1. Find: max d [ft] Qin d ψ = 0.1 [ft] Δθ = 0.3 time inflow 3 4 5 6
retention
time
[hr]
0-1
1-2
2-3
3-4
4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
sand 3 4 5 6
Find the maximum depth d, in feet. [pause] In this problem, ---
K=0.4 [ft/hr]
ψ = 0.1 [ft]
rectangular box
basin area = 1.0 [acre]
Δθ = 0.3
d=0 [ft] at t=0 [hr]

2. Find: max d [ft] Qin d ψ = 0.1 [ft] Δθ = 0.3 time inflow retention
[hr]
0-1
1-2
2-3
3-4
4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
stormwater runoff, with a given inflow hydrograph enters a retention basin.
sand
K=0.4 [ft/hr]
rectangular box
ψ = 0.1 [ft]
basin area = 1.0 [acre]
Δθ = 0.3
d=0 [ft] at t=0 [hr]

3. Find: max d [ft] Qin d ψ = 0.1 [ft] Δθ = 0.3 time inflow retention
[hr]
0-1
1-2
2-3
3-4
4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
The basin shape is a rectangular box, or more specifically, a rectangular parallelepiped, with a bottom area of 1.0 acres.
sand
K=0.4 [ft/hr]
rectangular box
ψ = 0.1 [ft]
basin area = 1.0 [acre]
Δθ = 0.3
d=0 [ft] at t=0 [hr]

4. Find: max d [ft] Qin d ψ = 0.1 [ft] Δθ = 0.3 time inflow retention
[hr]
0-1
1-2
2-3
3-4
4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
Water percolates down through the sand over this 1 acres area. The properties of the ----
sand
K=0.4 [ft/hr]
rectangular box
ψ = 0.1 [ft]
basin area = 1.0 [acre]
Δθ = 0.3
d=0 [ft] at t=0 [hr]

5. Find: max d [ft] Qin d ψ = 0.1 [ft] Δθ = 0.3 time inflow retention
[hr]
0-1
1-2
2-3
3-4
4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
sand are provided, and the depth of water in the retention basin is 0 feet, at time = 0 hours. [pause]
sand
K=0.4 [ft/hr]
rectangular box
ψ = 0.1 [ft]
basin area = 1.0 [acre]
Δθ = 0.3
d=0 [ft] at t=0 [hr]

6. Find: max d [ft] Qin d ψ = 0.1 [ft] Δθ = 0.3 time inflow retention
[hr]
0-1
1-2
2-3
3-4
4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
To solve this problem, the strategy will be to calculate the depth of water ---
sand
K=0.4 [ft/hr]
rectangular box
ψ = 0.1 [ft]
basin area = 1.0 [acre]
Δθ = 0.3
d=0 [ft] at t=0 [hr]

7. Find: max d [ft] time depth inflow infiltration depth [hr] [ft] [ft]
0-1
0.000
0.583
0.409
1-2
0.409
0.841
1.882
2-3
1.882
1.110
3.581
3-4
3.581
1.253
4.064
4-5
4.064
1.247
3.478
for each time period, starting at 0 feet, and accounting for the ---
5-6
3.478
1.151
2.575
K=0.4 [ft/hr]
basin shape = rectangular box
basin area = 1.0 [acre]
d=0 [ft] at t=0 [hr]

8. Find: max d [ft] time depth inflow infiltration depth [hr] [ft] [ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.841
1.882
2-3
1.882
2.810
1.110
3.581
3-4
3.581
1.736
1.253
4.064
4-5
4.064
0.661
1.247
3.478
the water entering the basin, shown in the inflow hydrograph, and the ---
5-6
3.478
0.248
1.151
2.575
K=0.4 [ft/hr]
basin shape = rectangular box
basin area = 1.0 [acre]
d=0 [ft] at t=0 [hr]

9. Find: max d [ft] time depth inflow infiltration depth [hr] [ft] [ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.841
1.882
2-3
1.882
2.810
1.110
3.581
3-4
3.581
1.736
1.253
4.064
4-5
4.064
0.661
1.247
3.478
water exiting the basin, by infiltration. The water depth at the end of an hour ---
5-6
3.478
0.248
1.151
2.575
K=0.4 [ft/hr]
basin shape = rectangular box
basin area = 1.0 [acre]
d=0 [ft] at t=0 [hr]

