Solution Manual for Hydrology and Floodplain Analysis, 6th Edition

Chapter 2-ANSWER KEY Concept Check 2.1 What characterizes a watershed? ANSWER: A watershed is the basic unit used in most hydrologic calculations relating to the water balance or computation of rainfall-runoff. It is characterized by one main channel and by tributaries that drain into a main channel at one or more confluence points. Larger watersheds can have many subareas that contribute runoff to a single outlet. The drainage area is another characteristic that reflects the volume of water that can be generated from rainfall. 2-1 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.2 Describe the four assumptions made that are inherent to the unit hydrograph. ANSWER: 1. Rainfall excesses of equal duration are assumed to produce hydrographs with equivalent time bases regardless of the intensity of the rain. 2. Direct runoff ordinates for a storm if given duration are assumed directly proportional to rainfall excess volumes. Thus, twice the rainfall produces a doubling hydrograph ordinates. 3. The time distribution of direct runoff is assumed independent of antecedent precipitation. 4.Rainfall distribution is assumed to be the same for all storms of equl duration, both spatially and temporally. These assumptions can sometimes limit the application of a unit hydrograph in a given watershed. 2-2 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.3 What physical factors affect the shape and timing of the unit hydrograph? ANSWER: The shape and timing of the hydrograph are related to duration and intensity of rainfall as well as the various factors governing the watershed area. The meteorological factors that influence the hydrograph shape and volume of runoff include: 1.Rainfall intensity and pattern 2.Areal distribution of rainfall over the basin The various factors that govern the watershed include: 1.Size and shape of the drainage area 2.Slope of the land surface and the main channel 3.Channel morphology and drainage type 4.Soil types and distribution 5.Storage detention in the watershed. 2-3 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.4 Discuss the parameters that describe the main timing aspects of the hydrograph. ANSWER: The physical factors that will affect the shape and timing of the unit hydrograph are: a. Duration of the rainfall excess: the time from start to finish of rainfall excess b. Lag Time: the time from the center of mass of rainfall excess to the peak of the hydrograph. c. Time of rise: the time from the start of rainfall excess to the peak of the hydrograph. d. Time of concentration: the time for a wave (of water) to propagate from the most distant point in the watershed to the outlet. One estimate is the time from the end of net rainfall to the inflection point of the hydrograph. 2-4 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.5 What two characteristics of the hydrograph do most methods for synthetic unit hydrographs relate? ANSWER: They relate hydrograph peak flow and timing to watershed characteristics. 2-5 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.6 In order to determine which of the three excavation sites could have potentially contributed to the sedimentation of the three lakes, it is necessary to determine the watershed boundaries and the contributing drainage areas of the system of lakes. Catchment basins (watersheds) can be determined by connecting ridge lines and dividing lines. Ridge lines will follow the elevation isocontours and dividing lines are perpendicular to the elevation isocontours at the end and start at the confluences of streams.Using the map on the textbook website displaying the elevation data, determine the following: a. How many confluences are displayed on the map? b. Delineate the watersheds using the elevation isocontours. The total drainage area of the study area is approximately 238 acres. c. Determine which of the three lakes may be affected by accelerated sedimentation. d. Determine which of the three excavation sites may be responsible for the accelerated sedimentation of the lakes. ANSWER: a. There are three confluence points b. See picture below c. Lakes 1 (lake 2 minimally) will be affected by the accelerated excavation. d. The sites that are responsible are sites 1 and 2. Three can be eliminated because it is outside the watershed. 2-6 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.6 cont’ HOMEWORK PROBLEMS 2.7 A watershed has the following characteristics: 2-7 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. A = 2500 ac L = 3.9 mi. S = 52 ft/mi. y = 1% The channel is lined with concrete. The watershed is a residential area with 1/4-ac lots. The soil is categorized as soil group B. Assume that the average watershed slope is the same as the channel slope. Determine the UH for this area for a storm duration of 1 hr using the SCS triangular UH method. ANSWER: From table 2-1 obtain the curve number with the given information. CN = 75 S = (1000/CN) – 10 = (1000/75) – 10 = 3.33 L = 3.9 mi. = (3.9mi.)(5280 ft/mi.) = 20592 ft. Plug into equation: tp  tp  L0.8 ( S  1) 0.7 1900 y (21,120) 0.8 (3.33  1) 0.7 1900 1 = 4.15 hr TR = D/2 + tp = 1/2 + 4.15 = 4.65 hr Qp  484 A  TR 484  2600ac  4.73 1mi.2 640ac  406 cfs 2.7 cont’ Vol = (1 in.)(2500 ac) = 2500 ac-in. ~ 2500 cfs-hr 2-8 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Vol = (1/2)(Qp)(TR + B) 2Vol  TR = 7.67 hr QP Flow (cfs) B 450 0 400 0.5 350 1 1.5 300 2 250 2.5 200 3 150 3.5 100 4 50 4.5 0 5 0 2 4 6 8 10 12 14 Time (hr) 2-9 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.8 Determine the storm hydrograph resulting from the rainfall pattern in Fig. P2–8(a) using the triangular 1-hr UH given in Fig. P2–8(b). Change Rainfall to 1, 2, 3, 2 Fig. P2-8a Fig. P2-8b ANSWER: Since the rainfall is given in net rainfall, we do not need to subtract any losses. The rainfall for each time period is multiplied by the unit hydrograph ordinates to obtain the storm hydrograph. This computational procedure is in the table below: 2.8 cont’ 2-10 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. In order to obtain R 1 U through R 4 U. You take the UH units and you multiply it by the intensity for that given hour. Once you proceed to the next hour the values are lagged one hour. To obtain Q the values are then added across the table. Time (hrs) 0 1 2 3 4 5 6 7 8 9 R2 U R1 U UH 0 