, point precipitation
Curve number (active construction)86TR-55, fallow / bare cover, HSG B
Time of concentration, tc22 minNRCS sheet + shallow-concentrated
Design particle size (capture)0.020 mmCoarse silt; standard NC criterion

Step 1 — RUSLE annual soil loss

Universal Soil Loss Equation (revised), tons per acre per year:

$A = R \cdot K \cdot LS \cdot C \cdot P$

The LS factor combines slope length L and slope steepness S. For a 4% slope < 9%:

$L = \left(\frac{\lambda}{72.6}\right)^m, \quad m = 0.4 \text{ (for } 3\% \le s < 5\%\text{)}$
$S = 10.8 \sin(\arctan(s/100)) + 0.03$

Plugging in:

$L = (200/72.6)^{0.4} = 2.755^{0.4} = 1.491$
$S = 10.8 \cdot \sin(\arctan(0.04)) + 0.03 = 10.8 \cdot 0.03998 + 0.03 = 0.462$
$LS = 1.491 \times 0.462 = 0.689$

Now A:

$A = 180 \times 0.48 \times 0.689 \times 0.90 \times 1.00 = 53.6 \text{ tons/acre/year}$

Across the 5-acre disturbed area, that's 268 tons/year of soil loss if the site stays bare for the full year. NCDEQ approval typically assumes ~9 months of active construction before stabilization, which scales the design yield to roughly:

$Y_{design} = 268 \times (9/12) = 201 \text{ tons over construction window}$

Converting tons to cubic feet at a typical sediment bulk density of 100 lb/ft³:

$V_{sed} = \frac{201 \times 2000}{100} = 4{,}020 \text{ ft}^3$
Note: RUSLE delivers gross erosion at the source; not all of it reaches the basin. A typical sediment delivery ratio for small construction watersheds is 0.4 – 0.6, which would reduce the design influent to ~2,000 – 2,400 ft³. The minimum NCDEQ basin volume usually controls anyway.

Step 2 — Required basin storage volume

NCDEQ Erosion and Sediment Control Planning & Design Manual §6.61 sets a minimum sediment storage volume per disturbed acre. The exact number depends on basin type (skimmer vs. principal-spillway) and the manual revision in force at permit issuance — typical values run 1,800 to 3,600 ft³ per disturbed acre. Use the larger of the two:

$V_{min} = (1{,}800 \text{ to } 3{,}600) \text{ ft}^3/\text{ac} \times 5.0 \text{ ac} = 9{,}000 \text{ to } 18{,}000 \text{ ft}^3$

This controls over the RUSLE-derived volume (4,020 ft³) in either case. Use the regulatory minimum, after confirming the current NCDEQ provision applicable to your basin type.

Step 3 — Peak runoff to the basin (10-yr, 24-hr)

SCS curve-number method. P = 5.4 in, CN = 86:

$S = \frac{1000}{CN} - 10 = \frac{1000}{86} - 10 = 1.628 \text{ in}$
$Q = \frac{(P - 0.2S)^2}{P + 0.8S} = \frac{(5.4 - 0.326)^2}{5.4 + 1.302} = \frac{25.75}{6.702} = 3.84 \text{ in}$

Time of concentration via NRCS sheet-flow + shallow-concentrated path returns tc = 22 min for these inputs. Peak discharge via TR-55 graphical (Type II, tc = 22 min, Ia/P = 0.06):

$q_p = q_u \cdot A \cdot Q = 740 \cdot 5.0 \cdot 3.84 \div 640 = 22.2 \text{ cfs}$

For surface-area sizing per Camp's criterion we use the inflow peak. Design Qp = 22.2 cfs.

