Engineering Hydrology Prof. Rajesh Bhagat Asst. Professor, CED, YCCE, Nagpur

1. Engineering Hydrology Prof. Rajesh Bhagat Asst. Professor, CED, YCCE, Nagpur B. E. (Civil Engg.) M. Tech. (Enviro. Engg.) GCOE, Amravati VNIT, Nagpur Experience & Achievement: Selected Scientist, NEERI-CSIR, Govt. of India. GATE Qualified Three Times. Selected Junior Engineer, ZP Washim. Three Times Selected as UGC Approved Assistant Professor. Assistant Professor, PCE, Nagpur. Assistant Professor, Cummins College of Engg. For Women (MKSSS, Nagpur) Mobile No.:- 8483002277 / 8483003474 Email ID :- rajeysh7bhagat@gmail.com Website:-www.rajeysh7bhagat.wordpress.com

2. UNIT-IV Floods: Causes and effects, Factors affecting peak flows and its estimation, Flood routing and Flood forecasting. Frequency analysis. Economic planning for flood control. Statistical Methods: Statistics in hydrological analysis, probability and probability distribution. (Gumbel’s Probability Distribution Method for Prediction of Flood Peak)

3. Flood Any flow which is relatively high and which overtops the natural or artificial banks in any reach of a river may be called a flood. Abnormal quantity of water arising from heavy rainfall. Flood is the most worldwide natural disaster. It occurs in every country and wherever there is rainfall or coastal hazards. Flood may be controlled by hydraulic structure. Floods are be estimated with reasonable accuracy for an economic and efficient design of hydraulic structure.

4. Different types of flood:- Local flood:- causeddue to high intensity of local rainfall which generate higher surface run-off. Urbanization, Surface sealing and loss of infiltration capacity increases the surface runoff causes local flood. River floods: - River flood occur when the river volume exceeds due to excess of run-off beyond the river capacity. The river levels rise slowly due to high intensity rainfall for long lasting duration. Flash floods:-Flash floods occur as a result of the rapid accumulation and release of runoff waters from upstream mountainous areas, which can be caused by very heavy rainfall, bursting of clouds, landslides, sudden break-up of an ice jam or failure of flood control works. Coastal floods:- High tides and storm caused by tropical depressions and cyclones can cause coastal floods in urban areas, low-lying land near the sea in general. Tidal effects can keep the river levels high for long periods of time and sustain flooding.

5. Major parameters factors responsible for the occurrences of flood: Climate:- Due to global warming or climate change many subsystems of the global water cycle are effected causing an increase of flood magnitude as well as flood frequency. Nature of collecting basin:- A collecting basin is an area where surface water from rain converges to a single point, it is usually exit of basin, where water join another water body, such as a river. Larger the area of basin more will be the water receiving capacity. Nature of the streams: -Water flowing through a channel, such as a stream, has the ability to transport sediment. Some percolates deep into the ground and replenishes the groundwater supply. Vegetative cover:- Trees reduce top soil erosion, prevent harmful land pollutants contained in the soil from getting into our waterways, slow down water run-off, and ensure that our groundwater supplies are continually being replenished. Rainfall: - It is difficult to predict and measure the precise changes in the hydrological cycle because of the processes such as evaporation, transport, and precipitation.

6. CAUSES OF FLOOD • Intense rainfall • Melting of snow • High Tides & Cyclone • CAUSES OF FLOODING • Culvert, drain and gully entry blockage (partial or total) • Dumped rubbish. • Floating debris, including leaves and branches. • Sediment deposition • Poorly maintained urban watercourses (silted, overgrown with vegetation, or blocked with rubbish.) • Bridge, culvert or tunnel waterway capacity exceeded. • Failure of drainage pump or jammed sluice gate. • High water level in receiving river that flows back up the drain or sewer.

