Hydrology And Irrigation Engineering

1. Runoff, Hydrograph

2. Module-III
• Runoff: Definition, concept of
catchment, factors affecting
runoff, rainfall – runoff
relationship using regression
analysis.
• Hydrographs: Definition,
components of hydrograph, base
flow separation, unit hydrograph,
assumption, application and
limitations, derivation from
simple storm hydrographs, S
curve and its computations,
Conversion of UH of different
durations

3. catchment

4. 4

5. factors affecting runoff hydrograph

6. Rainfall – Runoff relationship using regression analysis

7. 7
Hydrograph
 Hydrographs are graphical representations of “flow parameters
against time”. When a storm falls in a catchment, after initial
losses and loss due infiltration, rainfall excess eventually flows to
the main stream.
 “If the discharge ‘Q’, average velocity ’V’ and depth or stage ‘Y’ at
outlet point of the stream is measured from beginning of the
storm to the end of storm in the catchment and if they are plotted
against time, they may be called Hydrograph”.
i) If ‘Q’ is plotted against time ‘t’, it is called flow or “flood
Hydrograph”
ii) If ‘V’ is plotted against ‘t’, it is known as “Velocity
Hydrograph”
iii) If depth ‘Y’ or stage ‘H’ is plotted against time ‘t’, it is called
depth or “Stage Hydrograph”.
“But most commonly in Hydrology, Hydrograph means the
graph of flow rate ‘Q’ against time ‘t’”
HYDROGRAPH ANALYSIS

8. 8
Hydrograph of Single and Complex Storms (Picture-I)
HYDROGRAPH ANALYSIS

9. 9
Hydrograph of Single and Complex Storms (Picture-II)
HYDROGRAPH ANALYSIS

10. 10
Components of Hydrograph of Isolated Storm
i. AB is approach segment
ii. BC is rising limp or concentration curve
iii. CPD is crest segment or peak limb
iv. DE is falling limb
v. Lower portion DE is called “Ground Water Depletion” curve
vi. P is the point of crest or peak
vii. tB is time of rise
viii. Tp is the time of peak
ix. Area under ABE is the volume of base flow (shown by shaded
line). This volume is assumed to have no contribution of storm.
x. Area between BPE and BE is the volume of direct surface runoff
(DSR) shown by dots.
xi. OA is the value of discharge at initial time t=0 and this OA may
be assumed to be the base flow.
xii. ‘C’ is the first point of inflexion and ‘D’ is the second point of
inflexion.
xiii. OE1 is the base period ‘T’ of the hydrograph
HYDROGRAPH ANALYSIS

11. 11
Hydrograph Segment
HYDROGRAPH ANALYSIS

12. 12
Segments of Hydrograph of an Isolated Storm
Hydrograph is broadly divided into following three segments;
 Rising Limb/ Segment
In the initial time of rainfall, losses cause the discharge to rise
slowly. Building up of storage is evidently gradual.
However, if the storm is continuous, a accumulative discharge
from upstream reaches the outlet (as infiltration and absorption
loss decreases with time). “Of course, shape of basin and storm
characteristics affect the rising limp”.
 Peak Limb/ Crest Segment
It is the most important part of Hydrograph. Peak outflow is
essential for design of Hydraulic Structures. When runoffs from
all sources reach the outlet, peak flow occurs.
In large catchment, peak flow may even occur after the storm.
“Estimation of peak flow and its time of occurrence are very
important in Hydrology”.
HYDROGRAPH ANALYSIS

13. 13
Segments of Hydrograph of an Isolated Storm
 Recession Limb/ Depletion Curve
The recession limb represents the “withdrawal of water from
storage after all inflows to the channel have ceased. “Therefore, it
is time independent of rainfall or infiltration and is essentially
dependent on channel features alone”.
HYDROGRAPH ANALYSIS

14. Effective Rainfall (ER)

15. Base flow Separation
For the derivation of unit hydrograph, the base flow has to be separated from the total
runoff hydrograph
Base flow is the initial flow of the stream before the rain comes. “It is the sustained
or dry weather flow of the stream resulting from the outflow of perennial or almost
permanent ground water flow that reaches the channel”. This base flow of such
channel is more or less assumed to be constant.

