ENGINEERING HYDROLOGY

Transcription

1 ENGINEERING HYDROLOGY Prof. Rajesh Bhagat Asst. Professor Civil Engineering Department Yeshwantrao Chavan College Of Engineering Nagpur B. E. (Civil Engg.) M. Tech. (Enviro. Engg.) GCOE, Amravati VNIT, Nagpur Mobile No.: / ID:- rajeysh7bhagat@gmail.com Website:-

2 Unit-III 1) Runoff: Runoff, sources and component, classification of streams, factors affecting runoff, Estimation Methods. Measurement of discharge of a stream by Areaslope and Area-velocity methods. 2) Hydrograph: Flood hydrographs and its components, Base flow & Base flow separation, S-Curve technique, unit hydrograph, synthetic hydrograph. Instantaneous Unit hydrograph. 2

3 HYDROGRAPH:- 1) A plot of the discharge in stream against time chronologically. 2) Depending upon unit of time involved: 1) Annual hydrograph 2) Monthly hydrograph 3) Seasonal hydrograph 4) Flood hydrograph or storm hydrograph or hydrograph: it shows stream flow due to storm over catchment. It is used flooding characteristics of 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, drought studies. 3

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6 HYDROGRAPH storm of Duration D Precipitation P Discharge Q baseflow t l t p peak flow new baseflow w/o rainfall 6 DTEL 6

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8 Watershed Urbanization

9 Factors affecting Hydrograph: 1) Size 2) Shape 3) Slope 4) Drainage density 5) Land use or vegetation 6) Rainfall intensity 7) Rainfall duration 8) Direction of storm movement 9

10 Hydrograph: 1) Hydrograph is a graphical variation of discharge against time. 2) It is a response of a given catchment to a rainfall input. 3) The discharge noted in hydrograph is the combined effect of surface runoff, interflow & base flow. 4) If two storms occurs in a catchment such that the 2 nd one doesn t start before the direct runoff due to 1 st one has ceased, we get a singled peaked hydrograph. 1

11 Hydrograph: 1) If however, the second storm start before the direct runoff due to 1 st storm has ceased, (complex storm) then multipeak hydrograph are obtained. 11

12 1) A 1 ABCDEE 1 is called hydrograph due to isolated storm I 1. 2) AB is rising limb or concentration curve. 3) BCD is crest curve. 4) DE is falling curve or recession curve. 5) C is point of crest or peak. 6) E is end of direct runoff. 7) EA is the hydrograph in the period of ground water recession. 8) A is beginning of direct runoff due to 2 nd storm. 12

13 1) T is base period of 1 st storm hydrograph. 2) A 1 AEE 1 is the base flow contribution to total discharge. 3) ABCDE direct runoff contribution to total discharge. 4) G 1 is the centre of mass of rainfall. 5) G 2 is the centre of mass of hydrograph. 6) T L = lag time. 7) t pk = time of peak from starting point A 13

14 Hydrograph Separation: 1) In hydrological analysis it is necessary to obtain Direct Runoff Hydrograph (DRH) from Total Storm Hydrograph (TSH). 2) To separate DRH from TSH, various methods are available. 14

15 1) The flood data and base flow in a storm are estimated for a storm in a catchment area of 6 km2. calculate the effective rainfall. Time in Days Discharge (m 3 /s) Base flow (m 3 /s)

16 Ordinates of DRH after the separation of the base flow are: Time in Days Discharge (m 3 /s) Base flow (m 3 /s) Ordinates of DRH after the separation of the base flow Plot the DRH for given Ordinate. Volume of DRH = rainfall excess x catchment area Rainfall excess = (Volume of DRH / Catchment Area) 16

17 Ordinates of DRH after the separation of the base flow are: Time in Days Discharge (m 3 /s) Base flow (m 3 /s) Ordinates of DRH after the separation of the base flow Volume of DRH = rainfall excess x catchment area Rainfall excess = (Volume of DRH / Catchment Area) Volume of DRH = direct runoff = ( ) x 1 = 4 m 3 /s = 3456 m 3 /day = 3456 m3 Rainfall excess = (Volume of DRH / Catchment Area) Rainfall excess = (3456/ (6 x 1 6 )) =.576 m = 5.76 cm 17

