Solution Manual for Hydrology and Floodplain Analysis, 6th Edition

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1-1Chapter 1-Answer KeyCONCEPT CHECK1.1What is the hydrologic cycle? What are the pathways that precipitation falling onto the landsurface of the Earth is dispersed to the hydrologic cycle?ANSWER:The hydrologic cycle is a continuous process in which water is evaporated from water surfacesand the oceans, moves inland as moist are masses, and produces precipitation if the correctvertical lifting conditions exist.A portion of precipitation (rainfall) is retained in the soil near where it falls and returns to theatmosphere via evaporation (the conversion of water to water vapor from a water surface) andtranspiration (the loss of water vapor through plant tissue and leaves). Combined loss is calledevapotranspiration and is a maximum value if the water supply in the soil is adequate at all times.Some water enters the soil system in infiltration which is a function of soil moisture conditionsand soil and may reenter channels later as interflow or may percolate to recharge the shallowground water. The remaining portion of precipitation becomes overland flow or direct runoffwhich flows generally in a down-gradient direction to accumulate in local streams that then flowinto rivers.

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1-21.2Who is responsible for the first recorded rainfall measurements? Describe the techniqueused to obtain these measurements.ANSWER:The first recording was obtained in the seventeenth century by Perrault. He obtained his data bycomparing measured rainfall to the estimated flow in the Seine River to show how the two wererelated.

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1-31.3Explain the difference between humidity and relative humidity.ANSWER:Humidity is a measure of the amount of water vapor in the atmosphere and can be expressed inseveral ways. Specific humidity is a mass of water vapor in a unit mass of moist air whilerelative humidity is a ratio of the air’s actual water vapor content compared to the amount ofwater vapor at saturation for that temperature.

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1-41.4Explain how air masses are classified. Where are these types of air masses located?ANSWER:They are classified in two ways: the source from which they are generated, land (continental) orwater (maritime), and the latitude of generation (polar or tropical).These air masses are present in the United States. The Continental polar emanates from Canadaand passes over the northern United States. The maritime polar air mass also comes southwardfrom the Atlantic Coast of Canada and affects the New England states. Another maritime polarcomes from the Pacific and hits the extreme northwestern states. The maritime tropical airmasses come from the Pacific, the Gulf of Mexico and the Atlantic (these affect the entireSouthern United States). Continental tropical air masses from only during the summer. Theyoriginate in Texas and affect the states bordering the north.

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1-51.5List seven major factors that determine a watershed’s response to a given rainfall.ANSWER:Drainage AreaChannel SlopeSoil TypesLand UseLand CoverMain Channel and tributary characteristics-channel morphologyThe shape, slope and character of the floodplain

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1-6PROBLEMS1.6A lake with a surface area of 1260 acres was monitored over a period of time. During aone-month period, the inflow was 36 cfs, the outflow was 30 cfs, and a 1.6-in. seepage losswas measured. During the same month, the total precipitation was 4.3 in. Evaporation losswas estimated as 6.0 in. Estimate the storage change for this lake during the month.ANSWER:A = 1260 acresT = 1 monthI = 36 cfsO = 30 cfsG = 1.6 in.P = 4.3 in.E = 5 in.I-O+P-G-E =ΔSFirst convert inflow and outflow into inchesInflow =36𝑓𝑡3𝑠∗1𝑎𝑐43560𝑓𝑡2∗12𝑖𝑛1𝑓𝑡∗3600𝑠1ℎ𝑟∗24ℎ𝑟1𝑑𝑎𝑦∗30𝑑𝑎𝑦𝑠1𝑚𝑜𝑛𝑡ℎ∗11260 𝑎𝑐𝑟𝑒𝑠= 20.40 𝑖𝑛Outflow =30 𝑓𝑡3𝑠∗1𝑎𝑐43560𝑓𝑡2∗ 12𝑖𝑛1𝑓𝑡 ∗ 3600𝑠1ℎ𝑟∗ 24ℎ𝑟1𝑑𝑎𝑦 ∗ 30𝑑𝑎𝑦𝑠1𝑚𝑜𝑛𝑡ℎ ∗11260 𝑎𝑐𝑟𝑒𝑠 = 17.00 𝑖𝑛ΔS= 20.40 in.–17.00 in. + 4.3 in. - 1.6 in.–5 in. =1.1 in.ΔS in volume=1.1𝑖𝑛 ∗1𝑓𝑡12𝑖𝑛∗ 1260 𝑎𝑐𝑟𝑒𝑠=115.5 ac-ft

