1. Find: Qp [cfs] shed area tc C 1,050 1,200 1,300 1,450 A B C Q [acre]
[min] A
120
0.68
22.4 B
160
0.83
29.8
1,050
1,200
1,300
1,450 C
220
0.79
44.8
200
i [in/hr]=
Find the peak flow rate, Q sub p, in cubic feet per second. [pause] In this problem, ---
tc [min]+20
use Δt=5 [min]
use multiple triangle
hydrographs

2. Find: Qp [cfs] shed area tc C 1,050 1,200 1,300 1,450 A B C Q [acre]
[min] A
120
0.68
22.4 B
160
0.83
29.8
1,050
1,200
1,300
1,450 C
220
0.79
44.8
200
i [in/hr]=
3 watersheds, A, B and C, are adjacent to each other, and measure 120 acres, 160 acres, and 220 acres in size.
tc [min]+20
use Δt=5 [min]
use multiple triangle
hydrographs

3. Find: Qp [cfs] shed area tc C 1,050 1,200 1,300 1,450 A B C Q [acre]
[min] A
120
0.68
22.4 B
160
0.83
29.8
1,050
1,200
1,300
1,450 C
220
0.79
44.8
200
i [in/hr]=
The runoff coefficients and time of concentrations, for each watershed, are provided.
tc [min]+20
use Δt=5 [min]
use multiple triangle
hydrographs

4. Find: Qp [cfs] shed area tc C 1,050 1,200 1,300 1,450 A B C Q [acre]
[min] A
120
0.68
22.4 B
160
0.83
29.8
1,050
1,200
1,300
1,450 C
220
0.79
44.8
200
i [in/hr]=
All three watersheds discharge to a common point, and the intensity equation, for the region, is provided.
tc [min]+20
use Δt=5 [min]
use multiple triangle
hydrographs

5. Find: Qp [cfs] shed area tc C 1,050 1,200 1,300 1,450 A B C Q [acre]
[min] A
120
0.68
22.4 B
160
0.83
29.8
1,050
1,200
1,300
1,450 C
220
0.79
44.8
200
i [in/hr]=
Since the problem states to use multiple triangle hydrographs with a time step of 5 minutes, ---
tc [min]+20
use Δt=5 [min]
use multiple triangle
hydrographs

6. Find: Qp [cfs] shed area tc C 1,050 1,200 1,300 1,450 A B C Q [acre]
[min] A
120
0.68
22.4 B
160
0.83
29.8
1,050
1,200
1,300
1,450 C
220
0.79
44.8
200
i [in/hr]=
our strategy will be to determine the individual hydrographs, from each watershed, and then ---
tc [min]+20
use Δt=5 [min]
use multiple triangle
hydrographs

7. Find: Qp [cfs] shed area tc C 1,050 1,200 1,300 1,450 A B C Q [acre]
[min] A
120
0.68
22.4 B
160
0.83
29.8
1,050
1,200
1,300
1,450 C
220
0.79
44.8
200
i [in/hr]=
add their values together to calculate a hydrograph for all three watersheds combined. [pause]
tc [min]+20
use Δt=5 [min]
use multiple triangle
hydrographs

8. Find: Qp [cfs] tp Qp shed area tc C A B C Q [acre] [min] A 120 0.68
22.4 B
160
0.83
29.8 C
220
0.79
44.8 tp
The two characteristics which define a synthetic triangular hydrograph are the --- Qp

9. Find: Qp [cfs] tp Qp shed area tc C A B C Q [acre] [min] A 120 0.68
22.4 B
160
0.83
29.8 C
220
0.79
44.8 tp
peak flowrate, Q sub p, and the time to peak, --- Qp
peak
flowrate

10. Find: Qp [cfs] tp Qp shed area tc C A B C Q [acre] [min] A 120 0.68
22.4 B
160
0.83
29.8 C
220
0.79
44.8
time to tp
t sub p. These two values define one of the three ---
peak Qp
peak
flowrate

11. Find: Qp [cfs] tp Qp shed area tc C A B C Q [acre] [min] A 120 0.68
22.4 B
160
0.83
29.8 C
220
0.79
44.8 tp
points of the triangular hydrograph. A second point is at the origin of the hydrograph, where --- Qp

12. Find: Qp [cfs] tp Qp shed area tc C A B C Q [acre] [min] A 120 0.68
22.4 B
160
0.83
29.8 C
220
0.79
44.8 tp
time equals zero and flow equals zero. And the third point is located at flow equals zero, --- Qp

