1

QUESTION 1

(i) What do you understand hydrology?

(ii) Explain the hydraulic cycle in nature with the help of a neat sketch,

indicating its various phases.

(iii) Discuss clearly the various forms of precipitation

(iv) What are the functions of hydrology in water resources development?

(v) What are the basic data required for hydrological studies?

QUESTION 2

Explain hydrologic equation.

QUESTION 1

(i) What are the different types or(forms) of precipitation?

(ii) Discuss the three various types of recording rain gauges

(iii) What factors you consider in selecting a site for a rain-gauge station?

(iv) Distinguish between recording and non-recording rain-gauges,

(v) Explain A method for estimating the missing rainfall data at a basin,

QUESTION 2

Rain-gauge station D was inoperative for part of a month during which a storm

occured. The storm rainfall recorded in the three surrounding stations A, B and C

were 8.5, 6.7and 9.0 cm, respectively. If the a.a.r for the stations are 75, 84, 70 and

90 cm, respectively, estimate the storm rainfall at station D.

QUESTIONS ON CHAP 2 PRECIPITATION

QUESTIONS ON INTRODUCTION

2

QUESTION 3

(i) Define rain-gauge density

(ii) What do understand by

(iii) As the rainfall over a large area is not uniform, the average depth of

rainfall over the area is determined by three methods discuss them

briefly,

(iv) Example 2.3 Point rainfalls due to a storm at several rain-gauge

stations in a basin are shown in Fig. 2.10. Determine the mean areal

depth of rainfall over the basin by the three methods.

(v) Example 2.3 (a) The area shown in Fig. P (2.3a) is composed of a square

plus an equilateral triangular plot of side 10 km. The annual

precipitations at the rain-gauge stations located at the four corners and

centre of the square plot and apex of the traingular plot are indicated in

figure. Find the mean precipitation over the area by Thiessen polygon

method, and compare with the arithmetic mean.

3

QUESTION 4

(i) What do you understand by a hyetograph?

(ii) What do you understand by A mass curve of rainfall (or

precipitation)

(iii) Differentiate between Mean and Median

(iv) The annual rainfall at a place for a period of 10 years from 1961 to

1970 are respectively 30.3, 41.0, 33.5, 34.0, 33.3, 36.2, 33.6, 30.2,

35.5, 36.3. Determine the mean and median values of annual

rainfall for the place.

QUESTION 5

The following are the rain gauge observations during a storm. Construct:

(i) mass curve of precipitation,

(ii) hyetograph,

(iii) maximum intensity-duration curve and develop a formula, and

(iv) maximum depth-duration curve.

4

QUESTION 1

(i) write down the hydrologic equation of water losses

(ii) Enumerated the various water losses that occur in nature

(iii) What are the factors affecting evaporation?

(iv) State Daltons law of evaporation

(v) Discuss briefly methods used to estimate evaporation from a

lake.

QUESTION 2

(i) Differentiate between Floating pans and Land pan

(ii) What do you understand by Pan coefficient,

(iii) The following are the monthly pan evaporation data (Jan.-Dec.) at

Krishnarajasagara in a certain year in cm.

16.7, 14.3, 17.8, 25.0, 28.6, 21.4,

16.7,16.7, 16.7, 21.4, 16.7, 16.7

The water spread area in a lake nearby in the beginning of January in

that year was 2.80 km 2 and at the end of December it was measured as

2.55 km 2. Calculate the loss of water due to evaporation in that year.

Assume a pan coefficient of 0.7.

(iv) Compute the daily evaporation from a Class A pan if the amounts of

water added to bring the level to the fixed point are as follows:

Day: 1 2 3 4 5 6 7

Rainfall (mm): 14 6 12 8 0 5 6

Water 5 3 0 0 7 4 3

added (mm): (removed)

What is the evaporation loss of water in this week from a lake (surface

area = 640 ha) in the vicinity, assuming a pan coefficient of 0.75?

(v) The total observed runoff volume during a storm of 6-hr duration

with a uniform intensity of 15 mm/hr is 21.6 Mm 3 . If the area of the

Water losses (evaporation, infiltration, transpiration)

5

basin is 300 km 2 , find the average infiltration rate and the runoff

coefficient for the basin.

QUESTION 3

(i) List the recommendation measures to reduce reservoir evaporation.

