Hydrologic cycle,weather and temperature, Rain gauge, precipitation, discharge measurement,stream flow, hydrographs, flood and flood routing. unit hydrographs

1. Lecture 9
Annual runoff volume
Methods of estimation
Yield of the river:
The total quantity of water that can be expected from a stream in a
given period such as a year is called the yield of the river. Yield for
a period of a year means the total annual runoff volume = ∑R
There are four(4) methods for estimation of yield are listed below :
1. Correlation of stream flow and rainfall
2. Empirical equations
3. Watershed simulation
4. Rational method
1

2. Lecture 9(contd.)
Method- 1: Rainfall –Runoff Correlation
The relation between rainfall and resulting runoff is quite
complex due to paucity of data. However, one of the most
common method is to correlate runoff –R with rainfall- P values.
Plotting of values R against P and drawing a best-fit line can be
adopted for very rough estimates.
A better method is to fit a linear regression line between R and P
and to accept the result if the correlation coefficient is linear
unity.
The equation for straight line regression between runoff R and
rainfall P is given by
R = aP + b ...........................................Eq 6-2
P
R
2

3. Lecture 9(contd.)
and the values of a and b are given by
a = {N(∑ PR) – ( ∑P)( ∑R)} / {N(∑ P2 ) – (∑ P)2 } .........Eq 6-3
b = { (∑R) – a (∑P)}/N ....................Eq 6-4
where, N = number of observation sets R and P.
The coefficient of correlation (r) can be calculated as
r =[{N(∑ PR) – ( ∑P)( ∑R)}] / √ [{ N(∑ P2 ) – (∑ P)2 } × { N (∑ R2 ) – (∑ R)2 }]
3

4. 4
Lecture 9(contd.)
Example 6-2
The monthly rainfall P and the corresponding runoff
R values covering a period of 18 months for a
catchment given below in the table. Develop a
correlation between R and P. And hence also
calculate Correlation coefficient.
Month P R Month P R Month P R
1 5 0.5 7 5 0.1 13 2 0.0
2 35 10.0 8 31 12.0 14 22 6.5
3 40 138 9 36 16.0 15 30 9.4
4 30 8.2 10 30 8.0 16 25 7.6
5 15 3.1 11 10 2.3 17 8 1.5
6 10 3.2 12 8 1.6 18 6 0.5
Data Table

5. 5
Solution :
Given
Rainfall and Runoff data in the table.
To be computed correlation between R and P
The Equation :
R = aP + b .............. Eq-1
The values of a and b are calculated by Eq-6-3 & 6-4
as given below:
a = {N(∑ PR) – ( ∑P)( ∑R)} / {N(∑ P2 ) – (∑ P)2 }
b = { (∑R) – a (∑P)}/N
Lecture 9(contd.)

6. 6
Lecture 9(contd.)
Month P P2 R R2 PR
1 5 25 0.5 0.25 2.5
2 35 1225 10.0 100.0 350
3 40 1600 13.8 190.44 552
4 30 900 8.2 67.24 246
5 15 225 3.1 9.61 46.5
6 10 100 3.2 10.24 32.0
7 5 25 0.1 0.01 0.5
8 31 961 12.0 144.0 372.0
9 36 1296 16.0 256.0 576.0
10 30 900 8.0 64.0 240.0
11 10 100 2.3 5.29 23.0
12 8 64 1.6 2.56 12.8
13 2 4 0.0 0.0 0.0
14 22 484 6.5 42.25 143.0
15 30 900 9.4 88.36 282.0
16 25 625 7.6 57.76 190.0
17 8 64 1.5 2.25 12.0
18 6 36 0.5 0.25 3.0
Total= 348 9534 104.3 1040.51 3083.3
From the given data we can calculate the required values
as provided in the calculated Table below for a & b:

7. 7
Lecture 9(contd.)
Putting the respective values for calculation a & b
a = {N(∑ PR) – ( ∑P)( ∑R)} / {N(∑ P2 ) – (∑ P)2 }
a = 18( 3083.3) – ( 348)( 104.3)/ {18(9534 ) – 121104}
a = 0.380
b = { (∑R) – a (∑P)}/N = {104.3 – 0.380× 348} / 18
b = (-) 1.55
Now putting the values of a & b in Eq- 1 above
R = aP + b = 0.380 P + (-1.55)
R = 0.380 P – 1.55
This equation is the Co-relation between R &P

