1. Hydrological Forecasting

2. Introduction: How to use knowledge to predict from existing data, what will happen in future?. This is a fundamental problem of all engineering design. Water events are usually random in nature The hydrologist is frequently asked what the maximum possible discharge of a particular river will be. The only answer that can be given is that : From the data available, and making various assumptions, it would appear that a certain value will not be exceeded on average more than once in a specific number of years

3. Flood Formulae Catchment area formulae. The particular random variable of river flood discharge have been proposed to define the 'maximum flood' that could occur for a particular catchment. The formulae are empirical by nature, derived from observed floods on particular catchments and usually of the form: Q = CA n WhereQ =flood discharge in m 3 /s (or ft 3 /s). A =catchment area in km 2 (or mile 2 ). n =an index usually between 0.5 and 1.25 C =a coefficient depending on climate, catchment and units.

4. The Rational Method. The introduction of rainfall into a formula might be expected to improve it, bearing in mind the type of relationship which exists between rainfall and runoff This kind of direct relationship of runoff to rainfall depths has been used in the past to determine flood discharges. Mulvaney was the first to propose the idea in his work on Irish arterial drainage. It was also the basis of the Lloyd-Davis method of sewer designs and the Bransby-Williams estimating method for floods in India. Its use has persisted to the present because of its sim­plicity. The formulae are all of the form Q p = CiA

5. Where i = rainfall intensity in a time t A = catchment area C =a dimensionless runoff coefficient, whose value depends on the catchment characteristics. Q P = peak discharge due to the particular rainstorm, and assumed to occur after time t c when the whole catchment area is contributing t c =time of concentration, i.e. the time taken for rain falling on the catchment farthest from the gauging station to arrive there. Bransby-Williams gave a formula for the design rainfall duration D (in hours):

6. Where L = greatest distance from the edge of catchment to the outfall D =is the diameter of a circle of area equal to catchment area (L/d is, therefore, a dimensionless coefficient of circularity) A = catchment area in square miles h = the channel slope (as a percentage) along its greatest length t c =time of concentration in hours. Probability of the N-year event. The term recurrence interval (also called the return period period), denoted by T r, is the time that, on average, elapses between two events that equal or exceed a particular level. Putting it another way, t he N-year event, the event that is expected to be equalled or exceeded, on average, every N years, has a recurrence interval, T r of N years.

7. Let the probability P(X <x) represent the probability that x will not be equaled or exceeded in a certain period of time. Then P(X < x) n will represent the probability that x will not be equalled or exceeded in n such periods. For an independent series and from the multiple probability rule : P(X < x) n =[ P(X < x)] n = [1 - P(X > x)] n Therefore : P(X > x) n = 1 – [ 1 – P(X > x)] n Now:

8. Then : So, for example, the probability of X > x, where x is the value of a flood with return period of 20 years, occurring in a particular 3-year period is : P(X > 20 yr flood) 3 =1 - [ 1- ] 3 =1 – [0.95] 3 =1 – 0.857 =0.143 or 14.3 %

9. Table 1 Shows the probability (%) of the N-year flood occurring in a particular period Number of years in period N = Average return period T r (years ) 51020501002005001000 120105210.50.20.1 23619104210.40.2 3492714631.50.60.3 56741231052.510.5 108965401810521 2099886433181042 3099.9967845261463 60—99.895704326116 100——99.48763391810 200———98.287633318 500————99.3926339 1000—————99.38663

10. To find design period of a temporary project (n) : Since : 1 - P(X > x) n = (1 – ) n = ( )n )n Log (1 - P(X > x) n ) = n log ( ) Therefore: n= Example: 1 How long may a cofferdam remain in a river, with an even chance of not being overtopped, if it is designed to be secure against a 10-year flood? Here, the policy ruling is that there should be an even chance, so P(X > x) n = 0.50 and T r = 10 then

11. n = = 6.5 years Determining the magnitude of the N ‑ year) event by plotting. Having listed a series of events (for example, maximum floods) accorded a ranking m, starting with; m = 1 for the highest value, m = 2 for the next highest and so on in descending order. The recurrence interval T r can now be computed from one of a number of formulae:

12. Where: m = event ranking ; and n = number of events. But there are objections to its use because of the bias it introduces to the largest events in a short series. The probability P of an N-year event of return period T r is: (%) Cases of Q-T r Graphs: 1. Q against T r using linear co-ordinates (Figure 1). Extrapolation of the curve to the T r values of T r depends critically on the few highest points.

