Engineering Hydrology; Lecture Note

CHAPTER FOUR

FLOOD ROUTING

4.1 General
At a river gauging station, the stage and discharge hydrographs represent the passage of waves of
river depth and stream flow during flood, respectively. As this wave moves down the river, the
shape of the wave gets modified due to various factors, such as channel storage, resistance,
lateral addition or withdrawal of flows etc. when a flood wave passes through a reservoir, its
peak is attenuated and the time base is enlarged (translated) due to the effect of storage. Flood
waves passing down a river have their peaks attenuated due to friction if there is no lateral
inflow. In both reservoir and channel conditions the time to peak is delayed, and hence the peak
discharge is translated.

Flood routing is the technique of determining the flood hydrograph at a section of a river by
utilizing the data of flood flow at one or more upstream sections. The hydrologic analysis of
problems such as flood forecasting, flood protection, reservoir and spillway design invariably
include flood routing. In these applications two broad categories of routing can be recognized.
These are:
i) Reservoir routing and
ii) Channel routing

In reservoir routing the effect of a flood wave entering a reservoir is studied. Knowing the
volume-elevation characteristics of the reservoir and the out flow elevation relationship for
spillways and other outlet structures in the reservoir; the effect of a flood wave entering the
reservoir is studied to predict the variation of reservoir elevation and out flow discharge with
time. This form of routing is essential (i) in the design of the capacity of spillways and other
reservoir outlet structures and (ii) in the location and sizing of the capacity of reservoirs to meet
specific requirements.

In channel routing the changes in the shape of a hydrograph as it travels down a channel is
studied. By considering a channel reach and an input hydrograph at the upstream end, this form
of routing aims to predict the flood hydrograph at a various sections of the reach. Information on
the flood-peak attenuation and the duration of high-water levels obtained by channel routing is
utmost importance in flood forecasting operations and flood protection works.

A variety of flood routing methods are available and they can be broadly classified in to two
categories as: (i) hydraulic routing and (ii) hydrologic routing. Hydrologic routing methods
employ essentially the equation of continuity and a storage function, indicated as lumped
routing. Hydraulic methods, on the other hand, employ the continuity equation together with the
equation of motion of unsteady flow. The basic differential equations used in the hydraulic
routing, known as St. Venant equations afford a better description of unsteady flow than
hydrologic methods.

A flood hydrograph is modified in two ways as the storm water flows downstream. Firstly, and
obviously, the time of the peak rate of flow occurs later at downstream points. This is known as

~1~
Lecture Note
 Engineering Hydrology; Lecture Note

translation. Secondly, the magnitude of the peak rate of flow is diminished at downstream
points, the shape of the hydrograph flattens out, and the volume at the floodwater takes longer to
pass a lower section. This modification of the hydrograph is called attenuation.

Time
Figure 4.1: Flood translation and attenuation

4.2 Simple Non-storage Routing

Relationship between flood events and stages at upstream and downstream points in a single
river reach can be established by correlating known floods and stages at certain conditions. The
information could be obtained from flood marks on river banks and bridge sides.
Measurements/estimates of floods can then be related to known the level of the flood at the
upstream and downstream locations. With such curves it is possible to give satisfactory forecasts
of the downstream peak stage from an upstream peak stage measurement

.
Peak H downstream (m)
Figure 4.2: Peak stage relationship

The time of travel of the hydrograph crest (peak flow) also need to be determined to know the
complete trace of modification of the wave. Curves of upstream stage plotted against time travel
to the required downstream point can be compiled from the experience of several flood events.

~2~
Lecture Note
 Engineering Hydrology; Lecture Note

Peak travel time to downstream point

Figure 4.3: Flood Peak travel time

The complexities of rainfall-runoff relationships are such that these simple methods allow only
for average conditions. Flood events can have very many different causes that produce flood
hydrographs of different shapes.

The principal advantages of these simple methods are that they can be developed for stations
with only stage measurements and no rating curve, and they are quick and easy to apply
especially for warning of impending flood inundations when the required answers are
immediately given in stage heights.

