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- 1. Chapter 11 Hydrology• Hydrologic cycle• Infiltration• Rainfall hyetograph, Duration, Intensity, return period• Effective or Excess rain (ph index, Horton)• Runoff hydrograph• Runoff hydrograph for gauged stations (unit hydrograph)• Synthetic hydrograph for ungauged stations – Rational method – SCS curve number method
- 2. Rainfall Availability and AssociatedGrowth in Water Resources EngineeringProjects Worldwide
- 3. 11.1 The Hydrologic Cycle Atmospheric Moisture 39 1 0 0 Moisture over land P r e c i pdi t a t i o n o n l a n 61 385 P Evaporation from land Precipitation on ocean Snow melt Runoff Evap Surface runoff Precipitation ET 424 Evap Evaporation from ocean Infiltration Streams Groundwater Wat er t ab l Recharge e Runoff 38 Surface discharge Groundwater flow Impervious 1 GroundwaterLake strata GW discharge Reservoir
- 4. AtmosphereEvaporation Evaporation Precipitation Water on Surface Overland Flow Channel Reservoir FlowEvapotranspiration Ground Water Ground Water Flow Ocean The Hydrologic Cycle
- 5. Major Hydrologic Processes• Precipitation (measured by radar or rain gage)• Evaporation or ET (loss to atmosphere)• Infiltration (loss to subsurface soils)• Overland flow (sheet flow toward nearest stream)• Streamflow (measured flow at stream gage)• Ground water flow and well mechanics• Water quality and contaminant transport (S & GW)
- 6. Diagram Showing Two Watersheds (Catchments)
- 7. The Watershed or Basin• Area of land that drains to a single outlet and is separated from other watersheds by a drainage divide.• Rainfall that falls in a watershed will generate runoff to that watershed outlet.• Topographic elevation is used to define a watershed boundary (land survey or LIDAR)• Scale is a big issue for analysis
- 8. Figure 11.3 Watershed delineation: (a) peak point identification
- 9. Figure 11.3 (continued) Watershed delineation: (b) circumscribing boundary. Source: Adapted fromthe Natural Resource Conservation Service(www.nh.nrcs.usda.gov/technical/WS_delineation.html).
- 10. Watershed Response Tributary Precipitation over the area Portion Infiltrates the soil Portion Evaporates or ET back Reservoir Remainder - Overland Flow Natural stream Overland flow - Channel flow Urban Final Hydrograph at Outlet Concrete channel Q T
- 11. 11.2 Precipitation Watershed area = 240 km2
- 12. Arithmetic Mean Method• The simplest of all is the Arithmetic Mean Method, which taken an average of all the rainfall depths
- 13. The Theissen polygon method• This method, first proposed by Thiessen in 1911, considers the representative area for each rain gauge:• 1. Joining the rain gauge station locations by straight lines to form triangles• 2. Bisecting the edges of the triangles to form the so- called “Thiessen polygons”• 3. Calculate the area enclosed around each rain gauge station bounded by the polygon edges (and the catchment boundary, wherever appropriate) to find the area of influence corresponding to the rain gauge. (areas of influence of each rain gauge)
- 14. Calculation
- 15. The Isohyetal method• This is considered as one of the most accurate methods, but it is dependent on the skill and experience of the analyst. The method requires the plotting of isohyets as shown in the figure and calculating the areas enclosed either between the isohyets or between an isohyet and the catchment boundary. The areas may be measured with a planimeter if the catchment map is drawn to a scale.• Area I = 40 km2• Area II = 80 km2• Area III = 70 km2• Area IV = 50 km2• Total catchment area = 240 km2• The areas II and III fall between two isohyets each. Hence, these areas may be thought of as corresponding to the following rainfall depths:• Area II : Corresponds to (10 + 15)/2 = 12.5 mm rainfall depth• Area III : Corresponds to (5 + 10)/2 = 7.5 mm rainfall depth• For Area I, we would expect rainfall to be more than 15mm but since there is no record, a rainfall depth of 15mm is accepted. Similarly, for Area IV, a rainfall depth of 5mm has to be taken.