7. Design of culvert, bridges, drainage work and irrigation diversion works, needs a reliable estimates of the flood. The maximum flood that any structure can safely pass is called the design flood. As the magnitude of the design flood increase, the capital cost of the structure also increases but the probability of annual damages will decrease. Standard Project Flood (SPF) or Standard Probable Flood is the estimate of the flood likely to occur from the severe combination of the meteorological & hydrological conditions, which are reasonably characteristics of the drainage basin being considered, but excluding extremely rare combination. Floods with return period of 100 years are accounted. Maximum Probable Flood (MPF) differs from the SPF in that it includes the extremely rare / severe combination of the meteorological & hydrological conditions. It is a catastrophic floods and is usually confined to spillway design of very high dams. The SPF is usually around 80% of the MPF for the basin.

8. DESIGN FLOOD:- Actual value of flood for which structure is designed. Flood adopted for the design of hydraulic structure like spillways, bridges, flood banks, etc. Floods are be estimated with reasonable accuracy for an economic and safe or efficient design. It may be MPF or SPF or flood of any desired recurrence interval. As the magnitude of design flood increases, the capital cost of structure also increases but the probability of annual damage will decrease. Design flood in many cases is less than MPF because MPF is not economical. Several design flood may be adopted in a single structure for example MPF is used for spillway.

9. Importance of Statistics and Probability in Hydrology: If the outcome of a process is precisely predicted then it is a deterministic process. But in some cases, there is uncertainty or unpredictability regarding outcome. Average weather condition are same from year to year, but yearly rainfall at a given place is not same every year. In all such a cases, process is treated as a probabilistic process and outcome governed by some probability law.

10. DESIGN FLOOD:- • The methods used in the estimation of the design flood can be grouped as under: • Physical indication of past flood • Envelop curves • Empirical formulae • Rational method • Unit hydrograph application • Frequency analysis or statistical method • Methods 1 to 4 of above only give the flood peak. • Method 5 give the complete flood hydrograph but it can not be used for very large areas. • In method 6, flood hydrograph may be constructed by making its peak discharge equal to the estimated design flood. (Gumbel’s Distribution)

11. DESIGN FLOOD:- • Physical indication of past flood: • When flood data is not available or when the available data is short, the maximum flood that occurred during the past, and the number of years during which this flood was the highest are found out by local enquiries or from the observation of past records. • This is then increased by a certain percentage and adopted as the design period. • The percentage increase naturally depends on the number of years over which the observed flood was maximum. • If it is over a longer period the percentage increase may be less and for shorter periods the increase may have to be more.

12. DESIGN FLOOD:- • Envelope Curve: • When the observed peak floods in different catchments of a hydro meteorologically homogeneous region are plotted against their respective areas on a log-log sheet it will usually show that all the points lie below an enveloping curve which can be drawn by eye. • After the envelope curve is drawn the maximum flood for the catchment under consideration can be read by entering the plot with known area of the catchment. • It is based on past recorded floods and there is always a possibility that still higher flood may occur in future. As more and more data accumulates the curve are to be redrawn. The results from envelope curve should not be relied upon wholly.

14. Empirical Formulae: • Formulae considering area of the basin only into consideration. • Dickens Formula: Q = C x A (3/4) • A is in km2 and C is a constant depending on other catchment parameter which may 11.4 – 25 and Q is in m3/s. • Ryves Formula: Q = C x A (2/3) • A is in km2 and Q is in m3/s. Widely used in south. Values of C varies from 30 - 6.75. • Inglis Formula: Q = (124 A) / (A + 10.4)(1/2) • A is in km2 and and Q is in m3/s. • Ali NawazBahadur Formula: Q = C A (0.993 – (1/14(Log A))) • Developed for Hyderabad. C is 48 – 60, A is in km2 and and Q is in m3/s. • Formulae may consider one or more basin parameters apart from area and also rainfall characteristics . • Formulae may consider recurrence interval either explicitly into consideration.

