Lecture 6 _ Synthetic Unit Hydrograph.pdf

  • Uploaded by: Davis Nyaribo Mogambi
  • 0
  • 0
  • January 2021
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Lecture 6 _ Synthetic Unit Hydrograph.pdf as PDF for free.

More details

  • Words: 1,359
  • Pages: 17
Loading documents preview...
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

Related Documents

Lecture 6
February 2021 3
Ace 6 Unit 2
February 2021 0
Unit 6: 5 Photocopiable 1
February 2021 0
Lecture
February 2021 3

More Documents from "Aashique Rock"