1. Surface Water

2. WATER SEDIMENT The Lane Diagram

3. I. Events During Precipitation A. Interception B. Stem Flow C. Depression Storage D. Hortonian Overland Flow E. Interflow F. Throughflow -> Return Flow G. Baseflow

4. II. Hydrograph A. General

5. II. Hydrograph A. General B. Storm Hydrograph

9. II. Hydrograph A. General B. Storm Hydrograph 1. Shape and Distribution of “events”

10. direct ppt., runoff, baseflow, interflow

12. II. Hydrograph A. General B. Storm Hydrograph 1. Shape and Distribution of 2. Hydrograph Separation

14. II. Hydrograph A. General B. Storm Hydrograph 1. Shape and Distribution of 2. Hydrograph Separation C. Predicting the rate of Baseflow Recession after a storm

15. vs.

17. Why care?

19. Predicting the rate of Baseflow Recession after a storm

21. An example problem….

22. Gaining and Losing Streams…..

23. III. Rainfall-Runoff Relationships

24. III. Rainfall-Runoff Relationships A. Time of Concentration

25. III. Rainfall-Runoff Relationships A. Time of Concentration “The time required for overland flow and channel flow to reach the basin outlet from the most distant part of the catchment”

26. tc = L 1.15 7700 H 0.38

27. III. Rainfall-Runoff Relationships A. Time of Concentration “The time required for overland flow and channel flow to reach the basin outlet from the most distant part of the catchment” tc = L 1.15 7700 H 0.38 tc = time of concentration (hr) L = length of catchment (ft) along the mainstream from basin mouth to headwaters (most distant ridge) H = difference in elevation between basin outlet and headwaters (most distant ridge)

28. III. Rainfall-Runoff Relationships A. Time of Concentration example problem L = 13,385 ft H = 380 ft tc = L 1.15 7700 H 0.38

29. tc = (13,385) 1.15 7700 (380) 0.38 tc = time of concentration (hr) L = length of catchment (ft) along the mainstream from basin mouth to headwaters (most distant ridge) H = difference in elevation (ft) between basin outlet and headwaters (most distant ridge) L = 31,385 ft H = 380 ft Tc = 0.75 hrs, or 45 minutes

30. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation

31. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation If the period of ppt exceeds the time of concentration, then the Rational Equation applies

32. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation Q=CIA

33. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation Q=CIA Where Q=peak runoff rate (ft3/s) C= runoff coeffic. I = ave ppt intensity (in/hr) A = drainage area (ac)

35. First: solve for time of concentration (“Duration”); THEN: solve for rainfall intensity for a given X year storm.

36. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation example problem The drainage basin that ultimately flows past the JMU football stadium is dominated by an industrial park with flat roofed buildings, parking lots, shopping malls, and very little open area. The drainage basin has an area of 90 acres. Find the peak discharge during a storm that has a 25 year flood return interval.

37. First: solve for time of concentration (“Duration”); THEN: solve for rainfall intensity for a given X year storm. “45 minutes from previous exercise”

39. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation example problem Q = ciA The drainage basin that ultimately flows past the JMU football stadium is dominated by an industrial park with flat roofed buildings, parking lots, shopping malls, and very little open area. The drainage basin has an area of 90 acres. Find the peak discharge during a storm that has a 25 year flood return interval.

40. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation example problem Q = ciA Q = (0.85)*(2.5 in/hr)*(90 acres) The drainage basin that ultimately flows past the JMU football stadium is dominated by an industrial park with flat roofed buildings, parking lots, shopping malls, and very little open area. The drainage basin has an area of 90 acres. Find the peak discharge during a storm that has a 25 year flood return interval.

41. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation example problem Q = 191.3 ft 3 /s The drainage basin that ultimately flows past the JMU football stadium is dominated by an industrial park with flat roofed buildings, parking lots, shopping malls, and very little open area. The drainage basin has an area of 90 acres. Find the peak discharge during a storm that has a 25 year flood return interval.

42. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation example problem Calculate the mean velocity if the cross sectional area of the channel is 40 ft 2. The drainage basin that ultimately flows past the JMU football stadium is dominated by an industrial park with flat roofed buildings, parking lots, shopping malls, and very little open area. The drainage basin has an area of 90 acres. Find the peak discharge during a storm that has a 25 year flood return interval.

43. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation example problem An industrial park with flat roofed buildings, parking lots, and very little open area has a drainage basin area of 90 acres. The 25 year flood has an intensity of 2 in/hr. Find the peak discharge during the storm. Calculate the mean velocity if the cross sectional area of the channel is 40 ft 2. Discharge = Velocity x Area

44. III. Rainfall-Runoff Relationships A. Time of Concentration B. Rational Equation example problem An industrial park with flat roofed buildings, parking lots, and very little open area has a drainage basin area of 90 acres. The 25 year flood has an intensity of 2 in/hr. Find the peak discharge during the storm. Calculate the mean velocity if the cross sectional area of the channel is 40 ft 2. Discharge = Velocity x Area 191.3 ft 3 /s = 40ft 2 * V V = 4.8 ft/s

45. Calculate the mean velocity if the cross sectional area of the channel is 40 ft 2. Discharge = Velocity x Area 191.3 ft 3 /s = 40ft 2 * V V = 4.8 ft/s or 146.3 cm/s If the channel is made of fine sand, will it remain stable?

46. Hjulstrom Diagram 146.3 cm/s 0.10-0.25 mm (fine sand) size range

47. III. Measurement of Streamflow

48. III. Measurement of Streamflow A. Direct Measurements B. Indirect Measurements

49. III. Measurement of Streamflow A. Direct Measurements

50. III. Measurement of Streamflow A. Direct Measurements 1. Price /Gurley/Marsh-McBirney Current Meters

51. III. Measurement of Streamflow A. Direct Measurements 1. Price or Gurley Current Meter 2. Weirs

52. Weirs rectangular Q = 1.84 (L – 0.2H)H 3/2 Where L = length of weir crest (m), H = ht of backwater above weir crest (m), Q = m 3 /s *note: eq. 2.16B in Fetter is incorrect (exponent is 3/2 as shown above)

53. Weirs V notch Q=1.379 H 5/2 Where H = ht of backwater above weir crest (m) Q = m 3 /s

54. III. Measurement of Streamflow B. Indirect Measurements

55. III. Measurement of Streamflow B. Indirect Measurements 1.Manning Equation V = R 2/3 S ½ n Where V = average flow velocity (m/s) R = hydraulic radius (m) S = channel slope (unitless) n = Manning roughness coefficient

56. 1.Manning Equation V = R 2/3 S ½ n Where V = average flow velocity (m/s) R = hydraulic radius (m) S = channel slope (unitless) n = Manning roughness coefficient R = A/P A = Area (m 2 ) P = Wetted Perimeter (m)

57. 1.Manning Equation V = 1.49 * R 2/3 S ½ n Where V = average flow velocity (ft/s) R = hydraulic radius (ft) S = channel slope (unitless) n = Manning roughness coefficient R = A/P A = Area (ft 2 ) P = Wetted Perimeter (ft) If using English units…..

58. Q = A R 2/3 S ½ n Where Q = average flow discharge (m 3 /s) A = area of channel (m 2 ) R = hydraulic radius (m) S = channel slope (unitless) n = Manning roughness coefficient R = A/P A = Area P = Wetted Perimeter If Q = V * A, then V = R 2/3 S ½ n

59. Q = A R 2/3 S ½ N Where Q = average flow discharge A = area of channel r = hydraulic radius s = channel slope (unitless) n = Manning roughness coefficient R = A/P A = Area P = Wetted Perimeter Example Problem: A flood that occurred in a mountain stream comprised of cobbles, pebbles, and few boulders creates a high water mark of 3 meters above the bottom of the channel, and temporarily expands the channel width to 6 m. The slope of the water surface is 100 meters of drop per 1 km of distance. Determine V in m/s Determine Q in m 3 /s

61. B. Indirect Measurements 1.Manning Equation 2. SuperElevation Method 3. Measurement of Cobbles

62. B. Indirect Measurements 1.Manning Equation 2. SuperElevation Method

63. Q = A(R c *g*cosS * tanΘ) ½ Q = discharge, A = average radial cross section in the bend, Rc= radius of curvature S = slope of channel (degrees) Θ = angle between high water marks on opposite banks (degrees) Example Problem:

64. B. Indirect Measurements 3. Measurement of Cobbles

65. V = 0.18d 0.49 Where V =m/s, d=mm where 50 < d < 3200 mm Measure the 5 largest boulders, intermediate axis, take the average The “Costa Equation”

