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Area - Advanced Hydrology - Problems, Exercises of Engineering Dynamics

Dhirubhai Ambani Institute of Information and Communication TechnologyEngineering Dynamics

These are the Problems of Advanced Hydrology which includes Excess Ordinate, Data, Area Method, Histogram, Calculation, Contribution etc.Key important points are: Area, Method, Appropriate Shape, Duration Rainfall, Duration Of Rainfall, Small Watershed, Distributed

Typology: Exercises

2012/2013

Uploaded on 03/28/2013

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ekanath 🇮🇳

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Use Snyder’s method to develop aUH for the area of 100mi2described 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
Module 3
Example Problem
Synthetic unit hydrograph
Calculate Qp
Qp= 640(cp)(A)/tp
= 640(0.6)(100)/8.6
= 4465 cfs
Since this is a small watershed,
Tb ≈4tp = 4(8.6)
= 34.4 hr
Duration of rainfall
D= tp/5.5 hr
= 8.6/5.5 hr
= 1.6 hr
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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

Synthetic unit hydrograph

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

0

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0 5 10 15 20 25 30 35 40

Q (cfs

)

Time (hr)

Qp

W (^75)

W (^50) Area drawn to represent 1 in. of runoff over the watershed

W 75 = 440(Q (^) P/A) -1. W 50 = 770(Q (^) P/A) -1. (widths are distributed 1/3 before Q (^) p and 2/3 after)

Synthetic unit hydrograph

Example Problem Contd…

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Find TR using eq.

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

. Hence Qp = 1.110 cfs

Synthetic unit hydrograph

R

p

T

A

Q

R t p

D

T = +

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. = (10mi 2 ) (5280ft/mi) 2 (ac/43560ft 2 ) (1 in.) = 6400 ac-in Hence from eq.

B = 7.17 hr

Q T Q B

Vol

p R p

Example Problem Contd…

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0

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0 2 4 6 8 10 12 14

Q (cfs

)

Time (hr)

Qp = 1110 (cfs)

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

Synthetic unit hydrograph

Example Problem Contd…

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  1. 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 km 2

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. 24 0

Exercise problems Contd…

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  1. 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…

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