This PPT give the brief introduction about Water Resources Engineering

1. Lecture-1
Introduction to Hydrology
Prepared By,
Daundkar Ramesh Kailas,
Assistant Professor,
Department of Civil
Engineering,
RGUKT- R K Valley
Water Resources Engineering by Daundkar Ramesh Kailas is licensed under
a Creative Commons Attribution 4.0 International License.

2. What is Hydrology ?
• Hydrology means the science of water.
• It is the science that deals with the occurrence,
circulation and distribution of water of the earth
and earths atmosphere.
• As a branch of earth science, it is concerned with
the water in streams and lakes, rainfall and
snowfall on the land and water occurring below the
earth surface in the pores of the soil and rocks.

3. Hydrology is basically an applied science. It
deals with
(i) estimation of water resources
(ii) the study of processes such as
precipitation, runoff, evapotranspiration and
their interaction, and
(iii)The study of problems such as floods and
droughts and strategies to combat them.

4. Hydrologic Cycle
• Evaporation of water from water bodies such
as oceans and lakes, formation and
movement of clouds, rain and snowfall,
stream flow and groundwater movement are
various dynamic aspects of water related to
earth.
• The various aspects of water related to earth
can be explained in terms of a cycle known as
hydrologic cycle

6. • A convenient starting point to describe the cycle is in the
oceans. Water in the oceans evaporates due to heat energy
provided by solar radiation.
• Ocean plays key role in this vital cycle of water. The ocean
holds 97% of the total water on planet; 78% of global
precipitation occurs over the ocean and it is the source of
86% of global evaporation.
• The water vapor moves upwards and forms clouds. While
much of the clouds condense and fall back to the oceans as
rain, a part of the cloud is driven to the land areas by winds.
• The clouds condense over the land areas and precipitate on
to the land mass as rain, snow, hail, sleet, etc.

7. • A part of precipitation may evaporate back to the
atmosphere even while falling. Another part may be
intercepted by vegetation, structures, etc and it may
get evaporated back to the atmosphere or move down
to the ground.
• A portion of the water that reaches the ground enters
the earth’s surface through infiltration, enhances the
moisture content of the soil and finally reaches to
groundwater body.
• Vegetation sends some portion of soil water and
ground water back to the atmosphere through the
process of transpiration.

8. • The precipitation reaching the earths surface after
meeting the needs of evaporation and infiltration
moves down the natural slope over the surface and
through a network of gullies, streams and rivers
called as stream flow.
• The groundwater may come to the surface through
springs and other outlets. The portion of
precipitation which by variety of paths above and
below the surface of the earth reaches the stream
channel is called as runoff. Once it enters the
stream channel, runoff becomes stream flow.

9. • The sequence of events as above is a simplistic picture of
very complex cycle that has been taking place since the
formation of the earth.
• It is seen that hydrologic cycle is a very vast and
complicated cycle in which there are large number of paths
of varying time scales.
• Further, it is a continuous recirculating cycle in the sense
that there is neither a beginning nor an end or pause.
• Each path of the hydrologic cycle involves one or more of
the following aspects :
1. Transportation of water, 2. Temporary storage and
3. Change of state.

10. Main components of the hydrologic cycle
Transportation
Components
Storage Components
1. Precipitation 1. Storage on land surface in
lakes, Rivers etc.
2. Evaporation 2. Soil moisture storage
3. Transpiration 3. Groundwater storage
4. Infiltration
5. Runoff

11. • A qualitative representation of the hydrologic cycle first introduced by
Horton. The below diagram is popularly known as Horton’s
representation of the hydrologic cycle, illustrates very clearly the
storage and transportation components and their relative positions in
the cycle.

12. Water Budget Equation
• Catchment Area : The area of land draining into a
stream or water course at a given location is
known as the catchment area. It is also called
drainage area or drainage basin. In the USA, it is
known as watershed.
• A catchment area is separated from it’s
neighboring areas by a ridge called divide in the
USA. The catchment area can be measured by
planimeter on a topographic map.