16. Base flow Separation Procedure
1. By simply drawing a line ‘AC’ tangential to both the limbs at
their lower portion. This method is very simple but is
approximate and can be used only for preliminary estimates.
2. Extending the recession curve existing prior to the occurrence of
the storm up to the point ‘D’ directly under the peak of the
hydrograph and then drawing a straight line DE. Where E is a
point on hydrograph ‘N’ days after the peak & N (in days) is
given by
N= 0.84A0.2
3. Simply by drawing a straight line AE, from the point of rise to
the point E on the hydrograph, ‘N’ days after the peak.

17. 17
Computation of Direct Surface Runoff (DSR) from a Storm
Enclosed figure shows the flood hydrograph of a storm with flood
ordinates at different time interval Δt. “From the flood hydrograph,
separate the base flow by any one of the methods described earlier”.
Let the “Base Flow” is separated by simple straight line method and
Direct Surface Runoff (DSR) Hydrograph is drawn (figure enclosed).
Let Q1, Q2, Q3, Q4, …., Qn-2, Qn-1, Qn are the ordinates of Direct Surface
Runoff (DSR) Hydrograph at different time interval Δt.
Find the area under the (DSR) Hydrograph, which will represent the
volume of (DSR);
Area under the curve APB = Volume of DSR
= A1+A2+A3+A4+….. + An-2 + An-1 + An
= 1Q1Δt + [Q1+Q2]Δt + [Q2+Q3]Δt + [Q3+Q4] Δt+ … + [Qn+2+ Qn+1]Δt
2 2 2 2 2
+ [Qn-1+ Qn]Δt + 1 Δt
2 2
HYDROGRAPH ANALYSIS

18. 18
= [1Q Δt+Q1 Δt]+[Q2Δt+Q2 Δt]+[Q3Δt] + [Q4Δt+Q4 Δt] +…+ [Qn-2Δt+Qn-2 Δt]
2 2 2 2 2 2 2 2 2
+ [Qn-1Δt+Q1Δt] + [QnΔt+Qn Δt]
2 2 2 2
= [(Q1 + Q2 + Q3 + Q4 + …. Qn-2 + Qn-1 + Qn)] Δt
= ∑QΔt
Q = Total volume of Direct Surface Runoff. If ‘Q’ in m3/sec, & Δt in hr.
Volume of DSR = ∑QΔt m3 x (60 x 60) sec
sec
= ∑QΔt m3 x (60 x 60) m3
If ‘A’ is the area of the catchment in km2, A = A x 106 m3
DSR in cm = {[(60 x 60) ∑QΔt] x 100} cm
A x 106
DSR = 0.36 ∑Q x t (cms)
A
Thus, DSR volume (depth in cm) = Rainfall excess or net rainfall
HYDROGRAPH ANALYSIS

19. 19
Flood Hydrograph of a Single Storm
HYDROGRAPH ANALYSIS

20. 20
Calculation of Direct Surface Runoff (DSR) Hydrograph
(Depth)
HYDROGRAPH ANALYSIS

21. 21
Example:-
Calculate Direct Surface Run-off for Rainfall excess from a Storm
Hydrograph from the following data. Assume Area of Catchment as
30 km2
HYDROGRAPH ANALYSIS
Date and Time Ordinate of Hydrograph
Total Q (Cumecs)
Base Flow
(Cumecs)
Direct Runoff
(Cms)
1 2 3 4
14.08.2012
0500 14 14 0
0800 25 12 13
1100 51 11 40
1400 65 10 55
1700 54 11 43
2000 28 13 15
2300 14 14 0
∑Q = 166

22. 22
HYDROGRAPH ANALYSIS
Solution
 Find the “ordinates of storm Hydrograph”, representing total
discharge Q at a given time interval, say ‘t’ hours.
 Separate the “Base Flow” by any of the methods. Find the
ordinates of the base flow at the same time interval.
 Find the ordinates of “Direct Run-off” by subtracting the
ordinates of base flow from total discharge ordinates.
 The Direct Surface Run-off (depth in cm) is found from the
expression.

