1. Find: Qpeak [cfs] 125 150 175 200 Time Unit Rainfall Infiltration
[hour]
Hydrograph
[in]
[in]
[cfs/in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
125
150
175
200
4-5 64
Find the peak total runoff, in cubic feet per second. [pause] In this problem, ---
5-6 22
6-7
7-8
base flow = 15 [cfs]

2. Find: Qpeak [cfs] 125 150 175 200 Time Unit Rainfall Infiltration
[hour]
Hydrograph
[in]
[in]
[cfs/in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
125
150
175
200
4-5 64
a unit hydrograph for a particular watershed and discharge point, is provided.
5-6 22
6-7
7-8
base flow = 15 [cfs]

3. Find: Qpeak [cfs] 125 150 175 200 Time Unit Rainfall Infiltration
[hour]
Hydrograph
[in]
[in]
[cfs/in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
125
150
175
200
4-5 64
A 3-hour storm event drops 2.5 inches of rain, some of which is absorbed by the soil. [pause]
5-6 22
6-7
7-8
base flow = 15 [cfs]

4. Find: Qpeak [cfs] 125 150 175 200 Time Unit Rainfall Infiltration
[hour]
Hydrograph
[in]
[in]
[cfs/in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
125
150
175
200
4-5 64
To calculate the runoff hydrograph, we’ll use the -----
5-6 22
6-7
7-8
base flow = 15 [cfs]

5. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Time Unit Rainfall Infiltration
[hour]
Hydrograph
[in]
[in]
[cfs/in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
4-5 64
discrete convolution equation, which discritizes the data into equal periods of time, such as, 1 hour. In this equation, Q sub n equals ---
n<M Σ
5-6 22
Qn=
Pm * Un-m+1
6-7
m=1
7-8
base flow = 15 [cfs]

6. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 runoff [cfs] Time Unit Rainfall
Infiltration
[hour]
Hydrograph
[in]
[in]
[cfs/in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
runoff [cfs]
4-5 64
the runoff values, which we’re trying to calculate, in units of cubic feet per second. P sub m is the ----
n<M Σ
5-6 22
Qn=
Pm * Un-m+1
6-7
m=1
7-8
base flow = 15 [cfs]

7. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 runoff [cfs] excess rainfall [in]
Time
Unit
Rainfall
Infiltration
[hour]
Hydrograph
[in]
[in]
[cfs/in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
runoff [cfs]
4-5 64
average excess precipitation in the watershed, in inches, and the U sub n-m+1 is the ---
n<M Σ
5-6 22
Qn=
Pm * Un-m+1
6-7
m=1
7-8
excess
rainfall [in]
base flow = 15 [cfs]

8. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 runoff [cfs] excess unit hydro-
Time
Unit
Rainfall
Infiltration
[hour]
Hydrograph
[in]
[in]
[cfs/in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
runoff [cfs]
4-5 64
unit hydrograph, specific to the watershed, in units of cubic feet per second, per inch of excess rainfall.
n<M Σ
5-6 22
Qn=
Pm * Un-m+1
6-7
m=1
7-8
excess
unit hydro-
rainfall [in]
base flow = 15 [cfs]
graph [cfs/in]

9. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 runoff [cfs] excess unit hydro-
Time
Unit
Rainfall
Infiltration
[hour]
Hydrograph
[in]
[in]
[cfs/in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
runoff [cfs]
4-5 64
The unit hydrograph for the watershed has been provided in the problem statement, and the excess rainfall ---
n<M Σ
5-6 22
Qn=
Pm * Un-m+1
6-7
m=1
7-8
excess
unit hydro-
rainfall [in]
base flow = 15 [cfs]
graph [cfs/in]

10. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 runoff [cfs] excess unit hydro-
Time
Unit
Rainfall
Infiltration
[hour]
Hydrograph
[in]
[in]
[cfs/in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
runoff [cfs]
4-5 64
is the rainfall which contributes to runoff, which equals the ---
n<M Σ
5-6 22
Qn=
Pm * Un-m+1
6-7
m=1
7-8
excess
unit hydro-
rainfall [in]
base flow = 15 [cfs]
graph [cfs/in]

11. - = Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 runoff [cfs] excess
Time
Unit
Rainfall -
Infilt- =
Excess
[hour]
Hydrograph
[in]
ration
Rainfall
[in]
[cfs/in]
[in]
0-1 5
0.7
0.2
1-2 20
1.0
0.1
2-3 45
0.8
0.1
3-4 80
runoff [cfs]
4-5 64
the recorded rainfall, minus the infiltration. So the excess rainfall equals, ---
n<M Σ
5-6 22
Qn=
Pm * Un-m+1
6-7
m=1
7-8
excess
unit hydro-
rainfall [in]
base flow = 15 [cfs]
graph [cfs/in]

12. - = Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 runoff [cfs] excess
Time
Unit
Rainfall -
Infilt- =
Excess
[hour]
Hydrograph
[in]
ration
Rainfall
[in]
[cfs/in]
[in]
0-1 5
0.7
0.2
0.5
1-2 20
1.0
0.1
0.9
2-3 45
0.8
0.1
0.7
3-4 80
runoff [cfs]
4-5 64
0.5 inches, 0.9 inches, and 0.7 inches, for hours 1, 2 and 3.
n<M Σ
5-6 22
Qn=
Pm * Un-m+1
6-7
m=1
7-8
excess
unit hydro-
rainfall [in]
base flow = 15 [cfs]
graph [cfs/in]

13. - = Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 runoff [cfs] excess
Time
Unit
Rainfall -
Infilt- =
Excess
[hour]
Hydrograph
[in]
ration
Rainfall
[in]
[cfs/in]
[in]
0-1 5
0.7
0.2
0.5
1-2 20
1.0
0.1
0.9
2-3 45
0.8
0.1
0.7
3-4 80
runoff [cfs]
4-5 64
With all values of P and U determined, ---
n<M Σ
5-6 22
Qn=
Pm * Un-m+1
6-7
m=1
7-8
excess
unit hydro-
rainfall [in]
base flow = 15 [cfs]
graph [cfs/in]

14. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Time Unit [cfs/in] 5 20 45 80 6422
Hydro
Excess Rainfall [in]
Runoff n
[hour]
[cfs]
0.5
0.9
0.7
(m=1)
(m=2)
(m=3) 1
0-1
2.5 2
1-2
14.5 3
2-3
44.0 4
3-4
94.5 5
4-5
135.5
the hydrograph is set up. In the convolution equation, --- 6
5-6
124.6 7
6-7
64.6
Qn=
Pm * Un-m+1 Σ
n<M
m=1 8
7-8
15.4
base flow = 15 [cfs]

15. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Time Unit [cfs/in] 5 20 45 80 6422
Hydro
Excess Rainfall [in]
Runoff n
[hour]
[cfs]
0.5
0.9
0.7
(m=1)
(m=2)
(m=3) 1
0-1
2.5 2
1-2
14.5 3
2-3
44.0 4
3-4
94.5 5
4-5
135.5
n refers to the time period, and little m refers to the --- 6
5-6
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

16. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Time Unit [cfs/in] 5 20 45 80 6422
Hydro
Excess Rainfall [in]
Runoff n
[hour]
[cfs]
0.5
0.9
0.7
(m=1)
(m=2)
(m=3) 1
0-1
2.5 2
1-2
14.5 3
2-3
44.0 4
3-4
94.5 5
4-5
135.5
hour of rainfall. [pause] Therefore, it may be easier ---- 6
5-6
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

17. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Time Unit Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
(1,1)
2.5 2
1-2 20
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
to equate each cell in the hydrograph to the product of a P and U value. For example, the first column, first row, --- 6
5-6 22
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

18. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Time Unit Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
2.5 2
1-2 20
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
refers to n=1, m=1. Using the convolution equation, this cell would be equal to --- 6
5-6 22
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

19. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Time Unit Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
P1*U1
2.5 2
1-2 20
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
P1 times U1. [pause] Using this formula we can assign a product to the remaining cells --- 6
5-6 22
124.6 7
6-7 1
64.6 1 1
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

20. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Time Unit Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
P1*U1
2.5 2
1-2 20
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
based on the row, n, and column, m. 6
5-6 22
124.6 7
6-7 ?
64.6 ? ?
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

21. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Time Unit Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
P1*U1
P2*U0
P3*U-1
2.5 2
1-2 20
P1*U2
P2*U1
P3*U0
14.5 3
2-3 45
P1*U3
P2*U2
P3*U1
44.0 4
3-4 80
P1*U4
P2*U3
P3*U2
94.5 5
4-5 64
P1*U5
P2*U4
P3*U3
135.5
Solving for runoff from the first hour, Q1, --- 6
5-6 22
P1*U6
P2*U5
P3*U4
124.6 7
6-7
P1*U7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
P1*U8
15.4
m=1
base flow = 15 [cfs]

22. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Q1=P1*U1+P2*U0+P3*U-1 Time Unit
[cfs/in] 5 20 45 80 64 22
Hydro
Excess Rainfall [in]
Runoff n
[hour]
[cfs]
0.5
0.9
0.7
(m=1)
(m=2)
(m=3) 1
0-1
P1*U1
P2*U0
P3*U-1 Q1
2.5 2
1-2
14.5 3
2-3
44.0 4
3-4
94.5 5
4-5
135.5
we notice there is no U0 or U-1 term, since the second and third hour of rainfall cannot contribute the first hour of runoff.
Q1=P1*U1+P2*U0+P3*U-1 6
5-6
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

23. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Q1=P1*U1+P2*U0+P3*U-1 Time Unit
[cfs/in] 5 20 45 80 64 22
Hydro
Excess Rainfall [in]
Runoff n
[hour]
[cfs]
0.5
0.9
0.7
(m=1)
(m=2)
(m=3) 1
0-1
P1*U1
P2*U0
P3*U-1 Q1
2.5 2
1-2
14.5 3
2-3
44.0 4
3-4
94.5 5
4-5
135.5
So the second and third terms of the equation, equal 0, and the runoff from the first hour is simply, ---
Q1=P1*U1+P2*U0+P3*U-1 6
5-6
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

24. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Q1=P1*U1 Time Unit
Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
P1*U1 Q1
2.5 2
1-2 20
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
0.5 inches of excess rainfall times 5 cubic feet per second, per excess inch of rainfall, which equals, ---
Q1=P1*U1 6
5-6 22
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

25. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Q1=P1*U1=2.5[cfs] Time Unit
Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
2.5
2.5
2.5 2
1-2 20
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
2.5 cubic feet per second. [pause] This process is repeated for the second hour ---
Q1=P1*U1=2.5[cfs] 6
5-6 22
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

26. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Q2=P1*U2+P2*U1+P3*U0 Time Unit
Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
2.5
2.5
2.5 2
1-2 20
P1*U2
P2*U1
P3*U0 Q2
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
and we notice the third term cancels out of the equation again.
Q2=P1*U2+P2*U1+P3*U0 6
5-6 22
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

27. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Q2=P1*U2+P2*U1 Time Unit
Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
2.5
2.5
2.5 2
1-2 20
P1*U2
P2*U1 Q2
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
The values for U and P are plugged in, and the second hour of runoff equals, ---
Q2=P1*U2+P2*U1 6
5-6 22
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

28. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Q2=P1*U2+P2*U1 Time Unit
Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
2.5
2.5
2.5 2
1-2 20
P1*U2
P2*U1
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
14.5 cubic feet per second.
Q2=P1*U2+P2*U1 6
5-6 22
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

29. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Q3=P1*U3+P2*U2+P3*U1 Time Unit
Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
2.5
2.5
2.5 2
1-2 20 10
4.5
14.5 3
2-3 45
P1*U3
P2*U2
P3*U1
44.0 4
3-4 80
94.5 5
4-5 64
135.5
Runoff from the third hour of the storm is calculated the same way, ---
Q3=P1*U3+P2*U2+P3*U1 6
5-6 22
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

30. Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Q3=P1*U3+P2*U2+P3*U1 Time Unit
Excess Rainfall [in]
Runoff n
[hour]
Hydro
[cfs]
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
2.5
2.5
2.5 2
1-2 20 10
4.5
14.5 3
2-3 45
P1*U3
P2*U2
P3*U1
44.0 4
3-4 80
94.5 5
4-5 64
135.5
and equals 44 cubic feet per second. The last 5 hours of calculations are completed, ---
Q3=P1*U3+P2*U2+P3*U1 6
5-6 22
124.6 7
6-7
64.6
n<M
Qn=
Pm * Un-m+1 Σ 8
7-8
15.4
m=1
base flow = 15 [cfs]

31. Find: Qpeak [cfs] Time Unit Excess Rainfall [in] Runoff n [hour] Hydro
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
2.5
2.5
2.5 2
1-2 20 10
4.5
14.5 3
2-3 45 45 18
3.5
44.0 4
3-4 80 80
40.5 14
94.5 5
4-5 64 64 72
31.5
135.5
and we’re left with the runoff hydrograph, resulting from the excess precipitation, of the storm. 6
5-6 22 22
57.6 56
124.6 7
6-7
19.8
44.8
64.6 8
7-8
15.4
15.4
base flow = 15 [cfs]

32. Find: Qpeak [cfs] Time Unit Excess Rainfall [in] Runoff n [hour] Hydro
0.5
0.9
0.7
[cfs/in]
(m=1)
(m=2)
(m=3) 1
0-1 5
2.5
2.5
2.5 2
1-2 20 10
4.5
14.5 3
2-3 45 45 18
3.5
44.0 4
3-4 80 80
40.5 14
94.5 5
4-5 64 64 72
31.5
135.5
To calculate the total runoff, --- 6
5-6 22 22
57.6 56
124.6 7
6-7
19.8
44.8
64.6 8
7-8
15.4
15.4
base flow = 15 [cfs]

33. Find: Qpeak [cfs] Time Unit Runoff n [hour] Hydro [cfs] [cfs/in] 1 0-15
2.5
2.5 2
1-2 20
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
we need to add the base flow, --- 6
5-6 22
124.6 7
6-7
64.6 8
7-8
15.4
base flow = 15 [cfs]

34. Find: Qpeak [cfs] Time Unit Runoff n [hour] Hydro [cfs] [cfs/in] 1 0-15
2.5
2.5 2
1-2 20
14.5 3
2-3 45
44.0 4
3-4 80
94.5 5
4-5 64
135.5
of 15 cubic feet per second, --- 6
5-6 22
124.6 7
6-7
64.6 8
7-8
15.4
base flow = 15 [cfs]

35. Find: Qpeak [cfs] Time Unit Runoff base flow Total n [hour] Hydro
[cfs/in] 1
0-1 5
2.5
+ 15
2.5 2
1-2 20
14.5
+ 15 3
2-3 45
44.0
+ 15 4
3-4 80
94.5
+ 15 5
4-5 64
135.5
+ 15
to the runoff hydrograph we just computed. 6
5-6 22
124.6
+ 15 7
6-7
64.6
+ 15 8
7-8
15.4
+ 15
base flow = 15 [cfs]