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1-71.7A 5-acre apartment complex receives 1.25 in./hr. of rain for 6 hours. During the storm event,0.1 in/hr. of water evaporates, and a volume of 15 acre-inches accumulates on the lot. Therest leaves the lot as runoff. Assume that there is no infiltration in this scenario.(a) Using a water balance, determine the total volume of runoff generated by the storm andcalculate the runoff coefficient.(b) Calculate the peak flow rate during the storm (Hint: use the Rational Method).ANSWER:(a)Set up the water balance:P-R-E=∆SPut all variables in the same units:P = 1.25 in./hrorP = 7.5 inE = 0.1 in/hrorE = 0.6 in.R=?∆S=(15ac-in)/(6 hr)=2.5 ac-in/hr(2.5 ac-in/hr)/(5 acres) = 0.5 in/hr∆S = 0.5 in/hror∆S=3 inPlug values into water balance:1.25 in/hr-R-0.1 in/hr = 0.5 in/hror7.5 in–R-0.6 in = 3 inR=1.25 in/hr-0.1in/hr-0.5in/hrR = 7.5 in–0.6 in–3 in= 0.65 in/hr=3.9 inR=(0.65in/hr)*(6 hr)=3.9 inR = (3.9 in)*(5 acres)(3.9 in)*(5 acre)=19.5 acre-inchesR = 19.5 ac-inR = 19.5 ac-inThe runoff coefficient is a ratio of the amount of water that actually becomes runoff versus theamount that falls as precipitation. The higher the amount of impervious surfaces the higher therunoff coefficient.Calculate the runoff coefficient:C=R/P C=0.65/1.25orC=R/P=3.9 in/7.5in

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1-8C = 0.52(b)Use the Rational Method to calculate Q:Q=CiAQ=(0.52)*(1.25in/hr)*(5 acres)Q = 3.25 cfs

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1-91.8In a given year, a watershed with an area of 30002500 km2received 110 cm of precipitation.The average rate of flow measured in a gage at the outlet of the watershed was 29 m3/sec.Estimate the water losses due to the combined effects of evaporation, transpiration, andinfiltration due to ground water. How much runoff reached the river for the year (in cm)?ANSWER:A = 30002500 km2P = 110 cmR = 29 m3/secAssumingΔS= 0 in the span of the yearET+G = P-RConvert R to cmR =29𝑚3𝑠∗100𝑐𝑚1𝑚∗3600𝑠1ℎ𝑟∗24ℎ𝑟1𝑑𝑎𝑦∗365𝑑𝑎𝑦𝑠1𝑦𝑒𝑎𝑟∗13000 𝑘𝑚2∗1𝑘𝑚2(1000𝑚)2=30.48 cm is runoffET + G = 110 cm–30.48 cm =79.52 cm

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1-101.9Using the data from problem 1.8, what is the runoff coefficient?ANSWER:Runoff coefficient = R/P =30.48𝑐𝑚110𝑐𝑚=0.277

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1-111.10In a given month, a watershed with an area of 7km2received 63cm of precipitation. Duringthe same month, the loss due to evaporation was 12cm. Ignore losses due to transpirationand infiltration due to ground water. What would be the average rate of flow measured in agage at the outlet of the watershed in m3/s?ANSWER:A = 7k2m= 7,000,0002mP = 63cmE = 12cmHere precipitation is the inflow I-E = OO = 63cm-12cm = 51cm in 1 month over an area of 7km2,soO =51𝑐𝑚1𝑚𝑜𝑛𝑡ℎ∗1𝑚𝑜𝑛𝑡ℎ30𝑑𝑎𝑦𝑠∗1𝑑𝑎𝑦24ℎ𝑟∗1ℎ𝑟3600𝑠∗1𝑚100𝑐𝑚∗7𝐸6𝑚2=1.38daym3

