Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow rain [hour] [hour] [m3/s] [cm] A B C D 0-4 14.8 1.2 1 8 2 4 4-8 21.2 2.4 2 20 4 8 8-12 40.0 3.8 4 40 12 18 12-16 72.0 2.2 10 64 18 28 16-20 121.2 0.0 16 108 30 40 20-24 174.0 0.0 12 50 14 20 Find the 4-hour unit hydrograph, in cubic meters per second, per centimeter of excess rainfall. [pause] In this problem, --- 24-28 188.8 0.0 2 16 4 6 28-32 126.4 watershed area = 180 [km2] 32-36 54.8 base flow = 14.0 [m3/s] 36-40 21.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow rain [hour] [hour] [m3/s] [cm] A B C D 0-4 14.8 1.2 1 8 2 4 4-8 21.2 2.4 2 20 4 8 8-12 40.0 3.8 4 40 12 18 12-16 72.0 2.2 10 64 18 28 16-20 121.2 0.0 16 108 30 40 20-24 174.0 0.0 12 50 14 20 Stream flow and rainfall data, from a given storm is provided, in 4 hour increments. 24-28 188.8 0.0 2 16 4 6 28-32 126.4 watershed area = 180 [km2] 32-36 54.8 base flow = 14.0 [m3/s] 36-40 21.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow rain [hour] [hour] [m3/s] [cm] A B C D 0-4 14.8 1.2 1 8 2 4 4-8 21.2 2.4 2 20 4 8 8-12 40.0 3.8 4 40 12 18 12-16 72.0 2.2 10 64 18 28 16-20 121.2 0.0 16 108 30 40 20-24 174.0 0.0 12 50 14 20 The watershed area, and base flowrate, are also provided. 24-28 188.8 0.0 2 16 4 6 28-32 126.4 watershed area = 180 [km2] 32-36 54.8 base flow = 14.0 [m3/s] 36-40 21.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow rain [hour] [hour] [m3/s] [cm] A B C D 0-4 14.8 1.2 1 8 2 4 4-8 21.2 2.4 2 20 4 8 8-12 40.0 3.8 4 40 12 18 12-16 72.0 2.2 10 64 18 28 16-20 121.2 0.0 16 108 30 40 20-24 174.0 0.0 12 50 14 20 Possible answers A, B, C and D are shown on the right, in units of meters cubed per second, per centimeter of excess rainfall. 24-28 188.8 0.0 2 16 4 6 28-32 126.4 watershed area = 180 [km2] 32-36 54.8 base flow = 14.0 [m3/s] 36-40 21.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow flowrate [hour] [hour] [m3/s] 0-4 14.8 4-8 21.2 8-12 40.0 12-16 72.0 16-20 121.2 20-24 174.0 Before developing the hydrograph, we’ll first need to account for the base flow. 24-28 188.8 28-32 126.4 time 32-36 54.8 base flow = 14.0 [m3/s] 36-40 21.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow flowrate [hour] [hour] [m3/s] base flow 0-4 14.8 4-8 21.2 8-12 40.0 12-16 72.0 16-20 121.2 20-24 174.0 The base flow rate in the stream plus the 24-28 188.8 28-32 126.4 time 32-36 54.8 base flow = 14.0 [m3/s] 36-40 21.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow flowrate [hour] [hour] [m3/s] base flow 0-4 14.8 + storm stream flow 4-8 21.2 8-12 40.0 12-16 72.0 16-20 121.2 20-24 174.0 excess stream flow, caused by a rainstorm, equals the --- 24-28 188.8 28-32 126.4 time 32-36 54.8 base flow = 14.0 [m3/s] 36-40 21.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow flowrate [hour] [hour] [m3/s] base flow 0-4 14.8 + storm stream flow 4-8 21.2 total stream flow 8-12 40.0 12-16 72.0 16-20 121.2 20-24 174.0 total recorded stream flow. [pause] To develop the hydrograph, were interested in the --- 24-28 188.8 28-32 126.4 time 32-36 54.8 base flow = 14.0 [m3/s] 36-40 21.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow flowrate [hour] [hour] [m3/s] base flow 0-4 14.8 + storm stream flow 4-8 21.2 total stream flow 8-12 40.0 12-16 72.0 16-20 121.2 20-24 174.0 stream flow generated by the storm, only. Therefore we can subtract the --- 24-28 188.8 28-32 126.4 time 32-36 54.8 base flow = 14.0 [m3/s] 36-40 21.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow flowrate [hour] [hour] [m3/s] base flow 0-4 14.8 + storm stream flow 4-8 21.2 total stream flow 8-12 40.0 12-16 72.0 total stream flow 16-20 121.2 - base flow 20-24 174.0 base flow, from the measured stream flow, to determine the stream flow data, from the storm only. [pause] Graphically, --- storm stream flow 24-28 188.8 28-32 126.4 time 32-36 54.8 base flow = 14.0 [m3/s] 36-40 21.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow flowrate [hour] [hour] [m3/s] 0-4 14.8 4-8 21.2 8-12 40.0 12-16 72.0 16-20 121.2 20-24 174.0 we can plot the total stream flow, from our given data set, then plot --- 24-28 188.8 time 28-32 126.4 total stream flow 32-36 54.8 - base flow 36-40 21.2 storm stream flow