13. Figure 1: Annual maximum mean daily flows of the River Thames at Teddington, 1883 - 1988

14. 2. Q (linear) against T r (logarithmic) (Figure 2). This yields a straight line fitted to all but the lowest values. Although extrapolation is simpler, unless T r follows a logarithmic law, extrapolation is not necessarily more accurate than for Figure 1. Figure 2 : Annual maximum mean daily flows of the River Thames at Teddington, 1883 – 1988 [semi-log]

15. 3. Q (linear) against probability (per cent) (Figure 3). As often happens, flood series points lie on a shallow curve on probability paper (where a normal distribution of probability is assumed). Figure 3 : Annual maximum mean daily flows of the River Thames at Teddington, 1883-1988 (normal probability)

16. 4. Q (logarithmic) against probability (per cent) (Figure 4). The curve of figure 3 is now transformed to a straight line. A variation of the approach in figure 3 is to assume that the logarithm of the variate Q is normally distributed, leading to the use of logarithmic-normal distribution (or log- normal paper). Figure 4 : Annual maximum mean daily flows of the River Thames at Teddington, 1883-1988 (log-normal)

17. Analytical Methods : Other investigators have proposed methods assuming other theoretical frequency distributions. Gumbel used extreme-value theory (EVl) to show that in a series of extreme values X x, X 2 … X n where the samples are of equal size and X is an exponentially distributed variable (for example, the maximum discharge observed in a year's gauge readings), then the cumulative probability P' that any of the n values will be less than a particular value X (of return period T) approaches the value P' = e -e-y where e is the natural logarithm base, and y = - ln

18. That is, P' is the probability of non-occurrence of an event X in T years, or T= (Note that this argument refers to Gumbel's method. (The reader should not confuse this with the normal usage of Tr = 1/P where P - probability of occurrence.) The event X, of return period T years, is now defined as Q T, and : Q T = Q av +  (0.78y – 0.45) Where Q av = average of all values of 'annual flood' Q max  = standard deviation of the series.

19. Or :  = where n = number of years of record = number of Q max values  Q max = sum of the squares of n values of Q max

20. Table 2 : values of y as a function of T. T(years)y y 1.01-1.531004.60 1.580.002005.30 2.00.373005.70 5.01.504005.99 10.02.255006.21 202.9710006.91 503.9010009.21

21. Plotting on Gumbel Paper : Q 200 = Q av + 124.56( 0.78 x5.30 – 0.45) = 778 m 3 /s Q 100 = Q av + 124.56( 0.78 x4.60 – 0.45) = 710 m 3 /s

22. Determining the magnitude of the N-year event by calculation. Although the use of a normal probability distribution has been used above to plot events, and hence to extrapolate for rare values that may be used in design, values of particular probabilities can be calculated since a normal distribution curve is defined by only two parameters( the mean and standard deviation). Accordingly, to determine the specific discharge associated with a particular probability of occurrence r in an annual series that is normally distributed, it is necessary to compute only Q r = Q av + K  where  = Standard deviation and K is listed in Table 3.

23. Example 2 : Determine by calculation the mean daily discharge of the River Thames at Teddington with a 100-year return period, assuming the annual series normally distributed. Since : Q av = 329.7 m 3 /s and  = 133.8 m 3 /s. For T r = 100 years, P is 1.0 per cent; and from Table 2: K = 2.33 Therefore : Q 100 =329.7 + (2.33 x 133.8) = 641 m 3 /s

24. Table 3 : Values of a normal distribution Probability of exceedance (%) K K 0.13.09500.00 0.52.5855-0.13 1.02.3360-0.25 2.51.9665-0.385 51.64570-0.52 101.2875-0.67 151.0480-0.84 200.8485-1.04 250.6790-1.28 300.5295-1.645 350.38597.5-1.96 400.2599.0-2.33 450.1399.5-2.58 500.0099.9-3.09