Flood analysis

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Flood analysis

  1. 1. 12-1 GEOG415 Lecture 12: Flood AnalysisFlood probability analysisIt aims at estimating the magnitude of floods that will occur ata given probability, for example once in 100 year.Usefulness?Methods of analysis1. Obtain the record of flood. What kind? HYDAT data base (Extreme Flow).2. Rank the record and assign the exceedence probability. → see page 2-7 and 2-8.3. Plot them on a probability chart (e.g. Gumbel)4. Draw a straight line that represents the data set. Dunne and Leopold (1978, Fig. 10-14)
  2. 2. 12-2A map of the areas flooded by the 70-year flood of Bow River. (Montreal Engineering Co., 1973.City of Calgary flood study)
  3. 3. 12-3In Fig. 10-14 why does the straight line ignore the highestpoint?Is it possible that a 100-year flood occurs in a 30-yearobservation period? How will it show up on the Gumbel plot?Is it OK to estimate 100-year flood from 30-year data byextrapolation?Error estimationSuppose a 30-year annual maximum series having a standarddeviation (SD) of 16,000 cfs. The upper bound of the 90 %confidence limit of the 10-year flood is given by (see Table10-12, next page): SD × 0.50 = 8,000 cfs 8,000 cfs Dunne and Leopold (1978, Fig. 10-17)
  4. 4. 12-4 Dunne and Leopold (1978, Table 10-12)Every year, there is a 5 % chance of having a 20-year flood .What is the probability of having a 20-year flood in the nextfive years? → Equation (2-5) in Dunne and Leopold (1978).We have not had a 70-year flood for 71 years. What is theprobability of having a 70-year flood next year?
  5. 5. 12-5Mean annual floodGumbel extreme probability distribution is designed so thatthe average flood (arithmetic mean of all floods in the record)has a theoretical return period of 2.33 years.Using this property, mean annual flood can be determinedgraphically from a Gumbel chart.Homogeneity of flood recordsThe probability theory commonly used in hydrologicalanalysis assumes that the data are homogeneous.What does it mean?Partial-duration flood seriesHow is it different from annual Dunne and Leopold (1978, Table 10-13)maximum series?What is the actual return periodof bankfull discharge?
  6. 6. 12-6Regional flood-frequency analysisPlanners often require the flood frequency for ungaugedbasins. → Need for regional frequency curves.Assumption:For large regions of homogeneous climate, vegetation, andtopography, individual basins covering a wide range ofdrainage areas have similar flood-frequency characteristics Dunne and Leopold (1978, Fig. 10-19) Ratio of flood to the mean annual flood 1 1.5 2.33 5 10 25Dunne and Leopold (1978, Fig. 10-20a) Recurrence interval (years)Floods having specified recurrence interval can be estimatedfrom the mean annual flood.
  7. 7. 12-7How is mean annual flood estimated?Is this method applicablein mountainous regions? Dunne and Leopold (1978, Fig. 10-21)Effects of urbanizationWhat are expected effects? tp: lag to peak tp lc: centroid lag lc Dunne and Leopold (1978, Fig. 10-25)
  8. 8. 12-8Urban sewer systems reduce centroid lag time. Consequences? S: slope Dunne and Leopold (1978, Fig. 10-26) Dunne and Leopold (1978, Fig. 10-28)
  9. 9. 12-9The Unit HydrographWhen the total amount of runoff is given, how do we estimatethe temporal distribution of runoff? e.g time lag to peak, duration, etc.The unit hydrograph is the hydrograph of one inch of stormrunoff generated by a rainstorm of fairly uniform intensityoccurring within a specific period of time (e.g. one hour).The UH theory assumes that temporal distribution depends onbasin size, shape, slope, etc., but is fixed from storm to storm.Is this reasonable?Once the UH is established for a basin, hydrographs resultingfrom any amounts of runoff may be computed from the UH. Hydrograph that would result from 50.8 mm of runoff. 25.4 mm of runoff generated over the basin. Dunne and Leopold (1978, Fig. 10-30)