13. Find: Qp [cfs] tp Qp shed area tc C 2.67*tp A B C Q [acre] [min] A 120
0.68
22.4 B
160
0.83
29.8 C
220
0.79
44.8 tp
time equals 2.67 times the time to peak. Often times, these two red lines are called the --- Qp
2.67*tp

14. Find: Qp [cfs] tp Qp shed area tc C rising limb receding limb 2.67*tp
[acre]
[min] A
120
0.68
22.4 B
160
0.83
29.8 C
220
0.79
44.8 tp
rising limb and falling limb, or receding limb, of the hydrograph. Unless a time to peak is provided, ---
rising limb Qp
receding limb
2.67*tp

15. Find: Qp [cfs] tp = 0.67 * tc tp Qp shed area tc C rising limb
[acre]
[min] A
120
0.68
22.4 B
160
0.83
29.8 C
220
0.79
44.8
tp = 0.67 * tc tp
it can be approximated to equal 0.67 times the time of concentration. The times of concentrations are ---
rising limb Qp
receding limb
2.67*tp

16. Find: Qp [cfs] tp = 0.67 * tc tp Qp shed area tc C rising limb
[acre]
[min] A
120
0.68
22.4 B
160
0.83
29.8 C
220
0.79
44.8
tp = 0.67 * tc tp
plugged into the equation, and the time to peaks, for each watershed, ----
rising limb Qp
receding limb
2.67*tp

17. Find: Qp [cfs] tp = 0.67 * tc tp Qp shed area tc tp C rising limb
[acre]
[min]
[min] A
120
0.68
22.4
15.0 B
160
0.83
29.8
20.0 C
220
0.79
44.8
30.0
tp = 0.67 * tc tp
are calculated, and rounded to the nearest minute. [pause] The peak flowrate, Q sub p, is calculated using the ---
rising limb Qp
receding limb
2.67*tp

18. Find: Qp [cfs] tp = 0.67 * tc tp Qp shed area tc tp C flowrate
[acre]
[min]
[min] A
120
0.68
22.4
15.0 B
160
0.83
29.8
20.0 C
220
0.79
44.8
30.0
flowrate
tp = 0.67 * tc tp
rational method, where the peak flowrate equals the product of ---
ft3 s
Qp = C * i * A Qp
2.67*tp

19. Find: Qp [cfs] tp = 0.67 * tc tp Qp shed area tc tp C flowrate
[acre]
[min]
[min] A
120
0.68
22.4
15.0 B
160
0.83
29.8
20.0 C
220
0.79
44.8
30.0
flowrate
tp = 0.67 * tc tp
the runoff coefficient, C, ---
ft3
Qp = C * i * A Qp s
2.67*tp
runoff
coefficient

20. Find: Qp [cfs] tp = 0.67 * tc tp Qp shed area tc tp C flowrate
[acre]
[min]
[min] A
120
0.68
22.4
15.0 B
160
0.83
29.8
20.0 C
220
0.79
44.8
30.0
flowrate
tp = 0.67 * tc tp
the intensity, i, in inches per hour, ---
ft3
Qp = C * i * A Qp s in
intensity
2.67*tp hr
runoff
coefficient

21. Find: Qp [cfs] tp = 0.67 * tc tp Qp shed area tc tp C area flowrate
[acre]
[min]
[min] A
120
0.68
22.4
15.0 B
160
0.83
29.8
20.0
area C
220
0.79
44.8
30.0
[acre]
flowrate
tp = 0.67 * tc tp
and the area, A, in acres. The problem statement provides the ---
ft3
Qp = C * i * A Qp s in
intensity
2.67*tp hr
runoff
coefficient

22. Find: Qp [cfs] tp = 0.67 * tc tp Qp shed area tc tp C area flowrate
[acre]
[min]
[min] A
120
0.68
22.4
15.0 B
160
0.83
29.8
20.0
area C
220
0.79
44.8
30.0
[acre]
flowrate
tp = 0.67 * tc tp
runoff coefficient and the area. To find the intensity, we’ll use the given ---
ft3
Qp = C * i * A Qp s in
intensity
2.67*tp hr
runoff
coefficient

23. Find: Qp [cfs] tp Qp shed area tc tp C 2.67*tp [acre] [min] [min] A
120
0.68
22.4
15.0 B
160
0.83
29.8
20.0 C
220
0.79
44.8
30.0
200 tp
i [in/hr]=
Intensity, duration, frequency curve, which relates the time of concentration to intensity, for a given return period and location.
tc [min]+20 Qp
2.67*tp

