1. What you are given: (see example in notes, runoff 2, page 23-30)
3 inches of rainfall
12-acre site
Forest with good cover
Hydrological soil group B
What you need to do:
Estimate the depth (in.) and volume (ft3) of runoff using the SCS Curve number method for:
A. Antecedent moisture condition I
B. Antecedent moisture condition II
C. Antecedent moisture condition III
CN(I) = [4.2CN(II)] / [10 â 0.058CN(II)]
CN(III) = [23CN(II)] / [10 + 0.13CN(II)]
1. What you are given: (see example in notes, runoff 2, page 23-30)
3 inches of rainfall
12-acre site
Forest with good cover
Hydrological soil group B
What you need to do:
Estimate the depth (in.) and volume (ft3) of runoff using the SCS Curve number method for:
A. Antecedent moisture condition I
B. Antecedent moisture condition II
C. Antecedent moisture condition III
CN(I) = [4.2CN(II)] / [10 â 0.058CN(II)]
CN(III) = [23CN(II)] / [10 + 0.13CN(II)]
FOREST WITH GOOD COVER HYDROLOGIC GROUP B IS 55
POWER POINT PG 23-30
§No expression of time
ÃCan not account for rainfall duration or intensity
§0.2S based on data from agricultural watersheds
ÃMay not apply to urban areas with impervious areas
ÃEquations have been developed to adjust for urban areas
§Can not estimate runoff from:
à snowmelt
ÃRain on frozen ground
§CN procedure is less accurate for runoff < 0.5 in.
§CN procedure applies only to direct surface runoff and does not include:
ÃInterflow from forested areas
ÃRapid subsurface flow to streams (important in sandy soils)
ÃLateral subsurface flow in karst regions
ÃHigh groundwater levels that may come to the surface during rainfall events
Watershed with Land Use %
And HSGs Listed
HSG=D 30% bare soil
HSG=C 35% thin forest
HSG=B 35% grass
CN EXAMPLE:
Use HSG % CN (Table 3.16 Haan text)
Bare soil D 30 91 (bare soil w/o conservation)
Grass B 35 58 (grass in good condition)
Forest C 35 77 (poor cover)
Weighted CN = 0.30(91) + 0.35(58) + 0.35(77)
Weighted CN = 75
S = (1000 / CN) â 10 = (1000 / 75) â 10
S = 3.3 in.
Q = (P â 0.2S)2 / (P + 0.8S) for P > 0.2S
P > 0.2(3.3in.)
P > 0.66 in.
Q = [3 â 0.2*3.3]2 / [3+ 0.8(3.3)]
Q = 0.97in.
Q = 0.97 in. = 0.08 ft
5 acres = 217,800.9 ft2
Volume = 0.08ft x 217,800.9 ft2
Volume = 17424.4 ft3
CN(I) = 58 (dry conditions)
CN(II) = 77 (average conditions)
CN(III) = 89 (wet conditions)