Synthetic Unit Hydrograph

Module 3

Lecture 6: Synthetic unit hydrograph

Synthetic Unit Hydrograph •

In India, only a small number of streams are gauged (i.e., stream flows due to single and multiple storms, are measured)

•

There are many drainage basins (catchments) for which no stream flow records are available and unit hydrographs may be required for such basins

•

In such cases, hydrographs may be synthesized directly from other catchments, which are hydrologically and meteorologically homogeneous, or indirectly from other catchments through the application of empirical relationship

•

Methods for synthesizing hydrographs for ungauged areas have been developed from time to time by Bernard, Clark, McCarthy and Snyder. The best known approach is due to Snyder (1938)

Module 3

Synthetic unit hydrograph

Snyder’s method •

Snyder (1938) was the to develop a synthetic UH based on a study of watersheds in the Appalachian Highlands. In basins ranging from 10 – 10,000 mi.2 Snyder relations are tp = Ct(LLC)0.3 where tp= basin lag (hr) L= length of the main stream from the outlet to the divide (mi) Lc = length along the main stream to a point nearest the watershed centroid (mi) Ct= Coefficient usually ranging from 1.8 to 2.2 Module 3

Synthetic unit hydrograph

Snyder’s method

Contd…

Qp = 640 CpA/tp where Qp = peak discharge of the UH (cfs) A = Drainage area (mi2) Cp = storage coefficient ranging from 0.4 to 0.8, where larger values of cp are associated with smaller values of Ct

Tb = 3+tp/8

•

where Tb is the time base of hydrograph Note: For small watershed the above eq. should be replaced by multiplying tp by the value varies from 3-5 The above 3 equations define points for a UH produced by an excess rainfall of D= tp/5.5 duration

Snyder’s hydrograph parameter Module 3

Synthetic unit hydrograph

Example Problem Use Snyder’s method to develop a UH for the area of 100mi2 described below. Sketch the appropriate shape. What duration rainfall does this correspond to? Ct = 1.8, L= 18mi, Cp = 0.6, Lc= 10mi Calculate tp tp = Ct(LLC)0.3 = 1.8(18·10) 0.3 hr, = 8.6 hr

Since this is a small watershed, Tb ≈4tp = 4(8.6) = 34.4 hr

Calculate Qp Qp= 640(cp)(A)/tp = 640(0.6)(100)/8.6 = 4465 cfs Duration of rainfall D= tp/5.5 hr = 8.6/5.5 hr = 1.6 hr Module 3

Synthetic unit hydrograph

Example Problem

Contd…

5000

Qp

W 75 = 440(QP/A)-1.08 W 50 = 770(QP/A)-1.08 (widths are distributed 1/3 before Qp and 2/3 after)

4000

W 75 Q (cfs)

3000

W 50

Area drawn to represent 1 in. of runoff over the watershed

2000

1000

0 0

5

10

15

20

25

30

35

40

Time (hr)

Module 3

Synthetic unit hydrograph

SCS (Soil Conservation Service) Unit Hydrograph •

Unit = 1 inch of runoff (not rainfall) in 1 hour

tp

•

Can be scaled to other depths and times

•

Based on unit hydrographs from many watersheds

•

The

earliest

method

assumed

a

Qp

hydrograph as a simple triangle, with rainfall duration D, time of rise TR (hr), time of fall B. and peak flow Qp (cfs). TR

B

SCS triangular UH Module 3

Synthetic unit hydrograph

SCS Unit Hydrograph •

Contd…

The volume of direct runoff is

Vol =

Q pTR 2

+

Qp B 2

or

where B is given by

Qp =

2vol TR + B

B = 1.67TR Therefore runoff eq. becomes, for 1 in. of rainfall excess,

0.75vol Qp = TR 484 A Qp = TR

=

0.75(640) A(1.008) Qp = TR

where A= area of basin (sq mi) TR = time of rise (hr)

Module 3

Synthetic unit hydrograph

SCS Unit Hydrograph •

Contd…

Time of rise TR is given by

TR =

D + tp 2

where D= rainfall duration (hr) tp= lag time from centroid of rainfall to QP Lag time is given by 0.8 L

tp =

 1000  9 −     CN 19000y 0.5

0.7

where L= length to divide (ft) Y= average watershed slope (in present) CN= curve number for various soil/land use Module 3

Synthetic unit hydrograph

SCS Unit Hydrograph

Contd…

Runoff curve number for different land use (source: Woo-Sung et al.,1998)

Module 3

Synthetic unit hydrograph

Example Problem Use the SCS method to develop a UH for the area of 10 mi2 described below. Use rainfall duration of D = 2 hr Ct = 1.8, L= 5mi, Cp = 0.6, Lc= 2mi The watershed consist CN = 78 and the average slope in the watershed is 100 ft/mi. Sketch the resulting SCS triangular hydrograph .