8. 8
Calculation of Correlation coefficient (r)
r = { N(∑ PR) – ( ∑P)( ∑R)} / √ [{ N(∑ P2 ) – (∑ P)2 } × { N (∑ R2 )– (∑ R)2 }]
Putting the corresponding values in the above equation
r= { (18 × 3083.3) (348 × 104.3)}/ √ [ (18 ×9534) – 121104]
× [ (18 × 1040.51) – 10878.49] = 0.964
Hence Correlation coefficient (r) = 0.964
Lecture 9(contd.)

9. 9
Lecture 9(contd.)
Method -2 Empirical Equations : Khosla’ s Formula
Establishment of correlation between annual rainfall and runoff :
Khosla’s Formula:
Khosla(1960) analyzed the rainfall, runoff and temperature
data for various catchments in India and USA to arrive at an
empirical relationship between runoff and rainfall and
arrived a formula which is called Khosla’s Formula. The
time period is taken as a month. The relationship for
monthly runoff is
Rm = Pm –Lm .......................Eq 6-8
and
Lm = 0.48 Tm for Tm > 4.50C

10. 10
Lecture 9(contd.)
where Rm = monthly runoff in cm and Rm ≥ 0
Pm= monthly rainfall in cm
Lm = monthly losses in cm
Tm = mean monthly temperature of the catchment in 0C
For Tm ≤ 4.5 0C, the loss Lm may provisionally be assumed as
T0C 4.5 -1 - 6.5
Lm 2.17 1.78 1.52
(cm)
Important : Monthly loss is higher than monthly Rainfall or
precipitation then monthly Runoff will be equal to zero, i.e.
Rm = 0 when Lm > Pm
Monthly Annual runoff = ∑Rm

11. 11
Lecture 9(contd.)
Example 6-3
For a catchment,the mean monthly rainfall and
temperatures are given. Estimate the annual runoff and
annual runoff coefficient by Khosla’s formula.
Month Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec
Temp0C 12 16 21 27 31 34 31 29 28 29 19 14
Rainfall
(cm)
4 4 2 0 2 12 32 29 16 2 1 2
Solution:
Given ,
Rainfall & temp. of January to December is mentioned in the table.
To be estimated: Total annual Runoff ∑Rm & Runoff coefficient

12. 12
Lecture 9(contd.)
If the loss (Lm) is higher than the Pm then Rm is taken to be zero.
We know,
Rm = Pm –Lm
and Lm = 0.48 Tm for Tm > 4.50C
Estimation of Runoff for the month of January:
We know,
Rm = Pm –Lm
= 4 – Lm
Lm = 0.48 * 120C (Values from Data Table)
= 5.76 cm
Rm = 4 –Lm
Rm = 4 – 5.76 = - 1.76 ; As per condition Rm = 0

13. 13
Lecture 9(contd.)
Month Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec Total
cm
T0C 12 16 21 27 31 34 31 29 28 29 19 14 -
Lm = 0.48Tm
(cm)
5.76 7.68 10.08 12.96 17.76 16.32 14.88 13.92 13.44 13.92 9.12 6.72 -
Rainfall
(cm)
4 4 2 0 2 12 32 29 16.0 2 1 2 106
Runoff (cm)
Rm =Pm –Lm
0 0 0 0 0 0 17.22 15.08 2.56 0 0 0 34.9
Similarly estimation of Runoff for remaining months
Estimation Table
Annual Runoff coefficient
r = (Total Annual Runoff) / (Total Annual Rainfall)
r = 34.9/ 106 = 0.329 , say 0.33

14. 14
Lecture 9(contd.)
HYDROGRAPH
Definition of Hydrograph
A hydrograph is the graphical representation of the discharge
flowing in a river at the given location with the passage of time.
A hydrograph or a runoff : It is thus a plot between time (on (X-axis),
and discharge (on Y-axis) as shown in the following figure-1.Dischargeintheriver
Recessional curve of the hydrograph
Time
Figure-1 : Typical hydrograph

15. 15
A hydrograph represents
 discharge fluctuation in the river at a given site
over a given time period
 indicates the peak flow and can help to design the
hydraulic structure at site
The characteristics region (elements) of hydrographs
Peak flow
The maximum flow in the river due to any given storm is known as
the peak flow. The peak flow will be different for different kinds of
storms.
Base flow
The ground water inflow is called base flow or the flow in a
channel due to soil moisture or ground water is called base flow.
Lecture 9(contd.)

