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Solution Manual for Hydrology and Floodplain Analysis, 6th Edition

Solution Manual for Hydrology and Floodplain Analysis, 6th Edition is an essential tool for mastering your textbook, offering detailed solutions and helpful explanations.

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    1-1 Chapter 1-Answer Key CONCEPT CHECK 1.1 What is the hydrologic cycle? What are the pathways that precipitation falling onto the land surface of the Earth is dispersed to the hydrologic cycle? ANSWER: The hydrologic cycle is a continuous process in which water is evaporated from water surfaces and the oceans, moves inland as moist are masses, and produces precipitation if the correct vertical lifting conditions exist. A portion of precipitation (rainfall) is retained in the soil near where it falls and returns to the atmosphere via evaporation (the conversion of water to water vapor from a water surface) and transpiration (the loss of water vapor through plant tissue and leaves). Combined loss is called evapotranspiration 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 conditions and soil and may reenter channels later as interflow or may percolate to recharge the shallow ground water. The remaining portion of precipitation becomes overland flow or direct runoff which flows generally in a down-gradient direction to accumulate in local streams that then flow into rivers.

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    1-2 1.2 Who is responsible for the first recorded rainfall measurements? Describe the technique used to obtain these measurements. ANSWER: The first recording was obtained in the seventeenth century by Perrault. He obtained his data by comparing measured rainfall to the estimated flow in the Seine River to show how the two were related.

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    1-3 1.3 Explain 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 in several ways. Specific humidity is a mass of water vapor in a unit mass of moist air while relative humidity is a ratio of the air’s actual water vapor content compared to the amount of water vapor at saturation for that temperature.

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    1-4 1.4 Explain 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) or water (maritime), and the latitude of generation (polar or tropical). These air masses are present in the United States. The Continental polar emanates from Canada and passes over the northern United States. The maritime polar air mass also comes southward from the Atlantic Coast of Canada and affects the New England states. Another maritime polar comes from the Pacific and hits the extreme northwestern states. The maritime tropical air masses come from the Pacific, the Gulf of Mexico and the Atlantic (these affect the entire Southern United States). Continental tropical air masses from only during the summer. They originate in Texas and affect the states bordering the north.

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    1-5 1.5 List seven major factors that determine a watershed’s response to a given rainfall. ANSWER: Drainage Area Channel Slope Soil Types Land Use Land Cover Main Channel and tributary characteristics-channel morphology The shape, slope and character of the floodplain

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    1-6 PROBLEMS 1.6 A lake with a surface area of 1260 acres was monitored over a period of time. During a one-month period, the inflow was 36 cfs, the outflow was 30 cfs, and a 1.6-in. seepage loss was measured. During the same month, the total precipitation was 4.3 in. Evaporation loss was estimated as 6.0 in. Estimate the storage change for this lake during the month. ANSWER: A = 1260 acres T = 1 month I = 36 cfs O = 30 cfs G = 1.6 in. P = 4.3 in. E = 5 in. I-O+P-G-E = ΔS First convert inflow and outflow into inches Inflow = 36 𝑓𝑡 3 𝑠 ∗ 1𝑎𝑐 43560𝑓𝑡 2 ∗ 12𝑖𝑛 1𝑓𝑡 ∗ 3600𝑠 1ℎ𝑟 ∗ 24ℎ𝑟 1𝑑𝑎𝑦 ∗ 30𝑑𝑎𝑦𝑠 1𝑚𝑜𝑛𝑡ℎ ∗ 1 1260 𝑎𝑐𝑟𝑒𝑠 = 20.40 𝑖𝑛 Outflow = 30 𝑓𝑡 3 𝑠 ∗ 1𝑎𝑐 43560𝑓𝑡 2 ∗ 12𝑖𝑛 1𝑓𝑡 ∗ 3600𝑠 1ℎ𝑟 ∗ 24ℎ𝑟 1𝑑𝑎𝑦 ∗ 30𝑑𝑎𝑦𝑠 1𝑚𝑜𝑛𝑡ℎ ∗ 1 1260 𝑎𝑐𝑟𝑒𝑠 = 17.00 𝑖𝑛 ΔS = 20.40 in. – 17.00 in. + 4.3 in. - 1.6 in. – 5 in. = 1.1 in. ΔS in volume = 1.1𝑖𝑛 ∗ 1𝑓𝑡 12𝑖𝑛 ∗ 1260 𝑎𝑐𝑟𝑒𝑠 = 115.5 ac-ft

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    1-7 1.7 A 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. The rest 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 and calculate 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=∆S Put all variables in the same units: P = 1.25 in./hr or P = 7.5 in E = 0.1 in/hr or E = 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/hr or ∆S=3 in Plug values into water balance: 1.25 in/hr-R-0.1 in/hr = 0.5 in/hr or 7.5 in – R-0.6 in = 3 in R=1.25 in/hr-0.1in/hr-0.5in/hr R = 7.5 in – 0.6 in – 3 in = 0.65 in/hr =3.9 in R=(0.65in/hr)*(6 hr)=3.9 in R = (3.9 in)*(5 acres) (3.9 in)*(5 acre)= 19.5 acre-inches R = 19.5 ac-in R = 19.5 ac-in The runoff coefficient is a ratio of the amount of water that actually becomes runoff versus the amount that falls as precipitation. The higher the amount of impervious surfaces the higher the runoff coefficient. Calculate the runoff coefficient: C=R/P C=0.65/1.25 or C=R/P=3.9 in/7.5in