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1-121.11Plot Eq. (1–4) as a graph (𝑒𝑠vs.T) for a range of temperatures from−30℃to40℃anda range of pressures from 0 to 70 mb. The area below the curve represents the unsaturated aircondition. Using this graph, answer the following:(a) Select two saturated and two unsaturated samples of air from the dataset of pressureand temperature given below:Pressure (mb): {10, 20, 30}Temperature (℃): {10, 20, 30}(b) Let A and B be two air samples, where A: (T = 30℃, P = 25 mb) and B: ( T = 30℃, P= 30 mb). For each sample, determine the following:(i)Saturation vapor pressure(ii)Dew point(iii) Relative humidity(c) Suppose both samples A and B were cooled to 15℃What would be their relativehumidity? What would be their dew point temperature?

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1-131.11 cont’ANSWER:The graph can be made using a program like ExcelEquation (1–4): e =8107489.2exp))79.242(6.4278(dTa)Saturated samples of air:T = 10C, P = 20 mbT = 10C, P = 30 mbUnsaturated samples of air:T = 10C, P = 10 mbT = 20C, P = 20 mbb)A = (T = 30C, P = 25 mb)B = (T = 30C, P = 30 mb)i.Saturation Vapor Pressure is a function of temperature. For both samples A andB, T = 30C sosbsaee42.41 mbii.Dew point is a measure of water vapor pressureFor sample A, ea= 25mbTd=21.14CFor sample B, eb= 25mbTd=24.14CWater Vapor Pressure (mb)Temperature (C)ABTd

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1-141.11 cont’iii.Relative humidity is the ratio of the actual vapor pressure to the saturation vaporpressureSample A =10041.4225= 59%,Sample B =10041.4230=71%c)In 15C, both samples have become saturated:Their relative humidity is100%After condensation begins, the temperature and the dew point are the same soT = Td=15C

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1-151.12The gas constantRhas the value2.87 × 106𝑐𝑚2/𝑠2°𝐾for dry air, when pressure is inmb. Using the ideal gas law𝑃 = 𝜌𝑅𝑇find the density of dry air at20℃with a pressure of 1013mb. Find the density of moist air at the same pressure and temperature if the relative humidity is65 percent.ANSWER:R =2.87 × 106𝑐𝑚2/𝑠2°𝐾T =20℃P = 1013 mbNote: 1 mb = 100 Pa = 100 Nt/m2= 100 kg/m∙s2a)Rearrange for ρ gives𝜌 = 𝑃𝑅𝑇 =1013𝑚𝑏 ∙ (100 𝑁𝑡 𝑚2⁄1𝑚𝑏⁄) ∙ (1𝑘𝑔 𝑚 ∙ 𝑠2⁄1𝑁𝑡 𝑚2⁄⁄)2.87 × 106𝑐𝑚2∙ 𝑠−2∙ 𝐾−1∙ (20 + 273.15)𝐾∙ (100𝑐𝑚1𝑚)2= 1.20 𝑘𝑔 𝑚3⁄b)The density of moist air can be found using the equation ρm=)/378.01(PeRTPSincewe know the relative humidity, we can find e. (See Appendix C for values of esat 20C(need to convert from kN/m2to mb)At 20C es= 23.4 mbH = 100e/es65 = 100e/23.4e = 15.21 mbSubstituting this into equation givesρm=)/378.01(PeRTP=ρd)/378.01(Pe1.20 kg/m3(1 −0.378 ∙ 15.21 𝑚𝑏1013 𝑚𝑏)=1.19 kg/ m3Moist air is lighter than dry air!

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