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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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
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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
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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
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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
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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
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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
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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
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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]
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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 ---
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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]
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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]
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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]
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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
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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
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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]
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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
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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
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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
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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
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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
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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
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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
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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
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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] AnswerB 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 AnswerB
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
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