4.3 Storage Routing

When a storm event occurs, an increased amount of water flows down the river and in any one
short reach of the channel there is a greater volume of water than usual contained in temporary
storage. If at the beginning of the reach the flood hydrograph is (above normal flow) is given as
I, the inflow, then during the period of the flood, T1, the channel reach has received the flood
volume given by the area under the inflow hydrograph. Similarly, at the lower end of the reach,
with an outflow hydrograph O, the flood is given by the area under the curve. In a flood situation
relative quantities may be such that lateral and tributary inflows can be neglected, and thus by
the principle of continuity, the volume of inflow equals the volume of outflow, i.e. the flood
T1 T3

volume. V =∫ Idt=∫ Odt

0 T2
T1 T

At intermediate time T, an amount ∫ Idt has entered the reach and an amount∫ Odt has left the
0 0
reach. The difference must be stored within the reach, so the amount of storage, S, within the
T
reach at time t = S=∫ ( I −O)dtT is given by
0

The principle of hydrologic flood routings (both reservoir and channel) uses the continuity
equation in the form of “Inflow minus outflow equals rate of change of storage”.
ds
=I t −O t
dt
Where:
I = Inflow in to the reach

~3~
Lecture Note
 Engineering Hydrology; Lecture Note

O= Outflow from the reach

dS/dt =Rate of change of storage within the reach. Alternatively, the continuity (storage)
equation can be stated as in a small time interval ∆t the difference between the total inflow
volume and total outflow volume in a reach is equal to the change in storage in that reach, i.e.,
Í ∆ t−Ó ∆t=∆ S
Where, I = average inflow in time ∆t
O = average outflow in time ∆t
∆S = change in storage
∆t = routing period or the equation can be rewritten as:
S i+1−Si 1 1
= ( I i+1 + I i )− ( O i+1 +O i )
∆t 2 2

The time interval ∆t should be sufficiently short so that the inflow and out flow hydrographs can
be assumed to be straight line in that interval. As a rule of thumb ∆t ≤1/6 of the time to peak of
the inflow hydrograph is required.

The continuity equation (I-Q = dS/dt), forms basis for all the storage routing methods. The
routing problem consists of finding Q as a function of time, given I as a function of time, and
having information or making assumptions about storage, S.

4.4 Reservoir or level pool routing

The passage of the flood wave through a reservoir or a river reach is an unsteady flow
phenomenon. In hydraulics, it is classified as a gradually varied flow. The equation of continuity
used in all the hydrologic routing methods as the primary equation, states that the difference
between the inflow and outflow rate is equal to the rate of change of storage; i.e.
I – O = S---------- (1) Where: I = inflow rate
O = out flow rate
S = Storage

Alternatively, in a small time interval t, the difference between the total inflow volume and the
total outflow volume is equal to the change in storage volume;

I¯t - O¯t = S ----------- (2) Where I¯ = average inflow rate in time t

O¯ = average outflow rate in time t
S = change in storage during the time t
Since, I¯ = I1 + I 2 O¯ = O1 + O2
2 2

S = S2 - S1 Where, suffixes 1 and 2 denote the beginning and the end of the time interval t.

I 1 +I 2 0 1 +02
The above equation can be written as:
( ) (
2
∇ t−
2 ) ∇ t=S2 −S1 −−−−−−(3 )

~4~
Lecture Note
 Engineering Hydrology; Lecture Note

The time interval t should be sufficiently short, so that the inflow and outflow hydrograph can
be assumed to be in straight line in that time interval. Moreover, t must be shorter than the time
of transit of the flood wave through the reservoir or the given river reach.