Solution Find tp by the eq.

0.8 L

tp =

 1000  9 −    CN  19000y 0.5

0.7

Convert L= 5mi, or (5*5280 ft/mi) = 26400 ft. Slope is 100 ft/mi, so y = (100ft/mi) (1mi/5280 ft)(100%) = 1.9% Substituting these values in eq. of tp, we get tp = 3.36 hr Module 3

Synthetic unit hydrograph

Example Problem Find TR using eq.

Contd…

D TR = + tp 2

Given rainfall duration is 2 hr, TR = 4.36 hr, the rise of the hydrograph Then find Qp using the eq, given A= 10 mi2

484 A Qp = TR

. Hence Qp = 1.110 cfs

To complete the graph, it is also necessary to know the time of fall B. The volume is known to be 1 in. of direct runoff over the watershed. So, Vol. = (10mi2) (5280ft/mi)2 (ac/43560ft2) (1 in.) = 6400 ac-in Hence from eq.

Vol =

Q pTR 2

+

Qp B 2

B = 7.17 hr Module 3

Synthetic unit hydrograph

Example Problem

Contd…

1200

1000

Q (cfs)

800

600

Qp= 1110 (cfs) 400

200

0 0

2

TR=4.36 (hr)

4

6

8

Time (hr)

10

12

14

B=7.17 (hr)

Module 3

Exercise problems

1.

The stream flows due to three successive storms of 2.9, 4.9 and 3.9 cm of 6 hours duration each on a basin are given below. The area of the basin is 118.8 km2 . Assuming a constant base flow of 20 cumec, derive a 6-hour unit hydrograph for the basin. An average storm loss of 0.15 cm/hr can be assumed (Hint :- Use UH convolution method)

Time (hr)

0

3

6

9

12

15

Flow (cumec)

20 50 92 140 199 202

18

21

24

27

30

33

204

144

84

45

29

20

Module 3

Exercise problems

Contd…

2. The ordinates of a 4-hour unit hydrograph for a particular basin are given below. Derive the ordinates of (i) the S-curve hydrograph, and (ii) the 2-hour unit hydrograph, and plot them, area of the basin is 630 km2

Time (hr)

Discharge (cumec)

0

0

2

25

4

100

6

160

8

190

10

170

12

110

Time (hr)

Discharge (cumec)

14

70

16

30

18

20

20

6

22

1.5

24

0

Module 3

Exercise problems 3.

Contd…

The following are the ordinates of the 9-hour unit hydrograph for the entire catchment of the river Damodar up to Tenughat dam site: and the catchment characteristics are , A = 4480 km2, L = 318 km, Lca = 198 km. Derive a 3-hour unit hydrograph for the catchment area of river Damodar up to the head of Tenughat reservoir, given the catchment characteristics as, A = 3780km2, L = 284 km, Lca = 184km. Use Snyder’s approach with necessary modifications for the shape of the hydrograph.

Time (hr)

0

9

18

27

36

45

54

63

72 81

90

Flow (cumec)

0

69

1000

210

118

74

46

26

13 4

0

Module 3

Highlights in the Module  This module presents the concept of Rainfall-Runoff analysis, or the conversion of precipitation to runoff or streamflow, which is a central problem of engineering hydrology.  Gross rainfall must be adjusted for losses to infiltration, evaporation and depression storage to obtain rainfall excess, which equals Direct Runoff (DRO).  The concept of the Unit hydrograph allows for the conversion of rainfall excess into a basin hydrograph, through lagging procedure called hydrograph convolution.  The concept of synthetic hydrograph allows the construction of hydrograph, where no streamflow data are available for the particular catchment. Module 3