engineering hydrology - · pdf fileprof. rajesh bhagat asst. professor civil engineering...

Prof. Rajesh BhagatAsst. Professor

Civil Engineering Department

Yeshwantrao Chavan College Of Engineering

Nagpur

B. E. (Civil Engg.) M. Tech. (Enviro. Engg.)

GCOE, Amravati VNIT, Nagpur

Mobile No.:- 8483003474 / 8483002277

Email ID:- [email protected]

Website:- www.rajeysh7bhagat.wordpress.com

ENGINEERING HYDROLOGY

mailto:[email protected]

2

Unit-III

1) Runoff: Runoff, sources and component, classification of streams, factors

affecting runoff, Estimation Methods. Measurement of discharge of a stream by Area-

slope 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.

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.

DTEL

6

6

HYDROGRAPH

storm of Duration D

Precipitation

P

Discharge

Q baseflow

peak flow

new baseflow

tp

w/o rainfall

tl

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

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 2nd one doesn’t start before

the direct runoff due to 1st one has ceased, we get a singled peaked

hydrograph.

11

Hydrograph:

1) If however, the second storm start before the direct runoff due to 1st storm has

ceased, (complex storm) then multipeak hydrograph are obtained.

12

1) A1ABCDEE1 is called hydrograph due to isolated storm I1.

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 2nd storm.

13

1) T is base period of 1st storm hydrograph.

2) A1AEE1 is the base flow contribution to total discharge.

3) ABCDE direct runoff contribution to total discharge.

4) G1 is the centre of mass of rainfall.

5) G2 is the centre of mass of hydrograph.

6) TL = lag time.

7) tpk = time of peak from starting point A

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.

15

1) The flood data and base flow in a storm are estimated for a storm in a

catchment area of 600 km2. calculate the effective rainfall.

Time in Days 0 1 2 3 4 5 6 7 8 9

Discharge (m3/s) 20 63 151 133 90 63 44 29 20 20

Base flow (m3/s) 20 22 25 28 28 26 23 21 20 20

16

Ordinates of DRH after the separation of the base flow are:

Plot the DRH for given Ordinate.

Volume of DRH = rainfall excess x catchment area

Rainfall excess = (Volume of DRH / Catchment Area)

Time in Days 0 1 2 3 4 5 6 7 8 9

Discharge (m3/s) 20 63 151 133 90 63 44 29 20 20

Base flow (m3/s) 20 22 25 28 28 26 23 21 20 20

Ordinates of DRH

after the separation of

the base flow

0 41 126 105 62 37 21 8 0 0

17

Ordinates of DRH after the separation of the base flow are:

Volume of DRH = rainfall excess x catchment area


Volume of DRH = direct runoff = (41 + 126 + 105 + 62 + 37 + 21 + 8) x 1

= 400 m3/s = 34560000 m3/day = 34560000 m3


Rainfall excess = (34560000/ (600 x 106)) = 0.0576 m = 5.76 cm

Time in Days 0 1 2 3 4 5 6 7 8 9

Discharge (m3/s) 20 63 151 133 90 63 44 29 20 20

Base flow (m3/s) 20 22 25 28 28 26 23 21 20 20

Ordinates of DRH

after the separation of

the base flow

0 41 126 105 62 37 21 8 0 0

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.

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

20

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 as below.

Time from start of storm, Hr

0 2 4 6 8 10 12 14

Accumulated rainfall, cm 0 0.6 2.8 5.2 6.7 7.5 9.2 9.6

21

Hour Accumulatedrainfall, cm

Time interval, hour

Depth of rainfall,cm

Φ x (time

interval)

ER, cm

Intensity of ER, cm/hr

0 0 - - - - -

2 0.6 2 0.6 0.8 0 0

4 2.8 2 2.2 0.8 1.4 0.7

6 5.2 2 2.4 0.8 1.6 0.8

8 6.7 2 1.5 0.8 0.7 0.35

10 7.5 2 0.8 0.8 0 0

12 9.2 2 1.7 0.8 0.9 0.45

14 9.6 2 0.4 0.8 0 0

Plot the hyetograph for above Intensity of ER against time.

Area under ERH x Catchment area = Direct Runoff Volume

Total Effective Rainfall = (0.7 + 0.8 + 0.35 + 0.45) x 2 = 4.6 cm

Direct runoff volume = (4.6 / 100 ) x 5 x 1000000) = 230000 m3

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.

