1) The document describes deriving a unit hydrograph from a storm hydrograph for a catchment area that experienced two rainfall events - a 3 cm storm followed by a 2 cm storm. 2) The resulting direct runoff hydrograph (DRH) is calculated using the principle of superposition, by combining the individual DRHs from each rainfall event at each time step. 3) The summary provides an example calculation of the 5 cm DRH at 3 hours, which is the sum of the 3 cm DRH (75 cm) and 2 cm DRH (0 cm) at that time.

1. 1
Example 8-2
Two storms each of 6-h duration and having rainfall excess value
of 3.0 and 2.0 cm respectively. The 2-cm ER rain follows the 3 cm
rain. The 6-h unit hydrograph for the catchment is the same as in
given below. Calculate the resulting DRH.
Time(h) 0 3 6 9 12 15 18 24 30 36 42 48 54 60 69
UH
ordinate
(m3/s)
0 25 50 85 125 160 185 160 110 60 36 25 16 8 0
Solution
Given: Two Storms each of 6-h duration
Rainfall values: 3.0 cm and 2.0 cm
2.0 cm ER rain follows 3.0 cm
To be calculated DRH for Two storms
Presentation 11

2. 2
Plot 3 cm and 2 cm ERH in log – log paper/ mm graph
paper
First calculate the DRHs due to 3 cm and 2 cm ER.
Using TUH Method : Assumption — 1
The ER of 3.0 cm and 2.0 cm is calculated by multiplying 6-h
unit hydrograph respectively.
2-cm DRH occurs after 3-cm DRH, the ordinates of the 2-cm
DRH are lagged by 6 hours as shown in the Calculation
Table below.
Presentation 11(contd.)

3. 3
Using the method of superposition, the
ordinates of the resulting DRH are obtained by
combining the ordinates of the 3 cm and the 2 cm
DRHs at any instant, by this process the ordinates of
the 5 cm DRH are obtained (Assumption — 2)
Sample calculation:
2 cm DRH = 2 * THU ordinates = 2 * 25 = 50 cm
3 cm DRH = 3* THU ordinates = 3 * 25 = 75 cm
and so on
Presentation 11(contd.)

4. 4
0 6 12 18 24 30 36 ............ ... 54 60 66 72
900
800
700
600
500
400
300
200
100
0
Discharge(m3/s)
Time (hour)
3cm
2cm
0 6 12 h
2 cm & 3 cm ERH
A
B
A+B = 3 +2 = 5 cm DRH
A = DRH 3cm ER
B = DRH 2 cm ER
Figure11- 1 : Principle of superposition
Log-log paper/mm
graph paper
Rainfall(cm)
Time (h)
Presentation 11(contd.)

5. 5
Time
(h)
Ordinate
of 6-h UH
(m3/s)
Ordinate
of 3-cm
DRH
Ordinate
of 2-cm
DRH
Ordinate
of 5-cm
DRH
m3/s
Remark Time
(h)
Ordinate
of 6-h
UH
(m3/s)
Ordinat
e
of 3-cm
DRH
Ordinate
of 2-cm
DRH
Ordinate
of 5-cm
DRH
Remark
1 2 3 4 5 6 1 2 3 4 5 6
0 0 0 0 0 42 36 108 120 228
interpolat
ed value
3 25 75 0 75 48 25 75 72 147
6 50 150 0 150 54 16 48 50 98
9 85 255 50 305 60 8 24 32 56
12 125 375 100 475 (66) (2.7) (8.1) (16) (24.1)
15 160 480 170 650 69 0 0 (10.6) (10.6)
18 185 555 250 805 75 0 0 0 0
(21) (172.5) (517.5) (320)=
160×2
(837.5) Interp
olated
value
24 160 480 370 850
30 110 330 320 650
36 60 180 220 400
Calculation Table
5 cm DRH (at t = 3 hr) = 3* THU ordinates + 2 * THU ordinates
= 75 cm + 0 = 75 cm and so on
Presentation 11(contd.)

6. 6
Now again plot / Super impose 5 cm DRH over ERH of 2 cm &
3cm(A+B = 2+3=5cm)
DRH calculation for 21 hr and 66 hr as follows by interpolation:
(x1 –x2) / (x2 – x) = (y1 –y2) / (y2 – y)
x1 = 18 y1 = 185
x2 = 24 y2 = 160
x = 21 y = ?
(18- 24)/ ( 24-21) = ( 185- 160) / (160-y)
y = 172.5 cm and so on
Presentation 11(contd.)

7. 7
Presentation 11(contd.)
Derivation of unit hydrographs from hydrographs
of different depths:
A number of isolated storm hydrographs caused by short spells of
rainfall excess, each of approximately same duration (0.90 to 1.1 D h)
are selected from a study of the continuously gauged runoff of the
stream. For each of these surface hydrograph, the base flow is
separated by adopting one of the methods as mentioned earlier as
shown
1. Method I – Straight line method
2. Method II or
3. Method III

8. 8
The Unit hydrograph is obtained by dividing the
ordinates of hydrograph by its ER value .
i.e. THU = Storm Hydrographs / ER
Flood hydrographs should be selected for analysis
considering the following desirable features :
a. The storms should be isolated storms occurring individually.
b. The rainfall should be fairly uniform during the duration and
should cover the entire catchment area.
c. The duration of the rainfall should be 1/5 to 1/3 of the basin
lag
d. The rainfall excess of the selected storm should be high.
range of ER values of 1.0 to 4.0 cm is sometimes preferred.
Presentation 11(contd.)

