1. Find: Pe [in] N P=6 [in] Ia=0.20*S 4.0 4.2 4.4 4.6 Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
Industrial D
4.0
4.2
4.4
4.6
Find the excess precipitation, P sub e, in inches. [pause] In this problem, --- N

2. Find: Pe [in] N P=6 [in] Ia=0.20*S 4.0 4.2 4.4 4.6 Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
Industrial D
4.0
4.2
4.4
4.6
a lot is divided into 4 distinct land uses, whose percent area and soil type are provided. N

3. Find: Pe [in] N P=6 [in] Ia=0.20*S 4.0 4.2 4.4 4.6 Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
Industrial D
4.0
4.2
4.4
4.6
The total rainfall is 6 inches, and the initial abstraction, is 20% of the maximum retention. [pause] N

4. Find: Pe [in] N P=6 [in] Ia=0.20*S 4.0 4.2 4.4 4.6 Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
Industrial D
4.0
4.2
4.4
4.6
The SCS rainfall-runoff equation --- N

5. Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
Industrial D
(P-Ia)2
for excess precipitation, is, --- N
Pe=
P-Ia+S

6. Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
Industrial D
runoff
depth
(P-Ia)2
the excess precipitation, equals the quantity, --- N
Pe=
P-Ia+S

7. Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
initial
Industrial D
abstraction
runoff
depth
(P-Ia)2
the total rainfall depth, minus the initial abstraction, squared, --- N
Pe=
P-Ia+S
rainfall
depth

8. Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
initial
Industrial D
abstraction
runoff
depth
(P-Ia)2
divided by the rainfall depth, minus the initial abstraction, --- N
Pe=
P-Ia+S
rainfall
depth

9. Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
initial
Industrial D
abstraction
runoff
depth
(P-Ia)2
plus the maximum retention. [pause] N
Pe=
P-Ia+S
rainfall
maximum
depth
retention

10. Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
initial
Industrial D
abstraction
runoff
depth
(P-Ia)2
The problem statement provides the rainfall depth, as P = 6 inches, --- N
Pe=
P-Ia+S
rainfall
maximum
depth
retention

11. Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
initial
Industrial D
abstraction
runoff
depth
(P-Ia)2
and an initial abstraction of 0.2 times the maximum retention. N
Pe=
P-Ia+S
rainfall
maximum
depth
retention

12. Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
initial
Industrial D
abstraction
runoff
depth
(P-Ia)2
Substituting this value in 0.20 times S, in for the initial abstraction, --- N
Pe=
P-Ia+S
rainfall
maximum
depth
retention

13. Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S (P-0.2*S)2 Pe=
Land use
% Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
initial
Industrial D
abstraction
runoff
depth
(P-Ia)2
The equation reduces to a function of the rainfall depth, P, and the maximum retention, S. N
Pe=
P-Ia+S
(P-0.2*S)2
Pe=
P+0.8*S

14. Find: Pe [in] P=6 [in] Ia=0.20*S (P-Ia)2 1,000 S= -10 Pe= CN P-Ia+S
Land use
% Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
initial
Industrial D
abstraction
runoff
depth
(P-Ia)2
The maximum retention equals 1,000 divided by the curve number, minus 10, ---
1,000 S=
-10
Pe= CN
P-Ia+S
(P-0.2*S)2
Pe=
P+0.8*S

15. Find: Pe [in] P=6 [in] Ia=0.20*S (P-Ia)2 1,000 S= -10 Pe= CN P-Ia+S
Land use
% Area
Soil Type
Residential
40%
25%
20%
15% C
P=6 [in]
Open space D
Ia=0.20*S
Commercial C
initial
Industrial D
abstraction
runoff
depth
(P-Ia)2
and the curve number is a function of the land uses, percent areas, and soil types. [pause] Curve numbers based on various land uses,
1,000 S=
-10
Pe= CN
P-Ia+S
(P-0.2*S)2
Pe=
P+0.8*S

16. Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in] Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C
Open space D
Commercial C
Industrial D
for hydrologic soil types A through D, are provided in tables, and generally range between 40 and 98.
1,000
(P-0.2*S)2 S=
-10
Pe= CN
P+0.8*S
P=6 [in]

17. Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]
A (Sand)  Low CN
Find: Pe [in]
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C
Open space D
Commercial C
Industrial D
Soil type A refers to sandy soils which have a high infiltration potential, and low curve number.
1,000
(P-0.2*S)2 S=
-10
Pe= CN
P+0.8*S
P=6 [in]

18. Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C
Open space D
Commercial C
Industrial D
1,000
Soil type D refers to clayey soils which have low infiltration potential, and higher curve numbers. Soil types B and C are in between.
(P-0.2*S)2 S=
-10
Pe= CN
P+0.8*S
P=6 [in]

19. Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C
Open space D
Commercial C
Industrial D
1,000
For a medium density residential land use, with a soil type C, ---
(P-0.2*S)2 S=
-10
Pe= CN
P+0.8*S
P=6 [in]

20. Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
The curve number equals 83. [pause] The other three curve numbers are looked up as well ---
1,000
(P-0.2*S)2 S=
-10
Pe= CN
P+0.8*S
P=6 [in]

21. Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
[pause].
1,000
(P-0.2*S)2 S=
-10
Pe= CN
P+0.8*S
P=6 [in]

22. Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S P=6 [in]
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
The curve number to use in the retention equation, should be a composite curve number which account for all four land uses.
1,000
(P-0.2*S)2 S=
-10
Pe=
CNcomp
P+0.8*S
P=6 [in]

23. Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S CNcomp=
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
This value should be the weighted average of all curve numbers, based on the percent area of each land use.
1,000
(P-0.2*S)2 S=
-10
Pe=
CNcomp
P+0.8*S n Σ
CNcomp=
(% Area)i * CNi
P=6 [in]
i=1

24. Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S CNcomp=
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
Plugging in the values for percent areas and curve numbers, the composite curve number computes to ---
1,000
(P-0.2*S)2 S=
-10
Pe=
CNcomp
P+0.8*S n Σ
CNcomp=
(% Area)i * CNi
P=6 [in]
i=1

25. Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S CNcomp=
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
[pause] Substituting this value in, ---
1,000
(P-0.2*S)2 S=
-10
Pe=
CNcomp
P+0.8*S n Σ
CNcomp=
(% Area)i * CNi = 85.95
P=6 [in]
i=1

26. Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S CNcomp=
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
the maximum retention becomes, ---
1,000
(P-0.2*S)2 S=
-10
Pe=
CNcomp
P+0.8*S n Σ
CNcomp=
(% Area)i * CNi = 85.95
P=6 [in]
i=1

27. Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 = 1.634 [in] Pe= CNcomp
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
1.634 inches. [pause] Finally, the retention depth ---
1,000
(P-0.2*S)2 S=
-10 = [in]
Pe=
CNcomp
P+0.8*S n Σ
CNcomp=
(% Area)i * CNi = 85.95
P=6 [in]
i=1

28. Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 = 1.634 [in] Pe= CNcomp
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
and rainfall depth are substituted in for variables P, and S, and the excess rainfall equals, ---
1,000
(P-0.2*S)2 S=
-10 = [in]
Pe=
CNcomp
P+0.8*S n Σ
CNcomp=
(% Area)i * CNi = 85.95
P=6 [in]
i=1

29. Σ Find: Pe [in] Pe= 4.40 [in] 1,000 (P-0.2*S)2 S= -10 = 1.634 [in] Pe=
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
4.40 inches. [pause]
1,000
(P-0.2*S)2 S=
-10 = [in]
Pe=
CNcomp
P+0.8*S n
Pe= 4.40 [in] Σ
CNcomp=
(% Area)i * CNi
i=1

30. Σ Find: Pe [in] Pe= 4.40 [in] 4.0 4.2 4.4 4.6 1,000 (P-0.2*S)2 S=
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
4.0
4.2
4.4
4.6
1,000
When reviewing the possible solutions, ---
(P-0.2*S)2 S=
-10 = [in]
Pe=
CNcomp
P+0.8*S n
Pe= 4.40 [in] Σ
CNcomp=
(% Area)i * CNi
i=1

31. Σ Find: Pe [in] AnswerC Pe= 4.40 [in] 4.0 4.2 4.4 4.6 1,000
A (Sand)  Low CN
Find: Pe [in]
D (Clay)  High CN
Land use
% Area
Soil Type CN
Residential
40%
25%
20%
15% C 83
Open space D 80
Commercial C 94
Industrial D 93
4.0
4.2
4.4
4.6
1,000
the answer is C.
(P-0.2*S)2 S=
-10 = [in]
Pe=
CNcomp
P+0.8*S n
Pe= 4.40 [in] Σ
CNcomp=
(% Area)i * CNi
AnswerC
i=1

32. ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4