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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Find: Qp [cfs] shed tp Qp tb Q [cfs] 1,200 800 600 400 200 t [min] 20 A 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

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

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

? 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