- 16. calculation Area I = 40 km2 Area II = 80 km2 Area III = 70 km2 Area IV = 50 km2
- 17. Hydrologic Theory• One of the principal objectives in hydrology is to transform rainfall that has fallen over a watershed area into flows to be expected in the receiving stream.• Losses must be considered such as infiltration or evaporation (long-term)• Watershed characteristics are important
- 18. Watershed Characteristics Divide Size Slope Reservoir Shape 1 mile Natural stream Soil type Urban Storage capacity Concrete channel
- 19. 11.4 The Watershed Response - Hydrograph• As rain falls over a watershed area, a certain portion will infiltrate the soil. Some water will evaporate to atmosphere.• Rainfall that does not infiltrate or evaporate is available as overland flow and runs off to the nearest stream.• Smaller tributaries or streams then begin to flow and contribute their load to the main channel at confluences.• As accumulation continues, the streamflow rises to a maximum (peak flow) and a flood wave moves downstream through the main channel.• The flow eventually recedes or subsides as all areas drain out.
- 20. Measured Flow at Main St Gage 30,000 29,000 cfs 25,000 Jun 76Flow, cfs Apr 79 20,000 Sep 83 15,000 Mar 92 Mar 97 10,000 5,000 3 6 9 12 15 18 21 24 Time, hrs Time, hrs
- 21. Figure 11.11 Schematic of a stream gauging station. Source: USGS, 2008.
- 22. HOW IS FLOW MEASURED?
- 23. Figure 11.12 Procedure for discharge measurement with current meter. Source: USGS, 2008.
- 24. 11.5 Excess rainfall• Rainfall that is neither retained on the land surface nor infiltrated into the soil• Graph of excess rainfall versus time is called excess rainfall hyetograph• Direct runoff = observed streamflow - baseflow• Excess rainfall = observed rainfall - abstractions• Abstractions/losses – difference between total rainfall hyetograph and excess rainfall hyetograph
- 25. Infiltration and excess rain Excess rainfall rateRainfall or infiltration rate (mm/hr, in/hr) Actual Infiltration rate Infiltration capacity rate curve
- 26. f-index f-index: Constant rate of abstraction yielding excess rainfall hyetograph with depth equal to depth of direct runoff• Used to compute excess rainfall hyetograph when observed rainfall and streamflow data are available
- 27. Hydrologic Analysis RUNOFF HYDROGRAPHS VR   Q ( t ) dt VR R A
- 28. STORM WATER HYDROGRAPHS• Graphically represent runoff rates vs. time at watershed outlet or any selected points of interest in a watershed). and occurred as a result of rainfall.• Design Engineers are interested in calculating – Peak runoff rates (Qp) – Volume of runoff VR• Measured hydrographs for the maximum probable storm are best But not often available• The volume under the effective rainfall hyetograph is equal to the volume of surface-direct runoff.• Methods are available to develop a Unit hydrograph or “synthetic” hydrograph for watersheds that can be used to determine Qp and VR.
- 29. HYDROGRAPH COMPONENTS• Qp is the maximum flow rate on the hydrograph• tp (time to peak) is the time from the start of they hydrograph to qp.• tb (base time) is the total time duration of the hydrograph.
- 30. HYDROGRAPH COMPONENTS• Tc (time of concentration) time it takes water to flow from the hydraulically most remote point in a watershed to the watershed outlet• TL(lag time) is the average of the flow times from all locations in the watershed and can be estimated as the length of time from the center of mass of the first effective rainfall block, to the peak of the runoff hydrograph.• If each block of effective rainfall has a duration of D D Tp  TL  2
- 31. Typical hydrograph (separation of base flow)• Stream flow hydrographs have two components 1. Direct runoff resulted from a specific rainfall hyetograph 2. base flow from ground water seepage• Base or groundwater that flow can be separated from direct runoff
- 32. Rainfall-Runoff Relationships• Gauged and ungauged watersheds• Gauged watersheds – Watersheds where data on precipitation, streamflow, and other variables are available• Ungauged watersheds – Watersheds with no data on precipitation, streamflow or other variables.