0.5 1 0.75 0.5 0.25 0 0 0.5 1 0.75 0.5 0.25 0 1 2 1.5 1 0.5 0 R3U R4 U 0 1.5 3 2.25 1.5 0.75 0 Q 0 0.5 2 4.25 6 5.5 3.5 1.75 0.5 0 0 1 2 1.5 1 0.5 0 The hydrograph can then be plotted (below is new hydrograph) 7 6 Flow (cfs) 5 4 3 2 1 0 0 2 4 6 8 10 Time (hr) 2-11 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.9 (a) Given a triangular 1-hr UH with TB  12 hr, TR  4 hr, QP  200 cfs, where TB  time base of the UH, TR  time of rise, QP  peak flow, develop a storm hydrograph for hourly rainfall (in.) of P  0.1, 0.5, 1.2  . (b) Repeat the above problem for hourly rainfall (in.) of P  0.2, 1.0, 2.4  . ANSWER: The UH ordinates are derived using the triangular 1-hr UH graph in 1 hour intervals: 2.9 cont’ 2-12 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Time (hr) U (cfs) 0 0 1 50 2 100 3 150 4 200 5 175 6 150 7 125 8 100 9 75 10 50 11 25 12 0 The storm hydrograph is developed for both cases (a) and (b) with the use of the convolution equation. 2.9 cont’ 2-13 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. a. Time (hr) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 U (cfs) 0 50 100 150 200 175 150 125 100 75 50 25 0 P1*U 0 5 10 15 20 17.5 15 12.5 10 7.5 5 2.5 0 P2*U 0 25 50 75 100 87.5 75 62.5 50 37.5 25 12.5 0 P3*U 0 60 120 180 240 210 180 150 120 90 60 30 0 Q (cfs) 0 5 35 125 215 297.5 342.5 297.5 252.5 207.5 162.5 117.5 72.5 30 0 2-14 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.9 cont’ b. Time (hr) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 U (cfs) 0 50 100 150 200 175 150 125 100 75 50 25 0 P1*U P2*U 0 10 20 30 40 35 30 25 20 15 10 5 0 0 50 100 150 200 175 150 125 100 75 50 25 0 P3*U 0 120 240 360 480 420 360 300 240 180 120 60 0 Q 0 10 70 250 430 595 685 595 505 415 325 235 145 60 0 2-15 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.9 cont’ 2-16 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.10 A sketch of the Buffalo Creek Watershed is shown in Fig. P2–10. Areas A and B are identical in size, shape, slope, and channel length. UHs (1 hr) are provided for natural and fully developed conditions for both areas. Fig P2-10 (a) Assuming natural conditions for both areas, evaluate the peak outflow at point 1 if 2.5 in./hr of rain falls for 2 hr. Assume a total infiltration loss of 1 in. (b) Assume that area B has reached full development and area A has remained in natural conditions. Determine the outflow hydrograph at point 1 if a net rainfall of 2 in./hr falls for 1 hr. (c) Sketch the outflow hydrograph for the Buffalo Creek Watershed under complete development (A and B both urbanized) for the rainfall given in part (b). Time (hr) 0 1 2 3 4 5 6 7 8 UHdev (cfs) 0 40 196 290 268 185 90 30 0 Time (hr) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 UH (nat) 0 12 32 62 108 180 208 182 126 80 53 32 18 6 2.10 cont’ 2-17 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 0 ANSWER: a. Assuming uniform loss, there is a loss rate of: 1 in. for two hours = Net rainfall intensity = 2.5 – 0.5 = 2 in/hr for two hours The storm hydrograph at point 1 is found by combining the storm hydrographs for areas A and B. Existing conditions for both areas: Time (hr) (cfs) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 P1*U 0 12 32 62 108 180 208 182 126 80 53 32 18 6 0 0 0 24 64 124 216 360 416 364 252 160 106 64 36 12 0 0 P2*U 0 24 64 124 216 360 416 364 252 160 106 64 36 12 0 (cfs) 0 24 88 188 340 576 776 780 616 412 266 170 100 48 12 0 (cfs) Q (cfs) 0 0 24 48 88 176 188 376 340 680 576 1152 776 1552 780 1560 616 1232 412 824 266 532 170 340 100 200 48 96 12 24 0 0 2-18 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.10 cont’ Storm hydrograph at point 1: Existing conditions 1800 1600 Flow (cfs) 1400 1200 1000 800 600 400 200 0 0 2 4 6 8 10 12 14 16 Time (hr) b. In this case we have to determine the storm hydrograph at point 1 for a 1-hr duration rainfall. i = 2in/hr for 1 hr 2.10 cont’ 2-19 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Time (hr) Unat (cfs) Udevel Qa (cfs) Qb (cfs) Q (cfs) 0 0 0 0 0 0 1 12 40 24 80 104 2 32 196 64 392 456 3 62 290 124 580 704 4 108 268 216 536 752 5 180 185 360 370 730 6 208 90 416 180 596 7 182 30 364 60 424 8 126 0 252 0 252 9 80 160 160 10 53 106 106 11 32 64 64 12 18 36 36 13 6 12 12 14 0 0 0 15 0 0 0 2.10 cont’ 2-20 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. c. Areas A and B are fully developed. i = 2in/hr for 1 hr Time (hr) Udevel Qa (cfs) Qb (cfs) Q (cfs) 0 0 0 0 0 1 40 80 80 160 2 196 392 392 784 3 290 580 580 1160 4 268 536 536 1072 5 185 370 370 740 6 90 180 180 360 7 30 60 60 120 8 0 0 0 0 PROBLEM 2.11 2-21 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. (a) Examine the two hydrographs presented below for an area of 5000 acres. Which of the two is a unit hydrograph? Please show how you arrived at this conclusion. (b) Using the hydrograph that you chose from part (a), assuming it is a 1-hr UH, create a 3-hr UH. 1400 1200 Q (cfs 1000 800 Hydrograph 1 600 Hydrograph 2 400 200 0 0 2 4 6 8 10 12 Time (hr) 2-22 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. SOLUTION 2.11 (a) In order to determine this, you need to add the ordinates of each of the individual hydrographs Time (hr) Hydrograph 1 Hydrograph 2 1 0 0 2 75 200 3 250 500 4 450 700 5 600 1000 6 500 1300 7 350 700 8 225 400 9 50 200 10 0 0 2500 5000 Then multiply this cfs value by the time step, in this case 1 hour meaning that hydrograph 1 has a runoff volume of 2500 cfs-hrs and hydrograph 2 has a runoff volume of 5000 cfs-hr. Assuming that the conversion between cfs-hrs and acre-inches is a 1:1 relationship we now have the volumes in ac-in. To get the inches of direct runoff we divide these values by the size of the watershed, in this case 5000 acres to get the inches of direct runoff. Because hydrograph 2 corresponds to 1 inch of direct runoff while hydrograph 1 corresponds to ½ inch of direct runoff, hydrograph 2 is the unit hydrograph. (b) Using hydrograph two as our 1-hr UH we can develop the 3-hr UH. 2.11 cont’ 2-23 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Time (hr) 1-hr UH Lagged 1-hr UH 0 0 1 200 0 2 500 200 3 700 4 Sum 3-hr UH 0 0 200 66.67 0 700 233.33 500 200 1400 466.67 1000 700 500 2200 733.33 5 1300 1000 700 3000 1000 6 700 1300 1000 3000 1000 7 400 700 1300 2400 800 8 200 400 700 1300 433.33 9 0 200 400 600 200 0 200 200 66.67 0 0 0 10 Lagged 1-hr UH 11 3-hr UH 1200 Q (cfs) 1000 800 600 