2. HYDROGRAPH
It is a graph showing the rate of
flow (discharge) versus time past a
specific point in a river, or other
channel or conduit carrying flow.
It can also called as a graph
showing the volume of water
reaching a particular outfall.
3. Such hydrographs are commonly used in the
design of sewerage, more specifically, the
design of surface water sewerage systems
and combined sewers.
4. COMPONENTS OF A HYDROGRAPH
Rising limb:
The rising limb of hydro graph, also known as concentration
curve, reflects a prolonged increase in discharge from a catchment
area, typically in response to a rainfall event
Recession (or falling) limb:
The recession limb extends from the peak flow rate onward.
The end of stormflow (aka quickflow or direct runoff) and the
return to groundwater-derived flow (base flow) is often taken as
the point of inflection of the recession limb.
The recession limb represents the withdrawal of water from the
storage built up in the basin during the earlier phases of the
hydrograph.
5. Peak discharge:
The highest point on the hydro graph when the rate of
discharge is greatest
Lag time:
The time interval from the center of mass of rainfall excess
to the peak of the resulting hydrograph
Time to peak:
The time interval from the start of the resulting hydro graph
Discharge:
The rate of flow (volume per unit time) passing a specific
location in a river or other channel
6.
7. Hydrograph usually consists of a fairly regular
lower portion that changes slowly throughout the
year and a rapidly fluctuating component that
represents the immediate response to rainfall.
The lower, slowly changing portion is termed
base flow. The rapidly fluctuating component is
called direct runoff.
Unit Hydrograph
8.
9. UNIT HYDROGRAPH
The amount of run-off resulting from 1 unit (1cm,
1mm, 1ft, etc.) of rainfall excess.
is essentially a tool for determining the direct runoff
response to rainfall.
Once you know the watershed’s response to one
storm, you can predict what its response for another
will look like.
10. Basic Assumptions of UH
UH
1. The effective rainfall is uniformly distributed within its duration
2. The effective rainfall is uniformly distributed over the whole
drainage basin
3. The base duration of direct runoff hydrograph due to an effective
rainfall of unit duration is constant.
4. The ordinates of DRH are directly proportional to the total
amount of DR of each hydrograph
5. For a given basin, the runoff hydrograph due to a given period of
rainfall reflects all the combined physical characteristics of basin
(time-invariant)
11. Procedure for Derivation of UH from Hydrograph
Type-I
Derivation of Unit Hydrograph from Given
DRH or Flood Hydrograph
Time (hr) 0 6 12 18 24 30 36 42 48 54 60 66
Observed
hydrograph(m3
/s)
100 100 300 700 1000 800 600 400 300 200 100 100
In a typical 6-hr storm, 4 cm excess rainfall is
occurring. The flow recorded in the catchment as
shown below. Derive an unit hydrograph for 6-hr
storm. Assume Base flow is 100 cu.meter/sec.
12. Soln:
Step-1: Compute DRH (Col. 3)
DRH=Ordinates of flood hydrograph – Base flow
Step-2: Compute 6hr-UH (Col. 4)
Ordinates of UH = Ordinates of DRH / Excess
Rainfall
14. Type-II
Derivation of DRH or Flood Hydrograph
from Given Unit Hydrograph
Time (hr) 0 3 6 9 12 15 18 21 24 27 30 33 36 39
ordinates
of 3 hr
UH
(m3/s)
0 12 75 132 180 210 183 156 135 144 96 87 66 54
42 45 48 51 54 57 60 63
42 33 24 18 12 6 6 0
Q. The ordinates of 3 hr UH is given below. Assume
Base flow is 15 cu.meter/sec. Derive the DRH for 2
cm, 6 cm and 4 cm excess rainfall.
16. Type-III
Derivation of T-hr Unit Hydrograph from
Given D-hr unit Hydrograph
If UH of specified UH is available, then we can
derive of any other duration UH by using
superposition techniques.
But, the use of this technique is limited i.e. if
required duration is integral multiple of given
duration UH, then this technique is easy to use
and compute req. UH.
For ex. Derivation of 16 hr UH from given 4hr
UH, then superposition technique is applicable,
But when it is necessary to derive 4hr UH from
given 16hr UH, then S-curve technique should be
used.
17. Type-III (A): Derivation of req. T-hr UH from Given
D-hr UH (where T is the multiple integral of D)
Time (hr) 0 3 6 9 12 15 18 21 24 27 30 33 36 39
ordinates of
3 hr UH
(m3/s)
0 12 75 132 180 210 183 156 135 144 96 87 66 54
4
2
45 48 51 54 57 60 63
4
2
33 24 18 12 6 6 0
Q. The ordinates of 3 hr UH is given below. Assume
Base flow is 15 cu.meter/sec.
Derive the 9 hr UH.
18. Soln:
Step-1: See the given duration and req. duration.
Given Duration of UH = 3 hr
Req. duration = 9hr
i.e. 9 is the integral multiple of 3. Thus, superposition technique is
applicable.
Step-2: if we add 3 hr UH by 3 times then we got 9 hr UH
i.e. 3hr + 3hr + 3hr = 9 hr
but, at the same when we adding the 3 UH of 1 cm each, resulted
hydrograph will be DRH of 1+1+1= 3 cm excess rainfall.
Thus, in 2nd step, lag ordinates of given UH by 3 hr and then again
lag by 3 hr (Col. 3 & 4)
Step-3: Add these lagged UH i.e. Col.6, this will give us DRH of
3cm ER.
Step-4: Divide col.6/3cm (bcoz we want to plot UH)