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)

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

0 6 12 18 24 30 36 42 48 54 60 66 72 78 84

Discharge, m3/s

0 5 15 50 120 201 173 130 97 66 40 21 9 3.5 2

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.

3) 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 6hr-

unit hydrograph by corresponding rainfall amounts.

Time, Hrs 0 6 12 18

Effective rainfall, cm 0 2 4 3

29

Sample calculation for the example solved graphically is given table

Time, Hrs

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

0

6

12

18

24

30

36

42

48

54

60

66

72

78

84

90

96

0

5

15

50

120

201

173

130

97

66

40

21

9

3.5

2

0

10

30

100

240

402

346

260

194

132

80

42

18

7

4

0

0

0

0

20

60

200

480

804

692

520

388

264

160

84

36

14

8

0

0

0

0

15

45

150

360

603

519

390

291

198

120

63

27

10.5

6

0

10

50

175

485

1032

1510

1555

1233

910

635

400

222

106

45

18.5

6

30

3) The ordinates of 6 hr unit hydrograph of a catchment is given below:

Derive the flood hydrograph in the catchment due to the storm given below:

The storm loss rate for the catchment is estimated 0.25 cm/hr. The base flow

can be assumed to be 15 m3/s at the beginning and increasing by 2.0 m3/s

for every 12 hours till the end of the direct runoff hydrograph.

Time,Hr

0 3 6 9 12 15 18 24 30 36 42 48 54 60 69

Ordinatesof 6 hr UH

0 25 50 85 125 160 185 160 110 60 36 25 16 8 0

Time from start of storm (hr) 0 6 12 18

Accumulated Rainfall 0 3.5 11 16.5

31

Time interval of storm (hr) 6 12 18

Accumulated Rainfall 3.5 11 16.5

Rainfall 3.5 7.5 5.5

Loss @ 0.25cm/Hr for 6 Hrs 1.5 1.5 1.5

Effective Rainfall, cm 2 6 4

Due to unequal time interval of UH ordinates, a few entries have to be interpolated

to complete the table.

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)

0369121518212427303642485460666972757881

0255085125160185172.51601351106036251680

050100170250320370345320270220120725032160

0001503005107509601110103596066036021615096480

0000010020034050064074064044024014410064320

050100320550930132016451930194519201420872506326212112320

15151515171717171919192121232325252727

15651153355679471337166219491694193914418935293492371375927

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 780 km2 due to 6 hr storm. Derive the 6 hr unit hydrograph of the

basin.

Assume the base flow of 40 m3/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 = 1794 m3/s

DRH in depth = ((1794 x 6 x60 x 60) / (780 x 106)) x 100

= 4.968 cm (Rain Excess)

Time,Hr

6 12 18 24 30 36 42 48 54 60 66 72 78

Discharge, m3/s

40 64 215 360 405 350 270 205 145 100 70 50 40

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 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 of DRH

hydrograph by rain excess 4.968 cm to get ordinates of UH.

Time,Hr

6 12 18 24 30 36 42 48 54 60 66 72 78

Discharge, m3/s

40 64 215 360 405 350 270 205 145 100 70 50 40

Time,Hr

6 12 18 24 30 36 42 48 54 60 66 72 78

Direct Runoff

0 24 175 320 365 310 230 165 105 60 30 10 0

(Ordinates of UH)

Discharge, m3/s

0 4.83 35.22 64.42 74.47 62.4 46.29 33.21 21.13 12.077 6.04 2.01 0

35

5) The peak of flood hydrograph due to a 3 Hr duration isolated storm in a

catchment is 270 m3/s. The total depth of rainfall is 5.9cm. Assuming an

average infiltration losses of 0.3 cm/hr and a constant base flow of 20 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

Loss @ 0.3 cm/Hr for 3 hours = 0.3 x 3 = 0.9 cm

Rainfall Excess = 5 cm

Peak Flood Hydrograph = 270 m3/s

Base flow = 20 m3/s

Peak of DRH = 270 – 20 = 250 m3/s

Peak of 3 Hr Unit Hydrograph = (peak of DRH / Rainfall Excess) = 250 / 5 = 50 m3/s

36

5) The peak of flood hydrograph due to a 3 Hr duration isolated storm in a

catchment is 270 m3/s. The total depth of rainfall is 5.9cm. Assuming an

average infiltration losses of 0.3 cm/hr and a constant base flow of 20 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 = 50 m3/s

Area of catchment = 567 km2

(1/2) B x 50 x 60 x 60 = 567 x 106 x (1/100)

B = 63 Hours

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 = 0.25 cm/hr and base flow is 10 m3/sec.