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow flowrate [hour] [hour] [m3/s] 0-4 14.8 4-8 21.2 8-12 40.0 12-16 72.0 16-20 121.2 20-24 174.0 the constant base flow, as a negative value, shown in orange, and the the storm stream flow equals --- 24-28 188.8 time 28-32 126.4 total stream flow 32-36 54.8 - base flow 36-40 21.2 storm stream flow

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time stream flow flowrate [hour] [hour] [m3/s] 0-4 14.8 4-8 21.2 8-12 40.0 12-16 72.0 16-20 121.2 20-24 174.0 the stream flow, less the magnitude of the base flow, for all 10 measurements. [pause] To calculate the --- 24-28 188.8 time 28-32 126.4 total stream flow 32-36 54.8 - base flow 36-40 21.2 storm stream flow

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Flow rate [m3/s] [hour] [hour] flowrate total storm 0-4 14.8 4-8 21.2 8-12 40.0 12-16 72.0 16-20 121.2 20-24 174.0 stream flow from just the storm, --- 24-28 188.8 28-32 126.4 total stream flow 32-36 54.8 - base flow 36-40 21.2 storm stream flow

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Flow rate [m3/s] [hour] [hour] flowrate total storm 0-4 14.8 0.8 4-8 21.2 7.2 8-12 40.0 26.0 12-16 72.0 58.0 16-20 121.2 107.2 20-24 174.0 160.0 we can simply subtract 14 from the total flowrate. [pause] Also before developing the hydrograph, --- 24-28 188.8 174.8 28-32 126.4 112.4 total stream flow 32-36 54.8 40.8 - base flow 36-40 21.2 7.2 storm stream flow

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Flow rate [m3/s] [hour] [hour] total storm 0-4 14.8 0.8 4-8 21.2 7.2 δ F f= 8-12 40.0 26.0 δ t 12-16 72.0 58.0 16-20 121.2 107.2 20-24 174.0 160.0 the infiltration of rainfall into the soil, must be accounted for. 24-28 188.8 174.8 28-32 126.4 112.4 32-36 54.8 40.8 36-40 21.2 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Flow rate [m3/s] [hour] [hour] total storm 0-4 14.8 0.8 4-8 21.2 7.2 δ F f= 8-12 40.0 26.0 δ t 12-16 72.0 58.0 16-20 121.2 107.2 infiltration cm 20-24 174.0 160.0 rate The infiltration rate, in centimeters per 4 hour time period, equals, --- 4 hr 24-28 188.8 174.8 28-32 126.4 112.4 32-36 54.8 40.8 36-40 21.2 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Flow rate [m3/s] total [hour] [hour] total storm infiltration [cm] 0-4 14.8 0.8 4-8 21.2 7.2 δ F f= 8-12 40.0 26.0 δ t 12-16 72.0 58.0 16-20 121.2 107.2 infiltration cm 20-24 174.0 160.0 rate the total infiltration, in centimeters, divided by the --- 4 hr 24-28 188.8 174.8 28-32 126.4 112.4 32-36 54.8 40.8 36-40 21.2 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Flow rate [m3/s] total [hour] [hour] total storm infiltration [cm] 0-4 14.8 0.8 4-8 21.2 7.2 δ F storm f= duration 8-12 40.0 26.0 δ t [4 hr] 12-16 72.0 58.0 16-20 121.2 107.2 infiltration cm 20-24 174.0 160.0 rate storm duration, in 4 hour time increments. 4 hr 24-28 188.8 174.8 28-32 126.4 112.4 32-36 54.8 40.8 36-40 21.2 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Flow rate [m3/s] total [hour] [hour] total storm infiltration [cm] 0-4 14.8 0.8 4-8 21.2 7.2 Δ F storm f= duration 8-12 40.0 26.0 Δ t [4 hr] 12-16 72.0 58.0 16-20 121.2 107.2 infiltration cm 20-24 174.0 160.0 rate For this problem we’ll assume a constant infiltration rate, and use the capital Delta notation. 4 hr 24-28 188.8 174.8 28-32 126.4 112.4 32-36 54.8 40.8 36-40 21.2 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Flow rate [m3/s] total [hour] [hour] total storm infiltration [cm] 0-4 14.8 0.8 4-8 21.2 7.2 Δ F storm f= duration 8-12 40.0 26.0 Δ t [4 hr] 12-16 72.0 58.0 16-20 121.2 107.2 infiltration cm 20-24 174.0 160.0 rate We’ll first determine the total infiltration, delta F, which equals, --- 4 hr 24-28 188.8 174.8 28-32 126.4 112.4 32-36 54.8 40.8 36-40 21.2 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Flow rate [m3/s] total [hour] [hour] total storm infiltration [cm] 0-4 14.8 0.8 4-8 21.2 7.2 Δ F f= 8-12 40.0 26.0 Δ t 12-16 72.0 58.0 16-20 121.2 107.2 Δ F=drain-drunoff 20-24 174.0 160.0 the total rainfall depth, minus, the total runoff depth. 24-28 188.8 174.8 28-32 126.4 112.4 32-36 54.8 40.8 36-40 21.2 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t 12-16 58.0 2.2 16-20 107.2 Δ F=drain-drunoff 20-24 160.0 The depth of rainfall is the sum of the --- 24-28 174.8 28-32 112.4 32-36 40.8 36-40 7.2 watershed area = 180 [km2]