16. 16
Lecture 9(contd.)
DischargeQ(m3/s)
Figure 7-2 : Elements of flood hydrograph
Lag time TLTC
Storm
● A
C
B●
●
P●
● D
TB Time in hours.
tpk
QP
Elements
1. The rising limb-AB:
The joining point A ( the
starting point of the
rising curve) to the point
B (the point of inflection)
2. The crest segment-
BPC: between the two
points of inflection B with a
peak P to the deplection
point or the falling limb
Point C or depletion curve
CD– starting from the point
of depletion C.

17. 17
Lecture 9(contd.)
3. The falling limb or depletion curve CD – starting from the
point of depletion C.
4. Peak time( tpk)– The time to peak from the starting point A,
5. TL (Lag time) : The time interval from the centre of mass of
rainfall to the centre of mass of hydrograph called lag time(TL).
6. Qp- peak discharge : The highest discharge flowing in the
river is peak discharge.
7. The time base of hydrograph-TB -time from the starting
point A to the end of depletion point D .

18. 18
Lecture 9(contd.)
Factors Affecting hydrograph
Sl.No Physiographic factors Climatic factors
1. Basin characteristics: Storm characteristics:
a. shape
b. size
c. slope
d. nature of the valley
e. elevation
f. drainage density.
a. precipitation
b. duration
c. magnitude and
d. movement of storm
2. Infiltration characteristics: Infiltration
characteristics:
a. land use and cover
b. soil type and geological conditions
c. lakes, swamps and other storage
2. Initial loss
3. Channel characteristics: cross-section,
roughness and storage capacity.
3. Evapotranspiration.

19. 19
Components of hydrograph
The components of hydrograph are three(3).They are:
(1) the rising limb
(2) the crest segment
(3) the recession limb
(1) The Rising Limb
The rising limb of a hydrograph also called concentration curve
represents the increase in discharge due to the gradual building
up of storage in channels and over the catchments surface. The
initial losses and high infiltration losses during the early period of
a storm cause the discharge to rise rather slowly in initial periods.
Lecture 9(contd.)

20. 20
As the storm continues, more and more flow from distant parts
reach the basin outlet. Simultaneously the infiltration losses also
decrease with time. Thus under a uniform storm over the catchment
the runoff increases rapidly with time. Hence the basin and storm
characteristics control the rising limb of the hydrograph.
Crest segment
The most important part of hydrograph is the crest segment
because it contains peak flow. The peak flow occurs when the runoff
from various parts of the catchments simultaneously contribute the
maximum amount of flow at the outlet of the basin. Generally for
large basin/catchments, the peak flow occurs after the cessation of
rainfall, the time interval from the center of mass of rainfall to the
peak being essentially controlled by basin and storm characteristics.
Multiple- peaked complex hydrographs in a basin can occur when
two or more storms occur in close succession.
Lecture 9(contd.)

21. 21
Lecture 9(contd.)
(3) The Recession limb
The recession limb which extends from the point of inflection at the end
of the crest segment to the commencement of the natural groundwater
flow represents the withdrawal of water from the storage built up in the
basin during the earlier phases of the hydrograph. The starting point of
the recession limb, i.e. the point of inflection represents the condition of
maximum storage. Since the deflection of storage takes place after the
creation of rainfall, the shape of this part of the hydrograph is
independent of storm characteristics and depends entirely on the basin
characteristics.
The storage of water in the basin exists as :
 surface storage
interflow storage
ground water storage/base flow storage.

22. 22
Lecture 9(contd.)
Barnes (1940) showed that the recession of a storage can be expressed
as
Qt = Q0 Kr
t .............................Eq -1
Where, Qt = discharge at a time interval t days
and Q0 = initial discharge.
Kr = recession constant at time interval
t days value < unity
Eq- 1 can be represented as in the form of exponential decay as
Qt = Q0 e-at .........................Eq-2
a = - ln Kr

23. 23
Where the recession constant Kr can be considered to be made
up of three components to take care of the three types of
storages :
Kr = Krs . Kri . Krb .......................Eq-3
Where.
Krs = recession constant for surface storage
Kri = recession constant for interflow
Krb = recession constant for base flow
.
Lecture 9(contd.)