10. Find: max d [ft] time depth inflow infiltration depth d0 in0-1 out0-1
[hr]
[ft]
[ft]
[ft]
[ft] d0
in0-1
out0-1
0.583
0.409 d1
0-1
0.000
1-2
0.409
0.841
1.882
2-3
1.882
1.110
3.581
3-4
3.581
1.253
4.064
4-5
4.064
1.247
3.478
will equal the water depth at the beginning of the hour, plus the inflow, minus the infiltration. The depth at the end of the hour ---
5-6
3.478
1.151
2.575
dt+1=dt+inflowt-infiltrationt

11. Find: max d [ft] time depth inflow infiltration depth d0 in0-1 out0-1
[hr]
[ft]
[ft]
[ft]
[ft] d0
in0-1
out0-1
0.583 d1
0-1
0.409
1-2 d1
0.409
0.841
1.882
2-3
1.882
1.110
3.581
3-4
3.581
1.253
4.064
4-5
4.064
1.247
3.478
will equal the depth of water at the beginning of the subsequent hour, and this process will repeat for all ---
5-6
3.478
1.151
2.575
dt+1=dt+inflowt-infiltrationt

12. Find: max d [ft] time depth inflow infiltration depth d0 in0-1 out0-1
[hr]
[ft]
[ft]
[ft]
[ft] d0
in0-1
out0-1
0.583
0.409 d1
0-1
1-2
0.409 d1
0.841
1.882 d2
2-3 d2
1.882
1.110 d3
3.581
3-4
3.581 d3
1.253
4.064 d4
4-5
4.064 d4
1.247 d5
3.478
time periods, and the answer will be the maximum value between ---
5-6 d5
3.478
1.151
2.575 d6
dt+1=dt+inflowt-infiltrationt

13. Find: max d [ft] time depth inflow infiltration depth d0 in0-1 out0-1
[hr]
[ft]
[ft]
[ft]
[ft] d0
in0-1
out0-1
0.583 d1
0-1
0.409
1-2
0.409 d1
0.841
1.882 d2
2-3 d2
1.882
1.110 d3
3.581
3-4
3.581 d3
1.253
4.064 d4
4-5
4.064 d4
1.247 d5
3.478
all recorded water depths. [pause] Let’s begin by solving for the ---
5-6 d5
3.478
1.151
2.575 d6
dt+1=dt+inflowt-infiltrationt

14. Find: max d [ft] time depth inflow infiltration depth d0 in0-1 out0-1
[hr]
[ft]
[ft]
[ft]
[ft] d0
in0-1
out0-1
0.583 d1
0-1
0.409
1-2
0.409 d1
0.841
1.882 d2
2-3 d2
1.882
1.110 d3
3.581
3-4
3.581 d3
1.253
4.064 d4
4-5
4.064 d4
1.247 d5
3.478
inflow depth, in feet, for the first hour. The problem statement provides the ---
5-6 d5
3.478
1.151
2.575 d6
dt+1=dt+inflowt-infiltrationt

15. Find: max d [ft] Qin d time inflow retention [hr] 0-1 1-2 2-3 3-4 4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
inflow hydrograph, in units of cubic feet per second. Since we know the time period is ---
basin shape = rectangular box
basin area = 1.0 [acre]

16. Find: max d [ft] Qin d time inflow retention [hr] 0-1 1-2 2-3 3-4 4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
1 hour, we can determine the total volume of water which entered the basin that hour, and since we know the ---
Δt=1 [hr]
basin shape = rectangular box
basin area = 1.0 [acre]

17. Find: max d [ft] Qin d time inflow retention [hr] 0-1 1-2 2-3 3-4 4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
base area of the basin is 1 acre, we can compute the increase in water depth, due to the inflow, each hour. As an example, we’ll compute the ---
Δt=1 [hr]
basin shape = rectangular box
basin area = 1.0 [acre]

18. Find: max d [ft] Qin d time inflow retention [hr] 0-1 1-2 2-3 3-4 4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
inflow depth for the first time period, where the inflow is 12 cubic feet per second.
Δt=1 [hr]
basin shape = rectangular box
basin area = 1.0 [acre]