- 33. 03/02/2006 Unit Hydrograph (HU)• Unit Hydrograph are developed for gauged watersheds where data on precipitation, streamflow, and other variables are available• For Ungauged watersheds with no data on precipitation, streamflow or other variables we use synthetic hydrograph
- 34. Measured Flow at Main St Gage 30,000 29,000 cfs 25,000 Jun 76Flow, cfs Apr 79 20,000 Sep 83 15,000 Mar 92 Mar 97 10,000 5,000 3 6 9 12 15 18 21 24 Time, hrs Time, hrs
- 35. Unit Hydrograph Theory• Direct runoff hydrograph resulting from a unit depth of excess rainfall (1.0 in or 1.0 cm) occurring uniformly on a watershed at a constant rate for a specified duration.• Unit pulse response function of a linear hydrologic system• Can be used to derive runoff from any excess rainfall R on the watershed of similar duration D or series of excess rain.
- 36. Unit hydrograph assumptions• Assumptions – Excess rainfall has constant intensity during duration – Excess rainfall is uniformly distributed on watershed – Base time of runoff is constant – Ordinates of unit hydrograph are proportional to total runoff (linearity) – Unit hydrograph represents all characteristics of watershed (lumped parameter) and is time invariant (stationarity)
- 37. Typical Unit Hydrograph
- 38. Linearity and superposition of the Unit Hydrograph
- 39. Unit Hydrographs - ExampleObtain a Unit Hydrograph for a basin of 315 km2 of area using the grossrainfall hyetograph and streamflow hydrograph data tabulated below. Time Observed (h) Hydrograph Time Gross Precipitation (m3/s) (h) (GRH) 0 100 (cm/h) 1 100 0-1 0.5 2 300 1-2 2.5 3 700 2-3 2.5 4 1000 3-4 0.5 5 800 6 600 7 400 8 300 9 200 10 100 11 100
- 40. Cont…• Separate the baseflow from the observed streamflow hydrograph in order to obtain the Direct Runoff Hydrograph (DRH). For this example, use the horizontal line method to separate the baseflow. From observation of the hydrograph data, the streamflow at the start of the rising limb of the hydrograph is 100 m3/s.• Compute the volume of Direct Runoff. This volume must be equal to the volume of the Effective Rainfall Hyetograph (ERH).
- 41. Cont….• Volume of Direct Runoff Hydroraph VR = 200+600+900+700+500+300+200+100) m3/s (3600) s = 12600,000 m3• Express VR in equivalent units of depth: VDRH in equivalent units of depth = VDRH/Area of the basin = 12600,000 m3/(315000000 m2) = 0.04 m = 4 cm: R = 4 cm• Excess rain (4 cm) = Volume of runoff under the hydrograph• R = CP (C is the runoff constant =4/6=0.67)• Obtain a Unit Hydrograph by normalizing the DRH. Normalizing implies dividing the ordinates of the DRH by the VR in equivalent units of depth (4 cm).