13. Water-Budget Equation
For a given problem area, say a catchment , in an interval
of time ∆t, the continuity equation for water in its various
phases is written as,
Mass inflow – Mass outflow = Change in mass storage
If the density of the inflow, outflow and storage
volumes are the same,
∀𝑖 - ∀ 𝑜 = ∆S
Where ∀𝑖 = inflow volume of water into the problem area
during the time period,
∀ 𝑜= outflow volume of water from the problem area
during the time period and
∆S = change in the storage of the water volume over and
under the given area during the given period

14. In hydrologic calculations, the volumes are often expressed as average
depths over the catchment area. If an annual stream flow volume over 10
km2 catchment is 107 m3, it corresponds to the depth of
107
10 × 106 = 1m =
100 cm.
Rainfall, evaporation and runoff volumes are expressed in units of depth
over the catchment.
All the terms in a hydrological water budget may not be known to the same
degree of accuracy, an expression for the water budget of a catchment for a
time interval ∆t is written as:
P – R – G –E – T = ∆S
Where, P = Precipitation, R = Surface Runoff, G = Net ground water flow
out of the catchment, E = Evaporation, T = Transpiration and ∆S = Change
in storage.

15. The storage S consists of three components as
S = 𝑆𝑠 + 𝑆𝑠𝑚+ 𝑆 𝑔
Where, 𝑆𝑠 = surface water storage
𝑆𝑠𝑚 = water in storage as soil moisture
𝑆 𝑔 = water in storage as groundwater
Thus, ∆S = ∆𝑆𝑠 + ∆ 𝑆𝑠𝑚 + ∆ 𝑆 𝑔
In terms of rainfall-runoff relationship can be represented as R = P-
L
Where L = losses = water not available to runoff due to infiltration,
evaporation, transpiration and surface storage.

16. Ex.1. A lake had a water surface elevation of 103.200 m above datum at
the beginning of a certain month. In that month, the lake received an
average inflow of 6 m3/s from surface runoff sources. In the same period,
the outflow from the lake had an average value of 6.5 m3/s. Further in that
month, the lake received a rainfall of 145 mm and the evaporation from
the lake surface was estimated as 6.10 cm. Write the water budget
equation for the lake and calculate the water surface elevation of the lake
at the end of the month. The average lake-surface area can be taken as
5000 ha. Assume that there is no contribution to or from the groundwater
storage.
Ans: ⇒
Given: Water surface elevation at the beginning = 103.200 m
Average inflow of lake( 𝑰) = 6 m3/s
Average outflow from the lake(𝑸 ) = 6.5 m3/s
Rainfall during the month (𝑷) = 145 mm = 0.145 m
Evaporation from the lake surface (E) = 6.10 cm = 0.06 m
Average Lake Surface Area (A)= 5000 ha
Time Duration = ∆𝒕 = 1 month = 30 x 24 x 60 x 60 = 2.592 x 106 Seconds

17. In a time interval ∆t, the water budget for the lake can be written as :
Input volume – Output volume = Change in storage of the lake
𝑰 ∆𝒕 + 𝑷. 𝑨 − 𝑸 ∆𝒕 − E A = ∆S
In one month :
Inflow volume = 𝑰 ∆𝒕 = 6 x 2.592 x 106 = 15.552 x 106 m3
Outflow volume = 𝑸 ∆𝒕 = 6.5 x 2.592 x 106 = 16.848 x106 m3
Input due to precipitation = P A = 145 x 0.001 × 5000 × 104
=7.25 x 106
m3
Outflow due to evaporation = E A = 6.10 x 10-2 x 5000× 104= 3.05x106 m3
Hence,
∆S = 15.552 x 106 m3+ 7.25 x 106 m3− 16.848 x 106 m3 − 3.05x106 m3
∆S= 2.904 x 106 m3
Change in elevation = ∆ Z =
∆S
𝑨
=
2.904 x 106 m3
5000×104 𝑚2 =0.058 m
New water surface elevation at the end of the month = 103.200 + 0.058
= 103.258 m above the datum

