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Hydrographs
Response of a given catchment to rainfall input.
A hydrograph is a graph showing discharge versus time.
Hydrograph
Record of River Discharge over a period of time
River Discharge
= cross sectional area rivers mean (average) velocity
X
(at a particular point in its course)
Storm Hydrographs
Show the change in discharge caused by a period of rainfall
Hydrograph :applications
To find out discharge patterns of a particular drainage basin
Help predict flooding events, therefore used for flood prevention measures
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
Base flow
Through flow
Overland flow
Basin lag time
mm
4
3
2
Peak flow
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
2
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
mm
4
3
2
Rainfall shown in mm, as a bar graph
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
mm
4
3
2
Discharge in m3/s, as a line graph
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
mm
4
3
2
The rising flood water in
the river
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
mm
4
3
2
Peak flow
Peak flow
Maximum discharge in the river
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
mm
4
3
2
Peak flow Falling flood water in the
river
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
Basin lag time
mm
4
3
2
Peak flow
Basin lag time
Time difference
between the peak of the rain storm
and the peak flow of the
river
3
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
Base flow
Basin lag time
mm
4
3
2
Peak flow
Base flow
Normal discharge of the river
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
Base flow
Through flow
Overland flow
Basin lag time
mm
4
3
2
Peak flow
Overland flow
Through flow
+
= Storm Flow
Volume of water reaching the river from surface run off
Overland flow Through flow
Volume of water reaching the river
through the soil and underlying rock
layers
Storm Hydrographs
•Rainfall Intensity
•Rising Limb
•Recession Limb
•Lag time
•Peak flow compared to Base flow •Recovery rate, back to Base flow
Basin lag time
0 12 24 36 48 30 72
Hours from start of rain storm
3
2
1
Dis
char
ge (
m3/s
)
Base flow
Through flow
Overland flow
mm
4
3
2
Peak flow
Time Characteristics of hydrograph
1. Time to peak
– • From beginning of rising limb to peak discharge
– • Function of basin characteristics
– • Drainage density, slope channel size, roughness and soil
infiltration characteristics
2. Time of concentration
– • Time required for the farthermost rain to reach the outlet
3. Lag time or Basin lag time
– • Between centre of mass of rainfall and runoff hydrograph mass
– • Since it is very difficult to find the centre then it is the time
between centre of mass of effective rainfall to peak discharge
4. Duration of rainfall
5. Base time of hydrographs
Storm rainfall:
Hydrograph :
0
10
20
30
40
50
60
70
80
90
100
1 2 3 4 5 6 7 8 9
Series1
14 15 16 17 18 19 20 21 22
?
Precipitation and river response
4
Understanding hydrological
responses
Peak discharge?
Lag time?
Rate of recession?
Factors affecting flood hydrographs
Channel characteristics
– Cross section,
– roughness
– Storage capacity.
Infiltration characteristic
– (a) Land use and cover
– (b) Soil type and geological condition
– (c) Lakes, swamps and other storage
Climatic factors
– (a) Shape
– (b) Size
– ( c) slope
– (d) Nature of the valley
– (e) Elevation
– (f) Drainage density
Physiographic Factors
Factors affecting flood hydrographs
Storm characteristics
– precipitation,
– Intensity,
– duration,
– magnitude and
– movement of storm.
Initial loss
Evapotranspiration
Climatic Factors
Factors influencing Storm Hydrographs
• Area
• Shape
• Slope
• Rock Type
• Soil
• Land Use
• Drainage Density
• Precipitation / Temp
• Tidal Conditions
Stream order and flood response Regional
climate
and flood
response
5
Geological substrate and flood
response Basin geometry and flood
response
circular
elonga
ted
effect of
storm
path?
Land use and flood response
Area
Large basins receive more precipitation than small therefore have larger runoff
Larger size means longer lag time as water has a longer distance to travel to reach the trunk river
Area Rock Type Drainage Density
Shape Soil Precipitation / Temp
Slope Land Use Tidal Conditions
Shape
Elongated basin will produce a lower peak flow and longer lag time than a circular one of the same size
Area Rock Type Drainage Density
Shape Soil Precipitation / Temp
Slope Land Use Tidal Conditions
Slope
Channel flow can be faster down a steep slope therefore steeper rising limb and shorter lag time
Area Rock Type Drainage Density
Shape Soil Precipitation / Temp
Slope Land Use Tidal Conditions
6
Rock Type
Permeable rocks mean rapid infiltration and little overland flow therefore shallow rising limb
Area Rock Type Drainage Density
Shape Soil Precipitation / Temp
Slope Land Use Tidal Conditions
Soil
Infiltration is generally greater on thick soil, although less porous soils eg. clay act as impermeable layers
The more infiltration occurs the longer the lag time and shallower the rising limb
Area Rock Type Drainage Density
Shape Soil Precipitation / Temp
Slope Land Use Tidal Conditions
Land Use
Urbanisation - concrete and tarmac form impermeable surfaces, creating a steep rising limb and shortening the time lag
Afforestation - intercepts the precipitation, creating a shallow rising limb and lengthening the time lag
Area Rock Type Drainage Density
Shape Soil Precipitation / Temp
Slope Land Use Tidal Conditions
Drainage Density
A higher density will allow rapid overland flow
Area Rock Type Drainage Density
Shape Soil Precipitation / Temp
Slope Land Use Tidal Conditions
Precipitation & Temperature
Short intense rainstorms can produce rapid overland flow and steep rising limb
If there have been extreme temperatures, the ground can be hard (either baked or frozen) causing rapid surface run off
Snow on the ground can act as a store producing a long lag time and shallow rising limb. Once a thaw sets in the rising limb will become steep
Area Rock Type Drainage Density
Shape Soil Precipitation / Temp
Slope Land Use Tidal Conditions
Tidal Conditions
High spring tides can block the normal exit for the water, therefore extending the length of time the river basin takes to return to base flow
Area Rock Type Drainage Density
Shape Soil Precipitation / Temp
Slope Land Use Tidal Conditions
7
Base flow separation
To draw surface runoff hydrograph, it is
required to separate base flow
Methods
1. Straight line method
N=0.83A0.2 Line AB
2. Line ACB
3. Line AFE
Effective rainfall or Rainfall Excess
The direct runoff,
(initial losses and
infiltration losses are
separated).