19. Find: max d [ft] Qin d time inflow in0-1= 12 1 [hr] retention [hr] 0-1
1-2
2-3
3-4
4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
after converting the units, we find the the added depth from the inflow for the first hour is ---
acre
ft3 1
3,600 s
in0-1= 12
1 [hr] * * * hr s
43,560
ft2 1
1 [acre]

20. Find: max d [ft] Qin d time inflow in0-1= 0.992 [ft] in0-1= 12 1 [hr]
retention
time
[hr]
0-1
1-2
2-3
3-4
4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
0.992 feet. Returning to the table of depth values, ---
in0-1= [ft]
acre
ft3 1
3,600 s
in0-1= 12
1 [hr] * * * hr s
43,560
ft2 1
1 [acre]

21. Find: max d [ft] time depth inflow infiltration depth in0-1= 12 1 [hr]
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.841
1.882
2-3
1.882
2.810
1.110
3.581
3-4
3.581
1.736
1.253
4.064
4-5
4.064
0.661
1.247
3.478
the depth of feet is entered into the inflow for the first hour. [pause] Since the added depth from inflow is ---
5-6
3.478
0.248
1.151
2.575
acre
ft3 1
3,600 s
in0-1= 12
1 [hr] * * * hr s
43,560
ft2 1
1 [acre]

22. Find: max d [ft] time depth inflow infiltration depth constant inflow
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.841
1.882
2-3
1.882
2.810
1.110
3.581
3-4
3.581
1.736
1.253
4.064
constant
4-5
4.064
0.661
1.247
3.478
simply the inflow in cubic feet per second, multiplied a constant, ---
inflow
5-6
3.478
0.248
1.151
2.575
acre
ft3 1
3,600 s
in0-1= 12
1 [hr] * * * hr s
43,560
ft2 1
1 [acre]

23. Find: max d [ft] Qin d time inflow constant inflow int-t+1= Qin
retention
time
[hr]
0-1
1-2
2-3
3-4
4-5
5-6
inflow
[cfs] 12 28 34 21 8 3
basin
Qin d
The remaining 5 hours of inflow can be individually, ---
constant
inflow
ft3 1 s
int-t+1= Qin * s
12.1
ft2

24. Find: max d [ft] Qin d time inflow constant inflow int-t+1= Qin
retention
time
inflow
basin
Qin
[hr]
[cfs]
0-1 12
1-2 28
2-3 34 d
3-4 21
4-5 8
plugged into the equation, and the calculated values ---
constant
inflow
5-6 3
ft3 1 s
int-t+1= Qin * s
12.1
ft2

25. Find: max d [ft] time depth inflow infiltration depth int-t+1= Qin
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.841
1.882
2-3
1.882
2.810
1.110
3.581
3-4
3.581
1.736
1.253
4.064
4-5
4.064
0.661
1.247
3.478
can be entered into the table, for added depth, from inflow. [pause] Next, we’ll need to determine the depth of water ---
5-6
3.478
0.248
1.151
2.575
acre
ft3 1
3,600 s
int-t+1= Qin
1 [hr] * * * hr s
43,560
ft2 1
1 [acre]

26. Find: max d [ft] time depth inflow infiltration depth out0-1 out1-2
[hr]
[ft]
[ft]
[ft]
[ft]
out0-1
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
out1-2
0.841
1.882
2-3
1.882
2.810
out2-3
1.110
3.581
3-4
3.581
1.736
out3-4
1.253
4.064
4-5
4.064
0.661
out4-5
1.247
3.478
exiting the retention basin for each time period, due to infiltration. Let’s first focus on the infiltration ---
5-6
3.478
0.248
out5-6
1.151
2.575
acre
ft3 1
3,600 s
int-t+1= Qin
1 [hr] * * * hr s
43,560
ft2 1
1 [acre]

27. Find: max d [ft] time depth inflow infiltration depth out0-1 out1-2
[hr]
[ft]
[ft]
[ft]
[ft]
out0-1
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
out1-2
0.841
1.882
2-3
1.882
2.810
out2-3
1.110
3.581
3-4
3.581
1.736
out3-4
1.253
4.064
4-5
4.064
0.661
out4-5
1.247
3.478
during the first hour. [pause] The green-ampt equation for cumulative infiltration is, ---
5-6
3.478
0.248
out5-6
1.151
2.575
acre
ft3 1
3,600 s
int-t+1= Qin
1 [hr] * * * hr s
43,560
ft2 1
1 [acre]