15. Empirical Formulae: • Formulae may consider one or more basin parameters apart from area and also rainfall characteristics . • Craig’s Formula: Q = 10 k B v I ln ((4.97 L2) / B) • Q is in m3/s, L is greatest length of catchment in km, B is avg width of catchment in km, v is velocity of runoff in m/s, k is runoff coefficient & I is avg rainfall intensity in cm /hr. • Lillie’s Formula: Q = 0.058 V R k ∑ θ L • Catchment is divided in triangles, Q is m3/s, V is velocity of flow m/s, R is rainfall coefficient, k is coefficient, θ is apical angle and L is length in km. • Rhinds Formula: Q = 0.098 C S R (0.386 A ) p • Q is m3/s, S is avg fall of last 5 km, R is greatest annual rainfall in cm, p is variable ndex, L is greatest length in catchment, C is coefficient. • Chamiers Formula: Q = 3.5 I K A 3/4 • Iszkowski’s : Q = 0.03171 C M H A

16. Formulae may consider recurrence interval either explicitly into consideration. • Fullers Formula: Q = C x A (0.8) • Hortons Formula: Q = (114 Tr0.25) / A • Q is m3/s, Tr is return period & A is drainage area on km2. • Pettis Formula: Q = C (P B ) 1.25 • Q is flood discharge with return period of 100 years in m3/s, P is one day rainfall with 100 years return period, B is width of basin in km & C is coefficient.

17. Rational Method for Estimation of Design Flood:- • The time of concentration is the maximum time required by the surface runoff to reach the basin outlet. • When storms continues beyond time of concentration every part of the catchment would be contributing to the runoff at outlet and therefore it represents condition of peak runoff. • It is given by Q = C x A x I • Where, C is runoff coefficient or impermeability coefficient = (runoff/rainfall) • A = catchment area • I = intensity of rainfall. • If different portions of catchment have different runoff coefficient then C is calculated by • Ceq = (C1A1 + C2A2 + C3A3……) / (A1 + A2 + A3……)

18. Rational Method for Estimation of Design Flood:- Q = C x A x I The rational formula is found to be suitable for a peak flow prediction in small catchment up to 50 km2 in area. Application in urban drainage designs and in the design of small culverts and bridges. Rainfall Intensity, I = (K Tx) / (tc + a)m Probability exceedance P : (Return Period T = 1/P) K, a, x and m are constant. (K = 80 - 90, a =13 - 49 , x =0.2 & m = 0.46 - 0.98) Kirpich Equation, tc= 0.01947 L0.77 S-0.385 tc= Time of concentration in minutes L = Max. Length of traveller of water in meter S = Slope of catchment = ∆H / L ∆H = Difference in elevation

19. What is a return period? The probability that events such as floods, wind storms & earthquake will occur is often expressed as a return period. The inverse of probability (generally expressed in %), it gives the estimated time interval between events of a similar size or intensity. For example, the return period of a flood might be 100 years; otherwise expressed as its probability of occurring being 1/100, or 1% in any one year. This does not mean that if a flood with such a return period occurs, then the next will occur in about one hundred years' time - instead, it means that, in any given year, there is a 1% chance that it will happen, regardless of when the last similar event was. Or, put differently, it is 10 times less likely to occur than a flood with a return period of 10 years (or a probability of 10%). 

20. 1) An urban catchment has an area of 0.85 km2. the slope of the catchment is 0.006 and the maximum length of travel of water is 950 m. The maximum depth of rainfall with a 25 years return period is as below:- if culvert for drainage at the outlet of this area is to designed for a return period of 25 years, estimates the required peak flow rate by assuming the runoff coefficient as 0.3. Sol:- Time of concentration by Kirpich Equation, tc = 0.01947 L0.77 S-0.385 tc = 0.01947 x (9500.77 ) x (0.006)-0.385 = 27.39 minutes Max. Depth of rainfall for 27.39 minute duration (by interpolation): (50 - 40) / (30 - 20) = (50 - x) / (30 – 27.39) x = 47.39 mm = max. Depth of rainfall for 27.39 minutes. Avg intensity of rainfall = I = (47.39 / 27.39 ) 60 = 103.81 mm/hr Q = C x A x I = (0.3 x 0.85 x 1000000 x 103.81 ) / 1000 x 60 x 60 = 7.35 m3 / s