400 200 0 0 2 4 6 8 10 12 Time (hr) 2-24 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.12 A major storm event was recorded for Little Bear Creek. The incremental rainfall and measured hydrograph data for this storm are provided in Table P2–12 in 1-hr increments. The drainage area for the basin is 3.25 mi2. Assume base flow for Little Bear is zero. (See Fig. 1.26). (a) Using the storm hydrograph, estimate the volume of runoff that occurred in inches over the watershed. (b) Estimate the volume of infiltration in inches for this storm event based on the measured rainfall. (c) Estimate the time to peak t p for this watershed for the entire storm event. Table P2–12. Little Bear Creek Data Date/Time Incremental Precipitation (in.) Flow (cfs) 6/8/01 16:00 0 0.0 6/8/01 17:00 0.05 0.5 6/8/01 18:00 0.04 0.5 6/8/01 19:00 0 0.6 6/8/01 20:00 0.08 14 6/8/01 21:00 0.05 32 6/8/01 22:00 0.42 54 6/8/01 23:00 0.1 78 6/9/01 0:00 0.3 105 6/9/01 1:00 0.14 158 6/9/01 2:00 0.19 201 2-25 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 6/9/01 3:00 0.45 228 6/9/01 4:00 0.01 258 6/9/01 5:00 0 272 6/9/01 6:00 225 6/9/01 7:00 189 6/9/01 8:00 142 6/9/01 9:00 119 6/9/01 10:00 98 6/9/01 11:00 66 6/9/01 12:00 42 6/9/01 13:00 27 6/9/01 14:00 0 ANSWER: a. The volume can be determined be using the individual flows, timing it by the time interval and then adding them. The volume will be in cfs-hr. In this case since the time interval is 1 hr. 2309.6 cfs-hr 2309.6 ac-in We are given the area in square miles which can be converted to acres and can be used to find runoff in inches. 3.25 mi 2 b. Knowing the amount of direct runoff in inches we can then subtract it from total rainfall to find what was lost to infiltration: With the incremental precipitation we can find the total rainfall = . Thus, total precipitation – direct runoff = infiltration loss 2.12 cont’ 2-26 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. c. Time to peak is found knowing the center of mass of rainfall excess and the time to the peak of the hydrograph. The peak of the hydrograph happens 13 hrs after the storm starts (at 5 o’clock). The center of mass of rainfall can be calculated by finding the weighted average of the rain (multiplying rainfall by the time) and then dividing it by the total rainfall. See Excel table below. 13 13 i 0 i 0  Pi t i /  Pi  Time to Center of Mass of Rainfall 14.55in.  hr = 7.95 hr 1.83in. Time to Peak Flow – Time to CM of Rainfall = 13 – 7.95 = 5 hr = Time to Peak Note: “Time to Peak” is NOT equal to “Time to Peak Flow” Date/Time Time Increment, t (hr) Precipitation, P (in.) P*t Flow (cfs) 6/8/2001 16:00 0 0 0 0 6/8/2001 17:00 1 0.05 0.05 0.5 6/8/2001 18:00 2 0.04 0.08 0.5 6/8/2001 19:00 3 0 0 0.6 6/8/2001 20:00 4 0.08 0.32 14 6/8/2001 21:00 5 0.05 0.25 32 6/8/2001 22:00 6 0.42 2.52 54 6/8/2001 23:00 7 0.1 0.7 78 6/9/2001 0:00 8 0.3 2.4 105 6/9/2001 1:00 9 0.14 1.26 158 6/9/2001 2:00 10 0.19 1.9 201 6/9/2001 3:00 11 0.45 4.95 228 6/9/2001 4:00 12 0.01 0.12 258 6/9/2001 5:00 13 0 0 272 6/9/2001 6:00 225 6/9/2001 7:00 189 6/9/2001 8:00 142 6/9/2001 9:00 119 6/9/2001 10:00 98 6/9/2001 11:00 66 6/9/2001 12:00 42 6/9/2001 13:00 27 6/9/2001 14:00 0 SUM 1.83 14.55 2309.6 2-27 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.13 Plot the hydrograph for the storm data given in Problem 1.22 (flow rate vs. time). Label the following: (a) peak flow Q p , (b) time to peak t p (distance from center of mass of rainfall to peak flow), (c) time of rise TR (distance from start of discharge to peak flow), (d) time base TB (distance from start of discharge to end of discharge). ANSWER: (a) peak flow Q p =1380 cfs (b) time to peak t p = 1hr (c) time of rise TR = 1.25hrs (d) time base TB = 2.75 hrs PROBLEM 2.14 2-28 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Below you are provided with a 30-min UH and the rainfall and infiltration information for a storm event. Using this information, develop a storm hydrograph. Time (hr) U (m3/s) i (cm/hr) f (cm/hr) 0 0 0 0 0.5 15 0.75 0.25 1 40 1.5 0.2 1.5 60 3.0 0.2 2 85 1.75 0.1 2.5 105 0.5 0.1 3 120 3.5 110 4 95 4.5 80 5 60 5.5 45 6 30 6.5 10 7 0 30-min UH Q (m3/s) 150 100 50 0 0 2 4 6 8 Time (hr) 2-29 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. SOLUTION 2.14 We first need to determine the net rainfall for these periods: Time (hr) Rain (cm) Infiltration (cm) Net Rainfall (cm) 0 0 0 0 0.5 0.375 0.125 0.25 1.0 0.75 0.1 0.65 1.5 1.5 0.1 1.4 2.0 0.875 0.05 0.825 2.5 0.25 0.05 0.2 Now hydrograph convolution: Time (hr) U (m3/s) P1*UH P2*UH P3*UH P4*UH P5*UH Sum 0 0 0 0.5 15 0 1 40 3.75 0 1.5 60 10 9.75 0 2 85 15 26 21 0 2.5 105 21.25 39 56 12.375 0 128.625 3 120 26.25 55.25 84 33 3 201.5 3.5 110 30 68.25 119 49.5 8 274.75 4 95 27.5 78 147 70.125 12 334.625 4.5 80 23.75 71.5 168 86.625 17 366.875 5 60 20 61.75 154 99 21 0 3.75 19.75 62 2-30 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 355.75 5.5 45 15 52 133 90.75 24 314.75 6 30 11.25 39 112 78.375 22 262.625 6.5 10 7.5 29.25 84 66 19 205.75 7 0 2.5 19.5 63 49.5 16 150.5 0 6.5 42 37.125 12 97.625 0 14 24.75 9 47.75 0 8.25 6 14.25 0 2 2 0 0 7.5 8 8.5 9 9.5 Storm Hydrograph 400 350 300 Q (m3/s) 250 200 150 100 50 0 0 -50 1 2 3 4 5 6 7 8 9 10 Time (hr) 2.15 Using a program such as Excel, develop the S-curve for the given 30-min UH, and 2-31 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. then develop the 15-min UH from the 30-min UH. Time (hr) 0 0.25 0.5 0.75 1.0 1.25 1.5 U (cfs) 0 15 70.9 118.6 109.4 81.6 60.9 Time (hr) 1.75 2.0 2.25 2.5 2.75 3.0 3.25 U (cfs) 45.4 33.9 25.3 18.9 14.1 10.5 7.8 Time (hr) 3.5 3.75 4.0 4.25 4.5 4.75 5.0 5.25 U (cfs) 5.8 4.4 3.3 2.4 1.8 1.6 0.8 0 ANSWER: Create the S using the 30-min unit hydrograph. Lag by 30 min until you reach 0 Time 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 6 U (cfs) 0 15 70.9 118.6 109.4 81.6 60.9 45.4 33.9 25.3 18.9 14.1 10.5 7.8 5.8 4.4 3.3 2.4 1.8 1.6 0.8 0 30 min lagged UH 0 15 70.9 118.6 109.4 81.6 60.9 45.4 33.9 25.3 18.9 14.1 10.5 7.8 5.8 4.4 3.3 2.4 1.8 1.6 0.8 0 0 15 70.9 118.6 109.4 81.6 60.9 45.4 33.9 25.3 18.9 14.1 10.5 7.8 5.8 4.4 3.3 2.4 1.8 1.6 0.8 0 15 70.9 118.6 109.4 81.6 60.9 45.4 33.9 25.3 18.9 14.1 10.5 7.8 5.8 4.4 3.3 2.4 1.8 0 15 70.9 118.6 109.4 81.6 60.9 45.4 33.9 25.3 18.9 14.1 10.5 7.8 5.8 4.4 3.3 0 15 70.9 118.6 109.4 81.6 60.9 45.4 33.9 25.3 18.9 14.1 10.5 7.8 5.8 0 15 70.9 118.6 109.4 81.6 60.9 45.4 33.9 25.3 18.9 