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t 12-16 58.0 2.2 16-20 107.2 Δ F=drain-drunoff Σ 20-24 160.0 individual rainfall entries, which equals, 9.6 centimeters. [pause] The depth of runoff equals the --- 9.6 [cm] 24-28 174.8 28-32 112.4 32-36 40.8 36-40 7.2 watershed area = 180 [km2]

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t 12-16 58.0 2.2 16-20 107.2 Δ F=drain-drunoff Σ 20-24 160.0 volume of runoff, divided by the watershed area. 9.6 [cm] 24-28 174.8 Vrunoff 28-32 112.4 Awatershed 32-36 40.8 36-40 7.2 watershed area = 180 [km2]

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t 12-16 58.0 2.2 16-20 107.2 Δ F=drain-drunoff 20-24 160.0 The volume of runoff equals the sum of the flow rates, times the time step, delta t. 9.6 [cm] 24-28 174.8 Vrunoff ΣQ *Δ t 28-32 112.4 Awatershed 32-36 40.8 36-40 7.2 watershed area = 180 [km2]

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t 12-16 58.0 2.2 16-20 107.2 Δ F=drain-drunoff 20-24 160.0 The sum of the flow rates equals 694.4 meters cubed per second, and the time step is 4 hours. 9.6 [cm] 24-28 174.8 Vrunoff ΣQ *Δ t 28-32 112.4 Awatershed 32-36 40.8 m3 4 [hr] 36-40 7.2 694.4 area=180 [km2] s

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t 12-16 58.0 2.2 16-20 107.2 Δ F=drain-drunoff 20-24 160.0 The area of the watershed is 180 kilometers squared, and after a few conversions are made, --- 9.6 [cm] 24-28 174.8 Vrunoff ΣQ *Δ t 28-32 112.4 Awatershed 32-36 40.8 m3 4 [hr] 36-40 7.2 694.4 area=180 [km2] s

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t 5.56 [cm] 12-16 58.0 2.2 16-20 107.2 Δ F=drain-drunoff 20-24 160.0 the runoff depth computes to 5.56 centimeters. [pause] Therefore the total infiltration, delta F, equals, --- 9.6 [cm] 24-28 174.8 Vrunoff ΣQ *Δ t 28-32 112.4 Awatershed 32-36 40.8 m3 4 [hr] 36-40 7.2 694.4 area=180 [km2] s

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4.04 [cm] 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t 5.56 [cm] 12-16 58.0 2.2 16-20 107.2 Δ F=drain-drunoff 20-24 160.0 4.04 centimeters. [pause] Since the rainfall lasts for --- 9.6 [cm] 24-28 174.8 Vrunoff ΣQ *Δ t 28-32 112.4 Awatershed 32-36 40.8 m3 4 [hr] 36-40 7.2 694.4 area=180 [km2] s