21. If in the urban area, the land use of the area and the corresponding runoff coefficient are as given below calculate the equivalent runoff coefficient. Sol: Equivalent runoff coefficient Ceq = (C1A1 + C2A2 + C3A3 + C4A4) / (A1 + A2 + A3 +A4) Ceq = ((0.7x8) + (0.1x17) + (0.3x50) + (0.8x10)) / (8 + 17 + 50 +10) = 0.36

22. A small watershed consists of 1.5 km2 of cultivate area (C=0.2), 2.5 km2 under forest (C=0.1) and 1.0 km2 under grass cover (C=0.35). There is a fall of 22 m in a watercourse length of 1.8 km. The intensity frequency-duration relation for the area may be taken as I = (80 x Tr0.2) / (tc + 13)0.46 Where I is in cm/hr, tc is in minutes. Estimate the peak rate of runoff for a 25 year frequency. Sol: length of water course = 1.8 km = 1800 m & Fall in the elevation = 22 m S = 22 / 1800 = 0.0122 tc = 0.0195 x (L0.77 ) x (S)-0.385 tc = 0.0195 x (18000.77 ) x (0.0122)-0.385 = 34.15 minutes I = (80 x Tr0.2) / (tc + 13)0.46 I = (80 x 250.2) / (34.15 + 13)0.46 = 25.90 cm/hr Ceq = (C1A1 + C2A2 + C3A3) / (A1 + A2 + A3) Ceq = ((0.2x1.5) + (0.1x2.5) + (0.35x1)) / (1.5 + 2.5 + 1) = 0.18

23. A small watershed consists of 1.5 km2 of cultivate area (C=0.2), 2.5 km2 under forest (C=0.1) and 1.0 km2 under grass cover (C=0.35). There is a fall of 22 m in a watercourse length of 1.8 km. The intensity frequency-duration relation for the area may be taken as I = (80 x Tr0.2) / (tc + 13)0.46 Where I is in cm/hr, tc is in minutes. Estimate the peak rate of runoff for a 25 year frequency. Sol: S = 0.0122 tc = 34.15 minutes I = 25.90 cm/hr = 0.2590 m / hr Ceq= 0.18 Q = C x A x I = (0.18 x 5 x 1000000 x 0.2590 / 60 x 60 = 64.75 m3 / s

24. A bridge is proposed to constant across a river having catchment area of 2027 hectors. The catchment has a slope of 0.007 and length of travel of water is 20270m. Estimates 30 years flood, the intensity frequency duration relationship is given by I = (9876 x Tr0.2) / (tc + 45)0.98 Where I is in mm/hr, Tr is in years & tc is in minutes. Assume runoff coefficient as 0.35. Sol:- Time of concentration by Kirpich Equation, tc = 0.01947 L0.77 S-0.385 tc = 0.01947 x (202700.77 ) x (0.007)-0.385 = 272.64 minutes Max. intensity of rainfall = I = ((9876 x 300.2 )) / ((272.64 + 45)0.98) = 68.89 mm/hr = 6.889 cm/hr A = 2027 Ha = 20.27 Km2 1 Km2 = 100 Ha Q = C x A x I Peak Discharge, Q = C x A x I = (0.35 x 20.27 x 1000000 x 6.889 / (100 x 60 x 60) = 135.76 cumec. Or m3/s

29. Frequency Analysis: • Frequency analysis makes use of the observed data in the past to predict the future flood events along with their probabilities or return period. • It is based on the assumption that combination of numerous factors which produces floods are a matter of pure chance and therefore are subject to analysis according to the mathematical theory of probability. • The estimation of parameters and the selection of the distribution become unreliable when the observed data happen to be very small. • Now if the a plot is made between the value of the random variable as abscissa and the corresponding exceedence probability as the ordinate, the resulting graph is called a distribution graph, or probability plot or empirical distribution. • It must be based on accurate, homogeneous and adequate data. • If the sample is too small the predictions about the future about the future floods can not be expected reliable. • The record shorter than 20 years must not be used because it overestimates the design flood by 80%.