14.1 10.5 0 15 70.9 118.6 109.4 81.6 60.9 45.4 33.9 25.3 18.9 0 15 70.9 118.6 109.4 81.6 60.9 45.4 33.9 0 15 70.9 118.6 109.4 81.6 60.9 0 15 70.9 118.6 109.4 0 15 70.9 2.15 cont’ 2-32 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 0 S-curve 0 15 70.9 133.6 180.3 215.2 241.2 260.6 275.1 285.9 294 300 304.5 307.8 310.3 312.2 313.6 314.6 315.4 316.2 316.2 316.2 316.2 316.2 316.2 Time S-curve 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5 0 15 70.9 133.6 180.3 215.2 241.2 260.6 275.1 285.9 294 300 304.5 307.8 310.3 312.2 313.6 314.6 315.4 316.2 316.2 S-Curve lagged (15 min) Difference 15 min UH (D/D’)=2 0 15 70.9 133.6 180.3 215.2 241.2 260.6 275.1 285.9 294 300 304.5 307.8 310.3 312.2 313.6 314.6 315.4 316.2 0 15 55.9 62.7 46.7 34.9 26 19.4 14.5 10.8 8.1 6 4.5 3.3 2.5 1.9 1.4 1 0.8 0.8 0 0 30 111.8 125.4 93.4 69.8 52 38.8 29 21.6 16.2 12 9 6.6 5 3.8 2.8 2 1.6 1.6 0 2.15 cont’ 2-33 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2-34 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. PROBLEM 2.16 (a) Convert the following 1-hr UH into a 2-hr UH using Excel. Show table with time and ordinates as well as the plotted UH. Find the area of this watershed. (b) Given the net rainfall hyetograph, compute the storm hydrograph using hydrograph convolution and the unit hydrograph you develop ed in part, this time assuming it is a 1-hr UH (a). Show a table with times and ordinates as well as the plotted storm hydrograph. 1-hr UH: Time (hr) U (cfs) 0 0 1 100 2 300 3 400 4 325 5 250 6 175 7 100 8 50 9 0 2.16 cont’ 2-35 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Net Rainfall: Time (hr) Rainfall Intensity (in./hr) 0.25 0.8 0.5 0.8 0.75 0.8 1.0 0.6 1.25 1.0 1.5 1.2 1.75 0.8 2.0 0.6 2.25 0.6 2.5 0.2 2.75 0.2 3.0 0.2 3.25 0.1 3.50 0.4 3.75 0.4 4.0 0.1 4.25 0.1 4.5 0.1 4.75 0.04 5 0.04 2-36 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. SOLUTION 2.16 (a) Lag the 1-hr UH by a 1-hr increment. Then add the lagged hydrograph to the original and multiply the resulting ordinate values by the ratio of D/D’ where D is the original duration and D’ is the desired duration (1/2) Time (hr) 1-hr lagged UH’s 1-hr UH 0 1 2 3 4 5 6 7 8 9 10 0 100 300 400 325 250 175 100 50 0 Sum 2 hr UH 0 100 400 700 725 575 425 275 150 50 0 0 100 300 400 325 250 175 100 50 0 0 50 200 350 362.5 287.5 212.5 137.5 75 25 0 2 hr UH 400 350 Flow (cfs) 300 250 200 150 2 hr UH 100 50 0 0 2 4 6 8 10 12 Time (hr) 2.16 cont’ 2-37 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. The area of the watershed will be found as follows: Add the flows of the 2-hr UH: 1700 cfs Multiply by the incremental hour: (1700 cfs)*(1 hr) = 1700 cfs-hr Now convert this to acre-in. If they assumed that 1 cfs-hr = 1 acre-in then they should get 1700 ac-in. If they assumed 1 acre-in = 1.008 cfs-hr then they should get 1686.51 ac-in. Because this is a unit hydrograph and we are therefore dealing with 1 inch of rainfall, the area of the watershed is 1700 acres (or 1686.51 acres) (b) First we need to determine the hourly rainfall values from the table of intensities. The rainfall for 0.25, 0.5, 0.75 and 1.0 –> hour 1 1.25, 1.5, 1.75, and 2.0 –> hour 2, and so on. So the rainfall totals that will be used for hydrograph convolution are 2.16 cont’ 2-38 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Time (hr) Rainfall (in) 1 0.75 2 0.9 3 0.3 4 0.25 5 0.07 Now we do hydrograph convolution: Rainfall: [0.75, 0.9, 0.3, 0.25, 0.07] Time (hr) P1*U 0 0 1 50 0 2 150 60 0 3 250 180 20 0 4 262.5 300 60 16.67 0 639.17 5 225 315 99.99 50 4.67 694.67 6 150 270 105 83.33 14 622.33 7 112.5 180 90 87.5 23.33 493.33 8 50 135 60 75 24.5 344.5 9 25 60 45 50 21 201 10 0 30 20 37.5 14 101.5 0 9.99 16.67 10.5 37.17 0 8.33 4.67 13 11 12 P2*U P3*U P4*U P5*U SH 0 50 210 450 2-39 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 13 0 14 2.33 2.33 0 0 Problem 3 Storm Hydrograph 800 700 Flow (cfs) 600 500 400 300 200 100 0 0 2 4 6 8 10 12 14 Time (hr) 2.17 Given the following 2-hr unit hydrograph, calculate the 1-hr unit hydrograph. Then back calculate and find the 2-hr unit hydrograph to prove that the method of calculation is accurate. Graph both unit hydrographs against time on the same plot. Time (hr) 0 1 2 3 4 5 6 Flow (cfs) 0 25 125 250 400 500 450 Time (hr) 7 8 9 10 11 12 13 Flow (cfs) 350 300 225 150 100 25 0 2.17 cont’ 2-40 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 16 ANSWER: Procedure to obtain the 1 hr hydrograph: ~ Lag the 2-hr unit hydrograph by 2-hr increments to obtain the S curve ~ Then lag the S-curve by the time of duration of the new unit hydrograph (in this case, 1 hr) ~ Multiply the resulting ordinate values by the ratio of D/D’ where D is the original duration and D’ is the desired duration (2/1 =2) Time 2-Hr UH S-curve Lagged S-curve 0 0 0 1 25 25 2 125 0 3 250 25 4 400 125 5 500 6 Difference (cfs) 1 Hr UH 0 0 0 25 50 125 25 100 200 275 125 150 300 0 525 275 250 500 250 25 775 525 250 500 450 400 125 0 975 775 200 400 7 350 500 250 25 1125 975 150 300 8 300 450 400 125 0 1275 1125 150 300 9 225 350 500 250 25 1350 1275 75 150 10 150 300 450 400 125 0 1425 1350 75 150 11 100 225 350 500 250 25 1450 1425 25 50 12 25 150 300 450 400 125 0 1450 1450 0 0 13 0 100 225 350 500 250 25 1450 1450 0 0 The same procedure is done to convert to the 2 hr UH. 2-41 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.17 cont’ ~Lag the 1-hr unit hydrograph by 1-hr increments to obtain the S curve ~Then lag the S-curve by the time of duration of the new unit hydrograph (in this case, 2 hr) ~Multiply the resulting ordinate values by the ratio of D/D’ where D is the original duration and D’ is the desired duration (1/2 =.5) Time 0 1 2 3 4 5 6 7 8 9 10 11 12 13 1-Hr UH lagged 0 50 200 300 500 500 400 300 300 150 150 50 0 0 0 50 200 300 500 500 400 300 300 150 150 50 0 0 50 200 300 500 500 400 300 300 150 150 50 0 50 200 300 500 500 400 300 300 150 150 0 50 200 300 500 500 400 300 300 150 0 50 200 300 500 500 400 300 300 0 50 200 300 500 500 400 300 0 50 200 300 500 500 400 0 50 200 300 500 500 0 50 200 300 500 0 50 200 300 0 50 200 0 50 0 Scurve 0 50 250 550 1050 1550 1950 2250 2550 2700 2850 2900 2900 2900 Lagged Scurve 0 50 250 550 1050 1550 1950 2250 2550 2700 2850 2900 Blue = 1-hr UH, Red = 2-hr UH. 2-42 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Difference (cfs) 0 50 250 500 800 1000 900 700 600 450 300 200 50 0 1 Hr UH 0 25 125 