The above relationship seems to be very simple, but its evaluation is not easily possible without
drastic simplifying assumptions. This is because of the fact that the relations between time and
rate of inflow, elevation and storage of reservoir, and elevation and rate of outflow cannot be
expressed by simple algebraic equations. They are, respectively represented by the:
(i) inflow- flood hydrograph
(ii) the elevation - storage curve and
(iii) the outflow - elevation curve
The first two curves obviously cannot be represented by any simple equation and the third may
be represented by the spillway discharge equation (Q = 1.71 LH 3/2) only if, the discharge through
other outlets is neglected. If the discharge through the outlets is also not neglected, then all the
three curves will be unnameable to simple mathematical treatment, without drastic assumptions.
Several procedures however, been suggested by different investigators to solve the above basic
equation (Equ. (3)) by recharging the components in different manners. Depending upon the
different procedures adopted for solving the above basic equation, the following hydrologic
method may be used for reservoir routing.

Modified Pul's method of Reservoir Routing

In this method of reservoir routing, the basic equation

I 1+ I 2 0 +0
[ ] [ ]
2
Δt− 1 2 Δt=S 2−S1 −−−(1)
2 can be rearranged as :

I 1+ I 2 0 1 Λt 02 Δt
[ ] [ ][
2
Δt+ S 1 −
2
= S2 +
2 ]
−−−−−−−−−−2

I1, I2, S1 and O1 are known at the first time step t. Hence, the value of the function (S 2 + 02t/2) at
the end of the interval t can be calculated by evaluating the left hand side of equation (2).
All the factors arranged on the left hand side of this equation are known at the beginning of a time
Since storage (S) at different elevation (h) and outflow (0) at different elevations (h) are known,
the fraction (S2 +02t/2) at different elevations can be computed in the beginning.

A curve between (S2+02t/2)/ Vs. elevation (h) can thus be prepared and is used to read out the
reservoir elevation corresponding to the above computed values of (S 2 + 02t/2)/, so as to know
the value of the reservoir elevation at the end of the interval t.

~5~
Lecture Note
 Engineering Hydrology; Lecture Note

Corresponding to this computed reservoir elevation at the end of the interval, the outflow 0 2 at the
end of the interval can finally be read out from the outflow elevation curve. The procedure can be
repeated till the entire inflow hydrograph is analyzed to compute the outflow.

Steps involved in Computations

i. From the known storage (s) elevation data and of discharge (outflow) elevation data,
prepare curve I, between (S2 + O2t/2) Vs. Elevation. t may be any chosen time
interval, approximately 20 to 40 % of the time of rise of the inflow hydrograph.
ii. On the same plot, draw curve 2, between outflow discharge Vs, Elevation.
iii. The storage, elevation and outflow discharge at the start of routing (step I) are known to
help in computing the value of (I1 + I2)/2t + (S1 – O1t/2), which will indicate the
computed value of function (S2 + O2t/2)
iv. The computed value of function (S2 + O2t/2) is used to read out the corresponding value
of elevation from curve 1, and will indicate the reservoir elevation at the end of the
interval.
v. The above value f reservoir elevation (at the end of the interval) is used on curve 2 to
readout the corresponding outflow (at the end of the interval) is used on curve2, to read
out the corresponding out flow (at the end of the interval).
vi. Deducting O2t from the computed value of (S2 + O2t/2), we can compute the value of
(S2 - O2t/2) which can be used as the value of (S1 – O1t/2) for the start of the next time
step.
vii. The procedure is repeated till the entire inflow hydrograph is routed.

Figure 4.4 Reservoir routing

Example: A reservoir has the following elevation, discharge and storage relationships:

~6~
Lecture Note
 Engineering Hydrology; Lecture Note

Elevation (m) Storage (Mm3) mcm Outflow discharge (m3/s)

300.20 5.5 0 When the reservoir
300.70 6.0 15 level was at
301.20 6.6 40 300.20m, the
301.70 7.2 75 following flood
302.20 7.9 115 hydrograph entered
302.70 8.8 160 the reservoir:
Time (h) 0 3 6 9 12 15 18 21 24 27
i Discharge (m3/s) 10 20 52 60 53 43 32 22 16 10
Route the flood and obtain:
a) The out flow hydrograph; and
b) The reservoir elevation Vs time curve during the passage of the flood wave.
Solution: Let us choose a time interval of 3 hr, which is the interval given in inflow discharges.
This value is within 20 to 40% of the period of rise, which is 9 hrs.