Called as effective
rainfall hyetograph or
hyetograph of rainfall
excess
It is the rainfall which
contributed for direct
runoff
Example
A storm over a catchment of area 15 sq.km had a 12
hrs duration. The mass curve of rainfall is given. If φ
index is 0.3 cm/h, determine the effective rainfall
hyetograph and the volume of direct runoff from the
catchment.
Time from
start of the
storm
0 2 4 6 8 10 12
Accumulated
rainfall
(cm)
0 0.7 2.5 5.8 7.2 9.0 9.9
Solution Time from start of
the storm 0 2 4 6 8 10 12
Accumulated
0 0.7 2.5 5.8 7.2 9 9.9 rainfall
(cm)
φ index 0.6 0.6 0.6 0.6 0.6 0.6
depth of rainfall in cm 0.7 1.8 3.3 1.4 1.8 0.9
Effective rainfall (cm) 0.1 1.2 2.7 0.8 1.2 0.3
Rainfall intensity cm/hr 0.05 0.6 1.35 0.4 0.6 0.15 3.15
47250000
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1 2 3 4 5 6
Ra
infa
ll In
ten
sit
y (
cm
/hr)
Time from start of storm (hr)
Unit hydrograph
The measurement of runoff works out to be a costly and
more time consuming process
Hence various methods has been used like RR models,
empirical, rational, soft computing etc…
UNIT HYDROGRAPH METHOD – Introduced by
Sherman (1932)
The unit hydrograph represents the lumped response
of the catchment to a unit rainfall excess of D-h
duration to produce a direct-runoff hydrograph.
It relates only the direct runoff to the rainfall excess.
Hence the volume of water contained in the unit
hydrograph must be equal to the volume of rainfall
excess.
As 1 cm depth of rainfall excess is considered, the area
of the unit hydrograph is equal to a volume given by 1
cm over the catchment for that duration.
The rainfall is considered to have an average intensity of
excess rainfall (ER) of 1 cm/h for the D hr duration of the
storm.
The distribution of the storm is considered to be uniform
all over the catchment.
8
Basic assumptions of unit hydrograph
Time invariance
– First basic assumption is that direct-runoff response in a
catchment is time-invariant.
Linear Response
– The direct-runoff response to the rainfall excess is assumed
to be linear.
– This is the most important assumption of the unit-
hydrograph theory.
– The Rainfall –runoff obeys the principles of superposition
Application of unit hydrograph
It is useful in calculating DRH (flood as well as storm hydrograph) of a given storm occurred for D hr.
If a D hr duration UH is available, it is possible to derive multiples of D hr duration unit hydrograph
Unit hydrograph of different
durations
Method of superposition
– Suitable when m is an integer
Method of S- curve
– Suitable when m is a fraction
Method of superposition
suitable when m is an
integer
D hr Unit hydrograph
is available
Super impose m unit
hydrographs with
each hydrograph
lagged by D hr from
the previous unit
hydrograph
Method of S- curve Also suitable when m is a fraction
Develop a hydrograph produced by a continuous
effective rainfall of D hr for a infinite period.
Then get the summation hydrograph (S curve
ordinates)
Lag this S curve by mD hrs (SB) take the difference
between the two S curves, (SA-SB)
Example
Rainfall of magnitude 3.8 cm and 2.8 cm
occurring on two consecutive 4-h
durations on a catchment area 27km2
produced the following hydrograph of flow
at the outlet of the catchment. Estimate
the rainfall excess and φ-index
Time from start of rainfall -6 0 6 12 18 24 30 36 42 48 54 60 66
Observed flow (m3/s) 6 5 13 26 21 16 12 9 7 5 5 4.5 4.5
9
Solution
Time from start of rainfall -6 0 6 12 18 24 30 36 42 48 54 60 66
Observed flow (m3/s) 6 5 13 26 21 16 12 9 7 5 5 4.5 4.5
DRH ordinates 1 0 8 21 16 11 7 4 2 0 0 -0.5 -0.5
Example
The following are
ordinates of 4hr unit
hydrograph.
