Unit Hydrograph Model

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Unit Hydrograph Model

Response Functions of Linear Systems

Basic operational rules:

Principle of Proportionality: f(cQ ) = cf(Q)

Principle of Superposition: f(Q1+Q2) = f(Q1) + f(Q2)

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Schematic Diagrams of Linear System Responses

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Discrete Unit Impulse Response Function

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Discrete Unit Impulse Response Function I n p u t O u t p u t a t t = n t , Q n , C o n t r i b u t e d b y I m

m = 1 I 1

[ 0 , t ) I 1 u [ ( n - 1 + 1 ) t ] = I 1 u n

m = 2 I 2 [ t , 2 t ) I 2

u [ ( n - 2 + 1 ) t ] = I 2 u n - 1 m = 3 I 3

[ 2 t , 3 t ) I 3 u [ ( n - 3 + 1 ) t ] = I 3 u n - 2

. . .

. . .

. . . m = m I m

[ ( m - 1 ) t , m t ) I m u [ ( n - m + 1 ) t ] = I m u n - m + 1

B y l i n e a r s u p e r p o s i t i o n , t h e t o t a l o u t p u t a t t = n t i s

1mnu

M

1m

mI

nQ

D i m e n s i o n a l i t y o f u n i t i m p u l s e r e s p o n s e f u n c t i o n i s

in

s3ft

or mm

s3m

eg., , T2L

LT

3L u

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Rainfall-Runoff Modeling• In many hydrologic engineering designs, we need to predict peak

discharge or hydrograph resulting from a certain type of storm event. For this purpose, some kind of rainfall-runoff model is needed to translate rainfall input to produce discharge hydrograph.

• Hydrologic rainfall-runoff models range with various degrees of sophistication. The use of a particular model depends largely on the accuracy requirement of the results, importance of the project, data availability, and fiscal constraints.

• Among many rainfall-runoff models, the unit hydrograph (UH) method received considerable use and it is still being used widely by many water resources engineers & hydrologists.

• UH for a given watershed can be derived from historical rainfall-runoff data. For watersheds having no streamflow & rainfall records, the so-called synthetic UH which relates the properties of UH to basin characteristics, must be developed.

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Unit Hydrograph Model

• First proposed by Sherman (1932)

• Definition: The UH of a drainage basin is a direct runoff hydrograph (DRH) resulting from 1 unit of effective rainfall (rainfall excess) hyetograph (ERH) distributed uniformly over the entire basin at a uniform rate during a specified time period (or duration).

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Basic Assumptions of UH

1. The effective rainfall is uniformly distributed within its duration

2. The effective rainfall is uniformly distributed over the whole

drainage basin

3. The base duration of direct runoff hydrograph due to an effective

rainfall of unit duration is constant.

4. The ordinates of DRH are directly proportional to the total amount

of DR of each hydrograph (principles of linearity, superposition,

and proportionality)

5. For a given basin, the runoff hydrograph due to a given period of

rainfall reflects all the combined physical characteristics of basin

(time-invariant)

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Criteria for Selecting Storm Events to Derive UHs

• Storms are isolated and occur individually;

• Storm coverage should be uniform over the entire watershed - watershed area should not be too large, say < 5000 km2 ;

• Storms should be flood-producing storms – ER is high, 10mm < ER < 50mm is suggested;

• Duration of rainfall should be approx. 1/5 to 1/3 of basin lag;

• The number of storm events should be at least 5.

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Derivation of UH (For simple ERH)

1. Analyze hydrograph and perform baseflow separation.

2. Measure the total volume of DRH in equivalent

uniform depth (EUD)

3. Find the effective rainfall such that VDRH = VERH.

4. Assume that ERHs are uniform, the UH can be derived

by dividing the ordinates of DRH by VDRH

5. The duration of the UH is the duration of ERH.

6. In rainfall-runoff analysis, the times of occurrence for DRH and ERH are commonly made identical.

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Derivation of UH (Figure)

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Determination of DRH from known UH and ERH

F r o m t h e p r i n c i p l e o f l i n e a r i t y o f U H , t h e D R H c a n b e d e r i v e d b y a n y e f f e c t i v e r a i n f a l l i n p u t w i t h t h e s a m e d u r a t i o n a s t h a t o f U H

M

1m1mn

u m

Pn

Q

w h e r e P m = E f f e c t i v e r a i n f a l l a m o u n t ; M = T o t a l n o . o f e f f e c t i v e r a i n f a l l s ; u j = j - t h o r d i n a t e o f t h e U H ; a n d Q n = n - t h o r d i n a t e o f t h e D R H .

I n a l g e b r a i c f o r m

1Mnu

MP...

2nu

3P

1nu

2P

nu

1P

nQ

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Determination of DRH

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Determination of DRH (Example)

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Effect of Storm Duration on UH (1)

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For Complex Storm Events

• How to determine the UH corresponding to the duration t?