23. 23
HYDROGRAPH ANALYSIS

24. Unit Hydrograph
 Unit Hydrograph (UH) is “defined as the Hydrograph of surface runoff of a
catchment area resulting from unit depth (usually 1 cm) of rainfall
excess or net rainfall occurring uniformly over the basin and at uniform rate
for a specified duration”.
 Effective Rainfall : also known as excess rainfall is that part of the rainfall that becomes
direct runoff at the outlet of the watershed. It is thus the total rainfall in a given duration
from which abstractions such as infiltration and initial losses are subtracted.
 Unit Hydrograph is a linear model of the catchment which is used to find out
the volume of DSR due to 1 cm of direct surface runoff or 1 cm of rainfall
excess. This is always constant for the catchment since area of the
catchment is constant.
 If rainfall comes to the catchment producing 2 cm of rainfall excess, “the
ordinates of the DSR will be twice as great of the Unit Hydrograph (UH)
ordinates and volume of DSR will be two time the volume of DSR of unit
hydrograph”.
HYDROGRAPH ANALYSIS

25. Assumptions in Derivation of UH Theory
i. The rainfall is of “uniform intensity” within its specified duration.
ii. The effective rainfall is “uniformly distributed” throughout the
whole area of drainage basin.
iii. “The base of time duration of Hydrograph of direct runoff due to
effective rainfalls of unit duration is constant”. Base period of
Hydrograph of different rainfall intensities remain approximately
same. As represented in enclosed figure.
iv. The ordinates of DSR Hydrograph “due to net rains of different
intensities but same duration are proportional”.
v. A Unit Hydrograph reflects all the combined physical
characteristics of the basin.
HYDROGRAPH ANALYSIS

26. Limitations
• For unit hydrograph analysis storms of short duration
should be selected since uniform intensity over long
duration is less likely to occur.
• UH’s can be effectively applied to drainage basins with
small area, since uniform areal distribution of rainfall over
large areas is less likely to occur.
• Basins with odd shapes particularly those which are long
and narrow will commonly have uneven rainfall distribution
and hence unit hydrograph are not well adopted to such
basins.
• The unit hydrograph method can be applied when the
major portion of the precipitation is in the form of snow

27. Derivation of UH from a simple Flood Hydrograph –
(Isolated Storm)
Step 1 From the given flood Hydrograph, “separate the base flow” by
any one of the methods. Most commonly used method to
draw a straight line without much error for simplicity (Figure
enclosed)
Step 2 Determine the volume of DSR Hydrograph by the formula;
“Volume of DSR = ∑QΔt
”
Step 3 Divide this volume by “known Area of Catchment” to get DSR
volume (depth in cm) i.e., “net rainfall or rainfall excess”.
Step 4 Divide the ordinates of DSR by the depth of DSR Hydrograph
to obtain ordinates of UH.
Step 5 Plot the ordinates of UH of the catchment as per enclosed
figure.
HYDROGRAPH ANALYSIS

28. Unit Hydrograph Derivation
HYDROGRAPH ANALYSIS

29. Example-I
HYDROGRAPH ANALYSIS
Calculate and Draw a “Unit Hydrograph” from a Storm Hydrograph, resulting
from a Catchment Area of 25 km2. The data is shown in the following table:

30. HYDROGRAPH ANALYSIS
Direct Runoff = 0.36 (∑Q x t) = 0.36 x 822 x 2 = 23.7 cm
A 25

31. HYDROGRAPH ANALYSIS
Unit Hydrograph from a Single Flood Hydrograph

32. Example-II:-(UH) from a Single Flood Hydrograph (Isolated)
HYDROGRAPH ANALYSIS

33. Limitations and Uses of UH Theory
Limitations
i. Similar rainfall distribution from storm over a large area is rare.
Hence UH theory is suited to catchment area under about 500 km3
(=2000 sq. miles)
ii. Odd shaped basins particularly those which are long and narrow,
commonly have very uneven rainfall distribution and hence UH
theory for such basins is not much suitable.
iii. In mountainous regions, subject to orographic rainfall, aerial
distribution is very uneven, but the pattern tends to remain the
same from storm to storm, and UH theory may not be
successfully applied.
iv. The unit hydrograph method cannot be applied when an
appreciable portion of the storm precipitation falls as snow.
v. The catchment having large storage like reservoir, lake, low areas,
etc. affect the linear relationship and hence theory cannot be
applied.
vi. If the variation of base period and peak flow vary more than +/-
10%, theory applied is not generally accepted.
HYDROGRAPH ANALYSIS