- 42. Table of calculationTime (h) Observed Direct Runoff Unit Hydrograph Hydrograph Hydrograph (m3/s/cm) (m3/s) (DRH) (m3/s) 0 100 0 0 1 300 200 50 2 700 600 150 3 1000 900 225 4 800 700 175 5 600 500 125 6 400 300 75 7 300 200 50 8 200 100 25 9 100 0 0 10 100 0 0
- 43. Duration, f-index , Excess or Effective rainfall heytograph (ERH)• Determine the duration D of the ERH associated with the UH obtained in . In order to do this: – Determine the volume of losses, VLosses which is equal to the difference between the gross depth of rainfall, VGRH, and the depth of the direct runoff hydrograph, VR .• VLosses = VGRH - VR = (0.5 + 2.5 + 2.5 +0.5) cm/h 1 h - 4 cm = (6-4)= 2 cm – Compute the f -index equal to the ratio of the volume of losses to the rainfall duration, tr. Thus, f-index = VLosses/tr = 2 cm / 4 h = 0.5 cm/h – Determine the ERH by subtracting the infiltration (f -index) from the GRH:
- 44. Duration of the Unit hydrograph• As observed in the table, the duration of the effective rainfall hyetograph is 2 hours. Thus, D = 2 hours, and the Unit Hydrograph obtained above is a 2-hour Unit Hydrograph. Time Gross f-index Excess rainfall Precipitatio (cm/h) n (GRH) (cm/h) (h) (cm/h) 0-1 0.5 -0.5 0 1-2 2.5 -0.5 2 2-3 2.5 -0.5 2 3-4 0.5 -0.5 0
- 45. Application of UH• Once a UH is derived, it can be used/applied to find direct runoff and stream flow hydrograph from other storm events whose effective rainfall hyetographs can be represented as a sequence of uniform intensity (rectangular) pulses each of duration D.• using the principles of superposition and proportionality
- 46. Example Application of UH• Using the UH obtained in A., predict the total streamflow that would be observed as a result of the following ERH• Determine the volume of each ERH pulse, Pm, expressed in units of equivalent depth:• Use superposition and proportionality principles: Time Effective Rainfall (h) Precipitation depth (ERH) (cm/h) Pm (cm) 0-2 0.5 1.0 2-4 1.5 3.0 4-6 2.0 4.0 6-8 1.0 2.0
- 47. 1 2 3 4 5 6 7Time UH P1*UH P2*UH P3*UH P4*UH DRH Total (h) (m3/s/cm) (m3/s) (m3/s) (m3/s) (m3/s) (m3/s) (m3/s) 0 0 0 0 100 1 50 50 50 150 2 150 150 0 150 250 3 225 225 150 375 475 4 175 175 450 0 625 725 5 125 125 675 200 1000 1100 6 75 75 525 600 0 1200 1300 7 50 50 375 900 100 1425 1525 8 25 25 225 700 300 1250 1350 9 0 0 150 500 450 1100 1200 10 75 300 350 725 825 11 0 200 250 450 550 12 100 150 250 350 13 0 100 100 200 14 50 50 150 15 0 0 100
- 48. Solution…Cont…• Columns 2 - 5: Apply the proportionality principle to scale the UH by the actual volume of the corresponding rectangular pulse, Pm. Observe that the resulting hydrographs are lagged so that their origins coincide with the time of occurrence of the corresponding rainfall pulse.• Column 6: Apply the superposition principle to obtain the DRH by summing up Columns 2 - 5.• Column 7: Add back the baseflow in order to obtain the Total Streamflow Hydrograph.
- 49. Hydrograph
- 50. Need for synthetic UH• UH is applicable only for gauged watershed and for the point on the stream where data are measured• For other locations on the stream in the same watershed or for nearby (ungauged) watersheds, synthetic procedures are used.• Synthetic unit hydrographs provide ordinates of the unit hydrograph as a function of tp, Qp and a mathematical or empirical shape description.
- 51. Synthetic UH• Synthetic hydrographs are derived by – Relating hydrograph characteristics such as peak flow, base time etc. with watershed characteristics such as area and time of concentration. – Using dimensionless unit hydrograph – Based on watershed storage
- 52. 11.6 SCS procedures Excess Rainfall• When stream flow hydrograph is not available, the SCS Curve Number Approach is used to determine excess rain that becomes runoff from only rainfall data.• It also used to construct runoff hydrograph or Unit hydrograph called synthetic hydrograph• Qp and VR can then be determined for engineering design.