250 400 500 450 350 300 225 150 100 25 0 2.18 Develop storm hydrographs from UHs of subareas 1 and 2 shown in figure P2-18 for the given rainfall and infiltration. Fig P2-18 t i f (hr) (in./hr) (in./hr) 1 0.5 0.4 2 1.1 0.3 3 3 0.2 4 0.9 0.1 Time (hr) 0 1 2 3 4 5 6 7 UH1 (cfs) 0 200 450 650 450 300 150 0 UH 2 (cfs) 0 150 300 500 350 250 125 100 8 9 50 0 2-43 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.18 cont’ ANSWER: Time (hr) 1 2 3 4 i (in./hr) 0.5 1.1 3 0.9 f (in./hr) 0.4 0.3 0.2 0.1 Net Rainfall Intensity (in/hr) 0.1 0.8 2.8 0.8 Subarea 1 Time UH1 (hr) P1*UH1 P2*UH1 P3*UH1 P4*UH1 Q1 (cfs) (cfs) (cfs) (cfs) (cfs) 0 0 0 0 1 200 20 0 2 450 45 160 0 3 650 65 360 560 0 985 4 450 45 520 1260 160 1985 5 300 30 360 1820 360 2570 6 150 15 240 1260 520 2035 7 0 0 120 840 360 1320 8 0 0 420 240 660 9 0 0 120 120 10 0 0 0 20 205 2-44 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.18 cont’ Subarea 2 Time UH2 (hr) P1*UH2 P2*UH2 P3*UH2 P4*UH2 Q2 (cfs) (cfs) (cfs) (cfs) (cfs) 0 0 0 0 1 150 15 0 2 300 30 120 0 3 500 50 240 420 0 710 4 350 35 400 840 120 1395 5 250 25 280 1400 240 1945 6 125 12.5 200 980 400 1592.5 7 100 10 100 700 280 1090 8 50 5 80 350 200 635 9 0 0 40 280 100 420 10 0 0 140 80 220 11 0 0 40 40 12 0 0 0 15 150 2-45 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.18 cont’ 2.19 Develop a combined storm hydrograph at point A in the watershed (Fig. P2–18) and lag route (shift in time only) assuming that travel time from point A to B is exactly 2 hours. ANSWER: We add the ordinates Q1 and Q2 from the previous problem to develop the combined storm hydrograph at point A. We then account for a 2-hr lag between points A and B to create a combined hydrograph for A and B. 2-46 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.19 cont’ Time (hr) Q1 (cfs) Q2 (cfs) QA (cfs) 0 0 0 0 1 20 15 35 2 225 165 390 0 3 1030 740 1770 35 4 2030 1430 3460 390 5 2570 1950 4520 1770 6 2000 1568 3568 3460 7 1290 1068 2358 4520 8 630 620 1250 3568 9 105 412.5 517.5 2358 10 0 210 210 1250 11 35 35 517.5 12 0 0 210 QAB (cfs) 13 35 14 0 2-47 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.20 Develop a storm hydrograph for subarea 3 from the given UH, add to the combined hydrograph from Problem 2.19, and produce a final storm hydrograph at the outlet of the watershed, B. Time (hr) UH3 (cfs) I (in/hr) f (in/hr) 0 0 0.5 0.4 1 140 1.1 0.3 2 420 3 0.2 3 630 0.9 0.1 4 490 5 350 6 210 7 130 8 70 9 0 ANSWER: The net rainfall intensity was obtained in problem 2.18 time (hr) I (in/hr) f (in/hr) Net Rainfall Intensity (in/hr) 1 0.5 0.4 0.1 2 1.1 0.3 0.8 3 3 0.2 2.8 4 0.9 0.1 0.8 2-48 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.20 cont’ Time (hr) UH3 (cfs) P1*UH3 P2*UH3 P3*UH3 P4*UH3 Q3 0 0 0 1 140 14 0 2 420 42 112 0 3 630 63 336 392 0 4 490 49 504 1176 112 5 350 35 392 1764 336 6 210 21 280 1372 504 7 130 13 168 980 392 8 70 7 104 588 280 9 0 0 56 364 168 10 0 196 104 11 0 56 12 0 0 14 154 791 1841 2527 2177 1553 979 588 300 56 0 2.20 cont’ 2-49 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Time (hr) Q3 (cfs) QAB (cfs) Qtotal (cfs) 0 0 0 1 14 0 14 2 154 0 154 3 791 35 826 4 1841 355 2196 5 2527 1695 4222 6 2177 3380 5557 7 1553 4515 6068 8 979 3627.5 4606.5 9 588 2410 2998 10 300 1295 1595 11 56 540 596 12 0 220 220 13 40 40 14 0 0 2-50 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.21 Redo Example 2–6 if the watershed is soil type B in good cover forest land. How does the forested area compare to the meadow UH? ANSWER: From Table 2.1 SCS curve number is found to be 55. Therefore, S = (1000 / CN) – 10 = (1000 / 55) – 10 = 8.18 in. Convert miles into feet. L = (5 mi)(5280 ft/mi) = 26,400 ft. The slope is 100 ft/mi, so y = (100 ft/mi)(1 mi / 5280 ft)(100%) = 1.9% Thus, time to peak is  ( L) 0.8 ( S  1) 0.7   (26,400) 0.8 (8.18  1) 0.7  tp =   =   = 6.21 hr 1900 1.9   1900 y   With rainfall duration D = 2 hr, the time to rise is TR = D/2 + tp = 2/2 + 6.21 = 7.21 hr Peak flow is Qp = (484A / TR) = (484*10 / 6.21) = 779.4 cfs where A is the area of 10 mi2 To complete the graph it is necessary to know the time of fall B. B = 1.67TR = 12.04 2-51 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.21 cont’ Forested Area (Group B) Meadow (Group D) TR 7.2 hr 4.36 hr B 11.84 hr 7.17 hr Qp 672.2 cfs 1110 cfs S 8.18 2.82 in. PROBLEM 2.22 A developer has recently purchased a 7 mi2 area of land north of Houston. She is interested in determining the impact that development will have on the hydrologic response of the area. There is one main channel running through the area and the length along it from the outlet to the divide is 6 miles. The average slope in this area is 100 ft/mi. To determine the comparison hydrographs, a storm with duration of 3 hours will be used. Pre-development, the area is 40% wooded (good condition), 30% good condition meadow, and 30% cultivated land with conservation treatment. Post-development the watershed will be 10% wooded (good condition), 5% good condition meadow, 30% cultivated land with conservation treatment, 30% residential with an average lot size of 1/8 acre, and 25% commercial area. The area has 40% soil type A and 60% soil type C. Using the SCS method of unit hydrograph construction and Excel to aid in your calculations and graphing, plot the resulting triangular hydrographs on the same graph for comparison and note the differences in a few short sentences. How does development affect the hydrograph? What do the resulting hydrographs tell you about the effects of high or low curve numbers? 2-52 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. SOLUTION 2.22 Pre-Development 1. Determine the curve number of the area based on the soil type and land use. Table with values given on page 76 of textbook. Land Use Soil Group Fraction of Area CN Wooded (Good Condition) A (0.4)*(0.4) = 0.16 25 Wooded (Good Condition) C (0.6)*(0.4) = 0.24 70 Meadow (Good Condition) A (0.4)*(0.3) = 0.12 30 Meadow (Good Condition) C (0.6)*(0.3) = 0.18 71 Cultivated Land (w/ conservation) A (0.4)*(0.3) = 0.12 62 Cultivated Land (w/ conservation) C (0.6)*(0.3) = 0.18 78 Weighted pre-development curve number: CNpre = (0.16)*(25) + (0.24)*(70) + (0.12)*(30) + (0.18)*(71) + (0.12)*(62) + (0.18)*(78) = 58.66 2. Develop the pre-development SCS hydrograph parameters S = (1000/CNpre) – 10 = (1000/58.66) – 10 = 7.047 in 2.22 cont’ 2-53 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. L = 6 mi = (6 mi)*(5280 ft/mi) = 31,680 ft y = 100 ft/mi = (100 ft/mi)*(1 mi/5280 ft)*(100) = 1.9 % ?? = ?0.8 (?