(ii) Differentiate between transpiration and transpiration ratio

(iii) What is the difference between Evapotranspiration (Et) and Potential

evapotranspiration (Ept)

(iv) List some methods used for estimating evapotranspiration

(v) Determine the evapotranspiration and irrigation requirement for wheat, if

the water application efficiency is 65% and the consumptive use

coefficient for the growing season is 0.8 from the following data :

QUESTION 4

(i) Discuss briefly the factors that affects Evapotranspiration,

(ii) What do you understand by infiltration,

(iii) List down the various methods used to determine infiltration,

(iv) For a given basin, the following are the infiltration capacity rates at

various time intervals after the beginning of the storm. Make a plot

of the f-curve and establish an equation of the form developed by

Horton. Also determine the total rain and the excess rain (runoff).

6

(v) For a small catchment, the infiltration rate at the beginning of rain

was observed to be 90 mm/hr and decreased exponentially to

a constant rate of 8 mm/hr after

QUESTION 5

(i) Write down the water balance equation

7

QUESTION 1

(i) With a neat sketch discuss briefly various components of catchment

characteristics,

(ii) What are the physical characteristics of a drainage ?

(iii) Discuss briefly two types of stream classification,

(iv) Dicuss the various factors, which affect the runoff from a basin;

(v) What are the methods of estimating runoff from a catchment?

QUESTION 6

A 4-hour rain of average intensity 1 cm/hr falls over the fern leaf type catchment

as shown in Fig. 4.10. The time of concentration from the lines AA, BB, CC and

DD are 1, 2, 3 and 4 hours, respectively, to the site 0 where the discharge

measurements are made. The values of the runoff coefficient C are 0.5, 0.6, and 0.7

for the 1st, 2nd and 3rd hours of rainfail respectively and attains a constant value

of 0.8 after 3 hours. Determine the discharge at site 0.

CHAP4: RUNOFF

8

QUESTION 1

(i) What do you understand hydrology?

SOLUTION

Hydrology is the science, which deals with the occurrence,

distribution and disposal of water on the planet earth; it is the science

which deals with the various phases of the hydrologic cycle.

(ii) Explain the hydraulic cycle in nature with the help of a neat sketch,

indicating its various phases.

SOLUTION

Hydrologic cycle is the water transfer cycle, which occurs continuously in

nature; the three important phases of the hydrologic cycle are:

(a) Evaporation and evapotranspiration

(b) Precipitation and

(c) Runoff and is shown in Fig. bellow.

The globe has one-third land and Two- thirds Ocean. Evaporation from the

surfaces of ponds, lakes, reservoirs. Ocean surfaces, etc. and transpiration

from surface vegetation i.e., from plant leaves of cropped land and forests,

etc. take place. These vapours rise to the sky and are condensed at higher

altitudes by condensation nuclei and form clouds, resulting in droplet

growth. The clouds melt and sometimes burst resulting in precipitation of

different forms like rain, snow, hail, sleet, mist, dew and frost. A part of this

precipitation flows over the land called runoff and part infilters into the soil

which builds up the ground water table. The surface runoff joins the streams

and the water is stored in reservoirs. A portion of surface runoff and ground

water flows back to ocean.

Again evaporation starts from the surfaces of lakes, reservoirs and ocean,

and the cycle repeats. Of these three phases of the hydrologic cycle, namely,

QUESTIONS &ANSWERS ON INTRODUCTION

9

evaporation, precipitation and runoff, it is the runoff phase, which is

important to a civil engineer since he is concerned with the storage of

surface runoff in tanks and reservoirs for the purposes of irrigation,

municipal water supply hydroelectric power etc.

The hydrological budget

(iii) Discuss clearly the various forms of precipitation

SOLUTION

Drizzle a light steady rain in fine drops (0.5 mm) and intensity 0.5 mm, maximum size6 mm) from the cloud

Glaze Freezing of drizzle or rain when they come in contact with

cold objects.

10

Sleet Frozen rain drops while falling through air at subfreezing

temperature.

Snow ice crystals resulting from sublimation (i.e., water vapour

condenses to ice)

Snow flakes ice crystals fused together.

Hail small lumps of ice (>5 mm in diameter) formed by alternate

freezing and melting, when they are carried up and down in highly

turbulent air currents.

Dew moisture condensed from the atmosphere in small drops upon

cool surfaces.

Frost a feathery deposit of ice formed on the ground or on the

surface of exposed objects by dew or water vapour that has frozen

Fog a thin cloud of varying size formed at the surface of

the earth by condensation of atmospheric vapour (interfering with

visibility)

Mist a very thin fog

(iv) What are the functions of hydrology in water resources development?