28. Find: max d [ft] Δθ * (d+ψ) Δθ = 0.3 ψ = 0.1 [ft] time depth inflow
infiltration
depth
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583
cumulative
infiltration [ft] F
F- Δθ * (d+ψ) * ln 1+
- K * t = 0
Δθ * (d+ψ)
the cumulative infiltration, capital F, in feet, minus ----
Δt=1 [hr]
Δθ = 0.3
ψ = 0.1 [ft]
K=0.4 [ft/hr]

29. Find: max d [ft] Δθ * (d+ψ) Δθ = 0.3 ψ = 0.1 [ft] time depth inflow
infiltration
depth
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583
cumulative
infiltration [ft] F
F- Δθ * (d+ψ) * ln 1+
- K * t = 0
Δθ * (d+ψ)
the change in moisture content, delta theta, as a decimal, times the quantity, ---
change in
moisture
content
Δt=1 [hr]
Δθ = 0.3
ψ = 0.1 [ft]
K=0.4 [ft/hr]

30. Find: max d [ft] Δθ * (d+ψ) Δθ = 0.3 ψ = 0.1 [ft] time depth inflow
infiltration
depth
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583
cumulative
infiltration [ft] F
F- Δθ * (d+ψ) * ln 1+
- K * t = 0
Δθ * (d+ψ)
the depth of water in the basin, d, in feet, plus ---
change in
moisture
depth [ft]
content
Δt=1 [hr]
Δθ = 0.3
ψ = 0.1 [ft]
K=0.4 [ft/hr]

31. Find: max d [ft] Δθ * (d+ψ) Δθ = 0.3 ψ = 0.1 [ft] time depth inflow
infiltration
depth
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583
cumulative
infiltration [ft] F
F- Δθ * (d+ψ) * ln 1+
- K * t = 0
Δθ * (d+ψ)
the suction head, psi, in feet, times the natural logarithm of ---
change in
suction head [ft]
moisture
depth [ft]
content
Δt=1 [hr]
Δθ = 0.3
ψ = 0.1 [ft]
K=0.4 [ft/hr]

32. Find: max d [ft] Δθ * (d+ψ) Δθ = 0.3 ψ = 0.1 [ft] time depth inflow
infiltration
depth
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583
cumulative
infiltration [ft] F
F- Δθ * (d+ψ) * ln 1+
- K * t = 0
Δθ * (d+ψ)
the quantity 1 plus the cumulative infiltration divided by the ---
change in
suction head [ft]
moisture
depth [ft]
content
Δt=1 [hr]
Δθ = 0.3
ψ = 0.1 [ft]
K=0.4 [ft/hr]

33. Find: max d [ft] Δθ * (d+ψ) Δθ = 0.3 ψ = 0.1 [ft] time depth inflow
infiltration
depth
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583
cumulative
infiltration [ft] F
F- Δθ * (d+ψ) * ln 1+
- K * t = 0
Δθ * (d+ψ)
Delta theta times d plus psi, minus, ---
change in
suction head [ft]
moisture
depth [ft]
content
Δt=1 [hr]
Δθ = 0.3
ψ = 0.1 [ft]
K=0.4 [ft/hr]

34. Find: max d [ft] Δθ * (d+ψ) Δθ = 0.3 ψ = 0.1 [ft] time depth inflow
infiltration
depth
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583
cumulative
hydraulic
conductivity
infiltration [ft]
[ft/hr] F
F- Δθ * (d+ψ) * ln 1+
- K * t = 0
Δθ * (d+ψ)
the hydraulic conductivity, capital K, in feet per hour, times ----
change in
suction head [ft]
moisture
depth [ft]
content
Δt=1 [hr]
Δθ = 0.3
ψ = 0.1 [ft]
K=0.4 [ft/hr]

35. Find: max d [ft] Δθ * (d+ψ) Δθ = 0.3 ψ = 0.1 [ft] time depth inflow
infiltration
depth
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583
cumulative
hydraulic
conductivity
infiltration [ft]
[ft/hr] F
F- Δθ * (d+ψ) * ln 1+
- K * t = 0
Δθ * (d+ψ)
the time, t, in hours. [pause] The values for change in moisture content, ---
change in
suction head [ft]
time
moisture
depth [ft]
content
Δt=1 [hr]
Δθ = 0.3
ψ = 0.1 [ft]
K=0.4 [ft/hr]