+1)0.7 (31,680)0.8 (8.047)0.7 = = 6.553 hr 1900√? 1900√1.9 ? 3 ?? = 2 + ?? = (2) + 6.553 = 8.053 hr ?? = 484∗? 484∗7 = 8.053 = 420.71 cfs ?? Vol = (7 mi2)*(5280 ft/mi)2*(ac/43,560 ft)*(1 in) = 4480 ac-in ?= 2∗??? 2∗4480 − ?? = 420.71 − 8.053 = 13.244 hr ?? Post-Development 1. Determine the curve number of the area based on the soil type and land use. Table with values given on page 76 of textbook. 2.22 cont’ 2-54 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Land Use Soil Group Fraction of Area CN Wooded (Good Condition) A (0.4)*(0.1) = 0.04 25 Wooded (Good Condition) C (0.6)*(0.1) = 0.06 70 Meadow (Good Condition) A (0.4)*(0.05) = 0.02 30 Meadow (Good Condition) C (0.6)*(0.05) = 0.03 71 Cultivated Land (w/ conservation) A (0.4)*(0.3) = 0.12 62 Cultivated Land (w/ conservation) C (0.6)*(0.3) = 0.18 78 Residential (1/8 acre lots) A (0.4)*(0.3) = 0.12 77 Residential (1/8 acre lots) C (0.6)*(0.3) = 0.18 90 Commercial A (0.4)*(0.25) = 0.1 89 Commercial C (0.6)*(0.25) = 0.15 94 Weighted post-development curve number: CNpost = (0.04)*(25) + (0.06)*(70) + (0.02)*(30) + (0.03)*(71) + (0.12)*(62) + (0.18)*(78) + (0.12)*(77) + (0.18)*(90) + (0.1)*(89) + (0.15)*(94) = 77.85 2. Develop the post-development SCS hydrograph parameters S = (1000/CNpost) – 10 = (1000/77.85) – 10 = 2.845 in L = 6 mi = (6 mi)*(5280 ft/mi) = 31,680 ft 2-55 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.22 cont’ y = 100 ft/mi = (100 ft/mi)*(1 mi/5280 ft)*(100) = 1.9 % ?? = ?0.8 (?+1)0.7 (31,680)0.8 (3.845)0.7 = = 3.908 hr 1900√? 1900√1.9 ? 3 ?? = 2 + ?? = (2) + 3.908 = 5.40774 hr ?? = 484∗? 484∗7 = 5.40774 = 626.509 cfs ?? Vol = (7 mi2)*(5280 ft/mi)2*(ac/43,560 ft)*(1 in) = 4480 ac-in ?= 2∗??? 2∗4480 − ?? = 626.509 − 5.40774 = 8.894 hr ?? Graphs The graphs may be either hand-drawn or developed in EXCEL. To plot, each graph will need three points, connected in straight lines. Pre-Development points: (0 , 0) , (8.053 , 420.71) , (21.297 , 0) Post-Development points: (0 , 0) , (5.40774 , 626.509) , (14.302 , 0) Notes on the graph: Because of the development of the area and the changes in the curve number that are associated with an increase in imperviousness and other factors, the peak flow in the developed hydrograph occurs earlier and is also greater than in is pre-development. These graphs demonstrate that a higher curve number will translate to higher peak runoff. 2.22 cont’ 2-56 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. Direct Comparison of Pre and Post Development SCS Hydrographs 700 600 Flow (cfs) 500 400 300 Pre-Development 200 Post-Development 100 0 0 5 10 15 20 25 Time (hr) 2-57 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.23 A small watershed has the characteristics given below. Find the peak discharge Qp, the basin lag time tp, and the time base of the unit hydrograph Tb, using Snyder’s method. A = 150 mi.2; Ct = 1.70; L = 27 mi.; Lc = 15 mi.; Cp = 0.7 ANSWER: tp = Ct(LLc)0.3 = (1.70)(27*15 mi.2)0.3 = 10.3 hr Qp = 640(C p )( A) tp = 640(0.7)(150) = 6524 cfs 10.3 Since this is a small watershed, Tb  4tp = 41.2 hr 2-58 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.24 For a 55 mi2 watershed with Ct = 2.2, L = 15 mi., Lc = 7 mi., and Cp = 0.5, find tp, Qp, Tb, and D. Plot the resulting Snyder UH. ANSWER: tp = Ct(LLc)0.3 = 2.2(15*7)0.3 = 8.9 hr Qp = 640(C p )( A) tp = 640(0.5)(55) = 1978 cfs 8.9 Since this is a small watershed, Tb  4tp = 35.6 hr D = tp/5.5 = 1.6 hr TR = tp + D/2 = 8.9 + 1.6/2 = 9.7 hr W75 = 440(Qp/A)-1.08 = 440(1978/55)-1.08 = 9.2 hr W50 = 770(Qp/A)-1.08 = 770(1978/55)-1.08 = 16.1 hr It is necessary to obtain more flow points in order to be able to draw the hydrograph. We are aware the widths are distributed 1/3 before Qp and 2/3 after Qp so, W75 *(1/3) *(2/3) W50 3.1 6.1 5.4 10.7 For W75, Q75 = 0.75Qp = 1484 cfs t75,1 = 9.7 – W75(1/3) = 6.6 hr t75,2 = 9.7 + W75(2/3) = 15.8 hr For W50, Q50 = 0.50Qp = 989 cfs t50,1 = 9.7 – W50(1/3) = 4.3 hr t50,2 = 9.7 + W50(2/3) = 20.4 hr Therefore, the following points should be plotted: Time (hr) Flow (cfs) 0 4.3 6.6 9.7 0 989 1484 1978 2-59 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 15.8 20.4 35.6 1484 989 0 2.24 cont’ 2-60 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.25 Assume the watershed in Example 2–5 has gone through extensive commercial and industrial growth on the wooded area. Now 50% of the formerly wooded areas have become urbanized, so of that portion, 40% is commercial and business and 60% is fair condition lawn space. Assume the soil is 50% group B and 50% group C for all areas. Using Figure 2–8, determine the runoff volume for a rainfall of 6 in. ANSWER: Originally, the watershed was 40% wooded. The new watershed is now 20% wooded since 50% of the wooded area got developed. The other 20% is divided in the following way: Commercial and Business: 0.4*0.2 = 0.08 Fair Condition Lawn Space: 0.6*0.2 = 0.12 Land Use Soil Group Fraction of Area CN Wooded B 0.2*0.5 = 0.1 55 C 0.2*0.5 = 0.1 70 B 0.08*0.5 = 0.04 92 C 0.08*0.5 = 0.04 94 B 0.12*0.5 = 0.06 69 C 0.12*0.5 = 0.06 79 B 0.6*0.5 = 0.3 75 C 0.6*0.5 = 0.3 83 Commercial and Business Fair Condition Lawn Space Residential Weighted CN: CN = 55(0.1) + 70(0.1) + 92(0.04) + 94(0.04) + 69(0.06) + 79(0.06) + 75(0.3) + 83(0.3) = 76 From Fig. 2-8 we get 3.4 in. direct runoff from 6 in. of rainfall. 2-61 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.26 Make sure the unit hydrograph for Subbasin C is a unit hydrograph (Subbasin C area = 770 ac). The unit hydrograph for Subbasin C is graphed in figure P2-27. Subbasin C (Unit Hydrograph) 280 300.0 Q (cfs) 250.0 200.0 150.0 100.0 50.0 0.0 0 2 4 6 8 10 12 Time (0.5 hr) Fig 2-27 ANSWER: Area of Subbasin C = 770 ac Qp = 280 cfs DRO = Area under the hydrograph DRO = (280 cfs)(5.5 hr)(1/2) = 770 cfs-hr = 770 ac-in. 770 ac-in./770 ac = 1 in. This is a unit hydrograph. 2-62 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.27 Using Fig 2-14, find the daily evaporation from a shallow lake with the following characteristics: Mean daily temperature = 25.6°C, Daily solar radiation = 550 cal/cm2, Mean daily dew point = 4.4°C, Wind movement (6 in. above pan) = 5.5 ft/s. ANSWER: The chart in Fig. 2-14 is with different units of ˚F, cal/cm 2 , mi/day, and in. The given measurements given need to be converted to the proper units 25.6°C = (25.6)(1.8) + 32 = 78°F 4.4°C = (4.4)(1.8) + 32 = 40°F 5. 