t = 3 hrs = 3x60x60 sec = 1.08 x 104 sec = 0.0108 x 106 sec.
To use the modified pull's method, we have to draw curve between elevation (h) V s outflow, say
curve 2; and (S2+O2t/2) vs. Elevation (h) say curve 1. To draw curve we have to compute the
value of function (S2+O2t/2) for different elevations (h) as shown in col.5 of the following table.
Table: 1
Elevation Storage O, Outflow discharge O.t = 0.0108 x 106 (S +O.t)
(m) (m cm) (m3/s) 2 2 2 (mcm)
ie. col (3) x 0.0054 col. (2) + col. (4)
(mcm)
1 2 3 4 5
300.20 5.5 0 0.0 5.50
300.70 6.0 15 0.08 6.08
301.20 6.6 40 0.22 6.82
301.70 7.2 75 0.41 7.61
302.20 7.9 115 0.62 8.52
302.70 8.8 160 0.86 9.66

Table: 2 Flood routing through reservoir

Time(hr) Inflow I1+I2 (I1+I2)t Ot S1 - Ot S2+ Ot Elevation Outflow
(m3) 2 2 (mcm) 2 2 (m) (m3/s)
(m3/s) (mcm) (mcm) (mcm)
0 10 300.20 0
15 0.162 0x0.0105=0 5.50+0.162=5.662
3 20 300.35 4
36 0.589 4x.00108=0432 5.662-00432=5.6188 5.6188+0.389=6.007
6 52 300.60 13
56 0.605 13x0.0108=0.1404 6.007-0.1404=5.866 5.866+0.605=6.471
9 60 300.98 28
56.5 0.610 28x0.0108=0.3024 6.471-0.3024=6.168 6.168+0.610=6.778
12 53 301.18 38
48 0.518 38x0.0108=0.4104 6.778-0.4104=6.3680 6.368+0.518=6.886
15 43 301.25 44
37.5 0.405 44x0.00108=0.4752 6.886-0.4752=6.411 6.411+0.405=6.815

~7~
Lecture Note
 Engineering Hydrology; Lecture Note

18 32 301.20 40
27 0.292 40x0.0108=0.4320 6.815-0.432=6.383 6.383+0.292=6.675
21 22 301.10 34
19 0.205 34x0.0108=0.3672 6.675-0.3672=6.308 6.308+0.205=6.513
24 16 300.98 28
13 0.140 28x0.0105=0.3024 6.513-0.3024=6.211 6.211+0.140=6.350
27 10 300.88 24
A graph is now plotted between the value col (1) and col (5) of the above table as curve 1, as
shown in Fig. 1. Similarly, a curve (curve 2) is plotted between the values of col (1) and col (3)
as to give outflow elevation curve. The requisite routing calculations are now carried out in table 2.
From the computed values of outflow discharge in col (9) of table 2, outflow hydrograph can be
plotted over inflow hydrograph as shown in Fig. 2.
A graph between the reservoir elevation Vs time from values of col. 8 and 1 can also be plotted to
obtain the reservoir elevation Vs time curve, as desired in the question and as shown in Fig. 3

Elevation
303
302.5
302
301.5
301
300.5
300 Elevation
299.5
299
298.5
t) )
5.
5 08 82 61 52 66
O .D cm 6. 6. 7. 8. 9.
+ (m
(S 2

Figure 1 Elevation vs S curve

Outflow hydrograph
50
45
40
35
30 Outflow hydrograph
25
20
15
10
5
0
0 2 4 6 8 10 12 14

~8~
Lecture Note
 Engineering Hydrology; Lecture Note

Figure 2 outflow inflow hydrograph curve

Elevation
350

300

250

200 Elevation

150

100

50

0
0 2 4 6 8 10 12 14

Figure 3 Elevation Vs time curve

Example - 2: A reservoir has the following elevation, discharge and storage relationships:

