Module 3
Lecture 5: Derivation of S-curve and
discrete convolution equations
Unit hydrograph
S – Curve method
•
It is the hydrograph of direct surface discharge that
would result from a
continuous succession of unit storms producing 1cm(in.)in tr –hr
•
If the time base of the unit hydrograph is Tb hr, it reaches constant outflow
(Qe) at T hr, since 1 cm of net rain on the catchment is being supplied and
removed every tr hour and only T/tr unit graphs are necessary to produce an
S-curve and develop constant outflow given by,
Qe = (2.78·A) / tr
where
Qe = constant outflow (cumec)
tr = duration of the unit graph (hr)
A = area of the basin (km2 or acres)
Module 3
Unit hydrograph
S – Curve method
Contd…
Intensity (cm/hr)
Changing the duration of UG by S-curve technique
Successive unit storms of Pnet = 1 cm
To obtain tr-hr UG
multiply the S-curve
difference by tr/trI
Discharge Q (Cumec)
S-curve
hydrograph
Unit hydrograph in
Succession produce
Constant outflow
Qe cumec
Lagged s-curve
Lagged by tr-hr
trI
Lagged
Time t (hr)
Constant flow Qe (Cumec)
Module 3
Unit hydrograph
Example Problem
•
Convert the following 2-hr UH to a 3-hr UH using the S-curve method
Time (hr)
2-hr UH ordinate (cfs)
0
0
Solution
1
75
Make a spreadsheet with the 2-hr
2
250
UH ordinates, then copy them in
3
300
the next column lagged by D=2
4
275
hours. Keep adding columns until
5
200
6
7
100
75
8
50
9
25
10
0
the row sums are fairly constant.
The sums are the ordinates of your
S-curve
Module 3
Unit hydrograph
Example Problem
Contd…
Time
(hr)
2-hr
UH
0
0
0
1
75
75
2
250
0
250
3
300
75
375
4
275
250
0
525
5
200
300
75
575
6
100
275
250
0
625
7
75
200
300
75
650
8
50
100
275
250
0
675
9
25
75
200
300
75
675
10
0
50
100
275
250
0
675
25
75
200
300
75
675
11
2-HR lagged UH’s
Sum
Module 3
Unit hydrograph
Example Problem
Contd…
Draw your S-curve, as shown in figure below
800
S-curve
700
600
Q (cfs)
500
2 hr UH
Lagged by 2 hr
400
300
200
100
0
0
2
4
6 Time (hr) 8
10
12
14
Make a spreadsheet with the 2-hr UH ordinates, then copy them in the next column
lagged by D=2 hours. Keep adding columns until the row sums are fairly constant.
The sums are the ordinates of your S-curve.
Module 3
Unit hydrograph
Example Problem
Time (hr)
Contd…
S-curve
ordinate
S-curve
lagged 3hr
Difference
3-HR UH
ordinate
0
0
0
0
1
75
75
50
2
250
250
166.7
3
375
0
375
250
4
525
75
450
300
5
575
250
352
216
6
625
375
250
166.7
7
650
525
125
83.3
8
675
575
100
66.7
9
675
625
50
33.3
10
675
650
25
16.7
11
675
675
0
0
Unit hydrograph
Example Problem
Find the one hour unit hydrograph using the excess rainfall hyetograph and
direct runoff hydrograph given in the table
Time (1hr)
Excess Rainfall (in)
Direct Runoff (cfs)
1
1.06
428
2
1.93
1923
3
1.81
5297
4
9131
5
10625
6
7834
7
3921
8
1846
9
1402
10
830
11
313
Module 3
Contd….
Unit hydrograph
Discrete Convolution Equation
Suppose that there are M pulses of excess rainfall.
If N pulses of direct runoff are considered, then N equations can be written Qn in
terms of N-M+1unknown values of unit hydrograph ordinates, where n=
1, 2, …,N.
m*
Qn = ∑ PmUn−m+1
m=1
m* = min(n,M)
Where Qn = Direct runoff
Pm = Excess rainfall
Un-m+1 = Unit hydrograph ordinates
Contd….
Unit hydrograph
Input Pn
P1
P2
Output Qn
m*
Output Qn = ∑ PmUn−m+1
P3
m=1
Unit pulse response applied to
P1
Un-m+1
Combination of 3 rainfall
UH
U1
U2
U3
U4
U5
n-m+1
Un-m+1
Unit pulse response applied to
P2
n-m+1
Unit hydrograph
Contd….
The set of equations for discrete time convolution
Q1 = PU
11
=
Q
2 P2U1 + PU
1 2
Q3 =P3U1 + P2U2 + PU
1 3
QM = PMU1 + PM−1U2 + ..... + PU
1 M
QM+1 =
0 + PMU2 + ..... + P2UM + PU
1 M+1
QN−1 = 0 + 0 + ..... + 0 + 0 + ..... + PMUN−M + PM−1UN−M+1
QN = 0 + 0 + ..... + 0 + 0 + ..... + 0 + PM−1UN−M+1
m*
Qn = ∑ PmUn−m+1
m=1
n = 1, 2,…,N
Unit hydrograph
Example Problem
Contd…
Solution
•
The ERH and DRH in table have M=3 and N=11 pulses respectively.
•
Hence, the number of pulses in the unit hydrograph is N-M+1=11-3+1=9.
•
Substituting the ordinates of the ERH and DRH into the equations in table
yields a set of 11 simultaneous equations
=
U1
Q2 − P2U1 1,928 − 1.93x404
=
= 1,079 cfs / in
P1
1.06
Similarly calculate for remaining ordinates and the final UH is tabulated below
n
Un (cfs/in)
1
2
3
4
5
6
7
8
9
404
1,079
2,343
2,506
1,460
453
381
274
173
Module 3