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4.04 [cm] 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t storm 12-16 58.0 2.2 duration 16-20 107.2 [4 hr] 20-24 160.0 16 hours of measurement, the storm duration is -- 24-28 174.8 28-32 112.4 32-36 40.8 36-40 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4.04 [cm] 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t storm 12-16 58.0 2.2 duration 4 [4 hr] 16-20 107.2 [4 hr] 20-24 160.0 4, 4-hour time periods. [pause] The infiltration rate, f, equals --- 24-28 174.8 28-32 112.4 32-36 40.8 36-40 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4.04 [cm] 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t storm 12-16 58.0 2.2 duration 4 [4 hr] 16-20 107.2 [4 hr] 20-24 160.0 1.01 centimeters, ever 4 hours, which we will --- cm f=1.01 24-28 174.8 4 hr 28-32 112.4 32-36 40.8 36-40 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain total [hour] [hour] flow[m3/s] [cm] infiltration [cm] 0-4 0.8 1.2 4.04 [cm] 4-8 7.2 2.4 Δ F f= 8-12 26.0 3.8 Δ t storm 12-16 58.0 2.2 duration 4 [4 hr] 16-20 107.2 [4 hr] 20-24 160.0 round to 1.0 centimeters, ever 4 hours. [pause] Similar to the stream flow, --- 4 hr cm f=1.01 24-28 174.8 round 28-32 112.4 down f=1.0 cm 32-36 40.8 4 hr 36-40 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain [hour] [hour] flow[m3/s] [cm] 0-4 0.8 1.2 4-8 7.2 2.4 8-12 26.0 3.8 12-16 58.0 2.2 16-20 107.2 4 hr cm f=1.0 depth 20-24 160.0 the depth used to develop the unit hydrograph --- 24-28 174.8 28-32 112.4 32-36 40.8 36-40 7.2 time

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain [hour] [hour] flow[m3/s] [cm] 0-4 0.8 1.2 4-8 7.2 2.4 8-12 26.0 3.8 12-16 58.0 2.2 16-20 107.2 cm f=1.0 depth 20-24 160.0 4 hr should only be the rainfall which becomes runoff, --- Rain 24-28 174.8 - Infiltration 28-32 112.4 32-36 40.8 36-40 7.2 time

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain [hour] [hour] flow[m3/s] [cm] 0-4 0.8 1.2 4-8 7.2 2.4 8-12 26.0 3.8 12-16 58.0 2.2 16-20 107.2 cm f=1.0 depth 20-24 160.0 4 hr which we’ll call, the runoff depth. [pause] Graphically, --- Rain 24-28 174.8 - Infiltration 28-32 112.4 32-36 40.8 Runoff Depth 36-40 7.2 time

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain [hour] [hour] flow[m3/s] [cm] 0-4 0.8 1.2 4-8 7.2 2.4 8-12 26.0 3.8 4 hr cm 1.0 12-16 58.0 2.2 16-20 107.2 Rain depth 20-24 160.0 the rain, infiltration and runoff depth would plot as shown, and the runoff depths would tabulate to - Infiltration 24-28 174.8 Runoff 28-32 112.4 Depth 32-36 40.8 time 36-40 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Time Storm Rain Runoff [hour] [hour] flow[m3/s] [cm] depth[cm] 0-4 0.8 1.2 0.2 4-8 7.2 2.4 1.4 8-12 26.0 3.8 2.8 4 hr cm 1.0 12-16 58.0 2.2 1.2 16-20 107.2 Rain depth 20-24 160.0 1 centimeter less than the recorded rainfall, for each 4-hour time period. [pause] At this point, we can start calculating the unit hydrograph. - Infiltration 24-28 174.8 Runoff 28-32 112.4 Depth 32-36 40.8 time 36-40 7.2

Qn 1.0 Rain - Infiltration Runoff Depth Time Storm Rain Runoff [hour] flow[m3/s] [cm] depth[cm] 0-4 0.8 1.2 0.2 4-8 7.2 2.4 1.4 8-12 26.0 3.8 2.8 4 hr cm 1.0 12-16 58.0 2.2 1.2 16-20 107.2 Rain depth 20-24 160.0 For sake of convenience, lets the designate the storm flows, with Q sub n, and --- - Infiltration 24-28 174.8 Runoff 28-32 112.4 Depth 32-36 40.8 time 36-40 7.2