SOLUTION

The study of hydrology helps us to know

(i) The maximum probable flood that may occur at a given site and its

frequency; this is required for the safe design of drains and culverts,

dams and reservoirs, channels and other flood control structures.

(ii) The water yield from a basinits occurence, quantity and

frequency, etc; this is necessary for the design of dams, municipal

water supply, water power, river navigation, etc.

(iii) The ground water development for which a knowledge of

the hydrogeology of the area, i.e., of the formation soil, recharge

facilities like streams and reservoirs, rainfall pattern, climate,

cropping pattern, etc. are required.

(iv) The maximum intensity of storm and its frequency for the design

of a drainage project in the area.

(v) What are the basic data required for hydrological studies?

SOLUTION

11

(i)Climatological data

(ii) Hydrometeorological data like temperature, wind velocity,

humidity, etc.

(iii) Precipitation records

(iv) Stream-flow records

(v) Seasonal fluctuation of ground water table or piezometric heads

(vi) Evaporation data

(vii) Cropping pattern, crops and their consumptive use

(viii) Water quality data of surface streams and ground water

(ix) Geomorphologic studies of the basin, like area, shape and slope of

the basin, mean and median elevation, mean temperature (as well as

highest and lowest temperature recorded) and other physiographic

characteristics of the basin; stream density and drainage density; tanks

and reservoirs.

(x) Hydrometeorological characteristics of basin

QUESTION 2

Explain hydrologic equation.

SOLUTION

HYDROLOGIC EQUATION

The hydrologic equation is simply the statement of the law of conservation of

matter and is given by

I = O + S ...

Where I = inflow

O = outflow

S = change in storage

This equation states that during a given period, the total inflow into a given area

must equal the total outflow from the area plus the change is storage. While

solving this equation, the ground water is considered as an integral part of the

12

surface water and it is the subsurface inflow and outflow that pose problems in the

water balance studies of a basin.

QUESTION 1

(i) What are the different types or(forms) of precipitation?

SOLUTION

The precipitation may be due to

(i) Thermal convection (convectional precipitation)This type of

precipiation is in the form of local whirling thunder storms and is typical of

the tropics. The air close to the warm earth gets heated and rises due to its

low density, cools adiabatically to form a cauliflower shaped cloud, which

finally bursts into a thunder storm. When accompanied by destructive

winds, they are called tornados.

(ii) Conflict between two air masses (frontal precipitation)When two

air masses due to contrasting temperatures and densities clash with each

other, condensation and precipitation occur at the surface of contact, Fig.

2.1. This surface of contact is called a front or frontal surface. If a cold

air mass drives out a warm air mass it is called a cold front and if a warm

air mass replaces the retreating cold air mass, it is called a warm front. On

the other hand, if the two air masses are drawn simultaneously towards a low

pressure area, the front developed is stationary and is called a stationary

front. Cold front causes intense precipitation on comparatively small areas,

while the precipitation due to warm front is less intense but is spread

over a comparatively larger area. Cold fronts move faster than warm fronts

and usually over-take them, the frontal surfaces of cold and warm air sliding

against each other. This phenomenon is called occlusion and the resulting

frontal surface is called an occluded front.

(ii) Orographic lifting (orographic precipitation)The mechanical

lifting of moist air over mountain barriers, causes heavy precipitation on

the windward side (Fig. 2.2). For example Cherrapunji in the Himalayan

QUESTIONS & ANSWERS ON CHAP 2

PRECIPITATION

13

range and Agumbe in the western Ghats of south India get very heavy

orographic precipitation of 1250 cm and 900 cm (average annual rainfall),

respectively.

(iv) Cyclonic (cyclonic precipitation)This type of precipitation is due to

lifting of moist air converging into a low pressure belt, i.e., due to pressure

differences created by the unequal heating of the earths surface. Here the

winds blow spirally inward counterclockwise in the northern hemisphere and

clockwise in the southern hemisphere. There are two main types of

cyclonestropical cyclone (also called hurricane or typhoon) of

comparatively small diameter of 300-1500 km causing high wind

velocity and heavy precipitation, and the extra-tropical cyclone of large

diameter up to 3000 km causing wide spread frontal type precipitation.

14

(ii) Discuss the three various types of recording rain gauges

SOLUTION

Tipping bucket rain gauge

a)Tipping bucket rain gauge. This consists of a cylindrical receiver 30

cm diameter with a funnel inside (Fig. 2.4). Just below the funnel a pair

of tipping buckets is pivoted such that when one of the bucket receives a

rainfall of 0.25 mm it tips and empties into a tank below, while the other

bucket takes its position and the process is repeated. The tipping of the

bucket actuates on electric circuit which causes a pen to move on a chart

wrapped round a drum which revolves by a clock mechanism. This type

cannot record snow.