36. Find: max d [ft] Δθ * (d+ψ) Δθ = 0.3 ψ = 0.1 [ft] time depth inflow
infiltration
depth
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583
cumulative
hydraulic
conductivity
infiltration [ft]
[ft/hr] F
F- Δθ * (d+ψ) * ln 1+
- K * t = 0
Δθ * (d+ψ)
suction head, hydraulic conductivity, and time step, are plugged into the equation, and ---
change in
suction head [ft]
time
moisture
depth [ft]
content
Δt=1 [hr]
Δθ = 0.3
ψ = 0.1 [ft]
K=0.4 [ft/hr]

37. Find: max d [ft] Δθ * (d+ψ) ψ = 0.1 [ft] Δθ = 0.3 time depth inflow
infiltration
depth
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
a revised equation is calculated. [pause] At this point we’ll guess the water depth at 1 hour --- F
F- Δθ * (d+ψ) * ln 1+
- K * t = 0
Δθ * (d+ψ)
ψ = 0.1 [ft]
Δθ = 0.3
K=0.4 [ft/hr]
Δt=1 [hr]

38. Find: max d [ft] time depth inflow infiltration depth F d1 F
[hr]
[ft]
[ft]
[ft]
[ft] F d1
0-1
0.000
0.992
0.583 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d1=0.5 [ft]
guess
to be, 0.5 feet, and assuming constant infiltration during the hour, the average depth during the hour would equal ---

39. Find: max d [ft] time depth inflow infiltration depth F d1 F
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583 d1 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d1=0.5 [ft]
guess
0.25 feet. This value of 0.25 feet is plugged into the equation for d, ---
d0-1=0.25 [ft]

40. Find: max d [ft] time depth inflow infiltration depth F d1 F
[hr]
[ft]
[ft]
[ft]
[ft] F
0-1
0.000
0.992
0.583 d1 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d1=0.5 [ft]
guess
and the resulting cumulative infiltration equals ----
d0-1=0.25 [ft]

41. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992
0.600
0-1
0.583 d1 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d1=0.5 [ft]
guess
0.600 feet of infiltration, during the first hour.
d0-1=0.25 [ft]

42. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992
0.600
0-1
0.583 d1 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d1=0.5 [ft]
guess
Solving for the depth at 1 hour, ---
d0-1=0.25 [ft]
d1=d0+inflow0-infiltration0

43. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992
0.600
0-1
0.583 d1 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d1=0.5 [ft]
guess
the initial depth, the inflow and the infiltration are plugged in and ---
d0-1=0.25 [ft]
d1=d0+inflow0-infiltration0

44. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992
0.600 d1
0-1
0.583 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d1=0.5 [ft]
guess
the depth at the first hour calculates to feet. [pause] Now we check to see if the calculated depth ---
d0-1=0.25 [ft]
d1=d0+inflow0-infiltration0
d1=0.392 [ft]

45. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992 F d1
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992 F d1
0-1
0.583 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d1=0.5 [ft]
guess
at the first hour, feet, equals the guessed depth at the first hour, 0.5 feet, which it does not. We’ll use as our second approximation of the ---
d0-1=0.25 [ft]
d1=d0+inflow0-infiltration0
d1=0.392 [ft]
calculated

46. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992 F d1
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992 F d1
0-1
0.583 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d1=0.5 [ft]
guess
depth after the first hour, this would make the average depth during the first hour ---
d0-1=0.25 [ft]
d1=d0+inflow0-infiltration0
d1=0.392 [ft]
calculated

47. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992 F d1
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992 F d1
0-1
0.583 F
F- 0.3 * (d+0.1[ft]) * ln 1+
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d1=0.5 [ft]
guess
equals feet, and a second round of calculations is undertaken to find the resulting cumulative infiltration and calculated depth. If we set up a table -----
d0-1=0.196 [ft]
d0-1=0.25 [ft]
average
d1=d0+inflow0-infiltration0
d1=0.392 [ft]
calculated

48. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992 F d1
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992 F
0-1
0.583 d1
(guess)
(average)
(calculated)
(calculated) #
d1 [ft]
d0-1 [ft]
F [ft]
d1 [ft] 1
0.500
0.250
0.600
0.392
it is easier to track the guessed depth approach the calculated depth. For the second iteration, --- 2
0.392
0.196
.600
0.392

49. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992 F d1
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992 F
0-1
0.583 d1
(guess)
(average)
(calculated)
(calculated) #
d1 [ft]
d0-1 [ft]
F [ft]
d1 [ft] 1
0.500
0.250
0.600
0.392
the infiltration equals [feet] and the calculated depth equals feet. If we use feet --- 2
0.392
0.196
0.579
0.413

50. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992 F d1
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992 F
0-1
0.583 d1
(guess)
(average)
(calculated)
(calculated) #
d1 [ft]
d0-1 [ft]
F [ft]
d1 [ft] 1
0.500
0.250
0.600
0.392
For the next guess, the average depth for the first hour would equal --- 2
0.392
0.196
0.579
0.413 3
0.413
0.196
0.579
0.413

51. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992 F d1
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992 F
0-1
0.583 d1
(guess)
(average)
(calculated)
(calculated) #
d1 [ft]
d0-1 [ft]
F [ft]
d1 [ft] 1
0.500
0.250
0.600
0.392
0.207 feet, the cumulative infiltration would equal feet, and the calculated depth, at 1 hour, would equal feet. On the fourth iteration, --- 2
0.392
0.196
0.579
0.413 3
0.413
0.207
0.584
0.408

52. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992 F d1
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992 F
0-1
0.583 d1
(guess)
(average)
(calculated)
(calculated) #
d1 [ft]
d0-1 [ft]
F [ft]
d1 [ft] 1
0.500
0.250
0.600
0.392
the depth converges at feet or feet. 2
0.392
0.196
0.579
0.413 3
0.413
0.207
0.584
0.408 4
0.408
0.204
0.583
0.409

53. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992
0.583
0.409
0-1
(guess)
(average)
(calculated)
(calculated) #
d1 [ft]
d0-1 [ft]
F [ft]
d1 [ft] 1
0.500
0.250
0.600
0.392
We’ll use [pause] In review, at the beginning of the first hour, --- 2
0.392
0.196
0.579
0.413 3
0.413
0.207
0.584
0.408 4
0.408
0.204
0.583
0.409

54. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992
0.583
0.409
0-1
(guess)
(average)
(calculated)
(calculated) #
d1 [ft]
d0-1 [ft]
F [ft]
d1 [ft] 1
0.500
0.250
0.600
0.392
the depth equaled 0 feet. The inflow from 12 cubic feet per second--- 2
0.392
0.196
0.579
0.413 3
0.413
0.207
0.584
0.408 4
0.408
0.204
0.583
0.409

55. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992
0.583
0.409
0-1
(guess)
(average)
(calculated)
(calculated) #
d1 [ft]
d0-1 [ft]
F [ft]
d1 [ft] 1
0.500
0.250
0.600
0.392
added a depth of feet of water, while over the same hour, --- 2
0.392
0.196
0.579
0.413 3
0.413
0.207
0.584
0.408 4
0.408
0.204
0.583
0.409

56. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992
0.583
0.409
0-1
(guess)
(average)
(calculated)
(calculated) #
d1 [ft]
d0-1 [ft]
F [ft]
d1 [ft] 1
0.500
0.250
0.600
0.392
0.583 feet of water infiltrated into the sand, resulting in --- 2
0.392
0.196
0.579
0.413 3
0.413
0.207
0.584
0.408 4
0.408
0.204
0.583
0.409

57. Find: max d [ft] time depth inflow infiltration depth 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0.000
0.992
0.583
0.409
0-1
(guess)
(average)
(calculated)
(calculated) #
d1 [ft]
d0-1 [ft]
F [ft]
d1 [ft] 1
0.500
0.250
0.600
0.392
a final depth of feet, at 1 hour. The second hour is --- 2
0.392
0.196
0.579
0.413 3
0.413
0.207
0.584
0.408 4
0.408
0.204
0.583
0.409

58. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314 2 # 1 3 4
0.579
0.196
0.413
0.392
0.584
0.250
d1 [ft]
d0-1 [ft]
(guess)
(average)
F [ft]
(calculated)
0.500
0.600
0.207
0.408
0.583
0.204
0.409
calculated the same way. The initial depth is carried over from the ---

59. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314 2 # 1 3 4
0.579
0.196
0.413
0.392
0.584
0.250
d1 [ft]
d0-1 [ft]
(guess)
(average)
F [ft]
(calculated)
0.500
0.600
0.207
0.408
0.583
0.204
0.409
ft3 1 s
in1-2= 28 *
from the previous time period and the inflow was previously calculated. The infiltration during the second hour is determined --- s
12.1
ft2

60. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314 F
F- 0.3 * (d+0.1[ft]) * ln 1+
using the same iterative approach we used in the first hour.
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0

61. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314 F
F- 0.3 * (d+0.1[ft]) * ln 1+
If we guess a depth of 2 feet at the end of the second hour, ---
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d2=2.0 [ft]
guess

62. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314 F
F- 0.3 * (d+0.1[ft]) * ln 1+
then the average depth during the second hour would be feet, ---
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d2=2.0 [ft]
guess
d1-2=1.205 [ft]
0.409 [ft] [ft] 2

63. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314 F
F- 0.3 * (d+0.1[ft]) * ln 1+
which will be substituted in for the variable d, ---
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d2=2.0 [ft]
guess
d1-2=1.205 [ft]
0.409 [ft] [ft] 2

64. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314 F F
F- 0.3 * (d+0.1[ft]) * ln 1+
we solve for the corresponding infiltration, F, ----
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d2=2.0 [ft]
guess
d1-2=1.205 [ft]
0.409 [ft] [ft] 2

65. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314 F d F
F- 0.3 * (d+0.1[ft]) * ln 1+
then solve for the calculated depth, d, to be used as the next guess. [pause]
0.3 * (d+0.1[ft])
- 0.4 [ft] = 0
d2=2.0 [ft]
guess
d1-2=1.205 [ft]
0.409 [ft] [ft] 2

66. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
If we use 2.0 feet as our initial guess for the depth of water in the basin after 2 hours, ---
d2=2.0 [ft]
guess

67. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
(guess)
(average)
(calculated)
(calculated) #
d2 [ft]
d1-2 [ft]
F [ft]
d2 [ft]
We only need 3 iterations to determine the actual depth is feet, and the infiltration is feet during the second hour. 1
2.000
1.205
0.853
1.870 2
1.870
1.139
0.839
1.884 3
1.884
1.146
0.840
1.883

68. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.840
1.883
These values are plugged into the table, and the remaining values of ---

69. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.840
1.883
2-3
1.883
2.810
1.110
3.581
3-4
3.581
1.736
1.253
4.064
infiltration and depth are calculated.
4-5
4.064
0.661
1.247
3.478
5-6
3.478
0.248
1.151
2.575

70. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.840
1.883
2-3
1.883
2.810
1.110
3.581
3-4
3.581
1.736
1.253
4.064
The problem asks to find the maximum depth, d, ---
4-5
4.064
0.661
1.247
3.478
5-6
3.478
0.248
1.151
2.575

71. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.840
1.883
2-3
1.883
2.810
1.110
3.581
3-4
3.581
1.736
1.253
4.064
which is identify to be 4.064, feet, which occurs at 4 hours.
4-5
4.064
0.661
1.247
3.478
5-6
3.478
0.248
1.151
2.575

72. Find: max d [ft] time depth inflow infiltration depth 0-1 0.000 0.992
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.840
1.883
2-3
1.883
2.810
1.110
3.581 3 4 5 6
3-4
3.581
1.736
1.253
4.064
When reviewing the possible solutions, ---
4-5
4.064
0.661
1.247
3.478
5-6
3.478
0.248
1.151
2.575

73. Find: max d [ft] AnswerB time depth inflow infiltration depth 0-1
[hr]
[ft]
[ft]
[ft]
[ft]
0-1
0.000
0.992
0.583
0.409
1-2
0.409
2.314
0.840
1.883
2-3
1.883
2.810
1.110
3.581 3 4 5 6
3-4
3.581
1.736
1.253
4.064
the answer is B.
4-5
4.064
0.661
1.247
3.478
5-6
3.478
0.248
1.151
2.575
AnswerB

74. ? 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