Elevation Storage Outflow discharge

(m) (mcm) (m3/s)
50.00 4.000 0.0
50.50 4.062 2.2
51.0 4.230 5.2
51.50 4.434 8.5
52.00 4.640 12.2
52.50 4.900 16.8
53.00 5.152 20.8
53.50 5.400 26.6
54.00 5.700 33.0
54.50 5.944 40.0
55.00 6.200 47.2
55.50 6.280 55.2
56.00 6400 64.0
56.50 6.565 72.8
57.00 6.730 82.4
When the reservoir level was 50.50m, the following flood hydrograph entered the reservoir:
Time(hr) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
Inflow 2.0 6.8 37.0 53.8 57.7 53.3 46.5 36.0 18.0 10.75 7.0 4.7 3.3 2.7 2.4 2.4
m3/s
Route the flood and obtain the outflow hydrograph and the reservoir elevation V s time curve
during the passage of the flood wave.

~9~
Lecture Note
 Engineering Hydrology; Lecture Note

Solution: Let us choose a time interval t of 2 hrs which is the interval given in inflow discharges.
This value is within 20 to 40% of the period of rise, which is 8 hrs.

 t = 2 hrs = 2x60x60 = 7200 sec= 0.0072 x 106 sec.

Table-1 Data on elevation, discharge and storage

Elevation (m) Storage (mcm) 0. t = 0x0.0072x106/2 S+0xt/2

Outflow, 0 (m3/s)
2 (mcm)
ie. col 3x0.0036 (mcm) col.2+col.4
1 2 3 4 5
50.00 4.000 0.0 0 4.000
50.50 4.062 2.2 0.00792 4.070
51.00 4.230 5.2 0.01872 4.249
51.50 4.434 8.5 0.03060 4.465
52.00 4.640 12.2 0.04392 4.484
52.50 4.900 16.8 0.06048 4.960
53.00 5.152 20.8 0.07488 5.227
53.50 5.400 26.6 0.09576 5.496
54.00 5.700 33.0 0.11880 5.819
54.50 5.944 40.0 0.14400 6.088
55.00 6.200 47.2 0.16992 6.370
55.50 6.280 55.2 0.19872 6.479
56.00 6.400 64.0 0.23040 6.630
56.50 6.565 72.8 0.26208 6.827
57.00 6.730 82.4 0.29664 7.027
To use the modified Pul's method, we have to draw curve between elevation (h) V s outflow, say
curve 2, and (S+ 0. t/2) Vs elevation (h) say curve 1. To draw curve 1, we have to compute the
value of function (S+ 0. t/2) for different elevation (h) as shown in col. 5 of the above table.
Table - 2 Flood Routing through a reservoir
Time Inflow I1+I2 I1+I2 x. t 0 x t S1- 0 x. t S2+ 0 x. t Elevation Outflow
(hr) I 2 2 (mcm) 2 2 (m)
(m3/s) (m3/s) (mcm) (mcm) (mcm)

0 2.00 50.50 2.20

4.40 0.032 2.2 x t/2 4.062 - . 4.054+0.032=4.0
=.00792 00792=4.054 86
2 6.80 50.68 3.30
21.90 0.237 3.3 x t 4.086- 4.062+0.237=4.2
=.02376 0.02376=4.062 99
4 37.00 51.15 6.60
45.40 0.490 6.6 x t 4.299- 4.256+0.490=4.7
=0.04752 0.04752=4.256 46
6 53.80 52.10 13.30
55.75 0.602 13.3xt 4.746- 4.650+0.602=5.2
=0.09576 0.09576=4.650 52