Step 4 — Camp's surface-area criterion

Camp's overflow-rate equation: a particle settles to the bottom of an idealized rectangular basin if its settling velocity vs is greater than or equal to the surface overflow rate Q/As. Solving for required surface area:

$A_s = \frac{Q}{v_s}$

For the 0.020 mm silt design particle in 20°C water (sediment specific gravity 2.65), Stokes' law:

$v_s = \frac{g(\rho_s - \rho)d^2}{18\mu} = \frac{9.81 \cdot (2650 - 1000) \cdot (0.000020)^2}{18 \cdot 0.001} = 3.6 \times 10^{-4} \text{ m/s}$
$v_s = 3.6 \times 10^{-4} \text{ m/s} \times 3.281 \text{ ft/m} = 1.18 \times 10^{-3} \text{ ft/s}$

Required surface area, with 1.2 short-circuiting factor for non-ideal flow distribution:

$A_{s,design} = 1.2 \times \frac{22.2}{1.18 \times 10^{-3}} = 22{,}570 \text{ ft}^2$

Use 22,500 ft² (rounded for siting). A 2:1 length-to-width basin geometry gives roughly 212 ft × 106 ft footprint.

Sanity check: NCDEQ's rule-of-thumb minimum is 325 ft²/cfs of peak inflow; at 22.2 cfs that's 7,215 ft² — far less than Camp's. Camp controls. When Camp gives a larger number than the rule-of-thumb, use Camp.

Step 5 — Multi-bin trap efficiency (Stokes/Camp)

Real soil isn't a single particle size. The 7-bin Stokes/Camp method computes trap efficiency for each grain-size class separately and weights by the influent particle-size distribution. With As = 22,500 ft² and Q = 22.2 cfs, ηi = min(1, vs,i·As/Q):

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%29.9100%
3 (coarse silt)0.0507.38 × 10⁻³18%7.48100%
4 (medium silt)0.0201.18 × 10⁻³22%1.20100%
5 (fine silt)0.0102.95 × 10⁻⁴15%0.29929.9%
6 (very fine silt)0.0057.38 × 10⁻⁵12%0.07487.5%
7 (clay)0.0021.18 × 10⁻⁵13%0.01201.2%
Weighted overall trap efficiency100%65.5%

Sum of (mass fraction × trapped fraction) = 60.0% (bins 1–4) + 4.49% (bin 5) + 0.90% (bin 6) + 0.16% (bin 7) = 65.5% overall.

This basin doesn't meet NCDEQ's 80% trap-efficiency target. Two real fixes (a third common one barely helps):
  • Larger basin: size for the 0.010 mm bin instead of 0.020 mm. vs drops 4× (Stokes' law, vs ∝ d²), so As required quadruples to ~90,000 ft². Often the only choice on permits where chemical addition isn't allowed.
  • PAM dosing: polyacrylamide flocculation moves the effective particle size from the 0.005 mm range up to ~0.05 mm. Used on linear-utility, steep-slope, and infill projects with footprint constraints.
  • Forebay (operational, not capture): a forebay does not meaningfully raise trap efficiency on its own — both cells see the same Q, and the coarse particles a forebay catches were going to be caught by the main basin anyway. Worked counter-example: the forebay myth.

What changes if you tweak the inputs

If you change…The result moves…
Slope from 4% to 8%LS doubles (~1.4); RUSLE yield → 533 tons/ac/yr
Soil from silt loam to sandy loamK drops 0.48 → 0.27; yield drops 44%
Design particle 0.020 → 0.010 mmvs drops 4×; required As increases 4× to ~90,000 ft² (often impractical without forebay)
Storm 10-yr → 25-yr (P = 6.3 in)Q peak → ~27 cfs; required As → ~27,500 ft²
Add temporary seeding (C 0.90 → 0.45)Yield halves; basin still controls per NCDEQ minimum

Open this exact scenario in HydroComplete

Every input and intermediate computation above lives in the SEDCAD4 engine. Tweak the slope, swap the soil, add a forebay — see how the basin re-sizes in real time.

Sources and further reading

— Michael Flynn, PE
This worked example uses HydroComplete's SEDCAD4 engine values for R, K, C, P, and the 7-bin trap-efficiency calculation. Open the scenario in the app to verify or modify any input.