66. B. Indirect Measurements 3. Measurement of Cobbles V = 0.18d 0.49 Where V =m/s, d=mm and 50 < d < 3200 mm Measure the 5 largest boulders, intermediate axis And h c = {V } 1.5 {4.5*{(S + 0.001) 0.17 }} Where V = velocity, in m/s S energy slope (decimal form) hc = competent flood depth (m) Example Problem: Average of five largest boulders: 3.2m x 2.3m x 1.6 m Average slope = 5.5 degrees Find: average velocity and depth of flow

67. V. Hydraulic Geometry A. The relationships Q = V*A Q = V * w * d w = aQ b d = cQ f v = kQ m

68. V. Hydraulic Geometry A. The relationships B. “at a station” C. “distance downstream”

69. A. Hydraulic Geometry “at a station trends” M = 0.26 M = 0.4 M = 0.34

70. A. Hydraulic Geometry “distance downstream trends” M = 0.5 M = 0.1 M = 0.4

71. Distance Downstream

75. VI. Flood Frequency A. Flood Frequency Analysis B. Flow Duration Curves

76. VI. Flood Frequency A. Flood Frequency Analysis Flood recurrence interval (R.I.) use Weibull Method - calculates the R.I. by taking the average time between 2 floods of equal or greater magnitude. RI = (n + 1) / m where n is number of years on record, m is magnitude of given flood

77. VI. Flood Frequency A. Flood Frequency Analysis What does this mean??? the curve estimates the magnitude of a flood that can be expected within a specified period of time The probability that a flow of a given magnitude will occur during any year is P = 1/RI. EX: a 50 year flood has a 1/50 th chance, or 2 percent chance of occurring in any given year.

78. VI. Flood Frequency A. Flood Frequency Analysis For multiple years: q = 1- ( 1-1/RI) n where q = probability of flood with RI with a specified number of years n

79. VI. Flood Frequency A. Flood Frequency Analysis For multiple years: q = 1- ( 1-1/RI) n where q = probability of flood with RI with a specified number of years n EX: a 50 year flood has an 86% chance of occurring over 100 years Example Problem: Determine the water height during a 100 year storm at the Harrison Gaging Station near Grottoes, Virginia.

80. VI. Flood Frequency A. Flood Frequency Analysis Example Problem: Determine the water height during a 100 year storm at the Harrison Gaging Station near Grottoes, Virginia. Method: Access data at www.usgs.gov; select water tabwww.usgs.gov Select “water watch” under ‘streams, lakes, rivers’ option Choose the current stream flow map, your state and the respective station location Open station page by clicking on the station number Select “surface water - peak streamflow” option Choose ‘tab separated file’ format Highlight, copy, and paste (special) your data to Excel for analysis.

81. VI. Flood Frequency A. Flood Frequency Analysis Example Problem: Determine the water height during a 100 year storm at the Harrison Gaging Station near Grottoes, Virginia. Method (continued): Clean up data so that only ‘Year’, ‘Q’, and ‘Gage Ht.’ are present Sort data based on Q in descending order Add magnitude (m) ranking (highest = 1) Add RI formula, where RI = (n+1)/m Create graph depicting RI vs. Q Create graph of Q vs. Gage Ht. Determine Gage Height with respect to the given RI

84. YearQ (cfs)Gage Ht (ft) Magnitud eRI (yrs) 9/6/19962890015.57173.0 11/4/19852810015.47236.5 10/15/19422310017.2324.3 9/19/20032200014.41418.3 6/21/19722130015.25514.6 1924-05-002100016.6612.2 9/6/19791620013.47710.4 10/5/19721530013.2489.1 3/18/19361260013.0798.1 3/19/19751240012.2107.3 9/28/20041230012.26116.6 8/16/19401210012.91126.1 4/17/20111190012.15135.6 4/26/19371170013145.2 9/18/19451130012.8154.9 8/20/19691110012.72164.6 1/25/20101110011.9174.3 11/29/20051090011.84184.1 3/19/19831030011.44193.8 9/20/19281010011.9203.7 2/17/19981000011.59213.5 4/22/1992984011.8223.3 5/30/1971946011.93233.2 10/17/1932870011.5243.0 12/1/1934834011.3252.9 9/19/1944834011.33262.8 10/9/1976825010.62272.7 2/14/1984825010.6282.6 4/17/1987812011.08292.5 6/18/1949798011.06302.4 1/26/1978780010.38312.4

87. VI. Flood Frequency B. Flow Duration Curves

88. VI. Flood Frequency B. Flow Duration Curves “shows the percentage of time that a given flow of a stream will be equaled or exceeded.”