Qn Pm 1.0 Rain - Infiltration Runoff Depth Time Storm Rain Runoff [hour] [hour] flow[m3/s] [cm] depth[cm] 0-4 0.8 1.2 0.2 4-8 7.2 2.4 1.4 8-12 26.0 3.8 2.8 4 hr cm 1.0 12-16 58.0 2.2 1.2 16-20 107.2 Rain depth 20-24 160.0 the runoff depth,with P sub m. [pause] And the goal is to solve for the --- - Infiltration 24-28 174.8 Runoff 28-32 112.4 Depth 32-36 40.8 time 36-40 7.2

Σ Qn Pm Qn= Pm * Un-m+1 unit hydrograph ordinates discrete convolution Time Storm Rain Runoff [hour] [hour] flow[m3/s] [cm] depth[cm] 0-4 0.8 1.2 0.2 4-8 7.2 2.4 1.4 8-12 26.0 3.8 2.8 unit 12-16 58.0 2.2 1.2 hydrograph 16-20 107.2 ordinates 20-24 160.0 unit hydrograph ordinates, U sub n-m+1, as defined in the discrete convolution equation. We set up the hydrograph, so it reads --- 24-28 174.8 n<m Σ Qn= Pm * Un-m+1 28-32 112.4 m=1 32-36 40.8 discrete convolution 36-40 7.2 equation

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 5 16-20 107.2 6 20-24 160.0 similar to our convolution equation, where the storm flow --- 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph = Σ m3 Qn= Pm * Un-m+1 s*cm Qn Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 5 16-20 107.2 6 20-24 160.0 equals the the sum of the product of the runoff depths, times the unit hydrograph ordinates. [pause] The variable n, --- 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 5 16-20 107.2 6 20-24 160.0 represent time periods, and the variable m, represents --- 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 5 16-20 107.2 6 20-24 160.0 the quantity of flow each single hour of rainfall contributes to the total runoff, --- 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 2 4-8 7.2 3 8-12 26.0 m=1 m=2 m=3 m=4 4 12-16 58.0 5 16-20 107.2 6 20-24 160.0 where the four columns equal m’s 1 through 4. [pause] Beginning at the first row, --- 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 5 16-20 107.2 6 20-24 160.0 Since n equals 1, the second, third and fourth terms --- 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 5 16-20 107.2 6 20-24 160.0 are zero, since the rainfall has not yet begun for those time periods. 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 5 16-20 107.2 6 20-24 160.0 From our data table, we recall the runoff depth for the first four hours is 0.2 centimeters, therefore, --- 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 0.8 = 0.2 * U1 5 16-20 107.2 6 20-24 160.0 0.8 meter cubed per second equals 0.2 centimeters times U sub 1. Temporarily dropping the units, --- 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 0.8 = 0.2 * U1 5 16-20 107.2 0.8 6 20-24 160.0 U1 = and solving for U sub 1, it equals, ---- 0.2 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 0.8 = 0.2 * U1 5 16-20 107.2 0.8 m3 6 20-24 160.0 U1 = = 4 4 meters cubed per second, of runoff, per centimeter of excess rainfall. 0.2 s*cm 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 3 8-12 26.0 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 5 16-20 107.2 6 20-24 160.0 The second time period is solved the same way. But this time, the first two hours of runoff --- 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 U1 =4 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 0.2*U2 1.4*U1 3 8-12 26.0 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 5 16-20 107.2 6 20-24 160.0 contribute to the stream flow, because n equals 2. 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 U1 =4 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 0.2*U2 1.4*U1 3 8-12 26.0 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 7.2 = 0.2*U2 + 1.4*U1 5 16-20 107.2 6 20-24 160.0 Solving the convolution equation for the second ordinate, --- 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 U1 =4 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 0.2*U2 1.4*U1 3 8-12 26.0 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 7.2 = 0.2*U2 + 1.4*U1 5 16-20 107.2 7.2- 1.4*U1 U2 = 6 20-24 160.0 and plugging in the first ordinate, U sub 1, --- 0.2 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 U1 =4 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 0.2*U2 1.4*U1 3 8-12 26.0 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 7.2 = 0.2*U2 + 1.4*U1 5 16-20 107.2 7.2- 1.4*U1 m3 U2 = 6 20-24 160.0 = 8 U sub 2 equals 8 meters cubed per second, per centimeter of excess rainfall. 0.2 s*cm 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 U1 =4 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 0.2*U2 1.4*U1 3 8-12 26.0 0.2*U3 1.4*U2 2.8*U1 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 26.0 = 0.2*U3 + 1.4*U2 + 2.8*U1 5 16-20 107.2 6 20-24 160.0 For the third time period, we can solve for U sub 3 by writing out the convolution equation, 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 U1 = 4 U2 = 8 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 0.2*U2 1.4*U1 3 8-12 26.0 0.2*U3 1.4*U2 2.8*U1 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 26.0 = 0.2*U3 + 1.4*U2 + 2.8*U1 5 16-20 107.2 26- 1.4*U2 - 2.8*U1 6 20-24 160.0 U3= plugging in U sub 1 and U sub 2, and we calculate the third unit ordinate as, --- 0.2 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 U1 = 4 U2 = 8 9 32-36 40.8 10 36-40 7.2