24. 24
The values of recession constants are as follows for time t in days:
Krs = 0.05 to 0.20
Kri = 0.50 to 0.85
Krb = 0.85 to 0.99
If the interflow is not significant Kri can be assumed to be unity.
When Eq-1 or Eq-2 is plotted on a semi-log paper with
discharge on the log-scale, it plots as a straight line and
from this the value of Kr can be found. The values of Kr
can be calculated from storage recession curve
Lecture 9(contd.)

25. 25
Example 7-1
The recession portion of a flood hydrograph is given below.
The time is indicated from the arrival of peak. Assuming the
interflow component to be negligible, calculate the base flow
and surface flow recession coefficients,
Lecture 9(contd.)
Time from
peak (days)
Discharge
(m3/s)
Time from
peak (days)
Discharge
(m3/s)
0.0 90 3.5 5.0
0.5 66 4.0 3.8
1.0 34 4.5 3.0
1.5 20 5.0 2.6
2.0 13 5.5 2.2
2.5 9 6.0 1.8
3.0 6.7 6.5 1.6
7.0 1.5

26. 26
DischargeQinm3/s
Recession curve
Figure 7-5: Storage recession curve –for the
above example
Solution:
Given
Assumption- interflow
component negligible.
To be calculated base
flow and surface flow
recession coefficients,
The problem will be
solved by using
recession curve as
Figure 7-5.
Lecture 9(contd.)

27. 27
Lecture 9(contd.)
The given data are plotted on a semi-log paper with discharge on the log-
scale as per the above figure 7-5. The straight line AB of the curve
indicates base flow. The surface flow terminates at B – 5 days after peak.
Since we know, at the time elapse t , discharge Qt
Qt = Q0 Kr
t,
Hence base flow recession coefficient Krb
t
Krb
t = Qt / Q0
Taking log in both sides;
log Krb
t = log (Qt / Q0)
t log Krb = log (Qt / Q0)
log Krb = 1/t log (Qt / Q0)

28. 28
Lecture 9(contd.)
The base flow recession is shown by line ABM figure 7-5. From the
figure initial discharge at 1 day after the peak ; from curve
Q0= 6.60 m3 / s
and time interval of t = 2 days, i.e. 3 days after the peak, the discharge
Qt from the curve = 4.0 m3/s . Thus putting the values: we get
log Krb = 1/t log(Qt / Q0)
log Krb = log(Qt / Q0) 1/t
Krb = (Qt / Q0) 1/t
= ( 4.0/6.6) 1/2 = 0.778, say Krb = 0.78

29. 29
Lecture 9(contd.)
Surface runoff is calculated by subtracting base flow from the
storage recession curve. Then plot surface runoff curve PA with the
surface runoff values as calculated shown in the following table.
The above curve shows the surface runoff depletion plot as a straight line.
Time
after
Peak
Values from storage
recession curve PA
(m3/sec)
Values of Base flow
from base flow
recession ABM line
(m3/sec)
Surface runoff =
col 2 –col 3
(m3/sec)
1 2 3 4
0.5 66.0 8.3 57.7
1.0 34.0 8.0 26.0
1.5 20.0 5.5 14.5
2.0 13.0 4.5 8.5
2.5 9.0 4.3 3.7
3.0 6.7 4.35 2.25
3.5 5.0 2.8 2.2
4.0 3.8 2.4 1.6
4.5 3.0 2.3 0.70
5.0

30. 30
Lecture 9(contd.)
From the surface runoff curve, initial discharge- Q0 = 26
m3/s at time t=1.0 h and the corresponding Qt = 2.25
m3/s at t = 3.0 h i.e. at an time interval 2 hours.
Hence the recession constant for surface storage Krs is
given by
log Krs = ½ log(2.25 / 26)
= log(2.25 / 26) ½l
Krs = 0.29
Ans :
Krb = 0.78 and Krs = 0.29