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Derivation of UH from Complex Storm EventsBasic Equation

N ..., 2, 1,n ,1mn

uM

1mmP

nQ

Since N=J + M –1, therefore, J=N-M+1. See an example. In matrix form, the above convolution relation can be expressed as PNJ uJ1 = QN1 Use the above example, in which N=6, M=3, and J=4, the matrix P and vectors u and Q are:

P =

3

23

123

123

12

1

P000

PP00

PPP0

0PPP

00PP

000P

; u =

4

3

2

1

u

u

u

u

; and Q =

6

5

4

3

2

1

Q

Q

Q

Q

Q

Q

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UH Determination from Complex Storms

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Methods for deriving UH from complex storm events

• Successive Approximation

• System Transformation

• Least squares & its variations

• Optimization techniques (LP, others)

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Successive Approximation

Q1 = P1 u1 Q2 = P2 u1 + P1 u2 Q3 = P3 u1 + P2 u2 + P1 u3 . . . . . . Qn-2 = . . . . . P3 uJ-3 + P2 uJ-1 + P1 uJ Qn-1 = . . . . . . P3 uJ-1 + P2 uJ Qn = . . . . . . . P3 uJ

1. Forward Procedure: Start from u1 uJ

2. Backward procedure: Start solving UH from uJ u1

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Least Squares & Its Variations1 . O r d i n a r y L e a s t S q u a r e s : S o l v e P u = Q b y m i n i m i z i n g P u - Q 2 . T h e r e s u l t i n g

n o r m a l e q u a t i o n i s

( P ’ P ) u = P ’ Q a n d t h e l e a s t s q u a r e s U H c a n b e d e t e r m i n e d a s

u O L S = ( P ’ P ) - 1 P ’ Q

T h e d e r i v e d U H d e p e n d s o n t h e c o n d i t i o n o f m a t r i x P o r ( P ’ P ) r e p r e s e n t e d b y t h e c o n d i t i o n n u m b e r , minmax / w i t h minmax and b e i n g t h e l a r g e s t

a n d s m a l l e s t e i g e n v a l u e s o f P ’ P .

R i d g e R e g r e s s i o n : ( o b t a i n s t a b l e & s m o o t h U H )

u R i d g e = ( P ’ P + k I ) - 1 P ’ Q

T h e a b o v e U H m i n i m i z e s P u - Q 2 + k u 2 w h i c h r e p r e s e n t s t h e m e a n -

s q u a r e d e r r o r i n p r e d i c t i n g f u t u r e d i s c h a r g e h y d r o g r a p h . T h e i s s u e i s n o w

t o d e t e r m i n e t h e o p t i m a l k ( r i d g e c o n s t a n t ) .

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Basic Requirements for Selecting Methods to Derive UH

• The resulting UH ordinates are positive;• The UH shape is preserved;• Errors in the input data are not amplified during

UH derivation;• The method is capable for admitting a number of

events simultaneously for UH derivation;• The method is computationally simple, efficient,

and easily programmable

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Single-Event vs. Multiple-Event Analysis • How to Determine Representative UH?• UH derived from the single-event analysis could vary from

one storm to the other • In multiple-event analysis, rainfall-runoff data of different

storms are stacked as

where Pk and Qk are ERH matrix and DRH vector for the k-

th event and u is the vector of multi-event UH.

Q

Q

Q

u

P

P

P

K

2

1

K

2

1

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Single-Event Analysis

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Multiple-Event Analysis

• By ordinary least squares method

uOLS = ( P’P )-1 ( P’Q )

• By ridge least squares method

uRidge = ( P’P + kI )-1 ( P’Q )

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Derivation of UH of Different Durations• How to determine a UH of one duration from the UH of a

different duration?

• Lagging Method

Referring to Fig., if a t-hr UH, say from 1cm of ERH, is known and we wish to derive a 2t-hr UH. Since the sum of two t-hr UHs has the volume of 2 cm (or 2”), the 2t-hr UH can be obtained simply by dividing the sum of 2t-hr UHs (lagged by t-hr) by 2.

Similarly, the procedure can be used to derive the UH with a duration which is a integer multiplier of the duration of the original UH.

t’ = k t

where k = integers such as 2, 3, 4, …; and t = original UH duration. The lagging method is restrictive to develop UH whose duration is the integer multiplier of the original UH duration.

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Lagging Method

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S-Curve Analysis (1)

• S-curve is a DRH (also called S-hydrograph), resulting from an infinite series of rainfall excess of one unit, say 1 cm or 1 inch with a specified duration, t.

• After te hours, the continuous rainfall producing 1 cm (or 1 inch)

of runoff every t-hr would reach an equilibrium discharge, Qe.

• The equilibrium discharge, Qe, can be computed as

where Qe – [ft3/s] ; i – [in/hr]; A – [mi2]; and t – [hr], or

where Qe – [m3/s]; i – [cm/hr]; A – [km2]; and t – [hr].

t

645.6AiA

eQ

t

2.778AiA

eQ

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S-Curve Analysis (2)

• To derive a UH of duration t’-hr from the S-curve, g(t), obtained from t-hr UH, the S-curve is shifted to the right by t’-hr. Then, the difference between the two S-curves represents the direct runoff hydrograph resulting from a rainfall excess of t’/t cm or inches. The t’-hr UH then can be easily obtained by dividing the ordinates of S-S’ by t’/t.

• Note: The S-curve tends to fluctuate about Qe. This means that the

initial UH does not represent actually the runoff at a uniform rate over time. Such fluctuations usually occur because of lack of precision in selecting UH duration. That is, the duration of the UH may differ slightly from the duration used in calculation. Nevertheless, an average S-curve can usually be drawn through the points without too much difficulty.

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Construction of S-Curve

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Determine UH by S-Curve

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Adjustment of S-Curve

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S-Curve (Example 1)

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S-Curve (Example 2)

Documents

Unit Hydrograph Model