Download Synthetic Unit Hydrograph - Advanced Hydrology - Lecture Slides and more Slides Aeronautical Engineering in PDF only on Docsity!

Lecture 6: Synthetic unit hydrograph

Module 3

Docsity.com

• 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)

Docsity.com Module 3

Q p = 640 C pA/t p

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

T b = 3+t p/

where Tb is the time base of hydrograph Note: For small watershed the above eq. should be replaced by multiplying t (^) p by the value varies from 3-

• The above 3 equations define points for a UH produced by an excess rainfall of

duration D= t p/5.5 Snyder’s hydrograph parameter

Snyder’s method Contd…

Docsity.com

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, L (^) c= 10mi

Calculate t (^) p t (^) p = Ct (LL (^) C) 0. = 1.8(18·10) 0.3^ hr, = 8.6 hr

Example Problem

Calculate Qp Qp = 640(c (^) p )(A)/t (^) p = 640(0.6)(100)/8. = 4465 cfs

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

Duration of rainfall D= t (^) p /5.5 hr = 8.6/5.5 hr = 1.6 hr Docsity.com

• Unit = 1 inch of runoff (not rainfall) in 1 hour
• Can be scaled to other depths and times
• Based on unit hydrographs from many watersheds
• The earliest method assumed a hydrograph as a simple triangle, with rainfall duration D, time of rise TR (hr), time of fall B. and peak flow Qp (cfs).

tp

Qp

T (^) R B SCS triangular UH

SCS (Soil Conservation Service) Unit Hydrograph

Docsity.com

• The volume of direct runoff is

or where B is given by

Therefore runoff eq. becomes, for 1 in. of rainfall excess,

T B

vol

Q

R

p =^ +

B = 1. 67 T R

R

p T

vol Q

1. 75 =

R

p T

A Q

484

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

R

p T

A Q

1. 75 ( 640 ) ( 1. 008 ) =

Q T Q B

Vol = p^ R + p

SCS Unit Hydrograph Contd…

Docsity.com

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

Module 3

SCS Unit Hydrograph Contd…

Docsity.com

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

Example Problem

Solution

Find t (^) p by the eq.

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 t (^) p = 3.36 hr

L

19000y

9 CN

1000  

  

 − tp =

Docsity.com

0

200

400

600

800

1000

1200

0 2 4 6 8 10 12 14

Q (cfs

)

Time (hr)

Qp = 1110 (cfs)

T (^) R=4.36 (hr) B=7.17 (hr)

Example Problem Contd…

Docsity.com

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. km 2. 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 18 21 24 27 30 33

Flow (cumec)

Docsity.com Module 3

3. 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 km^2 , L = 318 km, L (^) ca = 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 = 3780km^2 , L = 284 km, L (^) ca = 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)

Exercise problems Contd…

Docsity.com

 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.

Highlights in the Module

Docsity.com