31. Estimation of Design Flood For A Particular Return Period: The return period is calculated by using Weibull’s formula: T = ( n + 1) / m T = return period in years n = number of years of record m = order number Return period represents the average no. of years within which a given event will be equalled or exceeded. Probability of Exceedance, p = 1 / T If the probability of occurrence of an event is p, then the probability of its no-occurrence is q = 1- p Probability of an event not occurring at all in ‘n’ successive years would be equal to qn qn= (1- p)n Probability of an event occurring at least once in ‘n’ successive years, = 1 – qn = 1- (1 - p)n This probability is called risk.

32. 5) A flood of a certain magnitude has a return period of 20 years. What is the probability that a flood of this magnitude may occurs in next 15 years. Return period represents the avg. no. of years within which a given event will be equalled or exceeded. Probability of Exceedance, p = 1 / T p = 1 / 20 = 0.05 5% Probability of an event not occurring at all in ‘n’ successive years would be equal to qn qn= (1- p)n = (1 – 0.05)15 = 0.46 46 % Probability of an event occurring at least once in ‘n’ successive years, Risk = 1 – qn = 1- (1 - p)n Risk = 1 – 0.46 = 0.54 54 %

34. 6) The flood data analysis at river site yielded mean of 12000 m3/s & standard deviation of 650 m3/s for what discharge the structure should be designed so as to provide 95% assurance that the structure will not fail in next 50 years. Percentage of assurance = 95 % Percentage risk = 5% Risk = 1 – qn = 1- (1 - p)n Probability of Exceedance, p = 1 / T 0.05 = 1- (1 - 1 / T)50 T = 975.286 years YT = - lnln ( YT = - lnln ( YT = 6.88 X = X + K . σX= 12000 + K . 650 Frequency factor, K = (YT – yn) / Sn K = ( K = ( K= 4.92 X = X + K . σX = 12000 + (4.92 x 650)X = 15198 m3/s

35. Estimation of Design Flood For A Particular Return Period: The return period is calculated by using Weibull’s formula: T = ( n + 1) / m T = return period in years n = number of years of record m = order number or rank of highest observed flood Return period represents the avg no. Of years within which a given event will be equalled or exceeded. Probability of Exceedance = 1 / T = P If a graph plotted between flood magnitude and its return period in simple plane co-ordinates, the plot is called probability or empirical distribution plot. Extrapolation for large return period may yield more result hence following method are used: Gumble’s method Log Pearson Type III Method

36. Gumbel’s Distribution Method: X = X + K . σ OR X = X + K . σn-1 X = Flood Magnitude of Return Period T years X = Mean Value of Past Flood Record = ( ∑X) / n K = Frequency Factor, K = (YT – yn) / sn K = ( YT = - lnln ( yn = Reduced Mean Variable = 0.577 Sn= Reduced Standard Deviation Variable = 1.2825 No. of years of records, n > 50 we use yn = 0.577 & sn= 1.2825 σ or σn-1 = Standard deviation for previous data = √ ( ( ∑ ( Xi – X )2 ) / ( n – 1 ) ) σn-1

37. Gumbel’s Probability Distribution Method for Prediction of Flood Peak: This method of estimating the flood peak was given by Gumbel and is known as Gumbel’s Distribution method. This is a flood frequency method which is most widely used in hydrological analysis & meteorological studies.

38. Procedure for Estimating Peak Flood: From the given discharge data & for sample size n, X & Sn are to be computed. σ or σn-1 = Standard deviation = √ ( ( ∑ ( Xi – X )2 ) / ( n – 1 ) ) σn-1 yn & Snare to be computed for given sample size, n. (assumed if not given 0.577 & 1.2825) For given T, YT is to be calculated by YT= - lnln ( Frequency factor (K ) is calculated by K = (YT – yn) / SnK = ( Now the magnitude of flood X can be obtained from, X = X + K . σ OR X = X + K . σn-1