18 Excess Rainfall & Effective Rainfall:- 1) Excess rainfall: if the initial losses and infiltration subtracted from the total rainfall, the remaining portion of rainfall is called rainfall excess. Surface runoff occurs only when there is rainfall excess. 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 interflow, the effective rainfall is slightly greater than rainfall excess. Effective rainfall = (direct runoff volume / area of catchment) Interflow is small, so direct runoff is equal to surface runoff & therefore they are used synonymously. 18

19 Effective Rainfall Hyetograph:- 1) When initial losses and filtration losses are subtracted from the rainfall hyetograph, we get Effective Rainfall Hyetograph (ERH). 2) It is also known as Hyetograph of rainfall excess. 3) Direct Runoff Hydrograph (DRH) is the result of Effective Rainfall Hyetograph (ERH). 4) Area under ERH x Catchment area = Runoff Volume = Area under direct DRH 19

20 2) A storm over catchment of area 5 km 2 had a duration of 14 hours. If the Φ index for the catchment is.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 as below. Time from start of storm, Hr Accumulated rainfall, cm

21 Hour Accumulated rainfall, cm Time interval, hour Depth of rainfall, cm Φ x (time interval) ER, cm Intensity of ER, cm/hr Plot the hyetograph for above Intensity of ER against time. Area under ERH x Catchment area = Direct Runoff Volume Total Effective Rainfall = ( ) x 2 = 4.6 cm Direct runoff volume = (4.6 / 1 ) x 5 x 1) = 23 m 3 21

22 Unit Hydrograph: 1) 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). 2) Thus we can have 6-Hr Unit Hydrograph, 12-Hr Unit Hydrograph, etc. 3) 6-Hr unit hydrograph will have an effective rainfall intensity of 1/6 cm/hr. 22

23 Unit Hydrograph: 1) The D-hr Unit Hydrograph, D should not be more than any of the following: 1) Time of concentration 2) Lag time 3) Period of rise 2) Volume of water contained inside the unit hydrograph (ie area of unit of hydrograph) is equal to (1cm x catchment area) 23

24 Unit Hydrograph: Assume that a 6-hour unit hydrograph(uh) of a catchment has been derived, whose ordinates are given in the following table and a corresponding graphical representation is shown in Figure. Time, Hr Discharge, m3/s

25 Unit Hydrograph: 2) Assume further that the effective rainfall hyetograph(erh) for a given storm on the region has been given as in the following table. Time, Hrs Effective rainfall, cm ) This means that in in the first 6 hours, 2cm excess rainfall has been recorded, 4cm in the next 6 hour & 3cm in the next. 4) Direct runoff hydrograph can then be calculated by the three separate hydrograph for three excess rainfalls by multiplying the ordinates of the 6hrunit hydrograph by corresponding rainfall amounts. 25

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29 Time, Hrs Sample calculation for the example solved graphically is given table UH Ordinates, m3/s Direct runoff due to 2cm excess rainfall in first 6hrs Direct runoff due to 4cm excess rainfall in second 6hrs Direct runoff due to 3cm excess rainfall in third 6hrs Direct runoff hydrograph, m3/s

30 3) The ordinates of 6 hr unit hydrograph of a catchment is given below: Time, Hr Ordinates of 6 hr UH Derive the flood hydrograph in the catchment due to the storm given below: Time from start of storm (hr) Accumulated Rainfall The storm loss rate for the catchment is estimated.25 cm/hr. The base flow can be assumed to be 15 m 3 /s at the beginning and increasing by 2. m 3 /s for every 12 hours till the end of the direct runoff hydrograph. 3

31 Time interval of storm (hr) Accumulated Rainfall Rainfall for 6 Hrs Effective Rainfall, cm Due to unequal time interval of UH ordinates, a few entries have to be interpolated to complete the table. 31