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

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    1-9 1.8 In a given year, a watershed with an area of 3000 2500 km 2 received 110 cm of precipitation. The average rate of flow measured in a gage at the outlet of the watershed was 29 m 3 /sec. Estimate the water losses due to the combined effects of evaporation, transpiration, and infiltration due to ground water. How much runoff reached the river for the year (in cm)? ANSWER: A = 3000 2500 km 2 P = 110 cm R = 29 m 3 /sec Assuming ΔS = 0 in the span of the year ET+G = P-R Convert R to cm R = 29 𝑚 3 𝑠 ∗ 100𝑐𝑚 1𝑚 ∗ 3600𝑠 1ℎ𝑟 ∗ 24ℎ𝑟 1𝑑𝑎𝑦 ∗ 365𝑑𝑎𝑦𝑠 1𝑦𝑒𝑎𝑟 ∗ 1 3000 𝑘𝑚 2 ∗ 1𝑘𝑚 2 (1000𝑚) 2 = 30.48 cm is runoff ET + G = 110 cm – 30.48 cm = 79.52 cm

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    1-10 1.9 Using the data from problem 1.8, what is the runoff coefficient? ANSWER: Runoff coefficient = R/P = 30.48𝑐𝑚 110𝑐𝑚 = 0.277

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    1-11 1.10 In a given month, a watershed with an area of 7km 2 received 63cm of precipitation. During the same month, the loss due to evaporation was 12cm. Ignore losses due to transpiration and infiltration due to ground water. What would be the average rate of flow measured in a gage at the outlet of the watershed in m 3 /s? ANSWER: A = 7 k 2 m = 7,000,000 2 m P = 63cm E = 12cm Here precipitation is the inflow I-E = O O = 63cm-12cm = 51cm in 1 month over an area of 7km 2 , so O = 51 𝑐𝑚 1 𝑚𝑜𝑛𝑡 ℎ ∗ 1 𝑚𝑜𝑛𝑡 ℎ 30 𝑑𝑎𝑦𝑠 ∗ 1 𝑑𝑎𝑦 24 ℎ 𝑟 ∗ 1ℎ 𝑟 3600 𝑠 ∗ 1 𝑚 100 𝑐𝑚 ∗ 7 𝐸 6 𝑚 2 = 1.38 day m 3

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    1-12 1.11 Plot Eq. (1 – 4) as a graph ( 𝑒 𝑠 vs. T ) for a range of temperatures from −30℃ to 40℃ and a range of pressures from 0 to 70 mb. The area below the curve represents the unsaturated air condition. Using this graph, answer the following: (a) Select two saturated and two unsaturated samples of air from the dataset of pressure and 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 relative humidity? What would be their dew point temperature?

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    1-13 1.11 con t’ ANSWER : The graph can be made using a program like Excel Equation (1 – 4): e = 8 10 7489 . 2  exp ) ) 79 . 242 ( 6 . 4278 (   d T a) Saturated samples of air: T = 10  C, P = 20 mb T = 10  C, P = 30 mb Unsaturated samples of air: T = 10  C, P = 10 mb T = 20  C, P = 20 mb b) 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 and B, T = 30  C so   sb sa e e 42.41 mb ii. Dew point is a measure of water vapor pressure For sample A, e a = 25mb  T d = 21.14  C For sample B, e b = 25mb  T d = 24.14  C Water Vapor Pressure (mb) Temperature (C) A B T d

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    1-14 1.11 cont’ iii. Relative humidity is the ratio of the actual vapor pressure to the saturation vapor pressure Sample A = 100 41 . 42 25  = 59%, Sample B = 100 41 . 42 30  = 71% c) In 15  C, both samples have become saturated: Their relative humidity is 100% After condensation begins, the temperature and the dew point are the same so T = T d = 15  C

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    1-15 1.12 The gas constant R has the value 2.87 × 10 6 𝑐𝑚 2 /𝑠 2 °𝐾 for dry air, when pressure is in mb. Using the ideal gas law 𝑃 = 𝜌𝑅𝑇 find the density of dry air at 20℃ with a pressure of 1013 mb. Find the density of moist air at the same pressure and temperature if the relative humidity is 65 percent. ANSWER: R = 2.87 × 10 6 𝑐𝑚 2 /𝑠 2 °𝐾 T = 20℃ P = 1013 mb Note: 1 mb = 100 Pa = 100 Nt/m 2 = 100 kg/m ∙ s 2 a) Rearrange for ρ gives 𝜌 = 𝑃 𝑅𝑇 = 1013𝑚𝑏 ∙ (100 𝑁𝑡 𝑚 2 ⁄ 1𝑚𝑏 ⁄ ) ∙ ( 1𝑘𝑔 𝑚 ∙ 𝑠 2 ⁄ 1𝑁𝑡 𝑚 2 ⁄ ⁄ ) 2.87 × 10 6 𝑐𝑚 2 ∙ 𝑠 −2 ∙ 𝐾 −1 ∙ (20 + 273.15)𝐾 ∙ ( 100𝑐𝑚 1𝑚 ) 2 = 1.20 𝑘𝑔 𝑚 3 ⁄ b) The density of moist air can be found using the equation ρ m = ) / 378 . 0 1 ( P e RT P  Since we know the relative humidity, we can find e. (See Appendix C for values of e s at 20  C (need to convert from kN/m 2 to mb) At 20  C e s = 23.4 mb H = 100e/e s 65 = 100e/23.4 e = 15.21 mb Substituting this into equation gives ρ m = ) / 378 . 0 1 ( P e RT P  = ρ d ) / 378 . 0 1 ( P e  1.20 kg/m 3 (1 − 0.378 ∙ 15.21 𝑚𝑏 1013 𝑚𝑏 ) = 1.19 kg/ m 3 Moist air is lighter than dry air!

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