1. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
rain
[hour]
[hour]
[m3/s]
[cm] A B C D
0-4
14.8
1.2 1 8 2 4
4-8
21.2
2.4 2 20 4 8
8-12
40.0
3.8 4 40 12 18
12-16
72.0
2.2 10 64 18 28
16-20
121.2
0.0 16
108 30 40
20-24
174.0
0.0 12 50 14 20
Find the 4-hour unit hydrograph, in cubic meters per second, per centimeter of excess rainfall. [pause] In this problem, ---
24-28
188.8
0.0 2 16 4 6
28-32
126.4
watershed area = 180 [km2]
32-36
54.8
base flow = 14.0 [m3/s]
36-40
21.2

2. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
rain
[hour]
[hour]
[m3/s]
[cm] A B C D
0-4
14.8
1.2 1 8 2 4
4-8
21.2
2.4 2 20 4 8
8-12
40.0
3.8 4 40 12 18
12-16
72.0
2.2 10 64 18 28
16-20
121.2
0.0 16
108 30 40
20-24
174.0
0.0 12 50 14 20
Stream flow and rainfall data, from a given storm is provided, in 4 hour increments.
24-28
188.8
0.0 2 16 4 6
28-32
126.4
watershed area = 180 [km2]
32-36
54.8
base flow = 14.0 [m3/s]
36-40
21.2

3. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
rain
[hour]
[hour]
[m3/s]
[cm] A B C D
0-4
14.8
1.2 1 8 2 4
4-8
21.2
2.4 2 20 4 8
8-12
40.0
3.8 4 40 12 18
12-16
72.0
2.2 10 64 18 28
16-20
121.2
0.0 16
108 30 40
20-24
174.0
0.0 12 50 14 20
The watershed area, and base flowrate, are also provided.
24-28
188.8
0.0 2 16 4 6
28-32
126.4
watershed area = 180 [km2]
32-36
54.8
base flow = 14.0 [m3/s]
36-40
21.2

4. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
rain
[hour]
[hour]
[m3/s]
[cm] A B C D
0-4
14.8
1.2 1 8 2 4
4-8
21.2
2.4 2 20 4 8
8-12
40.0
3.8 4 40 12 18
12-16
72.0
2.2 10 64 18 28
16-20
121.2
0.0 16
108 30 40
20-24
174.0
0.0 12 50 14 20
Possible answers A, B, C and D are shown on the right, in units of meters cubed per second, per centimeter of excess rainfall.
24-28
188.8
0.0 2 16 4 6
28-32
126.4
watershed area = 180 [km2]
32-36
54.8
base flow = 14.0 [m3/s]
36-40
21.2

5. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
flowrate
[hour]
[hour]
[m3/s]
0-4
14.8
4-8
21.2
8-12
40.0
12-16
72.0
16-20
121.2
20-24
174.0
Before developing the hydrograph, we’ll first need to account for the base flow.
24-28
188.8
28-32
126.4
time
32-36
54.8
base flow = 14.0 [m3/s]
36-40
21.2

6. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
flowrate
[hour]
[hour]
[m3/s]
base flow
0-4
14.8
4-8
21.2
8-12
40.0
12-16
72.0
16-20
121.2
20-24
174.0
The base flow rate in the stream plus the
24-28
188.8
28-32
126.4
time
32-36
54.8
base flow = 14.0 [m3/s]
36-40
21.2

7. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
flowrate
[hour]
[hour]
[m3/s]
base flow
0-4
14.8 +
storm stream flow
4-8
21.2
8-12
40.0
12-16
72.0
16-20
121.2
20-24
174.0
excess stream flow, caused by a rainstorm, equals the ---
24-28
188.8
28-32
126.4
time
32-36
54.8
base flow = 14.0 [m3/s]
36-40
21.2

8. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
flowrate
[hour]
[hour]
[m3/s]
base flow
0-4
14.8 +
storm stream flow
4-8
21.2
total stream flow
8-12
40.0
12-16
72.0
16-20
121.2
20-24
174.0
total recorded stream flow. [pause] To develop the hydrograph, were interested in the ---
24-28
188.8
28-32
126.4
time
32-36
54.8
base flow = 14.0 [m3/s]
36-40
21.2

9. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
flowrate
[hour]
[hour]
[m3/s]
base flow
0-4
14.8 +
storm stream flow
4-8
21.2
total stream flow
8-12
40.0
12-16
72.0
16-20
121.2
20-24
174.0
stream flow generated by the storm, only. Therefore we can subtract the ---
24-28
188.8
28-32
126.4
time
32-36
54.8
base flow = 14.0 [m3/s]
36-40
21.2

10. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
flowrate
[hour]
[hour]
[m3/s]
base flow
0-4
14.8 +
storm stream flow
4-8
21.2
total stream flow
8-12
40.0
12-16
72.0
total stream flow
16-20
121.2 -
base flow
20-24
174.0
base flow, from the measured stream flow, to determine the stream flow data, from the storm only. [pause] Graphically, ---
storm stream flow
24-28
188.8
28-32
126.4
time
32-36
54.8
base flow = 14.0 [m3/s]
36-40
21.2

11. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
flowrate
[hour]
[hour]
[m3/s]
0-4
14.8
4-8
21.2
8-12
40.0
12-16
72.0
16-20
121.2
20-24
174.0
we can plot the total stream flow, from our given data set, then plot ---
24-28
188.8
time
28-32
126.4
total stream flow
32-36
54.8 -
base flow
36-40
21.2
storm stream flow

12. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
flowrate
[hour]
[hour]
[m3/s]
0-4
14.8
4-8
21.2
8-12
40.0
12-16
72.0
16-20
121.2
20-24
174.0
the constant base flow, as a negative value, shown in orange, and the the storm stream flow equals ---
24-28
188.8
time
28-32
126.4
total stream flow
32-36
54.8 -
base flow
36-40
21.2
storm stream flow

13. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
stream flow
flowrate
[hour]
[hour]
[m3/s]
0-4
14.8
4-8
21.2
8-12
40.0
12-16
72.0
16-20
121.2
20-24
174.0
the stream flow, less the magnitude of the base flow, for all 10 measurements. [pause] To calculate the ---
24-28
188.8
time
28-32
126.4
total stream flow
32-36
54.8 -
base flow
36-40
21.2
storm stream flow

14. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Flow rate [m3/s]
[hour]
[hour]
flowrate
total
storm
0-4
14.8
4-8
21.2
8-12
40.0
12-16
72.0
16-20
121.2
20-24
174.0
stream flow from just the storm, ---
24-28
188.8
28-32
126.4
total stream flow
32-36
54.8 -
base flow
36-40
21.2
storm stream flow

15. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Flow rate [m3/s]
[hour]
[hour]
flowrate
total
storm
0-4
14.8
0.8
4-8
21.2
7.2
8-12
40.0
26.0
12-16
72.0
58.0
16-20
121.2
107.2
20-24
174.0
160.0
we can simply subtract 14 from the total flowrate. [pause] Also before developing the hydrograph, ---
24-28
188.8
174.8
28-32
126.4
112.4
total stream flow
32-36
54.8
40.8 -
base flow
36-40
21.2
7.2
storm stream flow

16. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Flow rate [m3/s]
[hour]
[hour]
total
storm
0-4
14.8
0.8
4-8
21.2
7.2
δ F f=
8-12
40.0
26.0
δ t
12-16
72.0
58.0
16-20
121.2
107.2
20-24
174.0
160.0
the infiltration of rainfall into the soil, must be accounted for.
24-28
188.8
174.8
28-32
126.4
112.4
32-36
54.8
40.8
36-40
21.2
7.2

17. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Flow rate [m3/s]
[hour]
[hour]
total
storm
0-4
14.8
0.8
4-8
21.2
7.2
δ F f=
8-12
40.0
26.0
δ t
12-16
72.0
58.0
16-20
121.2
107.2
infiltration cm
20-24
174.0
160.0
rate
The infiltration rate, in centimeters per 4 hour time period, equals, ---
4 hr
24-28
188.8
174.8
28-32
126.4
112.4
32-36
54.8
40.8
36-40
21.2
7.2

18. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Flow rate [m3/s]
total
[hour]
[hour]
total
storm
infiltration [cm]
0-4
14.8
0.8
4-8
21.2
7.2
δ F f=
8-12
40.0
26.0
δ t
12-16
72.0
58.0
16-20
121.2
107.2
infiltration cm
20-24
174.0
160.0
rate
the total infiltration, in centimeters, divided by the ---
4 hr
24-28
188.8
174.8
28-32
126.4
112.4
32-36
54.8
40.8
36-40
21.2
7.2

19. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Flow rate [m3/s]
total
[hour]
[hour]
total
storm
infiltration [cm]
0-4
14.8
0.8
4-8
21.2
7.2
δ F
storm f=
duration
8-12
40.0
26.0
δ t
[4 hr]
12-16
72.0
58.0
16-20
121.2
107.2
infiltration cm
20-24
174.0
160.0
rate
storm duration, in 4 hour time increments.
4 hr
24-28
188.8
174.8
28-32
126.4
112.4
32-36
54.8
40.8
36-40
21.2
7.2

20. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Flow rate [m3/s]
total
[hour]
[hour]
total
storm
infiltration [cm]
0-4
14.8
0.8
4-8
21.2
7.2
Δ F
storm f=
duration
8-12
40.0
26.0
Δ t
[4 hr]
12-16
72.0
58.0
16-20
121.2
107.2
infiltration cm
20-24
174.0
160.0
rate
For this problem we’ll assume a constant infiltration rate, and use the capital Delta notation.
4 hr
24-28
188.8
174.8
28-32
126.4
112.4
32-36
54.8
40.8
36-40
21.2
7.2

21. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Flow rate [m3/s]
total
[hour]
[hour]
total
storm
infiltration [cm]
0-4
14.8
0.8
4-8
21.2
7.2
Δ F
storm f=
duration
8-12
40.0
26.0
Δ t
[4 hr]
12-16
72.0
58.0
16-20
121.2
107.2
infiltration cm
20-24
174.0
160.0
rate
We’ll first determine the total infiltration, delta F, which equals, ---
4 hr
24-28
188.8
174.8
28-32
126.4
112.4
32-36
54.8
40.8
36-40
21.2
7.2

22. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Flow rate [m3/s]
total
[hour]
[hour]
total
storm
infiltration [cm]
0-4
14.8
0.8
4-8
21.2
7.2
Δ F f=
8-12
40.0
26.0
Δ t
12-16
72.0
58.0
16-20
121.2
107.2
Δ F=drain-drunoff
20-24
174.0
160.0
the total rainfall depth, minus, the total runoff depth.
24-28
188.8
174.8
28-32
126.4
112.4
32-36
54.8
40.8
36-40
21.2
7.2

23. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
12-16
58.0
2.2
16-20
107.2
Δ F=drain-drunoff
20-24
160.0
The depth of rainfall is the sum of the ---
24-28
174.8
28-32
112.4
32-36
40.8
36-40
7.2
watershed area = 180 [km2]

24. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
12-16
58.0
2.2
16-20
107.2
Δ F=drain-drunoff Σ
20-24
160.0
individual rainfall entries, which equals, 9.6 centimeters. [pause] The depth of runoff equals the ---
9.6 [cm]
24-28
174.8
28-32
112.4
32-36
40.8
36-40
7.2
watershed area = 180 [km2]

25. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
12-16
58.0
2.2
16-20
107.2
Δ F=drain-drunoff Σ
20-24
160.0
volume of runoff, divided by the watershed area.
9.6 [cm]
24-28
174.8
Vrunoff
28-32
112.4
Awatershed
32-36
40.8
36-40
7.2
watershed area = 180 [km2]

26. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
12-16
58.0
2.2
16-20
107.2
Δ F=drain-drunoff
20-24
160.0
The volume of runoff equals the sum of the flow rates, times the time step, delta t.
9.6 [cm]
24-28
174.8
Vrunoff
ΣQ *Δ t
28-32
112.4
Awatershed
32-36
40.8
36-40
7.2
watershed area = 180 [km2]

27. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
12-16
58.0
2.2
16-20
107.2
Δ F=drain-drunoff
20-24
160.0
The sum of the flow rates equals meters cubed per second, and the time step is 4 hours.
9.6 [cm]
24-28
174.8
Vrunoff
ΣQ *Δ t
28-32
112.4
Awatershed
32-36
40.8 m3
4 [hr]
36-40
7.2
694.4
area=180 [km2] s

28. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
12-16
58.0
2.2
16-20
107.2
Δ F=drain-drunoff
20-24
160.0
The area of the watershed is 180 kilometers squared, and after a few conversions are made, ---
9.6 [cm]
24-28
174.8
Vrunoff
ΣQ *Δ t
28-32
112.4
Awatershed
32-36
40.8 m3
4 [hr]
36-40
7.2
694.4
area=180 [km2] s

29. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
5.56 [cm]
12-16
58.0
2.2
16-20
107.2
Δ F=drain-drunoff
20-24
160.0
the runoff depth computes to 5.56 centimeters. [pause] Therefore the total infiltration, delta F, equals, ---
9.6 [cm]
24-28
174.8
Vrunoff
ΣQ *Δ t
28-32
112.4
Awatershed
32-36
40.8 m3
4 [hr]
36-40
7.2
694.4
area=180 [km2] s

30. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4.04 [cm]
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
5.56 [cm]
12-16
58.0
2.2
16-20
107.2
Δ F=drain-drunoff
20-24
160.0
4.04 centimeters. [pause] Since the rainfall lasts for ---
9.6 [cm]
24-28
174.8
Vrunoff
ΣQ *Δ t
28-32
112.4
Awatershed
32-36
40.8 m3
4 [hr]
36-40
7.2
694.4
area=180 [km2] s

31. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4.04 [cm]
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
storm
12-16
58.0
2.2
duration
16-20
107.2
[4 hr]
20-24
160.0
16 hours of measurement, the storm duration is --
24-28
174.8
28-32
112.4
32-36
40.8
36-40
7.2

32. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4.04 [cm]
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
storm
12-16
58.0
2.2
duration
4 [4 hr]
16-20
107.2
[4 hr]
20-24
160.0
4, 4-hour time periods. [pause] The infiltration rate, f, equals ---
24-28
174.8
28-32
112.4
32-36
40.8
36-40
7.2

33. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4.04 [cm]
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
storm
12-16
58.0
2.2
duration
4 [4 hr]
16-20
107.2
[4 hr]
20-24
160.0
1.01 centimeters, ever 4 hours, which we will --- cm
f=1.01
24-28
174.8
4 hr
28-32
112.4
32-36
40.8
36-40
7.2

34. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
total
[hour]
[hour]
flow[m3/s]
[cm]
infiltration [cm]
0-4
0.8
1.2
4.04 [cm]
4-8
7.2
2.4
Δ F f=
8-12
26.0
3.8
Δ t
storm
12-16
58.0
2.2
duration
4 [4 hr]
16-20
107.2
[4 hr]
20-24
160.0
round to 1.0 centimeters, ever 4 hours. [pause] Similar to the stream flow, ---
4 hr cm
f=1.01
24-28
174.8
round
28-32
112.4
down
f=1.0 cm
32-36
40.8
4 hr
36-40
7.2

35. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
[hour]
[hour]
flow[m3/s]
[cm]
0-4
0.8
1.2
4-8
7.2
2.4
8-12
26.0
3.8
12-16
58.0
2.2
16-20
107.2
4 hr cm
f=1.0
depth
20-24
160.0
the depth used to develop the unit hydrograph ---
24-28
174.8
28-32
112.4
32-36
40.8
36-40
7.2
time

36. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
[hour]
[hour]
flow[m3/s]
[cm]
0-4
0.8
1.2
4-8
7.2
2.4
8-12
26.0
3.8
12-16
58.0
2.2
16-20
107.2 cm
f=1.0
depth
20-24
160.0
4 hr
should only be the rainfall which becomes runoff, ---
Rain
24-28
174.8 -
Infiltration
28-32
112.4
32-36
40.8
36-40
7.2
time

37. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
[hour]
[hour]
flow[m3/s]
[cm]
0-4
0.8
1.2
4-8
7.2
2.4
8-12
26.0
3.8
12-16
58.0
2.2
16-20
107.2 cm
f=1.0
depth
20-24
160.0
4 hr
which we’ll call, the runoff depth. [pause] Graphically, ---
Rain
24-28
174.8 -
Infiltration
28-32
112.4
32-36
40.8
Runoff Depth
36-40
7.2
time

38. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
[hour]
[hour]
flow[m3/s]
[cm]
0-4
0.8
1.2
4-8
7.2
2.4
8-12
26.0
3.8
4 hr cm
1.0
12-16
58.0
2.2
16-20
107.2
Rain
depth
20-24
160.0
the rain, infiltration and runoff depth would plot as shown, and the runoff depths would tabulate to -
Infiltration
24-28
174.8
Runoff
28-32
112.4
Depth
32-36
40.8
time
36-40
7.2

39. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph
Time
Storm
Rain
Runoff
[hour]
[hour]
flow[m3/s]
[cm]
depth[cm]
0-4
0.8
1.2
0.2
4-8
7.2
2.4
1.4
8-12
26.0
3.8
2.8
4 hr cm
1.0
12-16
58.0
2.2
1.2
16-20
107.2
Rain
depth
20-24
160.0
1 centimeter less than the recorded rainfall, for each 4-hour time period. [pause] At this point, we can start calculating the unit hydrograph. -
Infiltration
24-28
174.8
Runoff
28-32
112.4
Depth
32-36
40.8
time
36-40
7.2

40. Qn 1.0 Rain - Infiltration Runoff Depth Time Storm Rain Runoff [hour]
flow[m3/s]
[cm]
depth[cm]
0-4
0.8
1.2
0.2
4-8
7.2
2.4
1.4
8-12
26.0
3.8
2.8
4 hr cm
1.0
12-16
58.0
2.2
1.2
16-20
107.2
Rain
depth
20-24
160.0
For sake of convenience, lets the designate the storm flows, with Q sub n, and --- -
Infiltration
24-28
174.8
Runoff
28-32
112.4
Depth
32-36
40.8
time
36-40
7.2

41. Qn Pm 1.0 Rain - Infiltration Runoff Depth Time Storm Rain Runoff
[hour]
[hour]
flow[m3/s]
[cm]
depth[cm]
0-4
0.8
1.2
0.2
4-8
7.2
2.4
1.4
8-12
26.0
3.8
2.8
4 hr cm
1.0
12-16
58.0
2.2
1.2
16-20
107.2
Rain
depth
20-24
160.0
the runoff depth,with P sub m. [pause] And the goal is to solve for the --- -
Infiltration
24-28
174.8
Runoff
28-32
112.4
Depth
32-36
40.8
time
36-40
7.2

42. Σ Qn Pm Qn= Pm * Un-m+1 unit hydrograph ordinates discrete convolution
Time
Storm
Rain
Runoff
[hour]
[hour]
flow[m3/s]
[cm]
depth[cm]
0-4
0.8
1.2
0.2
4-8
7.2
2.4
1.4
8-12
26.0
3.8
2.8
unit
12-16
58.0
2.2
1.2
hydrograph
16-20
107.2
ordinates
20-24
160.0
unit hydrograph ordinates, U sub n-m+1, as defined in the discrete convolution equation. We set up the hydrograph, so it reads ---
24-28
174.8
n<m Σ
Qn=
Pm * Un-m+1
28-32
112.4
m=1
32-36
40.8
discrete convolution
36-40
7.2
equation

43. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0 5
16-20
107.2 6
20-24
160.0
similar to our convolution equation, where the storm flow --- 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

44. Σ Find: 4-hr Unit Hydrograph = Σ m3 Qn= Pm * Un-m+1 s*cm Qn
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0 5
16-20
107.2 6
20-24
160.0
equals the the sum of the product of the runoff depths, times the unit hydrograph ordinates. [pause] The variable n, --- 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

45. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0 5
16-20
107.2 6
20-24
160.0
represent time periods, and the variable m, represents --- 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

46. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0 5
16-20
107.2 6
20-24
160.0
the quantity of flow each single hour of rainfall contributes to the total runoff, --- 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

47. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8 2
4-8
7.2 3
8-12
26.0
m=1
m=2
m=3
m=4 4
12-16
58.0 5
16-20
107.2 6
20-24
160.0
where the four columns equal m’s 1 through 4. [pause] Beginning at the first row, --- 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

48. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0 5
16-20
107.2 6
20-24
160.0
Since n equals 1, the second, third and fourth terms --- 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

49. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0 5
16-20
107.2 6
20-24
160.0
are zero, since the rainfall has not yet begun for those time periods. 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

50. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2 5
16-20
107.2 6
20-24
160.0
From our data table, we recall the runoff depth for the first four hours is 0.2 centimeters, therefore, --- 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

51. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2
0.8 = 0.2 * U1 5
16-20
107.2 6
20-24
160.0
0.8 meter cubed per second equals 0.2 centimeters times U sub 1. Temporarily dropping the units, --- 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

52. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2
0.8 = 0.2 * U1 5
16-20
107.2
0.8 6
20-24
160.0
U1 =
and solving for U sub 1, it equals, ----
0.2 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

53. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2
0.8 = 0.2 * U1 5
16-20
107.2
0.8 m3 6
20-24
160.0
U1 =
= 4
4 meters cubed per second, of runoff, per centimeter of excess rainfall.
0.2
s*cm 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1 9
32-36
40.8 10
36-40
7.2

54. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2 3
8-12
26.0 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2 5
16-20
107.2 6
20-24
160.0
The second time period is solved the same way. But this time, the first two hours of runoff --- 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1
U1 =4 9
32-36
40.8 10
36-40
7.2

55. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2
0.2*U2
1.4*U1 3
8-12
26.0 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2 5
16-20
107.2 6
20-24
160.0
contribute to the stream flow, because n equals 2. 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1
U1 =4 9
32-36
40.8 10
36-40
7.2

56. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2
0.2*U2
1.4*U1 3
8-12
26.0 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2
7.2 = 0.2*U *U1 5
16-20
107.2 6
20-24
160.0
Solving the convolution equation for the second ordinate, --- 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1
U1 =4 9
32-36
40.8 10
36-40
7.2

57. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2
0.2*U2
1.4*U1 3
8-12
26.0 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2
7.2 = 0.2*U *U1 5
16-20
107.2
*U1
U2 = 6
20-24
160.0
and plugging in the first ordinate, U sub 1, ---
0.2 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1
U1 =4 9
32-36
40.8 10
36-40
7.2

58. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2
0.2*U2
1.4*U1 3
8-12
26.0 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2
7.2 = 0.2*U *U1 5
16-20
107.2
*U1 m3
U2 = 6
20-24
160.0
= 8
U sub 2 equals 8 meters cubed per second, per centimeter of excess rainfall.
0.2
s*cm 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1
U1 =4 9
32-36
40.8 10
36-40
7.2

59. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2
0.2*U2
1.4*U1 3
8-12
26.0
0.2*U3
1.4*U2
2.8*U1 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2
26.0 = 0.2*U *U *U1 5
16-20
107.2 6
20-24
160.0
For the third time period, we can solve for U sub 3 by writing out the convolution equation, 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1
U1 = 4
U2 = 8 9
32-36
40.8 10
36-40
7.2

60. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2
0.2*U2
1.4*U1 3
8-12
26.0
0.2*U3
1.4*U2
2.8*U1 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2
26.0 = 0.2*U *U *U1 5
16-20
107.2
*U *U1 6
20-24
160.0
U3=
plugging in U sub 1 and U sub 2, and we calculate the third unit ordinate as, ---
0.2 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1
U1 = 4
U2 = 8 9
32-36
40.8 10
36-40
7.2

61. Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2
0.2*U2
1.4*U1 3
8-12
26.0
0.2*U3
1.4*U2
2.8*U1 4
12-16
58.0
depth[cm]
Runoff
P1 = 0.2
P2 = 1.4
P3 = 2.8
P4 = 1.2
26.0 = 0.2*U *U *U1 5
16-20
107.2
*U *U1 m3 6
20-24
160.0
U3=
= 18
18 meters cubed per second, per centimeter of excess rainfall. If we continue this process for the entire hydrograph, ---
0.2
s*cm 7
24-28
174.8 8
28-32
112.4
Qn=
Pm * Un-m+1 Σ
n<M
m=1
U1 = 4
U2 = 8 9
32-36
40.8 10
36-40
7.2

62. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph Qn
Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2
0.2*U2
1.4*U1 3
8-12
26.0
0.2*U3
1.4*U2
2.8*U1 4
12-16
58.0 5
16-20
107.2 6
20-24
160.0
we can complete the table, --- 7
24-28
174.8 8
28-32
112.4 9
32-36
40.8 10
36-40
7.2

63. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph Qn
Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1 2
4-8
7.2
0.2*U2
1.4*U1 3
8-12
26.0
0.2*U3
1.4*U2
2.8*U1 4
12-16
58.0
0.2*U4
1.4*U3
2.8*U2
1.2*U1 5
16-20
107.2
0.2*U5
1.4*U4
2.8*U3
1.2*U2 6
20-24
160.0
0.2*U6
1.4*U5
2.8*U4
1.2*U3
and solve for all unit hydrograph ordinate values. --- 7
24-28
174.8
0.2*U7
1.4*U6
2.8*U5
1.2*U4 8
28-32
112.4
1.4*U7
2.8*U6
1.2*U5 9
32-36
40.8
2.8*U7
1.2*U6 10
36-40
7.2
1.2*U7

64. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph Qn
Pm * Un-m+1
Time n
[hour]
[m3/s]
P1*Un
P2*Un-1
P3*Un-2
P4*Un-3 1
0-4
0.8
0.2*U1
U1 = 4
U2 = 8
U3=18
U4=28
U5=40
U6=20
U7= 6 2
4-8
7.2
0.2*U2
1.4*U1 3
8-12
26.0
0.2*U3
1.4*U2
2.8*U1 4
12-16
58.0
0.2*U4
1.4*U3
2.8*U2
1.2*U1 5
16-20
107.2
0.2*U5
1.4*U4
2.8*U3
1.2*U2 6
20-24
160.0
0.2*U6
1.4*U5
2.8*U4
1.2*U3
[pause] 7
24-28
174.8
0.2*U7
1.4*U6
2.8*U5
1.2*U4 8
28-32
112.4
1.4*U7
2.8*U6
1.2*U5 9
32-36
40.8
2.8*U7
1.2*U6 10
36-40
7.2
1.2*U7

65. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph A B C D 1 8 2 4
U1 = 4
U2 = 8
U3=18
U4=28
U5=40
U6=20
U7= 6 2 20 4 8 4 40 12 18 10 64 18 28 16
108 30 40 12 50 14 20
When reviewing the possible solutions, --- 2 16 4 6

66. Find: 4-hr Unit Hydrograph
s*cm m3
Find: 4-hr Unit Hydrograph A B C D 1 8 2 4
U1 = 4
U2 = 8
U3=18
U4=28
U5=40
U6=20
U7= 6 2 20 4 8 4 40 12 18 10 64 18 28 16
108 30 40 12 50 14 20
The answer is D. 2 16 4 6
AnswerD

67. ? 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