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

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.

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

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  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. Hydrograph: • A plot of the discharge in stream against timechronologically. • Depending upon unit of timeinvolved: • Annualhydrograph • Monthlyhydrograph • Seasonalhydrograph • Flood hydrograph or storm hydrograph or hydrograph: it showsstream • flow due to storm over catchment. It is used flooding characteristicsof • stream. • Above Hydrograph 1,2,3 are called long term hydrograph and are used for longed term studies like calculating the surface potential of stream,reservoir studies, droughtstudies.

  3. Factors affectingHydrograph: Size Shape Slope Drainagedensity Land use orvegetation Rainfallintensity Rainfallduration Direction of stormmovement

  4. Hydrograph: Hydrograph is a graphical variation of discharge againsttime. It is a response of a given catchment to a rainfallinput. The discharge noted in hydrograph is the combined effect of surfacerunoff, interflow & baseflow. If two storms occurs in a catchment such that the 2nd one doesn’t start before the direct runoff due to 1st one has ceased, we get a singled peaked hydrograph.

  5. Hydrograph: 1) If however, the second storm start before the direct runoff due to 1st stormhas ceased, (complex storm) then multipeak hydrograph areobtained.

  6. A1ABCDEE1 is called hydrograph due to isolated stormI1. AB is rising limb or concentrationcurve. BCD is crestcurve. DE is falling curve or recessioncurve. C is point of crest orpeak. E is end of directrunoff. EA’ is the hydrograph in the period of ground waterrecession. A’ is beginning of direct runoff due to 2ndstorm.

  7. T is base period of 1st stormhydrograph. A1AEE1 is the base flow contribution to totaldischarge. ABCDE direct runoff contribution to totaldischarge. G1 is the centre of mass ofrainfall. G2 is the centre of mass ofhydrograph. TL = lagtime. tpk = time of peak from starting pointA

  8. HydrographSeparation: In hydrological analysis it is necessary to obtain Direct RunoffHydrograph (DRH) from Total Storm Hydrograph(TSH). To separate DRH from TSH, various methods areavailable.

  9. 1) Theflooddataandbaseflowinastormareestimatedforastormina catchment area of 600 km2. calculate the rainfallexcess.

  10. Ordinates of DRH after the separation of the base floware: Volume of DRH = rainfall excess x catchment area Rainfall excess = (Volume of DRH / CatchmentArea) Volume of DRH = direct runoff = (41 + 126 + 105 + 62 + 37 + 21 + 8) x1 = 400 m3/s = 34560000 m3/day = 34560000m3 Rainfall excess = (Volume of DRH / CatchmentArea) Rainfall excess = (34560000/ (600 x 106)) = 0.0576 m = 5.76cm

  11. Rainfall Excess & EffectiveRainfall:- 1) Rainfall Excess: if the initial losses and infiltration subtracted from thetotal rainfall, the remaining portion of rainfall is called rainfall excess. Surface runoffoccurs only when there is rainfallexcess. Rainfall excess = Total rainfall –Φ.t 1) Effective rainfall: it is that portion of rainfall which cause direct runoff. As direct runoff includes both surface runoff and prompt interflow, theeffective rainfall is slightly greater than rainfallexcess. Effective rainfall = (direct runoff volume / area ofcatchment) Rapid interflow is small, so direct runoff is equal to surface runoff & thereforethey are used synonymously. (Effective rainfall = Rainfallexcess)

  12. Effective RainfallHyetograph:- When initial losses and filtration losses are subtracted from therainfall hyetograph, we get Effective Rainfall Hyetograph(ERH). It is also known as Hyetograph of rainfallexcess. Direct Runoff Hydrograph (DRH) is the result of EffectiveRainfall Hyetograph(ERH). Area under ERH x Catchment area = Runoff Volume = Area under directDRH

  13. 2) A storm over catchment of area 5 km2 had a duration of 14 hours. If the Φ index for the catchment is 0.4 cm/hr, determine the effective rainfall hyetograph and the volume of direct runoff from the catchment due to the storm. The mass curve of rainfall of the storm are asbelow.

  14. Volume of DRH = Effective Rainfall x Catchment Area Effective Rainfall = (0.7 + 0.8 + 0.35 + 0.45) x 2 = 4.6cm Direct Runoff Volume = Rainfall Excess x Catchment Area Direct Runoff Volume = (4.6 / 100 ) x 5 x 1000000) = 230000m3

  15. UnitHydrograph: The Unit Hydrograph of the catchment is defined as hydrograph of direct runoff (DRH) results from 1cm depth of effective rainfall occurring uniformly over the catchment at a uniform rate during a specified period of time(D-hr). Thus we can have 6-Hr Unit Hydrograph, 12-Hr Unit Hydrograph,etc. 6-Hr unit hydrograph will have an effective rainfall intensity of 1/6cm/hr.

