Question and Answers on Engineering Hydrology

QUESTIONS ON INTRODUCTION 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’.

QUESTIONS ON CHAP 2 PRECIPITATION QUESTION 1 (i) (ii) (iii) (iv) (v)

What are the different types or(forms) of precipitation? Discuss the three various types of recording rain gauges What factors you consider in selecting a site for a rain-gauge station? Distinguish between recording and non-recording rain-gauges, 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.

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QUESTION 3 (i) (ii) (iii)

(iv)

(v)

Define ‘rain-gauge density’ What do understand by 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, 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.

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.

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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.

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Water losses (evaporation, infiltration, transpiration) QUESTION 1 (i) (ii) (iii) (iv) (v)

write down the hydrologic equation of water losses Enumerated the various water losses that occur in nature What are the factors affecting evaporation? State Dalton’s law of evaporation Discuss briefly methods used to estimate evaporation from a lake.

QUESTION 2 (i) (ii) (iii)

(iv)

(v)

Differentiate between Floating pans and Land pan What do you understand by Pan coefficient, 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. 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? 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

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basin is 300 km 2 , find the average infiltration rate and the runoff coefficient for the basin. QUESTION 3 (i) (ii) (iii) (iv) (v)

List the recommendation measures to reduce reservoir evaporation. Differentiate between transpiration and transpiration ratio What is the difference between Evapotranspiration (Et) and Potential evapotranspiration (Ept) List some methods used for estimating evapotranspiration 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).

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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

QUESTION 5 (i)

Write down the water balance equation

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CHAP4: RUNOFF QUESTION 1 (i) (ii) (iii) (iv) (v)

With a neat sketch discuss briefly various components of catchment characteristics, What are the physical characteristics of a drainage ? Discuss briefly two types of stream classification, Dicuss the various factors, which affect the runoff from a basin; 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.

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QUESTIONS &ANSWERS ON INTRODUCTION 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, 8

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 size—6 mm) from the cloud Glaze — Freezing of drizzle or rain when they come in contact with cold objects. 9

(iv)

(v)

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 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 basin—its 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.

What are the basic data required for hydrological studies? SOLUTION 10

(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 11

surface water and it is the subsurface inflow and outflow that pose problems in the water balance studies of a basin.

QUESTIONS & ANSWERS ON CHAP 2 PRECIPITATION 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 12

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 earth’s surface. Here the winds blow spirally inward counterclockwise in the northern hemisphere and clockwise in the southern hemisphere. There are two main types of cyclones—tropical 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.

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(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. 14

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

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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

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(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 Symon’s 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.

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(v) (a) Explain: 1) A method for estimating the missing rainfall data at a station in a basin. SOLUTION a) Station-year method—In 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 analysis—The 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 18

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 rainfall—It 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 method—It 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 method—This 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 19

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 method—In 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.

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SOLUTION (a) Arithmetic average method

ΣP1 = 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:

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(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 = 22

Difference = 50 km 2 Area of each corner triangle in the square plot =

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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.

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(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 Σx = 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.

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(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.

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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)

(ii)

mass curve of precipitation, is the plot of ‘accumulated rainfall (cm) vs. time (min)’ gives the ‘mass curve of rainfall’ Fig. 2.19 (a). 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).

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(iii)

maximum intensity-duration curve and develop a formula, and Maximum depth–duration curve. By inspection of time (t) and accumulated rainfall(cm) the maximum rainfall depths during 5, 10, 28

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

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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

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QUESTIONS & ANSWERS ON Water losses (evaporation, infiltration, transpiration) QUESTION 1 (i)

(ii)

(iii)

(iv)

write down the hydrologic equation of water losses SOLUTION The hydrologic equation states that Rainfall–Losses = Runoff 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) Transpiration—from plant leaves. (d) Evapotranspiration for consumptive use—from irrigated or cropped land. (e) Infiltration—into the soil at the ground surface. (f) Watershed leakage—ground water movement from one basin to another or into the sea. 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, State Dalton’s law of evaporation

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SOLUTION The Dalton’s 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 Dalton’s law (d) Humidity and wind velocity gradients (e) The energy budget—this method involves too many hydrometeorological factors (variables) with too much sophisticated instrumentation and hence it is a specialist approach (f) The water budget—similar to (i) (g) Combination of aerodynamic and energy balance equations—Penman’s equation (involves too many variables)

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QUESTION 2 (i)

(ii)

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. What do you understand by Pan coefficient SOLUTION Pan coefficient—Evaporation 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.

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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

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(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)

(v)

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, 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 38

(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 Horton’s 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 & ANSWERS CHAP4: RUNOFF 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 Discuss briefly two types of stream classification SOLUTION Streams may be classified as:  Influent and Effluent streams  Intermittent and perennial streams

(iii)

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.

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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 44

 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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