- 53. SCS CURVE NUMBER APPROACH • By far the most popular method. • Combines initial abstractions and infiltration losses and estimates rainfall excess as:R P  0.2S 2 when P  0.2S P: Cumulative rainfall (mm, in) P  0.8S S: soil moisture storage deficit 1000 at time of rainfall (mm, in)S  10 for R, P, S in inches CN 25400 R: runoff (mm, in); begins onlyS  254 for R, P, S in mm. when P > 0.2 S) CN CN: Curve number
- 54. CN CURVE NUMBER• A parameter that combines soil type and land use to estimate runoff potential.• Based on the Hydrologic Soil Group (HSG), land use and condition.• Range between 0 and 100. The greater the curve number, the greater the potential for RO.• SCS classified more than 4000 soils into four general HSG (A, B, C, and D)• Based on soils minimum infiltration rate when the soil is bare and after prolonged wetting.• In general A have the highest infiltration capacity and lowest runoff potential (sandy soils) and D have lowest infiltration rates and highest runoff potential (clay soils)• Curve numbers for various land uses ranging from cultivated land to industrial and residential districts.• Impervious areas and water surfaces are assigned curve numbers of 98-100.
- 55. Antecedent Moisture Conditions
- 56. Table 11.4 SCS Runoff Curve Numbers for AMC II
- 57. CURVE NUMBERS
- 58. weighted CN for Mixed Land Uses and Soil group• An area weighted CN is used when the area considered is for mixed land uses and Hydrological Soil Group. CN   AiCNi  Ai
- 59. CREATING AN EFFECTIVE RAINFALL HYETOGRAPH• Calculate the accumulated P for each time step from a rainfall hyetograph.• Calculate the appropriate weighted CN.• Calculate S using Equation (11.4).• Find 0.2S.• For each time step where the accumulated P > 0.2 S calculate the accumulated R using Equation (11.3).• Find the incremental R at each time step.• Plot the incremental R vs. time.
- 60. EXAMPLE PROBLEM• Given: – Precipitation (P) = 4.04 in. – A watershed that has: • 35% cultivated with a D soil group • 30% meadow with a B soil group • 35% thin forest with a C soil group• Required: – Calculate the surface runoff (excess rainfall)
- 61. Watershed with Land Use % and HSGsListed 30% Meadow 35% Cultivated HSG = B HSG = D 35% Thin Forest HSG = C
- 62. Cont…EXAMPLE PROBLEM1. Find the curve numbers Use HSG % CN* Cultivated D 35 91 Meadow B 30 58 Thin Forest C 35 77 *Table 5.1 text (reference is important)2. Calculate a weighted CN Weights based on % area CNavg = 0.35(91) + 0.30(58) + 0.35(77) CN avg = 76.2 = 76
- 63. Cont…EXAMPLE PROBLEM3. Calculate the S term S = 1000 / CN – 10 = (1000 / 76) – 10 S = 3.16 in.4. Check to see if P > 0.2S 0.2S = 0.2(3.16) = 0.63 in.  P > 0.2S5. Calculate surface runoff (R) R = [(P - 0.2S)^2] / (P + 0.8S) R = [(4.04 – 0.2(3.16)]2 / [4.04 + ((0.8)3.16)] R = 1.77 in. For a rainfall event = 4.04 in. on the given watershed with average soil moisture conditions
- 64. SCS Triangle hydrograph• If R is calculated for a given storm in A watershed of area A, then a synthetic triangular hydrograph can be constructed• Only two parameters are needed – Peak discharge Qp, and – Time to peak Tp and Recession time Tr which usually depend on Tp.