Time (T)

(Hrs.)

Ordinates of 4

hr. UH

I

storm

II

storm

III

storm

DRH

O

Base

flow

Ordinates Flood

hydrograph

R=2

cm

R=3

cm

R=1

cm

0 0 0 0 10 10

2 16 32 32 10 42

4 22 44 0 44 10 54

6 43 86 48 134 10 144

8 64 128 66 0 194 10 204

10 49 98 129 16 243 10 253

12 32 64 192 22 278 10 288

14 20 40 147 43 230 10 240

16 0 0 96 64 160 10 170

18 60 49 109 10 119

20 0 32 32 10 42

22 20 20 10 30

24 0 0 10 10

6) 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.

Time (T)

(Hrs.)

A

Ordinates of 4

hr. UH

B

I

Storm

C

II

storm

D

III

storm

E

DRHO

F=B+C

+D+E

Base

flow

Ordinates

Flood

hydrograph

R=2

cm

R=8

cm

R=4

cm

G H=F+G

0 0 0 0 10 10

2 12.52 25.04 25.04 10 35.04

4 21.31 42.62 0 42.62 10 52.62

6 23.54 47.08 100.16 147.24 10 157.24

8 14.79 29.58 170.48 0 200.06 10 210.06

10 12.18 24.36 188.32 50.08 262.76 10 272.76

12 0 0 118.32 85.24 203.56 10 213.56

14 97.44 94.16 191.6 10 201.6

16 0 59.16 59.16 10 69.16

18 48.72 48.72 10 58.72

20 0 0 10 10

Time (T) (Hrs.) Ordinates of Flood hydrograph, m3/s

0 0

2 0.3

4 1.7

6 2.6

8 5.4

10 4

12 2.6

14 1.1

16 0.6

18 0

7) Determine the ordinates of unit hydrograph from flood hydrograph.

Neglect base flow. Area= 405 hectare.



Excess rainfall x Catchment area = runoff volume = Area of Hydrograph

Excess rainfall = (runoff volume) / Catchment area

Excess rainfall = (131760) / 4050000 = 0.0325m = 3.25 cm

Time (T) (Hrs.)

A

Ordinates of Flood hydrograph, m3/s

B

Ordinates of 2 hr. UH

C = B /3.25

0 0 0.000

2 0.3 0.092

4 1.7 0.523

6 2.6 0.800

8 5.4 1.662

10 4 1.231

12 2.6 0.800

14 1.1 0.338

16 0.6 0.185

18 0 0.000



Time (T) (Hrs.)

A

Ordinates of 3 hr UH

B

Ordinates of Flood hydrograph

C = B x 11.5

0 0 0.00

6 3 34.50

12 5 57.50

18 9 103.50

24 11 126.50

30 7 80.50

36 5 57.50

42 4 46.00

48 2 23.00

54 1 11.50

60 0 0.00

8) Determine the ordinates of flood hydrograph of 3 hr rainfall resulting

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) = 11.5 cm

S-CURVE METHODS-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.

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)

0 0 0 0 0.00

4 20 0 20 20 6.67

8 80 20 100 100 33.33

12 130 100 230 0 230 76.67

16 150 230 380 20 360 120.00

20 130 380 510 100 410 136.67

24 90 510 600 230 370 123.33

28 52 600 652 380 272 90.67

32 27 652 679 510 169 56.33

36 15 679 694 600 94 31.33

40 5 694 699 652 47 15.67

44 0 699 699 679 20 6.67

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.

(m3/s)

Ordinates of

4-hr U.H.

(m3/s)

lagged by 4-

hr

Ordinates

of 4-hr

U.H. (m3/s)

lagged by

8-hr

S-Curve

ordinates

(m3/s)

S-Curve

ordinates

(m3/s)

lagged by

3-hr

DifferenceOrdinates of 3-

hr U.H. (m3/s)

A B C D E=B+C+D FG=E-F

H=(4/3)*(G)

0 0 0 0.00 0.00

1 6 6 6.00 8.00

2 36 36 36.00 48.00

3 66 66 0 66.00 88.00

4 91 0 91 6 85.00 113.33

5 106 6 112 36 76.00 101.33

6 93 36 129 66 63.00 84.00

7 79 66 145 91 54.00 72.00

8 68 91 0 159 112 47.00 62.67

9 34 106 6 146 129 17.00 22.67

10 27 93 36 156 145 11.00 14.67

11 13 79 66 158 159 -1.00 -1.33

12 0 68 91 159 146 13.00 17.33

Q.10 The 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-Curve technique.