9. 9
Example 8-3
Following table shows the ordinates of a storm hydrograph of a
river draining a catchment area of 423 km2 due to a 6-h isolated
storm. Derive the ordinates of a 6-h unit hydrograph for the
catchment. A storm of 6-h duration rainfall excess value 3.0 cm.(ER)
Time from
start of
storm(h)
-6 0 6 12 18 24 30 36 42 48
Discharge(m3/s 10 10 30 87.5 115.5 102.5 85.0 71.0 59.0 47.5
Time from
start of
storm(h)
54 60 66 72 78 84 90 96 102
Discharge(m3/s) 39.0 31.5 26.0 21.5 17.5 15.0 12.5 12.0 12.0
Solution:
Given : Discharge in the Data table
ER = 3 cm ; Catchment Area, A = 423 km2
Presentation 11(contd.)

10. 10
To be estimated the ordinates of 6-h Unit Hydrograph
and plot the same.
The storm hydrograph is plotted to scale as shown
below (Figure 11-2). Denoting the time from beginning
of storm as t, by inspection of the figure 11-2
A = beginning of DRH, t = 0
B = end of DRH , t = 90 h
Pm = Mean peak t = 20 h
Presentation 11(contd.)

11. 11
Lecture 11(contd.)
-6 0 6 12 24 36 48 60 72 84 96 108
Time in hours
020406080100120
Dischargem3/s
6 h
3 cm ER
A B
Storm hydrograph
Figure-11-2 : Derivation of unit hydrograph from a storm hydrograph
A = beginning of DRH, t = 0
B = end of DRH , t = 90 h
Pm= peak t = 20 h
Pm= peak
Base Flow
Surface
Runoff

12. 12
We know, Unit hydrograph = DRH / ER
where DRH = Storm hydrograph – Base flow
Base flow may be calculated by Using Method-I (Straight
line method)
Lag time N (days) = Peak time to depletion point
(From graph )
N = (90-20) /24 = 70h/24 = 2.91 days
We know by empirical equation:
N = 0.83 A0.2 = 0.83 x (423)0.2 = 2.78 days
where A = catchment in km2
Presentation 11(contd.)

13. 13
For convenience N = 2.91 days is adopted instead of
N = 2.78 days.
Calculation of Base flow
Join inflection point A and depletion point B by straight line
as in the graph (Figure-11-2 )
Then the below the straight line AB is base flow and values
above the straight line is DRH ( Direct Runoff Hydrograph) .
The ordinates of DRH are obtained by subtracting the base
flow from the ordinates of the storm hydrograph as shown in
the followingTable.11-1.
Presentation 11(contd.)

14. 14
Time from
beginning of
storm(h)
Ordinate of storm
hydrograph
(m3/s)
Base
flow(m3/s)
Ordinate of
DRH(m3/s) ER
Ordinate of 6-h
unit hydrograph
-6 10.0 10.0 0
3 cm
0
0 10.0 10.0 0 0
6 30.0 10.0 20.0 6.7
12 87.5 10.5 77.0 25.7
18 111.5 10.5 101.0 33.7
24 102.5 10.5 92.0 30.7
30 85.0 11.0 74.0 24.7
36 71.0 11.0 60.0 20.0
42 59.0 11.0 48.0 16.0
48 47.5 11.5 36.0 12.0
54 30.0 11.5 27.5 9.2
60 31.5 11.5 20.0 6.6
66 26.0 12.0 14.0 4.6
72 21.5 12.0 9.5 3.2
78 17.5 12.0 5.5 1.8
84 15.0 12.0 2.5 0.8
90 12.5 12.0 0 0
96 12.0 12.0 0 0
102 12.0 12.0 0 0
Table : 11-1
Presentation 11(contd.)

15. 15
Use and limitations of unit hydrograph
Use
The use of unit hydrographs establish a relationship between
the ERH and DRH for a catchment, they are of the following:
 the development of flood hydrographs for extreme rainfall
magnitudes for use in the design of hydraulic structures.
 extension of flood- flow records based on rainfall
records and
 the development of flood forecasting and warning systems based
on rainfall.
Presentation 11(contd.)

16. 16
Limitations of unit hydrograph:
1. Precipitation must be from rainfall only
2. snow-melt runoff cannot be satisfactory
represented by unit hydrograph
3. The catchment should not have unusually large
storages in terms of tanks, ponds, large flood-bank
storages, etc. which affect the linear relationship
between storage and discharge.
4. If the precipitation is directly non uniform, unit
hydrographs cannot be expected to give results.
Presentation 11(contd.)