24. Find: Qp [cfs] tp Qp shed area tc tp C 2.67*tp [acre] [min] [min] A
120
0.68
22.4
15.0 B
160
0.83
29.8
20.0 C
220
0.79
44.8
30.0
200 tp
i [in/hr]=
Plugging in the time of concentration, t sub c, for each watershed, the intensities calculate to be ---
tc [min]+20 Qp
2.67*tp

25. Find: Qp [cfs] tp Qp shed area tc tp i C 2.67*tp [acre] [min] [min]
[in/hr] A
120
0.68
22.4
15.0
4.72 B
160
0.83
29.8
20.0
4.02 C
220
0.79
44.8
30.0
3.09
200 tp
i [in/hr]=
4.72 inches per hour, 4.02 inches per hour, and 3.09 inches per hour, respectively. Returning to Q=CiA, ----
tc [min]+20 Qp
2.67*tp

26. Find: Qp [cfs] tp Qp shed area tc tp i C Qp = C * i * A 2.67*tp [acre]
[min]
[min]
[in/hr] A
120
0.68
22.4
15.0
4.72 B
160
0.83
29.8
20.0
4.02 C
220
0.79
44.8
30.0
3.09
200 tp
i [in/hr]=
the runoff coefficients, intensities and areas ---
tc [min]+20 Qp
Qp = C * i * A
2.67*tp

27. Find: Qp [cfs] tp Qp shed area tc tp i C Qp = C * i * A 2.67*tp [acre]
[min]
[min]
[in/hr] A
120
0.68
22.4
15.0
4.72 B
160
0.83
29.8
20.0
4.02 C
220
0.79
44.8
30.0
3.09
200 tp
i [in/hr]=
are plugged into the equation, and the resulting flowrates equal ---
tc [min]+20 Qp
Qp = C * i * A
2.67*tp

28. Find: Qp [cfs] tp Qp shed area tp i Qp C Qp = C * i * A 2.67*tp [acre]
[min]
[in/hr]
[cfs] A
120
0.68
15.0
4.72
385 B
160
0.83
20.0
4.02
534 C
220
0.79
30.0
3.09
537
200 tp
i [in/hr]=
385, 534, and 537 cubic feet per second, for basins A, B and C, respectively. Lastly, we’ll want to calculate the ---
tc [min]+20 Qp
Qp = C * i * A
2.67*tp

29. Find: Qp [cfs] tp Qp shed area tp i Qp C Qp = C * i * A 2.67*tp [acre]
[min]
[in/hr]
[cfs] A
120
0.68
15.0
4.72
385 B
160
0.83
20.0
4.02
534 C
220
0.79
30.0
3.09
537
200 tp
i [in/hr]=
total duration of the hydrograph, 2.67 times the time to peak. Often times this value is assigned ---
tc [min]+20 Qp
Qp = C * i * A
2.67*tp

30. Find: Qp [cfs] tp tb Qp shed area tp i Qp C Qp = C * i * A 2.67*tp
[acre]
[min]
[in/hr]
[cfs] A
120
0.68
15.0
4.72
385 B
160
0.83
20.0
4.02
534 C
220
0.79
30.0
3.09
537
200 tp
i [in/hr]=
the variable t sub b, for time of base. Keeping our peak flowrate, --- tb
tc [min]+20 Qp
Qp = C * i * A
2.67*tp

31. Find: Qp [cfs] tp tb Qp tb shed Qp tp 2.67*tp [min] 40.1 53.4 80.1 A B
385
533
536
15.0
20.0
30.0 tp tp
time to peak, and time to base values, for each watershed, --- tb Qp
2.67*tp

32. Find: Qp [cfs] tb shed Qp tp Q [cfs] 600 500 400 300 200 100 t [min]
40.1
53.4
80.1 tb A B C
shed
[cfs] Qp
385
533
536
15.0
20.0
30.0 tp
Find: Qp [cfs]
Q [cfs]
600
500
400
300
the triangular hydrographs can be plotted.
200
100
t [min] 20 40 60 80

33. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 500 400 300 200 100 t [min]A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
500
400
300
Watershed A, will be blue, ---
200
100
t [min] 20 40 60 80

34. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 500 400 300 200 100 t [min]A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
500
400
300
Watershed B, will be orange, ---
200
100
t [min] 20 40 60 80

35. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 500 400 300 200 100 t [min]A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
500
400
300
and Watershed C, will be green.
200
100
t [min] 20 40 60 80

36. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 500 400 300 200 100 t [min]A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
500
use Δt=5 [min]
400
300
Since the problem states to use a time step of 5 minutes, we’ll need to interpolate to find the flowrate every 5 minutes.
200
100
t [min] 20 40 60 80

37. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 500 400 300 200 100 t [min]A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
500
use Δt=5 [min]
400
300
Using Watershed A as the example, the peak flowrate is ---
200
100
t [min] 20 40 60 80

38. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 385 500 400 300 200 100
[min]
[cfs]
[min]
Q [cfs] A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
385
500
use Δt=5 [min]
400
300
385 cubic feet per second, at 15 minutes, and the beginning and ending ---
200
100
t [min] 20 40 60 80

39. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 385 500 400 300 200 100
[min]
[cfs]
[min]
Q [cfs] A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
385
500
use Δt=5 [min]
400
300
flowrates are 0, cubic feet per second, ---
200
100
t [min] 20 40 60 80

40. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 385 500 400 308 257 300 231
[min]
[cfs]
[min]
Q [cfs] A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
385
500
use Δt=5 [min]
400
308
257
300
231
the remaining flowrates can all be calculated, by interpolation. This process is repeated for Watersheds B and C, ---
200
155
128 78
100
t [min] 20 40 60 80

41. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 500 400 300 200 100 t [min]A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
500
use Δt=5 [min]
400
300
and the flow rates for each time period for all hydrographs are added together. For example, ---
200
100
t [min] 20 40 60 80

42. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 500 400 300 128 200 100
[min]
[cfs]
[min]
Q [cfs] A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
500
use Δt=5 [min]
400
300
128
at 5 minutes, Watershed A generates 128 cubic feet per second, ---
200
100
t [min] 20 40 60 80

43. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 500 400 134 300 128 200 100
[min]
[cfs]
[min]
Q [cfs] A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
500
use Δt=5 [min]
400
134
300
128
Watershed B generates 134 cubic feet per second, ---
200
100
t [min] 20 40 60 80

44. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 500 400 134 300 128 200 100
[min]
[cfs]
[min]
Q [cfs] A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
500
use Δt=5 [min]
400
134
300
128
and Watershed C generates 90 cubic feet per second. These values sum to ----
200
100 90
t [min] 20 40 60 80

45. Find: Qp [cfs] shed tp Qp tb Q [cfs] 600 500 352 400 300 200 100
[min]
[cfs]
[min]
Q [cfs] A
15.0
385
40.1 B
20.0
533
53.4
600 C
30.0
536
80.1
500
352
use Δt=5 [min]
400
300
352 cubic feet per second. This summation repeats for time equal to ----
200
100
t [min] 20 40 60 80

46. Find: Qp [cfs] shed tp Qp tb Q [cfs] 704 600 500 352 400 162 300 269
[min]
[cfs]
[min]
Q [cfs] A
15.0
385
40.1
704 B
20.0
533
53.4
600 C
30.0
536
80.1
500
352
use Δt=5 [min]
400
162
300
269
10 minutes, 15 minutes, 20 minutes, and so on until no more flow. If we rescale the flow axis, ---
200 55
216
100
t [min]
109 20 40 60 80

47. Find: Qp [cfs] shed tp Qp tb Q [cfs] 1,200 800 600 400 200 t [min] 20A
15.0
385
40.1
1,200 B
20.0
533
53.4
1200 C
30.0
536
80.1
1000
use Δt=5 [min]
800
600
The peak flow rate equals 1,200 cubic feet per second, at a time of 20 minutes.
400
200
t [min] 20 40 60 80

48. Find: Qp [cfs] shed tp Qp tb Q [cfs] 1,050 1,200 1,300 1,450 1,200 800
[min]
[cfs]
[min]
Q [cfs] A
15.0
385
40.1
1,050
1,200
1,300
1,450
1,200 B
20.0
533
53.4
1200 C
30.0
536
80.1
1000
use Δt=5 [min]
800
600
When reviewing the possible solutions, ---
400
200
t [min] 20 40 60 80

49. Find: Qp [cfs] AnswerB shed tp Qp tb Q [cfs] 1,050 1,200 1,300 1,450
[min]
[cfs]
[min]
Q [cfs] A
15.0
385
40.1
1,050
1,200
1,300
1,450
1,200 B
20.0
533
53.4
1200 C
30.0
536
80.1
1000
AnswerB
use Δt=5 [min]
800
600
the answer is B.
400
200
t [min] 20 40 60 80

50. ? 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