8 57.70 52.96 20.80

55.50 0.599 20.8Xt 5.252- 5.102+0.599=5.7
=0.14976 0.14976=5.102 01

~ 10 ~
Lecture Note
 Engineering Hydrology; Lecture Note

10 53.30 53.82 30.60

49.90 0.539 30.6xt 5.701- 5.481+0.539=6.0
=0.22032 0.22032=5.481 20
12 46.50 54.36 38.00
41.25 0446 38.0xt 6.020- 5.746+0.446=6.1
=0.2736 0.2736=5.746 92
14 36.00 54.70 43.00
27.00 0.292 43.0xt 6.192- 5.882+0.292=6.1
=0.3096 0.3096=5.882 74
16 18.00 54.67 42.50
14.38 0.155 42.5xt 6.174- 5.868+0.155=6.0
=0.3060 0.3060=5.868 23
18 10.75 54.38 39.00
8.88 0.096 39.0xt 6.023- 5.742+0.096=5.8
=0.2808 0.2808=5.742 38
20 7.00 54.05 33.50
5.85 0.063 33.5xt 5.838- 5.597+0.063=5.6
=0.2412 0.2412=5.597 60
22 4.70 53.73 29.70
4.00 0.043 29.7xt 5.660- 5.446+0.043=5.4
=0.21384 0.21384=5.446 89
24 3.30 53.45 26.00
3.00 0.032 26.0xt 5.489- 5.302+0.032=5.3
= 0.18720 0.18720=5.302 34
26 2.70 53.13 22.20
2.55 0.028 22.2xt 5.334- 5.174+0.028=5.2
=0.15984 0.15984=5.174 02

29 2.40 52.93 20.20

2.40 0.026 20.2xt 5.202- 5.057+0.026=5.0
=0.14544 0.14544=5.057 83
30 2.40 52.70 18.20

4.5 Channel or River Routing

In reservoir routing, the storage was a unique function of the outflow discharge, S= f(Q).
However, in channel or river routing the storage is a function of both outflow and inflow
discharge and hence a different routing method is needed. The flow in a river during a flood
belongs to the category of gradually varied unsteady flow. The water surface in channel reach is
not parallel to the channel bottom only, but also varies with time. Considering a channel reach
having a flood flow, the total volume in storage can be considered under two categories as:
1) Prism storage and
2) Wedge storage

Prism Storage
It is the prism that would exist if uniform flow occurred at the downstream depth, i.e. the volume
formed by an imaginary plane parallel to the channel bottom at the outflow section water surface.

Wedge storage
It is the wedge like volume formed between the actual water surface profile and the top surface
of the prism storage.

~ 11 ~
Lecture Note
 Engineering Hydrology; Lecture Note

At a fixed depth at a stream section of river reach the prism storage is constant while the wedge
storage changes from a positive value at an advancing flood to a negative value during a receding
flood. The prism storage Sp = f(q). The wedge storage can be accounted for expressing it as Sw
= f (I). The total storage in the channel reach can then be expressed as:

S = k x Im+(1-x)Qm ------------------------------- (1)

Where k and x are coefficients and m = a constant exponent. It has been formed that the value of
m varies from 0.6 for rectangular channels to a value of about 1.0 for natural channels.

Maskingum Equation
Using m = 1. Eq (1) reduces to a linear relationship for S in terms of I and Q as:
S = k [xI + (1-x)Q] ------------------------------------------ (2)
and this relationship is known as the Maskingum equation. In this the parameter x is known as
weight factor and takes a value between 0 and 0.5 when x = 0 obviously the storage is the
friction of discharge and the eq (2) reduces to:
S = KQ
such storage is known as linear storage or linear reservoir when x = 0.5, both the inflow and
outflow are equally important in determining the storage.
The coefficient k is known as storage time constant and has the dimension of time. It is
approximately equal to the time of travel of a flood wave through the channel reach.