34. Limitations and Uses of UH Theory
Uses
i. As the UH is a “linear model of the catchment”, it is used to
determine runoff Hydrograph of the Catchment even for
extreme magnitude for “the calculation of peak discharge to
design the Hydraulic Structures”.
ii. It can also be used for flood forecasting and warning.
iii. Based on rainfall records, “it is used for extension of flood
records”.
HYDROGRAPH ANALYSIS

35. CALCULATION OF STORM
HYDROGRAPH
FROM
THE GIVEN UNIT HYDROGRAPHS
OF VARIOUS RAINFALL
INTENSITIES
HYDROGRAPH ANALYSIS

36. Application of the (UH) Theory for the “Construction of a Flood
Hydrograph Resulting from Rainfall of Single Duration”
 The Unit Hydrograph can be used to construct a “Flood Hydrograph“
resulting from rainfall of the “same unit duration for which the Unit
Hydrograph is available”.
 The number of Unit Hydrographs for a given drainage basin is
theoretically infinite since there may be one for every possible
duration of rainfall and every possible distribution pattern in a basin.
 Practically only a limited number of Unit Hydrographs can be used
for a given basin. “The Unit Hydrograph selected for computing the
flood Hydrograph should be corresponding to the storm of like
duration and pattern”.
 However, a tolerance of as much as 25% of the Unit Hydrograph
duration can ordinarily be accepted without much serious error. Thus
a 6-hours Unit Hydrograph must be applied for storms of 4.5 to 7.5
hours duration.
 Alternatively, Unit Hydrograph having a similar range in duration may
be averaged to obtain an average Unit Hydrograph.
HYDROGRAPH ANALYSIS
Case-I

37. Procedure
The calculations for the storm Hydrograph corresponding to ‘n’ cm of
rainfall-excess are done in the tabular form as per enclosed table;
Direct runoff ordinates=(ordinates of Unit Hydrograph) x‘n’ cm
Step-I : Write down the “given hourly ordinates” (Discharge) of
Unit Hydrograph.
Step-II : Multiply the ordinates of Unit Hydrograph with ‘n’ cm of
Rainfall.
Step-III : List out the given hourly “base flow (m3/sec)”
Step-IV : Find out the total storm ordinates i.e. = Step-II + Step-III
The above procedure is elaborated by the enclosed table. Through an
example (Calculation of Storm Hydrograph Resulting from 8 cm
Rainfall.
HYDROGRAPH ANALYSIS

38. HYDROGRAPH ANALYSIS
4 5
Example:- Calculate Storm Hydrograph of 8 cm Rainfall

39. Case-II
Application of Unit Hydrograph Theory “for the Construction of a
Flood Hydrograph Resulting from two or more Periods of
Rainfall (Multi Duration Rainfall)” from a Single UH
 A Unit Hydrograph of some specific unit duration can also be
utilized for construction of flood Hydrograph resulting from the
“rainfall lasting for a longer duration”.
 The essential condition, however, is that the storm pattern should
be the same as that for the Unit Hydrograph.
 As an example, let a 3-hours Unit Hydrograph is available, and it
is required “to compute the flood Hydrograph resulting from a
rainfall lasting for 9-hours with variable intensities of rainfall”.
 The intensity rates having “n1 cm/3-hours” for the first period of
3-hours, “n2 cm/3-hours” for the second 3-hours, and that of
“n3 cm/3-hours” for the last 3-hours.
HYDROGRAPH ANALYSIS

40. Step-I
 The storm is divided into 3 parts, and the flood Hydrograph of
each part is computed separately and added.
Step-II
 The Hydrographs for the second part of the storm starts 3 hours
later than that for the first part.
Step-III
 Similarly, the Hydrograph for the third part of the storm starts 6
hours later than that for the first, or 3 hours later than that for the
second part.
 The above procedure is illustrated through the enclosed example.
HYDROGRAPH ANALYSIS
Procedure

