Hydrology
Dr. Ashfaque Ahmed Memon
Department of Civil Engineering
• Syllabus and Reference Books
• Introduction
• Catchment area
• Water budget equation
• Examples
Syllabus:
Recommended Books:
Hydrology
Hydrology is the science that encompasses the occurrence, distribution,
movement and properties of the waters of the earth and their
relationship with the environment within each phase of the hydrologic
cycle.
Hydrology: Science of water’s:
• Occurrence
• Distribution
• Movement, and
• Properties
Surface water hydrology
Groundwater hydrology
An iceberg is a large piece of freshwater ice that has broken off a glacier or
an ice shelf and is floating freely in open water.
As it drifts into shallower waters, it may come into contact with the seabed.
The word "iceberg" is a partial loan translation from Dutch ijsberg, literally
meaning ice mountain.
Catchment Area
The area of land draining into a stream or a water course at a
given location is known as catchment area.
Also called as drainage area or drainage basin or watershed
(USA).
Diagram showing two Catchments
Catchment of River A at station M
Estimating the Concentration Time of a Catchment
Izzard’s Formula:
For small plots having no defined flow channels from which run off occurs as
laminar overland flow
111 .b.L1o/ 3
To minutes
K . p 2/3
Types of Surface Value of Cr
Where
Smooth asphalt surface 0.007
Lo = Length of overland flow (m)
Concrete pavement 0.012
K = Runoff coefficient (Refer Table)
Tar and gravel pavement 0.017
p = Rainfall intensity (cm/hr)
Closely clipped soil 0.046
0.000275 p Cr
b a coefficien t Dense blue grass turf 0.060
S o1/ 3
Cr = Retardance coefficient (Refer Table)
So = Slope of the surface
To = Over land Flow Time / Inlet Time (min)
These equations are applicable only when, p.Lo < 387.
0.385
L3o
To 0.885
H
Example:
An area of 5 hectares in a single family residence
district has an average length of overland flow of 40
meters, average slope of the plots of 0.003, and the
design rainfall is given by p = 64/To1/2 , where p is in
cm/hr and To is in minutes. Ignoring the intensity
term in the retardance coefficient equation and
using Cr = 0.05; find the time of concentration for the
overland flow from this area. Assuming gutter flow
time to add 10 min., find the peak rate of runoff to
be expected.
Solution:
We have the relationship for over land flow time,
111 .b.L1o/ 3 0.000275 p Cr
To minutes where b
Kp 2 / 3 S o1/ 3
Cr
Ingnoring p term and substitution of values yields, b 1/ 3
0.347
So
Also, for the Single family residence, from the Table, K = 0.3
Substitution of values in the equation for over land flow time, we have
111(0.347 ).(40)1 3
To which yields To = 75 minutes
[0.3(64 To1 2 )]2 3
Here,
Inlet time or overland flow time, To = 75 min
Gutter flow time, Tf = 10 min
Hence,
Total time of concentration, Tc = To + Tf = 75 + 10 = 85 min
Now,
p 64 / T01 2 64 751 2 7.4 cm/hr
and,
1 1
Qmax K . pc . A 0.37.45 0.31 cumec
36 36
Water Budget Equation
From the continuity equation for water, i.e.
Mass inflow - mass outflow = change in mass storage
Water budget of a catchment for a time interval Δt is written as
P–R–G–E-T=ΔS
Where
P = precipitation,
R = surface runoff,
G = net ground water flow out of the catchment,
E = evaporation,
T = transpiration and
Δ S = change in storage.
All terms in the equation may have the dimensions of volume or depth over
the catchment area.
EXAMPLE # 1
A lake had a water surface elevation of 103.2 m above datum at the
beginning of certain month. In that month the lake received an
average inflow of 6.0 m3/sec from surface runoff sources. In the
same period the outflow from the lake had an average value of 6.5
m3/sec. Further, in that month, the lake received a rainfall of 145
mm and the evaporation from the lack surface was estimated as
6.10 cm. Write the water budget equation for the lake and calculate
water surface elevation of the lake at the end of month. The
average lake surface area can be taken as 5000 he. Assume that
there is no contribution to or from the ground water storage.
DATA:
WSEstart = 103.2 m;
Surface Inflow = 6.0 m3/sec = 6.0 x 30 x 24 x 60 x 60 m3;
Surface Outflow = 6.5 m3/sec = 6.5 x 30 x 24 x 60 x 60 m3;
Area = 5000 he = 5000 x 104 m2;
Precipitation = 145 mm = 0.145 m = 0.145 x 5000 x 104 m3;
Evaporation = 6.10 cm = 6.10 x 10-2 m = 0.061 x 5000 x 104 m3;
Water budget equation = ?
WSEend = ?
Solution:
For the given lake, Water budget equation will be written as
Inflow volume – Outflow volume = Change in storage (∆S)
Inflow volume = Surface Inflow + Precipitation
= (6.0 x 30 x 24 x 60 x 60) + (0.145 x 5000 x 104)
= 22.802 x 106 m3
Outflow volume = Surface Outflow + Evaporation
= (6.5 x 30 x 24 x 60 x 60 ) + (0.061 x 5000 x 104)
= 19.898 x 106 m3
Therefore, ∆S = 22.802 x 106 - 19.898 x 106 = 2.904 x 106 m3
Change in elevation, ∆Z = ∆S/A = 2.904 x 106 / 5000 x 104
= 0.05808 m
Hence,
WSEend = WSEstart + ∆Z = 103.2 + 0.05808 = 103.25808 m
Example # 2
A small catchment of area 150 he received a rainfall of 10.5 cm in
90 min duration due to a storm. At the outlet of the catchment, the
stream draining the catchment was dry before the storm and
experienced a runoff lasting for 10 hr with an average value of 2.0
m3/sec. The stream was again dry after the runoff event. What is
the amount of water which was not available to runoff due to
combined effect of infiltration, evaporation and transpiration? Also
compute the ratio of runoff to precipitation.
Data:
Area = 150 he = 150 x 104 m2
Rainfall, P = 10.5 cm = 10.5/100 m = 0.105 m
Storm duration, ∆tp = 90 min
Runoff, R = 2.0 m3/sec = 2.0 x 10 x 60 x 60 m3
Runoff duration, ∆tr = 10 hr
Solution:
The general Water budget equation in time ∆t,
ΔS = P – R – G – E - T (1)
Amount of water not available for runoff, i.e.
Losses, L = G + E + T
Change in storage, ΔS = 0
(a) Hence Eq. (1) becomes, L = P – R (2)
Therefore, L = (0.105 x 150 x 104) – (2.0 x 10 x 60 x 60)
= 157,500 – 72,000
= 85,500 m3
(b) Runoff Coefficient = R/P = 72,000/157,500 = 0.457