For a given channel reach by selecting a routing interval of t and using the Maskingum
equation, the change

S2 - S1 = k (x(I2-I1)+ (1-x) (Q2- Q1) ----------------------------- (3)

Where, suffixes1 and 2 refer to the conditions before and after the time interval t. The
continually equation for the reach is :

S2 - S1 = I2+I1 t - Q2 + Q1 t ------------------------------ (4)

2 2
From Eq (3) and (4), Q2 is evaluated as:
Q2 = CoI2 + C1I1 + C2Q1 ------------------------------------- (5)

Where, Co = - kx + 0.5 t --------------------------------------- (5a)

k-kx + 0.5t
C1 = kx + 0.5 t -------------------------------------- (5b)
k-kx + 0.5t
C2 = k-kx - 0.5t -------------------------------------- (5c)
k-kx + 0.5t
Note that Co + C1+ C2 = 1.0. Eq. (5) can be written in a general form for the nth time step as:
Qn = CoIn+ C1In-1 + C2Qn-1 (6)

~ 12 ~
Lecture Note
 Engineering Hydrology; Lecture Note

Eq (5) is known as Muskingum Routing Equation and provides a simple linear equation for
channel routing. It has been found that for the best results the routing interval t should be so
chosen that kt2kx. if t2kx, then coefficient Co will be negative. Generally negative values
of coefficients are avoided by choosing appropriate values of t.
To use the Muskingum equation to route a given inflow hydrograph through a reach, the values
of k and x for the reach and the value of outflow, Q 1 from the reach at the start are needed. The
procedure is indeed simple.

a) Knowing k and x, select- an appropriate value of t.

b) Calculate Co, C1 and C2
c) Starting from the initial condition I1, Q1 and known I2 at the end of first time step t.,
calculate Q2 by equation (5)
d) The outflow calculated in step (c) becomes the known initial outflow for the next time
step. Repeat the calculations for the entire inflow hydrograph.
Example: Route the following hydrograph through a river reach for which K = 12.0h and x =
0.20. At the start of the inflow flood, the outflow discharge is 10m3/s.

Time (h) 0 6 12 18 24 30 36 42 48 54
inflow (m3/s 10 20 50 60 55 45 35 27 20 15
Solution: Since K = 12h and 2kx = 2x12x0.2=4.8h, t should be such that 12ht4.8h. In the
present case t = 6h is selected to suit the given inflow hydrograph ordinate interval. Using eqs
(5-a, b & c) the coefficient Co, C1 and C2 are calculated as:
Co = -12x0.2 + 0.5x6 = 0.6 = 0.048
12 - 12x0.2 + 0.5x6 12.6
C1 = 12x0.2 + 0.5x6 = 5.4 = 0.429
12-12x0.2+0.5x6 12.6
C2 = 12-12x0.2+0.5x6 = 6.6 = 0.523
12-12x0.2+0.5x6 12.6
For the first time interval, 0 to 6 h
I1 = 10.0  C1I1 = 0.429x10 = 4.29
I2 = 20.0 CoI2 = 0.048x20 = 0.96
Q1 = 10.0 C2Q1 = 0.523x10 = 5.23
From equation (5)
Q2 = CoI2 + C1I1+C2Q1
= 0.96 +4.29 + 5.23 = 10.48 m3/s
For the next step, 6 to 12 h, Q= 10.48 m 3/s. The procedure is repeated for the entire duration of
the inflow hydrograph. The computation is done in table form. Try plotting the inflow and
outflow hydrographs, the attenuation and peak lag are found to be 10m3/s and 12h respectively.

~ 13 ~
Lecture Note
 Engineering Hydrology; Lecture Note

Time (h) I(m3/s) 0.048I2 0.429I1 0.523 Q1 Q (m3/s)

1 2 3 4 5 6
0 10 10.00
0.96 4.29 5.23
6 20 10.48
2.40 8.58 5.48
12 50 16.46
2.88 21.45 8.61
18 60 32.94
2.64 25.74 17.23
24 55 45.61
2.16 23.60 23.85
30 45 49.61
1.68 19.30 25.95
36 35 46.93
1.30 15.02 24.55
42 27 40.87
0.96 11.58 21.38
48 20 33.97
0.72 8.58 17.74
54 15 27.04

~ 14 ~
Lecture Note