15

b) Weighing type rain gauge

Weighing type rain gauge. In this type of rain-gauge, when a certain

weight of rainfall is collected in a tank, which rests on a spring-lever

balance, it makes a pen to move on a chart wrapped round a clock driven

drum (Fig. 2.5). The rotation of the drum sets the time scale while the

vertical motion of the pen records the cumulative precipitation.

c) Float type rain gauge

16

Float type rain gauge. In this type, as the rain is collected in a float

chamber, the float moves up which makes a pen to move on a chart

wrapped round a clock driven drum (Fig. 2.6). When the float chamber

fills up, the water siphons out automatically through a siphon tube kept in

an interconnected siphon chamber. The clockwork revolves the drum

once in 24 hours. The clock mechanism needs rewinding once in a

week when the chart wrapped round the drum is also replaced. This

type of gauge is used by IMD.

(iii) What factors you consider in selecting a site for a rain-gauge station?

SOLUTION

Distance from the ocean

Direction of the prevailing winds

The mean annual temperature

Altitude of the place and

Its topography

17

(iv) (a) Distinguish between recording and non-recording rain-gauges,

SOLUTION

Recording Rain Gauge

This is also called self-recording, automatic or integrating rain gauge.

This type of rain gauge has an automatic mechanical arrangement

consisting of a clockwork, a drum with a graph paper fixed around it

and a pencil point, which draws the mass curve of rainfall From this

mass curve, the depth of rainfall in a given time, the rate or intensity

of rainfall at any instant during a storm, time of onset and cessation of

rainfall, can be determined.

The non-recording rain gauge

Is used in India is the Symons rain gauge (Fig. 2.3). It consists of a

funnel with a circular rim of 12.7 cm diameter and a glass bottle as a

receiver. The cylindrical metal casing is fixed vertically to the

masonry foundation with the level rim 30.5

cm above the ground surface. The rain falling into the funnel is

collected in the receiver and is measured in a special measuring glass

graduated in mm of rainfall; when full it can measure 1.25 cm of rain.

18

(v) (a) Explain:

1) A method for estimating the missing rainfall data at a station in a

basin.

SOLUTION

a) Station-year methodIn this method, the records of two or more stations are combined into one long record provided station records are

independent and the areas in which the stations are located are

climatologically the same. The missing record at a station in a particular

year may be found by the ratio of averages or by graphical comparison.

b)

By simple proportion (normal ratio method)This method is illustrated by the following example.

C) Double-mass analysisThe trend of the rainfall records at a station may slightly change after some years due to a change in the environment (or exposure)

of a station either due to coming of a new building, fence, planting of trees or

cutting of forest nearby, which affect the catch of the gauge due to change in the

wind pattern or exposure.

QUESTION 2

Rain-gauge station D was inoperative for part of a month during which a storm

occured. The storm rainfall recorded in the three surrounding stations A, B and C

were 8.5, 6.7and 9.0 cm, respectively. If the a.a.r for the stations are 75, 84, 70 and

90 cm, respectively, estimate the storm rainfall at station D.

Solution By equating the ratios of storm rainfall to the a.a.r. at each station, the

storm rainfall at station D (PD) is estimated as

Eg:pg 25

QUESTION 3

(i) Define rain-gauge density

SOLUTION

19

RAIN-GAUGE DENSITY

The following figures give a guideline as to the number of rain-gauges to be

erected in a given area or what is termed as rain-gauge density Area Rain-gauge density

Plains 1 in 520 km2

Elevated regions 1 in 260-390 km2

Hilly and very heavy 1 in 130 Km2 preferably with 10% of the

rainfall areas rain-gauge stations equipped with the self

recording type

(ii) What do understand by

SOLUTION

Point rainfallIt is the rainfall at a single station. For small areas less than 50 km2, point rainfall may be taken as the average depth over the area.

(iii) As the rainfall over a large area is not uniform, the average depth of

rainfall over the area is determined by three methods discuss them briefly

SOLUTION

Arithmetic average methodIt is obtained by simply averaging arithmetically the amounts of rainfall at the individual rain-gauge stations in

the area, i.e.,

This method is fast and simple and yields good estimates in flat country if the

gauges are uniformly distributed and the rainfall at different stations do not vary

very widely from the mean. These limitations can be partially overcome if

topographic influences and aerial representativity are considered in the selection of

gauge sites.

