Hydrological Forecasting
Introduction:
How to use knowledge to predict from existing data,
what will happen in future?.
This is a fundamental problem of all engineering design.
Water events are usually random in nature
The hydrologist is frequently asked
what the maximum possible discharge of a particular river will be.
The only answer that can be given is that :
From the data available, and making various assumptions,
it would appear that a certain value will not be exceeded
on average more than once in a specific number of years
Flood Formulae
Catchment area formulae.
The particular random variable of river flood discharge have been
proposed to define the 'maximum flood' that could occur for
a particular catchment.
The formulae are empirical by nature, derived from
observed floods on particular catchments and usually of the form:
Q = CAn
Where Q =
A=
n=
C=
flood discharge in m3/s (or ft3/s).
catchment area in km2 (or mile2).
an index usually between 0.5 and 1.25
a coefficient depending on climate, catchment and units.
The Rational Method.
The introduction of rainfall into a formula might be expected to improve it,
bearing in mind the type of relationship which exists between
rainfall and runoff
This kind of direct relationship of runoff to rainfall depths has been used
in the past to determine flood discharges.
Mulvaney was the first to propose the idea in his work on Irish
arterial drainage.
It was also the basis of the Lloyd-Davis method of sewer
designs and the Bransby-Williams estimating method for floods in India.
Its use has persisted to the present because of its simplicity.
The formulae are all of the form
Qp = CiA
Where
i=
A=
C=
QP =
tc =
rainfall intensity in a time t
catchment area
a dimensionless runoff coefficient, whose
value depends on the catchment characteristics.
peak discharge due to the particular rainstorm,
and assumed to occur after time tc when the whole catchment area
is contributing
time of concentration, i.e. the time taken for
rain falling on the catchment farthest from the gauging
station to arrive there.
Bransby-Williams gave a formula for the design rainfall
duration D (in hours):
D 
L
a 

5 

d
 h 
2
tc
Where
L=
D=
A=
greatest distance from the edge of catchment to the outfall
is the diameter of a circle of area equal to catchment
area (L/d is, therefore, a dimensionless coefficient of circularity)
catchment area in square miles
h=
tc =
the channel slope (as a percentage) along its greatest length
time of concentration in hours.
Probability of the N-year event.
The term recurrence interval (also called the return period period),
denoted by Tr, is the time that,
on average, elapses between two events that equal or exceed a particular level.
Putting it another way, t
he N-year event, the event that is expected to be equalled or exceeded,
on average,
every N years, has
a recurrence interval, Tr of N years.
Let the probability P(X <x) represent the probability that x will not be equaled
or
exceeded in a certain period of time.
Then P(X < x)n will represent the probability that x will not be equalled
or
exceeded in n such periods.
For an independent series and from the multiple probability rule :
P(X < x)n = [P(X < x)]n
=
[1 - P(X > x)]n
Therefore :
P(X > x)n = 1 – [ 1 – P(X > x)]n
Now:
Tr

1
P(X  x)
Then :
P(X  x) n

1
 1 - 1 

Tr 

n
So, for example, the probability of X > x, where x is the value of a flood
with
return period of 20 years, occurring in a particular 3-year period is :
P(X > 20 yr flood)3 =
=
=
=
1 - [ 1-
1 3
]
20
1 – [0.95]3
1 – 0.857
0.143 or 14.3 %
Table 1 Shows the probability (%) of the N-year flood occurring in
a particular period
Number of
years in
period
1
2
3
5
10
20
30
60
100
200
500
1000
5
20
36
49
67
89
99
99.9
—
—
—
—
—
N = Average return period Tr (years)
10
20
50
100
200
500
10
19
27
41
65
88
96
99.8
—
—
—
—
5
10
14
23
40
64
78
95
99.4
—
—
—
2
4
6
10
18
33
45
70
87
98.2
—
—
1
2
3
5
10
18
26
43
63
87
99.3
—
0.5
1
1.5
2.5
5
10
14
26
39
63
92
99.3
0.2
0.4
0.6
1
2
4
6
11
18
33
63
86
1000
0.1
0.2
0.3
0.5
1
2
3
6
10
18
39
63
To find design period of a temporary project
(n) :
)n = (
Since :
1 - P(X > x)n = (1 –
1 n
)
Tr
Log (1 - P(X > x)n) = n log (
Therefore:
n=
Tr - 1
)
Tr
 log (1 - P (X  x) n ) 




log( Tr  1)


Example: 1
How long may a cofferdam remain in a river, with an even chance of not being
overtopped, if it is designed to be secure against a 10-year flood?
Here, the policy ruling is that there should be an even chance, so P(X > x)n =
0.50 and Tr= 10 then
n=
log (1 - 0.5)
log 0.5 1.699 0.301