Σ Find: 4-hr Unit Hydrograph m3 Qn= Pm * Un-m+1 s*cm Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 0.2*U2 1.4*U1 3 8-12 26.0 0.2*U3 1.4*U2 2.8*U1 4 12-16 58.0 depth[cm] Runoff P1 = 0.2 P2 = 1.4 P3 = 2.8 P4 = 1.2 26.0 = 0.2*U3 + 1.4*U2 + 2.8*U1 5 16-20 107.2 26- 1.4*U2 - 2.8*U1 m3 6 20-24 160.0 U3= = 18 18 meters cubed per second, per centimeter of excess rainfall. If we continue this process for the entire hydrograph, --- 0.2 s*cm 7 24-28 174.8 8 28-32 112.4 Qn= Pm * Un-m+1 Σ n<M m=1 U1 = 4 U2 = 8 9 32-36 40.8 10 36-40 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 0.2*U2 1.4*U1 3 8-12 26.0 0.2*U3 1.4*U2 2.8*U1 4 12-16 58.0 5 16-20 107.2 6 20-24 160.0 we can complete the table, --- 7 24-28 174.8 8 28-32 112.4 9 32-36 40.8 10 36-40 7.2

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 2 4-8 7.2 0.2*U2 1.4*U1 3 8-12 26.0 0.2*U3 1.4*U2 2.8*U1 4 12-16 58.0 0.2*U4 1.4*U3 2.8*U2 1.2*U1 5 16-20 107.2 0.2*U5 1.4*U4 2.8*U3 1.2*U2 6 20-24 160.0 0.2*U6 1.4*U5 2.8*U4 1.2*U3 and solve for all unit hydrograph ordinate values. --- 7 24-28 174.8 0.2*U7 1.4*U6 2.8*U5 1.2*U4 8 28-32 112.4 1.4*U7 2.8*U6 1.2*U5 9 32-36 40.8 2.8*U7 1.2*U6 10 36-40 7.2 1.2*U7

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph Qn Pm * Un-m+1 Time n [hour] [m3/s] P1*Un P2*Un-1 P3*Un-2 P4*Un-3 1 0-4 0.8 0.2*U1 U1 = 4 U2 = 8 U3=18 U4=28 U5=40 U6=20 U7= 6 2 4-8 7.2 0.2*U2 1.4*U1 3 8-12 26.0 0.2*U3 1.4*U2 2.8*U1 4 12-16 58.0 0.2*U4 1.4*U3 2.8*U2 1.2*U1 5 16-20 107.2 0.2*U5 1.4*U4 2.8*U3 1.2*U2 6 20-24 160.0 0.2*U6 1.4*U5 2.8*U4 1.2*U3 [pause] 7 24-28 174.8 0.2*U7 1.4*U6 2.8*U5 1.2*U4 8 28-32 112.4 1.4*U7 2.8*U6 1.2*U5 9 32-36 40.8 2.8*U7 1.2*U6 10 36-40 7.2 1.2*U7

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph A B C D 1 8 2 4 U1 = 4 U2 = 8 U3=18 U4=28 U5=40 U6=20 U7= 6 2 20 4 8 4 40 12 18 10 64 18 28 16 108 30 40 12 50 14 20 When reviewing the possible solutions, --- 2 16 4 6

Find: 4-hr Unit Hydrograph s*cm m3 Find: 4-hr Unit Hydrograph A B C D 1 8 2 4 U1 = 4 U2 = 8 U3=18 U4=28 U5=40 U6=20 U7= 6 2 20 4 8 4 40 12 18 10 64 18 28 16 108 30 40 12 50 14 20 The answer is D. 2 16 4 6 AnswerD

? 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