Thiessen polygon methodThis method attempts to allow for non-uniform distribution of gauges by providing a weighting factor for each gauge. The

stations are plotted on a base map and are connected by straight lines.

Perpendicular bisectors are drawn to the straight lines, joining adjacent

stations to form polygons, known as Thiessen polygons

20

if P1, P2, P3, ....are the rainfalls at the individual stations, and A1, A2, A3,

.... are the areas of the polygons surrounding these stations, (influence areas)

respectively, the average depth of rainfall for the

entire basin is given by

The isohyetal methodIn this method, the point rainfalls are plotted on a suitablebbase map and the lines of equal rainfall (isohyets) are drawn giving

consideration to orographic effects and storm morphology, Fig. 2.11.

Figure at pag 28

This method if analyzed properly gives the best results.

(iv) Example 2.3 Point rainfalls due to a storm at several rain-gauge

stations in a basin are shown in Fig. 2.10. Determine the mean areal

depth of rainfall over the basin by the three methods.

21

SOLUTION

(a) Arithmetic average method

P1 = sum of the 15 station rainfalls.

(b) Thiessen polygon method

The Thiessen polygons are constructed as shown in

Fig. 2.10 and the polygonal areas are planimetered and the mean areal

depth of rainfall is worked out below:

22

(c) Isohyetal method

The isohyets are drawn as shown in Fig. 2.11 and the mean areal

depth of rainfall is worked out below:

(v) Example 2.3 (a) The area shown in Fig. P (2.3a) is composed of a square

plus an equilateral triangular plot of side 10 km. The annual

precipitations at the rain-gauge stations located at the four corners and

centre of the square plot and apex of the traingular plot are indicated in

figure. Find the mean precipitation over the area by Thiessen polygon

method, and compare with the arithmetic mean.

SOLUTION

The Thiessen polygon is constructed by drawing perpendicular bisectors

to the lines joining the rain-gauge stations as shown in Fig. P (2.3a). The

weighted mean precipitation is computed in the following table:

Area of square plot = 10 10 = 100 km 2

Area of inner square plot =

23

Difference = 50 km 2

Area of each corner triangle in the square plot =

24

Arithmetic mean =

Which compares fairly with the weighted mean

QUESTION 4

(i) What do you understand by a hyetograph?

SOLUTION

A hyetograph is a bar graph showing the intensity of rainfall

with respect to time(Fig. 2.16) and is useful in determining the

maximum intensities of rainfall during a particular storm as is

required in land drainage and design of culverts.

25

(ii) What do you understand by A mass curve of rainfall (or

precipitation)

SOLUTION

Is a plot of cumulative depth of rainfall against time (Fig. 2.17). From the mass

curve, the total depth of rainfall and intensity of rainfall at any instant of time can

be found.

(iii) Differentiate between Mean and Median

SOLUTION

The sum of all the items in a set divided by the number of items

gives the mean value

,

where x = the mean value

x = sum of all the items

n = total number of items.

To find the median, the items are arranged in the ascending order;

if the number of items is odd, the middle item gives the median; if

the number of items is even, the average of the central two items

gives the median.

26

(iv) The annual rainfall at a place for a period of 10 years from 1961 to

1970 are respectively 30.3, 41.0, 33.5, 34.0, 33.3, 36.2, 33.6, 30.2,

35.5, 36.3. Determine the mean and median values of annual

rainfall for the place.

SOLUTION

a)

b) Median: Arrange the samples in the ascending order 30.2, 30.3 33.3,

33.5, 33.6, 34.0, 35.5, 36.2, 36.3, 41.0

No. of items = 10, i.e., even

Median =

Note the difference between the mean and the median values. If 11 years

of record, say1960 to 1970, had been given, the median would have been

the sixth item (central value) when arranged in the ascending order.

27

QUESTION 5

The following are the rain gauge observations during a storm. Construct:

(i) mass curve of precipitation,

(ii) hyetograph,

(iii) maximum intensity-duration curve and develop a formula, and

(iv) maximum depth-duration curve.

SOLUTION

(i) mass curve of precipitation,

is the plot of accumulated rainfall (cm) vs. time (min) gives the

mass curve of rainfall Fig. 2.19 (a).

(ii) hyetograph,

Hyetograph. The intensity of rainfall at successive 5 min interval is

calculated and a bar-graph of i (cm/hr) vs. t (min) is constructed; this

depicts the variation of the intensity of rainfall with respect to time

and is called the hyetograph; 2.19 (b).