  16. UnitHydrograph: • The D-hr Unit Hydrograph, D should not be more than any of thefollowing: • Time ofconcentration • Lag time • Period ofrise • Volume of water contained inside the unit hydrograph (ie area of unit of hydrograph) is equal to (1cm x catchmentarea)

  17. UnitHydrograph: Assume that a 6-hour unit hydrograph(UH) of a catchment has beenderived, whose ordinates are given in the following table and acorresponding graphical representation is shown inFigure.

  18. UnitHydrograph: 2) Assumefurtherthattheeffectiverainfallhyetograph(ERH)foragivenstorm on the region has been given as in the followingtable. This means that in in the first 6 hours, 2cm excess rainfall has been recorded, 4cm in the next 6 hour & 3cm in thenext. Direct runoff hydrograph can then be calculated by the three separate hydrograph for three excess rainfalls by multiplying the ordinates of the 6hr- unit hydrograph by corresponding rainfallamounts.

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  22. Sample calculation for the example solved graphically is giventable

  23. 4) Determine the ordinates of flood hydrograph of 3 hr rainfallresulting into total rainfall of 15 cm. initial loss is 0.5 cm and Φ-index = 1 cm/hr. Sol: Excess rainfall = 15 – 0.5 – (1 x 3) = 11.5cm

  24. 5) Determine the ordinates of unit hydrograph from floodhydrograph. Neglect base flow. Area= 405hectare. Excess rainfall x Catchment area = runoff volume = Area ofHydrograph Excess rainfall = (runoff volume) / Catchmentarea Excess rainfall = (131760) / 4050000 = 0.0325m = 3.25cm

  25. Derivation Unit Hydrograph from Flood Hydrograph ofIsolated Storm: 6) The following are the ordinates of the flood hydrograph from a catchment area of 780 km2due to 6 hr storm. Derive the 6 hrunit hydrograph of the basin. Assume base flow 40 m3/s. Assume the base flow of 40m3/s. Direct Surface runoff = (64-40) + (215-40) + (360-40) + (405-40) + (350-40)+ (270-40) + (205-40) + (145-40) + (100-40) + (70-40) +(50-40) Direct Surface runoff = 1794m3/s Effective Rainfall = Direct Runoff / Catchment area Effective Rainfall= ((1794 x 6 x60 x 60) / (780 x 106)) x100 = 4.968 cm (RainfallExcess)

  26. 6) The following are the ordinates of the flood hydrograph from a catchment area of 780 km2 due to 6 hr storm. Derive the 6 hr unit hydrograph of the basin. Therefore the ordinates of UH are obtained by dividing the ordinates ofDRH hydrograph by rainfallexcess 4.968 cm to get ordinates ofUH.

  27. 7) The ordinates of 6 hr unit hydrograph of a catchment is givenbelow: Derive the flood hydrograph in the catchment due to the storm givenbelow: The storm loss rate for the catchment is estimated 0.25 cm/hr. The baseflow can be assumed to be 15 m3/s at the beginning and increasing by 2.0m3/s for every 12 hours till the end of the direct runoffhydrograph.

  28. Due to unequal time interval of UH ordinates, a few entries have to be interpolated to complete thetable.

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  30. 8) Determine the ordinates of flood hydrograph of 3 successive storms of 4 hr duration, each producing rainfall of 3 cm, 4 cm and 2 cm respectively. Φ-index = 0.25 cm/hr and base flow is 10m3/sec.

  31. 9) A storm produces rainfall intensities of 0.75, 2.25 and 1.25 cm/hr on a drainage area in 3 successive time period of 4 hr. Φ-index = 0.25 cm/hr and base flow is 10 m3/sec.

  32. Unit Hydrograph of Different Duration:1) Method of Superposition:2) S Curve Method:S-curve or the summation curve is the hydrograph of direct surface discharge that would result from a continuous succession of unit storms producing 1 cm in time (T)hrs.It is called 'S' hydrograph because the shape of the hydrograph comes out like alphabet 'S‘ though slightly deformed.S-hydrograph or S-curve is a hydrograph that is produced by a continuous effective rainfall at a constant rate for indefinite period. It is a continuous rising curve, in the form of letter S, till equilibrium is reached.

  33. Method of Superposition:

  34. S Curve Hydrograph:

  35. Q.11The ordinates of 4-hr unit hydrograph are given below. Determine the ordinates of 3-hr UH using S-Curvetechnique.

  36. Q.12The ordinates of surface runoff of 4-hr duration from a catchment area of 357 km2 are measured at 1 hr interval are given below. Determine the ordinates of 6-hr UH using S-Curvetechnique. 810 R = (810 x 60 x 60) / (357 x 1000000) = 0.816cm

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