Determine ordinates of
12 hr unit hydrograph by
S-curve method.
Time 4hr. UH
0 0 4 50 8 100 12 110 16 140 20 90 24 80 28 60 32 40 36 20 40 0
time 4hr. UH
S-curve addition
S-
Curve
Ordina
tes
S-Curve
lagged
by 12 hr. Diff. (Y) Y*T/t
0 0 0 0 0 0.00
4 50 0 50 0 50 16.67
8 100 50 0 150 0 150 50.00
12 110 100 50 0 260 0 260 86.67
16 140 110 100 50 0 400 50 350 116.67
20 90 140 110 100 50 0 490 150 340 113.33
24 80 90 140 110 100 50 0 570 260 310 103.33
28 60 80 90 140 110 100 50 0 630 400 230 76.67
32 40 60 80 90 140 110 100 50 0 670 490 180 60.00
36 20 40 60 80 90 140 110 100 50 0 690 570 120 40.00
40 0 20 40 60 80 90 140 110 100 50 0 690 630 60 20.00
690 670 20 6.67
690 690 0 0
690 690 0 0
Solution :
10
Hydrograph
0.00
20.00
40.00
60.00
80.00
100.00
120.00
140.00
0 10 20 30 40 50 60
Dis
cha
rge
Q (
cum
ec)
Time (Hours)
Example
Re-gernerate the ordinates of 4hr
UH by the 12 hr UH ordinates.
Solution :
time 12hr. UH S-Curve Addition
S-Curve
Ordinate
s
S-Curve
lagged
by 4 hr. Diff. (Y) Y*T/t
0 0 0 0 0
4 16.66 16.66 0 16.66 49.98
8 50 50 16.66 33.34 100.02
12 86.66 0 86.66 50 36.66 109.98
16 116.66 16.66 133.32 86.66 46.66 139.98
20 113.33 50 163.33 133.32 30.01 90.03
24 103.33 86.66 0 189.99 163.33 26.66 79.98
28 76.66 116.66 16.66 209.98 189.99 19.99 59.97
32 60 113.33 50 223.33 209.98 13.35 40.05
36 40 103.33 86.66 0 229.99 223.33 6.66 19.98
40 20 76.66 116.66 16.66 229.98 229.99 -0.01 -0.03
44 6.66 60 113.33 50 229.99 229.98 0.01 0.03
48 0 40 103.33 86.66 229.99 229.98 0.01 0.03
229.99
Hydrograph
0
20
40
60
80
100
120
140
160
0 10 20 30 40 50
Dis
charg
e Q
(cu
mec
)
Time (Hours)
Example
For the given
ordinates of 8hr unit
hydrograph
Obtain the ordinates
of 24 hr unit
hydrograph
time 8hr. UH
0 0
4 5
8 13
12 26
16 45
20 82
24 162
28 240
32 231
36 165
40 112
44 79
48 57
52 42
56 31
60 32
64 14
68 9.5
72 6.6
76 4
80 2
84 1
88 0
Solution :
time 8hr. UH
8hr UH
lagged by 8
hr
8hr UH
lagged by 16
hr Addirion 24 hr. UH
col. 1 col. 2 col. 3 col. 4 col5 = 2+3+4 col5/3
0 0 0 0.0
4 5 5 1.7
8 13 0 13 4.3
12 26 5 31 10.3
16 45 13 0 58 19.3
20 82 26 5 113 37.7
24 162 45 13 220 73.3
28 240 82 26 348 116.0
32 231 162 45 438 146.0
36 165 240 82 487 162.3
40 112 231 162 505 168.3
44 79 165 240 484 161.3
48 57 112 231 400 133.3
52 42 79 165 286 95.3
56 31 57 112 200 66.7
60 32 42 79 153 51.0
64 14 31 57 102 34.0
68 9.5 32 42 83.5 27.8
72 6.6 14 31 51.6 17.2
76 4 9.5 32 45.5 15.2
80 2 6.6 14 22.6 7.5
84 1 4 9.5 14.5 4.8
88 0 2 6.6 8.6 2.9
11
Hydrograph
0.0
20.0
40.0
60.0
80.0
100.0
120.0
140.0
160.0
180.0
0 20 40 60 80 100
Dis
cha
rge
Q (
cum
ec)
Time (Hours)
Time(min) Rainfall
(cm/hr) Rainfall
(cm) ϕ Index
(cm/hr) ϕ Index (cm) Effective
rainfall
0-20 2.5 0.83 3.20 1.07 0.00 20-40 2.5 0.83 3.20 1.07 0.00 40-60 10 3.33 3.20 1.07 2.27 60-80 7.5 2.50 3.20 1.07 1.43
80-100 1.25 0.42 3.20 1.07 0.00 100-120 1.25 0.42 3.20 1.07 0.00 120-140 5 1.67 3.20 1.07 0.60
Total Rainfall 10 Total Runoff 4.30
12