32 Time, Hr Ordinates of UH DRH due to 2cm ER DRH due to 6cm ER DRH due to 4cm ER Ordinates of Final DRH Base Flow, m3/s Ordinates of Flood Hydrograph, m3/s A B C =(B x 2) D = (B x 6) E = (Bx4) F=(C+D+E) G H=(G+F)

33 Derivation Unit Hydrograph from Flood Hydrograph of Isolated Storm: 4) The following are the ordinates of the flood hydrograph from a catchment area of 78 km2 due to 6 hr storm. Derive the 6 hr unit hydrograph of the basin. Time, Hr Discharge, m3/s Assume the base flow of 4 m3/s. Direct Surface runoff = (64-4) + (215-4) + (36-4) + (45-4) + (35-4) + (27-4) + (25-4) + (145-4) + (1-4) + (7-4) + (5-4) Direct Surface runoff = 1794 m3/s DRH in depth = ((1794 x 6 x6 x 6) / (78 x 1 6 )) x 1 = cm (Rain Excess) 33

34 Derivation Unit Hydrograph from Flood Hydrograph of Isolated Storm: 4) The following are the ordinates of the flood hydrograph from a catchment area of 78 km2 due to 6 hr storm. Derive the 6 hr unit hydrograph of the basin. Time, Hr Discharge, m3/s Therefore the ordinates of UH are obtained by dividing the ordinates of DRH hydrograph by rain excess cm to get ordinates of UH. Time, Hr Direct Runoff (Ordinates of UH) Discharge, m3/s

35 5) The peak of flood hydrograph due to a 3 Hr duration isolated storm in a catchment is 27 m3/s. The total depth of rainfall is 5.9cm. Assuming an average infiltration losses of.3 cm/hr and a constant base flow of 2 m3/s estimates the peak of the 3 hr unit hydrograph of this catchment. If the area of catchment is 567 km2 determine the base width of 3 hr unit hydrograph by assuming it to be triangular in shape. Duration of Rainfall Excess = 3 Hr Total Depth of Rainfall = 5.9 cm cm/hr for 3 hours =.3 x 3 =.9 cm Rainfall Excess = 5 cm Peak Flood Hydrograph = 27 m3/s Base flow = 2 m3/s Peak of DRH = 27 2 = 25 m3/s Peak of 3 Hr Unit Hydrograph = (peak of DRH / Rainfall Excess) = 25 / 5 = 5 m3/s 35

36 5) The peak of flood hydrograph due to a 3 Hr duration isolated storm in a catchment is 27 m3/s. The total depth of rainfall is 5.9cm. Assuming an average infiltration losses of.3 cm/hr and a constant base flow of 2 m3/s estimates the peak of the 3 hr unit hydrograph of this catchment. If the area of catchment is 567 km2 determine the base width of 3 hr unit hydrograph by assuming it to be triangular in shape. Let B = base width of 3 hr Unit Hydrograph Volume represented by the area of UH = Volume of 1 cm depth over the catchment Area of UH = Area of catchment x 1 cm Peak of 3 Hr Unit Hydrograph = 5 m3/s Area of catchment = 567 km2 (1/2) B x 5 x 6 x 6 = 567 x 1 6 x (1/1) B = 63 Hours 36

37 6) 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 =.25 cm/hr and base flow is 1 m3/sec. Time (T) (Hrs.) Ordinates of 4 hr. UH I storm R=2 cm II storm R=3 cm III storm R=1 cm DRH O Base flow Ordinates Flood hydrograph

38 6) A storm produces rainfall intensities of.75, 2.25 and 1.25 cm/hr on a drainage area in 3 successive time period of 4 hr. Φ-index =.25 cm/hr and base flow is 1 m3/sec. Time (T) (Hrs.) A Ordinates of 4 hr. UH B I Storm C R=2 cm II storm D R=8 cm III storm E R=4 cm DRHO F=B+C +D+E Base flow G Ordinates Flood hydrograph H=F+G

39 7) Determine the ordinates of unit hydrograph from flood hydrograph. Neglect base flow. Area= 45 hectare. Time (T) (Hrs.) Ordinates of Flood hydrograph, m3/s