41. Example:-
Find the ordinates of a “storm Hydrograph resulting from a 3 hours
storm with rainfall of 2.0, 6.75 and 3.75 cms during subsequent 3
hours intervals”. The ordinates of Unit Hydrograph are given in the
following table. Assume an initial loss of 5 mm, infiltration index of
2.5 mm/hour and base flow of 10 cumecs.
HYDROGRAPH ANALYSIS
Ordinates of given Unit Hydrograph
Hours 03 06 09 12 15 18 21 24 03 06 09 12 15 18 21 24
Ordinate of
UH
0 110 365 500 390 310 250 235 175 130 95 65 40 22 10 0

42. Solution:-
i) Rainfall excess during the first three hours
= 20 – (2.5 x 3) – 5 = 7.5 mm = 0.75 cm
ii) Rainfall excess during the second three hours
= 67.5 – (3 x 2.5) = 60 mm = 6 cm.
iii) Rainfall excess during the last three hours.
= 37.5 – (3 x 2.5) = 30 mm = 3 cm
 The Rainfall excess as ratio of the unit Rainfall of 1 cm during the
subsequent 3 hours intervals are 0.75, 6 and 3.
 The computations of Runoff due to 0.75 cm rainfall excess will
start from 03 hours. The computations of runoff due to 6 cm
rainfall excess will start from 6 hours. Lastly, the computations of
runoff due to 3 cm rainfall excess will start from 9 hours.
 “The above procedure is illustrated in the enclosed table”.
HYDROGRAPH ANALYSIS

43. Table:-
HYDROGRAPH ANALYSIS

44. Flood Hydrograph Resulting from a Storm of Longer Duration
HYDROGRAPH ANALYSIS

45. Example:-
Calculation of Storm Hydrograph Resulting from Single
Unit Hydrograph of Various Rainfall Intensities
HYDROGRAPH ANALYSIS
Contd…..

46. HYDROGRAPH ANALYSIS

47. Case-III
S-Hydrograph (S-Curve)
 S-Hydrograph or S-Curve is a Hydrograph which is produced by a
“continuous effective rainfall at a constant rate for indefinite
period”.
 It is a “continuous rising curve”, in the form of letter ‘S’, till
equilibrium is reached. At the time of equilibrium, “it will
represent a constant rate of continuous effective rainfall, say Ro
cm per hour”.
 At the time of equilibrium, “the S-Curve will represent a runoff
discharge as under”;
Qo = (A x 100 x 100) Ro
100 x 3600
= A Ro cumec
36
(Where A is the catchment area in hectares)
HYDROGRAPH ANALYSIS
Contd…

48.  If the catchment area ‘A’ is in “square kilometers”, the discharge
represented by S-Curve, at the time of equilibrium is given by;
Qo = (A x 1000 x 1000) Ro
100 x 3600
=(AR0) x 100 (cms)
36
= 2.778 A Ro (cms)
(‘A’ = Area of catchment in km2 and Ro = Constant rate of
continues effective Rainfall.)
 The S-Hydrograph or S-Curve is constructed by “adding together
number of Unit Hydrographs of unit time duration (T0) spaced at a
unit time duration (T0) (i.e., duration of effective rainfall).
 “This is illustrated in the enclosed example were in S-Curve has
been drawn for 6 hours Unit Hydrograph. Area of basin is 311
km2”.
HYDROGRAPH ANALYSIS

49. HYDROGRAPH ANALYSIS
Computation of S-Hydrograph from Successive Unit Hydrograph
Contd…

50. HYDROGRAPH ANALYSIS
This discharge of 144 cumecs will be achieved in the above table at 36
hours (which is equal to base period – T0 hours.)