  10. 10. 12-10Construction of the UH from discharge data1. Select a few storm hydrographs resulting from fairlyuniform rain having similar duration, and separate baseflow.2. Compute the total volume of stormflow (m3) for eachhydrograph and divide it by the basin area to obtain totalstormflow in depth unit R (mm).3. Multiply each discharge measurement by 25.4/R so that thetotal stormflow of the reduced hydrograph is equal to 25.4mm.4. Plot the reduced hydrographs and superimpose them, eachhydrograph beginning at the same time.5. By trial and error, draw the unit hydrograph representingthe shape of all reduced hydrographs and having a totalstormflow of 25.4 mm.
  11. 11. 12-11UH’s for storms of various durationsPrinciples of superposition:Four-hour UH consists of the superposition of two two-hourUH. It needs to be divided by a factor of two to adjust for thetotal stormflow (25.4 mm).Is this reasonable? Under what condition? Dunne and Leopold (1978, Fig. 10-32)
  12. 12. 12-12S-curve methodCommonly used to derive hydrographs resulting fromarbitrary-duration storms.In this example, theoriginal UH was derivedfor 2.5-hr storm.It is superimposedsuccessively at 2.5-hrinterval, resulting in a S-shaped curve. Dunne and Leopold (1978, Fig. 10-33)What does it represent?Two S-curves are now plotted with an offset of 1 hour (a).The difference between the two S-curves represents anotherhydrograph (b), which is adjusted for the total stormflow toyield the UH of 1-hour storm (c). Dunne and Leopold (1978, Fig. 10-34)
  13. 13. 12-13Synthetic UH’sHow can we obtain the UH for an ungauged basin?Assumptions? Dunne and Leopold (1978, Fig. 10-35)Snyder methodCorrelation between the lag to peak (tp, hours) and basinlength. tp = Ct (LLc)0.3 L: Length of main stream from outlet to divide (miles) Lc: Distance from the outlet to a point on the stream nearest the centroid of the basin. Ct: Empirical constant (0.3-10)Duration of rainstorm (Dr) and tp were correlated by: Dr = 0.18tpin Snyder’s study. → may not be true in other regions.
  14. 14. 12-14The peak discharge (Qpk, ft3 s-1) is given by: Qpk = CpA/tp A: Basin area (mile2) Cp: Empirical constant (370-405)The duration (tb) of the UH could be given by: tb = 72 + 3tp or tb = 5(tp + 0.5Dr)The width of the UH at 75 % (W75, hour) of the peak flow isgiven by: W75 = 440A/Qpk1.08Similarly, W50 is given by: W50 = 770A/Qpk1.08From tp, Dr, tb, Qpk, W75, and W50, the UH can now besynthesized. Dunne and Leopold (1978, Fig. 10-38)
  15. 15. 12-15Triangular UH by Soil Conservation ServiceAn alternative methodassumes a triangular shape ofthe UH. The detailed methodof construction is given inDunne and Leopold (1978,p.342-343).In this case, the time of riseto peak (Tr) is given by: Dunne and Leopold (1978, Fig. 10-39) Tr = Dr/2 + tpDimensionless UH by Soil Conservation ServiceThis method uses dimensionless time (T/Tp) and discharge(Q/Qpk) to plot the UH.Q and T for each basin canbe calculated from Fig.10-40once Tp and Qpk are given.Tp is dependent on thegeometry and dimension ofthe basin. Qpk is dependenton the amount of runoff,which is estimated from the Dunne and Leopold (1978, Fig. 10-40)curve number method.
  16. 16. 12-16Both methods by SCS were developed for small agriculturalwatersheds.ExampleA 5-km2, reasonably flat watershed has a curve number of 88.Estimated time of concentration (tc, see page 11-14) is 30minutes. Generate a synthetic hydrograph resulting from 51mm of rain applied uniformly over a two-hour period. Runoff (R) = 25 mm using the curve number method Tp = 0.5Dr + 0.6tc = 78 min = 1.3 hr from DL, Eq.(10-20)The dimensionless UH is designed so that Qpk ≅ 0.5R/Tp = 0.5 × 25 / 1.3 = 9.6 mm hr-1In terms of discharge, Qpk = 15.6 mm hr-1 × 5 km2 = 48000 m3 hr-1 = 13 m3 s-1 13 Discharge (m3 s-1) 0 1.3 Time (hour)

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