40 7) Determine the ordinates of unit hydrograph from flood hydrograph. Neglect base flow. Area= 45 hectare. Excess rainfall x Catchment area = runoff volume = Area of Hydrograph Excess rainfall = (runoff volume) / Catchment area Excess rainfall = (13176) / 45 =.325m = 3.25 cm

41 7) Determine the ordinates of unit hydrograph from flood hydrograph. Neglect base flow. Area= 45 hectare. Time (T) (Hrs.) A Ordinates of Flood hydrograph, m3/s B Ordinates of 2 hr. UH C = B /

42 8) Determine the ordinates of flood hydrograph of 3 hr rainfall resulting into total rainfall of 15 cm. initial loss is.5 cm and Φ-index = 1 cm/hr. Sol: Excess rainfall = 15.5 (1 x 3) = 11.5 cm Time (T) (Hrs.) A Ordinates of 3 hr UH B Ordinates of Flood hydrograph C = B x

43 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.

44 Time (T) (Hrs.) Ordinates of 4 hr UH Ordinates of 4 hr UH lagged by 4hr S-Curve ordinate S-Curve lagged by 12 hr Difference Ordinates of 12 hr UH A B C D=B+C E F = D - E G=(4/12)*(F)

45 Q.9 The ordinates of 4-hr unit hydrograph are given below. Determine the ordinates of 3-hr UH using S- Curve technique. Time (Hours) Ordinates of 4-hr U.H. (m 3 /s) Ordinates of Ordinates 4-hr U.H. of 4-hr (m 3 /s) U.H. (m 3 /s) lagged by 4- lagged by hr 8-hr S-Curve ordinates (m 3 /s) S-Curve ordinates (m 3 /s) lagged by 3-hr Difference Ordinates of 3- hr U.H. (m 3 /s) A B C D E=B+C+D F G=E-F H=(4/3)*(G)

46 Q.1 The ordinates of surface runoff of 4-hr duration from a catchment area of 357 km 2 are measured at 1 hr interval are given below. Determine the ordinates of 6-hr UH using S-Curve technique. Time (Hour s) Surface Runoff Ordinates (m 3 /s) Ordinates of 4-hr U.H. (m 3 /s) Ordinates of 4-hr U.H. (m 3 /s) lagged by 4-hr Ordinates of 4-hr U.H. (m 3 /s) lagged by 8-hr Ordinates of 4-hr U.H. (m 3 /s) lagged by 12-hr S-Curve ordinates (m 3 /s) S-Curve ordinates (m 3 /s) lagged by 6-hr Difference Ordinates of 6-hr U.H. (m 3 /s) A B C= B/.816 D E F G=C+D+E+F H I=H-G J= (4/6)*(I)

47 Q.11. The ordinates of 4-hr UH are given below. Determine the ordinates of 2-hr UH using S-Curve technique and plot the same. Time (Hours) Ordinates of 4-hr U.H. (m 3 /s) Ordinates of 2-hr U.H. (m 3 /s) A B A=B*(4/2)

48 Q.12 The ordinates of S-Curve Hydrograph are given below. Determine the ordinates of 3-hr UH. Effective rainfall is 1 cm/hr. Time (Hours) A Ordinates of S- Curve (m 3 /s) B Ordinates of S- Curve (m 3 /s) lagged by 3-hr C Difference D Ordinates of 3-hr U.H. (m3/s) E = D x (1/3)

49 Q.13 The ordinates of 6-hr UH are given below. Determine the ordinates of 4-hr UH using S-Curve technique and plot the same. Time (Hours ) Ordinates of 6-hr U.H. (m3/s) S-Curve addition S- Curve ordinat es (m 3 /s) Time S-Curve ordinates (m 3 /s) 4- hr duration S-Curve ordinates (m 3 /s) lagged by 4-hr Difference Ordinates of 4-hr U.H. (m 3 /s) 6=4*( 4/6) 7 8=6-7 9=8*(6/4) =