51. HYDROGRAPH ANALYSIS
Derivation of S-Hydrograph from Series of Unit Hydrograph

52. CALCULATION OF UNIT
HYDROGRAPH
FROM
THE GIVEN UNIT HYDROGRAPHS
OF VARIOUS INTENSITIES
HYDROGRAPH ANALYSIS

53. Construction of Unit Hydrograph of Different Unit Duration
from the a Unit Hydrograph of given Unit Duration
 Unit Hydrograph may be of different duration ‘D’. If a unit
hydrograph of 2 hrs. duration available, unit hydrograph of 4 hrs.
duration, 6 hrs. duration, 8 hrs. duration, etc. may be obtained i.e.,
the unit hydrograph to be derived is an integral multiple. In such
cases, durations from 2 hrs. to 4 hrs., 6 hrs. etc. can be done by
method of superposition.
 However, if the Duration of Unit Hydrograph to be derived is less
than duration of known Unit Hydrograph, “this method of
superposition will not hold good”. Thus, Unit Hydrograph of
different duration may be derived by the following two methods.
i) Method of superposition when duration of UH to be derived is
2 times, 3 times or 4 times of the duration of known UH.
ii) S-Curve method (Summation hydrograph method) when
duration of UH to be derived is less than the duration of the
known UH.
HYDROGRAPH ANALYSIS
Case-I

54. Unit Hydrograph of Different Durations
Phase-A:-
Construction of Longer period Unit Hydrograph from a given
Unit Hydrograph of Shorter Unit Period/ Duration.
Method of Superposition
i) As shown (figure enclosed) D-hr UH is plotted (Curve-1)
ii) Also plot the UH lagged by D hrs. (Curve-2)
iii) Add the two Unit Hydrographs and plot it (Curve-3)
iv) Divide Curve-III, i.e. DSR Hydrograph by 2D duration to get
the curve shown by dotted line, i.e. (Curve-4)
It is noted that base period of 2-D-hr UH shown by dotted line
increases by D-hrs from the given UH of T hrs base period.
Therefore, the peak of 2-D-hrs UH will decrease as the base
period is increased from T to (T + D hrs)
HYDROGRAPH ANALYSIS

55. Example – Unit Hydrograph of Different Durations
A 4-hr UH of different durations is shown in the following table. Derive
the 8 hr Unit Hydrograph.
HYDROGRAPH ANALYSIS
Column (5) gives the ordinates of 8-hr unit Hydrograph derived from
column (4) and unit UH in column (2). The peak of the 8-hr UH is
reduced to 16m3/sec from 20m3/sec as the base period is increased
from 20 to 24 hrs.

56. Method of Superposition
HYDROGRAPH ANALYSIS

57. Example-II:- Method of Superposition
HYDROGRAPH ANALYSIS

58. Case-II
Unit Hydrograph of Different Durations - S- Curve Method
 S-Curve (Summation Curve), also known as S-Hydrograph is a
Hydrograph produced by a “continuous effective rainfall of unit
depth at a constant rate for indefinite period”. It is a curve
obtained by summation of an infinite series of Unit Hydrographs
each lagged by D-hrs with respect to preceding one.
 It is shown that when a UH of ‘n’ ‘D’ duration is required where ‘n’
is whole number like 2,3,4 etc., method of superposition serves
the purpose. If ‘n’ is in-fraction, then S-Curve method is required.
 As an example, from UH of 4 hrs. duration, to develop UH of 2 hrs.
duration cannot be developed by the pervious method of
superposition. In such situation, “S-Hydrograph is constructed.
Which is used to find the UH of desired duration”.
 As shown (figure enclosed), S-Hydrograph is continuously a
rising curve which ultimately attains a constant value when
equilibrium discharge is reached.
HYDROGRAPH ANALYSIS

59. Unit Hydrograph of Different Durations
Unit Hydrograph can be derived as under;
Step 1 “Construct the S-Curve” from the given Unit Hydrograph of
known time duration D-hrs.
Step 2 “Then advance or offset” the position of all the ordinates of
“S-Hydrograph” for a period equal to the desired duration of
D0 hrs. of unknown UH. “Name this S-Hydrograph as offset
Hydrograph”.
Step 3 Find the “Difference” of the ordinates of Original S-Curve and
“Offset Hydrograph”.
Step 4 “Divide this each difference” by (D0/D) to get the ordinates of
new UH of D0 hrs. duration.
HYDROGRAPH ANALYSIS