18. Ex.2) Estimate the constant rate of withdrawal from a 1375 ha in a month
of 30 days during which reservoir level dropped by 0.75 m inspite of an
average inflow into the reservoir of 0.5 x 10 6 m3/day. During the month,
the average seepage loss from the reservoir was 2.5 cm, total precipitation
on the reservoir was 18.5 cm and the total evaporation was 9.5 cm.
Ans :⇒
Given : Reservoir Surface Area = 1375 ha
Time Period = 30 days = 30 x 24 x 3600 =2.592 x 106Seconds
Change in reservoir water level = − 0.75 m
∴ Change in reservoir storage volume = − 0.75 m x 1375 x 104
∆S= − 0.1031 x 106 m3
∆S = [𝑰 ∆𝒕 + 𝑷. 𝑨] − [𝑸 ∆𝒕 +E A+G.A]
Total input volume = 𝑰 ∆𝒕 + 𝑷. 𝑨
Total output volume = 𝑸 ∆𝒕 +E A+G.A
𝑸 ∆𝒕 ={ [𝑰 ∆𝒕 + 𝑷. 𝑨] − [E A+G.A]} − ∆S

19. 𝑸 ∆𝒕= { [𝑰 ∆𝒕 + 𝑷. 𝑨] − [E A+G.A]} − ∆S
𝑰 ∆𝒕 + 𝑷. 𝑨 = [0.5 x 10 6m3/day x 30 ] + [18.5 x 10-2 x 1375 x 104 ]
= 17.5438 x 106 m3
E A+G.A = [9.5 x10-2 x 1375 x 104] +[ 2.5x 10-2 x 1375 x 104]
= 1.60 x 106 m3
𝑸 ∆𝒕 = {[17.5438 x 106 ] − [1.60 x 106m3]} −(− 0.1031 x 106)
= 16.047 × 106 𝑚3
Constant rate of withdrawal = 𝑸 =
16.047 × 106
2.592 x1 0 6 = 6.191 m3/s

20. World Water Balance:
The total quantity of water in the world is estimated to be about 1386 M
km3. About 96.5% of this water is contained in the oceans as saline water.
Some of the water on the land amounting to about 1% of the total water is
also saline. Thus only about 35 M km3 of fresh water is available. Out of
this about 10.6 M km3 is both liquid and fresh and the remaining 24.4 M
km3 is contained in frozen state as ice in the polar regions and on
mountain tops and glaciers. An estimated distribution of water on the
earth is given in below table:
It is seen that the annual evaporation from the world’s oceans and inland
areas are 0.505 and 0.072 M km3 respectively. Thus over the oceans about
9% more water evaporates than that falls back as precipitation. i.e. 91% of
evaporated water from sea falls back into the sea. The total runoff of
0.047 M km3 is the runoff and groundwater outflow from the land mass to
oceans. It is interesting to know that less than 4% of this total river flow is
used for irrigation and rest flows down to sea.

22. Global Freshwater Resources

23. Applications in Engineering
• Hydrology finds it’s greatest application in the design and operation of
water-resources engineering projects, such as those for (i) Irrigation
(ii) Water Supply (iii) Flood Control (iv) Water Power and (v)
Navigation
• In all these projects, hydrological investigations for the proper
assessment of the following factors are necessary:
1. The capacity of storage structure such as reservoirs.
2. The magnitude of flood flows to enable safe disposal of the excess
flow.
3. The minimum flow and quantity of flow available at various seasons.
4. The interaction of the flood wave and hydraulic structures, such as
levees, reservoirs, barrages and bridges.

24. • The hydrological study of a project should necessarily precede
structural and other detailed design studies. It involves the collection
of relevant data and analysis of the data by applying the principles and
theories of hydrology to seek solutions to practical problems.
• Many important projects in the past have failed due to improper
assessment of hydrological factors. Some typical failures of hydraulic
structures are:
1. Overtopping and consequent failure of an earthen dam due to an
inadequate spillway capacity
2. Failure of bridges and culverts due to excess flood flow and
3. Inability of large reservoir to fill up with water due to overestimation
of the stream flow.
Such a failure is often called as hydrologic failures. Practically , all
hydrologic phenomenon are complex and at the present level of
knowledge, they can at best be interpreted with the aid of probability
concepts. Hydrological events are treated as random processes and
historical data relating to the event are analyzed by statistical methods to
obtain information on probabilities of occurrence of various events.