Time

(Hour

s)

Surface

Runoff

Ordinates

(m3/s)

Ordinates

of 4-hr

U.H. (m3/s)

Ordinates

of 4-hr

U.H.

(m3/s)

lagged by

4-hr

Ordinates

of 4-hr

U.H.

(m3/s)

lagged by

8-hr

Ordinates

of 4-hr

U.H.

(m3/s)

lagged by

12-hr

S-Curve

ordinates

(m3/s)

S-Curve

ordinates

(m3/s)

lagged by

6-hr

DifferenceOrdinates of 6-hr

U.H. (m3/s)

A B C= B/0.816 D E F G=C+D+E+F H I=H-G J= (4/6)*(I)

0.00 0.00 0.00 0.00 0.00 0.001.00 15.00 18.36 18.36 18.36 12.242.00 25.00 30.61 30.61 30.61 20.403.00 36.00 44.07 44.07 44.07 29.384.00 38.00 46.52 0.00 46.52 46.52 31.025.00 48.00 58.77 18.36 77.13 77.13 51.426.00 69.00 84.48 30.61 115.08 0.00 115.08 76.727.00 91.00 111.41 44.07 155.48 18.36 137.12 91.418.00 113.00 138.34 46.52 0.00 184.87 30.61 154.26 102.849.00 101.00 123.65 58.77 18.36 200.78 44.07 156.71 104.47

10.00 88.00 107.74 84.48 30.61 222.82 46.52 176.30 117.5311.00 71.00 86.92 111.41 44.07 242.41 77.13 165.28 110.1912.00 54.00 66.11 138.34 46.52 0.00 250.98 115.08 135.90 90.6013.00 31.00 37.95 123.65 58.77 18.36 238.74 155.48 83.25 55.5014.00 21.00 25.71 107.74 84.48 30.61 248.53 184.87 63.66 42.4415.00 9.00 11.02 86.92 111.41 44.07 253.43 200.78 52.64 35.10

810

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. (m3/s) Ordinates of 2-hr U.H. (m

3/s)

A B A=B*(4/2)

0 0 0

2 12.5 25

4 62.5 125

6 130 260

8 175 350

10 180 360

12 140 280

14 90 180

16 50 100

18 35 70

20 13 26

22 3 6

24 0 0

Time

(Hours)

A

Ordinates of S-

Curve (m3/s)

B

Ordinates of S-

Curve (m3/s) lagged

by 3-hr

C

Difference

D

Ordinates of 3-hr U.H. (m3/s)

E = D x (1/3)

0 0 0 0.001 55 55 18.332 141 141 47.003 251 0 251 83.674 344 55 289 96.335 413 141 272 90.676 463 251 212 70.677 501 344 157 52.338 523 413 110 36.679 538 463 75 25.00

10 546 501 45 15.00

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

)

Ordinates of

6-hr U.H.

(m3/s)

S-Curve

addition

S-

Curve

ordinat

es

(m3/s) Time

S-Curve

ordinates

(m3/s) 4-

hr

duration

S-Curve

ordinates

(m3/s)

lagged by

4-hr

Difference Ordinates

of 4-hr

U.H.

(m3/s)

1 2 3 4=2+3 56=4*( 4/6) 7 8=6-7 9=8*(6/4)

0 0 0 0 0.00 0.00 0

6 40 0 40 4 26.67 0.00 26.67 40

12 90 40 130 8 86.67 26.67 60.00 90

18 100 130 230 12 153.33 86.67 66.67 100

24 130 230 360 16 240.00 153.33 86.67 130

30 80 360 440 20 293.33 240.00 53.33 80

36 70 440 510 24 340.00 293.33 46.67 70

42 50 510 560 28 373.33 340.00 33.33 50

48 30 560 590 32 393.33 373.33 20.00 30

54 10 590 600 36 400.00 393.33 6.67 10

600 600 40 400.00 400.00 0.00 0

44 0.00 400.00 -400.00 -600

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.

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