89. B. Flow Duration Curves “shows the percentage of time that a given flow of a stream will be equaled or exceeded.” P = * m n+1 * (100)

90. B. Flow Duration Curves “shows the percentage of time that a given flow of a stream will be equaled or exceeded.” P = * m * n+1 (100)

91. VI. Flood Frequency B. Flow Duration Example Problem: Determine the discharge that can be expected 80% of the time at the Harrison Gaging Station near Grottoes, Virginia. Method: Access data at www.usgs.gov; select water tabwww.usgs.gov Select “water watch” under ‘streams, lakes, rivers’ option Choose the current stream flow map, your state and the respective station location Open station page by clicking on the station number Select “daily data” option, Then click ‘mean discharge’ option Choose the earliest date of record through ‘present’ Choose ‘tab separated file’ format, and select ‘go’ Highlight, copy, and paste (special) your data to Excel for analysis.

92. VI. Flood Frequency A. Flood Frequency Analysis Example Problem: Determine the discharge that can be expected 80% of the time at the Harrison Gaging Station near Grottoes, Virginia. Method (continued): Clean up data so that only ‘Year’, ‘Q’, and ‘Gage Ht.’ are present Sort data based on Q in descending order Add magnitude (m) ranking (highest = 1) Add P formula, where P = m/(n+1) Pick out desired probability value, and record the respective discharge

94. P = * m * 100 n+1

96. How much water would this value of discharge yield for a full day?

97. How much water would this value of discharge yield for a full day? 81 ft 3 * 3600 s * 24 hr = 6,998,400 ft 3 of water in one day s 1 hr 1 d

98. V. Sediment Transport A. Shear Stress τ c = critical boundary shear stress hc = minimal water depth required for flow ρ w = water density (assume 1.00 g/cm 3 ) g = gravitational acceleration (981 cm/s 2 ) S = slope (decimal e.g., meters per meters)

99. V. Sediment Transport A. Shear Stress τ c = h c ρ w g S τ c = critical boundary shear stress (force per unit area) (g/cm-s 2 ) hc = minimal water depth required for flow (cm) ρ w = water density (assume 1.00 g/cm 3 ) g = gravitational acceleration (981 cm/s 2 ) S = slope (decimal, e.g., meters per meters)

100. V. Sediment Transport 1. Shear Stress 2. The Shields Equation τ c = h c ρ w g S τ c = τ *c (ρ s – ρ w )gD 50 τ c = critical boundary shear stress h c = minimal water depth required for flow ρ s, ρ w = grain density (assume 2.65 g/cm 3 ) and water density g = gravitational acceleration (981 cm/s 2 ) D 50 = median bed material grain size τ *c = dimensionless critical shear stress (the Shields number) 0.03 for sand, 0.047 for gravel

102. V.Sediment Transport 1. Shear Stress 2. Shields Equation τ c = h c ρ w g S τ c = τ *c (ρ s – ρ w )gD 50 τ c = critical boundary shear stress hc = minimal water depth required for flow ρ s, ρ w = grain density (assume 2.65 g/cm 3 ) and water density g = gravitational acceleration (981 cm/s 2 ) D 50 = median bed material grain size τ *c = dimensionless critical shear stress (the Shields number) 0.03 for sand, 0.047 for gravel h c =(ρ s – ρ w ) τ *c D 50 ρ w S OR

103. τ c = h c ρ w g S τ c = τ *c (ρ s – ρ w )gD 50 τ c = critical boundary shear stress hc = minimal water depth required for flow ρ s, ρ w = grain density (assume 2.65 g/cm 3 ) and water density g = gravitational acceleration (981 cm/s 2 ) D 50 = median bed material grain size τ *c = dimensionless critical shear stress (the Shields number) 0.03 for sand, 0.047 for gravel h c =(ρ s – ρ w ) τ *c D 50 ρ w S OR Problem: A gravel bed stream of slope 2 m per 1 km has a median grain size of 60 mm. Caculate: 1) the critical shear stress required for bedload mobilization; 2) The critical water depth to initiate motion