Find: Qpeak [cfs] Time Unit Rainfall Infiltration

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Σ Find: Qpeak [cfs] Qn= Pm * Un-m+1 Time Unit [cfs/in] 5 20 45 80 64 22 Hydro Excess Rainfall [in] Runoff n [hour] [cfs] 0.5 0.9 0.7 (m=1) (m=2) (m=3) 1 0-1 2.5 2 1-2 14.5 3 2-3 44.0 4 3-4 94.5 5 4-5 135.5 hour of rainfall. [pause] Therefore, it may be easier ---- 6 5-6 124.6 7 6-7 64.6 n<M Qn= Pm * Un-m+1 Σ 8 7-8 15.4 m=1 base flow = 15 [cfs]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

? 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