28

(iii) maximum intensity-duration curve and develop a formula, and

Maximum depthduration curve. By inspection of time (t) and

accumulated rainfall(cm) the maximum rainfall depths during 5, 10,

29

15, 20, 25, 30, 35, 40, 45 and 50 min durations are 0.7, 1.3, 1.6, 1.8,

2.3, 2.5, 2.7, 2.9, 3.0 and 3.1 cm respectively. The plot of the

maximum rainfall depths against different durations on a log-log

paper gives the maximum depth-duration curve, which is a straight

line, Fig. 2.20 (a).

(iv) maximum depth-duration curve.

Maximum intensity-duration curve. Corresponding to the maximum depths

obtained in (c) above, the corresponding maximum intensities can be obtained

60, i.e., 8.4, 7.8, 6.4,5.4, 5.52, 5.0, 4.63, 4.35, 4.0 and 3.72 cm/hr,

respectively. The plot of the maximum intensities against the different duration on

a log-log paper gives the maximum intensity-duration curve which is a straight

line, Fig. 2.20 (b). The equation for the maximum intensity duration curve is of the

form

30

which can now be verified as

t = 10 min, i = 7.2 cm/hr

t = 40 min, i = 4.25 cm/hr

which agree with the observed data.

Example 2 in the book

31

QUESTION 1

(i) write down the hydrologic equation of water losses

SOLUTION

The hydrologic equation states that

RainfallLosses = Runoff

(ii) Enumerated the various water losses that occur in nature

SOLUTION

WATER LOSSES IN NATURE

(a) Interception loss-due to surface vegetation, i.e., held by

plant leaves.

(b) Evaporation:

from water surface, i.e., reservoirs, lakes, ponds, river

channels, etc.

from soil surface, appreciably when the ground water

table is very near the soil surface.

(c) Transpirationfrom plant leaves.

(d) Evapotranspiration for consumptive usefrom irrigated or

cropped land.

(e) Infiltrationinto the soil at the ground surface.

(f) Watershed leakageground water movement from one

basin to another or into the sea.

(iii) What are the factors affecting evaporation?

SOLUTION

air and water temperature,

relative humidity,

wind velocity,

surface area (exposed),

barometric pressure and

salinity of the water,

(iv) State Daltons law of evaporation

QUESTIONS & ANSWERS ON Water losses

(evaporation, infiltration, transpiration)

32

SOLUTION

The Daltons law states that the evaporation is proportional to

the difference in vapour pressures ew and ea. A more general

form of the Eq. (3.2) is given by

(v) Discuss briefly methods used to estimate evaporation from a

lake

SOLUTION

Evaporation from water surfaces can be determined from the

following methods:

(a) The storage equation

P + I Og = E + O S where

P = Precipitation

I = surface inflow

Og = subsurface inflow or outflow

E = evaporation

O = surface outflow

S = change in surface water storage

(b) Auxiliary pans like land pans, floating pans, colarado

sunken pans, etc.

(c) Evaporation formula like that of Daltons law

(d) Humidity and wind velocity gradients

(e) The energy budgetthis method involves too many

hydrometeorological factors (variables) with too much

sophisticated instrumentation and hence it is a specialist

approach

(f) The water budgetsimilar to (i)

(g) Combination of aerodynamic and energy balance

equationsPenmans equation (involves too many variables)

33

QUESTION 2

(i) Differentiate between Floating pans and Land pan

SOLUTION

Floating pans (made of GI) of 90 cm square and 45 cm deep are

mounted on a raft floating in water. The volume of water lost due

to evaporation in the pan is determined by knowing the volume of

water required to bring the level of water up to the original mark

daily and after making allowance for rainfall, if there has been

any.

Land pan. Evaporation pans are installed in the vicinity of the

reservoir or lake to determine the lake evaporation.

Colarado sunken pan. This is 92 cm square and 42-92 cm deep

and is sunk in the ground such that only 5-15 cm depth projects

above the ground surface and thus the water level is maintained

almost at the ground level. The evaporation is measured by a point

gauge.

(ii) What do you understand by Pan coefficient

SOLUTION

Pan coefficientEvaporation pan data cannot be applied to free water

surfaces directly but must be adjusted for the differences in physical and

climatological factors. For example, a lake is larger and deeper and may

be exposed to different wind speed, as compared to a pan.