60. S-Hydrograph or S-Curve
HYDROGRAPH ANALYSIS

61. HYDROGRAPH ANALYSIS
 Let it be required to obtain a “Unit Hydrograph” of unit period t0 from a
given Unit Hydrograph of unit period. T0, “where t0 can be either greater
or smaller than T0”. For this, the S-Hydrograph method will be used.
 From the given Unit Hydrograph of unit period T0, “S-Curve is derived”.
This S-Curve will represent a constant effective rainfall of R0 = 1/ T0
cm/hour.
 “An offset curve is then drawn by advancing or offsetting the position of
original S-Curve” for a period equal to the desired unit period to hours.
 “The difference between the ordinates of original S-Curve and the offset
S-Curve divide by the R0 t0“ will give the ordinate of the desired Unit
Hydrograph.
 Thus, for any time period t, if the difference ordinates of the two S-
Curves is Δy, then the ordinate of the desired Unit Hydrograph of t0 unit
period.
Δy = Δy = Δy. T0
R0t0 (1/T0)t0 t0
Phase-B:
Construction of Shorter or Longer Period Unit Hydrograph
from a given Unit Hydrograph

62. HYDROGRAPH ANALYSIS
Example-I
Derivation of Unit Hydrograph of t0 (= 3 hours) unit duration from the
given Unit Hydrograph of duration T0 (= 6 hours)

63. HYDROGRAPH ANALYSIS
Construction of Shorter or Longer Period of Unit Hydrograph
from a given Unit Hydrograph

64. Example-II:- of S-Curve Method
Given below are the ordinates of 4 hrs UH of a basin.
Derive 2 hrs UH from it using S-Curve Method.
HYDROGRAPH ANALYSIS

65. HYDROGRAPH ANALYSIS

66. HYDROGRAPH ANALYSIS

67. Case-III
Derivation of UH from Complex Storm
 Practically, “it is not always possible “to have such isolated
storms”. Storms having different rainfall excess like R1, R2 and R3
may occur. “Such storms of different rainfall excess are called
complex storm”.
 Assume 3 storms which can produce rainfall excess R1, R2 and
R3. The Hydrograph of these 3 storms (after deducting base flow)
is shown in figure enclosed.
 Let Q1, Q2, Q3,….. Qn are the known ordinates of complex storm as
per figure enclosed and U1, U2, U3, ….. Un are ordinates of Unit
Hydrograph which are to be determined. R1, R2 and R3 are known
rainfall excesses due to complex storm, which are also known.
Then;
HYDROGRAPH ANALYSIS

68. Derivation of UH from Complex Storm
Qn = RnU1 + Rn-1U2 + Rn-2U3 + ….,
When
n = 1, Q1 = R1U1
(i.e. Q1 and R1 are known, U1 is determined)
n = 2, Q2= R2U1 + R1U2
(i.e. Q2,R2,U1,R1 are known, U2 is determined)
n=3, Q3=R3U1+R2U2+R1U3
(i.e. Q3,R3,U1,U2,R1,R2 are known, U3 is determined)
n=4, Q4=R4U1+R3U2+R2U3+R1U4,
R4U1 is zero, since no 4th rainfall is considered
Since Q4, R3, R2, R1, U2, U3 are known, U4 can be determined
Thus, all ordinate of Unit Hydrograph U1, U2, U3, U4, …… Un can be
determined and resulting Unit Hydrograph (UH) can be obtained.
HYDROGRAPH ANALYSIS

69. Derivation of UH from Complex Storm
HYDROGRAPH ANALYSIS

70. Derivation (UH) from Complex Storm
Example
Draw the Unit Hydrograph from the complex storm having the data
given below;
HYDROGRAPH ANALYSIS
Solution
Net Rainfall are DSR in depth 1st hr = 4 – 5 Φ Index = 4 – 2 x 1 = 2 cm
2nd hr = 3 – 2 x 1 = 1 cm
3rd hr = 2.5 – 2 x 1 = 0.5 cm
Time (Hrs) 0 1 2 3 4 5 6 7 8 9 10 11 12 13
UH Ordinates
m3/sec
0 58 110 96 53 26 14 8 5 4 3 1.5 1 0
Compute the storm Hydrograph resulting from three hour storm rainfall
as under:-
Time 1st hr 2nd hr 3rd hr
Rainfall depth (cm) 4 3 2.5
Take Φ-index as 2 cm/hr and assume a base flow of 2 m3/sec.

71. Example of Derivation (UH) from Complex Storm
HYDROGRAPH ANALYSIS