Unit Hydrograph Model
Response Functions of Linear Systems
Basic operational rules:
- Principle of Proportionality: f(cQ ) = cf(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
Input
m=1
m=2
m=3
.
.
.
m=m
Output at t=nt, Qn, Contributed by Im
I1[0, t)
I2[t, 2t)
I3[2t, 3t)
.
.
.
Im[(m-1)t, mt)
I1 u[(n-1+1)t] = I1 un
I2 u[(n-2+1)t] = I2 un-1
I3 u[(n-3+1)t] = I3 un-2
.
.
.
Im u[(n-m+1)t] = Im un-m+1
By linear superposition, the total output at t=nt is
M
Q   I u
n
m
n - m 1
m 1
Dimensionality of unit impulse response function is
 L3
u    T
L



ft 3
m3
 L2 
s
s
   T  , eg., mm or in

 

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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 socalled 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
From the principle of linearity of UH, the DRH can be derived by any
effective rainfall input with the same duration as that of UH
M
Q   P u
n
m n - m 1
m 1
where Pm = Effective rainfall amount; M = Total no. of effective rainfalls; uj =
j-th ordinate of the UH; and Qn = n-th ordinate of the DRH.
In algebraic form
Q Pu P u
P u
 ...  P u
n
1 n
2 n -1
3 n-2
M n - M 1
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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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Effect of Storm Duration on UH (2)
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Effect of Storm Duration on UH (3)
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Effect of Storm Duration on UH (4)
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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 Events
Basic Equation
M
Q   P u
, n  1, 2, ..., N
n
m n - m 1
m 1
Since N=J + M –1, therefore, J=N-M+1. See an example.
In matrix form, the above convolution relation can be expressed as
PNJ uJ1 = QN1
Use the above example, in which N=6, M=3, and J=4, the matrix P and
vectors u and Q are:
 P1
P
 2
 P3
P= 
0
0

 0
0
0
P1
0
P2
P1
P3
P2
0
P3
0
0
0
 Q1 
Q 
0 
 u1 
 2


Q 3 
u2 
0

; and Q =  
;u=
u 3 
P1 
Q 4 


Q 5 
P2 
u 4 

 
P3 
Q 6 
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UH Determination from Complex Storms
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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 = P 1 u 1
Q2 = P 2 u 1 + P 1 u 2
Q3 = P 3 u 1 + P 2 u 2 + P 1 u 3
.
.
.
.
.
.
Qn-2 =
.
.
.
Qn-1 =
.
.
.
Qn =
.
.
.
.
.
.
.
.
.
1. Forward Procedure: Start from u1  uJ
2. Backward procedure: Start solving UH from uJ  u1
UH
P3 uJ-3 + P2 uJ-1 + P1 uJ
.
P3 uJ-1 + P2 uJ
.
.
P 3 uJ
UH Determination from Complex Storms
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Least Squares & Its Variations
1. Ordinary Least Squares: Solve Pu=Q by minimizing Pu-Q2. The resulting
normal equation is
(P’P) u = P’Q
and the least squares UH can be determined as
uOLS = (P’P)-1P’Q
The derived UH depends on the condition of matrix P or (P’P) represented
 /  min
 and  min
by the condition number, max
with max
being the largest
and smallest eigenvalues of P’P.
- Ridge Regression: (obtain stable & smooth UH)
uRidge = (P’P + kI)-1P’Q
The above UH minimizes Pu-Q2 + ku2 which represents the meansquared error in predicting future discharge hydrograph. The issue is now
to determine the optimal k (ridge constant).
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UH Determination from Complex Storms
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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  P 
Q 
1
1
P 
Q 
2
 
 2

  
  u 

  

  
 
 
PK 
Q K 
where Pk and Qk are ERH matrix and DRH vector for the
k-th event and u is the vector of multi-event UH.
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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
645.6A
Q  iA 
e
t
where Qe – [ft3/s] ; i – [in/hr]; A – [mi2]; and t – [hr], or
Q  iA 
e
2.778A
t
where Qe – [m3/s]; i – [cm/hr]; A – [km2]; and t – [hr].
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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)
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