- 65. SCS Triangular Hydrograph• RA = volume under Excess rain (R) of duration D the curve D 1 TL RA  (2.67Tp)Q p Q (m3/s; cfs) Qp 2 2 RA Qp  2.67T p 0.75 RA Qp  Tp Tr = 1.67 Tp Tp Time (hr)
- 66. Tp time to peak• Now R is find from the SCS procedure• Tp is usually a function of the longest travel time in a watershed, called time of concentration Tc Tp = 0.67 Tc
- 67. Time of Concentration (tc)• The time it takes flow to move from the most hydraulically remote point in a watershed to the watershed outlet – The distance from the hydraulically most remote point to the outlet is called the hydraulic length• tc is the sum of flow times for the various flow segments as the water travels to the watershed outlet – Overland flow + shallow channel flow + open channel flow• Travel time for each segment depends on length of travel and flow velocity
- 68. Hydraulically mostremote point in thewatershed
- 69. TIME of Concentration, Tc• SCS Equation to calculate time of Concentration • L = hydraulic length of watershed (feet) • S = curve number parameter (inches) • Y = average land slope of the watershed (%) • Tc = time lag (hours) L ( S  1) 0.8 0.7 Tc  0.5 1140Y
- 70. SCS Unit Hydrograph Time to Peak and Duration• Duration of rainfall excess should be taken 1/5 to 1/3 Tp.• Base time, Tb T T  D p L – Tb=2.67 Tp. 2 – Some use Tb = 5Tp D  0.133Tc – Some use Tb = ∞ TL  0.6Tc T p  0.67Tc Tb  2.67T p
- 71. SCS Synthetic Unit hydrograph• Synthetic UH in which the discharge is expressed by the ratio of Q to Qp and time by the ratio of T to Tp• If peak discharge and lag time are known, UH can be estimated. R 1 0.75 A Qp  tp
- 72. SCS unit hydrograph
- 73. Example • Construct a 10-min SCS UH. A = 3.0 km2 and Tc = 1.25 h 0.833 h DT p  TL   0.6(1.25)  0.166 / 2  0.833h 2 qTb  2.67(0.833)  2.22h 7.49 m3/s.cm 2.08(3)Qp   7.49 m3 / s 0.833Multiply y-axis of SCShydrograph by Qp and x-axisby Tp to get the required UH, t 2.22 hor construct a triangular UH
- 74. EXAMPLESolution: – HSG = D / Commercial  T. 5.1  CN = 95 • S = 0.53 in. – Assume AMC = II  R = 1.96 in. of runoff – Find points to develop the unit hydrograph • tl = 0.75 hr (45 min) • tp = 1.25 hr (75 min) • tb = 3.33 hr (200 min) • qp = 302 cfs / 1 in. of runoff – Plot unit hydrograph – Check area under the triangle  1 in.
- 75. 600 400Q (cfs) qp = 302.5 cfs 300 200 50 tp = 75 100 150 tb = 200 T(min)
- 76. Check runoff depth? Volume under triangle = (302.5 cfs x 4,500 sec) / 2 + [(302.5 x (12,000 – 4,500 sec)] / 2 = 1,812,000 ft3Surface runoff depth = 1,812,000 ft3 / 21,780,000 ft2 =0.08 ft = 1.0 in. ok
- 77. EXAMPLESolution: qp 2.5” rain = 302.5 cfs x 1.96 in. of SRO qp 2.5” rain = 592.9 cfs Plot storm hydrograph Check area under the triangle  1.96 in.
- 78. Q (cfs) qp = 592.9 cfs600 Surface runoff depth = 1.96 in. ok400 Volume under triangle300 = 3,557,400 ft3200 T(min) 50 tp = 75 100 150 tb = 200
- 79. 11.8 The Rational Method• Empirical method developed in 1851 to estimate Qp (m3/s, cfs). many limitations but still widely used Q p  0.0028CiA Q p  1.008CiA • Qp = peak flow (m3/s; cfs) • C = coefficient (dimensionless) • i = rainfall intensity (mm/hr; in/hr) with duration (D) = tc – tc = time of concentration • A = drainage area (ha; ac) • 1.008 is conversion factor from ac-in/hr to cfs. • 0.0028 is conversion factor ha-mm/hr to m3/s
- 80. Rational Method• Rationale of rational method: – If rain falls steadily across the entire watershed long enough, Peak Q will increase until it equal the average rate of rain times the basin area (adjust by a coefficient to account for infiltration) – the entire watershed contribution to runoff occurs when rainfall duration (D) = Time of Concentration tc – If D ≠ tc , runoff less than when D = tc – Qp is at a maximum when D = tc
- 81. Qp = CiA “Rational Method” Limitations• Reasonable for small watersheds < 80 ha; 200 acre• Assumes constant and uniform rainfall and constant infiltration• The runoff coefficient is not constant during a storm• No ability to predict flow as a function of time (only peak flow)• Only applicable for storms with duration longer than the time of concentration• Runoff frequency = rainfall frequency used• Peak flow equation only• Cannot be used to predict time to the peak (tp) and should not be used to develop hydrographs
- 82. Time of Concentration (Tc): Kirpich• Tc = time of concentration [min]• L = “stream” or “flow path” length [ft]• h = elevation difference between basin ends [ft] 0.385  3.35 x 10 L  6 3 tc       h  Watch those units!