39. Gumbel’s Distribution Method: X = X + K . σ OR X = X + K . σn-1 X = Flood Magnitude of Return Period T years X = Mean Value of Past Flood Record = ( ∑X) / n K = Frequency Factor, K = (YT – yn) / sn K = ( YT = - lnln ( yn = Reduced Mean Variable = 0.577 Sn= Reduced Standard Deviation Variable = 1.2825 No. of years of records, n > 50 we use yn = 0.577 & sn= 1.2825 σ or σn-1 = Standard deviation for previous data = √ ( ( ∑ ( Xi – X )2 ) / ( n – 1 ) ) σn-1

40. 7) The annual flood data for a river shows mean of 10000 m3/s with the standard deviation of 3000 m3/s. If a hydraulic structure is designed for a flood of 15000 m3/s. Determine its return period. Sol: X = 10000 m3/s σ = 3000 m3/s X = 15000 m3/s T = ? YT = - lnln ( K = ( X = X + K . σ X = X + K . σ 15000 = 10000 + (K) 3000 K = 1.67 K = ( 1.67 = ( YT = 2.714 YT = - lnln ( 2.714 = - lnln ( T = 15.59 Years

41. 8) The flood data analysis at river site yielded mean of 12000 m3/s and standard deviation of 650 m3/s for what discharge the structure should be so as to provide 95 % assurance that the structure will not fail in next 50 years. Sol: X = 12000 m3/s σ = 650 m3/s n = 50 yrs Assurance = 95% X = ? Probability of occurring in ‘n’ years, Risk = 1 – ( 1 – ( 1 / T )) n 0.05 = 1 – ( 1 – ( 1 / T )) 50 T = 975.286 Yrs YT= - lnln ( K = ( X = X + K . σ YT= - lnln ( YT = - lnln ( YT = 6.88 K = ( K = ( K = 4.915 X = X + K . σ X = 12000 + (4.915 x 650) X = 15195 m3/s

42. 9) Flood frequency analysis carried out at river site yielded flood of magnitude 40809 m3/s for return period of 50 years, 46300 m3/s for return period of 100 years what is the design flood for return period of 500 years. X50 = 40809 m3/s T50 = 50Yrs X100= 46300 m3/sT100= 100Yrs X500 = ? YT = - lnln ( K = ( X = X + K . σ CASE-I: T50 = 50Yrs YT = - lnln ( YT = - lnln ( Y50= 3.9 K = ( K50 = ( K50 =2.59 X = X + K . σ40809 = X + 2.59 . σ Eq. (1) CASE-I: T100= 100Yrs YT = - lnln ( YT = - lnln ( Y100= 4.6 K = ( K100 = ( K100 = 3.13 X = X + K . σ46300 = X + 3.13 . σ Eq. (2) Solve Eqn (1) & (2) simultaneously:

43. 9) Flood frequency analysis carried out at river site yielded flood of magnitude 40809 m3/s for return period of 50 years, 46300 m3/s for return period of 100 years what is the design flood for return period of 500 years. X50 = 40809 m3/s T50 = 50Yrs X100= 46300 m3/sT100= 100Yrs X500 = ? CASE-I: T50 = 50Yrs X = X + K . σ 40809 = X + 2.59 . σ Eq. (1) CASE-I: T100= 100Yrs X = X + K . σ46300 = X + 3.13 . σ Eq. (2) Solve Eqn (1) & (2) simultaneously: X = 14472.5 m3/s σ = 10168.52 CASE-III: T500 = 500 Yrs Y500= - lnln ( Y500= - lnln ( Y500= 6.21 K = ( K = ( X = X + K . σ X = 14472.5 + (4.39 x 10168.52 ) X = 59144 m3/s

44. Flood Routing:- It is the technique of determining the flood hydrograph at a section of a river by utilising the data of flood flow at one or more upstream sections. It is also used in analysis of a flood forecasting, flood protection, reservoir design, spillway design etc. The broad categories of flood routing are reservoir routing and channel routing. In reservoir routing the effect of reservoir storage on a flood hydrograph is analysed. This is used in design, location and sizing of the capacity of reservoir. In channel routing the effect of storage of a specified channel reach on the flood hydrograph is studied. It is used in flood forecasting and flood protection.