9
log 0.9 1.954 0.046
log
10
= 6.5 years
Determining the magnitude of the N-year) event by plotting.
Having listed a series of events (for example, maximum floods) accorded a
ranking m,
starting with; m = 1 for the highest value, m = 2 for the next highest and so
on in descending order.
The recurrence interval Tr can now be computed from one of a number of
formulae:
n 1
Tr 
m
Where:
m = event ranking ; and
n = number of events.
But there are objections to its use because of the bias it
introduces to the largest events in a short series.
The probability P of an N-year event of return period Tr is:
100
P 
Tr
(%)
Cases of Q-Tr Graphs:
1. Q against Tr using linear co-ordinates (Figure 1).
Extrapolation of the curve to the Tr values of Tr depends
critically on the few highest points.
Figure 1: Annual maximum mean daily flows of the
River Thames at Teddington, 1883 - 1988
2. Q (linear) against Tr (logarithmic) (Figure 2).
This yields a straight line fitted to all but the lowest values. Although
extrapolation is simpler, unless Tr follows a logarithmic law,
extrapolation is not necessarily more accurate than for Figure 1.
Figure 2 : Annual maximum mean daily flows of the
River Thames at Teddington, 1883 – 1988 [semi-log]
3. Q (linear) against probability (per cent) (Figure 3).
As often happens, flood series points lie on a shallow curve on
probability paper (where a normal distribution of probability is
assumed).
Figure 3 : Annual maximum mean daily flows of the River
Thames at Teddington, 1883-1988 (normal probability)
4. Q (logarithmic) against probability (per cent) (Figure 4).
The curve of figure 3 is now transformed to a straight line. A variation of the
approach in figure 3 is to assume that the logarithm of the variate Q is normally
distributed, leading to the use of logarithmic-normal distribution (or lognormal paper).
Figure 4 : Annual maximum mean daily flows of the
River Thames at Teddington, 1883-1988 (log-normal)
Analytical Methods :
Other investigators have proposed methods assuming other theoretical
frequency distributions.
Gumbel used extreme-value theory (EVl) to show that in a series of extreme
values Xx, X2… Xn where the samples are of equal size and X is an
exponentially distributed variable (for example, the maximum discharge
observed in a year's gauge readings), then the cumulative probability P' that
any of the n values will be less than a particular value X (of return period T)
approaches the value
P' = e-e-y
where e is the natural logarithm base,
and

y = - ln  ln 

 1 
1   
 Tr  
That is, P' is the probability of non-occurrence of an
event X in T years, or
T=
1
1  P'
(Note that this argument refers to Gumbel's method.
(The reader should not confuse this with the normal
usage of Tr = 1/P
where P - probability of occurrence.)
The event X, of return period T years, is now defined as QT, and :
QT = Qav + (0.78y – 0.45)
Where
Qav = average of all values of 'annual flood' Qmax
 = standard deviation of the series.
Or :
 =
n   Q2max 2 

- Qav 
n 1 n

where
n = number of years of record = number of Qmax values
Qmax = sum of the squares of n values of Qmax
Table 2 : values of y as a function of T.
T(years)
y
T(years)
y
1.01
-1.53
100
4.60
1.58
0.00
200
5.30
2.0
0.37
300
5.70
5.0
1.50
400
5.99
10.0
2.25
500
6.21
20
2.97
1000
6.91
50
3.90
1000
9.21
Plotting on Gumbel Paper :
Q = Qav + 124.56( 0.78 x5.30 – 0.45) = 778 m3/s
Q = Qav + 124.56( 0.78 x4.60 – 0.45) = 710 m3/s
200
100
Determining the magnitude of the N-year event by
calculation.
Although the use of a normal probability distribution has been used above to
plot events, and hence to extrapolate for rare values that may be used in design,
values of particular probabilities can be calculated since a normal distribution
curve is defined by only two parameters( the mean and standard deviation).
Accordingly, to determine the specific discharge associated with a particular
probability of occurrence r in an annual series that is normally distributed, it is
necessary to compute only
Qr = Qav + K
where
 = Standard deviation
and
K
is listed in Table 3.
Example 2 :
Determine by calculation the mean daily discharge of the River
Thames at Teddington with a 100-year return period, assuming the
annual series normally distributed.
Since :
Qav = 329.7 m3/s and  = 133.8 m3/s.
For Tr = 100 years, P is 1.0 per cent;
and from Table 2: K = 2.33
Therefore :
Q100 =329.7 + (2.33 x 133.8) = 641 m3/s
Table 3 :
Probability of
exceedance (%)
0.1
0.5
1.0
2.5
5
10
15
20
25
30
35
40
45
50
Values of a normal distribution
K
Probability of
exceedance (%)
K
3.09
2.58
2.33
1.96
1.645
1.28
1.04
0.84
0.67
0.52
0.385
0.25
0.13
0.00
50
55
60
65
70
75
80
85
90
95
97.5
99.0
99.5
99.9
0.00
-0.13
-0.25
-0.385
-0.52
-0.67
-0.84
-1.04
-1.28
-1.645
-1.96
-2.33
-2.58
-3.09