5 ft 3600s 24hr 1mi. mi. ft   = 5.5  = 90 s s hr day 5280 ft day Draw a bullet point at the intersection of T = 78°F and radiation = 550 cal/cm2 (upper left quadrant). Project a horizontal line to Td = 40°F (upper right quadrant) and draw a second point. Next project a vertical line to v = 90 mi./day (bottom right quadrant) and draw a third point. Then, project a vertical line down from the first point and a horizontal line to the left from the third point. The intersection of these two lines will be the point of daily evaporation. E = 0.25 in. 2-63 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.28 A class A pan is maintained near a small lake to determine daily evaporation (see table). The level in the pan is observed at the end of everyday. Water is added if the level falls near 7 in. For each day the difference in height level is calculated between the current and previous day, and the precipitation value is from the current day. Determine the daily lake evaporation if the pan coefficient is 0.70. Rainfall Water Level (in.) (in.) 1 0 8.00 2 0.23 7.92 3 0.56 7.87 4 0.05 7.85 5 0.01 7.76 6 0 7.58 7 0.02 7.43 8 0.01 7.32 9 0 7.25 10 0 7.19 11 0 7.08* 12 0.01 7.91 13 0 7.86 14 0.02 7.8 Day *Refilled at this point to 8 inches ANSWER: Pan evaporation = Change in water level in the pan + Amount of Rainfall Lake Evaporation = (Pan evaporation)(coefficient) 2-64 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.28 cont’ Rainfall (in) Day Water Level (in.) Pan Evaporation (in) Lake Evaporation (in) 1 0 8 0 0 2 0.23 7.92 0.31 0.217 3 0.56 7.87 0.61 0.427 4 0.05 7.85 0.07 0.049 5 0.01 7.76 0.1 0.070 6 0 7.58 0.18 0.126 7 0.02 7.43 0.17 0.119 8 0.01 7.32 0.12 0.084 9 0 7.25 0.07 0.049 10 0 7.19 0.06 0.042 11 0 7.08* 0.11 0.077 12 0.01 7.91 0.10* 0.070 13 0 7.86 0.05 0.035 14 0.02 7.8 0.08 0.056 * On Day 11, the pan is refilled. Thus, the pan evaporation that occurs on Day 12 (before the measurement is taken) starts from 8 in, not 7.08 in. So, 8.00 – 7.91 = 0.09 in. = ∆ water level Pan evaporation = ∆ water level + amount of rainfall = 0.09 in. + 0.01 in. = 0.10 in. 2-65 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.29 Given an initial rate of infiltration equal to 1.4 in./hr and a final capacity of 0.6 in./hr, use Horton’s equation [Eq. (2–42)] to find the infiltration capacity at the following times: t  10 min, 15 min, 30 min, 1 hr, 2 hr, 4 hr, and 6 hr. You may assume a time constant k = 0.24/hr. ANSWER: Horton’s equation: f = fc + (fo – fc)e-kt f = 0.6 in./hr + (1.4 – 0.6 in./hr) e 0.24t f = 0.6 + .8 e 0.24t plugging in given times (have to be in hours) t (hr) f(in/hr) 1 6 1.37 ¼ 1.35 ½ 1.31 1 1.22 2 1.10 4 0.91 6 0.79 2-66 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.30 Determine a Horton equation to fit the following times and infiltration capacities. Time f (hr) (in./hr) 1 6.34 2 5.20 6.5 2.50  1.20 ANSWER: Horton’s Equation: f = fc + (fo – fc)e-kt At t = ∞, f = 1.20 = fc + 0 fc = 1.20 (in./hr) At t = 1 hr, f = 6.34 = 1.20 + (fo – 1.20)e-k At t = 2 hr, f = 5.20 = 1.20 + (fo – 1.20)e-2k Solve for f0 from t = 1 hr equation: fo = 6.34  1.20  1.20 e k Substitute into the t = 2 hr equation f  5.20  1.20  (( 5.20 = 1.20  5.14  1.20)  1.20)e  2 k k e 5.14  2 k e e k 2.30 cont’ 2-67 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 5.20 = 1.20  5.14e  k 4.00 = 5. 14e  k e  k = 0.78 k = 0.25 Solve for f0 f = 6.34 = 1.20 + (fo – 1.20)e-0.25 5.14 = (fo – 1.20)e-0.25 fo = 7.8 in./hr Plug known values into original equation: f = 6.34 = 1.20 + (7.8 – 1.20)e-0.25t 2-68 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.31 A 5-hr storm over a 15-ac basin produces a 5-in. rainfall: 1.2 in./hr for the first hour, 2.1 in./hr for the second hour, 0.9 in./hr for the third hour, and 0.4 in./hr for the last two hours. Determine the infiltration that would result from the Horton model with k = 1.1/hr, fc = 0.2 in./hr, and fo = 0.9 in./hr. Plot the overland flow for this condition in in./hr vs. t. ANSWER: Time Horton’s Equation: f = fc + (f0 – fc)e-kt Infiltration Rainfall 0 0.90 f = 0.2 + (0.9 – 0.2)e-1.1t 1 0.43 1.2 f = 0.2 + 0.7e-1.1t 2 0.28 2.1 3 0.23 0.9 4 0.21 0.4 5 0.20 0.4 Plug in the values of t to solve for f. To determine the volume of infiltration, take the integral of Horton’s equation: 5 1.1t  (0.2  0.7e )dt 0 0.2t 0  5 =   0.7 1.1t 5 e 0  1.1 = 1  (0.636)  (0.004  1) =1.633 in. over the watershed 2-69 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.32 A plot of the infiltration curve obtained using Horton’s equation is shown in Fig. P2–33. Prove that k  ( fo  fc ) if F´ is the area between the curve and the fc line. F´ Find the area by integration over time, as time approaches infinity. Fig 2.33 ANSWER: A point is chosen on the curve such that at t = t 1 , f  fc. The area F´ is equal to the area under the curve above the line fc from t = 0 to t = to . This area may be found by the integration as follows: Let ‘ta’ be some arbitrary time. ta ta F ‘   [ f c  ( f o  f c )e ]dt   f c dt 0  kt 0 ta   ( f o  f c )e kt dt 0  ( f o  f c )(1 / k )(e  kta )  ( f o  f c )(1 / k ) As ta  ∞, e-kta  0. Therefore, 2-70 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.32 cont’ F ‘  (1 / k )( f o  f c ) k  ( fo  fc ) / F ‘ 2.33 Determine the  index of Figure P2–34 if the runoff depth was a)5.6 in. of rainfall over the watershed area and b) 6.5 in. ANSWER: a. This is trial and error problem. From looking at the graph it is possible to set up the equation: (0.4 – φ)6 + (0.5 – φ)2 + (0.7 – φ)4 + (0.9 – φ)2 + (0.3 – φ)4 + (0.2 – φ)2 = 5.6 Assume φ = 0.2 in./hr, (0.4 – 0.2)6 + (0.5 – 0.2)2 + (0.7 – 0.2)4 + (0.9 – 0.2)2 + (0.3 – 0.2)4 + (0.2 – 0.2)2 = 5.6 Both sides of the equation agree, so 2-71 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.33 cont’ So φ index = 0.2 in./hr. b. Again, this is trial and error problem. From looking at the graph it is possible to set up the equation: (0.4 – φ)6 + (0.5 – φ)2 + (0.7 – φ)4 + (0.9 – φ)2 + (0.3 – φ)4 + (0.2 – φ)2 = 6.5 With φ index = 0.1 in./hr = 7.1 so the index needs to be between 0.2 and 0.1 Trying φ index =.15 the answer is 6.6 with index = .155 the answer is 6.5 so the φ index for a rainfall of 6.5 in. is .155 in./hr. 2.34 A sandy loam has an initial moisture content of 0.18, hydraulic conductivity of 7.8 mm/hr, and average