Thus the pan evaporation data have to be corrected to obtain the actual

evaporation from water surfaces of lakes and reservoirs, i.e., by

multiplying by a coefficient called pan coefficient and is defined as

Pan coefficient =

(iii) The following are the monthly pan evaporation data (Jan.-Dec.)

at Krishnarajasagara in a certain year in cm.

34

16.7, 14.3, 17.8, 25.0, 28.6, 21.4,

16.7,16.7, 16.7, 21.4, 16.7, 16.7

The water spread area in a lake nearby in the beginning of January in

that year was 2.80 km 2 and at the end of December it was measured as

2.55 km 2. Calculate the loss of water due to evaporation in that year.

Assume a pan coefficient of 0.7.

SOLUTION

Mean water spread area of lake

Annual loss of water due to evaporation (adding up the monthly values)

= 228.7 cm

Annual volume of water lost due to evaporation

(iv) Compute the daily evaporation from a Class A pan if the amounts of

water added to bring the level to the fixed point are as follows:

Day: 1 2 3 4 5 6 7

Rainfall (mm): 14 6 12 8 0 5 6

Water 5 3 0 0 7 4 3

added (mm): (removed)

What is the evaporation loss of water in this week from a lake (surface

area = 640 ha) in the vicinity, assuming a pan coefficient of 0.75?

SOLUTION

Pan evaporation

35

(v) The total observed runoff volume during a storm of 6-hr duration

with a uniform intensity of 15 mm/hr is 21.6 Mm 3 . If the area of the

basin is 300 km 2 , find the average infiltration rate and the runoff

coefficient for the basin.

SOLUTION

(i) Infiltration loss Fp = Rainfall (P) Runoff (R)

QUESTION 3

(i) List the recommendation measures to reduce reservoir evaporation.

SOLUTION

Storage reservoirs of more depth and less surface area, i.e., by

choosing a cross section of the reservoir like a deep gorge.

By growing tall trees like Causerina on the windward side of the

reservoirs to act as wind breakers.

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By spraying certain chemicals or fatty acids and formation of films. By

spreading a manomolecular layer of acetyl alcohol (hexadecanol) C16

H33OH over the reservoir surface (from boats).

By allowing flow of water, temperature is reduced and evaporation is

reduced; i.e.,by designing the outlet works so that the warmer surface

water can be released.

By removing the water loving weeds and plants like Phreatophytes from

the periphery of the reservoir.

By straightening the stream-channels the exposed area of the water

surface (along the length) is reduced and hence evaporation is reduced.

By providing mechanical coverings like thin polythene sheets to small

agricultural ponds and lakes.

By developing underground reservoirs, since the evaporation from a

ground water table is very much less than the evaporation from a water

surface.

If the reservoir is surrounded by huge trees and forest, the evaporation

loss will be less due to cooller environment.

(ii) Differentiate between transpiration and transpiration ratio

SOLUTION

Transpiration is the process by which the water vapour escapes from the

living plant leaves and enters the atmosphere.

Transpiration ratio is the ratio of the weight of water absorbed (through

the root system), conveyed through and transpired from a plant during the

growing season to the weight of the dry matter produced exclusive of

roots.

Transpiration ratio =

(iii) What is the difference between Evapotranspiration (Et) and Potential

evapotranspiration (Ept)

SOLUTION

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Evapotranspiration (Et) or consumptive use (U) is the total water lost from a

cropped (or irrigated) land due to evaporation from the soil and transpiration by the

plants or used by the plants in building up of plant tissue. Potential

evapotranspiration (Ept) is the evapotranspiration from the short green

vegetation when the roots are supplied with unlimited water covering the soil. It is

usually expressed as a depth (cm, mm) over the area.

(iv) List some methods used for estimating evapotranspiration

SOLUTION

Tanks and lysimeter experiments

Field experimental plots

Installation of sunken (colarado) tanks

Evapotranspiration equations as developed by Lowry-Johnson,

Penman,Thornthwaite, Blaney-Criddle, etc.

Evaporation index method,

(v) Determine the evapotranspiration and irrigation requirement for wheat, if

the water application efficiency is 65% and the consumptive use

coefficient for the growing season is 0.8 from the following data :

SOLUTION

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QUESTION 4

(i) Discuss briefly the factors that affects Evapotranspiration

SOLUTION

Factors affect the evapotranspiration:

Climatological factors like percentage sunshine hours, wind speed,

mean monthly temperature and humidity.

Crop factors like the type of crop and the percentage growing

season.

The moisture level in the soil.