36. Find: Qpeak [cfs] Time Unit Runoff base flow Total n [hour] Hydro
[cfs/in] 1
0-1 5
2.5
+ 15 =
2.5 2
1-2 20
14.5
+ 15 = 3
2-3 45
44.0
+ 15 = 4
3-4 80
94.5
+ 15 = 5
4-5 64
135.5
+ 15 =
--- 6
5-6 22
124.6
+ 15 = 7
6-7
64.6
+ 15 = 8
7-8
15.4
+ 15 =
base flow = 15 [cfs]

37. Find: Qpeak [cfs] Time Unit Runoff base flow Total n [hour] Hydro
[cfs/in] 1
0-1 5
2.5
+ 15 =
17.5
2.5 2
1-2 20
14.5
+ 15 =
29.5 3
2-3 45
44.0
+ 15 =
59.0 4
3-4 80
94.5
+ 15 =
109.5 5
4-5 64
135.5
+ 15 =
150.5
[pause] Since the problem asks to find the --- 6
5-6 22
124.6
+ 15 =
139.6 7
6-7
64.6
+ 15 =
79.6 8
7-8
15.4
+ 15 =
30.4
base flow = 15 [cfs]

38. Find: Qpeak [cfs] Time Unit Runoff base flow Total n [hour] Hydro
[cfs/in] 1
0-1 5
2.5
+ 15 =
17.5
2.5 2
1-2 20
14.5
+ 15 =
29.5 3
2-3 45
44.0
+ 15 =
59.0 4
3-4 80
94.5
+ 15 =
109.5 5
4-5 64
135.5
+ 15 =
150.5
the peak total runoff, we identify this value as --- 6
5-6 22
124.6
+ 15 =
139.6 7
6-7
64.6
+ 15 =
79.6 8
7-8
15.4
+ 15 =
30.4
base flow = 15 [cfs]

39. Find: Qpeak [cfs] Time Unit Runoff base flow Total n [hour] Hydro
[cfs/in] 1
0-1 5
2.5
+ 15 =
17.5
2.5 2
1-2 20
14.5
+ 15 =
29.5 3
2-3 45
44.0
+ 15 =
59.0 4
3-4 80
94.5
+ 15 =
109.5 5
4-5 64
135.5
+ 15 =
150.5
150.5 cubic feet per second. 6
5-6 22
124.6
+ 15 =
139.6 7
6-7
64.6
+ 15 =
79.6 8
7-8
15.4
+ 15 =
30.4
base flow = 15 [cfs]

40. Find: Qpeak [cfs] 125 150 175 200 Time Unit Runoff base flow Total n
[hour]
Hydro
[cfs]
[cfs]
Runoff
[cfs]
[cfs/in] 1
0-1 5
2.5
+ 15 =
17.5
2.5 2
1-2 20
14.5
+ 15 =
29.5
125
150
175
200 3
2-3 45
44.0
+ 15 =
59.0 4
3-4 80
94.5
+ 15 =
109.5 5
4-5 64
135.5
+ 15 =
150.5
When reviewing the possible solutions, --- 6
5-6 22
124.6
+ 15 =
139.6 7
6-7
64.6
+ 15 =
79.6 8
7-8
15.4
+ 15 =
30.4
base flow = 15 [cfs]

41. Find: Qpeak [cfs] AnswerB 125 150 175 200 Time Unit Runoff base flow
Total n
[hour]
Hydro
[cfs]
[cfs]
Runoff
[cfs]
[cfs/in] 1
0-1 5
2.5
+ 15 =
17.5
2.5 2
1-2 20
14.5
+ 15 =
29.5
125
150
175
200 3
2-3 45
44.0
+ 15 =
59.0 4
3-4 80
94.5
+ 15 =
109.5 5
4-5 64
135.5
+ 15 =
150.5
the answer is B. 6
5-6 22
124.6
+ 15 =
139.6 7
6-7
64.6
+ 15 =
79.6 8
7-8
15.4
+ 15 =
30.4
AnswerB
base flow = 15 [cfs]

42. ? 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