capillary suction of 100 mm. Rain falls at 2.9 cm/hr, and the final moisture content is measured to be 0.45. When does surface saturation occur? Plot the infiltration rate vs. the infiltration volume, using the Green and Ampt method of infiltration. ANSWER θi = 0.18 θf = 0.45 Ks = 78 mm/hr = .78 cm/hr Ψ = –100 mm = -10 cm i = 2.9 cm/hr (for 6 hours) Md = θf – θi = 0.45 – 0.18 = 0.27 FS  MD  100mm  0.27 = 9.93 mm   i   2.9cm / hr   1    1    0.78cm / hr   KS  Until 9.93 mm has infiltrated, the rate of infiltration is equal to the rainfall rate (2.9 cm/hr). This will happen at: 2-72 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.33 cont’ .993 cm  1hr = 0.34hr 2.9cm After surface saturation, the rate of infiltration is calculated using:  M  f  K S 1  d  F   F (cm) f (cm/hr) 1 2 3 4 5 6 7 8 2.89 1.83 1.48 1.31 1.20 1.13 1.08 1.04 2-73 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.35 Use the parameters given to graph the infiltration rate vs. the infiltration volume for the same storm for both types of soil. Prepare a graph using the Green–Ampt method, comparing all the curves calculated with both the lower-bound and upperbound porosity parameters. The rainfall intensity of the storm was 1.5 cm/hr for several hours, and the initial moisture content of all the soils was 0.15. Capillary Suction Hydraulic Conductivity Soil Porosity (cm) (cm/hr) Silt loam 0.42 – 0.58 16.75 0.65 Sandy clay 0.37 – 0.49 23.95 0.10 ANSWER: θi = 0.15 Ψ = – capillary suction i = 1.5 cm/hr Md = θ f – θ i SL 0.65 SC 0.10 Ψ (cm) -16.75 -23.95 θf 0.42 to 0.58 0.37 to 0.49 K s (cm/hr) Silty Loam (SL): FS   16.75cm  M d MD   12.81M d  i   1.5cm / hr   1    1    0.65cm / hr   KS  For low porosity: Md = 0.42 – 0.15 = 0.27 Fs = 3.46 cm Saturation time = Fs/i = 2.31 h 2-74 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.35 cont’ For high porosity: Md = 0.58 – 0.15 = 0.43 Fs = 5.51 cm Saturation time = Fs/i = 3.67 hr Sandy Clay (SC): FS   23.95cm  M d MD   1.71M d  i   1.5cm / hr   1    1    0.10cm / hr   KS  For low porosity: Md = 0.37 – 0.15 = 0.22 Fs = .38 cm Saturation time = Fs/i = 0.25 hr For high porosity: Md = 0.49 – 0.15 = .34 Fs = .58 cm Saturation time = Fs/i = 0.39 hr  M d  Plot the infiltration volume vs. the infiltration rate using: f  K S 1   F   SL SC f lp F 0.25 0.5 1 2 3 4 5 6 7 8 9 10 3.6 3.6 1.6 1.4 1.2 1.1 1.1 1.0 1.0 0.9 f hp 5.3 3.0 2.2 1.8 1.6 1.4 1.3 1.2 1.2 1.1 f lp graph f hp graph 1.5 1.5 1.5 1.5 1.5 1.4 1.2 1.1 1.1 1.0 1.0 0.9 f lp F 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.4 1.3 1.2 1.2 1.1 0.25 0.5 1 2 3 4 5 6 7 8 9 10 2.2 1.2 0.6 0.4 0.3 0.2 0.2 0.2 0.2 0.2 0.2 0.2 f hp f lp graph 3.4 1.7 0.9 0.5 0.4 0.3 0.3 0.2 0.2 0.2 0.2 0.2 2-75 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. f hp graph 1.5 1.2 0.6 0.4 0.3 0.2 0.2 0.2 0.2 0.2 0.2 0.2 1.5 1.5 0.9 0.5 0.4 0.3 0.3 0.2 0.2 0.2 0.2 0.2 2.35 cont’ 2-76 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.36 The Green and Ampt infiltration equation is a loss function used to compute the cumulative infiltration, F (cm) for a given infiltration rate, f (cm/hr). Recall that, f=Ks·(1-(M·wf )/F). For the given soil properties and infiltration rate, answer the following. K 1 cm/hr wf -11 cm M 0.2 t(hr) f(cm/hr) F(cm) 0 0 0 0.01 15.34 0.15 0.25 3.32 0.95 0.50 2.42 1.55 0.75 2.06 2.07 a. Compute the infiltration rate, f (cm/hr) for F=2.07 cm, and show your computations. f=Ks·(1-(M·wf )/F) = 1(1-0.2*(-11)/2.07) =2.06 cm/hr b. Compute the cumulative infiltration resulting from a constant rain rate of 0.5 cm/hr for 1 hr. F=0.5 * 1 = 0.5 cm c. At saturation, what is the infiltration rate in cm/hr? Justify. At saturation, f=K=1 cm/hr 2-77 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.37 Please refer to the Green and Ampt Infiltration and Runoff Example posted on the textbook website along with the associated Excel Spreadsheet. Complete the problem in the example, and then repeat the procedure with a constant rainfall rate of 50 mm/hr for 1.8 hrs. Determine the new cumulative runoff and runoff coefficient. Infiltration Rate, f (mm/hr) ANSWER: Time, t (hr) Rainfall (mm/hr) Infiltration Rate (mm/hr) Ks = wf = Md = 6.5 mm/hr -166.8 mm 0.75 Moisture Deficit Cumulative Rainfall (mm) = Cumulative Runoff (mm) = Cumulative Infiltration (mm) = Runoff Coefficient = 92.5 32.1 60.4 34.7% 78 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.38 A constant rain rate storm (2 cm/hr for 4 hr) falls on a silt loam soil with an effective saturation of 30%. Determine the infiltration rate (f) and cumulative infiltration (F) at time t = 1 hr. Determine the time (t) and infiltration rate (f) when the cumulative infiltration F = 6 cm. Solution Values obtained from problem statement and table 2-4 for silt loam. ?? = ?? − (? − ?? ) , ?? ?? = ?? − (? − ?? ) ?? 0.3 ∗ 0.486 = ?? − (0.501 − 0.486), ?? = 0.1608 ?? = ?? (1 − ?? ) = 0.486(1 − 0.3) = 0.3402 ? = ?? (?? ∗ ? ∗ψ ? ?? = 1−?/? = ψ 0.3402 3.688 + 1) = 0.65 (16.68 ∗ + 1) = + 0.65 F ? ? 0.3402∗−16.68 1−2/0.65 = 2.732?? ?? = 1.366 ℎ? 2 a) When t=1, f= 2cm/hr, F= i*t= 2 cm/hr ?? = b) When F= 6cm, ? = 3.688 6 + 0.65 = 1.265 ??/ℎ? ψ ∗ ?? − ? ?? (? − ?? ) = ? − ?? + ψ ∗ ?? ∗ ln ( ) ψ ∗ ?? − ?? −16.68 ∗ 0.3402 − 6 0.65(? − 1.366) = 6 − 2.732 − 16.68 ∗ 0.3402 ∗ ln ( ) 16.68 ∗ 0.3402 − 2.732 ? = 3.527ℎ? 79 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. 2.39 Repeat Example 2–11 in the chapter with the same rainfall, except use Horton’s equation for the infiltration with f0 = 1.2 in/hr, fc = 0.2 in/hr, and k = 0.35hr−1. Solution (Assuming the runoff is unknown, and we are trying to find it.) f0 fc k A 1.2 0.2 0.35 0.875 in/hr in/hr 1/hr mi^2 First, put the area into units of acres: 0.875 mi2 * 640 ac/mi2 = 560 acres Then add the rainfall values for each time period to get total rainfall: time i 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1.4 1.4 2.3 2.3 2.3 1.1 1.1 0.7 0.7 0.7 0.3 0.3 14.6 in total rainfall Then use equation (2-43) with the given parameters to solve for the total infiltration: ? −? ?(?) = ?? ∗ ? + 0 ? ? ∗ (1 − ? −?? ), 1.2 − 0.2 ∗ (1 − ? −0.3∗12 ) = 5.21 in infiltrated 0.35 Then subtract the rain infiltrated from the rain fallen ?(?) = 0.2 ∗ 12 + 14.6-5.21 = 9.29 in, 9.29 in*1ft/12 in = 0.78 ft Multiply this amount of rain by the area of the watershed to get total runoff 0.78 ft*560 acres = 438 acre-ft runoff 80 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.