(ii) What do you understand by infiltration

SOLUTION

INFILTRATION

Water entering the soil at the ground surface is called infiltration. It

replenishes the soil moisture deficiency and the excess moves

downward by the force of gravity called deep seepage or

Percolation and builds up the ground water table.

(iii) List down the various methods used to determine infiltration

SOLUTION

The methods of determining infiltration are:

Infiltrometers

Observation in pits and ponds

Placing a catch basin below a laboratory sample

Artificial rain simulators

Hydrograph analysis

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(iv) For a given basin, the following are the infiltration capacity rates at

various time intervals after the beginning of the storm. Make a plot

of the f-curve and establish an equation of the form developed by

Horton. Also determine the total rain and the excess rain (runoff).

SOLUTION

The precipitation and infiltration rates versus time are plotted as

shown in Fig. 3.10.In the Hortons equation, the Hortons constant

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(v) For a small catchment, the infiltration rate at the beginning of rain

was observed to be 90 mm/hr and decreased exponentially to

a constant rate of 8 mm/hr after

SOLUTION

QUESTION 5

(ii) Write down the water balance equation

SOLUTION

The input items into a basin are essentially precipitation (P) and

subsurface inflow (Gi) while the water losses are evaporation (E),

evapotranspiration (Et) and subsurface outflow (Go). The

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Balance goes to recharge ground water (Gr), increase the soil moisture

(SMA) and as surface runoff (streamflow, R).

The water balance equation can be written as

P + Gi = E + Et+ G0 + SMA + Gr + R

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QUESTION 1

(i) With a neat sketch discuss briefly various components of catchment

characteristics

SOLUTION

(ii) What are the physical characteristics of a drainage ?

SOLUTION

The characteristics of the drainage net may be physically described by:

The number of streams

The length of streams

Stream density

Drainage density

(iii) Discuss briefly two types of stream classification

SOLUTION

Streams may be classified as:

Influent and Effluent streams

Intermittent and perennial streams

Influent and Effluent streams. If the GWT is below the bed of the stream, the

seepage from the stream feeds the ground-water resulting in the build up of water

mound (Fig. 4.6).Such streams are called influent streams. Irrigation channels

function as influent streams andmany rivers which cross desert areas do so.

QUESTION & ANSWERS CHAP4: RUNOFF

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When the GWT is above water surface elevation in the stream, the ground water

feeds the stream, Fig. 4.7. Such streams are called effluent streams. The base flow

of surface streams is the effluent seepage from the drainage basin. Most of

the perennial streams are mainly effluent streams.

Intermittent and perennial streams. If the GWT lies above the bed of the

stream during the wet season but drops below the bed during the dry season, the

stream flows during wet season (due to surface runoff and ground water

contribution) but becomes dry during dry seasons. Such streams are called

intermittent streams.

(iv) Dicuss the various factors, which affect the runoff from a basin

SOLUTION

The various factors, which affect the runoff from a drainage basin depend

upon the following characteristics:

(a) Storm characteristics : Type or nature of storm and

season

Intensity

Duration

Areal extent (distribution)

Frequency

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Antecedent precipitation

Direction of storm movement

(b) Meteorological characteristics :

Temperature

Humidity

Wind velocity,

Pressure variation

(c) Basin characteristics :

Size

Shape

Slope

Altitude (elevation)

Topography

Geology (type of soil)

Land use /vegetation

Orientation

Type of drainage net

Proximity to ocean and mountain

Ranges

(d) Storage characteristics :

Depressions

Pools and ponds / lakes

Stream

Channels

Check dams (in gullies)

Upstream reservoir /or tanks

Flood plains, swamps

Ground water storage in pervious

deposits (aquifers)

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(v) What are the methods of estimating runoff from a catchment?

SOLUTION

The runoff from rainfall may be estimated by the following methods:

Empirical formulae, curves and tables

Infiltration method

Rational method

Overland flow hydrograph

Unit hydrograph method

Coaxial Graphical Correlation

QUESTION 6

A 4-hour rain of average intensity 1 cm/hr falls over the fern leaf type catchment

as shown in Fig. 4.10. The time of concentration from the lines AA, BB, CC and

DD are 1, 2, 3 and 4 hours, respectively, to the site 0 where the discharge

measurements are made. The values of the runoff coefficient C are 0.5, 0.6, and 0.7

for the 1st, 2nd and 3rd hours of rainfail respectively and attains a constant value

of 0.8 after 3 hours. Determine the discharge at site 0.

SOLUTION

The discharge computations are made in Table 4.4.

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