- 83. Time of Concentration Contd.• Another name for Tc is gathering time. Tc can be related to catchment area, slope etc. using the Kirpich equation: Tc = 0.02 L 0.77 S – 0.385• Tc is the time of concentration (min);• L is the maximum length of flow (m);• S is the watershed gradient (m/m).
- 84. Time of Concentration (Tc): Hatheway• Tc = time of concentration [min]• L = “stream” or “flow path” length [ft]• S = mean slope of the basin• N = Manning’s roughness coefficient (0.02 smooth to 0.8 grass overland) 0.47  2nL  tc    3 S   
- 85. ESTIMATING THE TIME PARAMETERS• Time of concentration (tc)• For some areas, we can sum the time for various flow segments as the water flows toward the watershed outlet.• Segments – Overland flow nL – Shallow channel flow t  c  i – Flow in open channels. i 1 v i
- 86. Time of Concentration Contd.• With Tc obtained for the catchment, decide on a return period.• For small conservation works, return period is assumed as 10 years.• With the Tc and assumed return period, get an intensity value from the Intensity-Duration curve derived for the area.
- 87. Intensity-Durtaion-Frequency Curve for Madaba
- 88. Rational Method• Runoff Coefficient (C) – Difficult to accurately determine – Must reflect factors such as: interception, infiltration, surface detention, antecedent moisture conditions – Studies have shown that C is not a constant • C increases with wetter conditions – Table 5.5 contains a range of values for C – Compute average C for composite areas • Area weighted basis • Cavg = SC A / SA (same method used with Curve i i i Numbers)
- 89. Runoff Coefficients for the Rational Method
- 90. Qp = CiA “Rational Formula” - Review• Estimate tc• Pick duration of storm = tc Why is this the max flow?• Estimate point rainfall intensity based on synthetic storm• Convert point rainfall intensity to average area intensity• Estimate runoff coefficient based on land use• Calculate Qp
- 91. “Rational Formula” - Fall Creek 10 Year Storm• Area = 126 mi2 = 126 x 640= 80640 acre = 326 km2• L - 15 miles - 80,000 ft• H - 800 ft (between Beebe lake and hills)• tc = 274 min = 4.6 hours 3.35 x 10 L  6 3 0.385 t c     h • 6 hr storm = 2.5” or 0.42”/hr• Area factor = 0.87 therefore i = 0.42 x 0.87 = 0.36 in/hr Area correction
- 92. “Rational Formula” - Fall Creek 10 Year Storm• C - 0.25 (moderately steep, grass-covered clayey soils, some development)• Qp = 1.008CiA  0.36in  2 640acre   Q p  1.0080.25 126mi   hr   2 mi• QP = 7300 ft3/s (200 m3/s)• Empirical 10 year flood is approximately 150 m3/s
- 93. Rational Formula Example• Suppose it rains 0.25” in 30 minutes (i = 0.5 in/hr) on Fall Creek watershed (A = 126 mi2)and runoff coefficient is 0.25. What is the peak flow?  0.25in  2 640acre   Q p  1.0080.25 126mi   0.5h   2 miQ p  40,650cfs  1150m3 / s Peak flow in record was 450 m3/s. What is wrong? Method not valid for storms with duration less than tc.

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